Numerical Solution of the KdV - Revision history

Levi: /* Example Calculations */

2025-09-24T16:50:28Z

Example Calculations

← Older revision		Revision as of 16:50, 24 September 2025
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	<center><math>		<center><math>
	u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)		u(x,0) = 8\,\mathrm{sech}^{2}\left(2(x+8)\right)
	+ \, 2 \mathrm{sech}^{2}\left(~~\left(~~		+ 2\, \mathrm{sech}^{2}\left(x+1\right)</math></center>
	x +1~~\right)~~ \right)</math></center>

	{\| class="wikitable"		{\| class="wikitable"
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	\|}		\|}


	== Lecture Videos ==		== Lecture Videos ==

2025-09-24T16:45:58Z

Numerical Solution of the KdV

← Older revision		Revision as of 16:45, 24 September 2025
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	spectral methods. We begin with the KdV equation written as		spectral methods. We begin with the KdV equation written as
	<center><math>		<center><math>
	\partial _{t}u+3\partial _{x}\left( u~~\right)~~ ^{2}+\partial _{x}^{3}u=0.		\partial _{t}u+3\partial _{x}\left( u^{2}\right)+\partial _{x}^{3}u=0.
	</math></center>		</math></center>
	The Fourier transform of the KdV is therefore		The Fourier transform of the KdV is therefore

2025-09-24T15:48:47Z

Numerical Implementation Using the FFT

← Older revision		Revision as of 15:48, 24 September 2025
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	==Numerical Implementation Using the FFT==		==Numerical Implementation Using the FFT==

	The Fast Fourier Transform (FFT) is a method to calculate the ~~fourier~~		The Fast Fourier Transform (FFT) is a method to calculate the Fourier
	transform efficiently for discrete sets of points. These points need to be		transform efficiently for discrete sets of points. These points need to be
	evenly spaced in the <math>x</math> plane and are given by		evenly spaced in the <math>x</math> plane and are given by

2025-09-24T15:47:34Z

Solution for the Linearized KdV

← Older revision		Revision as of 15:47, 24 September 2025
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	u\left( x,0\right) =f\left( x\right)		u\left( x,0\right) =f\left( x\right)
	</math></center>		</math></center>
	We can solve this equation by taking the Fourier transform. ~~and~~ obtain		We can solve this equation by taking the Fourier transform. We obtain
	<center><math>		<center><math>
	\partial _{t}\hat{u}-ik^{3}\hat{u}=0		\partial _{t}\hat{u}-ik^{3}\hat{u}=0

2023-08-17T02:58:20Z

Fourier Transform

← Older revision		Revision as of 02:58, 17 August 2023
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	where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm		where we have assumed that the function <math>f\left( x\right) </math> vanishes at <math>\pm
	\infty .</math> This means that the Fourier transform converts differentiation to		\infty .</math> This means that the Fourier transform converts differentiation to
	multiplication by <math>k</math>.		multiplication by <math>i k</math>.

	==Solution for the Linearized KdV==		==Solution for the Linearized KdV==

2020-08-31T02:49:58Z

Lecture Videos

2020-08-31T02:46:13Z

Example Calculations

2017-03-23T00:42:17Z

Solution for the Linearized KdV

← Older revision		Revision as of 00:42, 23 March 2017
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	equation		equation
	<center><math>		<center><math>
	\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty <x<\infty		\partial _{t}u+\partial _{x}^{3}u=0,\ \ -\infty < x <\infty
	</math></center>		</math></center>
	subject to solve initial conditions		subject to solve initial conditions

2015-01-14T22:22:21Z

Fourier Transform

← Older revision		Revision as of 22:22, 14 January 2015
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	\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)		\mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x &=&-\int_{-\infty }^{\infty }f\left( x\right)
	\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\		\left( \partial_{x}\mathrm{e}^{-\mathrm{i}kx}\right) \mathrm{d}x \\
	&=&-k\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x		&=&ik\int_{-\infty }^{\infty }f\left( x\right) \mathrm{e}^{-\mathrm{i}kx}\mathrm{d}x
	\end{matrix}		\end{matrix}
	</math></center>		</math></center>

2010-11-09T03:58:40Z

Numerical Solution of the KdV

← Older revision		Revision as of 03:58, 9 November 2010
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	\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\		\mathrm{e}^{\mathrm{i}k^{3}\Delta t} \\
	\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)		\hat{u}\left( k,t+\Delta t\right) &=&\hat{u}_{1}\left( k,t+\Delta t\right)
	-3ik \Delta t\widehat{\left( u^{2}\right)}		-3ik \Delta t\widehat{\left( u_{1}^{2}\right)}
	\end{matrix}</math></center>		\end{matrix}</math></center>
	Note that we are using Euler's method to time step and that the solution		Note that we are using Euler's method to time step and that the solution