Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 328 / 3

The formal adjoint of $\mathcal{L}=\Delta -k^2$ is itself. With the bilinear convention used in Green second identity, take any complex-valued $v$ satisfying $\mathcal{L}v=0$. The product rule for divergence gives the familyEquivalently, the differential one-formis a closed differential form. Arbitrary linear combinations of adjoint solutions give further members of this family. One can also write it as $\nabla \cdot [v^2\nabla (u/v)]=0$ wherever $v\ne 0$, but the first formula has no division and remains valid at zeros of $v$. For a sesquilinear convention replace $v$ by $\overline{v}$; no such conjugation is built into the bilinear identity used below.

Write $z=x+iy$ and $\overline{z}=x-iy$. A convenient spectral parameter for a linear boundary value problem is any $\lambda \in \mathbb{C}\setminus \{0\}$. Define the modified Helmholtz adjoint plane wavewhereIndeed $A^2+B^2=k^2$, so $(\Delta -k^2)v_{\lambda }=0$. Equivalently, $4{\partial }_z{\partial }_{\overline{z}}{v}_{\lambda }=k^2{v}_{\lambda }$. The exclusion of zero is only the coordinate singularity of this spectral parametrization; it still supplies a whole complex one-parameter family. The conjugate companion isIt also solves the adjoint equation.

Let $\Omega$ be a bounded piecewise smooth domain, $f=u{|}_{\partial \Omega }$ and $q={\partial }_{n}u$, with $n$ the outward normal vector. Integrating the divergence form from part (i) and using the divergence theorem gives two global relations for a linear boundary value problem:These conjugate global relations for the modified Helmholtz equation are identities between the prescribed Dirichlet boundary condition and its unknown normal derivative; neither is a pointwise boundary equation. Both hold for every nonzero boundary spectral parameter under the usual trace regularity assumptions.

If $u$ is real, its two boundary traces $f,q$ are real. Since ${\tilde{v}}_{\lambda }=\overline{v_{\overline{\lambda }}}$ and the outward normal vector is real,Thus apply complex conjugation to the first global relation for a linear boundary value problem, and replace its parameter by $\overline{\lambda }$, to get the second. For complex $u$, conjugation would also replace its traces by their conjugates, so that argument requires the stated reality assumption. There is also an algebraic redundancy in this particular parametrization: ${\tilde{v}}_{\lambda }=v_{1/\lambda }$, hence $\tilde{\mathcal{G}}(\lambda )=\mathcal{G}(1/\lambda )$ even for complex data. This does not change the requested conjugation identity.

Label the sides by $L,R,B,T$, and parametrize the two vertical sides by $s=y\in [-1,1]$, the horizontal sides by $s=x\in [-1,1]$. Their outward normal derivatives are $q_L=-u_x(-1,s)$, $q_R=u_x(1,s)$, $q_B=-u_y(s,-1)$ and $q_T=u_y(s,1)$. Let $f_j$ denote the prescribed side values. Assume compatible, sufficiently regular boundary traces; the square's corners have zero arclength measure and do not require separate normal values.

For sine collocation of square modified Helmholtz global relations, use Legendre polynomials for the known Dirichlet boundary condition and a Fourier sine series for the unknown normal derivative:The known coefficients are $d_{j\ell }=(2\ell +1){\int }_{-1}^1{f}_j{P}_{\ell }ds/2$. The sine functions have ${\int }_{-1}^1{\varphi }_m{\varphi }_{n}ds={\delta }_{mn}$ and form a complete basis in $L^2(-1,1)$. Choosing them for the normal derivative does not impose zero flux at a corner: the expansion is an $L^2$ representation, and endpoint values are not determined by it. If preserving corner values of the approximated Dirichlet trace is necessary, subtract its endpoint-interpolating line before polynomial approximation, and add that line back. All subsequent known-data integrals can alternatively be evaluated with the exact $f_j$.

Define the entire function basis transformsAt $\tau =\pm i{p}_m$ the quotient is evaluated by its removable limit or by the defining integral. For example, ${\mathsf{P}}_0=2\sinh \tau /\tau$, and polynomial expansion of $P_{\ell }$ expresses every ${\mathsf{P}}_{\ell }$ in derivatives of ${\mathsf{P}}_0$. PutFor compactness write $A=A(\lambda )$ and $B=B(\lambda )$. Substituting these expansions into the first global relation for a linear boundary value problem givesThe signs are fixed by the outward normal vectors, rather than by a choice of traversal direction. The second approximate global relation for a linear boundary value problem replaces $A$ by $-A$, leaving $B$ unchanged:Here $\simeq 0$ records omitted boundary-expansion tails. In a finite spectral method the selected equations are set equal to zero to solve for the $4N$ unknown real coefficients $c_{jm}$. For real data the two families obey the same complex conjugation relation as their exact counterparts.

A particularly simple constructive choice uses paired collocation points for a global relation. For $n=1,\dots ,N$, putFour suitable sets, one for each side, areThe braces in each set range over all $n$. For large $p_n/k$, compute $r_n^{-1}=k/({\omega }_n+p_n)$ to avoid cancellation. The corresponding $(A,B)$ pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these $4N$ equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.

The key advantage is that $v_{Bn},v_{Tn}$ vanish on the vertical sides, while $v_{Ln},v_{Rn}$ vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with $f_{jn}={\int }_{-1}^1{f}_j^{(M)}(s){\varphi }_n(s)ds$, the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.

Scale each row by $e^{-{\omega }_n}$. Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled $4N\times 4N$ system is strictly diagonally dominant, with row margin $1-{\delta }_n>1-e^{-\pi }$. This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$ gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact $f_j$, these are exactly the first $N$ sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.