Laplace's Equation — Question 2

Question 2 – Solution

Strategy. Use the boundary Fourier coefficients for construction, a harmonic barrier for bounds, and orthogonality for energy.

Step 1: Compute the boundary expansion. Two integrations by parts give $$b_n=2\int_0^1x(1-x)\sin(n\pi x)\,dx =\frac{4(1-(-1)^n)}{(n\pi)^3}.$$ Thus $b_n=8/(n\pi)^3$ for odd $n$, and $b_n=0$ for even $n$. The solution is $$\begin{gathered} \boxed{u(x,y)=\sum_{\substack{n\ge 1\\n\ {\mathrm{odd}}}} \frac{8}{(n\pi)^3}\frac{\sinh(n\pi y)}{\sinh(n\pi)}\sin(n\pi x).} \end{gathered}$$

Step 2: Establish the appropriate convergence. The hyperbolic ratio is between zero and one for $0\le y\le 1$. Since $\sum|b_n|<\infty$, the series converges uniformly on the closed square. Its top trace is the sine series of $x(1-x)$, and its other traces vanish. On any strip $y\le 1-\epsilon$, derivatives of every fixed order are bounded by a polynomial in $n$ times $e^{-n\pi\epsilon}$. They converge uniformly there, so the Laplacian vanishes term by term. Uniform convergence of the original series alone would not justify taking two derivatives.

Step 3: Compare with a harmonic barrier. The boundary values are nonnegative. The function $h(x,y)=y/4$ is harmonic, is nonnegative on the sides, vanishes on the bottom, and dominates $x(1-x)\le 1/4$ on the top. Applying the maximum principle to $u$ and $u-h$ gives $$\boxed{0\le u(x,y)\le y/4.}$$ Both inequalities are strict in the interior: either equality there would make the corresponding harmonic difference constant, contrary to its boundary values.

Step 4: Evaluate energy without losing boundary terms. For a finite partial sum, Green’s identity leaves only the top contribution $\int_0^1u(x,1)u_y(x,1)\,dx$. Sine orthogonality gives $$\begin{gathered} \mathcal E_N=\frac 12\sum_{\substack{n\le N\\n\ {\mathrm{odd}}}} b_n^2\,n\pi\coth(n\pi). \end{gathered}$$ The same identity applied to a difference of partial sums proves their gradients are Cauchy in $L^2$, since the summands are $O(n^{-5})$. The uniform limit has that weak gradient and is the harmonic field already constructed. Therefore $$\begin{gathered} \boxed{\mathcal E=\frac{32}{\pi^5} \sum_{\substack{n\ge 1\\n\ {\mathrm{odd}}}}\frac{\coth(n\pi)}{n^5}.} \end{gathered}$$ Every term is positive; truncating this energy series gives a lower bound.