Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 9 / 5

Use the Fourier coefficientsThey are continuous for $0\le y\le 1$ and bounded by $2\Vert u{\Vert }_{\infty }$. On compact subintervals of $0<y<1$, the equation is uniformly elliptic with smooth coefficients. Local elliptic regularity at the flat vertical sides with zero boundary values permits differentiation and integration by parts there. ThusFor completeness, this identity also follows in the distributional sense without assuming boundary derivatives initially. Integrate on $(\epsilon ,\pi -\epsilon )$ against $\sin (nx)$ and a test function of $y$ with support away from zero and one. The interior gradient estimate near a zero Dirichlet side gives $|u_x|\le C{\epsilon }^{-1}\sup |u|$ on a nearby ball of radius comparable to $\epsilon$; continuity and the zero side value make that supremum $o(1)$ uniformly over the support in $y$. Therefore the boundary products $u_x\sin (nx)$ and $u\cos (nx)$ tend to zero. Move the $y$ derivatives to the test function before taking $\epsilon \downarrow 0$. This proves the ordinary differential equation, which makes each $b_n$ smooth in $0<y<1$ and yields the same classical identity.

The Cauchy-Euler differential equation has indicial rootsFor $n\ge 1$ the negative root is strictly negative. Boundedness as $y\downarrow 0$ forces $B_n=0$. Continuity at $y=1$ identifiesBecause $p_n^{+}>n$, the series ${\sum }_{n\ge 1}{A}_n{y}^{p_n^{+}}\sin (nx)$ converges absolutely and uniformly for $0\le y\le \rho <1$. On compact subsets with $y>0$, differentiated series converge as well. At each fixed $0<y<1$, its Fourier coefficients equal those of $u(\cdot ,y)$; completeness of the Fourier sine basis in $L^2(0,\pi )$ and continuity in $x$ make the two functions equal. ThereforeThis is also the requested sum over integers: take all coefficients with $n\le 0$ to be zero. Negative indices give redundant Fourier modes, and the $n=0$ sine vanishes. No pointwise convergence of the ordinary Fourier series on the top boundary is needed.

An important consequence is a forced zero trace at a quadratically degenerate boundary. Indeed the absolute sum is bounded by $2\Vert u{\Vert }_{\infty }{\sum }_{n\ge 1}{y}^n=2\Vert u{\Vert }_{\infty }y/(1-y)$ for $y<1$, which tends to zero as $y\downarrow 0$, uniformly in $x$. Thus $u(x,0)=0$ on the entire lower side. It follows also from $b_n(0)=0$ for every sine coefficient and completeness.

Choose continuous boundary values equal to $\sin x$ on the lower side and zero on the other three sides. These values agree at all four corners, so they define a continuous function on the boundary of the rectangle. Any solution would satisfy the zero vertical-side conditions of part (a), whose Fourier sine series forces its lower boundary trace to be zero. This contradicts the prescribed value $\sin x$ for $0<x<\pi$. Hence these continuous Dirichlet data admit no solution in $C^0(\overline{R})\cap C^2(R)$.

The missing hypothesis is uniform ellipticity up to the boundary. The principal coefficient matrix and its quadratic form areAlthough this is positive-definite at every interior point, no single positive uniform ellipticity constant works on $R$: taking $\xi =(0,1)$ requires $y^2\ge \lambda$ for arbitrarily small positive $y$. The normal second-derivative coefficient vanishes on the lower side. This degenerate ellipticity allows the equation and boundedness to force a boundary trace, instead of allowing arbitrary continuous Dirichlet data. The coefficient matrix is positive-definite at every interior point, but lacks the uniform lower bound used by the usual bounded-domain existence theorem.