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| | {{nonlinear waves course | | {{nonlinear waves course |
| | | chapter title = Introduction to the Inverse Scattering Transform | | | chapter title = Introduction to the Inverse Scattering Transform |
| − | | next chapter = [[Reaction-Diffusion Systems]] | + | | next chapter = [[Properties of the Linear Schrodinger Equation]] |
| | | previous chapter = [[Conservation Laws for the KdV]] | | | previous chapter = [[Conservation Laws for the KdV]] |
| | }} | | }} |
| | | | |
| − |
| |
| − | == Introduction ==
| |
| | | | |
| | The inverse scattering transformation gives a way to solve the KdV equation | | The inverse scattering transformation gives a way to solve the KdV equation |
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| | Fourier transformation, except it works for a non linear equation. We want to | | Fourier transformation, except it works for a non linear equation. We want to |
| | be able to solve | | be able to solve |
| − | <center><math>\begin{align} | + | <center><math>\begin{matrix} |
| | \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ | | \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ |
| | u(x,0) & =f\left( x\right) | | u(x,0) & =f\left( x\right) |
| − | \end{align}</math></center> | + | \end{matrix}</math></center> |
| | with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> | | with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> |
| | | | |
| | The Miura transformation is given by | | The Miura transformation is given by |
| | <center><math> | | <center><math> |
| − | u=v^{2}+v_{x} \, | + | u=-v^{2}-\partial_x v\, |
| | </math></center> | | </math></center> |
| | and if <math>v</math> satisfies the mKdV | | and if <math>v</math> satisfies the mKdV |
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| | then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura | | then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura |
| | transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This | | transformation as being a nonlinear ODE solving for <math>v</math> given <math>u.</math> This |
| − | nonlinear ODE is also known as the Riccati equation and there is a well know | + | nonlinear ODE is also known as the Riccati equation and there is a well known |
| | transformation which linearises this equation. It we write | | transformation which linearises this equation. It we write |
| | <center><math> | | <center><math> |
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| | in the inverse scattering transformation. Note that this is Schrodinger's equation. | | in the inverse scattering transformation. Note that this is Schrodinger's equation. |
| | | | |
| − | ==Properties of the eigenfunctions== | + | == Lecture Videos == |
| − | | |
| − | The equation
| |
| − | <center><math>
| |
| − | \partial_{x}^{2}w+uw=-\lambda w
| |
| − | </math></center>
| |
| − | has two kinds of solutions for <math>u\rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> The
| |
| − | first are waves and the second are bound solutions. It is well known that
| |
| − | there are at most a finite number of bound solutions (provided <math>u\rightarrow0</math>
| |
| − | as <math>x\pm\infty</math> sufficiently rapidly) and a continum of solutions for the
| |
| − | incident waves.
| |
| − | | |
| − | ===Example: Scattering by a Well===
| |
| − | | |
| − | The properties of the eigenfunction is prehaps seem most easily through the
| |
| − | following example
| |
| − | <center><math>
| |
| − | u\left( x\right) =\left\{
| |
| − | \begin{matrix}
| |
| − | | |
| − | 0, & x\notin\left[ -1,1\right] \\
| |
| − | b, & x\in\left[ -1,1\right]
| |
| − | \end{matrix}
| |
| − | \right.
| |
| − | </math></center>
| |
| − | where <math>b>0.</math>
| |
| − | | |
| − | ==Case when <math>\lambda<0</math>==
| |
| − | | |
| − | If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
| |
| − | get
| |
| − | <center><math>
| |
| − | w\left( x\right) =\left\{
| |
| − | \begin{matrix}
| |
| − | | |
| − | a_{1}\mathrm{e}^{kx}, & x<-1\\
| |
| − | b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
| |
| − | a_{2}\mathrm{e}^{-kx} & x>1
| |
| − | \end{matrix}
| |
| − | \right.
