WikiWaves - User contributions [en]

Method of Characteristics for Linear Equations

2025-11-11T11:52:24Z

Levi: /* Example 1 */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(X,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(X,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = X</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>

</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|zXEvCJyHQSg}}

=== Part 2 ===

{{#ev:youtube|vRRVU4VrxV0}}

=== Part 3 ===

{{#ev:youtube|xkGdvFGit_c}}

=== Part 4 ===

{{#ev:youtube|rW-voMoG0KI}}

Method of Characteristics for Linear Equations

2025-11-11T11:52:07Z

Levi: /* General Form */

{{nonlinear waves course
| chapter title = Method of Characteristics for Linear Equations
| next chapter = [[Traffic Waves]]
| previous chapter =
}}

{{complete pages}}

We present here a brief account of the method of characteristic for linear waves.

== Introduction ==

The method of characteristics is an important method for hyperbolic PDE's which
applies to both linear and nonlinear equations.

We begin with the simplest wave equation
<center>
<math>
\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - 1 \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = 1</math> <math>u(x,t)</math> must be a constant.
These are nothing but the straight lines <math>x = t+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t+c,t) = u(c,0) = f(c) = f(x-t)\,
</math></center>
Therefore the solution is <math>u(x,t) = f(x-t)\,</math>.

== General Form ==

If we consider the equation
<center>
<math>
\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
then we can apply the method of characteristics.
We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(X,t) \right).
</math>
</center>
This gives us the following o.d.e. for the characteristic curves (along which the solution is a constant)
<center>
<math>
\frac{\mathrm{d} X}{\mathrm{d} t} = a(X,t) .
</math>
</center>

== Example 1 ==

[[Image:Characteristic_linear1.jpg|thumb|right|350px|Characteristic for Example 1]]
Consider the equation
<center>
<math>
\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d}X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - x \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = ce^t</math>
This means that we have
<center>
<math>
u(x,t) = u(ce^t,t) = u(c,0) = f(c) = f(xe^{-t})\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(xe^{-t})\,</math>.

[[Image:Waterfall_linear1.jpg|thumb|right|350px|Solution for Example 1 with <math>f(x) = e^{-x^2}</math>]]

== Example 2 ==

Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - t \right)
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> <math>u(x,t)</math> must be a constant.
These are the curves <math>x = t^2/2+c</math>
This means that we have
<center>
<math>
u(x,t) = u(t^2/2 + c,t) = u(c,0) = f(c) = f(x - t^2/2)\,
</math></center>
Therefore the solution is given <math>u(x,t) = f(x - t^2/2)\,</math>.

== Non-homogeneous Example ==

We can also use the method of characteristics in the non-homogeneous case. We show this through an example
Consider the equation
<center>
<math>
\partial_t u + t \partial_x u = xt,\,\,-\infty < x < \infty,\,\,t>0,
</math></center>
subject to the initial conditions
<center>
<math>
\left. u \right|_{t=0} = f(x)
</math></center>

We consider the solution along the curve <math>(x,t) = (X(t),t)</math>. We then have
<center>
<math>
\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
</math></center>

Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = t</math> which are the curves <math>x = t^2/2+c</math>
<center>
<math>
\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
</math>

</center>
Therefore
<center>
<math>
u(t^2/2+c,t) = t^4/8 + c t^2/2 + c_2\,
</math>
</center>
Now
<center>
<math>
u(c,0) = c_2 = f(c)
</math></center>

Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|zXEvCJyHQSg}}

=== Part 2 ===

{{#ev:youtube|vRRVU4VrxV0}}

=== Part 3 ===

{{#ev:youtube|xkGdvFGit_c}}

=== Part 4 ===

{{#ev:youtube|rW-voMoG0KI}}

Reaction-Diffusion Systems

2025-11-06T10:37:46Z

Levi: /* The discrete Fourier transform */

{{nonlinear waves course
| chapter title = Reaction-Diffusion Systems
| next chapter = [[Burgers Equation]]
| previous chapter = [[Example Calculations for the KdV and IST]]
}}

{{complete pages}}

We present here a brief theory of reaction diffusion waves.

== Law of Mass Action ==

The law of mass action states that equation rates are proportional to the concentration
of reacting species and the ratio in which they combined. It is discussed in detail in
[[Billingham and King 2000]]. We will present here a few simple examples.

=== Example 1: Simple Decay ===

Suppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.
<center>
<math>P \to A</math>
</center>
with rate <math>k[P]</math> where <math>[P]</math> denotes concentration. Then if we
set <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations
<center>
<math> \frac{\mathrm{d}p}{\mathrm{d}t} = -kp\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}a}{\mathrm{d}t} = kp</math>
</center>
which has solution
<center>
<math>
p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt})
</math>
</center>
where <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>
<math>p</math> repectively at <math>t=0</math>.

=== Example 2: Quadratic Autocatalysis ===

This example will be important when we consider reaction diffusion problems.
We consider the reaction
<center>
<math>A + B \to 2B</math>
</center>
with rate proportional to <math>k[A][B]</math>. If we define <math>a = [A]</math>
and <math>b = [B]</math> we obtain the following equations
<center>
<math> \frac{\mathrm{d}a}{\mathrm{d}t} = -kab\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}b}{\mathrm{d}t} = kab</math>
</center>
We can solve these equations by observing that
<center>
<math>
\frac{\mathrm{d}(a+b)}{\mathrm{d}t} = 0
</math>
</center>
so that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain
<center>
<math>
\frac{\mathrm{d}b}{\mathrm{d}t} = k(a_0 + b_0 - b)b
</math>
</center>
which is separable with solution
<center>
<math>
b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
and
<center>
<math>
a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
Note that <math>a\to 0</math> and <math>b\to a_0 + b_0</math>
as <math>t\to \infty.</math>

== Diffusion ==

The equation for spatially homogeneous diffusion of a chemical with concentration
<math>c</math> is
<center>
<math>
\partial_t c = D\nabla^2 c
</math>
</center>
which is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in
only one spatial dimension. Consider it on the boundary <math>-\infty < x < \infty</math>. In this case
we can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain
<center>
<math>
\partial_t \hat{c} = -D k^2 \hat{c}
</math>
</center>
where <math>\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution
<center>
<math>
\hat{c} = \hat{c}_0 e^{-D k^2 t}
</math>
</center>
We can find the inverse transform using convolution and obtain
<center>
<math>
c(x,t) = \mathcal{F}^{-1} \left\{ e^{-k^2 D t} \mathcal{F} \left\{ c_0(x) \right\} \right\} = c_0(x) * \frac{1}{\sqrt{4\pi D t}} e^{-x^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{-(x-s)^2/4Dt}\mathrm{d}s
</math>
</center>

=== Solution of the dispersion equation using FFT ===

We can solve the dispersion equation using the discrete Fourier transform and
its closely related numerical implementation the FFT (Fast Fourier Transform).
We have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail.
We consider the concentration
on the finite domain <math>-L \leq x \leq L</math> and use a
[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(t) e^{\mathrm{i} k_n x}
</math>
</center>
where <math>k_n = \pi n /L </math>. If we substitute this into the diffusion equation we obtain
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x}
</math>
</center>
Note that this is not the same solution as we obtained on the infinite domain because
of the boundary conditions on the finite domain. The coefficients <math>\hat{c}_n(0)</math>
are found using the initial conditions so that
<center>
<math>
\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
</math>
</center>

The key to the numerical solution of this equation is the use of the FFT. We begin by discretising the
domain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to
approximate the integral above and obtain
<center>
<math>
\hat{c}_n(0) = \frac{1}{N} \sum_{m=0}^{N-1} e^{-\mathrm{i} k_n x_m} c_0(x_m)
</math>
</center>
<center>
<math>
= \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} e^{\mathrm{i} \pi n} c_0(x_m)
</math>
</center>

We also get
<center>
<math>
c(x_m,t) = \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
</math>
</center>
<center>
<math>
= \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
</math>
</center>
but we know that
<center>
<math>
\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
</math>
</center>
So
<center>
<math>
c(x_m,t) = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{l=0}^{N-1} c_0(x_l) e^{-2\mathrm{i} \pi nl/N} e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N}
</math>
</center>

=== The discrete Fourier transform ===
The
[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
of a sequence of ''N'' complex numbers ''c''0, ..., ''c''''N''−1 is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math>0, ..., <math>\hat{c}</math>''N''−1 by the DFT according to the formula:
<center>
<math>\hat{c}_n = \sum_{m=0}^{N-1} c_m e^{-2\pi \mathrm{i}mn/N} \quad \quad n = 0, \dots, N-1</math>
</center>

We denote the transform by the symbol <math>\mathcal{F}</math>, as in <math>\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \} </math> or <math>\mathcal{F} \left ( \mathbf{x} \right )</math> or <math>\mathcal{F} \mathbf{x}</math>.

The '''inverse discrete Fourier transform (IDFT)''' is given by
<center>
<math>c_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{c}_n e^{2\pi \mathrm{i}mn/N} \quad \quad m = 0,\dots,N-1.</math>
</center>

Therefore we can write again, similar to the continuous transform case:
<center>
<math>
c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\} = \frac{1}{\sqrt{4\pi D t}} \sum_{l=0}^{N-1} c_0(x_{(m-l)\bmod N}) \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}_l
</math>
</center>
Unlike it, though, now <math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}\ne\frac{1}{\sqrt{4\pi D t}} e^{-x_m^2/4Dt}</math>. It can be proven that for our previously defined <math>x_m = -L + 2Lm/N</math> and <math>k_{n}=\left\{
\begin{matrix}
n\pi/L,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right)\pi/L,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right. </math> we have:
<center><math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}_m=\dfrac{2L}{N\sqrt{4\pi DT}}\sum_{l=-\infty}^\infty \operatorname{exp}\left[ -\dfrac{(x_m+(2l+1)L)^2}{4Dt} \right]\approx\dfrac{2L}{N\sqrt{4\pi DT}}\left( \operatorname{exp}\left[ -\dfrac{(x_m+L)^2}{4Dt} \right] + \operatorname{exp}\left[ -\dfrac{(x_m-L)^2}{4Dt} \right] \right)</math></center>

The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]
or '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse)
will involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to order <math>N \log(N)</math>. This is an incredible speed up, for example
if N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT
algorithms are used so extensively.

== Reaction Diffusion Equations ==

We consider an auto catalytic reaction where the chemical species also diffuse. In this
case the equations are
<center>
<math>\partial_t a = D\partial_x^2 a - kab</math>
</center>
<center>
<math>\partial_t b = D\partial_x^2 b + kab</math>
</center>
We can non-dimensionalise these equations scaling the variables as
<center>
<math>
z = x/x^*\,\,\,\tau = t/t^*\,\,\,\alpha = a/a_0\,\,\,\beta = b/a_0
</math>
</center>
So that the equations become
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \alpha = \frac{a_0}{(x^*)^2}D\partial_z^2 \alpha
- k a_0^2 \alpha\beta
</math>
</center>
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \beta = \frac{a_0}{(x^*)^2}D\partial_z^2 \beta
+ k a_0^2 \alpha\beta
</math>
</center>
If we choose
<center>
<math>
x^* = \sqrt{\frac{D}{ka_0}}\,\,\,t^*=\frac{1}{ka_0}
</math>
</center>
then we obtain the system
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
-\alpha\beta
</math>
</center>
<center>
<math>
\partial_\tau \beta =\partial_z^2 \beta
+ \alpha\beta
</math>
</center>

=== Solution via split step method ===

[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\alpha(z,0) =1
\, \beta(z,0) = \exp(-10z^2)</math>]]

We can solve this equations numerically using a
[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume
that at time <math>\tau</math> we know <math>\alpha(z,\tau)</math>
and <math>\beta(z,\tau)</math>. We then solve first the following equation
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
</math>
</center>
from <math>\tau</math> to <math>\tau + \Delta\tau</math>
(which we can do exactly using the spectral methods just discussed for
the dispersion equation). We write this solution as
<math>\tilde{\alpha}(z,\tau + \Delta\tau)</math>
Then we solve
<center>
<math>
\partial_\tau \alpha = -\alpha\beta
</math>
</center>
by assuming that <math>\beta</math> is constant and subject to the boundary
condition that <math>\alpha(z,\tau) = \tilde{\alpha}(z,\tau + \Delta\tau)</math>.
This gives
<center>
<math>
\alpha(z,\tau + \Delta\tau) = e^{-\beta(z,\tau) \Delta\tau} \tilde{\alpha}(z,\tau+ \Delta\tau)\,
</math>
</center>
and we do likewise for the equation for <math>\beta</math>. Note that
while both steps are exact the result from the split step method is an
approximation with error which becomes smaller as the step size becomes
smaller.

We can easily implement this split step method in matlab and we obtain
a pair of travelling waves.

== Travelling Waves solution ==

When we solve the equations we found the solution formed travelling waves and
we now consider this phenomena in detail.

We define a new coordinate <math>y = z - v\tau</math> (so we will consider only
waves travelling to the right, although we could analyse waves travelling to
the left in a similar fashion). We seek stationary solutions in
<math>\alpha(y)</math> and <math>\beta(y)</math> which satisfy
<center>
<math>
\frac{\mathrm{d}^2 \alpha}{\mathrm{d}y^2} + v \frac{\mathrm{d} \alpha}{\mathrm{d}y} = \alpha\beta
</math>
</center>
and
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} = -\alpha\beta
</math>
</center>
If we add these equations we obtain
<center>
<math>
\frac{\mathrm{d}^2 (\alpha+\beta)}{\mathrm{d}y^2} + v \frac{\mathrm{d} (\alpha+\beta)}{\mathrm{d}y} = 0
</math>
</center>
so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
<math>\beta = 0</math> and if <math>y\to-\infty</math> then <math>\alpha = 0</math> and
<math>\beta = 1</math>. Therefore <math>\alpha + \beta = 1</math>.
This means that, since <math>\alpha \geq 0</math>, we must have
<math>0\leq \beta \leq 1</math>.
We can then obtain the following equation
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} + \beta(1-\beta)= 0
</math>
</center>
which we can write as the system of first order equations.

[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing
the equilibrium points and the heteroclinic connection]]

We define the variable <math>\gamma = \frac{\mathrm{d}\beta}{\mathrm{d}y}</math>
and we obtain
<center><math>
\begin{align}
\frac{\mathrm{d}\beta}{\mathrm{d} y} &= \gamma&\\
\frac{\mathrm{d}\gamma}{\mathrm{d} y} &= -v\gamma + \beta(\beta -1)& \\
\end{align}
</math></center>
This dynamical system has equilibrium points at <math>(0,0)</math>
and <math>(1,0)</math>. We can analyse these equilibrium points by
linearization. The Jacobian matrix is
<center>
<math>
J =\begin{pmatrix}
0 & 1 \\
-1 + 2\beta & -v
\end{pmatrix}
</math>
</center>

We can easily see that the Jacobian evaluated at our first equilibrium point is
<center>
<math>
J_{(0,0)} =\begin{pmatrix}
0 & 1 \\
-1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\mu_{\pm} = -1/2 (v \mp \sqrt{v^2-4})</math>. Therefore
this point is a nodal sink (possibly a spiral)

[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on
the heteroclinic connection]]

Additionally,
<center>
<math>
J_{(1,0)} =\begin{pmatrix}
0 & 1 \\
1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\lambda_{\pm} = -1/2 (v \mp \sqrt{v^2+4})</math>.
This is a a saddle point. The unstable and stable
separatrices leave the equilibrium point at <math>(1,0)</math> in the directions
<math> \begin{pmatrix}\lambda_{\pm} \\ 1\end{pmatrix}</math>. The only path on which <math>\beta</math> is bounded
as <math>y\to-\infty</math> are the unstable separatrices. Also, only the
unstable separatrix which enters the region <math>\beta<1</math> is physically meaningful.

To find a travelling wave we need to find a heteroclinic connection
between the two equilibrium points which also has to satisfy the conditions
that <math>0\leq \beta \leq 1</math>.

We need to show that the heteroclinic connection does not cross the <math>\beta</math> axis.
Consider the region
<center>
<math>
R = \left\{(\beta,\gamma)\,|\, \beta<1,\,-k\beta<\gamma<0\right\}
</math>
</center>
On the line <math>\beta = 1,d\beta/dy<0</math> and hence all flow it into <math>R</math>.
On the line <math>\gamma = 0, d\gamma/dy < 0</math> for <math>0<\beta<1</math>. On the line
<math>\gamma = -k \beta</math> we know that <math>d\beta/dy < 0</math> so that integral paths
enter the region if and only if <math>d\gamma/d\beta < \gamma/\beta</math>. We know that
<center>
<math>
d\gamma/d\beta - \gamma/\beta = -v - \frac{\beta(1-\beta)}{\gamma} -\frac{\gamma}{\beta}
= \frac{1}{k} (k^2 - vk +1 -\beta)
</math>
</center>
when <math>\gamma = -k\beta</math>. Therefore we need to find a value of <math>k</math>
so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
<math>k = \dfrac{v}{2}</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|rF4X42jP0v8}}

=== Part 2 ===

{{#ev:youtube|D0NwYlM-uOg}}

=== Part 3 ===

{{#ev:youtube|5IEZJtJaDHk}}

=== Part 4 ===

{{#ev:youtube|t_OjTSwVgdo}}

=== Part 5 ===

{{#ev:youtube|kTHVZaYezLk}}

=== Part 6 ===

{{#ev:youtube|MVuSg5_sfYI}}

=== Part 7 ===

{{#ev:youtube|hxKMOHyy6Bw}}

=== Part 8 ===

{{#ev:youtube|yQ-O2KIqu44}}

[[Category:Simple Nonlinear Waves]]

Burgers Equation

2025-11-06T10:35:03Z

Levi: /* Numerical Solution of Burgers equation */

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u</math> and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime} =c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} .
We note that the wave speed
is
<center><math>
c=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik \widehat{\left( u^{2}\right)} -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>
Or, in convolution form:
<center><math>u(x, t+\Delta t)=\left(u(x, t)-\dfrac{\Delta t}2\mathcal F^{-1}\left\{ik\right\}*u^2(x, t)\right)*\mathcal F^{-1}\left\{e^{-\nu k^2\Delta t}\right\}</math></center>
And for our previously defined <math>x_m</math> and <math>k_n</math> it can be proven that <math>\mathcal F^{-1}\left\{ik\right\}_m=\dfrac\pi{2L}(-1)^m\left(\cot\dfrac{\pi m}N+i\right)</math> (<math>\cot 0</math> taken to be <math>0</math>).

{| class="wikitable"
|-
! Phase plane for a travelling wave solution
! Numerical solution of Burgers equation
|-
| [[Image:Burgers_phase.jpg|thumb|right|500px|Phase plane for a travelling wave solution of Burgers equation]]
| [[Image:File-Burgers2.gif|thumb|right|500px| Numerical solution of Burgers equation]]

|}

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}
-2\nu^2 \frac{\partial_x^2\phi}{\phi}
-\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) \mathrm{d}s\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\Phi \left( x\right) * \dfrac1{\sqrt{4\pi\nu t}} e^{-x^2 / 4\nu t} \\
&=&\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&-2\nu \dfrac{\partial _{x}\left( \Phi \left( x\right) * e^{-x^2 / 4\nu t} \right) }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } \\
&=&-2\nu \dfrac{\partial _{x} \Phi \left( x\right) * e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } = \dfrac{\Phi \left( x\right) * x e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * t e^{-x^2 / 4\nu t} } \\
&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] \mathrm{d}y}
\end{matrix}</math></center>
as <math>\partial_x(f * y)=\partial_x f * y = f * \partial_x y</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|tVXQmxOG_6Y}}

=== Part 2 ===

{{#ev:youtube|hzgpMM_wWts}}

=== Part 3 ===

{{#ev:youtube|uH4B1XsGB-0}}

=== Part 4 ===

{{#ev:youtube|h6aDmCtJygM}}

=== Part 5 ===

{{#ev:youtube|CsnUKrLjtyQ}}

Reaction-Diffusion Systems

2025-11-04T18:12:32Z

Levi: /* The discrete Fourier transform */ fix

{{nonlinear waves course
| chapter title = Reaction-Diffusion Systems
| next chapter = [[Burgers Equation]]
| previous chapter = [[Example Calculations for the KdV and IST]]
}}

{{complete pages}}

We present here a brief theory of reaction diffusion waves.

== Law of Mass Action ==

The law of mass action states that equation rates are proportional to the concentration
of reacting species and the ratio in which they combined. It is discussed in detail in
[[Billingham and King 2000]]. We will present here a few simple examples.

=== Example 1: Simple Decay ===

Suppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.
<center>
<math>P \to A</math>
</center>
with rate <math>k[P]</math> where <math>[P]</math> denotes concentration. Then if we
set <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations
<center>
<math> \frac{\mathrm{d}p}{\mathrm{d}t} = -kp\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}a}{\mathrm{d}t} = kp</math>
</center>
which has solution
<center>
<math>
p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt})
</math>
</center>
where <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>
<math>p</math> repectively at <math>t=0</math>.

=== Example 2: Quadratic Autocatalysis ===

This example will be important when we consider reaction diffusion problems.
We consider the reaction
<center>
<math>A + B \to 2B</math>
</center>
with rate proportional to <math>k[A][B]</math>. If we define <math>a = [A]</math>
and <math>b = [B]</math> we obtain the following equations
<center>
<math> \frac{\mathrm{d}a}{\mathrm{d}t} = -kab\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}b}{\mathrm{d}t} = kab</math>
</center>
We can solve these equations by observing that
<center>
<math>
\frac{\mathrm{d}(a+b)}{\mathrm{d}t} = 0
</math>
</center>
so that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain
<center>
<math>
\frac{\mathrm{d}b}{\mathrm{d}t} = k(a_0 + b_0 - b)b
</math>
</center>
which is separable with solution
<center>
<math>
b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
and
<center>
<math>
a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
Note that <math>a\to 0</math> and <math>b\to a_0 + b_0</math>
as <math>t\to \infty.</math>

== Diffusion ==

The equation for spatially homogeneous diffusion of a chemical with concentration
<math>c</math> is
<center>
<math>
\partial_t c = D\nabla^2 c
</math>
</center>
which is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in
only one spatial dimension. Consider it on the boundary <math>-\infty < x < \infty</math>. In this case
we can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain
<center>
<math>
\partial_t \hat{c} = -D k^2 \hat{c}
</math>
</center>
where <math>\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution
<center>
<math>
\hat{c} = \hat{c}_0 e^{-D k^2 t}
</math>
</center>
We can find the inverse transform using convolution and obtain
<center>
<math>
c(x,t) = \mathcal{F}^{-1} \left\{ e^{-k^2 D t} \mathcal{F} \left\{ c_0(x) \right\} \right\} = c_0(x) * \frac{1}{\sqrt{4\pi D t}} e^{-x^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{-(x-s)^2/4Dt}\mathrm{d}s
</math>
</center>

=== Solution of the dispersion equation using FFT ===

We can solve the dispersion equation using the discrete Fourier transform and
its closely related numerical implementation the FFT (Fast Fourier Transform).
We have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail.
We consider the concentration
on the finite domain <math>-L \leq x \leq L</math> and use a
[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(t) e^{\mathrm{i} k_n x}
</math>
</center>
where <math>k_n = \pi n /L </math>. If we substitute this into the diffusion equation we obtain
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x}
</math>
</center>
Note that this is not the same solution as we obtained on the infinite domain because
of the boundary conditions on the finite domain. The coefficients <math>\hat{c}_n(0)</math>
are found using the initial conditions so that
<center>
<math>
\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
</math>
</center>

The key to the numerical solution of this equation is the use of the FFT. We begin by discretising the
domain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to
approximate the integral above and obtain
<center>
<math>
\hat{c}_n(0) = \frac{1}{N} \sum_{m=0}^{N-1} e^{-\mathrm{i} k_n x_m} c_0(x_m)
</math>
</center>
<center>
<math>
= \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} e^{\mathrm{i} \pi n} c_0(x_m)
</math>
</center>

We also get
<center>
<math>
c(x_m,t) = \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
</math>
</center>
<center>
<math>
= \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
</math>
</center>
but we know that
<center>
<math>
\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
</math>
</center>
So
<center>
<math>
c(x_m,t) = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{l=0}^{N-1} c_0(x_l) e^{-2\mathrm{i} \pi nl/N} e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N}
</math>
</center>

=== The discrete Fourier transform ===
The
[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
of a sequence of ''N'' complex numbers ''c''0, ..., ''c''''N''−1 is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math>0, ..., <math>\hat{c}</math>''N''−1 by the DFT according to the formula:
<center>
<math>\hat{c}_n = \sum_{m=0}^{N-1} c_m e^{-2\pi \mathrm{i}mn/N} \quad \quad n = 0, \dots, N-1</math>
</center>

We denote the transform by the symbol <math>\mathcal{F}</math>, as in <math>\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \} </math> or <math>\mathcal{F} \left ( \mathbf{x} \right )</math> or <math>\mathcal{F} \mathbf{x}</math>.

The '''inverse discrete Fourier transform (IDFT)''' is given by
<center>
<math>c_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{c}_n e^{2\pi \mathrm{i}mn/N} \quad \quad m = 0,\dots,N-1.</math>
</center>

Therefore we can write again, similar to the continuous transform case:
<center>
<math>
c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\} = \frac{1}{\sqrt{4\pi D t}} \sum_{l=0}^{N-1} c_0(x_{(m-l)\bmod N}) \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}_l
</math>
</center>
Unlike it, though, now <math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}\ne\frac{1}{\sqrt{4\pi D t}} e^{-x_m^2/4Dt}</math>. It can be proven that for our previously defined <math>x_m = -L + 2Lm/N</math> and <math>k_{n}=\left\{
\begin{matrix}
n\pi/L,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right)\pi/L,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right. </math> we have:
<center><math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}=\dfrac{2L}{N\sqrt{4\pi DT}}\sum_{l=-\infty}^\infty \operatorname{exp}\left[ -\dfrac{(x_m+(2l+1)L)^2}{4Dt} \right]\approx\dfrac{2L}{N\sqrt{4\pi DT}}\left( \operatorname{exp}\left[ -\dfrac{(x_m+L)^2}{4Dt} \right] + \operatorname{exp}\left[ -\dfrac{(x_m-L)^2}{4Dt} \right] \right)</math></center>

The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]
or '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse)
will involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to order <math>N \log(N)</math>. This is an incredible speed up, for example
if N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT
algorithms are used so extensively.

