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Consider the Unit Disk. can we solve for a harmonic function in the unit disk such that: $\triangle u = 0 $ in D and $ u = f $ on $\partial D$ where $ f = 1$ for $|\theta| \leq \epsilon$ and $ |\theta - \pi| \leq \epsilon$ and $f=0$ everywhere else?

Equivalently, can we solve the following Dirichlet problem in the unit square $\Omega$ with vertices at (0,0),(1,0),(0,1),(1,1): $\triangle u = 0$ in $\Omega$ and $u=f$ on $\partial \Omega$ where $f(x,y)=M_{\epsilon}$ when $ |x-\frac{1}{2}|\leq \epsilon$ and $f=0$ everywhere else?

I know the answer of course in both cases is yes and in particular let me expand on the second question:

One can give the following direct formula for the solution $u(x,y)$ in the unit square: $ u(x,y)= \sum_{n=1,3,5,...} \frac{b_n}{sinh n\pi}sin(n\pi x) (sinh n\pi y + sinh n\pi (1-y))$ where $ b_n=\frac{2M_{\epsilon}}{n\pi} (-1)^{(n-1)/2} sin(n\pi \epsilon/2)$ the very interesting question that I want to pose now is this: what happens to the energy $\int_\Omega |du|^2$ as $\epsilon$ approaches zero? Let's assume for the sake of simplicity that $M_{\epsilon} = \frac{1}{\epsilon}$ but more generally can we pick M such that the energy doesn't blow up in the limit?

PS. based on elliptic pde's ofcourse one can solve for $\triangle u = 0$ and $u=f$ on $\partial \Omega$ given that $ f \in H^{\frac{1}{2}} (\partial \Omega)$ Notice that the step function is not however in this space and I can only think of it being in $L^2$ so how would this change the elliptic regularity results? would the solution still be smooth inside?

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  • $\begingroup$ I have actually written this out and it seems to me that the energy blows up indeed. No matter how one goes about choosing the M value the dirichlet energy always blows up... I would think this means that for the laplacian operator it costs less to smear as opposed to concentrate around a minimal surface. $\endgroup$
    – Ali
    Mar 4, 2014 at 3:20

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