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, [math]\displaystyle{ \epsilon = \frac{h}{\lambda} }[/math] and [math]\displaystyle{ \alpha=\frac{a}{h} }[/math], are involved. By choosing appropriate magnitudes for [math]\displaystyle{ \epsilon }[/math] and [math]\displaystyle{ \alpha }[/math], 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,

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

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

[math]\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} }[/math]

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

[math]\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} }[/math]

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

[math]\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} }[/math]

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

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

[math]\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} }[/math]

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

Hence,

[math]\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} }[/math]

Now we use asymptotic expansions of the form,

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

to derive an equation for each [math]\displaystyle{ H_i }[/math].

* Derivation of [math]\displaystyle{ H_1 }[/math]:

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

[math]\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} }[/math]