| |
| − | </math></center>
| |
| − | where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
| |
| − | no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
| |
| − | to solve for <math>a</math> and <math>b</math>. This leads to two system of equations, one for the
| |
| − | even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
| |
| − | (<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
| |
| − | <center><math>
| |
| − | \left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | \mathrm{e}^{-kx} & -\cos\kappa\\
| |
| − | k\mathrm{e}^{-kx} & \sin\kappa
| |
| − | \end{matrix}
| |
| − | \right) \left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | a_{1}\\
| |
| − | b_{1}
| |
| − | \end{matrix}
| |
| − | \right) =\left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | 0\\
| |
| − | 0
| |
| − | \end{matrix}
| |
| − | \right)
| |
| − | </math></center>
| |
| − | This has non trivial solutions when
| |
| − | <center><math>
| |
| − | \det\left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | \mathrm{e}^{-kx} & -\cos\kappa\\
| |
| − | k\mathrm{e}^{-kx} & - \kappa \sin\kappa
| |
| − | \end{matrix}
| |
| − | \right) =0
| |
| − | </math></center>
| |
| − | which gives us the equation
| |
| − | <center><math>
| |
| − | - \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
| |
| − | </math></center>
| |
| − | or
| |
| − | <center><math>
| |
| − | \kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
| |
| − | </math></center>
| |
| − | We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
| |
| − | finite number of solutions.
| |
| − | | |
| − | In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
| |
| − | Similarly we repeat the above process for the odd solutions.
| |
| − | | |
| − | ==Case when <math>\lambda>0</math>==
| |
| − | | |
| − | When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
| |
| − | <center><math>
| |
| − | w\left( x\right) =\left\{
| |
| − | \begin{matrix}
| |
| − | | |
| − | \mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
| |
| − | b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
| |
| − | a\mathrm{e}^{-\mathrm{i}kx} & x>1
| |
| − | \end{matrix}
| |
| − | \right.
| |
| − | </math></center>
| |
| − | where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
| |
| − | obtain
| |
| − | <center><math>
| |
| − | \left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | -\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
| |
| − | ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
| |
| − | 0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
| |
| − | 0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
| |
| − | \end{matrix}
| |
| − | \right) \left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | r\\
| |
| − | b_{1}\\
| |
| − | b_{2}\\
| |
| − | a
| |
| − | \end{matrix}
| |
| − | \right) =\left(
| |
| − | \begin{matrix}
| |
| − | | |
| − | \mathrm{e}^{ik}\\
| |
| − | ik\mathrm{e}^{-ik}\\
| |
| − | 0\\
| |
| − | 0
| |
| − | \end{matrix}
| |
| − | \right)
| |
| − | </math></center>
| |
| − | | |
| − | ==Connection with the KdV==
| |
| − | | |
| − | If we substitute the relationship
| |
| − | <center><math>
| |
| − | \partial_{x}^{2}w+uw=-\lambda w
| |
| − | </math></center>
| |
| − | into the KdV after some manipulation we obtain
| |
| − | <center><math>
| |
| − | \partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
| |
| − | _{x}wQ\right) =0
| |
| − | </math></center>
| |
| − | 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
| |
| − | <center><math>
| |
| − | \partial_{t}\lambda=0
| |
| − | </math></center>
| |
| − | provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
| |
| − | state eigenfunctions). This shows that the discrete eigenvalues are unchanged
| |
| − | and <math>u\left( x,t\right) </math> evolves according to the KdV.