== Reaction Diffusion Equations ==

We consider an auto catalytic reaction where the chemical species also diffuse. In this
case the equations are
<center>
<math>\partial_t a = D\partial_x^2 a - kab</math>
</center>
<center>
<math>\partial_t b = D\partial_x^2 b + kab</math>
</center>
We can non-dimensionalise these equations scaling the variables as
<center>
<math>
z = x/x^*\,\,\,\tau = t/t^*\,\,\,\alpha = a/a_0\,\,\,\beta = b/a_0
</math>
</center>
So that the equations become
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \alpha = \frac{a_0}{(x^*)^2}D\partial_z^2 \alpha
- k a_0^2 \alpha\beta
</math>
</center>
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \beta = \frac{a_0}{(x^*)^2}D\partial_z^2 \beta
+ k a_0^2 \alpha\beta
</math>
</center>
If we choose
<center>
<math>
x^* = \sqrt{\frac{D}{ka_0}}\,\,\,t^*=\frac{1}{ka_0}
</math>
</center>
then we obtain the system
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
-\alpha\beta
</math>
</center>
<center>
<math>
\partial_\tau \beta =\partial_z^2 \beta
+ \alpha\beta
</math>
</center>

=== Solution via split step method ===

[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\alpha(z,0) =1
\, \beta(z,0) = \exp(-10z^2)</math>]]

We can solve this equations numerically using a
[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume
that at time <math>\tau</math> we know <math>\alpha(z,\tau)</math>
and <math>\beta(z,\tau)</math>. We then solve first the following equation
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
</math>
</center>
from <math>\tau</math> to <math>\tau + \Delta\tau</math>
(which we can do exactly using the spectral methods just discussed for
the dispersion equation). We write this solution as
<math>\tilde{\alpha}(z,\tau + \Delta\tau)</math>
Then we solve
<center>
<math>
\partial_\tau \alpha = -\alpha\beta
</math>
</center>
by assuming that <math>\beta</math> is constant and subject to the boundary
condition that <math>\alpha(z,\tau) = \tilde{\alpha}(z,\tau + \Delta\tau)</math>.
This gives
<center>
<math>
\alpha(z,\tau + \Delta\tau) = e^{-\beta(z,\tau) \Delta\tau} \tilde{\alpha}(z,\tau+ \Delta\tau)\,
</math>
</center>
and we do likewise for the equation for <math>\beta</math>. Note that
while both steps are exact the result from the split step method is an
approximation with error which becomes smaller as the step size becomes
smaller.

We can easily implement this split step method in matlab and we obtain
a pair of travelling waves.

== Travelling Waves solution ==

When we solve the equations we found the solution formed travelling waves and
we now consider this phenomena in detail.

We define a new coordinate <math>y = z - v\tau</math> (so we will consider only
waves travelling to the right, although we could analyse waves travelling to
the left in a similar fashion). We seek stationary solutions in
<math>\alpha(y)</math> and <math>\beta(y)</math> which satisfy
<center>
<math>
\frac{\mathrm{d}^2 \alpha}{\mathrm{d}y^2} + v \frac{\mathrm{d} \alpha}{\mathrm{d}y} = \alpha\beta
</math>
</center>
and
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} = -\alpha\beta
</math>
</center>
If we add these equations we obtain
<center>
<math>
\frac{\mathrm{d}^2 (\alpha+\beta)}{\mathrm{d}y^2} + v \frac{\mathrm{d} (\alpha+\beta)}{\mathrm{d}y} = 0
</math>
</center>
so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
<math>\beta = 0</math> and if <math>y\to-\infty</math> then <math>\alpha = 0</math> and
<math>\beta = 1</math>. Therefore <math>\alpha + \beta = 1</math>.
This means that, since <math>\alpha \geq 0</math>, we must have
<math>0\leq \beta \leq 1</math>.
We can then obtain the following equation
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} + \beta(1-\beta)= 0
</math>
</center>
which we can write as the system of first order equations.

[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing
the equilibrium points and the heteroclinic connection]]

We define the variable <math>\gamma = \frac{\mathrm{d}\beta}{\mathrm{d}y}</math>
and we obtain
<center><math>
\begin{align}
\frac{\mathrm{d}\beta}{\mathrm{d} y} &= \gamma&\\
\frac{\mathrm{d}\gamma}{\mathrm{d} y} &= -v\gamma + \beta(\beta -1)& \\
\end{align}
</math></center>
This dynamical system has equilibrium points at <math>(0,0)</math>
and <math>(1,0)</math>. We can analyse these equilibrium points by
linearization. The Jacobian matrix is
<center>
<math>
J =\begin{pmatrix}
0 & 1 \\
-1 + 2\beta & -v
\end{pmatrix}
</math>
</center>

We can easily see that the Jacobian evaluated at our first equilibrium point is
<center>
<math>
J_{(0,0)} =\begin{pmatrix}
0 & 1 \\
-1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\mu_{\pm} = -1/2 (v \mp \sqrt{v^2-4})</math>. Therefore
this point is a nodal sink (possibly a spiral)

[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on
the heteroclinic connection]]

Additionally,
<center>
<math>
J_{(1,0)} =\begin{pmatrix}
0 & 1 \\
1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\lambda_{\pm} = -1/2 (v \mp \sqrt{v^2+4})</math>.
This is a a saddle point. The unstable and stable
separatrices leave the equilibrium point at <math>(1,0)</math> in the directions
<math> \begin{pmatrix}\lambda_{\pm} \\ 1\end{pmatrix}</math>. The only path on which <math>\beta</math> is bounded
as <math>y\to-\infty</math> are the unstable separatrices. Also, only the
unstable separatrix which enters the region <math>\beta<1</math> is physically meaningful.

To find a travelling wave we need to find a heteroclinic connection
between the two equilibrium points which also has to satisfy the conditions
that <math>0\leq \beta \leq 1</math>.

We need to show that the heteroclinic connection does not cross the <math>\beta</math> axis.
Consider the region
<center>
<math>
R = \left\{(\beta,\gamma)\,|\, \beta<1,\,-k\beta<\gamma<0\right\}
</math>
</center>
On the line <math>\beta = 1,d\beta/dy<0</math> and hence all flow it into <math>R</math>.
On the line <math>\gamma = 0, d\gamma/dy < 0</math> for <math>0<\beta<1</math>. On the line
<math>\gamma = -k \beta</math> we know that <math>d\beta/dy < 0</math> so that integral paths
enter the region if and only if <math>d\gamma/d\beta < \gamma/\beta</math>. We know that
<center>
<math>
d\gamma/d\beta - \gamma/\beta = -v - \frac{\beta(1-\beta)}{\gamma} -\frac{\gamma}{\beta}
= \frac{1}{k} (k^2 - vk +1 -\beta)
</math>
</center>
when <math>\gamma = -k\beta</math>. Therefore we need to find a value of <math>k</math>
so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
<math>k = \dfrac{v}{2}</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|rF4X42jP0v8}}

=== Part 2 ===

{{#ev:youtube|D0NwYlM-uOg}}

=== Part 3 ===

{{#ev:youtube|5IEZJtJaDHk}}

=== Part 4 ===

{{#ev:youtube|t_OjTSwVgdo}}

=== Part 5 ===

{{#ev:youtube|kTHVZaYezLk}}

=== Part 6 ===

{{#ev:youtube|MVuSg5_sfYI}}

=== Part 7 ===

{{#ev:youtube|hxKMOHyy6Bw}}

=== Part 8 ===

{{#ev:youtube|yQ-O2KIqu44}}

[[Category:Simple Nonlinear Waves]]

Burgers Equation

2025-11-04T08:06:38Z

Levi: /* Exact Solution of Burgers equations */ more beautiful solution form

{{nonlinear waves course
| chapter title = Burgers Equation
| next chapter =
| previous chapter = [[Reaction-Diffusion Systems]]
}}

==Introduction==

We have already met the conservation law for the traffic equations
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
</math></center>
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
<center><math>
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
</math></center>
where <math>\nu >0.</math>

The simplest equation of this type is to write
<center><math>
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
</math></center>
(changing variables to <math>u</math> and this equation is known as Burgers equation.

==Travelling Wave Solution==

We can find a travelling wave solution by assuming that
<center><math>
u\left( x,t\right) =u\left( x-ct\right) =u\left( \xi \right)
</math></center>
This leads to the equations
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
}</math> so that
<center><math>\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
\end{matrix}</math></center>
This is a degenerate system with the entire <math>u</math> axis being equilibria.

We can also solve this equation exactly as follows.
<center><math>
-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
</math></center>
can be integrated to give
<center><math>
-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime} =c_{1}
</math></center>
which can be rearranged to give
<center><math>
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu-2c_{1}\right)
</math></center>
We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
u-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
solution if we have two real roots and for the bounded solution
<math>u_{2} .
We note that the wave speed
is
<center><math>
c=\frac{1}{2}\left( u_{1}+u_{2}\right)
</math></center>
The equation can therefore be written as
<center><math>
2\nu u^{\prime }=\left( u-u_{1}\right) \left( u-u_{2}\right)
</math></center>
which has solution
<center><math>
u\left( \xi \right) =\frac{1}{2}\left( u_{1}+u_{2}\right) -\frac{1}{2}\left(
u_{1}-u_{2}\right) \tanh \left[ \left( \frac{\xi }{4\nu }\right) \left(
u_{1}-u_{2}\right) \right]
</math></center>

==Numerical Solution of Burgers equation==

We can solve the equation using our split step spectral method. The equation
can be written as
<center><math>
\partial _{t}u=-\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
</math></center>
We solve this by solving in Fourier space to give
<center><math>
\partial _{t}\hat{u}=-\frac{1}{2}ik \widehat{\left( u^{2}\right)} -\nu k^{2}\hat{u}
</math></center>
Then we solve each of the steps in turn
for a small time interval to give
<center><math>\begin{matrix}
\tilde{u}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right) -\frac{
\Delta t}{2}ik\mathcal{F}\left( \left[ \mathcal{F}^{-1}\hat{u}\left(
k,t\right) \right] ^{2}\right) \\
\hat{u}\left( k,t+\Delta t\right) &=&\tilde{u}\left( k,t+\Delta t\right)
\exp \left( -\nu k^{2}\Delta t\right)
\end{matrix}</math></center>

{| class="wikitable"
|-
! Phase plane for a travelling wave solution
! Numerical solution of Burgers equation
|-
| [[Image:Burgers_phase.jpg|thumb|right|500px|Phase plane for a travelling wave solution of Burgers equation]]
| [[Image:File-Burgers2.gif|thumb|right|500px| Numerical solution of Burgers equation]]

|}

==Exact Solution of Burgers equations==

We can find an exact solution to Burgers equation. We want to solve
<center><math>\begin{matrix}
\partial _{t}u+u\partial _{x}u &=&\nu \partial _{x}^{2}u \\
u\left( x,0\right) &=&F\left( x\right)
\end{matrix}</math></center>
Frist we write the equation as
<center><math>
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}-\nu \partial _{x}u\right)
=0
</math></center>
We want to find a function <math>\psi \left( x,t\right) </math> such that
<center><math>
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u-\frac{u^{2}}{2
}
</math></center>
Note that because <math>\partial _{x}\partial _{t}\psi =\partial _{t}\partial
_{x}\psi </math> we will satisfy Burgers equation. This gives us the following
equation for <math>\psi </math>
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
We introduce the ''Cole-Hopf '' transformation
<center><math>
\psi =-2\nu \log \left( \phi \right)
</math></center>
From this we can obtain the three results:
<center><math>
\begin{align}
\partial _{x}\psi &=-2\nu \frac{\partial _{x}\phi }{\phi } \\
\partial _{x}^{2}\psi &=2\nu \left( \frac{\partial _{x}\phi }{\phi }\right)
^{2}-\frac{2\nu }{\phi }\partial _{x}^{2}\phi \\
\partial _{t}\psi &=-2\nu \frac{\partial _{t}\phi }{\phi }
\end{align}
</math></center>
Therefore
<center><math>
\partial _{t}\psi =\nu \partial _{x}^{2}\psi -\frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
</math></center>
becomes
<center><math>
-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
_{x}\phi }{\phi }\right) ^{2}
-2\nu^2 \frac{\partial_x^2\phi}{\phi}
-\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
}{\phi }\right) ^{2}
</math></center>
or
<center><math>
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
</math></center>
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
<center><math>
F\left( x\right) =u\left( x,0\right) =-2\nu \frac{\partial _{x}\phi \left(
x,0\right) }{\phi \left( x,0\right) }
</math></center>
We can write this as
<center><math>
\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
x\right)
</math></center>
which has solution
<center><math>
\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
\int_{0}^{x}F\left( s\right) \mathrm{d}s\right)
</math></center>
We need to solve
<center><math>\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}</math></center>
We take the Fourier transform and obtain
<center><math>\begin{matrix}
\partial _{t}\hat{\phi} &=&-k^{2}\nu \hat{\phi} \\
\hat{\phi}\left( k,0\right) &=&\hat{\Phi}\left( k\right)
\end{matrix}</math></center>
which has solution
<center><math>
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{-k^{2}\nu t}
</math></center>
We can then use the convolution theorem to write
<center><math>\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
e^{-k^{2}\nu t}\right] \\
&=&\Phi \left( x\right) * \dfrac1{\sqrt{4\pi\nu t}} e^{-x^2 / 4\nu t} \\
&=&\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
\end{matrix}</math></center>
Which can be expressed as
<center><math>
\phi \left( x,t\right) =\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty
}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
</math></center>
where
<center><math>
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
\left( x-y\right) ^{2}}{2t}
</math></center>
To find <math>u</math> we recall that
<center><math>\begin{matrix}
u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&-2\nu \dfrac{\partial _{x}\left( \Phi \left( x\right) * e^{-x^2 / 4\nu t} \right) }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } \\
&=&-2\nu \dfrac{\partial _{x} \Phi \left( x\right) * e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } = \dfrac{\Phi \left( x\right) * x e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * t e^{-x^2 / 4\nu t} } \\
&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
2\nu }\right] \mathrm{d}y}
\end{matrix}</math></center>
as <math>\partial_x(f * y)=\partial_x f * y = f * \partial_x y</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|tVXQmxOG_6Y}}

=== Part 2 ===

{{#ev:youtube|hzgpMM_wWts}}

=== Part 3 ===

{{#ev:youtube|uH4B1XsGB-0}}

=== Part 4 ===

{{#ev:youtube|h6aDmCtJygM}}

=== Part 5 ===

{{#ev:youtube|CsnUKrLjtyQ}}

Reaction-Diffusion Systems

2025-11-04T07:23:12Z

Levi: /* Diffusion */ Fixes + discrete convolution

{{nonlinear waves course
| chapter title = Reaction-Diffusion Systems
| next chapter = [[Burgers Equation]]
| previous chapter = [[Example Calculations for the KdV and IST]]
}}

{{complete pages}}

We present here a brief theory of reaction diffusion waves.

== Law of Mass Action ==

The law of mass action states that equation rates are proportional to the concentration
of reacting species and the ratio in which they combined. It is discussed in detail in
[[Billingham and King 2000]]. We will present here a few simple examples.

=== Example 1: Simple Decay ===

Suppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.
<center>
<math>P \to A</math>
</center>
with rate <math>k[P]</math> where <math>[P]</math> denotes concentration. Then if we
set <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations
<center>
<math> \frac{\mathrm{d}p}{\mathrm{d}t} = -kp\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}a}{\mathrm{d}t} = kp</math>
</center>
which has solution
<center>
<math>
p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt})
</math>
</center>
where <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>
<math>p</math> repectively at <math>t=0</math>.

=== Example 2: Quadratic Autocatalysis ===

This example will be important when we consider reaction diffusion problems.
We consider the reaction
<center>
<math>A + B \to 2B</math>
</center>
with rate proportional to <math>k[A][B]</math>. If we define <math>a = [A]</math>
and <math>b = [B]</math> we obtain the following equations
<center>
<math> \frac{\mathrm{d}a}{\mathrm{d}t} = -kab\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}b}{\mathrm{d}t} = kab</math>
</center>
We can solve these equations by observing that
<center>
<math>
\frac{\mathrm{d}(a+b)}{\mathrm{d}t} = 0
</math>
</center>
so that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain
<center>
<math>
\frac{\mathrm{d}b}{\mathrm{d}t} = k(a_0 + b_0 - b)b
</math>
</center>
which is separable with solution
<center>
<math>
b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
and
<center>
<math>
a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
Note that <math>a\to 0</math> and <math>b\to a_0 + b_0</math>
as <math>t\to \infty.</math>

== Diffusion ==

The equation for spatially homogeneous diffusion of a chemical with concentration
<math>c</math> is
<center>
<math>
\partial_t c = D\nabla^2 c
</math>
</center>
which is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in
only one spatial dimension. Consider it on the boundary <math>-\infty < x < \infty</math>. In this case
we can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain
<center>
<math>
\partial_t \hat{c} = -D k^2 \hat{c}
</math>
</center>
where <math>\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution
<center>
<math>
\hat{c} = \hat{c}_0 e^{-D k^2 t}
</math>
</center>
We can find the inverse transform using convolution and obtain
<center>
<math>
c(x,t) = \mathcal{F}^{-1} \left\{ e^{-k^2 D t} \mathcal{F} \left\{ c_0(x) \right\} \right\} = c_0(x) * \frac{1}{\sqrt{4\pi D t}} e^{-x^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{-(x-s)^2/4Dt}\mathrm{d}s
</math>
</center>

=== Solution of the dispersion equation using FFT ===

We can solve the dispersion equation using the discrete Fourier transform and
its closely related numerical implementation the FFT (Fast Fourier Transform).
We have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail.
We consider the concentration
on the finite domain <math>-L \leq x \leq L</math> and use a
[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(t) e^{\mathrm{i} k_n x}
</math>
</center>
where <math>k_n = \pi n /L </math>. If we substitute this into the diffusion equation we obtain
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x}
</math>
</center>
Note that this is not the same solution as we obtained on the infinite domain because
of the boundary conditions on the finite domain. The coefficients <math>\hat{c}_n(0)</math>
are found using the initial conditions so that
<center>
<math>
\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
</math>
</center>

The key to the numerical solution of this equation is the use of the FFT. We begin by discretising the
domain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to
approximate the integral above and obtain
<center>
<math>
\hat{c}_n(0) = \frac{1}{N} \sum_{m=0}^{N-1} e^{-\mathrm{i} k_n x_m} c_0(x_m)
</math>
</center>
<center>
<math>
= \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} e^{\mathrm{i} \pi n} c_0(x_m)
</math>
</center>

We also get
<center>
<math>
c(x_m,t) = \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
</math>
</center>
<center>
<math>
= \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
</math>
</center>
but we know that
<center>
<math>
\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
</math>
</center>
So
<center>
<math>
c(x_m,t) = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{l=0}^{N-1} c_0(x_l) e^{-2\mathrm{i} \pi nl/N} e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N}
</math>
</center>

=== The discrete Fourier transform ===
The
[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
of a sequence of ''N'' complex numbers ''c''0, ..., ''c''''N''−1 is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math>0, ..., <math>\hat{c}</math>''N''−1 by the DFT according to the formula:
<center>
<math>\hat{c}_n = \sum_{m=0}^{N-1} c_m e^{-2\pi \mathrm{i}mn/N} \quad \quad n = 0, \dots, N-1</math>
</center>

We denote the transform by the symbol <math>\mathcal{F}</math>, as in <math>\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \} </math> or <math>\mathcal{F} \left ( \mathbf{x} \right )</math> or <math>\mathcal{F} \mathbf{x}</math>.

The '''inverse discrete Fourier transform (IDFT)''' is given by
<center>
<math>c_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{c}_n e^{2\pi \mathrm{i}mn/N} \quad \quad m = 0,\dots,N-1.</math>
</center>

Therefore we can write again, similar to the continuous transform case (with the only difficulty that we need to define carefully the values of
<math>k_n</math>):
<center>
<math>
c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \frac{1}{\sqrt{4\pi D t}} e^{-x_m^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \sum_{n=0}^{N-1} c_0(x_{(m-n)\bmod N}) e^{-x_{n}^2/4Dt}
</math>
</center>

The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]
or '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse)
will involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to order <math>N \log(N)</math>. This is an incredible speed up, for example
if N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT
algorithms are used so extensively.

== Reaction Diffusion Equations ==

We consider an auto catalytic reaction where the chemical species also diffuse. In this
case the equations are
<center>
<math>\partial_t a = D\partial_x^2 a - kab</math>
</center>
<center>
<math>\partial_t b = D\partial_x^2 b + kab</math>
</center>
We can non-dimensionalise these equations scaling the variables as
<center>
<math>
z = x/x^*\,\,\,\tau = t/t^*\,\,\,\alpha = a/a_0\,\,\,\beta = b/a_0
</math>
</center>
So that the equations become
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \alpha = \frac{a_0}{(x^*)^2}D\partial_z^2 \alpha
- k a_0^2 \alpha\beta
</math>
</center>
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \beta = \frac{a_0}{(x^*)^2}D\partial_z^2 \beta
+ k a_0^2 \alpha\beta
</math>
</center>
If we choose
<center>
<math>
x^* = \sqrt{\frac{D}{ka_0}}\,\,\,t^*=\frac{1}{ka_0}
</math>
</center>
then we obtain the system
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
-\alpha\beta
</math>
</center>
<center>
<math>
\partial_\tau \beta =\partial_z^2 \beta
+ \alpha\beta
</math>
</center>

=== Solution via split step method ===

[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\alpha(z,0) =1
\, \beta(z,0) = \exp(-10z^2)</math>]]

We can solve this equations numerically using a
[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume
that at time <math>\tau</math> we know <math>\alpha(z,\tau)</math>
and <math>\beta(z,\tau)</math>. We then solve first the following equation
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
</math>
</center>
from <math>\tau</math> to <math>\tau + \Delta\tau</math>
(which we can do exactly using the spectral methods just discussed for
the dispersion equation). We write this solution as
<math>\tilde{\alpha}(z,\tau + \Delta\tau)</math>
Then we solve
<center>
<math>
\partial_\tau \alpha = -\alpha\beta
</math>
</center>
by assuming that <math>\beta</math> is constant and subject to the boundary
condition that <math>\alpha(z,\tau) = \tilde{\alpha}(z,\tau + \Delta\tau)</math>.
This gives
<center>
<math>
\alpha(z,\tau + \Delta\tau) = e^{-\beta(z,\tau) \Delta\tau} \tilde{\alpha}(z,\tau+ \Delta\tau)\,
</math>
</center>
and we do likewise for the equation for <math>\beta</math>. Note that
while both steps are exact the result from the split step method is an
approximation with error which becomes smaller as the step size becomes
smaller.

We can easily implement this split step method in matlab and we obtain
a pair of travelling waves.

== Travelling Waves solution ==

When we solve the equations we found the solution formed travelling waves and
we now consider this phenomena in detail.

We define a new coordinate <math>y = z - v\tau</math> (so we will consider only
waves travelling to the right, although we could analyse waves travelling to
the left in a similar fashion). We seek stationary solutions in
<math>\alpha(y)</math> and <math>\beta(y)</math> which satisfy
<center>
<math>
\frac{\mathrm{d}^2 \alpha}{\mathrm{d}y^2} + v \frac{\mathrm{d} \alpha}{\mathrm{d}y} = \alpha\beta
</math>
</center>
and
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} = -\alpha\beta
</math>
</center>
If we add these equations we obtain
<center>
<math>
\frac{\mathrm{d}^2 (\alpha+\beta)}{\mathrm{d}y^2} + v \frac{\mathrm{d} (\alpha+\beta)}{\mathrm{d}y} = 0
</math>
</center>
so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
<math>\beta = 0</math> and if <math>y\to-\infty</math> then <math>\alpha = 0</math> and
<math>\beta = 1</math>. Therefore <math>\alpha + \beta = 1</math>.
This means that, since <math>\alpha \geq 0</math>, we must have
<math>0\leq \beta \leq 1</math>.
We can then obtain the following equation
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} + \beta(1-\beta)= 0
</math>
</center>
which we can write as the system of first order equations.

[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing
the equilibrium points and the heteroclinic connection]]

We define the variable <math>\gamma = \frac{\mathrm{d}\beta}{\mathrm{d}y}</math>
and we obtain
<center><math>
\begin{align}
\frac{\mathrm{d}\beta}{\mathrm{d} y} &= \gamma&\\
\frac{\mathrm{d}\gamma}{\mathrm{d} y} &= -v\gamma + \beta(\beta -1)& \\
\end{align}
</math></center>
This dynamical system has equilibrium points at <math>(0,0)</math>
and <math>(1,0)</math>. We can analyse these equilibrium points by
linearization. The Jacobian matrix is
<center>
<math>
J =\begin{pmatrix}
0 & 1 \\
-1 + 2\beta & -v
\end{pmatrix}
</math>
</center>

We can easily see that the Jacobian evaluated at our first equilibrium point is
<center>
<math>
J_{(0,0)} =\begin{pmatrix}
0 & 1 \\
-1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\mu_{\pm} = -1/2 (v \mp \sqrt{v^2-4})</math>. Therefore
this point is a nodal sink (possibly a spiral)

[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on
the heteroclinic connection]]

Additionally,
<center>
<math>
J_{(1,0)} =\begin{pmatrix}
0 & 1 \\
1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\lambda_{\pm} = -1/2 (v \mp \sqrt{v^2+4})</math>.
This is a a saddle point. The unstable and stable
separatrices leave the equilibrium point at <math>(1,0)</math> in the directions
<math> \begin{pmatrix}\lambda_{\pm} \\ 1\end{pmatrix}</math>. The only path on which <math>\beta</math> is bounded
as <math>y\to-\infty</math> are the unstable separatrices. Also, only the
unstable separatrix which enters the region <math>\beta<1</math> is physically meaningful.

To find a travelling wave we need to find a heteroclinic connection
between the two equilibrium points which also has to satisfy the conditions
that <math>0\leq \beta \leq 1</math>.

