=Travelling Wave Solutions of the KdV Equation=

The KdV equation has two qualitatively different types of permanent form travelling wave solution. These are referred to as cnoidal waves and solitary waves.

==KdV equation in <math>(z,\tau)</math> space ==

Assume we have wave travelling with speed <math> V_0 </math> without change of form,

<center><math> H(z,\tau)=H(z-V_0\tau) </math></center>

and substitute into KdV equation then we obtain

<center><math>
-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center>
where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate.

We integrate this equation twice with respect to <math>\xi</math> to give

<center><math> \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2) </math></center> where D_1 and D_2 are constants of integration.

==Standardization of KdV equation==

We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,
so <math>f(H)=\frac{1}{6}H_\xi^2 </math>

It turns out that we require 3 real roots to obtain periodic solutions.
Let roots be <math> H_1 \leq H_2 \leq H_3</math>.

We can imagine the graph of cubic function which has 3 real roots and we can now write a function
<center><math> f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)</math></center>

From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0.</math>

We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.

and now solve equation in terms of the roots <math>H_i,</math>

We define <math>X=\frac{H}{H_3}</math>, and obtain

<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
where <math>X_i=\frac{H_i}{H}</math>

crest to be at <math>\xi=0 and X(0)=0</math>

and a further variable Y via

<center><math> X=1+(X_2-1)sin^2(Y) </math></center>

<center><math>Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}sin(Y)^2 \right\}

...(1)</math></center>

so <math>Y(0)=0.</math>

and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))},</math></center>
which is separable.

In order to get this into a completely standard form we define

<center><math>k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1)

...(2) </math></center>

Clearly, <math>0 \leq k^2 \leq 1</math> and <math>l>0.</math>

==Solution of the KdV equation==

A simple quadrature of equation (1) subject to the condition (2) the gives us

<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form

<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>

or equivalently

<center><math> x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} </math></center>

Now we can write Y with fixed values of x,k as

<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>

<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
and hence
<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>

<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".

Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form

<center><math>H=H_2+(H_3-H_2)cn^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} </math></center>

User talk:Wheo001

2008-10-14T04:23:28Z

Wheo001:

=Travelling Wave Solutions of the KdV Equation=

The KdV equation has two qualitatively different types of permanent form travelling wave solution. These are referred to as cnoidal waves and solitary waves.

==KdV equation in <math>(z,\tau)</math> space ==

Assume we have wave travelling with speed <math> V_0 </math> without change of form,

<center><math> H(z,\tau)=H(z-V_0\tau) </math></center>

and substitute into KdV equation then we obtain

<center><math>
-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center>
where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate.

We integrate this equation twice with respect to <math>\xi</math> to give

<center><math> \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2) </math></center> where D_1 and D_2 are constants of integration.

==Standardization of KdV equation==

We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,
so <math>f(H)=\frac{1}{6}H_\xi^2 </math>

It turns out that we require 3 real roots to obtain periodic solutions.
Let roots be <math> H_1 \leq H_2 \leq H_3</math>.

We can imagine the graph of cubic function which has 3 real roots and we can now write a function
<center><math> f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)</math></center>

From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0.</math>

We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.

and now solve equation in terms of the roots <math>H_i,</math>

We define <math>X=\frac{H}{H_3}</math>, and obtain

<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
where <math>X_i=\frac{H_i}{H}</math>

crest to be at <math>\xi=0 and X(0)=0</math>

and a further variable Y via

<center><math> X=1+(X_2-1)sin^2(Y) </math></center>

<center><math>Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}sin(Y)^2 \right\}

...(1)</math></center>

so <math>Y(0)=0.</math>

and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))},</math></center>
which is separable.

In order to get this into a completely standard form we define

<center><math>k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1)

...(2) </math></center>

Clearly, <math>0 \leq k^2 \leq 1</math> and <math>l>0.</math>

==Solution for KdV equation==

A simple quadrature of equation (1) subject to the condition (2) the gives us

<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form

<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>

or equivalently

<center><math> x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} </math></center>

Now we can write Y with fixed values of x,k as

<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>

<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
and hence
<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>

<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".

Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form

<center><math>H=H_2+(H_3-H_2)cn^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} </math></center>

User talk:Wheo001

2008-10-14T04:22:57Z

Wheo001: /* Standardization of KdV equation */

User talk:Wheo001

2008-10-14T04:20:59Z

Wheo001: /* Rearranging KdV equation */

=Travelling Wave Solutions of the KdV Equation=

The KdV equation has two qualitatively different types of permanent form travelling wave solution. These are referred to as cnoidal waves and solitary waves.

==KdV equation in <math>(z,\tau)</math> space ==

Assume we have wave travelling with speed <math> V_0 </math> without change of form,

<center><math> H(z,\tau)=H(z-V_0\tau) </math></center>

and substitute into KdV equation then we obtain

<center><math>
-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center>
where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate.

