Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 70 / 2

Use the Wirtinger derivative $q_z=(q_x-i{q}_y)/2$ and write $z=x+iy$. Since $q$ solves the Laplace equation, $q_z$ is a holomorphic function. The transforms in the question select solutions with sufficient decay at infinity. In this class the solution, when it exists, is unique; hence reflection of the symmetric data about $y=\ell /2$ givesFor $\gamma >0$, this uniqueness follows directly from Green's first identity: the homogeneous problem has integral $\int |\nabla q{|}^2+\gamma {\int }_{y=0,\ell }{q}^2=0$. For arbitrary real $\gamma$, the decaying-class qualifications are explained in part (c). Without a condition at infinity, symmetry of the data alone would not force symmetry of every solution.

Put $c=q(0,0)$, $f(x)=q(x,0)$, and introduce a known transformThe bottom Robin boundary condition gives $q_y=\gamma f$, while the top one gives $q_y=-\gamma f$. The bottom side is traversed from infinity to zero. Integration by parts therefore givesThe top side is traversed from $i\ell$ towards infinity, and its exponential supplies $e^{k\ell }$, givingOn the vertical side, $dz=idy$ and the prescribed Neumann boundary condition is $q_x=g$. ThusBoth the side orientations and the factor $1/2$ from the Wirtinger derivative are essential to these signs.

The Cauchy integral theorem applied to $e^{-ikz}{q}_z$ yields the global relationInitially this follows for $\mathrm{Im}k\le 0$, where the far vertical closing segment vanishes. The identities below can be compared on the real axis, and then continued wherever the respective transforms are analytic.

Define the characteristic factor of the symmetric Robin strip transformSubstituting part (a) into the global relation givesReflection symmetry gives $H(-k)=e^{-k\ell }H(k)$ and $\varphi (-k)=-e^{-k\ell }\varphi (k)$, because $q_y(0,\ell -y)=-q_y(0,y)$. Also $D(-k)=e^{-k\ell }D(k)$. Use the global relation at $-k$ and multiply by $-e^{k\ell }$ to obtainComparison therefore provesAt a zero of $D$, this identity is interpreted through the necessary solvability condition and a removable limit when a decaying solution exists. For $\gamma >0$, $D$ has no real zeros, so no such qualification is needed on the two real integration rays.

Substitute part (b) into the spectral representation. The terms containing $\psi (ik)$ areThe transform $\psi (ik)$ is analytic for $\mathrm{Im}k>0$. Write $k=a+ib$. In the first quadrant,which decays since $x,y>0$. The Cauchy integral theorem in that quadrant equates the two integrals in the first parentheses. In the second quadrant,which decays since $a<0$, $b>0$, and $y<\ell$. The same contour integration equates the two integrals in the second parentheses. Therefore $\psi (ik)$ makes no contribution. This is upper-quadrant cancellation of reflected boundary transforms, and it uses the whole paired expression rather than attempting to discard an individual integral.

The terms involving $c$ cancel by the same two quadrant arguments with the analytic factors $k\pm \gamma$ and $\psi$ omitted. There is a sign defect in the printed reconstruction prefactor. The sides specified in part (a) form a clockwise boundary, so the Cauchy integral theorem gives a negative reconstruction sign. Indeed, integrating each spectral ray first yields $i/(z-\zeta )$, and hence the sum of the three ray integrals is $i{\oint }_{\mathrm{clockwise}}{q}_z(\zeta )/(z-\zeta )d\zeta =-2\pi q_z(z)$. Thus the printed positive prefactor reconstructs $-q_z$. The requested cancellation above holds with either overall sign, but the actual data-dependent derivative isDecay at infinity fixes the integration constant when recovering $q$ from its Wirtinger derivative.

For completeness, the missing condition at infinity and the unrestricted printed $\gamma$ deserve an explicit solvability of a decaying Robin strip check. Let $Y_n$ be orthonormal eigenfunctions from a Sturm-Liouville problem of $-d^2/d{y}^2$ with $Y_n^{\prime }(0)=\gamma Y_n(0)$ and $Y_n^{\prime }(\ell )=-\gamma Y_n(\ell )$, and eigenvalues ${\lambda }_n$. With $g_n={\int }_0^{\ell }g{Y}_{n}dy$, separation of variables givesIndeed, each coefficient obeys $q_n^{\prime \prime }-{\lambda }_n{q}_n=0$ and $q_n^{\prime }(0)=g_n$. A positive eigenvalue has one decaying exponential; a zero eigenvalue has only affine solutions and a negative eigenvalue only oscillatory solutions, neither of which decays unless its coefficient vanishes. This also proves uniqueness in the decay class and justifies reflection symmetry there. For $\gamma >0$, all eigenvalues are positive. For $\gamma =0$, the constant eigenfunction requires ${\int }_0^{\ell }gdy=0$; the apparent zero of $D$ at the origin is then removable. For $\gamma <0$, compatibility with every nonpositive eigenmode is necessary. These restrictions cannot be inferred from smoothness and reflection symmetry alone. The printed boundary value problem without any far-field condition permits additional growing or nondecaying homogeneous solutions, whereas the spectral transforms select the decaying branch.