Connection betwen KdV and the Schrodinger Equation - Revision history

Levi: /* Single Soliton Example */

2025-10-05T08:42:21Z

Single Soliton Example

← Older revision		Revision as of 08:42, 5 October 2025
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	u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\		u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
	& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}		& =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( e^{2k_{1}
	x-8k_{1}^{3}t-\alpha} + 1/2k_1\right) ^{2}}\\		x-8k_{1}^{3}t-\alpha} + 1/(2k_1)\right) ^{2}}\\
	& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt		& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
	{2k_{1}}\right) ^{2}}\\		{2k_{1}}\right) ^{2}}\\
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	^{3}t\right\}		^{3}t\right\}
	\end{align}</math></center>		\end{align}</math></center>
	where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-~~kx_~~{0}		where <math>\theta=k_{1}x-4k_{1}^{3}t-\alpha/2</math> and <math>\sqrt{2k_{1}}e^{-\alpha/2}=e^{-k_{1}x_{0}
	}</math>. This is of course the single soliton solution.		}</math>. This is of course the single soliton solution.

Levi: /* Reflectionless Potential */ this can actually be solved nicely (Dunajski 2009)

2025-09-30T06:50:39Z

Reflectionless Potential: this can actually be solved nicely (Dunajski 2009)

← Older revision		Revision as of 06:50, 30 September 2025
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	the elements of <math>\mathbf{C}</math> are given by		the elements of <math>\mathbf{C}</math> are given by
	<center><math>		<center><math>
	c_{mn} = \frac{c_{n}^2\left( t\right)}		c_{nm} = \frac{c_{n}^2\left( t\right)}
	{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}		{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x}
	</math></center>		</math></center>
	This ~~gives us~~		This means
	<center><math>		<center><math>
	K\left( x,y,t\right) =-\sum_{m=1}^{N} \left(		v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_m
	\mathbf{I}+\mathbf{C}\right) ^{-1}\~~vec~~{f}e^{-k_{m}y}		</math></center>
			Since <math>\partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n</math>, this gives us
			<center><math>\begin{align}
			K\left( x,x,t\right) =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\
			=-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\
			=&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\
			=\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\
			=\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\
			=\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
			\end{align}
			</math></center>
			And therefore
			<center><math>
			u\left( x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)
	</math></center>		</math></center>
	~~We then find <math>u(x,t)</math> from <math>K</math>.~~

	=== Single Soliton Example ===		=== Single Soliton Example ===
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	& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt		& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt
	{2k_{1}}\right) ^{2}}\\		{2k_{1}}\right) ^{2}}\\
	& =2k_{1}^{2}\~~mbox~~{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}		& =2k_{1}^{2}\operatorname{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}
	^{3}t\right\}		^{3}t\right\}
	\end{align}</math></center>		\end{align}</math></center>

Levi: /* Reflectionless Potential */ revert

2025-09-30T05:15:41Z

Reflectionless Potential: revert

← Older revision		Revision as of 05:15, 30 September 2025
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	v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0		v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
	</math></center>		</math></center>
	which is an algebraic (finite dimensional system) for the unknowns <math>v_{m}.</math>.		which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>.

	We can write this as		We can write this as

Levi: /* Reflectionless Potential */ typos

2025-09-30T05:11:45Z

Reflectionless Potential: typos

← Older revision		Revision as of 05:11, 30 September 2025
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	+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }		+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) }
	-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(		-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{ c_{n}^{2}\left(
	t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) e^{-k_{m}x}e^{-k_{n}\left(		t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left(
	y+x\right) }=0		y+x\right) }=0
	</math></center>		</math></center>
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	v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0		v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0
	</math></center>		</math></center>
	which is an algebraic (finite dimensional system) for the ~~unknows~~ <math>v_{n}.</math>.		which is an algebraic (finite dimensional system) for the unknowns <math>v_{m}.</math>.