We need to show that the heteroclinic connection does not cross the <math>\beta</math> axis.
Consider the region
<center>
<math>
R = \left\{(\beta,\gamma)\,|\, \beta<1,\,-k\beta<\gamma<0\right\}
</math>
</center>
On the line <math>\beta = 1,d\beta/dy<0</math> and hence all flow it into <math>R</math>.
On the line <math>\gamma = 0, d\gamma/dy < 0</math> for <math>0<\beta<1</math>. On the line
<math>\gamma = -k \beta</math> we know that <math>d\beta/dy < 0</math> so that integral paths
enter the region if and only if <math>d\gamma/d\beta < \gamma/\beta</math>. We know that
<center>
<math>
d\gamma/d\beta - \gamma/\beta = -v - \frac{\beta(1-\beta)}{\gamma} -\frac{\gamma}{\beta}
= \frac{1}{k} (k^2 - vk +1 -\beta)
</math>
</center>
when <math>\gamma = -k\beta</math>. Therefore we need to find a value of <math>k</math>
so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
<math>k = \dfrac{v}{2}</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|rF4X42jP0v8}}

=== Part 2 ===

{{#ev:youtube|D0NwYlM-uOg}}

=== Part 3 ===

{{#ev:youtube|5IEZJtJaDHk}}

=== Part 4 ===

{{#ev:youtube|t_OjTSwVgdo}}

=== Part 5 ===

{{#ev:youtube|kTHVZaYezLk}}

=== Part 6 ===

{{#ev:youtube|MVuSg5_sfYI}}

=== Part 7 ===

{{#ev:youtube|hxKMOHyy6Bw}}

=== Part 8 ===

{{#ev:youtube|yQ-O2KIqu44}}

[[Category:Simple Nonlinear Waves]]

Reaction-Diffusion Systems

2025-10-31T11:34:54Z

Levi: /* Reaction Diffusion Equations */

{{nonlinear waves course
| chapter title = Reaction-Diffusion Systems
| next chapter = [[Burgers Equation]]
| previous chapter = [[Example Calculations for the KdV and IST]]
}}

{{complete pages}}

We present here a brief theory of reaction diffusion waves.

== Law of Mass Action ==

The law of mass action states that equation rates are proportional to the concentration
of reacting species and the ratio in which they combined. It is discussed in detail in
[[Billingham and King 2000]]. We will present here a few simple examples.

=== Example 1: Simple Decay ===

Suppose we have of chemical <math>P</math> which decays to <math>A</math>, i.e.
<center>
<math>P \to A</math>
</center>
with rate <math>k[P]</math> where <math>[P]</math> denotes concentration. Then if we
set <math>p=[P]</math> and <math>a = [A] </math> we obtain the equations
<center>
<math> \frac{\mathrm{d}p}{\mathrm{d}t} = -kp\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}a}{\mathrm{d}t} = kp</math>
</center>
which has solution
<center>
<math>
p = p_0 e^{-kt}\,\,\,\textrm{and}\,\,\, a = a_0 + p_0(1-e^{-kt})
</math>
</center>
where <math>a_0</math> and <math>p_0</math> are the values of <math>a</math>
<math>p</math> repectively at <math>t=0</math>.

=== Example 2: Quadratic Autocatalysis ===

This example will be important when we consider reaction diffusion problems.
We consider the reaction
<center>
<math>A + B \to 2B</math>
</center>
with rate proportional to <math>k[A][B]</math>. If we define <math>a = [A]</math>
and <math>b = [B]</math> we obtain the following equations
<center>
<math> \frac{\mathrm{d}a}{\mathrm{d}t} = -kab\,\,\,\textrm{and}\,\,\, \frac{\mathrm{d}b}{\mathrm{d}t} = kab</math>
</center>
We can solve these equations by observing that
<center>
<math>
\frac{\mathrm{d}(a+b)}{\mathrm{d}t} = 0
</math>
</center>
so that <math>a + b = a_0 + b_0</math>. We can then eliminate <math>a</math> to obtain
<center>
<math>
\frac{\mathrm{d}b}{\mathrm{d}t} = k(a_0 + b_0 - b)b
</math>
</center>
which is separable with solution
<center>
<math>
b = \frac{b_0(a_0 + b_0)e^{k(a_0 + b_0)t}}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
and
<center>
<math>
a = \frac{a_0(a_0 + b_0)}{a_0 + b_0e^{k(a_0 + b_0)t}}
</math>
</center>
Note that <math>a\to 0</math> and <math>b\to a_0 + b_0</math>
as <math>t\to \infty.</math>

== Diffusion ==

The equation for spatially homogeneous diffusion of a chemical with concentration
<math>c</math> is
<center>
<math>
\partial_t c = D\nabla^2 c
</math>
</center>
which is the [http://en.wikipedia.org/wiki/Heat_equation heat equation]. We will consider this in
only one spatial dimension. Consider it on the boundary <math>-\infty < x < \infty</math>. In this case
we can solve by the [http://en.wikipedia.org/wiki/Fourier_transform Fourier transform] and obtain
<center>
<math>
\partial_t \hat{c} = -D k^2 \hat{c}
</math>
</center>
where <math>\hat{c}</math> is the Fourier transform of <math>c</math>. This has solution
<center>
<math>
\hat{c} = \hat{c}_0 e^{-D k^2 t}
</math>
</center>
We can find the inverse transform using convolution and obtain
<center>
<math>
c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}\mathrm{d}s
</math>
</center>

=== Solution of the dispersion equation using FFT ===

We can solve the dispersion equation using the discrete Fourier transform and
its closely related numerical implementation the FFT (Fast Fourier Transform).
We have already met the FFT [[Numerical Solution of the KdV]] but we consider it here in more detail.
We consider the concentration
on the finite domain <math>-L \leq x \leq L</math> and use a
[http://en.wikipedia.org/wiki/Fourier_series Fourier series] expansion
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(t) e^{\mathrm{i} k_n x}
</math>
</center>
where <math>k_n = \pi n /L </math>. If we substitute this into the diffusion equation we obtain
<center>
<math>
c(x,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0)e^{-k_n^2 D t} e^{\mathrm{i} k_n x}
</math>
</center>
Note that this is not the same solution as we obtained on the infinite domain because
of the boundary conditions on the finite domain. The coefficients <math>\hat{c}_n(0)</math>
are found using the initial conditions so that
<center>
<math>
\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
</math>
</center>

The key to the numerical solution of this equation is the use of the FFT. We begin by discretising the
domain into a series of <math>N</math> points <math>x_m = -L + 2Lm/N </math>. We then use this to
approximate the integral above and obtain
<center>
<math>
\hat{c}_n(0) = \frac{1}{N} \sum_{m=0}^{N-1} e^{-\mathrm{i} k_n x_m} c_0(x_m)
</math>
</center>
<center>
<math>
= \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} e^{\mathrm{i} \pi n} c_0(x_m)
</math>
</center>

We also get
<center>
<math>
c(x_m,t) = \sum_{n=-\infty}^{\infty} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
</math>
</center>
<center>
<math>
= \sum_{n=-\infty}^{\infty} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
</math>
</center>
but we know that
<center>
<math>
\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
</math>
</center>

=== The discrete Fourier transform ===
The
[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
of a sequence of ''2N'' complex numbers ''c''0, ..., ''c''''2N''−1 is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math>0, ..., <math>\hat{c}</math>''N''−1 by the DFT according to the formula:
<center>
<math>\hat{c}_m = \sum_{n=0}^{N-1} c_n e^{-2\pi \mathrm{i}mn/N} \quad \quad m = 0, \dots, N-1</math>
</center>

We denote the transform by the symbol <math>\mathcal{F}</math>, as in <math>\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \} </math> or <math>\mathcal{F} \left ( \mathbf{x} \right )</math> or <math>\mathcal{F} \mathbf{x}</math>.

The '''inverse discrete Fourier transform (IDFT)''' is given by
<center>
<math>c_n = \frac{1}{N} \sum_{m=0}^{N-1} \hat{c}_m e^{2\pi \mathrm{i}mn/N} \quad \quad n = 0,\dots,N-1.</math>
</center>

Therefore we can write
<center>
<math>
c(x_m,t) = \mathcal{F} \left\{ e^{-k_n^2 D t} \mathcal{F}^{-1} \left\{ c_0(x_m) \right\}
\right\}
</math>
</center>
Note that the choice of where to put the <math>1/N</math> is arbitrary in the definition of FFT and IFFT and
does not exactly match here. Of course since it appears once the formula above is correct regardless.
The only difficulty is that we need to define carefully the values of
<math>k_n</math>

The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]
or '''FFT''' algorithm. A naive implementation of the discrete Fourier transform above (or its inverse)
will involve order <math>N^2</math> operations. Using FFT algorithms, this can be reduced to order <math>N \log(N)</math>. This is an incredible speed up, for example
if N = 1024, FFT algorithms are more efficient by a factor of 147. This is the reason FFT
algorithms are used so extensively.

== Reaction Diffusion Equations ==

We consider an auto catalytic reaction where the chemical species also diffuse. In this
case the equations are
<center>
<math>\partial_t a = D\partial_x^2 a - kab</math>
</center>
<center>
<math>\partial_t b = D\partial_x^2 b + kab</math>
</center>
We can non-dimensionalise these equations scaling the variables as
<center>
<math>
z = x/x^*\,\,\,\tau = t/t^*\,\,\,\alpha = a/a_0\,\,\,\beta = b/a_0
</math>
</center>
So that the equations become
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \alpha = \frac{a_0}{(x^*)^2}D\partial_z^2 \alpha
- k a_0^2 \alpha\beta
</math>
</center>
<center>
<math>
\frac{a_0}{t^*}\partial_\tau \beta = \frac{a_0}{(x^*)^2}D\partial_z^2 \beta
+ k a_0^2 \alpha\beta
</math>
</center>
If we choose
<center>
<math>
x^* = \sqrt{\frac{D}{ka_0}}\,\,\,t^*=\frac{1}{ka_0}
</math>
</center>
then we obtain the system
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
-\alpha\beta
</math>
</center>
<center>
<math>
\partial_\tau \beta =\partial_z^2 \beta
+ \alpha\beta
</math>
</center>

=== Solution via split step method ===

[[Image:Reaction diffusion.gif|thumb|right|500px|Solutions for <math>\alpha(z,0) =1
\, \beta(z,0) = \exp(-10z^2)</math>]]

We can solve this equations numerically using a
[http://en.wikipedia.org/wiki/Split-step_method split step method]. We assume
that at time <math>\tau</math> we know <math>\alpha(z,\tau)</math>
and <math>\beta(z,\tau)</math>. We then solve first the following equation
<center>
<math>
\partial_\tau \alpha = \partial_z^2 \alpha
</math>
</center>
from <math>\tau</math> to <math>\tau + \Delta\tau</math>
(which we can do exactly using the spectral methods just discussed for
the dispersion equation). We write this solution as
<math>\tilde{\alpha}(z,\tau + \Delta\tau)</math>
Then we solve
<center>
<math>
\partial_\tau \alpha = -\alpha\beta
</math>
</center>
by assuming that <math>\beta</math> is constant and subject to the boundary
condition that <math>\alpha(z,\tau) = \tilde{\alpha}(z,\tau + \Delta\tau)</math>.
This gives
<center>
<math>
\alpha(z,\tau + \Delta\tau) = e^{-\beta(z,\tau) \Delta\tau} \tilde{\alpha}(z,\tau+ \Delta\tau)\,
</math>
</center>
and we do likewise for the equation for <math>\beta</math>. Note that
while both steps are exact the result from the split step method is an
approximation with error which becomes smaller as the step size becomes
smaller.

We can easily implement this split step method in matlab and we obtain
a pair of travelling waves.

== Travelling Waves solution ==

When we solve the equations we found the solution formed travelling waves and
we now consider this phenomena in detail.

We define a new coordinate <math>y = z - v\tau</math> (so we will consider only
waves travelling to the right, although we could analyse waves travelling to
the left in a similar fashion). We seek stationary solutions in
<math>\alpha(y)</math> and <math>\beta(y)</math> which satisfy
<center>
<math>
\frac{\mathrm{d}^2 \alpha}{\mathrm{d}y^2} + v \frac{\mathrm{d} \alpha}{\mathrm{d}y} = \alpha\beta
</math>
</center>
and
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} = -\alpha\beta
</math>
</center>
If we add these equations we obtain
<center>
<math>
\frac{\mathrm{d}^2 (\alpha+\beta)}{\mathrm{d}y^2} + v \frac{\mathrm{d} (\alpha+\beta)}{\mathrm{d}y} = 0
</math>
</center>
so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
<math>\beta = 0</math> and if <math>y\to-\infty</math> then <math>\alpha = 0</math> and
<math>\beta = 1</math>. Therefore <math>\alpha + \beta = 1</math>.
This means that, since <math>\alpha \geq 0</math>, we must have
<math>0\leq \beta \leq 1</math>.
We can then obtain the following equation
<center>
<math>
\frac{\mathrm{d}^2 \beta}{\mathrm{d}y^2} + v \frac{\mathrm{d} \beta}{\mathrm{d}y} + \beta(1-\beta)= 0
</math>
</center>
which we can write as the system of first order equations.

[[Image:R_d_phase_portrait.jpg|thumb|right|500px|Phase portrait for out system showing
the equilibrium points and the heteroclinic connection]]

We define the variable <math>\gamma = \frac{\mathrm{d}\beta}{\mathrm{d}y}</math>
and we obtain
<center><math>
\begin{align}
\frac{\mathrm{d}\beta}{\mathrm{d} y} &= \gamma&\\
\frac{\mathrm{d}\gamma}{\mathrm{d} y} &= -v\gamma + \beta(\beta -1)& \\
\end{align}
</math></center>
This dynamical system has equilibrium points at <math>(0,0)</math>
and <math>(1,0)</math>. We can analyse these equilibrium points by
linearization. The Jacobian matrix is
<center>
<math>
J =\begin{pmatrix}
0 & 1 \\
-1 + 2\beta & -v
\end{pmatrix}
</math>
</center>

We can easily see that the Jacobian evaluated at our first equilibrium point is
<center>
<math>
J_{(0,0)} =\begin{pmatrix}
0 & 1 \\
-1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\mu_{\pm} = -1/2 (v \mp \sqrt{v^2-4})</math>. Therefore
this point is a nodal sink (possibly a spiral)

[[Image:R_d_wave.gif|right|Travelling wave solution for v=2 and the position on
the heteroclinic connection]]

Additionally,
<center>
<math>
J_{(1,0)} =\begin{pmatrix}
0 & 1 \\
1 & -v
\end{pmatrix}
</math>
</center>
which has eigenvalues <math>\lambda_{\pm} = -1/2 (v \mp \sqrt{v^2+4})</math>.
This is a a saddle point. The unstable and stable
separatrices leave the equilibrium point at <math>(1,0)</math> in the directions
<math> \begin{pmatrix}\lambda_{\pm} \\ 1\end{pmatrix}</math>. The only path on which <math>\beta</math> is bounded
as <math>y\to-\infty</math> are the unstable separatrices. Also, only the
unstable separatrix which enters the region <math>\beta<1</math> is physically meaningful.

To find a travelling wave we need to find a heteroclinic connection
between the two equilibrium points which also has to satisfy the conditions
that <math>0\leq \beta \leq 1</math>.

We need to show that the heteroclinic connection does not cross the <math>\beta</math> axis.
Consider the region
<center>
<math>
R = \left\{(\beta,\gamma)\,|\, \beta<1,\,-k\beta<\gamma<0\right\}
</math>
</center>
On the line <math>\beta = 1,d\beta/dy<0</math> and hence all flow it into <math>R</math>.
On the line <math>\gamma = 0, d\gamma/dy < 0</math> for <math>0<\beta<1</math>. On the line
<math>\gamma = -k \beta</math> we know that <math>d\beta/dy < 0</math> so that integral paths
enter the region if and only if <math>d\gamma/d\beta < \gamma/\beta</math>. We know that
<center>
<math>
d\gamma/d\beta - \gamma/\beta = -v - \frac{\beta(1-\beta)}{\gamma} -\frac{\gamma}{\beta}
= \frac{1}{k} (k^2 - vk +1 -\beta)
</math>
</center>
when <math>\gamma = -k\beta</math>. Therefore we need to find a value of <math>k</math>
so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
<math>k = \dfrac{v}{2}</math>.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|rF4X42jP0v8}}

=== Part 2 ===

{{#ev:youtube|D0NwYlM-uOg}}

=== Part 3 ===

{{#ev:youtube|5IEZJtJaDHk}}

=== Part 4 ===

{{#ev:youtube|t_OjTSwVgdo}}

=== Part 5 ===

{{#ev:youtube|kTHVZaYezLk}}

=== Part 6 ===

{{#ev:youtube|MVuSg5_sfYI}}

=== Part 7 ===

{{#ev:youtube|hxKMOHyy6Bw}}

=== Part 8 ===

{{#ev:youtube|yQ-O2KIqu44}}

[[Category:Simple Nonlinear Waves]]

Connection betwen KdV and the Schrodinger Equation

2025-10-05T08:42:21Z

Levi: /* Single Soliton Example */

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{nm} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This means
<center><math>
v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_m
</math></center>
Since <math>\partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n</math>, this gives us
<center><math>\begin{align}
K\left( x,x,t\right) =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\
=-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\
=&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\
=\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\
=\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\
=\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
\end{align}
</math></center>
And therefore
<center><math>
u\left( x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
</math></center>

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/(2k_1)\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\operatorname{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k_{1}^{3}t-\alpha/2</math> and <math>\sqrt{2k_{1}}e^{-\alpha/2}=e^{-k_{1}x_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Conservation Laws for the KdV

2025-10-04T09:26:05Z

Levi: /* Proof of an Infinite Number of Conservation Laws */ fix

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting features of the KdV is the existence of
infinitely many conservation laws. Let's begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since we can rewrite the equation into the conservation form <math>\partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0</math>, then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u + 2 u^2)
</math></center>
<center><math>
w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just an <math>\partial _{x}</math> of some <math>X(u)</math> and therefore does not actually provide a conservation law.

As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>\int_{-\infty }^{\infty }u\mathrm{d} x</math> (mass), <math>\int_{-\infty }^{\infty }u^2\mathrm{d} x</math> (momentum), and <math>\int_{-\infty }^{\infty }(2 u^3 - (\partial_{x} u)^2)\mathrm{d} x</math> (energy).

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Connection betwen KdV and the Schrodinger Equation

2025-09-30T06:50:39Z

Levi: /* Reflectionless Potential */ this can actually be solved nicely (Dunajski 2009)

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{nm} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This means
<center><math>
v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_m
</math></center>
Since <math>\partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n</math>, this gives us
<center><math>\begin{align}
K\left( x,x,t\right) =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\
=-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\
=&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\
=\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\
=\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\
=\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
\end{align}
</math></center>
And therefore
<center><math>
u\left( x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
</math></center>

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\operatorname{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Example Calculations for the KdV and IST

2025-09-30T06:01:51Z

Levi: /* Example 2: Hat Function Potential */

{{nonlinear waves course
| chapter title = Example Calculations for the KdV and IST
| next chapter = [[Reaction-Diffusion Systems]]
| previous chapter = [[Connection betwen KdV and the Schrodinger Equation]]
}}

We consider here the two examples we treated in [[Properties of the Linear Schrodinger Equation]].

==Example1: <math>\delta</math> function potential==

We have already calculated the scattering data for the delta function
potential in [[Properties of the Linear Schrodinger Equation]]. The scattering data is
<center><math>
S\left( \lambda,0\right) =\left( k_{1},\sqrt{k_{1}},\frac{u_{0}}{2ik-u_{0}
},\frac{2ik}{2ik-u_{0}}\right)
</math></center>
The spectral data evolves as
<center><math>
k_{1}=k_{1}
</math></center>
<center><math>
c_{1}\left( t\right) =c_{1}\left( 0\right) e^{4k_{1}^{3}t}=\sqrt{k_{1}
}e^{4k_{1}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
so that
<center><math>
S\left( \lambda,t\right) =\left( k_{1},\sqrt{k_{1}}e^{4k_{1}^{3}t}
,\frac{u_{0}}{2ik-u_{0}}e^{8ik^{3}t},\frac{2ik}{2ik-u_{0}}\right)
</math></center>

==Example 2: Hat Function Potential==

We solve for the case when
<center><math>
u\left( x\right) =\left\{
\begin{matrix}
0, & x\notin\left[ -1,1\right] \\
20, & x\in\left[ -1,1\right]
\end{matrix}
\right.
</math></center>
We have already solved this case in [[Properties of the Linear Schrodinger Equation]].
For the even solutions we need to solve
<center><math>
\tan\kappa=\frac{k}{\kappa}
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math>.

For the odd solutions we need to solve and
<center><math>
\tan\kappa=-\frac{\kappa}{k}
</math></center>

Recall that the solitons have amplitude <math>2k_{n}^{2}</math> or <math>-2\lambda_{n}</math>. This
can be seen in the height of the solitary waves.

We cannot work with a hat function numerically, because the jump in <math>u</math> leads
to high frequencies which dominate the response. We can smooth our function
by a number of methods. We use here the function <math>\tanh\left( x\right) </math> so
we write
<center><math>
u\left( x\right) =\frac{20}{2}\left( \tanh\left( \nu\left( x+1\right)
\right) -\tanh\left( \nu\left( x-1\right) \right) \right)
</math></center>
where <math>\nu</math> is an appropriate constant to make the function increase in value
sufficiently rapidly but not too rapidly.

{| class="wikitable"
|-
! Animation
! Three-dimensional plot
|-
| [[Image:Wide_function.gif|thumb|right|500px|Evolution of <math>u(x,t)</math>.]]
| [[Image:Widefunction.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math>. ]]

|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|K0zeheguRKo}}

Introduction to KdV

2025-09-30T05:55:39Z

Levi: /* Formula for the solitary wave */

{{nonlinear waves course
| chapter title = Introduction to KdV
| next chapter = [[Numerical Solution of the KdV]]
| previous chapter = [[Nonlinear Shallow Water Waves]]
}}

{{complete pages}}

The KdV (Korteweg-De Vries) equation is one of the most important non-linear
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The derivation of
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
is written as
<center><math>
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.
</math></center>

More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]

==Travelling Wave Solution==

The KdV equation posesses travelling wave solutions. One particular
travelling wave solution is called a soltion and it was discovered
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
in 1834. However, it was not understood
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.

We begin with the assumption that the wave travels with constant form, i.e.
is of the form
<center><math>
u\left( x,t\right) =f\left( x-ct\right)
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math>
Only very special values of <math>c</math> and <math>f</math> will give travelling
waves.
We introduce the coordinate <math>\zeta = x - ct</math>.
If we substitute this expression into the KdV equation we obtain
<center><math>
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0
</math></center>
We can integrate this with respect to <math>\zeta</math> to obtain
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
where <math>A_1</math> is a constant of integration.

If we think about this equation as Newton's second law in a potential well <math>
V(f) </math> then the equation is
<center><math>
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}
</math></center>
then the potential well is given by
<center><math>
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}
</math></center>
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
in a cubic well.

The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
We can choose the constant <math>A_{1}=0</math> and then we have a
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
and then runs up the other side and finally returns to the maximum at <math>
t=\infty .</math> This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using <math>
f^{^{\prime }}=v.</math> This gives us
<center><math>\begin{matrix}
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2}
\end{matrix}</math></center>
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) </math> and <math>(c/3,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is a nonlinear center (it is neither repelling nor
attracting).
The Jacobian matrix for the saddle point is
<center><math>
J_{\left( 0,0\right) }=\left(
\begin{array}{cc}
0 & 1 \\
c & 0
\end{array}
\right)
</math></center>
which has eigenvalues at <math>\pm \sqrt{c}</math> and the incident directions are
<center><math>
\left(
\begin{array}{c}
1\\
\sqrt{c}
\end{array}
\right) \leftrightarrow \sqrt{c},\left(
\begin{array}{c}
-1 \\
\sqrt{c}
\end{array}
\right) \leftrightarrow -\sqrt{c}
</math></center>
There is a
homoclinic connection which connects the equilibrium point at the origin. This homoclinic
connection represents the solitary wave. Within this homoclinic connection
lie periodic orbits which represent the cnoidal waves.

{| class="wikitable"
|-
! Phase portrait
! Solitary Wave
! Cnoidal Wave
|-
| [[Image:Kdv phase portrait.jpg|thumb|350px|Phase portrait]]
| [[Image:Kdv wave solitary2.gif|thumb|350px|Solitary Wave]]
| [[Image:Kdv wave cn.gif|thumb|350px|Cnoidal Wave]]
|}

We can also integrate the equation
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
by multiplying by <math>f^{\prime }</math>and integrating. This gives us
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
f^{2}}{2}-f^{3} = -V(f)
</math></center>
It is no coincidence that the right hand side is the potential energy, because this
is nothing more that the equation for conservation of energy (or the first
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.

This is a separable equation and the only challenge is to integrate
<center><math>
\int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f.
</math></center>

==Formula for the solitary wave==

We know that the solitary wave solution is found when <math>
A_{0}=A_{1}=0.</math> This gives us
<center><math>
\left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right)
</math></center>
This can be solved by separation of variables to give
<center><math>
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
</math></center>
We then substitute
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)
</math></center>
and note that
<center><math>
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)
</math></center>
and that
<center><math>
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }
</math></center>
This means that
<center><math>\begin{matrix}
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt{c}}s
\end{matrix}</math></center>
This gives us
<center><math>
-\frac{2}{\sqrt{c}}s=\zeta+a
</math></center>
Therefore
<center><math>
f(\zeta)=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)
</math></center>
Of course we assumed that <math>\zeta=x-ct</math> so the formula for the solitary wave is
given by
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
x-ct+a\right) \right]
</math></center>
Note that a solution exists for each
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
discovered experimentally by
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].

