Paper 2, Section II, D

Integrable Systems
Part II, 2014

Let u=u(x)u=u(x) be a smooth function that decays rapidly as x|x| \rightarrow \infty and let L=x2+u(x)L=-\partial_{x}^{2}+u(x) denote the associated Schrödinger operator. Explain very briefly each of the terms appearing in the scattering data

S={{χn,cn}n=1N,R(k)},S=\left\{\left\{\chi_{n}, c_{n}\right\}_{n=1}^{N}, R(k)\right\},

associated with the operator LL. What does it mean to say u(x)u(x) is reflectionless?

Given SS, define the function

F(x)=n=1Ncn2eχnx+12πeikxR(k)dkF(x)=\sum_{n=1}^{N} c_{n}^{2} e^{-\chi_{n} x}+\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{\mathrm{i} k x} R(k) \mathrm{d} k

If K=K(x,y)K=K(x, y) is the unique solution to the GLM equation

K(x,y)+F(x+y)+xK(x,z)F(z+y)dz=0K(x, y)+F(x+y)+\int_{x}^{\infty} K(x, z) F(z+y) \mathrm{d} z=0

what is the relationship between u(x)u(x) and K(x,x)K(x, x) ?

Now suppose that u=u(x,t)u=u(x, t) is time dependent and that it solves the KdV equation ut+uxxx6uux=0u_{t}+u_{x x x}-6 u u_{x}=0. Show that L=x2+u(x,t)L=-\partial_{x}^{2}+u(x, t) obeys the Lax equation

Lt=[L,A], where A=4x33(ux+xu).L_{t}=[L, A], \quad \text { where } A=4 \partial_{x}^{3}-3\left(u \partial_{x}+\partial_{x} u\right) .

Show that the discrete eigenvalues of LL are time independent.

In what follows you may assume the time-dependent scattering data take the form

S(t)={{χn,cne4χn3t}n=1N,R(k,0)e8ik3t}.S(t)=\left\{\left\{\chi_{n}, c_{n} e^{4 \chi_{n}^{3} t}\right\}_{n=1}^{N}, R(k, 0) e^{8 \mathrm{i} k^{3} t}\right\} .

Show that if u(x,0)u(x, 0) is reflectionless, then the solution to the KdV equation takes the form

u(x,t)=22x2log[detA(x,t)]u(x, t)=-2 \frac{\partial^{2}}{\partial x^{2}} \log [\operatorname{det} A(x, t)]

where AA is an N×NN \times N matrix which you should determine.

Assume further that R(k,0)=k2f(k)R(k, 0)=k^{2} f(k), where ff is smooth and decays rapidly at infinity. Show that, for any fixed xx,

eikxR(k,0)e8ik3t dk=O(t1) as t\int_{-\infty}^{\infty} e^{\mathrm{i} k x} R(k, 0) e^{8 \mathrm{i} k^{3} t} \mathrm{~d} k=O\left(t^{-1}\right) \quad \text { as } t \rightarrow \infty

Comment briefly on the significance of this result.

[You may assume 1detAddx(detA)=tr(A1dA dx)\frac{1}{\operatorname{det} A} \frac{\mathrm{d}}{\mathrm{d} x}(\operatorname{det} A)=\operatorname{tr}\left(A^{-1} \frac{\mathrm{d} A}{\mathrm{~d} x}\right) for a non-singular matrix A(x)A(x).]