Paper 2, Section II, D

Integrable Systems | Part II, 2012

Consider the KdV\mathrm{KdV} equation for the function u(x,t)u(x, t)

ut=6uuxuxxxu_{t}=6 u u_{x}-u_{x x x}

(a) Write equation (1) in the Hamiltonian form

ut=xδH[u]δuu_{t}=\frac{\partial}{\partial x} \frac{\delta H[u]}{\delta u}

where the functional H[u]H[u] should be given. Use equation (1), together with the boundary conditions u0u \rightarrow 0 and ux0u_{x} \rightarrow 0 as x|x| \rightarrow \infty, to show that Ru2dx\int_{\mathbb{R}} u^{2} d x is independent of tt.

(b) Use the Gelfand-Levitan-Marchenko 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) d z=0

to find the one soliton solution of the KdV equation, i.e.

u(x,t)=4βχexp(2χx)[1+β2χexp(2χx)]2u(x, t)=-\frac{4 \beta \chi \exp (-2 \chi x)}{\left[1+\frac{\beta}{2 \chi} \exp (-2 \chi x)\right]^{2}}

[Hint. Consider F(x)=βexp(χx)F(x)=\beta \exp (-\chi x), with β=β0exp(8χ3t)\beta=\beta_{0} \exp \left(8 \chi^{3} t\right), where β0,χ\beta_{0}, \chi are constants, and tt should be regarded as a parameter in equation (2). You may use any facts about the Inverse Scattering Transform without proof.]

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