==Formula for the cnoidal wave==

If we consider the case when the solution oscillates between two values <math>F_2 < F_3</math>
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
we can integrate the equation to obtain
<center><math>
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)
</math></center>
where <math>cn</math> is a Jacobi Elliptic function and
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
Derivation of this equation is found [[KdV Cnoidal Wave Solutions]].
We can write this equation as
<center><math>
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)
</math></center>
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
small amplitude.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|SzZ-KhvvPio}}

=== Part 2 ===

{{#ev:youtube|QltlSQQBtrs}}

=== Part 3 ===

{{#ev:youtube|NJ7h3Z9QtvU}}

=== Part 4 ===

{{#ev:youtube|NictSlSgRbM}}

Connection betwen KdV and the Schrodinger Equation

2025-09-30T05:15:41Z

Levi: /* Reflectionless Potential */ revert

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{mn} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Connection betwen KdV and the Schrodinger Equation

2025-09-30T05:11:45Z

Levi: /* Reflectionless Potential */ typos

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknowns <math>v_{m}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{mn} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Connection betwen KdV and the Schrodinger Equation

2025-09-29T07:00:52Z

Levi: some clarifications, add book link

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknows <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{mn} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Connection betwen KdV and the Schrodinger Equation

2025-09-28T10:51:04Z

Levi:

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{-\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points,
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknows <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{mn} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Connection betwen KdV and the Schrodinger Equation

2025-09-28T10:49:03Z

Levi: some more explanations

{{nonlinear waves course
| chapter title = Connection betwen KdV and the Schrodinger Equation
| next chapter = [[Example Calculations for the KdV and IST]]
| previous chapter = [[Properties of the Linear Schrodinger Equation]]
}}

If we substitute the relationship
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
into the KdV after some manipulation we obtain
<center><math>
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
</math></center>
where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w.</math> If we integrate this equation we obtain
<center><math>
\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
_{x}wQ\right)\Bigg|^\infty_{\infty}=0
</math></center>
and, since <math>w^2>0</math> except for some isolated points
<center><math>
\partial_{t}\lambda=0
</math></center>
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
properties can be found.

==Scattering Data==

For the discrete spectrum the eigenfunctions behave like
<center><math>
w_{n}\left( x,t\right) =c_{n}\left( t\right) e^{-k_{n}x}
</math></center>
as <math>x\rightarrow\infty</math> with
<center><math>
\int_{-\infty}^{\infty}\left( w_{n}\left( x,t\right) \right) ^{2}\mathrm{d}x=1
</math></center>
The continuous spectrum looks like
<center><math>
w\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
</math></center>
<center><math>
w\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
</math></center>
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
coefficient. This gives us the scattering data at <math>t=0</math>
<center><math>
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
</math></center>
The scattering data evolves as
<center><math>
k_{n}(t)=k_{n}(0) = k_{n}
</math></center>
<center><math>
c_{n}\left( t\right) =c_{n}\left( 0\right) e^{4k_{n}^{3}t}
</math></center>
<center><math>
r\left( k,t\right) =r\left( k,0\right) e^{8ik^{3}t}
</math></center>
<center><math>
a\left( k,t\right) =a\left( k,0\right)
</math></center>
We can recover <math>u</math> from scattering data. We write
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}
x}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
Then solve
<center><math>
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
</math></center>
This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find <math>u</math> from
<center><math>
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
</math></center>

==Reflectionless Potential==

In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
<math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
<center><math>
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}x}
</math></center>
then
<center><math>
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
From the equation we can see that
<center><math>
K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}
</math></center>
If we substitute this into the equation,
<center><math>
-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}y}+\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}
-\sum_{m=1}^{N}v_{m}\left( x,t\right) e^{-k_{m}z}\sum_{n=1}^{N}c_{n}
^{2}\left( t\right) e^{-k_{n}\left( y+z\right) }\mathrm{d}z=0
</math></center>
which leads to
<center><math>
-\sum_{n=1}^{N} v_{n}\left( x,t\right) e^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(
y+x\right) }=0
</math></center>
and we can eliminate the sum over <math>n</math> and the
<math>e^{-k_{n}y}</math> to obtain
<center><math>
-v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
</math></center>
which is an algebraic (finite dimensional system) for the unknows <math>v_{n}.</math>.

We can write this as
<center><math>
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
</math></center>
where <math>f_{m}=c_{m}^2\left( t\right) e^{-k_{m}x}</math> and
the elements of <math>\mathbf{C}</math> are given by
<center><math>
c_{mn} = \frac{c_{n}^2\left( t\right)}
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
</math></center>
This gives us
<center><math>
K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y}
</math></center>
We then find <math>u(x,t)</math> from <math>K</math>.

=== Single Soliton Example ===

If <math>n=1</math> (a single soliton
solution) we get
<center><math>\begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}^2\left( t\right)
e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
{2k_{1}}e^{-2 k_{1} x}}\\
& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
\end{matrix}</math></center>
where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore
<center><math>\begin{align}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\
& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
{2k_{1}}\right) ^{2}}\\
& =2k_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
^{3}t\right\}
\end{align}</math></center>
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
}</math>. This is of course the single soliton solution.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|KHGoPCoyP28}}

=== Part 2 ===

{{#ev:youtube|8mVq6MQWO3I}}

=== Part 3 ===

{{#ev:youtube|meTgaaGKsfQ}}

=== Part 4 ===

{{#ev:youtube|AfAQyjzvJUU}}

Properties of the Linear Schrodinger Equation

2025-09-28T09:27:56Z

Levi: /* Case when \lambda>0 */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>

===Case when <math>\lambda<0</math>===

We consider the cases of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain (as <math>w\rightarrow0</math> as <math>x\rightarrow\pm\infty</math>)
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math>
<math>
2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg|_{0}^{\infty}=1
</math>
<math>\dfrac{2a^2}{u_{0}}=1</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>

=== Case when <math>\lambda>0</math> ===

The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
where <math>\mathrm{e}^{-\mathrm{i}kx}</math> is the incident wave, <math>r\mathrm{e}^{\mathrm{i}kx}</math> is the reflected wave, and <math>t\mathrm{e}^{-\mathrm{i}kx}</math> is the transmitted wave.

Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\zeta,\zeta\right] \\
b & x\in\left[ -\zeta,\zeta\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\zeta,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
a_{2}e^{-kx}, & x>\zeta,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\zeta</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\zeta,\\
b_{1}\cos\kappa x, & -\zeta< x <\zeta,\\
a_{1}e^{-kx}, & x>\zeta,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\zeta</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\zeta} & -\cos\kappa\zeta\\
ke^{-k\zeta} & -\kappa\sin\kappa\zeta
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\zeta} & -\cos\kappa\zeta\\
ke^{-k\zeta} & -\kappa\sin\kappa\zeta
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\zeta}\sin\kappa\zeta+k\cos\kappa\zeta
e^{-k\zeta}=0
</math></center>
or
<center><math>
\tan\kappa\zeta=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\zeta,\\
b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
-a_{1}e^{-kx} & x > \zeta,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\zeta</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\zeta} & \sin\kappa\zeta\\
ke^{-k\zeta} & -\kappa\cos\kappa\zeta
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\zeta} & \sin\kappa\zeta\\
ke^{-k\zeta} & -\kappa\cos\kappa\zeta
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\zeta}a\cos\kappa\zeta+k\sin\kappa\zeta e^{-k\zeta}=0
</math></center>
or
<center><math>
\tan\zeta\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\zeta\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\zeta< x <\zeta\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\zeta
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivatives at <math>x=\pm\zeta</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\zeta} & \cos\kappa\zeta & -\sin\kappa\zeta & 0\\
ik\mathrm{e}^{-\mathrm{i}k\zeta} & \kappa\sin\kappa\zeta & \kappa\cos\kappa
\zeta & 0\\
0 & \cos\kappa\zeta & \sin\kappa\zeta & -\mathrm{e}^{-\mathrm{i}k\zeta}\\
0 & -\kappa\sin\kappa\zeta & \kappa\cos\kappa\zeta &
ik\mathrm{e}^{-\mathrm{i}k\zeta}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Properties of the Linear Schrodinger Equation

2025-09-28T08:44:25Z

Levi: change varsigma to actual zeta

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>

===Case when <math>\lambda<0</math>===

We consider the cases of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain (as <math>w\rightarrow0</math> as <math>x\rightarrow\pm\infty</math>)
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math>
<math>
2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg|_{0}^{\infty}=1
</math>
<math>\dfrac{2a^2}{u_{0}}=1</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>

=== Case when <math>\lambda>0</math> ===

The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
where <math>\mathrm{e}^{-\mathrm{i}kx}</math> is the incident wave, <math>r\mathrm{e}^{\mathrm{i}kx}</math> is the reflected wave, and <math>t\mathrm{e}^{-\mathrm{i}kx}</math> is the transmitted wave.

Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\zeta,\zeta\right] \\
b & x\in\left[ -\zeta,\zeta\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\zeta,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
a_{2}e^{-kx}, & x>\zeta,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\zeta</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\zeta,\\
b_{1}\cos\kappa x, & -\zeta< x <\zeta,\\
a_{1}e^{-kx}, & x>\zeta,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\zeta</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\zeta} & -\cos\kappa\zeta\\
ke^{-k\zeta} & -\kappa\sin\kappa\zeta
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\zeta} & -\cos\kappa\zeta\\
ke^{-k\zeta} & -\kappa\sin\kappa\zeta
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\zeta}\sin\kappa\zeta+k\cos\kappa\zeta
e^{-k\zeta}=0
</math></center>
or
<center><math>
\tan\kappa\zeta=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\zeta,\\
b_{2}\sin\kappa x, & -\zeta< x <\zeta,\\
-a_{1}e^{-kx} & x > \zeta,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\zeta</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\zeta} & \sin\kappa\zeta\\
ke^{-k\zeta} & -\kappa\cos\kappa\zeta
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\zeta} & \sin\kappa\zeta\\
ke^{-k\zeta} & -\kappa\cos\kappa\zeta
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\zeta}a\cos\kappa\zeta+k\sin\kappa\zeta e^{-k\zeta}=0
</math></center>
or
<center><math>
\tan\zeta\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\zeta\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\zeta< x <\zeta\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\zeta
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm\zeta</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\zeta} & \cos\kappa\zeta & -\sin\kappa\zeta & 0\\
ik\mathrm{e}^{-\mathrm{i}k\zeta} & \kappa\sin\kappa\zeta & \kappa\cos\kappa
\zeta & 0\\
0 & \cos\kappa\zeta & \sin\kappa\zeta & -\mathrm{e}^{-\mathrm{i}k\zeta}\\
0 & -\kappa\sin\kappa\zeta & \kappa\cos\kappa\zeta &
ik\mathrm{e}^{-\mathrm{i}k\zeta}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Properties of the Linear Schrodinger Equation

2025-09-28T08:43:39Z

Levi: /* Case when \lambda>0 */

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>

===Case when <math>\lambda<0</math>===

We consider the cases of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain (as <math>w\rightarrow0</math> as <math>x\rightarrow\pm\infty</math>)
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math>
<math>
2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg|_{0}^{\infty}=1
</math>
<math>\dfrac{2a^2}{u_{0}}=1</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>

=== Case when <math>\lambda>0</math> ===

The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
where <math>\mathrm{e}^{-\mathrm{i}kx}</math> is the incident wave, <math>r\mathrm{e}^{\mathrm{i}kx}</math> is the reflected wave, and <math>t\mathrm{e}^{-\mathrm{i}kx}</math> is the transmitted wave.

Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
a_{2}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
a_{1}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\varsigma</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\varsigma,\\
b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
-a_{1}e^{-kx} & x > \varsigma,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\varsigma</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm\varsigma</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Conservation Laws for the KdV

2025-09-28T06:19:49Z

Levi:

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting features of the KdV is the existence of
infinitely many conservation laws. Let's begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since we can rewrite the equation into the conservation form <math>\partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0</math>, then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u + 2 u^2)
</math></center>
<center><math>
w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just an <math>\partial _{x}</math> of some <math>X(u)</math> and therefore does not actually provide a conservation law.

As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>u</math> (mass), <math>u^2</math> (momentum), and <math>2 u^3 - (\partial_{x} u)^2</math> (energy).

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Properties of the Linear Schrodinger Equation

2025-09-28T06:18:36Z

Levi: a bit more explanations

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>

===Case when <math>\lambda<0</math>===

We consider the cases of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain (as <math>w\rightarrow0</math> as <math>x\rightarrow\pm\infty</math>)
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math>
<math>
2\left( a^2\dfrac{e^{-u_{0}x}}{-u_{0}}\right)\Bigg|_{0}^{\infty}=1
</math>
<math>\dfrac{2a^2}{u_{0}}=1</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>

=== Case when <math>\lambda>0</math> ===

The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
where <math>\mathrm{e}^{-\mathrm{i}kx}</math> is the incident wave, <math>r\mathrm{e}^{\mathrm{i}kx}</math> is the reflected wave, and <math>t\mathrm{e}^{-\mathrm{i}kx}</math> is the transmitted wave.

Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
a_{2}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
a_{1}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\varsigma</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\varsigma,\\
b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
-a_{1}e^{-kx} & x > \varsigma,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\varsigma</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Conservation Laws for the KdV

2025-09-27T06:38:52Z

Levi:

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Let's begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since we can rewrite the equation into the conservation form <math>\partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0</math>, then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u + 2 u^2)
</math></center>
<center><math>
w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just an <math>\partial _{x}</math> of some <math>X(u)</math> and therefore does not actually provide a conservation law.

As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>u</math> (mass), <math>u^2</math> (momentum), and <math>2 u^3 - (\partial_{x} u)^2</math> (energy).

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Properties of the Linear Schrodinger Equation

2025-09-27T06:38:07Z

Levi:

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\rightarrow\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
a_{2}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
a_{1}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\varsigma</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\varsigma,\\
b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
-a_{1}e^{-kx} & x > \varsigma,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\varsigma</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Properties of the Linear Schrodinger Equation

2025-09-25T20:27:31Z

Levi:

{{nonlinear waves course
| chapter title = Properties of the Linear Schrodinger Equation
| next chapter = [[Connection betwen KdV and the Schrodinger Equation]]
| previous chapter = [[Introduction to the Inverse Scattering Transform]]
}}

The linear Schrödinger equation
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
incident waves. This is easiest seen through the following examples

==Example 1: <math>\delta</math> function potential==

We consider here the case when <math>u\left( x,0\right) = u_0 \delta\left( x\right)
.</math> Note that this function can be thought of as the limit as of the potential
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varepsilon,\varepsilon\right] \\
\frac{u_{0}}{2\varepsilon} & x\in\left[ -\varepsilon,\varepsilon\right]
\end{matrix}
\right.
</math></center>
In this case we need to solve
<center><math>
\partial_{x}^{2}w+ u_0\delta(x) w=-\lambda w
</math></center>
We consider the case of <math>\lambda<0</math> and <math>\lambda>0</math> separately. For the first
case we write <math>\lambda=-k^{2}</math> and we obtain
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

ae^{kx}, & x<0\\
be^{-kx}, & x>0
\end{matrix}
\right.
</math></center>
We have two conditions at <math>x=0,</math> <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_0 w\left( 0\right) =0.</math> This final condition is obtained by integrating `across' zero as follows
<center><math>\begin{align}
\int_{0^{-}}^{0^{+}} \left(\partial_x^2 w +u_0\delta(x) w + \lambda w \right) \ \mathrm{d}x = 0.
\end{align}
</math></center>

This gives the condition that <math>a=b</math> and
<math>k=u_{0}/2.</math> We need to normalise the eigenfunctions so that
<center><math>
\int_{-\infty}^{\infty}\left( w\left( x\right) \right) ^{2}\mathrm{d}x=1.
</math></center>
Therefore
<center><math>
2\int_{0}^{\infty}\left( ae^{-u_{0}x/2}\right) ^{2}\mathrm{d}x=1
</math></center>
which means that <math>a=\sqrt{u_{0}/2}.</math> Therefore, there is only one discrete
spectral point which we denote by <math>k_{1}=u_{0}/2</math>
<center><math>
w_{1}\left( x\right) =\left\{
\begin{matrix}

\sqrt{k_{1}}e^{k_{1}x}, & x<0\\
\sqrt{k_{1}}e^{-k_{1}x}, & x>0
\end{matrix}
\right.
</math></center>
The continuous eigenfunctions correspond to <math>\lambda=k^{2}>0</math> are of the form
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<0\\
t\mathrm{e}^{-\mathrm{i}kx}, & x>0
\end{matrix}
\right.
</math></center>
Again we have the conditions that <math>w</math> must be continuous at <math>0</math> and
<math>\partial_{x}w\left( 0^{+}\right) -\partial_{x}w\left( 0^{-}\right)
+u_{0}w\left( 0\right) =0.</math> This gives us
<center><math>\begin{matrix}
1+r & =t\\
-ikt+ik-ikr & =-tu_{0}
\end{matrix}</math></center>
which has solution
<center><math>\begin{matrix}
r & =\frac{u_{0}}{2ik-u_{0}}\\
t & =\frac{2ik}{2ik-u_{0}}
\end{matrix}</math></center>

==Example 2: Hat Function Potential==

The properties of the eigenfunction is perhaps seem most easily through the
following example
<center><math>
u\left( x\right) =\left\{
\begin{matrix}

0 & x\notin\left[ -\varsigma,\varsigma\right] \\
b & x\in\left[ -\varsigma,\varsigma\right]
\end{matrix}
\right.
</math></center>
where <math>b>0.</math>

===Case when <math>\lambda<0</math>===

If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
get
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
a_{2}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b-k^{2}}</math> which means that <math>0\leq k\leq\sqrt{b}</math> (there is
no solution for <math>k>\sqrt{b}).</math> We then match <math>w</math> and its derivative at
<math>x=\pm\varsigma</math> to solve for <math>a</math> and <math>b</math>. This leads to two system of
equations, one for the even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd
solutions (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions
is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x<-\varsigma,\\
b_{1}\cos\kappa x, & -\varsigma< x <\varsigma,\\
a_{1}e^{-kx}, & x>\varsigma,
\end{matrix}
\right.
</math></center>
If we impose the condition that the function and its derivative are continuous at
<math>x=\pm\varsigma</math> we obtain the following equation
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This has non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & -\cos\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\sin\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
-\kappa e^{-k\varsigma}\sin\kappa\varsigma+k\cos\kappa\varsigma
e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\kappa\varsigma=\frac{k}{\kappa}
</math></center>
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
finite number of solutions.

The solution for the odd solutions is
<center><math>
w\left( x\right) =\left\{
\begin{matrix}
a_{1}e^{kx}, & x <-\varsigma,\\
b_{2}\sin\kappa x, & -\varsigma< x <\varsigma,\\
-a_{1}e^{-kx} & x > \varsigma,
\end{matrix}
\right.
</math></center>
and again imposing the condition that the solution and its derivative is continuous
at <math>x=\pm\varsigma</math> gives
<center><math>
\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) \left(
\begin{matrix}

a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}

0\\
0
\end{matrix}
\right)
</math></center>
This can non trivial solutions when
<center><math>
\det\left(
\begin{matrix}

e^{-k\varsigma} & \sin\kappa\varsigma\\
ke^{-k\varsigma} & -\kappa\cos\kappa\varsigma
\end{matrix}
\right) =0
</math></center>
which gives us the equation
<center><math>
\kappa e^{-k\varsigma}a\cos\kappa\varsigma+k\sin\kappa\varsigma e^{-k\varsigma}=0
</math></center>
or
<center><math>
\tan\varsigma\kappa=-\frac{\kappa}{k}
</math></center>

=== Case when <math>\lambda>0</math> ===

When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
<center><math>
w\left( x\right) =\left\{
\begin{matrix}

\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x <-\varsigma\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -\varsigma< x <\varsigma\\
t\mathrm{e}^{-\mathrm{i}kx} & x>\varsigma
\end{matrix}
\right.
</math></center>
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
obtain
<center><math>
\left(
\begin{matrix}

-\mathrm{e}^{-\mathrm{i}k\varsigma} & \cos\kappa\varsigma & -\sin\kappa\varsigma & 0\\
ik\mathrm{e}^{-\mathrm{i}k\varsigma} & \kappa\sin\kappa\varsigma & \kappa\cos\kappa
\varsigma & 0\\
0 & \cos\kappa\varsigma & \sin\kappa\varsigma & -\mathrm{e}^{-\mathrm{i}k\varsigma}\\
0 & -\kappa\sin\kappa\varsigma & \kappa\cos\kappa\varsigma &
ik\mathrm{e}^{-\mathrm{i}k\varsigma}
\end{matrix}
\right) \left(
\begin{matrix}

r\\
b_{1}\\
b_{2}\\
t
\end{matrix}
\right) =\left(
\begin{matrix}

\mathrm{e}^{\mathrm{i}k}\\
ik\mathrm{e}^{-\mathrm{i}k}\\
0\\
0
\end{matrix}
\right)
</math></center>

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|anAThvCcpNw}}

=== Part 2 ===

{{#ev:youtube|SDPIx42VjLQ}}

=== Part 3 ===

{{#ev:youtube|OUmjeLZWr3M}}

=== Part 4 ===

{{#ev:youtube|hIfcO3a8_XU}}

=== Part 5 ===

{{#ev:youtube|z13lKSTficA}}

=== Part 6 ===

{{#ev:youtube|2XlQpEscxE4}}

=== Part 7 ===

{{#ev:youtube|iMMQ4NUdXNc}}

=== Part 8 ===

{{#ev:youtube|0F_dINNxMlw}}

Introduction to the Inverse Scattering Transform

2025-09-25T10:48:24Z

Levi:

{{nonlinear waves course
| chapter title = Introduction to the Inverse Scattering Transform
| next chapter = [[Properties of the Linear Schrodinger Equation]]
| previous chapter = [[Conservation Laws for the KdV]]
}}

The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
<center><math>\begin{matrix}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{matrix}</math></center>
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>

The Miura transformation is given by
<center><math>
u=-v^{2}-\partial_x v\,
</math></center>
and if <math>v</math> satisfies the mKdV
<center><math>
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
</math></center>
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This
nonlinear ODE is also known as the Riccati equation and there is a well known
transformation which linearises this equation. It we write
<center><math>
v=\frac{\left( \partial_{x}w\right) }{w}
</math></center>
then we obtain the equation
<center><math>
\partial_{x}^{2}w+uw=0
</math></center>
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
problem
<center><math>
\partial_{x}^{2}w+uw=-\lambda w
</math></center>
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrödinger equation.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|P3uMk9OS8p4}}

Conservation Laws for the KdV

2025-09-24T19:17:33Z

Levi: /* Proof of an Infinite Number of Conservation Laws */ some changes

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Lets begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since we can rewrite the equation into the conservation form <math>\partial _{t}w+\partial_{x}\left(3w^2-2\varepsilon ^{2}w^{3}+\partial _{x}^{2}w\right)=0</math>, then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u = \partial _{x}(\partial _{x}^2 u + 2 u^2)
</math></center>
<center><math>
w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}(\partial _{x}^3 u + 6 u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just an <math>\partial _{x}</math> of some <math>X(u)</math> and therefore does not actually provide a conservation law.

As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>u</math> (mass), <math>u^2</math> (momentum), and <math>2 u^3 - (\partial_{x} u)^2</math> (energy).

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Conservation Laws for the KdV

2025-09-24T18:04:10Z

Levi: /* Proof of an Infinite Number of Conservation Laws */ add w_4 and mass-momentum-energy

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Lets begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since the equation is in conservation form then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>
<center><math>
0=w_{4}-\partial _{x}w_{3}-2w_{0}w_{2}-w_{1}^2
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u
</math></center>
<center><math>
w_{4} = \partial _{x}^4 u + 6 u \partial_{x}^2 u + 5 (\partial_{x} u)^2 + 2 u^3 = \partial _{x}^4 u + 6 \partial_{x} (u \partial_{x} u) + 2 u^3 - (\partial_{x} u)^2
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just the derivative (with some modification) of the previous law and
therefore does not actually provide a new conservation law.

As <math>\int_{-\infty }^{\infty }\partial_{x} X(u) \mathrm{d} x = 0</math>, <math>w_0</math>, <math>w_2</math>, and <math>w_4</math> correspond to conservation of <math>u</math> (mass), <math>u^2</math> (momentum), and <math>2 u^3 - (\partial_{x} u)^2</math> (energy).

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Conservation Laws for the KdV

2025-09-24T17:51:22Z

Levi: some fixes

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Lets begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of mass. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of momentum. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since the equation is in conservation form then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>

We can solve recursively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just the derivative (with some modification) of the previous law and
therefore does not actually provide a new conservation law.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Conservation Laws for the KdV

2025-09-24T17:44:57Z

Levi:

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Lets begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>X(u)\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of momentum. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of energy. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since the equation is in conservation form then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>

We can solve recurrsively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just the derivative (with some modification) of the previous law and
therefore does not actually provide a new conservation law.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Conservation Laws for the KdV

2025-09-24T16:59:25Z

Levi:

{{nonlinear waves course
| chapter title = Conservation Laws for the KdV
| next chapter = [[Introduction to the Inverse Scattering Transform]]
| previous chapter = [[Numerical Solution of the KdV]]
}}

One of the most interesting freatures of the KdV is the existence of
infinitely many conservation laws. Lets begin with some basics of
conservation laws. If we can write our equation of the form
<center><math>
\partial _{t}T\left( u\right) +\partial _{x}X\left( u\right) =0
</math></center>
Then we can integrate this equation from <math>-\infty </math> to <math>\infty </math> to obtain
<center><math>
\int_{-\infty }^{\infty }\partial _{t}T\left( u\right) \mathrm{d} x = -\int_{-\infty
}^{\infty }\partial _{x}X\left( u\right) \mathrm{d} x
</math></center>
The second integral will be zero if <math>u\rightarrow 0</math> as <math>x\rightarrow \pm
\infty .</math> Therefore
<center><math>
\partial _{t}\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x=0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }T\left( u\right) \mathrm{d} x
</math></center>
must be conserved by the solution of the equation. For the KdV we can write
<center><math>
\partial _{t}u+\partial _{x}\left( 3u^{2}+\partial _{x}^{2}u\right) =0.
</math></center>
so that we immediately see that the quantity
<center><math>
\int_{-\infty }^{\infty }u\mathrm{d} x
</math></center>
is conserved. This corresponds to conservation of momentum. We can also
write the KdV equation as
<center><math>
\partial _{t}\left( u^{2}\right) +\partial _{x}\left( 4u^{3}+2u\partial
_{x}^{2}u-\left( \partial _{x}u\right) ^{2}\right) = 0
</math></center>
so that the quantity
<center><math>
\int_{-\infty }^{\infty }u^{2}\mathrm{d} x
</math></center>
must be conserved. This corresponds to conservation of energy. It turns out
that there is an infinite number of conserved quantities and we give here
the proof of this.

