Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 67 / 3

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the modified Helmholtz equation, including its Laplace equation case $\lambda =0$. Equivalently,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.

Fix counterclockwise arclength on each side and take the outward normal derivative. WriteThe unit tangent is $i{c}_j$ and the outward unit normal is $c_j$. On side $j$ this gives the chain-rule identityPut $w=c_{j}k$. The exponential in the closed differential one-form pulls back toAlso $-\lambda qd\bar{z}/(ik)=(\lambda /w)q^{(j)}ds$. Consequently the contribution of side $j$ is ${\rho }_j(c_{j}k)$, where the required local side spectra areThese are spectral integral transforms in the local coordinate of each side. Keeping the tangential-derivative term is important: premature integration by parts in a single side integral would leave corner values, which cannot simply be discarded. At the special samples used below, those endpoint terms do vanish for the rotationally invariant solution.

The Generalized Stokes theorem applied to the closed differential one-form givesEquivalently, the global relation isThe identical outward Neumann boundary data make the three known transforms equal:To reduce the unknown traces to a single function, use rotation invariance and uniqueness. For $\lambda >0$ the difference $u$ of two solutions has zero outward derivative andby Green's first identity. Thus $u=0$. Rotation by $a$ preserves the equation, the domain and the oriented boundary data, so it preserves $q$; hence $q^{(1)}=q^{(2)}=q^{(3)}=q$ and ${\Phi }_j=\Phi$. The global relation becomes the single equation with three spectral unknownsThe three unknowns are evaluations of the same boundary transform, not three freely specified boundary data.

The printed allowance of all real $\lambda$ requires a qualification. At $\lambda =0$, existence requires $3{\int }_{-l/2}^{l/2}f(s)ds=0$ by the divergence theorem, and solutions differ by a constant. Fixing a rotation-invariant normalization gives uniqueness and the same rotational conclusion. In fact the unnormalized solution is already invariant: its rotation difference is constant, and three successive rotations make three times that constant zero. For $\lambda <0$, uniqueness fails precisely when $-4\lambda$ is an eigenvalue of the Neumann Laplacian. At these parameters identical data do not force an arbitrary solution to be rotationally invariant. Averaging any solution over the three rotations produces a solution with the common trace used here; the difference is a homogeneous Neumann eigenfunction, or a sum of them. Part (e) describes both solvability and this remaining freedom rather than assuming uniqueness for every real parameter.

Let $\Phi$ denote the common trace transform of the unique solution, or of the rotational average at an exceptional parameter. Because $q$ and $f$ are real and $\lambda$ is real, $\overline{\Phi (\bar{k})}=\Phi (k)$ and $\overline{\Psi (\bar{k})}=\Psi (k)$. Taking the Schwarz conjugate of the global relation means replacing $k$ by $\bar{k}$ and conjugating the whole equation. It yields the second of the conjugate global relations for the modified Helmholtz equation:In this conjugation the rotated arguments interchange; after putting terms in the order $k,\bar{a}k,ak$, the coefficients have the displayed signs and arguments. DefineThe elementary cube-root identities giveMultiply the first global relation by $E(iak)$ and the conjugate one by $E(-iak)$. Then the coefficient of $\Phi (ak)$ is one in both equations:where all the quantities on the right are known,Subtracting eliminates $\Phi (ak)$. Ifthe resulting solution for the requested transform isAt a zero of the denominator this identity is interpreted in its undivided form. The transforms themselves remain finite for $k\ne 0$; consequently the numerator must vanish. That cancellation, rather than evaluation of a singular quotient, determines the Fourier coefficients below.

For $\lambda >0$, choose a sample $k_n$ satisfying $\chi (k_n)=i\pi n$ for each integer $n$. A convenient branch isThus $k_n+\lambda /k_n=2\pi in/l$. The undivided identity in part (d) givesRotation invariance makes all three vertex values equal, so $q(l/2)=q(-l/2)$. The exponential has the same endpoint value $(-1{)}^n$ at both ends. Integration by parts in the definition of $\Phi$ therefore givesThese are the Fourier coefficients with the positive-exponential convention. Hence the computed boundary trace isEvery quantity in this rotationally invariant Neumann reconstruction in an equilateral triangle is determined by the prescribed $f$. To check that no denominator was silently divided by zero, put $k_n=i{\kappa }_n$, ${\kappa }_n>0$. ThenNeither factor vanishes for $\lambda >0$. For sufficiently smooth compatible traces, the periodic extension of $q$ is continuous and piecewise smooth; the Fourier series recovers $q$, with the usual endpoint convention. Reality implies $Q_{-n}=\overline{Q_n}$ even though the sampling formula uses complex arguments.

For $\lambda =0$, first impose the zero-flux compatibility from part (c). For $n\ne 0$ take the nonzero root $k_n=2\pi in/l$ and use the same coefficient formula with $\lambda =0$. The coefficient $Q_0$ is an arbitrary real constant times $l$, expressing the additive constant for this problem with a Neumann boundary condition. The zero sample cannot determine it. For example, $f=0$ admits every constant solution, so no uniquely specified $q(s)$ is possible without a normalization.

For completeness, negative parameters can be handled without assuming away resonances. Let ${\varphi }_j$ be a real orthonormal eigenbasis of the Neumann Laplacian, with $-\Delta {\varphi }_j={\mu }_j{\varphi }_j$, and defineusing the corresponding oriented $f$ on each side. The weak equation and Green's first identity give the explicit coefficient relationAt a nonresonant parameter this determines the unique solution. Its common trace is the continuation of the displayed spectral Fourier coefficients from $\lambda >0$; choose either nonzero root of $k_n^2-(2\pi in/l)k_n+\lambda =0$. If an individual quotient is $0/0$ at a nonresonant parameter, use its removable limit in $\lambda$, rather than declare a new eigenvalue. For example, a double sampling root can make $k_n-\lambda /k_n=0$ without making the boundary problem singular. The weak coefficient relation proves that the solution is analytic in the parameter wherever no ${\mu }_j+4\lambda$ vanishes, so these nonresonant limits exist.

At a genuine resonance $\lambda =-{\mu }_{\ast }/4$, the Fredholm solvability condition is precisely $F_j=0$ for every ${\mu }_j={\mu }_{\ast }$. Necessity follows directly from the coefficient relation. For compatible smooth data, sufficiency follows by dividing all nonresonant coefficients and leaving the resonant ones free. The boundary functional is continuous on $H^1(D)$, so ${\sum }_j|F_j{|}^2/(1+{\mu }_j)<\infty$; away from the finite resonant eigenspace this implies ${\sum }_j(1+{\mu }_j)|F_j/({\mu }_j-{\mu }_{\ast }){|}^2<\infty$. Thus the resulting series converges in $H^1(D)$ and satisfies the weak equation, not just its individual coefficient identities. The full solution and its boundary values are thuswith convergence in the weak solution space and the associated trace sense. Rotation averaging selects the common-side component; a nonsymmetric homogeneous addition can have different traces on the three sides. Incompatible resonant data give no solution. For arbitrary real $\lambda$, compatibility and the free homogeneous components are essential; the source does not specify conditions that would select a unique trace at resonance.