KdV Cnoidal Wave Solutions - Revision history

Meylan at 10:20, 6 September 2010

2010-09-06T10:20:53Z

← Older revision		Revision as of 10:20, 6 September 2010
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	<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>		<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>

	[[Category:Nonlinear Water ~~Waves~~]]		[[Category:Nonlinear Water-Wave Theory]]

Meylan at 10:19, 6 September 2010

2010-09-06T10:19:51Z

← Older revision		Revision as of 10:19, 6 September 2010
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			{{complete pages}}

	==Introduction==		==Introduction==

Meylan at 10:19, 6 September 2010

2010-09-06T10:19:33Z

← Older revision		Revision as of 10:19, 6 September 2010
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	<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>		<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>

	[[Category:~~789~~]]		[[Category:Nonlinear Water Waves]]

Meylan: moved KdV Equation Solutions to KdV Cnoidal Wave Solutions

2010-09-06T10:19:01Z

moved KdV Equation Solutions to KdV Cnoidal Wave Solutions

← Older revision	Revision as of 10:19, 6 September 2010
(No difference)

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

2008-10-29T11:55:26Z

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

← Older revision		Revision as of 11:55, 29 October 2008
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	<center><math>\Longrightarrow		<center><math>\Longrightarrow
	\frac{1}{3} \int H_{\xi\xi}H_\xi d\xi = 2V_o \int H H_\xi d\xi - \frac{3}{2} \int H^2H\xi d\xi + \int D_1 H_\xi d\xi</math></center>		\frac{1}{3} \int H_{\xi\xi}H_\xi d\xi = 2V_o \int H H_\xi d\xi - \frac{3}{2} \int H^2H_\xi d\xi + \int D_1 H_\xi d\xi</math></center>

Wheo001: /* Solution of the KdV equation */

2008-10-29T11:53:48Z

Solution of the KdV equation

← Older revision		Revision as of 11:53, 29 October 2008
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	and <math>l>0.</math>		and <math>l>0.</math>

	~~==Solution of the KdV equation==~~

Meylan: /* Solution of the KdV equation */

2008-10-21T08:46:17Z

Solution of the KdV equation

← Older revision		Revision as of 08:46, 21 October 2008
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	<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>		<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>

			[[Category:789]]

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:31:04Z

Solution of the KdV equation

← Older revision		Revision as of 07:31, 17 October 2008
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	==Solution of the KdV equation==		==Solution of the 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.~~		==Solution of the KdV equation==
	~~Let roots be <math> H_1 \leq H_2 \leq H_3</math>.~~

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

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

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

	~~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>.		<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>,

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

	~~We define~~ <math>X=\frac{H}{~~H_3~~}</math>~~, and obtain~~		<center><math> x=\int_{0}^{\sin^{-1}y} \frac{dS}{\sqrt{1-k^2\sin^2(s)}} </math></center>

	~~<center>~~<math>~~X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)~~</math>~~</center>~~		Now we can write Y with fixed values of <math>x</math>,<math>k</math> as
	~~where~~ <math>~~X_i=\frac{H_i}{H}~~</math>

	~~crest to be at~~ <math>\xi=0</math> ~~and <math>X(0)=0~~</~~math~~>		<center><math> \bar{Y}=\sin^{-1}(\mathrm {sn} (\sqrt{l}\int_0^{\bar{Y}} d\xi,k)),</math></center>

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

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

	<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>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>

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

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

	~~and <center><math>\frac{dY~~}{d\xi~~}=\sqrt{l(1-~~k^2\~~sin^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~~>

Wheo001: /* Solution of the KdV equation */

2008-10-17T07:30:18Z

Solution of the KdV equation

← Older revision		Revision as of 07:30, 17 October 2008
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	==Solution of the KdV equation==		==Solution of the KdV equation==
	~~Use <math>sin</math> insead of <math>\sin</math> and use <math>{\rm sn}</math>~~

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

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

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

			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>

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

	<~~center~~><math> ~~x=\int_{0}^{\sin^{-1}y} \frac{dS}{\sqrt{1-k^2\sin^2(s)}}~~ </math></~~center~~>		We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.

	~~Now we can write Y with fixed values~~ of x,~~k as~~		and now solve equation in terms of the roots <math>H_i,</math>

	~~<center>~~<math> ~~\bar{Y}~~=~~sin^{-1}(sn(~~\~~sqrt~~{l}~~\int_0^{\bar~~{Y}~~} d\xi,k)),~~</math>~~</center>~~		We define <math>X=\frac{H}{H_3}</math>, and obtain

	<center><math> ~~sin~~(Y)~~=sn~~(~~\sqrt{l}\xi;k~~),</math></center>		<center><math>X_\xi^2=3H_3(X-X_1)(X-X_2)(1-X)</math></center>
	~~and hence~~		where <math>X_i=\frac{H_i}{H}</math>
	~~<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".~~		crest to be at <math>\xi=0</math> and <math>X(0)=0</math>

	~~Using the result <math>cn^2+sn^2=1</math>, our final result can be expressed in the form~~		and a further variable Y via

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

	~~[[Category:789]]~~
			<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^2\sin^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>

Wheo001: /* Standardization of KdV equation */

2008-10-17T07:29:36Z

Standardization of KdV equation

← Older revision		Revision as of 07:29, 17 October 2008
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	==Standardization of KdV equation==		==Standardization of KdV equation==


	We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,		We define <math>f(H)= V_oH^2-\frac{1}{2}H^3+D_1H+D2</math>,
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	We can imagine the graph of cubic function which has 3 real roots and we can now write a function		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>		<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>		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>.		We are only interested in solution for <math>H_2 < H < H_3</math> and we need <math>H_2 < H_3</math>.
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	We define <math>X=\frac{H}{H_3}</math>, and obtain		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>		<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>		where <math>X_i=\frac{H_i}{H}</math>

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

	and a further variable Y via ~~the relation~~		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)}s\~~in^2~~(Y) \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>

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

	and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2\sin^2(Y))},</math></center>		and <center><math>\frac{dY}{d\xi}=\sqrt{l(1-k^2\sin^2(Y))},</math></center>
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	In order to get this into a completely standard form we define		In order to get this into a completely standard form we define

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

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

	==Solution of the KdV equation==		==Solution of the KdV equation==

KdV Cnoidal Wave Solutions - Revision history

Meylan at 10:20, 6 September 2010

Meylan at 10:19, 6 September 2010

Meylan at 10:19, 6 September 2010

Meylan: moved KdV Equation Solutions to KdV Cnoidal Wave Solutions

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

Wheo001: /* Solution of the KdV equation */

Meylan: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Solution of the KdV equation */

Wheo001: /* Standardization of KdV equation */

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