Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 86 / 2 / a

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. In particular $q_{z\bar{z}}=\Delta q/4$. Let $E=e^{-ikz-\lambda \bar{z}/(ik)}$, $A=E(q_z+ikq)$ and $B=E(q_{\bar{z}}+\lambda q/(ik))$, so the printed differential form is $W=Adz-Bd\bar{z}$. Since $E_z=-ikE$ and $E_{\bar{z}}=-\lambda E/(ik)$,Thus the exterior derivative isBecause $E$ never vanishes and $k\ne 0$, this proves both directions: $W$ is a closed differential form if and only if $q_{z\bar{z}}=\lambda q$. The latter is the modified Helmholtz equation $\Delta q-4\lambda q=0$ for positive $\lambda$. The associated global relations on a bounded region $\Omega$ are, for every nonzero spectral parameter,The boundary orientation is positive, by Stokes theorem. Each admissible spectral parameter gives a separate identity between the traces. Interchanging $z$ and $\bar{z}$ gives the dual family, equivalently the same family at $k\mapsto -\lambda /k$ with an overall minus sign. No assumption that $q$ is real is needed.

For later use with $\lambda >0$, as in the remaining parts, make the global relation explicit in the first quadrant. PutLet $h_2(x)=q(x,0)$, $n_2(x)=q_y(x,0)$ and $h_1(y)=q(0,y)$, $n_1(y)=q_x(0,y)$. These normals are inward-coordinate derivatives. The bottom is traversed from zero to infinity and the left side from infinity to zero. Their oriented spectral integrals areHere ${\rho }_x$ is analytic for $\mathrm{Im}k<0$ and ${\rho }_y$ for $\mathrm{Re}k<0$, with suitable boundary limits. Indeed $\mathrm{Im}a=(r+\lambda /r)\sin \theta$ and $\mathrm{Re}b=(r+\lambda /r)\cos \theta$ for $k=r{e}^{i\theta }$. On their common third quadrant, both exponential weights decay. Apply the finite-domain relation to expanding rectangles; sufficient decay of the solution and traces eliminates the distant edges. This derivesThe limiting relations on the negative real and negative imaginary axes will remove the unknown derivative traces.