# KdV Equation Derivation

We consider the method of derivation of KdV Equation in the concept of Nonlinear Shallow Water Waves.

## Introduction

In the analysis of Nonlinear Shallow Water Waves equations we see that there are two important geometrical parameters, $\displaystyle{ \epsilon = \frac{h}{\lambda} }$ and $\displaystyle{ \alpha=\frac{a}{h} }$ involved. By choosing appropriate magnitudes for $\displaystyle{ \epsilon }$ and $\displaystyle{ \alpha }$, we can consider a theory in which dispersion and nonlinearity are in balance. The Korteweg-de Vries Equation verifies the relation between dispersion and nonlinearity properties.

## Derivation

We begin with the equations for waves on water,

$\displaystyle{ \begin{matrix} &\Phi_{xx} + \Phi_{yy} &= 0 \quad &-\infin\lt x\lt \infin, 0 \le y \le \eta(x,t) \\ \end{matrix} }$

Provided that at $\displaystyle{ y=\eta(x,t)=h+aH(x,t) }$ we have,

$\displaystyle{ \begin{matrix} &\Phi_{y} &= &\eta_t + \Phi_x \eta_x \\ &\Phi_t + \frac{1}{2}({\Phi_x}^2 + {\Phi_y}^2) + g\eta &= &B(t)\\ &\Phi_y = 0 &, &y = 0 \end{matrix} }$

To make these equations dimensionless, we use the scaled variables,

$\displaystyle{ \bar{x}=\frac{x}{\lambda}, \quad \bar{y}=\frac{y}{h}, \quad \bar{\Phi}=\frac{h\Phi}{\lambda a \sqrt{gh}}, \quad \bar{t}=\frac{t\sqrt{gh}}{\lambda} }$

where $\displaystyle{ \sqrt{gh} }$ is defined as linear wave speed in shallow water. Hence the dimensionless system is,

$\displaystyle{ \begin{matrix} &\epsilon^2 {\bar{\Phi}}_{\bar{x}\bar{x}} + {\bar{\Phi}}_{\bar{y}\bar{y}} &= &0 \\ \\ &{\bar{\Phi}}_{\bar{y}} &= &\epsilon^2(H_{\bar{t}}+\alpha {\bar{\Phi}}_{\bar{x}} H_{\bar{x}}) \\ \\ &{\bar{\Phi}}_{\bar{t}} + \frac{1}{2}\alpha ({{\bar{\Phi}}_{\bar{x}}}^2 + \epsilon^2 {{\bar{\Phi}}_{\bar{y}}}^2) + H &= &(B(t)-gh) / ag \\ \\ &{\bar{\Phi}}_{\bar{y}} = 0 &, &\bar{y} = 0 \end{matrix} }$

where $\displaystyle{ \epsilon = \frac{h}{\lambda} }$ and $\displaystyle{ \alpha=\frac{a}{h} }$ are two small parameters which are given in this problem.

In the next step we use the transform $\displaystyle{ \bar{\Phi} \to \bar{\Phi} + \int\limits_{0}^{\bar{t}}(\frac{B(s) - gh}{ag})\mathrm{d}s }$ and introduce further transformation to remove $\displaystyle{ \epsilon }$ from the equations,

$\displaystyle{ z = \frac{\alpha^{1 / 2}}{\epsilon}(\bar{x}-\bar{t}), \quad \tau = \frac{\alpha^{3/2}}{\epsilon}\bar{t}, \quad \Psi = \frac{\alpha^{1/2}}{\epsilon}\bar{\Phi} }$

The key idea is that $\displaystyle{ \frac{\alpha^{1 / 2}}{\epsilon} }$ is $\displaystyle{ O(1) }$.

