Difference between revisions of "KdV Equation Derivation"

From WikiWaves
Jump to navigationJump to search
Line 1: Line 1:
 
We consider the method of derivation of KdV Equation in the concept of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves].
 
We consider the method of derivation of KdV Equation in the concept of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves].
  
===Introduction===
+
== Introduction ==
  
 
In the analysis of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves] equations we see that there are two important geometrical parameters, <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math>, are involved. By choosing appropriate magnitudes for <math>\epsilon</math> and <math>\alpha</math>, we can consider a theory in which dispersion and nonlinearity are in balance. The <b>Korteweg-de Vries Equation</b> verifies the relation between dispersion and nonlinearity properties.
 
In the analysis of [http://www.wikiwaves.org/index.php/Nonlinear_Shallow_Water_Waves Nonlinear Shallow Water Waves] equations we see that there are two important geometrical parameters, <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math>, are involved. By choosing appropriate magnitudes for <math>\epsilon</math> and <math>\alpha</math>, we can consider a theory in which dispersion and nonlinearity are in balance. The <b>Korteweg-de Vries Equation</b> verifies the relation between dispersion and nonlinearity properties.
  
===Derivation===
+
== Derivation ==
  
 
We begin with the equations for waves on water,
 
We begin with the equations for waves on water,
Line 44: Line 44:
  
 
where <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math> are two small parameters which are given in this problem.
 
where <math>\epsilon = \frac{h}{\lambda}</math> and <math>\alpha=\frac{a}{h}</math> are two small parameters which are given in this problem.
 
----
 
  
 
In the next step we use the transform <math>\bar{\Phi} \to \bar{\Phi} + \int\limits_{0}^{\bar{t}}(\frac{B(s) - gh}{ag})ds</math> and introduce further transformation to remove <font size='3'><math>\epsilon</math></font> from the equations,
 
In the next step we use the transform <math>\bar{\Phi} \to \bar{\Phi} + \int\limits_{0}^{\bar{t}}(\frac{B(s) - gh}{ag})ds</math> and introduce further transformation to remove <font size='3'><math>\epsilon</math></font> from the equations,
Line 70: Line 68:
 
<center><math>
 
<center><math>
 
\begin{matrix}
 
\begin{matrix}
&\Psi &= &\Psi_0 + \alpha \Psi_1 + {\alpha}^2 \Psi_2 + o({\alpha}^2) \\ \\
+
&\Psi &= &\Psi_0 + \alpha \Psi_1 + {\alpha}^2 \Psi_2 + o({\alpha}^2)  &(5)\\ \\
&H &= &H_0 + \alpha H_1 + o(\alpha)
+
&H &= &H_0 + \alpha H_1 + o(\alpha) &(6)
 
\end{matrix}
 
\end{matrix}
 
</math></center>
 
</math></center>
  
to derive an equation for each <math>H_i</math>.
+
to derive an equation for each <font size='3'><math>H_i</math></font>.
 +
 
 +
 
 +
<b>* Derivation of <font size='3'><math>H_1</math></font>:</b>
 +
 
 +
Substituting (5) and (6), (1) must be true for all powers of <font size='3'><math>\alpha</math></font> therefore,
 +
 
 +
<center><math>
 +
\Psi_{0, \bar{y}\bar{y}} = 0 \quad \rArr \Psi_0 = B_0(z, \tau)
 +
</math></center>
  
----
 
  
===Summery===
+
== Summery ==
 
[[Category:789]]
 
[[Category:789]]

Revision as of 12:07, 14 October 2008

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{ \Psi_{0, \bar{y}\bar{y}} = 0 \quad \rArr \Psi_0 = B_0(z, \tau) }[/math]


Summery