==Modified KdV==

The modified KdV is
<center><math>
\partial _{t}v-3\partial _{x}\left( v^{3}\right) +\partial _{x}^{3}v=0
</math></center>
It is connected to the KdV by Miura's transformation
<center><math>
u=-\left( v^{2}+\partial _{x}v\right)
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial
_{x}^{3}u=-(2v+\partial _{x})\left( \partial _{t}v-3\partial _{x}\left(
v^{3}\right) +\partial _{x}^{3}v\right)
</math></center>
Note that this shows that every solution of the mKdV is a solution of the
KdV but not vice versa.

==Proof of an Infinite Number of Conservation Laws==

An ingenious proof of the exisitence of an infinite number of conservation
laws can be obtain from a generalization of Miura's transformation
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
If we substitute this into the KdV we obtain
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right) +\partial _{x}^{3}u=\left(
1-\varepsilon \partial _{x}-2\varepsilon ^{2}w\right) \left( \partial
_{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial _{x}w+\partial
_{x}^{3}w\right)
</math></center>
Therefore <math>u</math> solves the KdV equation provided that
<center><math>
\partial _{t}w+6\left( w-\varepsilon ^{2}w^{2}\right) \partial
_{x}w+\partial _{x}^{3}w=0
</math></center>
We write the solution to this equation as a formal power series
<center><math>
w\left( x,t,\varepsilon \right) =\sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right)
</math></center>
Since the equation is in conservation form then
<center><math>
\int_{-\infty }^{\infty }w\left( x,t,\varepsilon \right) \mathrm{d} x = \mathrm{constant}
</math></center>
and since this is true for all <math>\varepsilon </math> this implies that
<center><math>
\int_{-\infty }^{\infty }w_{n}\left( x,t\right) \mathrm{d} x = \mathrm{constant}
</math></center>
We then consider the expression
<center><math>
u=w-\varepsilon \partial _{x}w-\varepsilon ^{2}w^{2}
</math></center>
which implies that
<center><math>
u=\sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left( x,t\right) -\varepsilon
\partial _{x}\left( \sum_{n=0}^{\infty }\varepsilon ^{n}w_{n}\left(
x,t\right) \right) -\varepsilon ^{2}\left( \sum_{n=0}^{\infty }\varepsilon
^{n}w_{n}\left( x,t\right) \right) ^{2}
</math></center>
It we equate powers of <math>\varepsilon </math> we obtain
<center><math>
u=w_{0}
</math></center>
<center><math>
0=w_{1}-\partial _{x}w_{0}
</math></center>
<center><math>
0=w_{2}-\partial _{x}w_{1}-w_{0}^{2}
</math></center>
<center><math>
0=w_{3}-\partial _{x}w_{2}-2w_{0}w_{1}
</math></center>

We can solve recurrsively to obtain
<center><math>
w_{0}=u
</math></center>
<center><math>
w_{1}=\partial _{x}u
</math></center>
<center><math>
w_{2} = \partial_{x}^2 u + u^{2}
</math></center>
<center><math>
w_{3} = \partial _{x}^3 u + 4 u \partial_{x} u
</math></center>
Note that each of the odd conservation laws (<math>w_1, w_3</math> etc.) are just the derivative (with some modification) of the previous law and
therefore does not actually provide a new conservation law.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|9w3-IWX2b5g}}

=== Part 2 ===

{{#ev:youtube|QQjh14uguvg}}

Numerical Solution of the KdV

2025-09-24T16:50:28Z

Levi: /* Example Calculations */

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&ik\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>i k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty < x <\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. We obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the Fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right)+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u_{1}^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ 2\, \mathrm{sech}^{2}\left(x+1\right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|ZjeuLNdxcRc}}

=== Part 2 ===

{{#ev:youtube|C7gI5PCKvFQ}}

=== Part 3 ===

{{#ev:youtube|X7-p8nBmKw4}}

=== Part 4 ===

{{#ev:youtube|xevnpuOoks0}}

Numerical Solution of the KdV

2025-09-24T16:45:58Z

Levi: /* Numerical Solution of the KdV */

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&ik\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>i k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty < x <\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. We obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the Fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u^{2}\right)+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u_{1}^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ \, 2 \mathrm{sech}^{2}\left(\left(
x +1\right) \right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|ZjeuLNdxcRc}}

=== Part 2 ===

{{#ev:youtube|C7gI5PCKvFQ}}

=== Part 3 ===

{{#ev:youtube|X7-p8nBmKw4}}

=== Part 4 ===

{{#ev:youtube|xevnpuOoks0}}

Numerical Solution of the KdV

2025-09-24T15:48:47Z

Levi: /* Numerical Implementation Using the FFT */

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&ik\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>i k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty < x <\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. We obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the Fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u\right) ^{2}+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u_{1}^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ \, 2 \mathrm{sech}^{2}\left(\left(
x +1\right) \right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|ZjeuLNdxcRc}}

=== Part 2 ===

{{#ev:youtube|C7gI5PCKvFQ}}

=== Part 3 ===

{{#ev:youtube|X7-p8nBmKw4}}

=== Part 4 ===

{{#ev:youtube|xevnpuOoks0}}

Numerical Solution of the KdV

2025-09-24T15:47:34Z

Levi: /* Solution for the Linearized KdV */

{{nonlinear waves course
| chapter title = Numerical Solution of the KdV
| next chapter = [[Conservation Laws for the KdV]]
| previous chapter = [[Introduction to KdV]]
}}

== Introduction ==

We present here a method to solve the KdV equation numerically. There are
many different methods to solve the KdV and we use here a spectral method
which has been found to work well. Spectral methods work by using the
Fourier transform (or some varient of it) to calculate the derivative.

==Fourier Transform==

Recall that the Fourier transform is given by
<center><math>
\mathcal{F}\left[ f(x)\right] =\hat{f}\left( k\right) =\int_{-\infty
}^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
</math></center>
and the inverse Fourier transform is given by
<center><math>
f\left( x\right) =\mathcal{F}^{-1}\left[ \hat{f}\left( k\right) \right] =
\frac{1}{2\pi }\int_{-\infty }^{\infty }\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>
(note that there are other ways of writing this transform). The most
important property of the Fourier transform is that
<center><math>
\begin{matrix}
\int_{-\infty }^{\infty }\left( \partial _{x}f\left( x\right) \right)
\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
&=&ik\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
\end{matrix}
</math></center>
where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
\infty .</math> This means that the Fourier transform converts differentiation to
multiplication by <math>i k</math>.

==Solution for the Linearized KdV==

We begin with a simple example. Suppose we want to solve the linearized KdV
equation
<center><math>
\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty < x <\infty
</math></center>
subject to solve initial conditions
<center><math>
u\left( x,0\right) =f\left( x\right)
</math></center>
We can solve this equation by taking the Fourier transform. We obtain
<center><math>
\partial _{t}\hat{u}-ik^{3}\hat{u}=0
</math></center>
So that
<center><math>
\hat{u}\left( k,t\right) =\hat{f}\left( k\right) \mathrm{e}^{\mathrm{i}k^{3}t}
</math></center>
Therefore
<center><math>
u\left( x,t\right) = \frac{1}{2\pi}
\int_{-\infty}^{\infty }\hat{f}\left( k\right)
\mathrm{e}^{\mathrm{i}k^{3}t}\mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
</math></center>

==Numerical Implementation Using the FFT==

The Fast Fourier Transform (FFT) is a method to calculate the fourier
transform efficiently for discrete sets of points. These points need to be
evenly spaced in the <math>x</math> plane and are given by
<center><math>
x_{n}=x_{0}+n\Delta x,\ \ 0\leq n\leq N-1
</math></center>
(note they can start at any <math>x</math> value). For the FFT to be as efficient as
possible <math>N</math> should be a power of <math>2.</math> Corresponding to the discrete set of
points in the <math>x</math> domain is a discrete set of points in the <math>k</math> plane given
by
<center><math>
k_{n}=\left\{
\begin{matrix}
n\Delta k,\ \ 0\leq n\leq \frac{N}{2} \\
\left( n-N\right) \Delta k,\ \ \frac{N}{2}+1\leq n\leq N-1
\end{matrix}
\right.
</math></center>
where <math>\Delta k=2\pi /(N\Delta x).</math> Note that this numbering seems slighly
odd and is due to aliasing. We are not that interested in the frequency
domain solution but we need to make sure that we select the correct values
of <math>k</math> for our numerical code.

==Numerical Code for the Linear KdV==

Here is the code to solve the linear KdV using MATLAB

N = 1024;

t=0.1;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

f = exp(-x.^2);

f_hat = fft(f);

u = real(ifft(f_hat.*exp(i*k.^3*t)));

==Numerical Solution of the KdV==

It turns out that a method to solve the KdV equation can be derived using
spectral methods. We begin with the KdV equation written as
<center><math>
\partial _{t}u+3\partial _{x}\left( u\right) ^{2}+\partial _{x}^{3}u=0.
</math></center>
The Fourier transform of the KdV is therefore
<center><math>
\partial _{t}\hat{u}+3ik\widehat{\left( u^{2}\right)} -ik^{3}\hat{u}=0
</math></center>
We solve this equation by a split step method. We write the equation as
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)} +ik^{3}\hat{u}
</math></center>
We can solve
<center><math>
\partial _{t}\hat{u}=ik^{3}\hat{u}
</math></center>
exactly while the equation
<center><math>
\partial _{t}\hat{u}=-3ik\widehat{\left( u^{2}\right)}
</math></center>
needs to be solved by time stepping. The idea of the split step method is to
solve alternatively each of these equations when stepping from <math>t</math> to <math>
t+\Delta t.</math> Therefore we solve
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
-3ik \Delta t\widehat{\left( u_{1}^{2}\right)}
\end{matrix}</math></center>
Note that we are using Euler's method to time step and that the solution
could be improved by using a better method, such as the Runge-Kutta 4
method.

The only slighly tricky thing is that we have both <math>u</math> and <math>\hat{u}</math> in the
equation, but we can simply use the Fourier transform to connect these. The
equation then becomes
<center><math>\begin{matrix}
\hat{u}_{1}\left( k,t+\Delta t\right) &=&\hat{u}\left( k,t\right)
\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
- 3ik\Delta t\left( \mathcal{F}\left( \left( \mathcal{F}^{-1}\left[ \hat{u}_{1}\left( k,t+\Delta t\right)\right]
\right) ^{2}\right) \right)
\end{matrix}</math></center>

==Numerical Code to solve the KdV by the split step method==

Here is some code to solve the KdV using MATLAB

N = 256;

x = linspace(-10,10,N);

delta_x = x(2) - x(1);

delta_k = 2*pi/(N*delta_x);

k = [0:delta_k:N/2*delta_k,-(N/2-1)*delta_k:delta_k:-delta_k];

c=16;

u = 1/2*c*(sech(sqrt(c)/2*(x+8))).^2;

delta_t = 0.4/N^2;

tmax = 0.1; nmax = round(tmax/delta_t);

U = fft(u);

for n = 1:nmax

% first we solve the linear part

U = U.*exp(1i*k.^3*delta_t);

%then we solve the non linear part

U = U - delta_t*(3i*k.*fft(real(ifft(U)).^2));

end

== Example Calculations ==
We cosider the evolution of the KdV with two solitons as
initial condition as given below.
<center><math>
u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
+ \, 2 \mathrm{sech}^{2}\left(\left(
x +1\right) \right)</math></center>

{| class="wikitable"
|-
! Animation
! Three-dimensional plot.
|-
| [[Image:Two_soliton.gif|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.]]
| [[Image:Two_soliton.jpg|thumb|right|500px|Evolution of <math>u(x,t)</math> for two solitons.
Note the phase shift which occurs in the soliton interaction.]]

|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|ZjeuLNdxcRc}}

=== Part 2 ===

{{#ev:youtube|C7gI5PCKvFQ}}

=== Part 3 ===

{{#ev:youtube|X7-p8nBmKw4}}

=== Part 4 ===

{{#ev:youtube|xevnpuOoks0}}

Introduction to KdV

2025-09-24T08:27:30Z

Levi: /* Travelling Wave Solution */

{{nonlinear waves course
| chapter title = Introduction to KdV
| next chapter = [[Numerical Solution of the KdV]]
| previous chapter = [[Nonlinear Shallow Water Waves]]
}}

{{complete pages}}

The KdV (Korteweg-De Vries) equation is one of the most important non-linear
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The derivation of
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
is written as
<center><math>
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.
</math></center>

More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]

==Travelling Wave Solution==

The KdV equation posesses travelling wave solutions. One particular
travelling wave solution is called a soltion and it was discovered
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
in 1834. However, it was not understood
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.

We begin with the assumption that the wave travels with constant form, i.e.
is of the form
<center><math>
u\left( x,t\right) =f\left( x-ct\right)
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math>
Only very special values of <math>c</math> and <math>f</math> will give travelling
waves.
We introduce the coordinate <math>\zeta = x - ct</math>.
If we substitute this expression into the KdV equation we obtain
<center><math>
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0
</math></center>
We can integrate this with respect to <math>\zeta</math> to obtain
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
where <math>A_1</math> is a constant of integration.

If we think about this equation as Newton's second law in a potential well <math>
V(f) </math> then the equation is
<center><math>
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}
</math></center>
then the potential well is given by
<center><math>
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}
</math></center>
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
in a cubic well.

The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
We can choose the constant <math>A_{1}=0</math> and then we have a
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
and then runs up the other side and finally returns to the maximum at <math>
t=\infty .</math> This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using <math>
f^{^{\prime }}=v.</math> This gives us
<center><math>\begin{matrix}
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2}
\end{matrix}</math></center>
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) </math> and <math>(c/3,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is a nonlinear center (it is neither repelling nor
attracting).
The Jacobian matrix for the saddle point is
<center><math>
J_{\left( 0,0\right) }=\left(
\begin{array}{cc}
0 & 1 \\
c & 0
\end{array}
\right)
</math></center>
which has eigenvalues at <math>\pm \sqrt{c}</math> and the incident directions are
<center><math>
\left(
\begin{array}{c}
1\\
\sqrt{c}
\end{array}
\right) \leftrightarrow \sqrt{c},\left(
\begin{array}{c}
-1 \\
\sqrt{c}
\end{array}
\right) \leftrightarrow -\sqrt{c}
</math></center>
There is a
homoclinic connection which connects the equilibrium point at the origin. This homoclinic
connection represents the solitary wave. Within this homoclinic connection
lie periodic orbits which represent the cnoidal waves.

{| class="wikitable"
|-
! Phase portrait
! Solitary Wave
! Cnoidal Wave
|-
| [[Image:Kdv phase portrait.jpg|thumb|350px|Phase portrait]]
| [[Image:Kdv wave solitary2.gif|thumb|350px|Solitary Wave]]
| [[Image:Kdv wave cn.gif|thumb|350px|Cnoidal Wave]]
|}

We can also integrate the equation
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
by multiplying by <math>f^{\prime }</math>and integrating. This gives us
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
f^{2}}{2}-f^{3} = -V(f)
</math></center>
It is no coincidence that the right hand side is the potential energy, because this
is nothing more that the equation for conservation of energy (or the first
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.

This is a separable equation and the only challenge is to integrate
<center><math>
\int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f.
</math></center>

==Formula for the solitary wave==

We know that the solitary wave solution is found when <math>
A_{0}=A_{1}=0.</math> This gives us
<center><math>
\left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right)
</math></center>
This can be solved by separation of variables to give
<center><math>
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
</math></center>
We then substitute
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)
</math></center>
and note that
<center><math>
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)
</math></center>
and that
<center><math>
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }
</math></center>
This means that
<center><math>\begin{matrix}
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt{c}}s
\end{matrix}</math></center>
This gives us
<center><math>
-\frac{2}{\sqrt{c}}s=\zeta+a
</math></center>
Therefore
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)
</math></center>
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
given by
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
x-ct+a\right) \right]
</math></center>
Note that a solution exists for each
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
discovered experimentally by
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].

==Formula for the cnoidal wave==

If we consider the case when the solution oscillates between two values <math>F_2 < F_3</math>
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
we can integrate the equation to obtain
<center><math>
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)
</math></center>
where <math>cn</math> is a Jacobi Elliptic function and
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
Derivation of this equation is found [[KdV Cnoidal Wave Solutions]].
We can write this equation as
<center><math>
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)
</math></center>
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
small amplitude.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|SzZ-KhvvPio}}

=== Part 2 ===

{{#ev:youtube|QltlSQQBtrs}}

=== Part 3 ===

{{#ev:youtube|NJ7h3Z9QtvU}}

=== Part 4 ===

{{#ev:youtube|NictSlSgRbM}}

Introduction to KdV

2025-09-21T10:17:50Z

Levi: /* Travelling Wave Solution */

{{nonlinear waves course
| chapter title = Introduction to KdV
| next chapter = [[Numerical Solution of the KdV]]
| previous chapter = [[Nonlinear Shallow Water Waves]]
}}

{{complete pages}}

The KdV (Korteweg-De Vries) equation is one of the most important non-linear
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The derivation of
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
is written as
<center><math>
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.
</math></center>

More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]

==Travelling Wave Solution==

The KdV equation posesses travelling wave solutions. One particular
travelling wave solution is called a soltion and it was discovered
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
in 1834. However, it was not understood
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.

We begin with the assumption that the wave travels with constant form, i.e.
is of the form
<center><math>
u\left( x,t\right) =f\left( x-ct\right)
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math>
Only very special values of <math>c</math> and <math>f</math> will give travelling
waves.
We introduce the coordinate <math>\zeta = x - ct</math>.
If we substitute this expression into the KdV equation we obtain
<center><math>
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0
</math></center>
We can integrate this with respect to <math>\zeta</math> to obtain
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
where <math>A_1</math> is a constant of integration.

If we think about this equation as Newton's second law in a potential well <math>
V(f) </math> then the equation is
<center><math>
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}
</math></center>
then the potential well is given by
<center><math>
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}
</math></center>
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
in a cubic well.

The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
We can choose the constant <math>A_{1}=0</math> and then we have a
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
and then runs up the other side and finally returns to the maximum at <math>
t=\infty .</math> This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using <math>
f^{^{\prime }}=v.</math> This gives us
<center><math>\begin{matrix}
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2}
\end{matrix}</math></center>
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) </math> and <math>(c/3,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is a nonlinear center (it is neither repelling nor
attracting).
The Jacobian matrix for the saddle point is
<center><math>
J_{\left( 0,0\right) }=\left(
\begin{array}{cc}
0 & 1 \\
c & 0
\end{array}
\right)
</math></center>
which has eigenvalues at <math>\pm \sqrt{c}</math> and the incident directions are
<center><math>
\left(
\begin{array}{c}
1\\
\sqrt{c}
\end{array}
\right) \leftrightarrow \sqrt{c},\left(
\begin{array}{c}
-1 \\
\sqrt{c}
\end{array}
\right) \leftrightarrow -\sqrt{c}
</math></center>
There is a
homoclinic connection which connects the equilibrium point at the origin. This holoclinic
connection represents the solitary wave. Within this homoclinic connection
lie periodic orbits which represent the cnoidal waves.

{| class="wikitable"
|-
! Phase portrait
! Solitary Wave
! Cnoidal Wave
|-
| [[Image:Kdv phase portrait.jpg|thumb|350px|Phase portrait]]
| [[Image:Kdv wave solitary2.gif|thumb|350px|Solitary Wave]]
| [[Image:Kdv wave cn.gif|thumb|350px|Cnoidal Wave]]
|}

We can also integrate the equation
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
by multiplying by <math>f^{\prime }</math>and integrating. This gives us
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
f^{2}}{2}-f^{3} = -V(f)
</math></center>
It is no coincidence that the right hand side is the potential energy, because this
is nothing more that the equation for conservation of energy (or the first
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.

This is a separable equation and the only challenge is to integrate
<center><math>
\int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f.
</math></center>

==Formula for the solitary wave==

We know that the solitary wave solution is found when <math>
A_{0}=A_{1}=0.</math> This gives us
<center><math>
\left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right)
</math></center>
This can be solved by separation of variables to give
<center><math>
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
</math></center>
We then substitute
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)
</math></center>
and note that
<center><math>
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)
</math></center>
and that
<center><math>
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }
</math></center>
This means that
<center><math>\begin{matrix}
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt{c}}s
\end{matrix}</math></center>
This gives us
<center><math>
-\frac{2}{\sqrt{c}}s=\zeta+a
</math></center>
Therefore
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)
</math></center>
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
given by
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
x-ct+a\right) \right]
</math></center>
Note that a solution exists for each
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
discovered experimentally by
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].

==Formula for the cnoidal wave==

If we consider the case when the solution oscillates between two values <math>F_2 < F_3</math>
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
we can integrate the equation to obtain
<center><math>
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)
</math></center>
where <math>cn</math> is a Jacobi Elliptic function and
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
Derivation of this equation is found [[KdV Cnoidal Wave Solutions]].
We can write this equation as
<center><math>
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)
</math></center>
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
small amplitude.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|SzZ-KhvvPio}}

=== Part 2 ===

{{#ev:youtube|QltlSQQBtrs}}

=== Part 3 ===

{{#ev:youtube|NJ7h3Z9QtvU}}

=== Part 4 ===

{{#ev:youtube|NictSlSgRbM}}

Introduction to KdV

2025-09-21T10:15:54Z

Levi: /* Travelling Wave Solution */

{{nonlinear waves course
| chapter title = Introduction to KdV
| next chapter = [[Numerical Solution of the KdV]]
| previous chapter = [[Nonlinear Shallow Water Waves]]
}}

{{complete pages}}

The KdV (Korteweg-De Vries) equation is one of the most important non-linear
pde's. It was originally derived to model shallow water waves with weak
nonlinearities, but it has a wide variety of applications. The derivation of
the KdV is given in [[KdV Equation Derivation]]. The KdV equation
is written as
<center><math>
\partial _{t}u+6u\partial _{x}u+\partial _{x}^{3}u=0.
</math></center>

More information about it can be found at [http://en.wikipedia.org/wiki/Korteweg%E2%80%93de_Vries_equation Korteweg de Vries equation]

==Travelling Wave Solution==

The KdV equation posesses travelling wave solutions. One particular
travelling wave solution is called a soltion and it was discovered
experimentally by [http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell]
in 1834. However, it was not understood
theoretically until the work of [http://en.wikipedia.org/wiki/Diederik_Korteweg Korteweg] and
[http://en.wikipedia.org/wiki/Gustav_de_Vries de Vries] in 1895.

We begin with the assumption that the wave travels with constant form, i.e.
is of the form
<center><math>
u\left( x,t\right) =f\left( x-ct\right)
</math></center>
Note that in this equation the parameter <math>c</math> is an unknown as is the
function <math>f.</math>
Only very special values of <math>c</math> and <math>f</math> will give travelling
waves.
We introduce the coordinate <math>\zeta = x - ct</math>.
If we substitute this expression into the KdV equation we obtain
<center><math>
-c\frac{\mathrm{d}f}{\mathrm{d}\zeta}+6f\frac{\mathrm{d}f}{\mathrm{d}\zeta}+\frac{\mathrm{d}^{3}f}{\mathrm{d}\zeta^{3}}=0
</math></center>
We can integrate this with respect to <math>\zeta</math> to obtain
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
where <math>A</math> is a constant of integration.

If we think about this equation as Newton's second law in a potential well <math>
V(f) </math> then the equation is
<center><math>
\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}= -\frac{\mathrm{d}V}{\mathrm{d}f}
</math></center>
then the potential well is given by
<center><math>
V\left( f\right) =-A_{0}-A_{1}f-c\frac{f^{2}}{2}+f^{3}
</math></center>
Therefore our equation for <math>f</math> may be thought of as the motion of a particle
in a cubic well.

The constant <math>A_0</math> has no effect on our solution so we can set it to be zero.
We can choose the constant <math>A_{1}=0</math> and then we have a
maximum at <math>f=0</math>. There is a solution which rolls from this at <math>t=-\infty </math>
and then runs up the other side and finally returns to the maximum at <math>
t=\infty .</math> This corresponds to a solitary wave solution.

We can also think about the equation as a first order system using <math>
f^{^{\prime }}=v.</math> This gives us
<center><math>\begin{matrix}
\frac{\mathrm{d}f}{\mathrm{d}\zeta} &=&v \\
\frac{\mathrm{d}v}{\mathrm{d}\zeta} &=&A_{1}+cf-3f^{2}
\end{matrix}</math></center>
If we chose <math>A_{1}=0</math> then we obtain two equilibria at <math>(f,v)=\left(
0,0\right) </math> and <math>(c/3,0).</math> If we analysis these equilibria we find the
first is a saddle and the second is a nonlinear center (it is neither repelling nor
attracting).
The Jacobian matrix for the saddle point is
<center><math>
J_{\left( 0,0\right) }=\left(
\begin{array}{cc}
0 & 1 \\
c & 0
\end{array}
\right)
</math></center>
which has eigenvalues at <math>\pm \sqrt{c}</math> and the incident directions are
<center><math>
\left(
\begin{array}{c}
1\\
\sqrt{c}
\end{array}
\right) \leftrightarrow \sqrt{c},\left(
\begin{array}{c}
-1 \\
\sqrt{c}
\end{array}
\right) \leftrightarrow -\sqrt{c}
</math></center>
There is a
homoclinic connection which connects the equilibrium point at the origin. This holoclinic
connection represents the solitary wave. Within this homoclinic connection
lie periodic orbits which represent the cnoidal waves.

{| class="wikitable"
|-
! Phase portrait
! Solitary Wave
! Cnoidal Wave
|-
| [[Image:Kdv phase portrait.jpg|thumb|350px|Phase portrait]]
| [[Image:Kdv wave solitary2.gif|thumb|350px|Solitary Wave]]
| [[Image:Kdv wave cn.gif|thumb|350px|Cnoidal Wave]]
|}

We can also integrate the equation
<center><math>
-cf+3f^{2}+\frac{\mathrm{d}^{2}f}{\mathrm{d}\zeta^{2}}=A_{1}
</math></center>
by multiplying by <math>f^{\prime }</math>and integrating. This gives us
<center><math>
\frac{\left( f^{^{\prime }}\right) ^{2}}{2}=A_{0}+A_{1}f + c\frac{
f^{2}}{2}-f^{3} = -V(f)
</math></center>
It is no coincidence that the right hand side is the potential energy, because this
is nothing more that the equation for conservation of energy (or the first
integral of the Lagrangian system) which does not depend on <math>\zeta</math>.