Hence,

$\displaystyle{ \begin{matrix} &\alpha \Psi_{zz} + \Psi_{\bar{y}\bar{y}} = 0 & -\infin \lt z \lt \infin , 0 \le \bar{y} \le 1 + \alpha H(z,\tau) &(1) \\ \\ &\Psi_{\bar{y}} = \alpha (-H_z+\alpha H_{\tau} + \alpha \Psi_z H_z) & y=1+ \alpha H(z,\tau) &(2) \\ \\ &H - \Psi_z + \alpha \Psi_{\tau} + \frac{1}{2} ({\Psi_{\bar{y}}}^2+\alpha {\Psi_z}^2)=0 &y=1+ \alpha H(z,\tau) &(3) \\ \\ &\Psi_{\bar{y}} = 0 &\bar{y}=0 &(4) \end{matrix} }$

The boundary condition (4) expresses $\displaystyle{ \Psi }$ at the flat bed, $\displaystyle{ \bar{y}=0 }$. The boundary condition (3) is Bernoulli equation and (2) is kinematic boundary condition. Now we use asymptotic expansions of the form,

$\displaystyle{ \begin{matrix} &\Psi &= &\Psi_0 + \alpha \Psi_1 + {\alpha}^2 \Psi_2 + o({\alpha}^3) &(5)\\ \\ &H &= &H_0 + \alpha H_1 + o(\alpha^2) &(6) \end{matrix} }$

to derive an equation for each $\displaystyle{ H_i }$ according to the boundary conditions (2) to (4).

* Derivation of $\displaystyle{ H_i }$'s:

Substituting (5) and (6), (1) must be true for all powers of $\displaystyle{ \alpha }$. Therefore,

$\displaystyle{ \begin{matrix} &O(\alpha^0) &: &\Psi_{0, \bar{y}\bar{y}} = 0 &\rArr &\Psi_0 = B_0(z, \tau) \\ \\ &O(\alpha) &: &\Psi_{1, \bar{y}\bar{y}} = -\Psi_{0, zz} &\rArr &\Psi_1 = -\frac{1}{2}{\bar{y}}^2 B_{0, zz}+B_1(z, \tau) \\ \\ &O(\alpha^2) &: &\Psi_{2, \bar{y}\bar{y}} = -\Psi_{1, zz} &\rArr &\Psi_2 = \frac{1}{24}{\bar{y}}^4B_{0,zzzz}-\frac{1}{2}{\bar{y}}^2 B_{1,zz}+ B_2(z, \tau) \end{matrix} }$

Now at leading order the Bernoulli and kinematic equations, (3) and (2), gives,

$\displaystyle{ \begin{matrix} &H_0(z,\tau) = \Psi_{0,z} = B_{0,z} &(a) \\ \\ &H_1-B_{1,z}+\frac{1}{2}B_{0,zzz}+B_{0,\tau}+\frac{1}{2}B^2_{0,z} = 0 &(b) \\ \\ &-H_0B_{0,zz}+\frac{1}{6}B_{0,zzzz}-B_{1,zz} = -H_{1,z}+H_{0,\tau}+B_{0,z}H_{0,z} &(c) \end{matrix} }$

Differentiating (b) and eliminating $\displaystyle{ H_1 }$ and $\displaystyle{ B_1 }$ from (c) allow us to write,

$\displaystyle{ -H_0B_{0,zz}-\frac{1}{3}B_{0,zzzz}-B_{0,z\tau}-B_{0,z}B_{0,zz} = H_{0,\tau}+B_{0,z}H_{0,z} }$

Finally, (a) gives $\displaystyle{ B_0 }$ in terms of $\displaystyle{ H_0 }$ and hence

$\displaystyle{ 2H_{0,\tau}+3H_0H_{0,z}+\frac{1}{3}H_{0,zzz}=0 }$

which is named Korteweg-de Vries (KdV) equation.

## Interpretation

KdV equation includes dispersive effects through the term $\displaystyle{ H_{0,zzz} }$ and nonlinear effects through the term $\displaystyle{ H_0H_{0,z} }$ and governs the behavior of the small amplitude waves, with $\displaystyle{ \alpha\lt \lt 1 }$. It is reasonable to ask when and where the independent variables, $\displaystyle{ z }$ and $\displaystyle{ \tau }$, are of $\displaystyle{ O(1) }$ in order to determine more precisely the region in physical space where the KdV equation is valid as an approximation of the actual flow. According to the definition of $\displaystyle{ z }$ and $\displaystyle{ \tau }$, if $\displaystyle{ \alpha=O(\epsilon^2) }$, then $\displaystyle{ \bar{t}\gt \gt 1 }$ and $\displaystyle{ \bar{x}=\bar{t}+O(1) }$. This leads us to interpret any waveform that arises as a solution of the KdV equation as the large time limit of an initial value problem.

For solution of KdV equation please refer here.