User talk:Wheo001 - Revision history

Wheo001: /* Standardization of KdV equation */

2008-10-17T07:19:21Z

Standardization of KdV equation

← Older revision		Revision as of 07:19, 17 October 2008
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	and a further variable Y via		and a further variable Y via

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

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:18:23Z

Solution of the KdV equation

← Older revision		Revision as of 07:18, 17 October 2008
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	<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>		<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>
	for ~~'''~~ <math>0 < k^2 < 1 </math> ~~'''~~		for <math>0 < k^2 < 1 </math>,

	or equivalently		or equivalently
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	Now we can write Y with fixed values of <math>x</math>,<math>k</math> as		Now we can write Y with fixed values of <math>x</math>,<math>k</math> as

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

	<center><math> \sin(Y)=sn(\sqrt{l}\xi;k),</math></center>		<center><math> \sin(Y)=\mathrm {sn}(\sqrt{l}\xi;k),</math></center>
	and hence		and hence
	<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>		<center><math>X=1+(X_2-1)\mathrm {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".		<math>\mathrm {cn}(x;k)</math> is another Jacobi elliptic function with <math>\mathrm {cn}^2+\mathrm {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		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>		<center><math>H=H_2+(H_3-H_2)\mathrm {cn}^2\left\{ \left[ \frac{3}{4}(H_3-H_1) \right]^{\frac{1}{2}}\xi;k \right\} </math></center>

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:11:23Z

Solution of the KdV equation

← Older revision		Revision as of 07:11, 17 October 2008
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	<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>		<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>
	for ~~'''Bold text~~'''<math>0 < k^2 < 1 </math>		for ''' <math>0 < k^2 < 1 </math> '''

	or equivalently		or equivalently

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:10:56Z

Solution of the KdV equation

← Older revision		Revision as of 07:10, 17 October 2008
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	<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>		<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>
	<math> ~~for~~ 0 < k^2 < 1 </math>		for '''Bold text'''<math>0 < k^2 < 1 </math>

	or equivalently		or equivalently

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:10:13Z

Solution of the KdV equation

← Older revision		Revision as of 07:10, 17 October 2008
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	<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}} , for 0 < k^2 < 1</math~~></center~~>		<center><math> x= \int_{0}^y \frac{dt}{\sqrt{1-t^2}\sqrt{1-k^2t^2}}</math></center>
			<math> for 0 < k^2 < 1 </math>

	or equivalently		or equivalently

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:08:55Z

Solution of the KdV equation

← Older revision		Revision as of 07:08, 17 October 2008
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	Now we can write Y with fixed values of <math>x</math>,<math>k</math> as		Now we can write Y with fixed values of <math>x</math>,<math>k</math> as

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

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:08:14Z

Solution of the KdV equation

← Older revision		Revision as of 07:08, 17 October 2008
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	<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2\sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>		<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2\sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

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


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	Now we can write Y with fixed values of <math>x</math>,<math>k</math> as		Now we can write Y with fixed values of <math>x</math>,<math>k</math> as

	<center><math> \bar{Y}=sin^{-1}(\sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>		<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>		<center><math> \sin(Y)=sn(\sqrt{l}\xi;k),</math></center>
	and hence		and hence
	<center><math>X=1+(X_2-1)\sn^2(\sqrt{l}\xi;k)</math></center>		<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".		<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		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>		<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>

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:07:12Z

Solution of the KdV equation

← Older revision		Revision as of 07:07, 17 October 2008
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	A simple quadrature of equation (1) subject to the condition (2) the gives us		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>		<center><math>\int_{0}^\bar{Y} \frac{dS}{\sqrt{l(1-k^2\sin^2(Y))}} = \int_{0}^\bar{Y} dS</math></center>

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


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	or equivalently		or equivalently

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

	Now we can write Y with fixed values of <math>x</math>,<math>k</math> as		Now we can write Y with fixed values of <math>x</math>,<math>k</math> as

	<center><math> \bar{Y}=sin^{-1}(sn(\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>		<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>		<center><math> \sin(Y)=\sn(\sqrt{l}\xi;k),</math></center>
	and hence		and hence
	<center><math>X=1+(X_2-1)sn^2(\sqrt{l}\xi;k)</math></center>		<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".		<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		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>		<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>

Wheo001: /* KdV equation in $(z,\tau)$ space */

2008-10-17T07:05:24Z

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

← Older revision		Revision as of 07:05, 17 October 2008
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	<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 <math>D_1</math> and <math>D_2</math> are constants of integration.		<center><math> \Longrightarrow
			\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 <math>D_1</math> and <math>D_2</math> are constants of integration.

	==Standardization of KdV equation==		==Standardization of KdV equation==

Wheo001: /* Standardization of KdV equation */

2008-10-17T07:04:57Z

Standardization of KdV equation

← Older revision		Revision as of 07:04, 17 October 2008
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	and a further variable Y via		and a further variable Y via

	<center><math> X=1+(X_2-1)sin^2(Y) </math></center>		<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\}		<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>		...(1)</math></center>
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	so <math>Y(0)=0.</math>		so <math>Y(0)=0.</math>

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

User talk:Wheo001 - Revision history

Wheo001: /* Standardization of KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

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

Wheo001: /* Standardization of KdV equation */

Wheo001: /* KdV equation in $(z,\tau)$ space */