Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 6

Let $w$ be the difference of two solutions. It is a harmonic function in the open rectangle, continuous on its closure, withInterpret these Neumann boundary conditions classically on the open edges, with the corresponding first derivatives extending there. We use the strong maximum principle for harmonic functions together with even reflection across a homogeneous Neumann boundary. This avoids imposing differentiability at the two bottom corners, where the original data fail a corner compatibility condition for mixed boundary data.

By continuity on the compact rectangle, $w$ attains a maximum $M$. Suppose $M>0$. It cannot occur on the bottom edge, where $w=0$. If it occurs in the interior, the strong maximum principle for harmonic functions makes $w$ constant, contradicting its bottom value. If it occurs on an open vertical or top edge, the zero normal derivative allows an even reflection across that edge. The reflected harmonic function then has an interior maximum, giving the same contradiction.

The only remaining possibilities are the two top corners. Reflect evenly across both meeting edges. This gives a bounded harmonic function on a punctured neighbourhood of the corner. The removable singularity for a bounded harmonic function fills the centre with its continuous value. The corner is now an interior maximum, again impossible. Therefore $w\le 0$. Applying the identical argument to $-w$ gives $w\ge 0$.

Thus $\boxed{w=0}$, and the solution is unique among the continuous classical solutions with the stated edge conditions.

For $0<y<\pi$, define $J(y)={\int }_0^{\pi }{u}_y(x,y)dx$. Differentiate under the integral and use Laplace's equation:Thus $J$ is constant. The top Neumann boundary condition gives $J(\pi )=0$, soFor the included endpoint $y^{\ast }=0$, the integral is interpreted as an improper integral at the bottom corners if necessary. The Fourier series constructed in c gives, on $0<x<\pi$,The second series is absolutely and uniformly convergent; the first term has only integrable logarithmic singularities. Its integral is zero since ${\int }_0^{\pi }\log (2\sin x)dx=0$, while every cosine term also has zero integral. This proves the endpoint as well:For completeness, the logarithmic integral follows without any endpoint differentiation. Put $I={\int }_0^{\pi /2}\log (\sin x)dx$ and substitute $x\mapsto \pi /2-x$ to obtain $I={\int }_0^{\pi /2}\log (\cos x)dx$. Adding and using $\sin x\cos x=\sin (2x)/2$ gives $2I=I-(\pi /2)\log 2$, hence $I=-(\pi /2)\log 2$ and the required full-interval result.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.