| |
| − | | |
| − | ==Scattering Data==
| |
| − | | |
| − | For the discrete spectrum the eigenfunctions behave like
| |
| − | <center><math>
| |
| − | w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
| |
| − | </math></center>
| |
| − | as <math>x\rightarrow\infty</math> with
| |
| − | <center><math>
| |
| − | \int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
| |
| − | </math></center>
| |
| − | The continuous spectrum looks like
| |
| − | <center><math>
| |
| − | v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
| |
| − | ,\ \ \ x\rightarrow-\infty
| |
| − | </math></center>
| |
| − | <center><math>
| |
| − | v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
| |
| − | \infty
| |
| − | </math></center>
| |
| − | where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
| |
| − | coefficient. This gives us the scattering data at <math>t=0</math>
| |
| − | <center><math>
| |
| − | S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
| |
| − | \right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
| |
| − | </math></center>
| |
| − | The scattering data evolves as
| |
| − | <center><math>
| |
| − | k_{n}=k_{n}
| |
| − | </math></center>
| |
| − | <center><math>
| |
| − | c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
| |
| − | </math></center>
| |
| − | <center><math>
| |
| − | r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
| |
| − | </math></center>
| |
| − | <center><math>
| |
| − | a\left( k,t\right) =a\left( k,0\right)
| |
| − | </math></center>
| |
| − | We can recover <math>u</math> from scattering data. We write
| |
| − | <center><math>
| |
| − | F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
| |
| − | x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
| |
| − | </math></center>
| |
| − | Then solve
| |
| − | <center><math>
| |
| − | K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
| |
| − | x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
| |
| − | </math></center>
| |
| − | This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
| |
| − | }equation. We then find <math>u</math> from
| |
| − | <center><math>
| |
| − | u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
| |
| − | </math></center>
| |
| − | | |
| | | | |
| − | ==Reflectionless Potential== | + | === Part 1 === |
| | | | |
| − | In general the IST is difficult to solve. However, there is a simplification
| + | {{#ev:youtube|P3uMk9OS8p4}} |
| − | we can make when we have a reflectionless potential (which we will see gives
| |
| − | rise to the soliton solutions). The reflectionless potential is the case when
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| − | <math>r\left( k,0\right) =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
| |
| − | <center><math>
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| − | F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
| |
| − | </math></center>
| |
| − | then
| |
| − | <center><math>
| |
| − | K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
| |
| − | \mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
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| − | \sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
| |
| − | </math></center>
| |
| − | From the equation we can see that
| |
| − | <center><math>
| |
| − | K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
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| − | x\right) \mathrm{e}^{-k_{m}y}
| |
| − | </math></center>
| |
| − | If we substitute this into the equation
| |
| − | <center><math>
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| − | -\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
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| − | +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
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| − | +\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
| |
| − | \mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
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| − | y+z\right) }dz=0
| |
| − | </math></center>
| |
| − | which leads to
| |
| − | <center><math>
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| − | -\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
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| − | +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
| |
| − | -\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
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| − | t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
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| − | y+x\right) }=0
| |
| − | </math></center>
| |
| − | and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
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| − | <math>\mathrm{e}^{-k_{n}y}</math> to obtain
| |
| − | <center><math>
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| − | -v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
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| − | ^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
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| − | v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
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| − | </math></center>
| |
| − | which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
| |
| − | can write this as
| |
| − | <center><math>
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| − | \left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
| |
| − | </math></center>
| |
| − | where <math>f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x}</math> and
| |
| − | <center><math>
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| − | c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
| |
| − | }{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
| |
| − | </math></center>
| |
| − | <center><math>
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| − | K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
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| − | \mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
| |
| − | </math></center>
| |
| − | This leads to
| |
| − | <center><math>
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| − | u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
| |
| − | +\mathbf{C}\right) \right]
| |
| − | </math></center>
| |
| − | Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
| |
| − | solution) we get
| |
| − | <center><math>\begin{matrix}
| |
| − | K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
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| − | t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
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| − | t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
| |
| − | & =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
| |
| − | \end{matrix}</math></center>
| |
| − | where <math>\mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) .</math> Therefore
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| − | <center><math>\begin{matrix}
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| − | u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
| |
| − | & =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
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| − | x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
| |
| − | & =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
| |
| − | /\sqrt{2k_{1}}\right) ^{2}}\\
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| − | & =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
| |
| − | \end{matrix}</math></center>
| |
| − | where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
| |
| − | }</math>. This is of course the single soliton solution. | |