This is a separable equation and the only challenge is to integrate
<center><math>
\int \frac{1}{\sqrt{-2V(f)}} \mathrm{d}f.
</math></center>

==Formula for the solitary wave==

We know that the solitary wave solution is found when <math>
A_{0}=A_{1}=0.</math> This gives us
<center><math>
\left( f^{^{\prime }}\right) ^{2}=f^{2}\left( c-2f\right)
</math></center>
This can be solved by separation of variables to give
<center><math>
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}}=\int \mathrm{d}\zeta
</math></center>
We then substitute
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( s\right)
</math></center>
and note that
<center><math>
c-2f=c\left( 1-\mathrm{sech}^{2}\left( s\right) \right) =c\tanh ^{2}\left( f\right)
</math></center>
and that
<center><math>
\frac{\mathrm{d}f}{\mathrm{d}s}=-c\frac{\sinh \left( f\right) }{\cosh ^{3}\left( f\right) }
</math></center>
This means that
<center><math>\begin{matrix}
\int \frac{\mathrm{d}f}{f\sqrt{c-2f}} &=&-\frac{2}{\sqrt{c}}\int \frac{\sinh \left(
f\right) }{\mathrm{sech}^{2}\left( s\right) \tanh \left( f\right) \cosh ^{3}\left(
f\right) }\mathrm{d}s \\
&=&-\frac{2}{\sqrt{c}}s
\end{matrix}</math></center>
This gives us
<center><math>
-\frac{2}{\sqrt{c}}s=\zeta+a
</math></center>
Therefore
<center><math>
f=\frac{1}{2}c\,\mathrm{sech}^{2}\left( \frac{\sqrt{c}}{2}\left( \zeta+a\right) \right)
</math></center>
Of course we assumed that <math>x=x-ct</math> so the formula for the solitary wave is
given by
<center><math>
f\left( x-ct\right) =\frac{1}{2}c\,\mathrm{sech}^{2}\left[ \frac{\sqrt{c}}{2}\left(
x-ct+a\right) \right]
</math></center>
Note that a solution exists for each
<math>c</math>, and that the amplitude is proportional to <math>c.</math> All of this was
discovered experimentally by
[http://en.wikipedia.org/wiki/John_Scott_Russell John Scott Russell].

==Formula for the cnoidal wave==

If we consider the case when the solution oscillates between two values <math>F_2 < F_3</math>
(which we can assume are also roots of <math>V(f)</math> without loss of generality) then
we can integrate the equation to obtain
<center><math>
f\left( \zeta\right) =F_2+(F_3 - F_2) \mathrm{cn}^{2}\left( \gamma \zeta ;k\right)
</math></center>
where <math>cn</math> is a Jacobi Elliptic function and
<math>\gamma</math> and <math>k</math> are constants which depend on <math>c</math>.
Derivation of this equation is found [[KdV Cnoidal Wave Solutions]].
We can write this equation as
<center><math>
f\left( x-ct\right) =a+b \mathrm{cn}^{2}\left( \gamma (x-ct);k\right)
</math></center>
where <math>a=k^2\gamma^2</math> and <math>c = 6b + 4(2k^2 -1)\gamma^2</math>.
These waves are known as [http://en.wikipedia.org/wiki/Cnoidal_wave cnoidal waves].

In the limit the two solutions agree. We also obtain a sinusoidal solution in the limit of
small amplitude.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|SzZ-KhvvPio}}

=== Part 2 ===

{{#ev:youtube|QltlSQQBtrs}}

=== Part 3 ===

{{#ev:youtube|NJ7h3Z9QtvU}}

=== Part 4 ===

{{#ev:youtube|NictSlSgRbM}}

Nonlinear Shallow Water Waves

2025-08-24T21:41:01Z

Levi: /* Speed of the shock */ fix

{{nonlinear waves course
| chapter title = Nonlinear Shallow Water Waves
| next chapter = [[Introduction to KdV]]
| previous chapter = [[Traffic Waves]]
}}

{{complete pages}}

== Introduction ==

We assume that water is incompressible,
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
This allows us to assume [[:Category:Shallow Depth|Shallow Depth]]. We assume that the problem has not variation
in either the <math>y</math> or <math>z</math> direction. The fluid is governed by two parameters,
<math>u(x,t)</math>, the velocity of the water, and <math>h(x,t)</math> the water depth (note that this is not the still water depth since the problem
is nonlinear).

The theory we present here is discussed in [[Stoker 1957]], [[Billingham and King 2000]] and [[Johnson 1997]].

== Equations of Motion ==

The equation for the conservation of mass can derived by considering a a region <math>[x,x+\Delta x]</math>
Conservation of mass then implies that
<center>
<math>
\partial_t \int_x^{x + \Delta x} \rho h(s,t) \mathrm{d}s = \rho h(x,t)u(x,t) - \rho h(x+\Delta x,t)u(x+\Delta x,t)
</math>
</center>
If we take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0
</math>
</center>

A second equation comes from conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x + \Delta x}^{x}
+ \int_0^{h(x)} P(x,z,t) \mathrm{d}z -
\int_0^{h(x + \Delta x)} P(x+\Delta x,z,t) \mathrm{d}z
</math>
</center>
where <math>\rho \,</math> denotes density, and the pressure <math>P \,</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). This then gives us
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= -\left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \frac{1}{2}\rho g {h(x)}^2 -
\frac{1}{2}\rho g {h(x + \Delta x)}^2
</math>
</center>
If we then take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t \left( h u \right)
+ \partial_x \left(u^2 h + \frac{1}{2} gh^2\right) =0
</math>
</center>
We can simplify this using the equation derived from conservation of mass to
to obtain
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>

The equations
<center>
<math>
\partial_t h + u \partial_x h + h \partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>
are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

We can rewrite them in terms of the local wave speed <math>c(x, t) = \sqrt{gh(x, t)}</math> as follows:
<center>
<math>2\partial_t c + 2u\partial_x c + c\partial_x u = 0
</math>
</center>
<center>
<math>\partial_t u + u\partial_x u + 2c \partial_x c = 0
</math>
</center>
These equation are almost identical to those of compressible gas dynamics. Much of our understanding
of the equations for water have been found by researchers studying compressible gas dynamics.

== Linearized Equations ==

We can linearize these equations by assuming that <math>u</math> is small and that
<math>h=h_0 + \zeta </math> where <math>h_0</math> is the average water depth
and <math>\zeta</math> is also assumed small. This gives us
<center>
<math>
\partial_t \zeta + h_0\partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + g \partial_x \zeta = 0
</math>
</center>
These linear shallow water equations which can be
derived from the linear equations for water of finite
depth and taking the limit of small depth (see [[:Category:Shallow Depth|Shallow Depth]]).

== Characteristics ==

The equations possess characteristics.
Adding and subtracting the two equations above we obtain
<center>
<math>
\frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0
</math>
</center>
This means that
on the <math>\;C_+</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_+}{\mathrm{d} t} = u + c = u + \sqrt{gh}
</math>
</center>
the <math>\;C_+</math> invariant
<center>
<math>
R_+ = u + 2c = u + 2\sqrt{gh}
</math>
</center>
is a constant,
and on the <math>\;C_-</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - c = u - \sqrt{gh}
</math>
</center>
the <math>\;C_-</math> invariant
<center>
<math>
R_- = u - 2c = u - 2\sqrt{gh}
</math>
</center>
is a constant.

The functions <math>R_{\pm} (u ,c) = u \pm 2c</math>, are called the Riemannian invariants.

== Simple Waves ==

The problem as formulated can be solved by advancing the solution along the characteristics, but
this will in general be quite difficult analytically. However, there is a special class of problems,
called ''Simple Waves'' in which the solution only changes on one characteristic.
They are best illustrated through some examples. Note that the characteristic can meet forming
a shock, which is called a [http://en.wikipedia.org/wiki/Tidal_bore bore] or a
[http://en.wikipedia.org/wiki/Hydraulic_jump hydraulic jump]
when it occurs on the surface of the water.

=== The dam break problem ===
Assume the water occupies the region <math>{x < 0 ; 0 < z < h_0 }</math> initially held back by a dam at <math>x = 0</math>.
At <math>t = 0</math>, the dam is removed (breaks). What is the height of the water
<math>h(x,t) \,</math> for <math>t > 0? \,</math> The initial condition is therefore
<center>
<math>h(x,0) = \begin{cases}
h_0, & x < 0 \\
0, & x > 0
\end{cases}
</math>
</center>
<center>
<math>u(x ,0) = 0. \,
</math>
</center>
On the characteristic that originates at <math>t = 0</math> for <math>x < 0</math>,
<center>
<math>
R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_0} = \pm 2c_0
</math>
</center>
where <math>c_0 = \sqrt{gh_0}</math> is the initial (linear) wave speed.

Therefore, if a <math>C_+</math> and a <math>C_-</math> characteristic from this region intersect, then
<center>
<math>
u + 2\sqrt{gh} = 2c_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0
</math>
</center>
and hence, <math>u = 0</math> and <math>h = h_0</math>.
Moreover,
<center>
<math>
\frac{\mathrm{d} X_\pm}{\mathrm{d} t} = u \pm \sqrt{gh} = \pm c_0
</math>
</center>
so these characteristics are straight lines in the region <math>\big\{x < -c_0 t \big\}</math>
(the undisturbed region).

The <math>\;C_+</math> characteristic leave the region a<math>\big\{x < -c_0 t \big\}</math>
and enter <math>\big\{x > -c_0 t \big\}</math>. For now we will assume that these characteristics fill the domain
(and show that this is true shortly).
For <math>\big\{x > -c_0 t \big\}</math>, the <math>C_-</math> characteristics are given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - \sqrt{gh}
</math>
</center>
and on each of the <math>C_-</math> characteristics <math>R_- = u - 2\sqrt{gh}</math> is constant.
However, since this region is filled with <math>C_+</math> characteristics where <math>R_+ = u + 2\sqrt{gh} = 2c_0</math>, <math>u</math> and <math>h</math>
must be constant on each <math>C_-</math> characteristic.
This also means that the <math>C_-</math> characteristics must be straight lines.

Since the fluid occupies <math>\big\{x < 0 \big\}</math> at <math>t = 0</math>,
these <math>C_-</math> characteristics must start at the origin, with
<center>
<math>
X_-(t) = \left(u - \sqrt{gh}\right)t
</math>
</center>
which in turn implies that
<center>
<math>
u - \sqrt{gh} = \frac{x}{t}
</math>
</center>
We also have <math>R_+ = u + 2\sqrt{gh} = 2c_0</math> from the
<math>C_+</math> characteristics. We can solve these equations
at each point in <math>\big\{x > -c_0 t \big\}</math>. Solving for <math>u</math> and <math>h</math> gives
<center>
<math>
h(x, t) =
\begin{cases}
\frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2, \quad -c_0 t < x< 2 c_0 t,\\
h_0,\quad x <-c_0 t,
\end{cases}
</math>
</center>
<center>
<math>
u(x, t) =
\begin{cases}
\frac{2}{3} \left (c_0 + \frac{x}{t} \right ), \quad -c_0 t < x< 2 c_0 t,\\
0,\quad x <-c_0 t.
\end{cases}
</math>
</center>
Where we have assumed that, since <math>h = 0</math> for <math>x = 2c_0 t,</math>
the <math>C_+</math> characteristic only exist in the region <math>\big\{x < 2c_0 t \big\}</math>,
We will verify this by explicitly calculating them.

It remains to determine the <math>C_+</math> characteristic, which originated in <math>\big\{x < 0 \big\}</math>,
and show they
will fill the domain <math>\big\{x < 2c_0 t \big\}</math>. For <math>\big\{x < -c_0 t \big\}</math>,
the <math>\;C_+</math> characteristics are straight lines with slope <math>c_0</math> and are given by
<center>
<math>
X_+ (t) = -x_0 + c_0 t, \quad \left(x_0 > 0,\;\; t < \frac{x_0}{2c_0}\right)
</math>
</center>
When <math>t = \frac{x_0}{2c_0},\;\;\ X_{+} (t) = -c_0 t</math> so that for
<center>
<math>
t > \frac{x_0}{2c_0}, \quad \frac{\mathrm{d} X_{+} (t)}{\mathrm{d} t} = u + \sqrt{gh}
</math>
</center>
and substituting the solution we found for <math>h</math> and <math>u</math>
<center>
<math>
\frac{\mathrm{d}X_{+} (t)}{\mathrm{d} t} = \frac{4}{3}c_0 + \frac{X_{+} (t)}{3t}
</math>
</center>
Solving this ODE subject to <math>X_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2}</math> gives
<center>
<math>
X_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\;
</math>
</center>
the equation for a characteristic curve.
The curves indeed fill the domain <math>\big\{x < 2c_0 t \big\}</math>
and all satisfy <math>\big\{X_+ (t) < 2c_0 t \big\}</math>. To summarize, the
<math>C^{+}</math> characteristics are given by
<center>
<math>
X_+ (t) =
\begin{cases}
2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\quad t> x_0/2 c_0\\
-x_0 + c_0 t, \quad 0\ < t < x_0/2 c_0
\end{cases}
</math>
</center>

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Characteristics_dam_break.jpg|thumb|right|500px|Characteristics for the dam
break problem, blue for <math>C_{+}</math> and red for <math>C_{-}</math>. The solid red lines
show the curves <math>x=-c_0 t</math> and <math>x=2c_0 t</math> (note we have assumed
here that <math>c_0 =1</math>]]
| [[Image:Dambreak.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the Dam Break problem]]
|}

== Shocks ==

For a unique solution two exist there must be a single <math>C_+</math> and <math>C_-</math>
characteristic through each point. When two characteristics of the same kind meet we
have a shock forming.

=== Accelerating Piston ===

We now consider the problem of water initially at rest occupying the
half space <math>x>0</math> which is initially at rest. At <math>t=0</math>
the piston at <math>x=0</math> begins to move to the right with constant
acceleration <math>a</math> so that the position of the piston is given by
<math>\frac{1}{2}at^2</math>.

We assume that the <math>C_-</math> characteristics which originate in the
water at <math>t=0</math> fill the fluid. On these characteristics
<center>
<math>
R_- = u - 2 c = -2c_0 \,
</math>
</center>
and this condition must hold throughout the fluid.
On the <math>C_+</math> characteristics we know that
<math> R_+ = u + 2 c </math> must be a constant and hence on the
<math>C_+</math> characteristics <math>u</math> and <math>c</math>
must be constant and hence the <math>C_+</math> characteristics must
be straight lines. Note that this does not mean that the
<math>C_+</math> characteristics have the same slope and there is no
requirement that the <math>C_-</math> characteristics are straight lines.

The <math>C_+</math> characteristic originate from the fluid
or from the front of the piston. We consider those which originate from
the piston. The <math>C_+</math> characteristic which originates from
the piston at <math>t=t_0</math> must satisfy
<center>
<math>
u+2c = a t_0 + 2c_{\text{plate}} \,
</math>
</center>
where <math> a t_0</math> is the velocity of the piston at time <math>t=t_0</math>
and <math>c_{\text{plate}}</math> is the speed (related to height) at the plate.
We know that <math> R_- = u - 2 c =-2c_0</math> through out the fluid, so that if
we solve this at the plate (where <math>u=at_0</math> and <math>c=c_{\text{plate}}</math>)
then we get
<center><math>
c_{\text{plate}} = at_0/2 + c_0\,
</math></center>

On the <math>C_+</math> characteristics <math>u</math> and <math>c</math>
are constant and therefore
<center>
<math>
\frac{\mathrm{d}X_+}{\mathrm{d}t} = u+c = at_0 + \left( \frac{1}{2}at_0 + c_0 \right)
</math>
</center>
Hence
<center>
<math>
X_+(t,t_0) = \left( \frac{3}{2} a t_0 + c_0 \right) t - c_0 t_0 -a t_0^2
</math>
</center>
using the condition <math>X_+(t_0,t_0) = \frac{1}{2} a t_0^2</math> (the initial value which
comes from the position of the piston at <math>t=t_0</math>).
The slope of these lines increases and eventually meet to form a shock.
We find this point of intersection by considering neighboring characteristics
and determining when they first intersect.
<center>
<math>
X_+(t,t_0 + \Delta t) = X_+(t,t_0) + \Delta t \frac{\partial X_+}{\partial t_0} (t,t_0) = X_+(t,t_0)
</math>
</center>
It follows that neighbouring characteristics will meet when
<center>
<math>
\frac{\partial X_+}{\partial t_0} (t,t_0) = 0
</math>
</center>
which implies that
<center>
<math>
t = \frac{2c_0}{3a} + \frac{4}{3}t_0
</math>
</center>
The first time that a shock forms is the minimum value of this equation.
For this piston example, this occurs when <math>t_0 = 0</math> and the value
of <math>t</math> is <math>t = 2c_0/(3a)</math>. At this point
a shock is formed and we can no longer find a unique solution by following the
characteristics.

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Accelerating_piston.jpg|thumb|right|500px| <math>C_{+}</math> characteristics for the
accelerating piston, red undisturbed, blue from the piston and the green line shows the transition.]]
| [[Image:Accelerating_piston2.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the
accelerating piston.]]
|}

=== Piston Moving with Constant Velocity ===

This example is also known as the Moving Wall Problem, and is connected to Shallow Water Bores.

We consider the case of a piston, with positive constant velocity <math>V</math> (which is initially at <math>x=0</math>), advancing into a semi-infinite expanse of
fluid that is initially at rest with depth <math>h_0</math>.

The <math>C_+</math> characteristics which originate in the fluid
at <math>t=0</math> have slope
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0}</math>
</center>
and the <math>C_+</math> characteristic which originate at the piston at
<math>t=0</math> must satisfy
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0} + \frac{3}{2} V </math>
</center>
so that these two characteristics will intersect at <math>t=0</math>.
Therefore a shock forms immediately and we can track this by determining the
speed of the shock

=== Speed of the shock ===

We need the conservation equations in integral form to determine the speed
of the shock. Conservation of mass, written as an integral is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h \mathrm{d}x
+ \left. \rho u h \right|_{x_1}^{x_2} =0
</math>
</center>
If the shock is located at <math>s(t)</math>
which we assume is located between <math>x_1</math>
and <math>x_2</math>, then
<center>
<math>
\partial_t \int_{x_1}^{x_2} h \mathrm{d}x
= \partial_t \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) h \mathrm{d}x
= \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) \partial_t h \mathrm{d}x - h^{+} \partial_t s(t) + h^{-}\partial_t s(t),
</math>
</center>
where <math>h^+</math> is the height on the right (positive) side of
the jump and <math>h^-</math> is the height on the left (negative) side.
If we take the limit as <math>x_1\to x_2</math> we then obtain the following
identity
<center>
<math>
h^{+}\partial_t s(t) - h^{-}\partial_t s(t) -
u^{+} h^{+} + u^{-} h^{-} = 0
</math>
</center>
where <math>u^+</math> is the flow speed on the right (positive) side of
the jump and <math>u^-</math> is the flow speed on the left (negative) side.

We now need to consider the equation for conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x_2}^{x_1}
+ \int_0^{h(x_1)} P(x_1,z,t) \mathrm{d}z -
\int_0^{h(x_2)} P(x_2,z,t) \mathrm{d}z
</math>
</center>
where the pressure <math>P</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). We can apply a similar argument as before to obtain
<center>
<math>
h^{+}u^{+}\partial_t s(t) - h^{-}u^{-}\partial_t s(t) =
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2
</math>
</center>

=== Hydraulic Jump ===

For a hydraulic jump, <math>\dot{s}(t) = 0</math>, which means that we must solve
<center>
<math>
u^{+} h^{+} - u^{-} h^{-} = 0 \,
</math>
</center>
<center>
<math>
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2 =0
</math>
</center>
If we introduce the variables
<center>
<math>
H = \frac{h^{+}}{h^{-}}
</math>
</center>
and
<center>
<math>
\mathrm{Fr} = \frac{u^{-}}{\sqrt{gh^{-}}}
</math>
</center>
where <math>\mathrm{Fr}</math> is the ''Froude'' number
which is equivalent to the Mach number for gas dynamics.
Then we obtain
<center>
<math>
H^2 -1 = 2 \mathrm{Fr}^2 \left(1 - \frac{1}{H}\right)
</math>
</center>
This expression has the roots
<center>
<math>
H=1, \quad H=\frac{1}{2}\left(-1\pm\sqrt{1 + 8 \mathrm{Fr}^2}\right)
</math>
</center>
The only physically meaningful solution is the root which satisfies
<math>H>1</math>. This is only true providing <math>\mathrm{Fr} > 1</math>, which means
that we can only obtain a hydraulic jump if the flow is supercritical.

Below is a video of a hydraulic jump. You can clearly see the point where the flow is changing from supercritical to subcritical (look for the small turbulent region in the channel)
{{#ev:youtube|5etwhZ0d2GU}}

=== Shallow Water Bore ===
We now consider a bore, in which
the shock wave advances into still water.
We denote the fluid speed by <math>V = u^{-}</math>.
We denote the height on the wall side
by <math>h_1</math> and the height on the other side must be <math>h_0</math>, i.e.
<math>h^{+} = h_0</math>, <math>h^{-} = h_1</math>, <math>u^{+} = 0</math>.
This means that
<center>
<math>
h_{0}\partial_t s(t) - h_{1}\partial_t s(t)
+ V h_{1}= 0
</math>
</center>
and
<center>
<math>
- h_{1}V\partial_t s(t) = - \left(V\right)^2 h_{1}
+ \frac{1}{2} g \left(h_{0}\right)^2 - \frac{1}{2} g \left(h_{1}\right)^2
</math>
</center>
which can be solved to obtain the shock speed and the height of the moving fluid.

Below is a video of surfing on the [http://en.wikipedia.org/wiki/Severn Severn] bore, do not believe everything they
say. You might also want to check out the [http://en.wikipedia.org/wiki/Pororoca Pororoca]
a tidal bore on the Amazon.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|Yi8fytUszrQ}}

=== Part 2 ===

{{#ev:youtube|_668xmJ9DAQ}}

=== Part 3 ===

{{#ev:youtube|yTat11ERJMg}}

=== Part 4 ===

{{#ev:youtube|qG8bPJwX96k}}

=== Part 5 ===

{{#ev:youtube|38ZWW2dM1Qs}}

=== Part 6 ===

{{#ev:youtube|AZHlhA5pQBg}}

=== Part 7 ===

{{#ev:youtube|VhlY5lQcTlM}}

[[Category:Simple Nonlinear Waves]]

[[Category:Simple Nonlinear Waves]]
[[Category:Nonlinear Water-Wave Theory]]
[[Category:Shallow Depth]]

Nonlinear Shallow Water Waves

2025-08-23T23:12:11Z

Levi: /* Speed of the shock */ fix signs

{{nonlinear waves course
| chapter title = Nonlinear Shallow Water Waves
| next chapter = [[Introduction to KdV]]
| previous chapter = [[Traffic Waves]]
}}

{{complete pages}}

== Introduction ==

We assume that water is incompressible,
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
This allows us to assume [[:Category:Shallow Depth|Shallow Depth]]. We assume that the problem has not variation
in either the <math>y</math> or <math>z</math> direction. The fluid is governed by two parameters,
<math>u(x,t)</math>, the velocity of the water, and <math>h(x,t)</math> the water depth (note that this is not the still water depth since the problem
is nonlinear).

The theory we present here is discussed in [[Stoker 1957]], [[Billingham and King 2000]] and [[Johnson 1997]].

== Equations of Motion ==

The equation for the conservation of mass can derived by considering a a region <math>[x,x+\Delta x]</math>
Conservation of mass then implies that
<center>
<math>
\partial_t \int_x^{x + \Delta x} \rho h(s,t) \mathrm{d}s = \rho h(x,t)u(x,t) - \rho h(x+\Delta x,t)u(x+\Delta x,t)
</math>
</center>
If we take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0
</math>
</center>

A second equation comes from conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x + \Delta x}^{x}
+ \int_0^{h(x)} P(x,z,t) \mathrm{d}z -
\int_0^{h(x + \Delta x)} P(x+\Delta x,z,t) \mathrm{d}z
</math>
</center>
where <math>\rho \,</math> denotes density, and the pressure <math>P \,</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). This then gives us
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= -\left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \frac{1}{2}\rho g {h(x)}^2 -
\frac{1}{2}\rho g {h(x + \Delta x)}^2
</math>
</center>
If we then take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t \left( h u \right)
+ \partial_x \left(u^2 h + \frac{1}{2} gh^2\right) =0
</math>
</center>
We can simplify this using the equation derived from conservation of mass to
to obtain
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>

The equations
<center>
<math>
\partial_t h + u \partial_x h + h \partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>
are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

We can rewrite them in terms of the local wave speed <math>c(x, t) = \sqrt{gh(x, t)}</math> as follows:
<center>
<math>2\partial_t c + 2u\partial_x c + c\partial_x u = 0
</math>
</center>
<center>
<math>\partial_t u + u\partial_x u + 2c \partial_x c = 0
</math>
</center>
These equation are almost identical to those of compressible gas dynamics. Much of our understanding
of the equations for water have been found by researchers studying compressible gas dynamics.

== Linearized Equations ==

We can linearize these equations by assuming that <math>u</math> is small and that
<math>h=h_0 + \zeta </math> where <math>h_0</math> is the average water depth
and <math>\zeta</math> is also assumed small. This gives us
<center>
<math>
\partial_t \zeta + h_0\partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + g \partial_x \zeta = 0
</math>
</center>
These linear shallow water equations which can be
derived from the linear equations for water of finite
depth and taking the limit of small depth (see [[:Category:Shallow Depth|Shallow Depth]]).

