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I already to try ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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 $I_G=\int_{-1}^{1}u(x,0) G(x)\, dx$. 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).

I understand that this problem does not well-define. But all helpfull considerations will be very useful for me.

I already tried to ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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 $I_G=\int_{-1}^{1}u(x,0) G(x)\, dx$. 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 that can help solve this problem. We can go to a non-orthogonal coordinate system so that $D$ becomes a square. But after that the functions $f(x)$ and $f'(x)$ will appear in the operator (Laplacian in the new coordinate frame).

I understand that this problem is not well-defined. But all helpfull considerations will be very useful for me.

I already to try ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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).

I understand that this problem does not well-define. But all helpfull considerations will be very useful for me.

I already to try ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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 $I_G=\int_{-1}^{1}u(x,0) G(x)\, dx$. 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).

I understand that this problem does not well-define. But all helpfull considerations will be very useful for me.

I already to try ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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).

I already to try ask the same question here Unusual problem of calculus-of-variations, but I did a huge mistake and have no time to understand it. In the past post, I tried to simplify the problem, but simplified it to a trivial example. The difference is in the differential equation $\Delta u(x,y)=-\lambda u(x,y)$.

Full text of problem:

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 $\Delta u(x,y)=-\lambda u(x,y)$, $\forall (x,y)\in D$ 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).

I understand that this problem does not well-define. But all helpfull considerations will be very useful for me.