Consider the KdV equation for the function u(x,t)
ut=6uux−uxxx
(a) Write equation (1) in the Hamiltonian form
ut=∂x∂δuδH[u]
where the functional H[u] should be given. Use equation (1), together with the boundary conditions u→0 and ux→0 as ∣x∣→∞, to show that ∫Ru2dx is independent of t.
(b) Use the Gelfand-Levitan-Marchenko equation
K(x,y)+F(x+y)+∫x∞K(x,z)F(z+y)dz=0
to find the one soliton solution of the KdV equation, i.e.
u(x,t)=−[1+2χβexp(−2χx)]24βχexp(−2χx)
[Hint. Consider F(x)=βexp(−χx), with β=β0exp(8χ3t), where β0,χ are constants, and t should be regarded as a parameter in equation (2). You may use any facts about the Inverse Scattering Transform without proof.]