== Characteristics ==

The equations possess characteristics.
Adding and subtracting the two equations above we obtain
<center>
<math>
\frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0
</math>
</center>
This means that
on the <math>\;C_+</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_+}{\mathrm{d} t} = u + c = u + \sqrt{gh}
</math>
</center>
the <math>\;C_+</math> invariant
<center>
<math>
R_+ = u + 2c = u + 2\sqrt{gh}
</math>
</center>
is a constant,
and on the <math>\;C_-</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - c = u - \sqrt{gh}
</math>
</center>
the <math>\;C_-</math> invariant
<center>
<math>
R_- = u - 2c = u - 2\sqrt{gh}
</math>
</center>
is a constant.

The functions <math>R_{\pm} (u ,c) = u \pm 2c</math>, are called the Riemannian invariants.

== Simple Waves ==

The problem as formulated can be solved by advancing the solution along the characteristics, but
this will in general be quite difficult analytically. However, there is a special class of problems,
called ''Simple Waves'' in which the solution only changes on one characteristic.
They are best illustrated through some examples. Note that the characteristic can meet forming
a shock, which is called a [http://en.wikipedia.org/wiki/Tidal_bore bore] or a
[http://en.wikipedia.org/wiki/Hydraulic_jump hydraulic jump]
when it occurs on the surface of the water.

=== The dam break problem ===
Assume the water occupies the region <math>{x < 0 ; 0 < z < h_0 }</math> initially held back by a dam at <math>x = 0</math>.
At <math>t = 0</math>, the dam is removed (breaks). What is the height of the water
<math>h(x,t) \,</math> for <math>t > 0? \,</math> The initial condition is therefore
<center>
<math>h(x,0) = \begin{cases}
h_0, & x < 0 \\
0, & x > 0
\end{cases}
</math>
</center>
<center>
<math>u(x ,0) = 0. \,
</math>
</center>
On the characteristic that originates at <math>t = 0</math> for <math>x < 0</math>,
<center>
<math>
R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_0} = \pm 2c_0
</math>
</center>
where <math>c_0 = \sqrt{gh_0}</math> is the initial (linear) wave speed.

Therefore, if a <math>C_+</math> and a <math>C_-</math> characteristic from this region intersect, then
<center>
<math>
u + 2\sqrt{gh} = 2c_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0
</math>
</center>
and hence, <math>u = 0</math> and <math>h = h_0</math>.
Moreover,
<center>
<math>
\frac{\mathrm{d} X_\pm}{\mathrm{d} t} = u \pm \sqrt{gh} = \pm c_0
</math>
</center>
so these characteristics are straight lines in the region <math>\big\{x < -c_0 t \big\}</math>
(the undisturbed region).

The <math>\;C_+</math> characteristic leave the region a<math>\big\{x < -c_0 t \big\}</math>
and enter <math>\big\{x > -c_0 t \big\}</math>. For now we will assume that these characteristics fill the domain
(and show that this is true shortly).
For <math>\big\{x > -c_0 t \big\}</math>, the <math>C_-</math> characteristics are given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - \sqrt{gh}
</math>
</center>
and on each of the <math>C_-</math> characteristics <math>R_- = u - 2\sqrt{gh}</math> is constant.
However, since this region is filled with <math>C_+</math> characteristics where <math>R_+ = u + 2\sqrt{gh} = 2c_0</math>, <math>u</math> and <math>h</math>
must be constant on each <math>C_-</math> characteristic.
This also means that the <math>C_-</math> characteristics must be straight lines.

Since the fluid occupies <math>\big\{x < 0 \big\}</math> at <math>t = 0</math>,
these <math>C_-</math> characteristics must start at the origin, with
<center>
<math>
X_-(t) = \left(u - \sqrt{gh}\right)t
</math>
</center>
which in turn implies that
<center>
<math>
u - \sqrt{gh} = \frac{x}{t}
</math>
</center>
We also have <math>R_+ = u + 2\sqrt{gh} = 2c_0</math> from the
<math>C_+</math> characteristics. We can solve these equations
at each point in <math>\big\{x > -c_0 t \big\}</math>. Solving for <math>u</math> and <math>h</math> gives
<center>
<math>
h(x, t) =
\begin{cases}
\frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2, \quad -c_0 t < x< 2 c_0 t,\\
h_0,\quad x <-c_0 t,
\end{cases}
</math>
</center>
<center>
<math>
u(x, t) =
\begin{cases}
\frac{2}{3} \left (c_0 + \frac{x}{t} \right ), \quad -c_0 t < x< 2 c_0 t,\\
0,\quad x <-c_0 t.
\end{cases}
</math>
</center>
Where we have assumed that, since <math>h = 0</math> for <math>x = 2c_0 t,</math>
the <math>C_+</math> characteristic only exist in the region <math>\big\{x < 2c_0 t \big\}</math>,
We will verify this by explicitly calculating them.

It remains to determine the <math>C_+</math> characteristic, which originated in <math>\big\{x < 0 \big\}</math>,
and show they
will fill the domain <math>\big\{x < 2c_0 t \big\}</math>. For <math>\big\{x < -c_0 t \big\}</math>,
the <math>\;C_+</math> characteristics are straight lines with slope <math>c_0</math> and are given by
<center>
<math>
X_+ (t) = -x_0 + c_0 t, \quad \left(x_0 > 0,\;\; t < \frac{x_0}{2c_0}\right)
</math>
</center>
When <math>t = \frac{x_0}{2c_0},\;\;\ X_{+} (t) = -c_0 t</math> so that for
<center>
<math>
t > \frac{x_0}{2c_0}, \quad \frac{\mathrm{d} X_{+} (t)}{\mathrm{d} t} = u + \sqrt{gh}
</math>
</center>
and substituting the solution we found for <math>h</math> and <math>u</math>
<center>
<math>
\frac{\mathrm{d}X_{+} (t)}{\mathrm{d} t} = \frac{4}{3}c_0 + \frac{X_{+} (t)}{3t}
</math>
</center>
Solving this ODE subject to <math>X_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2}</math> gives
<center>
<math>
X_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\;
</math>
</center>
the equation for a characteristic curve.
The curves indeed fill the domain <math>\big\{x < 2c_0 t \big\}</math>
and all satisfy <math>\big\{X_+ (t) < 2c_0 t \big\}</math>. To summarize, the
<math>C^{+}</math> characteristics are given by
<center>
<math>
X_+ (t) =
\begin{cases}
2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\quad t> x_0/2 c_0\\
-x_0 + c_0 t, \quad 0\ < t < x_0/2 c_0
\end{cases}
</math>
</center>

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Characteristics_dam_break.jpg|thumb|right|500px|Characteristics for the dam
break problem, blue for <math>C_{+}</math> and red for <math>C_{-}</math>. The solid red lines
show the curves <math>x=-c_0 t</math> and <math>x=2c_0 t</math> (note we have assumed
here that <math>c_0 =1</math>]]
| [[Image:Dambreak.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the Dam Break problem]]
|}

== Shocks ==

For a unique solution two exist there must be a single <math>C_+</math> and <math>C_-</math>
characteristic through each point. When two characteristics of the same kind meet we
have a shock forming.

=== Accelerating Piston ===

We now consider the problem of water initially at rest occupying the
half space <math>x>0</math> which is initially at rest. At <math>t=0</math>
the piston at <math>x=0</math> begins to move to the right with constant
acceleration <math>a</math> so that the position of the piston is given by
<math>\frac{1}{2}at^2</math>.

We assume that the <math>C_-</math> characteristics which originate in the
water at <math>t=0</math> fill the fluid. On these characteristics
<center>
<math>
R_- = u - 2 c = -2c_0 \,
</math>
</center>
and this condition must hold throughout the fluid.
On the <math>C_+</math> characteristics we know that
<math> R_+ = u + 2 c </math> must be a constant and hence on the
<math>C_+</math> characteristics <math>u</math> and <math>c</math>
must be constant and hence the <math>C_+</math> characteristics must
be straight lines. Note that this does not mean that the
<math>C_+</math> characteristics have the same slope and there is no
requirement that the <math>C_-</math> characteristics are straight lines.

The <math>C_+</math> characteristic originate from the fluid
or from the front of the piston. We consider those which originate from
the piston. The <math>C_+</math> characteristic which originates from
the piston at <math>t=t_0</math> must satisfy
<center>
<math>
u+2c = a t_0 + 2c_{\text{plate}} \,
</math>
</center>
where <math> a t_0</math> is the velocity of the piston at time <math>t=t_0</math>
and <math>c_{\text{plate}}</math> is the speed (related to height) at the plate.
We know that <math> R_- = u - 2 c =-2c_0</math> through out the fluid, so that if
we solve this at the plate (where <math>u=at_0</math> and <math>c=c_{\text{plate}}</math>)
then we get
<center><math>
c_{\text{plate}} = at_0/2 + c_0\,
</math></center>

On the <math>C_+</math> characteristics <math>u</math> and <math>c</math>
are constant and therefore
<center>
<math>
\frac{\mathrm{d}X_+}{\mathrm{d}t} = u+c = at_0 + \left( \frac{1}{2}at_0 + c_0 \right)
</math>
</center>
Hence
<center>
<math>
X_+(t,t_0) = \left( \frac{3}{2} a t_0 + c_0 \right) t - c_0 t_0 -a t_0^2
</math>
</center>
using the condition <math>X_+(t_0,t_0) = \frac{1}{2} a t_0^2</math> (the initial value which
comes from the position of the piston at <math>t=t_0</math>).
The slope of these lines increases and eventually meet to form a shock.
We find this point of intersection by considering neighboring characteristics
and determining when they first intersect.
<center>
<math>
X_+(t,t_0 + \Delta t) = X_+(t,t_0) + \Delta t \frac{\partial X_+}{\partial t_0} (t,t_0) = X_+(t,t_0)
</math>
</center>
It follows that neighbouring characteristics will meet when
<center>
<math>
\frac{\partial X_+}{\partial t_0} (t,t_0) = 0
</math>
</center>
which implies that
<center>
<math>
t = \frac{2c_0}{3a} + \frac{4}{3}t_0
</math>
</center>
The first time that a shock forms is the minimum value of this equation.
For this piston example, this occurs when <math>t_0 = 0</math> and the value
of <math>t</math> is <math>t = 2c_0/(3a)</math>. At this point
a shock is formed and we can no longer find a unique solution by following the
characteristics.

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Accelerating_piston.jpg|thumb|right|500px| <math>C_{+}</math> characteristics for the
accelerating piston, red undisturbed, blue from the piston and the green line shows the transition.]]
| [[Image:Accelerating_piston2.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the
accelerating piston.]]
|}

=== Piston Moving with Constant Velocity ===

This example is also known as the Moving Wall Problem, and is connected to Shallow Water Bores.

We consider the case of a piston, with positive constant velocity <math>V</math> (which is initially at <math>x=0</math>), advancing into a semi-infinite expanse of
fluid that is initially at rest with depth <math>h_0</math>.

The <math>C_+</math> characteristics which originate in the fluid
at <math>t=0</math> have slope
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0}</math>
</center>
and the <math>C_+</math> characteristic which originate at the piston at
<math>t=0</math> must satisfy
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0} + \frac{3}{2} V </math>
</center>
so that these two characteristics will intersect at <math>t=0</math>.
Therefore a shock forms immediately and we can track this by determining the
speed of the shock

=== Speed of the shock ===

We need the conservation equations in integral form to determine the speed
of the shock. Conservation of mass, written as an integral is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h \mathrm{d}x
+ \left. \rho u h \right|_{x_1}^{x_2} =0
</math>
</center>
If the shock is located at <math>s(t)</math>
which we assume is located between <math>x_1</math>
and <math>x_2</math>, then
<center>
<math>
\partial_t \int_{x_1}^{x_2} h \mathrm{d}x
= \partial_t \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) h \mathrm{d}x
= \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) \partial_t h \mathrm{d}x - h^{+} \partial_t s(t) + h^{-}\partial_t s(t),
</math>
</center>
where <math>h^+</math> is the height on the right (positive) side of
the jump and <math>h^-</math> is the height on the left (negative) side.
If we take the limit as <math>x_1\to x_2</math> we then obtain the following
identity
<center>
<math>
h^{+}\partial_t s(t) - h^{-}\partial_t s(t) -
u^{+} h^{+} + u^{-} h^{-} = 0
</math>
</center>
where <math>u^+</math> is the height on the right (positive) side of
the jump and <math>u^-</math> is the height on the left (negative) side.

We now need to consider the equation for conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x_2}^{x_1}
+ \int_0^{h(x_1)} P(x_1,z,t) \mathrm{d}z -
\int_0^{h(x_2)} P(x_2,z,t) \mathrm{d}z
</math>
</center>
where the pressure <math>P</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). We can apply a similar argument as before to obtain
<center>
<math>
h^{+}u^{+}\partial_t s(t) - h^{-}u^{-}\partial_t s(t) =
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2
</math>
</center>

=== Hydraulic Jump ===

For a hydraulic jump, <math>\dot{s}(t) = 0</math>, which means that we must solve
<center>
<math>
u^{+} h^{+} - u^{-} h^{-} = 0 \,
</math>
</center>
<center>
<math>
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2 =0
</math>
</center>
If we introduce the variables
<center>
<math>
H = \frac{h^{+}}{h^{-}}
</math>
</center>
and
<center>
<math>
\mathrm{Fr} = \frac{u^{-}}{\sqrt{gh^{-}}}
</math>
</center>
where <math>\mathrm{Fr}</math> is the ''Froude'' number
which is equivalent to the Mach number for gas dynamics.
Then we obtain
<center>
<math>
H^2 -1 = 2 \mathrm{Fr}^2 \left(1 - \frac{1}{H}\right)
</math>
</center>
This expression has the roots
<center>
<math>
H=1, \quad H=\frac{1}{2}\left(-1\pm\sqrt{1 + 8 \mathrm{Fr}^2}\right)
</math>
</center>
The only physically meaningful solution is the root which satisfies
<math>H>1</math>. This is only true providing <math>\mathrm{Fr} > 1</math>, which means
that we can only obtain a hydraulic jump if the flow is supercritical.

Below is a video of a hydraulic jump. You can clearly see the point where the flow is changing from supercritical to subcritical (look for the small turbulent region in the channel)
{{#ev:youtube|5etwhZ0d2GU}}

=== Shallow Water Bore ===
We now consider a bore, in which
the shock wave advances into still water.
We denote the fluid speed by <math>V = u^{-}</math>.
We denote the height on the wall side
by <math>h_1</math> and the height on the other side must be <math>h_0</math>, i.e.
<math>h^{+} = h_0</math>, <math>h^{-} = h_1</math>, <math>u^{+} = 0</math>.
This means that
<center>
<math>
h_{0}\partial_t s(t) - h_{1}\partial_t s(t)
+ V h_{1}= 0
</math>
</center>
and
<center>
<math>
- h_{1}V\partial_t s(t) = - \left(V\right)^2 h_{1}
+ \frac{1}{2} g \left(h_{0}\right)^2 - \frac{1}{2} g \left(h_{1}\right)^2
</math>
</center>
which can be solved to obtain the shock speed and the height of the moving fluid.

Below is a video of surfing on the [http://en.wikipedia.org/wiki/Severn Severn] bore, do not believe everything they
say. You might also want to check out the [http://en.wikipedia.org/wiki/Pororoca Pororoca]
a tidal bore on the Amazon.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|Yi8fytUszrQ}}

=== Part 2 ===

{{#ev:youtube|_668xmJ9DAQ}}

=== Part 3 ===

{{#ev:youtube|yTat11ERJMg}}

=== Part 4 ===

{{#ev:youtube|qG8bPJwX96k}}

=== Part 5 ===

{{#ev:youtube|38ZWW2dM1Qs}}

=== Part 6 ===

{{#ev:youtube|AZHlhA5pQBg}}

=== Part 7 ===

{{#ev:youtube|VhlY5lQcTlM}}

[[Category:Simple Nonlinear Waves]]

[[Category:Simple Nonlinear Waves]]
[[Category:Nonlinear Water-Wave Theory]]
[[Category:Shallow Depth]]

Nonlinear Shallow Water Waves

2025-08-06T23:42:27Z

Levi: /* Equations of Motion */

{{nonlinear waves course
| chapter title = Nonlinear Shallow Water Waves
| next chapter = [[Introduction to KdV]]
| previous chapter = [[Traffic Waves]]
}}

{{complete pages}}

== Introduction ==

We assume that water is incompressible,
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
This allows us to assume [[:Category:Shallow Depth|Shallow Depth]]. We assume that the problem has not variation
in either the <math>y</math> or <math>z</math> direction. The fluid is governed by two parameters,
<math>u(x,t)</math>, the velocity of the water, and <math>h(x,t)</math> the water depth (note that this is not the still water depth since the problem
is nonlinear).

The theory we present here is discussed in [[Stoker 1957]], [[Billingham and King 2000]] and [[Johnson 1997]].

== Equations of Motion ==

The equation for the conservation of mass can derived by considering a a region <math>[x,x+\Delta x]</math>
Conservation of mass then implies that
<center>
<math>
\partial_t \int_x^{x + \Delta x} \rho h(s,t) \mathrm{d}s = \rho h(x,t)u(x,t) - \rho h(x+\Delta x,t)u(x+\Delta x,t)
</math>
</center>
If we take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t h(x ,t) + \partial_x (h(x ,t)u(x ,t)) = 0
</math>
</center>

A second equation comes from conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x + \Delta x}^{x}
+ \int_0^{h(x)} P(x,z,t) \mathrm{d}z -
\int_0^{h(x + \Delta x)} P(x+\Delta x,z,t) \mathrm{d}z
</math>
</center>
where <math>\rho \,</math> denotes density, and the pressure <math>P \,</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). This then gives us
<center>
<math>
\partial_t \int_{x}^{x + \Delta x} \rho h u \mathrm{d}x
= -\left. \rho u^2 h \right|_{x}^{x + \Delta x}
+ \frac{1}{2}\rho g {h(x)}^2 -
\frac{1}{2}\rho g {h(x + \Delta x)}^2
</math>
</center>
If we then take the limit as <math>\Delta x \to 0</math> we obtain
<center>
<math>
\partial_t \left( h u \right)
+ \partial_x \left(u^2 h + \frac{1}{2} gh^2\right) =0
</math>
</center>
We can simplify this using the equation derived from conservation of mass to
to obtain
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>

The equations
<center>
<math>
\partial_t h + u \partial_x h + h \partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + u \partial_x u + g \partial_x h = 0
</math>
</center>
are called the nonlinear shallow water equations. They determine the horizontal water velocity and the local water depth.

We can rewrite them in terms of the local wave speed <math>c(x, t) = \sqrt{gh(x, t)}</math> as follows:
<center>
<math>2\partial_t c + 2u\partial_x c + c\partial_x u = 0
</math>
</center>
<center>
<math>\partial_t u + u\partial_x u + 2c \partial_x c = 0
</math>
</center>
These equation are almost identical to those of compressible gas dynamics. Much of our understanding
of the equations for water have been found by researchers studying compressible gas dynamics.

== Linearized Equations ==

We can linearize these equations by assuming that <math>u</math> is small and that
<math>h=h_0 + \zeta </math> where <math>h_0</math> is the average water depth
and <math>\zeta</math> is also assumed small. This gives us
<center>
<math>
\partial_t \zeta + h_0\partial_x u = 0
</math>
</center>
and
<center>
<math>
\partial_t u + g \partial_x \zeta = 0
</math>
</center>
These linear shallow water equations which can be
derived from the linear equations for water of finite
depth and taking the limit of small depth (see [[:Category:Shallow Depth|Shallow Depth]]).

== Characteristics ==

The equations possess characteristics.
Adding and subtracting the two equations above we obtain
<center>
<math>
\frac{\partial (u \pm 2c)}{\partial t}+ (u \pm c)\frac{\partial (u \pm 2c)}{\partial x} = 0
</math>
</center>
This means that
on the <math>\;C_+</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_+}{\mathrm{d} t} = u + c = u + \sqrt{gh}
</math>
</center>
the <math>\;C_+</math> invariant
<center>
<math>
R_+ = u + 2c = u + 2\sqrt{gh}
</math>
</center>
is a constant,
and on the <math>\;C_-</math> characteristic, given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - c = u - \sqrt{gh}
</math>
</center>
the <math>\;C_-</math> invariant
<center>
<math>
R_- = u - 2c = u - 2\sqrt{gh}
</math>
</center>
is a constant.

The functions <math>R_{\pm} (u ,c) = u \pm 2c</math>, are called the Riemannian invariants.

== Simple Waves ==

The problem as formulated can be solved by advancing the solution along the characteristics, but
this will in general be quite difficult analytically. However, there is a special class of problems,
called ''Simple Waves'' in which the solution only changes on one characteristic.
They are best illustrated through some examples. Note that the characteristic can meet forming
a shock, which is called a [http://en.wikipedia.org/wiki/Tidal_bore bore] or a
[http://en.wikipedia.org/wiki/Hydraulic_jump hydraulic jump]
when it occurs on the surface of the water.

=== The dam break problem ===
Assume the water occupies the region <math>{x < 0 ; 0 < z < h_0 }</math> initially held back by a dam at <math>x = 0</math>.
At <math>t = 0</math>, the dam is removed (breaks). What is the height of the water
<math>h(x,t) \,</math> for <math>t > 0? \,</math> The initial condition is therefore
<center>
<math>h(x,0) = \begin{cases}
h_0, & x < 0 \\
0, & x > 0
\end{cases}
</math>
</center>
<center>
<math>u(x ,0) = 0. \,
</math>
</center>
On the characteristic that originates at <math>t = 0</math> for <math>x < 0</math>,
<center>
<math>
R_\pm = u \pm 2\sqrt{gh} = \pm 2\sqrt{gh_0} = \pm 2c_0
</math>
</center>
where <math>c_0 = \sqrt{gh_0}</math> is the initial (linear) wave speed.

Therefore, if a <math>C_+</math> and a <math>C_-</math> characteristic from this region intersect, then
<center>
<math>
u + 2\sqrt{gh} = 2c_0 , \;\mathrm{and}\; u - 2\sqrt{gh} = -2c_0
</math>
</center>
and hence, <math>u = 0</math> and <math>h = h_0</math>.
Moreover,
<center>
<math>
\frac{\mathrm{d} X_\pm}{\mathrm{d} t} = u \pm \sqrt{gh} = \pm c_0
</math>
</center>
so these characteristics are straight lines in the region <math>\big\{x < -c_0 t \big\}</math>
(the undisturbed region).

The <math>\;C_+</math> characteristic leave the region a<math>\big\{x < -c_0 t \big\}</math>
and enter <math>\big\{x > -c_0 t \big\}</math>. For now we will assume that these characteristics fill the domain
(and show that this is true shortly).
For <math>\big\{x > -c_0 t \big\}</math>, the <math>C_-</math> characteristics are given by
<center>
<math>
\frac{\mathrm{d} X_-}{\mathrm{d} t} = u - \sqrt{gh}
</math>
</center>
and on each of the <math>C_-</math> characteristics <math>R_- = u - 2\sqrt{gh}</math> is constant.
However, since this region is filled with <math>C_+</math> characteristics where <math>R_+ = u + 2\sqrt{gh} = 2c_0</math>, <math>u</math> and <math>h</math>
must be constant on each <math>C_-</math> characteristic.
This also means that the <math>C_-</math> characteristics must be straight lines.

Since the fluid occupies <math>\big\{x < 0 \big\}</math> at <math>t = 0</math>,
these <math>C_-</math> characteristics must start at the origin, with
<center>
<math>
X_-(t) = \left(u - \sqrt{gh}\right)t
</math>
</center>
which in turn implies that
<center>
<math>
u - \sqrt{gh} = \frac{x}{t}
</math>
</center>
We also have <math>R_+ = u + 2\sqrt{gh} = 2c_0</math> from the
<math>C_+</math> characteristics. We can solve these equations
at each point in <math>\big\{x > -c_0 t \big\}</math>. Solving for <math>u</math> and <math>h</math> gives
<center>
<math>
h(x, t) =
\begin{cases}
\frac{h_0}{9}\left(2 - \frac{x}{c_0 t}\right)^2, \quad -c_0 t < x< 2 c_0 t,\\
h_0,\quad x <-c_0 t,
\end{cases}
</math>
</center>
<center>
<math>
u(x, t) =
\begin{cases}
\frac{2}{3} \left (c_0 + \frac{x}{t} \right ), \quad -c_0 t < x< 2 c_0 t,\\
0,\quad x <-c_0 t.
\end{cases}
</math>
</center>
Where we have assumed that, since <math>h = 0</math> for <math>x = 2c_0 t,</math>
the <math>C_+</math> characteristic only exist in the region <math>\big\{x < 2c_0 t \big\}</math>,
We will verify this by explicitly calculating them.

It remains to determine the <math>C_+</math> characteristic, which originated in <math>\big\{x < 0 \big\}</math>,
and show they
will fill the domain <math>\big\{x < 2c_0 t \big\}</math>. For <math>\big\{x < -c_0 t \big\}</math>,
the <math>\;C_+</math> characteristics are straight lines with slope <math>c_0</math> and are given by
<center>
<math>
X_+ (t) = -x_0 + c_0 t, \quad \left(x_0 > 0,\;\; t < \frac{x_0}{2c_0}\right)
</math>
</center>
When <math>t = \frac{x_0}{2c_0},\;\;\ X_{+} (t) = -c_0 t</math> so that for
<center>
<math>
t > \frac{x_0}{2c_0}, \quad \frac{\mathrm{d} X_{+} (t)}{\mathrm{d} t} = u + \sqrt{gh}
</math>
</center>
and substituting the solution we found for <math>h</math> and <math>u</math>
<center>
<math>
\frac{\mathrm{d}X_{+} (t)}{\mathrm{d} t} = \frac{4}{3}c_0 + \frac{X_{+} (t)}{3t}
</math>
</center>
Solving this ODE subject to <math>X_+ \left(\frac{x_0}{2c_0}\right) = -\frac{x_0}{2}</math> gives
<center>
<math>
X_+ (t) = 2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\;\;
</math>
</center>
the equation for a characteristic curve.
The curves indeed fill the domain <math>\big\{x < 2c_0 t \big\}</math>
and all satisfy <math>\big\{X_+ (t) < 2c_0 t \big\}</math>. To summarize, the
<math>C^{+}</math> characteristics are given by
<center>
<math>
X_+ (t) =
\begin{cases}
2c_0 t - 3\left(\frac{x_0}{2}\right)^{2/3}(c_0 t)^{1/3},\quad t> x_0/2 c_0\\
-x_0 + c_0 t, \quad 0\ < t < x_0/2 c_0
\end{cases}
</math>
</center>

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Characteristics_dam_break.jpg|thumb|right|500px|Characteristics for the dam
break problem, blue for <math>C_{+}</math> and red for <math>C_{-}</math>. The solid red lines
show the curves <math>x=-c_0 t</math> and <math>x=2c_0 t</math> (note we have assumed
here that <math>c_0 =1</math>]]
| [[Image:Dambreak.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the Dam Break problem]]
|}

== Shocks ==

For a unique solution two exist there must be a single <math>C_+</math> and <math>C_-</math>
characteristic through each point. When two characteristics of the same kind meet we
have a shock forming.