We integrate this equation twice with respect to <math>\xi</math> to give

<center><math> \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2) </math></center> where D_1 and D_2 are constants of integration.

==Standardization of KdV equation==

We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,
so <math>f(H)=\frac{1}{6}H_\xi^2 </math>

It turns out that we require 3 real roots to obtain periodic solutions.
Let roots be <math> H_1 \leq H_2 \leq H_3</math>.

We can imagine the graph of cubic function which has 3 real roots and we can now write a function
<center><math> f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)</math></center>

From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0.</math>

We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.

and now solve equation in terms of the roots <math>H_i,</math>

We define <math>X=\frac{H}{H_3}</math>, and obtain

<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
where <math>X_i=\frac{H_i}{H}</math>

crest to be at <math>\xi=0 and X(0)=0</math>

and a further variable Y via

<center><math> X=1+(X_2-1)sin^2(Y) </math></center>

<center><math>Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}sin(Y)^2 \right\}

...(1)</math></center>

so <math>Y(0)=0.</math>

and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))},</math></center>
which is separable.

In order to get this into a completely standard form we define

<center><math>k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1)

...(2) </math></center>

Clearly, <math>0 \leq k^2 \leq 1</math> and <math>l>0.</math>

A simple quadrature of equation (1) subject to the condition (2) the gives us

<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form

<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>

or equivalently

<center><math> x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} </math></center>

Now we can write Y with fixed values of x,k as

<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>

<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
and hence
<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>

<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".

Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form

<center><math>H=H_2+(H_3-H_2)cn^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} </math></center>

User talk:Wheo001

2008-10-14T04:18:52Z

Wheo001: /* KdV equation in <math>(z,\tau)</math> space */

User talk:Wheo001

2008-10-14T04:17:38Z

Wheo001: /* KdV equation in <math>(\z,\tau)</math> space */

=Travelling Wave Solutions of the KdV Equation=

The KdV equation has two qualitatively different types of permanent form travelling wave solution. These are referred to as cnoidal waves and solitary waves.

==KdV equation in <math>(z,\tau)</math> space ==

Assume we have wave travelling with speed <math> V_0 </math> without change of form,

<center><math> H(z,\tau)=H(z-V_0\tau) </math></center>

and substitute into KdV equation then we obtain

<center><math>
-2V_oH_\xi+3HH_\xi+\frac{1}{3}H_{\xi\xi\xi}=0</math></center>
where <math>\xi=z-V_0\tau </math> is the travelling wave coordinate.

We integrate this equation twice with respect to <math>\xi</math> to give

<center><math> \frac{1}{6}H_\xi^2=V_oH^2-\frac{1}{2}H^3+D_1H+D_2=f(H,V_0,D_1,D_2) </math></center> where D_1 and D_2 are constants of integration.

We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,
so <math>f(H)=\frac{1}{6}H_\xi^2 </math>

It turns out that we require 3 real roots to obtain periodic solutions.
Let roots be <math> H_1 \leq H_2 \leq H_3</math>.

We can imagine the graph of cubic function which has 3 real roots and we can now write a function
<center><math> f(H)= \frac{-1}{2}(H-H_1)(H-H_2)(H-H_3)</math></center>

From the equation <math>f(H)=\frac{1}{6}H_\xi^2</math>, we require <math>f(H)>0.</math>

We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.

and now solve equation in terms of the roots <math>H_i,</math>

We define <math>X=\frac{H}{H_3}</math>, and obtain

<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
where <math>X_i=\frac{H_i}{H}</math>

crest to be at <math>\xi=0 and X(0)=0</math>

and a further variable Y via

<center><math> X=1+(X_2-1)sin^2(Y) </math></center>

<center><math>Y_\xi^2=\frac{3}{4}H_3(1-X_1)\left\{ 1-\frac{(1-X_2)}{(1-X_1)}sin(Y)^2 \right\}

...(1)</math></center>

so <math>Y(0)=0.</math>

and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2sin^2(Y))},</math></center>
which is separable.

In order to get this into a completely standard form we define

<center><math>k^2=\frac{1-X_2}{1-X_1}, l=\frac{3}{4}H_3(1-X_1)

...(2) </math></center>

Clearly, <math>0 \leq k^2 \leq 1</math> and <math>l>0.</math>

A simple quadrature of equation (1) subject to the condition (2) the gives us

<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

Jacobi elliptic function <math> y= sn(x,k)</math> can be written in the form

<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math></center>

or equivalently

<center><math> x=\int_{0}^{sin^{-1}y} \frac{dS}{\sqrt{1-k^2sin^2(s)}} </math></center>

Now we can write Y with fixed values of x,k as

<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>

<center><math> sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
and hence
<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>

<math>cn(x;k)</math> is another Jacobi elliptic function with <math>cn^2+sn^2=1</math>, and waves are called "cnoidal waves".

Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form

<center><math>H=H_2+(H_3-H_2)cn^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} </math></center>

==Rearranging KdV equation==