Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 61 / 3 / e / Solution

To first order in the potential, $a=A=1+O(q^2)$, while the spatial and temporal Volterra integral equations giveThus the linear limit of the global relation is exactlyfor the linear equation $i{q}_t+q_{xx}=0$. This ties the two boundary traces together, rather than allowing them to be independent data.

The count is made explicit by a temporal Laplace transform. For $\mathrm{Re}p>0$, the transformed equation is ${\tilde{q}}_{xx}+ip\tilde{q}=i{q}_0$. Put $\alpha =\sqrt{-ip}$ with positive real part. Of the two homogeneous solutions $e^{\pm \alpha x}$, only $e^{-\alpha x}$ decays at infinity. There is therefore exactly one free complex coefficient to be fixed at the endpoint. In particular, prescribed Dirichlet trace ${\tilde{g}}_0$ gives the definite Neumann traceThis follows by differentiating the decaying Dirichlet Green kernel $-(e^{-\alpha |x-y|}-e^{-\alpha (x+y)})/(2\alpha )$ at $x=0$. Thus one complex boundary condition is needed, such as Dirichlet, Neumann or a suitable Robin combination. Prescribing both traces arbitrarily would overdetermine the problem. In a Sobolev space with the required trace regularity, the cubic term is a locally Lipschitz perturbation, so the small-data perturbative problem retains the same boundary-condition count; corner compatibility still must be imposed for a classical solution.