	We can write this as		We can write this as
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	where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}		where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
	}</math>. This is of course the single soliton solution.		}</math>. This is of course the single soliton solution.


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

Levi: some clarifications, add book link

2025-09-29T07:00:52Z

some clarifications, add book link

← Older revision		Revision as of 07:00, 29 September 2025
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	\partial_{t}\lambda=0		\partial_{t}\lambda=0
	</math></center>		</math></center>
	provided that the eigenfunction <math>w</math> ~~is bounded~~ (which is true for the bound		provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound
	state eigenfunctions). This shows that the discrete eigenvalues are unchanged		state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged
	and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other		and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
	properties can be found.		properties can be found.

			Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3).

	==Scattering Data==		==Scattering Data==

Levi at 10:51, 28 September 2025

2025-09-28T10:51:04Z

← Older revision		Revision as of 10:51, 28 September 2025
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	<center><math>		<center><math>
	\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial		\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
	_{x}wQ\right)\Bigg\|^\infty_{\infty}=0		_{x}wQ\right)\Bigg\|^\infty_{-\infty}=0
	</math></center>		</math></center>
	and, since <math>w^2>0</math> except for some isolated points		and, since <math>w^2>0</math> except for some isolated points,
	<center><math>		<center><math>
	\partial_{t}\lambda=0		\partial_{t}\lambda=0

Levi: some more explanations

2025-09-28T10:49:03Z

some more explanations

← Older revision		Revision as of 10:49, 28 September 2025
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	</math></center>		</math></center>
	where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)		where <math>Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
	\partial_{x}w.</math> If we integrate this equation ~~then~~ we obtain ~~the result that~~		\partial_{x}w.</math> If we integrate this equation we obtain
			<center><math>
			\int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial
			_{x}wQ\right)\Bigg\|^\infty_{\infty}=0
			</math></center>
			and, since <math>w^2>0</math> except for some isolated points
	<center><math>		<center><math>
	\partial_{t}\lambda=0		\partial_{t}\lambda=0
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	state eigenfunctions). This shows that the discrete eigenvalues are unchanged		state eigenfunctions). This shows that the discrete eigenvalues are unchanged
	and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other		and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other
	properties can be found		properties can be found.

	==Scattering Data==		==Scattering Data==

Meylan: /* Scattering Data */

2023-09-14T05:30:01Z

Scattering Data

← Older revision		Revision as of 05:30, 14 September 2023
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	The scattering data evolves as		The scattering data evolves as
	<center><math>		<center><math>
	k_{n}=k_{n}		k_{n}(t)=k_{n}(0) = k_{n}
	</math></center>		</math></center>
	<center><math>		<center><math>

Meylan: /* Single Soliton Example */

2020-10-01T03:28:30Z

Single Soliton Example

← Older revision		Revision as of 03:28, 1 October 2020
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	where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}		where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
	}</math>. This is of course the single soliton solution.		}</math>. This is of course the single soliton solution.


			== Lecture Videos ==

			=== Part 1 ===

			{{#ev:youtube\|KHGoPCoyP28}}

			=== Part 2 ===

			{{#ev:youtube\|8mVq6MQWO3I}}

			=== Part 3 ===

			{{#ev:youtube\|meTgaaGKsfQ}}

			=== Part 4 ===

			{{#ev:youtube\|AfAQyjzvJUU}}

Meylan: /* Single Soliton Example */

2020-10-01T03:02:50Z

Single Soliton Example

← Older revision		Revision as of 03:02, 1 October 2020
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	e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }		e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) }
	{2k_{1}}e^{-2 k_{1} x}}\\		{2k_{1}}e^{-2 k_{1} x}}\\
	& =\frac{-1}{e^{-2k_{1}x + 8k_{1}^{3}t-\alpha} + 1/2k_1}		& =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1}
	\end{matrix}</math></center>		\end{matrix}</math></center>
	where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore		where <math>e^{-\alpha}=1/c_{0}^{2}\left( 0\right) .</math> Therefore