Laplace's Equation — Question 1

Question 1 – Solution

Strategy. Separate the two imposed spatial frequencies and track their normal derivatives as well as their temperatures.

Step 1: Extend each boundary mode. The zero side values give $\sin(n\pi x)$; the corresponding vertical equation is $Y''-(n\pi)^2Y=0$. The bottom value selects $\sinh(n\pi y)$ and the top value normalizes it: $$\boxed{u=\frac{\sinh(\pi y)}{\sinh\pi}\sin(\pi x) +\frac 12\frac{\sinh(3\pi y)}{\sinh(3\pi)}\sin(3\pi x).}$$ Each term has zero Laplacian. Substitution gives all four traces, including compatible corner values. The maximum principle gives uniqueness among continuous harmonic extensions.

Step 2: Compare penetration of the modes. Since $\sinh(a/2)/\sinh a=1/(2\cosh(a/2))$, the first and third modes are attenuated by $1/(2\cosh(\pi/2))$ and $1/(2\cosh(3\pi/2))$. Their third-to-first amplitude ratio is $$\boxed{\frac{\cosh(\pi/2)}{2\cosh(3\pi/2)}.}$$ It is smaller than the boundary ratio $1/2$: the higher spatial frequency loses amplitude faster away from the driven edge.

Step 3: Balance the four edge fluxes. Integrating $-\partial_nu$ along each unit edge gives $$F_{\mathrm{top}}=-2\coth\pi-\coth(3\pi),\qquad F_{\mathrm{bottom}}=2\,\operatorname{csch}\pi+\operatorname{csch}(3\pi),$$ $$F_{\mathrm{left}}=F_{\mathrm{right}} =\tanh(\pi/2)+\tfrac 12\tanh(3\pi/2).$$ For example, at the bottom the outward normal is $(0,-1)$, so the outward heat flux is $u_y$, not $-u_y$. The identity $\coth a-\operatorname{csch}a=\tanh(a/2)$ proves that the four totals sum to zero.

Step 4: Interpret the profiles and their sign. Writing $s=\sin(\pi x)\in[0,1]$, the top value is $s+\tfrac 12(3s-4s^3)=s(5/2-2s^2)\ge 0$, and it is positive for $0<x<1$. The weak maximum principle gives $u\ge 0$; the strong maximum principle gives $u>0$ inside because the boundary data are not constant. Individual modes need not be positive.