Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 69 / 3

Put $E_{\lambda }=e^{-i\beta (\lambda z-\bar{z}/\lambda )}$. The Wirtinger derivatives give $(E_{\lambda }{)}_z=-i\beta \lambda E_{\lambda }$ and $(E_{\lambda }{)}_{\bar{z}}=i\beta E_{\lambda }/\lambda$. Write $W=Adz+Bd\bar{z}$, whereUsing the exterior derivative, $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}$. The terms involving $u_z$ and $u_{\bar{z}}$ cancel, leavingBecause $E_{\lambda }$ never vanishes for $\lambda \ne 0$, $dW=0$ if and only if $u_{z\bar{z}}-{\beta }^{2}u=0$. In real coordinates this is the modified Helmholtz equation with mass parameter $2\beta$, not $\beta$. In fact any one nonzero spectral parameter suffices for the equivalence; the full family supplies many independent boundary tests.

Use a consistent number $n$ of vertices, with $z_{n+1}=z_1$, and set $m_j=(z_j+z_{j+1})/2$, $h_j=(z_{j+1}-z_j)/2$. Pull back the one-form to $z_j(s)=m_j+s{h}_j$. Since $dz=h_{j}ds$ and $d\bar{z}={\bar{h}}_{j}ds$, the side integrand isFor counterclockwise traversal, put ${\ell }_j=|h_j|$ and let $q_j={\partial }_{n}u$ be the outward normal derivative, while $g_j=u{|}_{S_j}$ is the Dirichlet boundary data. The outward unit normal is $-i{h}_j/{\ell }_j$ in complex notation. Therefore $h_j{u}_z-{\bar{h}}_j{u}_{\bar{z}}=i{\ell }_j{q}_j$, andThere is no tangential-derivative term: it cancels in this particular one-form. The Generalized Stokes theorem and $dW=0$ now give the polygonal global relationThe same zero identity holds with every side traversed clockwise, but then $h_j{u}_z-{\bar{h}}_j{u}_{\bar{z}}=-i{\ell }_j{q}_j$ for outward normals. One must change this sign consistently rather than mix the two orientations.

Interpret the printed coordinate notation as the square corners $1+i$, $1-i$, $-1-i$, $-1+i$. This listed order is clockwise, contrary to the counterclockwise convention in (ii). Keep the printed first side directed from top to bottom. Then $m_1=1$, $h_1=-i$, and $z(s)=1-is$. Put $g_1(s)=u(1,-s)$ and $q_1(s)=u_x(1,-s)$, where $q_1$ is the outward normal derivative on the right side. The pullback formula, without any orientation shortcut, givesIndeed $h_1{u}_z-{\bar{h}}_1{u}_{\bar{z}}=-i(u_z+u_{\bar{z}})=-i{u}_x$, and $i\beta (\lambda h_1+{\bar{h}}_1/\lambda )=\beta (\lambda -{\lambda }^{-1})$. If one reverses the side to match the counterclockwise convention, its parameter is $1+is$ and its integrand is $-W_1(-s,\lambda )$. Reversing every side multiplies the global relation by minus one, leaving its zero value unchanged.

The unheaded numerical reconstruction request. The four unknown Neumann boundary conditions are four functions on the sides, not four scalar values. The polygonal global relation is linear in their outward normal derivatives, and its remaining terms depend only on the prescribed Dirichlet boundary data.

Choose a finite approximation on each side, for example an expansion of $q_j(s)$ in Legendre polynomials or piecewise polynomials. Substitute those expansions into the consistently oriented global relation. At chosen nonzero spectral parameters for a linear boundary value problem, integrate the exponential kernels against each basis function to assemble a complex linear system; the known right-hand side is obtained by integrating the given Dirichlet boundary data. Use enough independent samples to resolve all side coefficients, and preferably oversample. The conjugate global relations for the modified Helmholtz equation provide useful companion tests; for complex data use the corresponding independent adjoint relation rather than assuming the traces real.

Solve the scaled system by least-squares solution using a stable factorization such as a singular value decomposition. Sampling directions should probe all sides, and exponential row scaling avoids overflow and poor conditioning. Refine the side approximation and spectral samples until the recovered traces and unused global relation residuals stabilize. Corner incompatibilities or limited corner regularity call for mesh refinement or enriched endpoint basis functions. This realizes a numerical Dirichlet-to-Neumann map without first discretizing the whole interior.

The underlying Dirichlet problem is uniquely solvable in the usual trace class for $\Delta u-4{\beta }^{2}u=0$ with $\beta >0$: the homogeneous problem has ${\int }_{\Omega }(|\nabla u{|}^2+4{\beta }^2|u{|}^2)dA=0$. This supports the boundary reconstruction, although uniqueness of the continuous problem alone does not guarantee that an arbitrary finite set of spectral samples is well conditioned.