=== Accelerating Piston ===

We now consider the problem of water initially at rest occupying the
half space <math>x>0</math> which is initially at rest. At <math>t=0</math>
the piston at <math>x=0</math> begins to move to the right with constant
acceleration <math>a</math> so that the position of the piston is given by
<math>\frac{1}{2}at^2</math>.

We assume that the <math>C_-</math> characteristics which originate in the
water at <math>t=0</math> fill the fluid. On these characteristics
<center>
<math>
R_- = u - 2 c = -2c_0 \,
</math>
</center>
and this condition must hold throughout the fluid.
On the <math>C_+</math> characteristics we know that
<math> R_+ = u + 2 c </math> must be a constant and hence on the
<math>C_+</math> characteristics <math>u</math> and <math>c</math>
must be constant and hence the <math>C_+</math> characteristics must
be straight lines. Note that this does not mean that the
<math>C_+</math> characteristics have the same slope and there is no
requirement that the <math>C_-</math> characteristics are straight lines.

The <math>C_+</math> characteristic originate from the fluid
or from the front of the piston. We consider those which originate from
the piston. The <math>C_+</math> characteristic which originates from
the piston at <math>t=t_0</math> must satisfy
<center>
<math>
u+2c = a t_0 + 2c_{\text{plate}} \,
</math>
</center>
where <math> a t_0</math> is the velocity of the piston at time <math>t=t_0</math>
and <math>c_{\text{plate}}</math> is the speed (related to height) at the plate.
We know that <math> R_- = u - 2 c =-2c_0</math> through out the fluid, so that if
we solve this at the plate (where <math>u=at_0</math> and <math>c=c_{\text{plate}}</math>)
then we get
<center><math>
c_{\text{plate}} = at_0/2 + c_0\,
</math></center>

On the <math>C_+</math> characteristics <math>u</math> and <math>c</math>
are constant and therefore
<center>
<math>
\frac{\mathrm{d}X_+}{\mathrm{d}t} = u+c = at_0 + \left( \frac{1}{2}at_0 + c_0 \right)
</math>
</center>
Hence
<center>
<math>
X_+(t,t_0) = \left( \frac{3}{2} a t_0 + c_0 \right) t - c_0 t_0 -a t_0^2
</math>
</center>
using the condition <math>X_+(t_0,t_0) = \frac{1}{2} a t_0^2</math> (the initial value which
comes from the position of the piston at <math>t=t_0</math>).
The slope of these lines increases and eventually meet to form a shock.
We find this point of intersection by considering neighboring characteristics
and determining when they first intersect.
<center>
<math>
X_+(t,t_0 + \Delta t) = X_+(t,t_0) + \Delta t \frac{\partial X_+}{\partial t_0} (t,t_0) = X_+(t,t_0)
</math>
</center>
It follows that neighbouring characteristics will meet when
<center>
<math>
\frac{\partial X_+}{\partial t_0} (t,t_0) = 0
</math>
</center>
which implies that
<center>
<math>
t = \frac{2c_0}{3a} + \frac{4}{3}t_0
</math>
</center>
The first time that a shock forms is the minimum value of this equation.
For this piston example, this occurs when <math>t_0 = 0</math> and the value
of <math>t</math> is <math>t = 2c_0/(3a)</math>. At this point
a shock is formed and we can no longer find a unique solution by following the
characteristics.

{| class="wikitable"
|-
! Characteristics
! Surface elevation
|-
| [[Image:Accelerating_piston.jpg|thumb|right|500px| <math>C_{+}</math> characteristics for the
accelerating piston, red undisturbed, blue from the piston and the green line shows the transition.]]
| [[Image:Accelerating_piston2.gif|thumb|right|500px|Evolution of the fluid surface <math>h(x,t)</math> for the
accelerating piston.]]
|}

=== Piston Moving with Constant Velocity ===

This example is also known as the Moving Wall Problem, and is connected to Shallow Water Bores.

We consider the case of a piston, with positive constant velocity <math>V</math> (which is initially at <math>x=0</math>), advancing into a semi-infinite expanse of
fluid that is initially at rest with depth <math>h_0</math>.

The <math>C_+</math> characteristics which originate in the fluid
at <math>t=0</math> have slope
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0}</math>
</center>
and the <math>C_+</math> characteristic which originate at the piston at
<math>t=0</math> must satisfy
<center>
<math>\frac{\mathrm{d} X_+}{\mathrm{d}t} = \sqrt{gh_0} + \frac{3}{2} V </math>
</center>
so that these two characteristics will intersect at <math>t=0</math>.
Therefore a shock forms immediately and we can track this by determining the
speed of the shock

=== Speed of the shock ===

We need the conservation equations in integral form to determine the speed
of the shock. Conservation of mass, written as an integral is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h \mathrm{d}x
+ \left. \rho u h \right|_{x_2}^{x_1} =0
</math>
</center>
If the shock is located at <math>s(t)</math>
which we assume is located between <math>x_1</math>
and <math>x_2</math>, then
<center>
<math>
\partial_t \int_{x_1}^{x_2} h \mathrm{d}x
= \partial_t \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) h \mathrm{d}x
= \left( \int_{x_1}^{s(t)} +
\int_{s(t)}^{x_2} \right) \partial_t h \mathrm{d}x + h^{+} \partial_t s(t) - h^{-}\partial_t s(t),
</math>
</center>
where <math>h^+</math> is the height on the right (positive) side of
the jump and <math>h^-</math> is the height on the left (negative) side.
If we take the limit as <math>x_1\to x_2</math> we then obtain the following
identity
<center>
<math>
h^{+}\partial_t s(t) - h^{-}\partial_t s(t) -
u^{+} h^{+} + u^{-} h^{-} = 0
</math>
</center>
where <math>u^+</math> is the height on the right (positive) side of
the jump and <math>u^-</math> is the height on the left (negative) side.

We now need to consider the equation for conservation of momentum. In integral form
this is
<center>
<math>
\partial_t \int_{x_1}^{x_2} \rho h u \mathrm{d}x
= \left. \rho u^2 h \right|_{x_1}^{x_2}
+ \int_0^{h(x_1)} P(x_1,z,t) \mathrm{d}z -
\int_0^{h(x_2)} P(x_2,z,t) \mathrm{d}z
</math>
</center>
where the pressure <math>P</math> is given by
<center>
<math>
P = \rho g \left(h - z\right)
</math>
</center>
(i.e. we have hydrostatic equilibrium). We can apply a similar argument as before to obtain
<center>
<math>
h^{+}u^{+}\partial_t s(t) - h^{-}u^{-}\partial_t s(t) =
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2
</math>
</center>

=== Hydraulic Jump ===

For a hydraulic jump, <math>\dot{s}(t) = 0</math>, which means that we must solve
<center>
<math>
u^{+} h^{+} - u^{-} h^{-} = 0 \,
</math>
</center>
<center>
<math>
\left(u^{+}\right)^2 h^{+} - \left(u^{-}\right)^2 h^{-}
+ \frac{1}{2} g \left(h^{+}\right)^2 - \frac{1}{2} g \left(h^{-}\right)^2 =0
</math>
</center>
If we introduce the variables
<center>
<math>
H = \frac{h^{+}}{h^{-}}
</math>
</center>
and
<center>
<math>
\mathrm{Fr} = \frac{u^{-}}{\sqrt{gh^{-}}}
</math>
</center>
where <math>\mathrm{Fr}</math> is the ''Froude'' number
which is equivalent to the Mach number for gas dynamics.
Then we obtain
<center>
<math>
H^2 -1 = 2 \mathrm{Fr}^2 \left(1 - \frac{1}{H}\right)
</math>
</center>
This expression has the roots
<center>
<math>
H=1, \quad H=\frac{1}{2}\left(-1\pm\sqrt{1 + 8 \mathrm{Fr}^2}\right)
</math>
</center>
The only physically meaningful solution is the root which satisfies
<math>H>1</math>. This is only true providing <math>\mathrm{Fr} > 1</math>, which means
that we can only obtain a hydraulic jump if the flow is supercritical.

Below is a video of a hydraulic jump. You can clearly see the point where the flow is changing from supercritical to subcritical (look for the small turbulent region in the channel)
{{#ev:youtube|5etwhZ0d2GU}}

=== Shallow Water Bore ===
We now consider a bore, in which
the shock wave advances into still water.
We denote the fluid speed by <math>V = u^{-}</math>.
We denote the height on the wall side
by <math>h_1</math> and the height on the other side must be <math>h_0</math>, i.e.
<math>h^{+} = h_0</math>, <math>h^{-} = h_1</math>, <math>u^{+} = 0</math>.
This means that
<center>
<math>
h_{0}\partial_t s(t) - h_{1}\partial_t s(t)
+ V h_{1}= 0
</math>
</center>
and
<center>
<math>
- h_{1}V\partial_t s(t) = - \left(V\right)^2 h_{1}
+ \frac{1}{2} g \left(h_{0}\right)^2 - \frac{1}{2} g \left(h_{1}\right)^2
</math>
</center>
which can be solved to obtain the shock speed and the height of the moving fluid.

Below is a video of surfing on the [http://en.wikipedia.org/wiki/Severn Severn] bore, do not believe everything they
say. You might also want to check out the [http://en.wikipedia.org/wiki/Pororoca Pororoca]
a tidal bore on the Amazon.

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|Yi8fytUszrQ}}

=== Part 2 ===

{{#ev:youtube|_668xmJ9DAQ}}

=== Part 3 ===

{{#ev:youtube|yTat11ERJMg}}

=== Part 4 ===

{{#ev:youtube|qG8bPJwX96k}}

=== Part 5 ===

{{#ev:youtube|38ZWW2dM1Qs}}

=== Part 6 ===

{{#ev:youtube|AZHlhA5pQBg}}

=== Part 7 ===

{{#ev:youtube|VhlY5lQcTlM}}

[[Category:Simple Nonlinear Waves]]

[[Category:Simple Nonlinear Waves]]
[[Category:Nonlinear Water-Wave Theory]]
[[Category:Shallow Depth]]

Traffic Waves

2025-08-05T06:52:42Z

Levi: Some fixes

{{nonlinear waves course
| chapter title = Traffic Waves
| next chapter = [[Nonlinear Shallow Water Waves]]
| previous chapter = [[Method of Characteristics for Linear Equations]]
}}

{{complete pages}}

We consider here some simple equations which model traffic flow. This problem is discussed in
[[Billingham and King 2000]].

[[Category:Reference]]

== Equations ==

We consider a single lane of road, and we measure distance along the road with
the variable <math>x</math> and <math>t</math> is time.
We define the following variables
<center><math>
\begin{matrix}
&\rho(x,t) &: &\mbox{car density (cars/km)} \\
& v(\rho) &: &\mbox{car velocity (km/hour)} \\
& q(x,t) =\rho v &: &\mbox{car flow rate (cars/hour)} \\
\end{matrix}
</math></center>

If we consider a finite length of road <math>x_1\leq x \leq x_2</math> then the net flow of cars
in and out must be balanced by the change in density. This means that
<center><math>
\frac{\partial}{\partial t} \int_{x_1}^{x_2} \rho(x,t) \mathrm{d}x = -q(x_2,t) + q(x_1,t)
</math></center>
We now consider continuous densities (which is obviously an approximation) and
set <math>x_2 = x_1 + \Delta x</math> and we obtain
<center><math>
\frac{\partial}{\partial t} \rho(x_1,t) = -\frac{q(x_2,t) + q(x_1,t)}{\Delta x}
</math></center>
and if we take the limit as <math>\Delta x \to 0</math> we obtain the differential equation
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} = 0
</math></center>
Note that this equation has been derived purely from the need to conserve cars (it is a conservation equation) and
is not possible to solve this equation until we have derived a connection between <math>\rho</math>
and <math>q</math>.

== Equation for <math>\rho</math> only ==

At the moment we assume that we have some expression for <math>v(\rho)</math>
If we substitute the expression for <math>q = v\rho</math> into our differential equation we obtain
<center><math>
\frac{\partial \rho}{\partial t} + \frac{\partial }{\partial x} \left(v(\rho)\rho\right) = 0
</math></center>
which gives us
<center><math>
\frac{\partial \rho}{\partial t} + \left(v^{\prime}(\rho)\rho + v(\rho)\right)
\frac{\partial \rho }{\partial x} = 0
</math></center>
or
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho)\frac{\partial \rho }{\partial x} = 0
</math></center>
where <math>c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right)</math>
is the '''kinematic wave speed'''. Note that this is not the speed of the cars, but
the speed at which disturbances in the density travel.

== A simple relationship between <math>\rho</math> and <math>q</math> ==

The relationship between <math>\rho</math> and <math>q</math> is an equation of state and
there is no ''exact'' equation since it depends on many unknowns. One of the
simplest relationship between <math>\rho</math> and <math>q</math> is derived from
the following assumptions

* When the density <math>\rho = 0</math> the speed is <math>v=v_0</math>
* When the density is <math>\rho = \rho_{\max} </math> the speed is <math>v=0</math>
* The speed is a linear function of <math>\rho</math> between these two values.

This also gives good fit with measured data. We will either consider the general case or use this simple
relationship. Using this we obtain
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
The flux of cars is given by
<center><math>
q = \rho v(\rho) = v_0\frac{\rho(\rho_{\max} - \rho)}{\rho_{\max}}
</math></center>
and the wave speed is
<center><math>
c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right) = -\frac{v_0}{\rho_{\max}}\rho + v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
= v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>

{| class="wikitable"
|-
! <math>v</math>
! <math>q</math>
! <math>c</math>
|-
| [[Image:Velocity.jpg|thumb|350px|<math>v(\rho)</math> versus <math>\rho</math>]]
| [[Image:Q_flux.jpg|thumb|350px|<math>q(\rho)</math> versus <math>\rho</math>]]
| [[Image:C_speed.jpg|thumb|350px|<math>c(\rho)</math> versus <math>\rho</math>]]
|}

== Small Amplitude Disturbances ==

We can linearise the model by assuming that the variation in density is small so
that we can write
<center><math>
\rho = \rho_0 + \tilde{\rho}
</math></center>
where we assume that <math>\tilde{\rho}</math> is small. This allows us to write the equations as
<center><math>
\frac{\partial \tilde{\rho}}{\partial t} + c(\rho_0) \frac{\partial \tilde{\rho}}{\partial x} = 0
</math></center>
where the main difference between this and the full equation is that the wave speed <math>c</math> is
a constant. This is the linearised equation. Note that this linearisation does not give a good model because
traffic density does not vary only a small amount about some mean (as is the case for accoustic waves where the
density of air is roughly constant).

Under these assumptions the solution to the equation is
<center><math>
\tilde{\rho} = f(x - c(\rho_0)t)
</math></center>
where <math>f</math> is determined by the initial condition. This represents
disturbances which travel with speed <math>c(\rho_0)</math> in the positive
<math>x</math> direction.

We now consider the '''characteristic curves''' which are curves along which the density
<math>\rho</math> is a constant. These are give by
<center><math>
x = X(t) = x_0 + c(\rho_0) t.\,
</math></center>
which are just straight lines of constant slope. We will see shortly that the full (nonlinear)
equations also possess characteristics.

== Nonlinear Initial Value Problem ==

We wish to solve
<center><math>
\frac{\partial \rho}{\partial t} + c(\rho) \frac{\partial \rho}{\partial x} = 0
</math></center>
subject to the initial conditions
<center><math>
\rho(x,0) = \rho_0(x) \,
</math></center>
It turns out that the concept of characteristic curves is very important for this problem.

If we want <math>\rho(X(t),t)</math> to be a constant then we require
<center><math>
\frac{\mathrm{d}}{\mathrm{d}t}\rho(X(t),t) = \frac{\mathrm{d} X}{\mathrm{d}t} \frac{\partial \rho}{\partial x} +
\frac{\partial \rho}{\partial t} = 0.
</math></center>
Comparing this to the governing partial differential equation we can see that we require
<center><math>
\frac{\mathrm{d} X}{\mathrm{d}t} = c(\rho)
</math></center>
This means that the characteristics are straight lines (since <math>\rho</math> is constant) with
slope given by <math> c(\rho_0(x_0))</math> so that the equation for the characteristics is
<center><math>
X(t) = x_0 + c(\rho_0(x_0))t \,
</math></center>

This does not allow us to write down a solution to the initial value problem,
all we can do is write
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho_0(x_0)\,
</math></center>
which allows us to calculate the solution stepping forward in time, but not to determine the solution given
a value of <math>(x,t)</math> (because we generally cannot solve the equation <math>x_0 + c(\rho_0(x_0))t = x</math> for <math>x_0</math>).

The characteristics are a family of straight lines which will all have different slopes. If two characteristics
meet, our solution method will break down because there will be two values of the density <math>\rho</math>.
This gives rise to a '''shock'''. It turns
out that this the formation of shocks is a product of the equations themselves and not with the solution method.
We will see shortly that special methods are required to treat these shocks.

=== Case when no shocks are formed ===

The characteristic curves will fill the space without meeting provided that the wave speed <math>c(\rho_0)</math>
is a monotonically increasing function of the distance <math>x</math>. If we work with our previous model we
have
<center><math>
v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}}
</math></center>
and
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
so that <math>c</math> is a monotonically decreasing function of density <math>\rho</math>. This means
that the wave speed <math>c(\rho_0)</math> will be
a monotonically increasing function of the distance <math>x</math> if an only if the density is a
monotonically decreasing function. In this case the solution can be calculated straightforwardly
by expansion of the initial density.

==== No shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1- \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example1 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1- \tanh(x))</math> ]]
| [[Image:Traffic_example1_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example1_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1- \tanh(x))</math>]]
| [[Image:Traffic_example11.gif|thumb|350px|<math>\rho(x,t) </math><math> for </math><math>\rho_0 = 1/2(1- \tanh(x))</math>]]
|}

==== Riemann problem and the expansion fan ====

We can consider a simple problem in which there is a jump in the initial density
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} > \rho_{R}</math> so that we do not form a shock. In this case
the characteristics on each side of <math>x=0</math> have a different slope and the
question is what happens between. It is easiest to think about the following problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < -\epsilon \\
\frac{\rho_{R}-\rho_{L}}{2\epsilon}x + \frac{\rho_{R}+\rho_{L}}{2} & -\epsilon \leq x \leq \epsilon \\
\rho_{R},& x > \epsilon
\end{matrix}
\right.
</math></center>
We can then see that we have lines of uniformly varying slope for <math>-\epsilon < x <\epsilon</math>
with slope between <math>c(\rho_L)</math> and <math>c(\rho_R)</math>. If we then take the limit
as <math>\epsilon \to 0</math> we obtain an expansion fan emanating from <math>x=0</math>.

If we assume that
<center><math>
c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}}
</math></center>
then we know that on the lines of the expansion fan (which all start at <math>x=0</math>) we have
<math>c(\rho) = x/t</math> (as the characteristic lines themselves are <math>x = c(\rho(x, t))t</math>). We can rearrange this and solve for <math>x</math> and obtain
<math>\rho(x, t) =\frac{ 1}{2} \rho_{\max} \left(1-\dfrac x{v_0 t}\right)</math>

The solution is then given by
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < c(\rho_L) t\\
\frac{ \rho_{\max}}{2} \left(1-\dfrac x{v_0 t}\right),& c(\rho_L) t \leq x \leq c(\rho_R) t\\
\rho_{R},& x > c(\rho_R) t
\end{matrix}
\right.
</math></center>
This solution is known as an '''expansion fan'''.

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.6,& x < 0 \\
0.3,& x > 0.
\end{matrix}
\right.
</math></center> The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Expansion_fan_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Expansion_fan_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Expansion_fan_characteristics.jpg|thumb|350px|Characterisitics]]
| [[Image:Expansion_fan2.gif|thumb|350px|<math>\rho(x,t)</math>]]
|}

=== Shocks ===

So far we have only considered the case when <math>g(x_0)=c(\rho_0(x_0))</math> is monotonically increasing so that
two characteristics never cross. We now consider the case when characteristics can meet.
A movie of this case is shown below.

{{#ev:youtube|Suugn-p5C1M}}

We can easily see that
the first characteristics to meet will be neighbouring characteristics. Consider two characteristics
with
<center><math>
X_1(t) = x_0 + g(x_0)t\,
</math></center>
<center><math>
X_2(t) = x_0 + \delta x + g(x_0+\delta x)t\,
</math></center>
Then these curves will meet at time <math>T</math> where
<center><math>
x_0 + g(x_0)T = x_0 + \delta x + g(x_0+\delta x)T\,
</math></center>
which implies that
<center><math>
T = -\frac{1}{g^{\prime}(x_0)}
</math></center>
Note the following
* If <math>g^{\prime}(x) > 0 </math> then no shock will form.
* The shock first forms at the minimum positive value of <math> - \frac{1}{g^{\prime}(x)} </math> for <math> -\infty < x <\infty </math>.

==== Shock Fitting ====

If we calculate the solution using our formula
<center><math>
\rho(x_0 + c(\rho_0(x_0))t,t) = \rho(x_0)\,
</math></center>
then we find that the solution becomes multivalued in the case when a shock forms.
We then have to fit a shock. One way to do this is by imposing the condition that equal
areas are removed and added when we chose the position of the shock.
This corresponds to the condition that
the number of cars must be conserved

==== Speed of the shock ====

If we consider the case when there is a shock at <math>s(t)</math> with
<math>\rho = \rho^{-}</math> at <math>x=s^{-}</math>
and
<math>\rho = \rho^{+}</math> at <math>x=s^{+}</math>
(where <math>s^{-}</math>
is just less than s(t) and <math>s^{+}</math>
is just greater than s(t) ). If we substitute
this into the governing integral equation we obtain
<center><math>
\frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} \rho(x,t)\mathrm{d}x + \int_{s(t)}^{x_2}
\rho(x,t)\mathrm{d}x \right) = q(x_1,t) - q(x_2,t)
</math></center>
and hence
<center><math>
\int_{x_1}^{x_2}
\frac{\partial \rho(x,t)}{\partial t} \mathrm{d}x + \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(x_1,t) - q(x_2,t)
</math></center>
If we now take the limit as <math>x_1\to x_2</math> we obtain
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t}\rho^{-}
- \frac{\mathrm{d}s}{\mathrm{d}t}\rho^{+} = q(\rho^{-}) - q(\rho^{+})
</math></center>
so that
<center><math>
\frac{\mathrm{d}s}{\mathrm{d}t} = \frac{q(\rho^{-}) - q(\rho^{+})}
{\rho^{-} - \rho^{+}}
</math></center>

==== Shock example ====

We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by <math>\rho_0 = 1/2(1 + \tanh(x))</math>. The figures below show the initial density, the initial speed,
the characteristics and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Traffic example2 rho.jpg|thumb|350px| <math>\rho_0 = 1/2(1+ \tanh(x))</math> ]]
| [[Image:Traffic_example2_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Traffic_example2_characteristics.jpg|thumb|350px|Characterisitics for <math>\rho_0 = 1/2(1+ \tanh(x))</math>]]
| [[Image:Traffic_example2.gif|thumb|350px|<math>\rho(x,t) </math><math> for </math><math>\rho_0 = 1/2(1+ \tanh(x))</math> Dotted solution is without shock fitting.]]
|}

==== Riemann problem ====

We now consider the Riemann problem
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
\rho_{L},& x < 0 \\
\rho_{R},& x > 0
\end{matrix}
\right.
</math></center>
where <math>\rho_{L} < \rho_{R}</math>. In this case a shock forms immediately and
the characteristics terminate at the shock. The shock moves with constant speed given by
the equation for the motion of the shock (or can be found by the equal areas rule). We obtain
<center><math>
\rho(x,t) =
\left\{
\begin{matrix}
\rho_{L},& x < \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t \\
\rho_{R},& x > \frac{1}{2} \left(c(\rho_{L}) + c(\rho_{R}) \right) t
\end{matrix}
\right.
</math></center>
We consider the case when <math>\rho_{\max} = v_0 = 1</math> and where the initial density is given
by
<center><math>
\rho(x,0) =
\left\{
\begin{matrix}
0.3,& x < 0 \\
0.6,& x > 0
\end{matrix}
\right.
</math></center>

The figures below show the initial density, the initial speed,
the characteristics, and <math>\rho(x,t)</math> for this case.

{| class="wikitable"
|-
! <math>\rho</math>
! <math>c</math>
|-
| [[Image:Shock_rho.jpg|thumb|350px| <math>\rho_0</math> ]]
| [[Image:Shock_c.jpg|thumb|350px|<math>c(\rho_0)</math>]]
|}

{| class="wikitable"
|-
! characteristics
! <math>\rho(x,t)</math>
|-
| [[Image:Shock_characteristics.jpg|thumb|350px|Characterisitics with shock shown in green.]]
| [[Image:Shock3.gif|thumb|350px|<math>\rho(x,t)</math> with the red dotted line showing the solution without shock fitting]]
|}

== Lecture Videos ==

=== Part 1 ===

{{#ev:youtube|SWIXL97H0fw}}

=== Part 2 ===

{{#ev:youtube|j652tvrgm4c}}

=== Part 3 ===

{{#ev:youtube|QVtYsUGtb7I}}

=== Part 4 ===

{{#ev:youtube|Ik2Q7H_aIV8}}

=== Part 5 ===

{{#ev:youtube|lNeFNHAqgEA}}

=== Part 6 ===

{{#ev:youtube|Yo0sW3Nh2FI}}

=== Part 7 ===

{{#ev:youtube|gOEMfOQZAAM}}

=== Part 8 ===

{{#ev:youtube|ZwvTZcWLZKM}}

[[Category:Simple Nonlinear Waves]]