Introduction
We have already met the conservation law for the traffic equations
[math]
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =0
[/math]
and seen how this leads to shocks. We can smooth this equation by adding
dispersion to the equation to give us
[math]
\partial _{t}\rho +c\left( \rho \right) \partial _{x}\rho =\nu \partial
_{x}^{2}\rho
[/math]
where [math]\nu \gt 0.[/math]
The simplest equation of this type is to write
[math]
\partial _{t}u+u\partial _{x}u=\nu \partial _{x}^{2}u
[/math]
(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
[math]
u\left( x,t\right) =u\left( xct\right) =u\left( \xi \right)
[/math]
This leads to the equations
[math]
cu^{\prime }+u^{\prime }u\nu u^{\prime \prime }=0
[/math]
We begin by looking at the phase plane for this system, writing [math]w=u^{\prime
}[/math] so that
[math]\begin{matrix}
\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( uc\right) \right)
\end{matrix}[/math]
This is a degenerate system with the entire [math]u[/math] axis being equilibria.
We can also solve this equation exactly as follows.
[math]
cu^{\prime }+u^{\prime }u\nu u^{\prime \prime }=0
[/math]
can be integrated to give
[math]
cu+\frac{1}{2}\left( u\right) ^{2}\nu u^{\prime} =c_{1}
[/math]
which can be rearranged to give
[math]
u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}2cu2c_{1}\right)
[/math]
We define the two roots of the quadratic [math]\left( u\right) ^{2}2\nu
u2c_{1}=0[/math] by [math]u_{1}[/math] and [math]u_{2}[/math]
and we assume that [math]u_{2} \lt 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} \lt u \lt u_{1}[/math].
We note that the wave speed
is
[math]
c=\frac{1}{2}\left( u_{1}+u_{2}\right)
[/math]
The equation can therefore be written as
[math]
2\nu u^{\prime }=\left( uu_{1}\right) \left( uu_{2}\right)
[/math]
which has solution
[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]
Numerical Solution of Burgers equation
We can solve the equation using our split step spectral method. The equation
can be written as
[math]
\partial _{t}u=\frac{1}{2}\partial _{x}\left( u^{2}\right) +\nu \partial
_{x}^{2}u
[/math]
We solve this by solving in Fourier space to give
[math]
\partial _{t}\hat{u}=\frac{1}{2}ik \widehat{\left( u^{2}\right)} \nu k^{2}\hat{u}
[/math]
Then we solve each of the steps in turn
for a small time interval to give
[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]
Phase plane for a travelling wave solution

Numerical solution of Burgers equation

Phase plane for a travelling wave solution of Burgers equation

Numerical solution of Burgers equation

Exact Solution of Burgers equations
We can find an exact solution to Burgers equation. We want to solve
[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]
Frist we write the equation as
[math]
\partial _{t}u+\partial _{x}\left( \frac{u^{2}}{2}\nu \partial _{x}u\right)
=0
[/math]
We want to find a function [math]\psi \left( x,t\right) [/math] such that
[math]
\partial _{x}\psi =u,\ \ \partial _{t}\psi =\nu \partial _{x}u\frac{u^{2}}{2
}
[/math]
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]
[math]
\partial _{t}\psi =\nu \partial _{x}^{2}\psi \frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
[/math]
We introduce the ColeHopf transformation
[math]
\psi =2\nu \log \left( \phi \right)
[/math]
From this we can obtain the three results:
[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]
Therefore
[math]
\partial _{t}\psi =\nu \partial _{x}^{2}\psi \frac{1}{2}\left( \partial
_{x}\psi \right) ^{2}
[/math]
becomes
[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]
or
[math]
\partial _{t}\phi =\nu \partial _{x}^{2}\phi
[/math]
which is just the diffusion equation. Note that we also have to transform the
boundary conditions. We have
[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]
We can write this as
[math]
\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =\frac{1}{2\nu }F\left(
x\right)
[/math]
which has solution
[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]
We need to solve
[math]\begin{matrix}
\partial _{t}\phi &=&\nu \partial _{x}^{2}\phi \\
\phi \left( x,0\right) &=&\Phi \left( x\right)
\end{matrix}[/math]
We take the Fourier transform and obtain
[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]
which has solution
[math]
\hat{\phi}\left( k,t\right) =\hat{\Phi}\left( k\right) e^{k^{2}\nu t}
[/math]
We can then use the convolution theorem to write
[math]\begin{matrix}
\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{1}\left[
e^{k^{2}\nu t}\right] \\
&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{\infty }^{\infty }\Phi \left( y\right)
\exp \left[ \frac{\left( xy\right) ^{2}}{4\nu t}\right] \mathrm{d}y
\end{matrix}[/math]
Which can be expressed as
[math]
\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{\infty }^{\infty
}\exp \left[ \frac{f}{2\nu }\right] \mathrm{d}y
[/math]
where
[math]
f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
\left( xy\right) ^{2}}{2t}
[/math]
To find [math]u[/math] we recall that
[math]\begin{matrix}
u\left( x,t\right) &=&2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
\left( x,t\right) } \\
&=&\dfrac{\int_{\infty }^{\infty }\left( \frac{xy}{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]