Laplace's Equation — Question 8

Question 8 – Solution

Strategy. Neumann side conditions select cosine modes, including a zero mode whose vertical factor is linear rather than hyperbolic.

Step 1: Keep the zero mode. The side eigenfunctions are $1,\cos(n\pi x)$ with eigenvalues $0,(n\pi)^2$. The mean mode satisfies $Y''=0$ with bottom value zero and top value one, so it is $y$. The second cosine mode gives $$\boxed{u(x,y)=y+\frac{\sinh(2\pi y)}{\sinh(2\pi)}\cos(2\pi x).}$$ Dropping the zero mode would lose the entire prescribed mean temperature.

Step 2: Verify and prove mixed uniqueness. The displayed terms are harmonic. The sine factor in $u_x$ vanishes at both vertical sides; the bottom and top traces are exactly zero and $1+\cos(2\pi x)$. For the difference $w$ of two solutions, Green’s identity yields $\int|\nabla w|^2=\int_{\partial\Omega}w\partial_nw=0$: $w=0$ on the horizontal edges and $\partial_nw=0$ on the vertical ones. Thus $w$ is constant, and its bottom value forces $w=0$.

Step 3: Track the mean and edge fluxes. The cosine has zero horizontal integral, so $\overline u(y)=\int_0^1u(x,y)\,dx=y$. The integrated outward conductive fluxes are $$\boxed{F_{\mathrm{top}}=-1,\qquad F_{\mathrm{bottom}}=1,\qquad F_{\mathrm{left}}=F_{\mathrm{right}}=0.}$$ They balance. Insulation forbids heat crossing the vertical sides; it does not forbid a vertical gradient or heat transfer between the horizontal edges.

Step 4: Compute energy and bound the field. Only the top contributes to $\int_{\partial\Omega}u\partial_nu$. At the top, $u_y=1+2\pi\coth(2\pi)\cos(2\pi x)$, so orthogonality gives $$\boxed{\mathcal E=1+\pi\coth(2\pi).}$$ For $0<y<1$, strict convexity of $\sinh(2\pi y)$ gives $0<\sinh(2\pi y)/\sinh(2\pi)<y$. Therefore $0<y-\sinh(2\pi y)/\sinh(2\pi)\le u \le y+\sinh(2\pi y)/\sinh(2\pi)<2y<2$. This direct estimate avoids incorrectly treating the unspecified side temperatures as Dirichlet data.