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] 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]

The boundary condition (4) expresses [math]\displaystyle{ \Psi }[/math] at the flat bed, [math]\displaystyle{ \bar{y}=0 }[/math]. The boundary condition (3) is Bernoulli equation and (2) is kinematic boundary condition. 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] according to the boundary conditions (2) to (4).

* Derivation of [math]\displaystyle{ H_i }[/math]'s:

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]

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

[math]\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} }[/math]

Differentiating (b) and eliminating [math]\displaystyle{ H_1 }[/math] and [math]\displaystyle{ B_1 }[/math] from (c) allow us to write,

[math]\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} }[/math]

Finally, (a) gives [math]\displaystyle{ B_0 }[/math] in terms of [math]\displaystyle{ H_0 }[/math] and hence

[math]\displaystyle{ 2H_{0,\tau}+3H_0H_{0,z}+\frac{1}{3}H_{0,zzz}=0 }[/math]

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

Interpretation

KdV equation includes dispersive effects through the term [math]\displaystyle{ H_{0,zzz} }[/math] and nonlinear effects through the term [math]\displaystyle{ H_0H_{0,z} }[/math] and governs the behavior of the small amplitude waves, with [math]\displaystyle{ \alpha\lt \lt 1 }[/math]. It is reasonable to ask when and where the independent variables, [math]\displaystyle{ z }[/math] and [math]\displaystyle{ \tau }[/math], are of [math]\displaystyle{ O(1) }[/math] 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 [math]\displaystyle{ z }[/math] and [math]\displaystyle{ \tau }[/math], if [math]\displaystyle{ \alpha=O(\epsilon^2) }[/math], then [math]\displaystyle{ \bar{t}\gt \gt 1 }[/math] and [math]\displaystyle{ \bar{x}=\bar{t}+O(1) }[/math]. 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.