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I have a domain $D:\quad$ $-1<x<1$ and $-f(x)<y<f(x)$.\ There is function $u(x,y)$ which is obey the equation $\Delta u(x,y)=0$, $\forall (x,y)\in D$ with the Dirichlet boundary condition $u(x,y)=0$, $\forall(x,y)\in \partial D$ and $\int_D u^2(x,y) dx dy=1$ Let us define $I_G=\int_{-1}^{1} dx u(x,0) G(x)$. The problem is as follows. How to find the function f so that integral $I_G$ is maximal?

Is it possible to solve such kind of problem?

I have an idea, which can help to solve this problem. We can go to a non-orthogonal coordinate system so that $D$ become a square. But after that the function $f(x)$ and $f'(x)$ appears in the operator (Laplacian in the new coordinate frame).

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  • $\begingroup$ I guess the buzz word is "free boundary problem" $\endgroup$
    – Dirk
    Commented Mar 21, 2018 at 13:06
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    $\begingroup$ Aren't solutions of the Dirichlet problem unique? How can the integral of u be 1? $\endgroup$
    – Michael Renardy
    Commented Mar 21, 2018 at 15:26
  • $\begingroup$ I made a typo. It should be $\int u^2(x,y)dxdy=1$. It is a usual normalization, which is fix a constant. $\endgroup$
    – Peter
    Commented Mar 21, 2018 at 17:12
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    $\begingroup$ Peter, have you paid attention to what Michael said? The only harmonic $u$ that satisfies the Dirichlet boundary condition you imposed is identically $0$, so editing the normalization condition won't help much: the problem is just nonsensical as posed. Perhaps, you meant something else (say no boundary condition, just the normalization itself). $\endgroup$
    – fedja
    Commented Mar 21, 2018 at 17:15
  • $\begingroup$ The maximum principle tells us that $u$ is everywhere zero, as both $u$ and $-u$ are harmonic, so achieve their maxima on the boundary, as Michael Renardy says. $\endgroup$
    – Ben McKay
    Commented Mar 22, 2018 at 11:39

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