3.II.31E

Integrable Systems | Part II, 2006

The solution of the initial value problem of the KdV\mathrm{KdV} equation is given by

q(x,t)=2ilimkkNx(x,t,k),q(x, t)=-2 i \lim _{k \rightarrow \infty} k \frac{\partial N}{\partial x}(x, t, k),

where the scalar function N(x,t,k)N(x, t, k) can be obtained by solving the following RiemannHilbert problem:

M(x,t,k)a(k)=N(x,t,k)+b(k)a(k)exp(2ikx+8ik3t)N(x,t,k),kR,\frac{M(x, t, k)}{a(k)}=N(x, t,-k)+\frac{b(k)}{a(k)} \exp \left(2 i k x+8 i k^{3} t\right) N(x, t, k), \quad k \in \mathbb{R},

M,NM, N and aa are the boundary values of functions of kk that are analytic for Imk>0\operatorname{Im} k>0 and tend to unity as kk \rightarrow \infty. The functions a(k)a(k) and b(k)b(k) can be determined from the initial condition q(x,0)q(x, 0).

Assume that MM can be written in the form

Ma=M(x,t,k)+cexp(2px+8p3t)N(x,t,ip)kip,Imk0,\frac{M}{a}=\mathcal{M}(x, t, k)+\frac{c \exp \left(-2 p x+8 p^{3} t\right) N(x, t, i p)}{k-i p}, \quad \operatorname{Im} k \geqslant 0,

where M\mathcal{M} as a function of kk is analytic for Imk>0\operatorname{Im} k>0 and tends to unity as k;ck \rightarrow \infty ; c and pp are constants and p>0p>0.

(a) By solving the above Riemann-Hilbert problem find a linear equation relating N(x,t,k)N(x, t, k) and N(x,t,ip)N(x, t, i p).

(b) By solving this equation explicitly in the case that b=0b=0 and letting c=2ipe2x0c=2 i p e^{-2 x_{0}}, compute the one-soliton solution.

(c) Assume that q(x,0)q(x, 0) is such that a(k)a(k) has a simple zero at k=ipk=i p. Discuss the dominant form of the solution as tt \rightarrow \infty and x/t=O(1)x / t=O(1).

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