Unusual problem of calculus-of-variations. Attempt 2

Question

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.

Answer 1

Since the OP states that $\lambda$ is not fixed from the outset, and assuming that $$J_G\equiv\int_{-1}^1 \cos(\tfrac{1}{2}\pi x)G(x)\,dx\neq 0,$$ we can make $I_G$ arbitrarily large by choosing $f(x)\equiv\epsilon>0$ an infinitesimal constant and $$\lambda=\frac{\pi^2}{4}(1+\epsilon^{-2}).$$ Then we have $$u(x,y)=({\rm sign}\,J_G)\sqrt{\frac{1}{\epsilon}}\cos(\tfrac{1}{2}\pi x)\cos(\tfrac{1}{2}\pi y/\epsilon)\Rightarrow I_G=\sqrt{\frac{1}{\epsilon}}\,|J_G|,$$ and $I_G\rightarrow\infty$ when $\epsilon\downarrow 0$.

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Update:

The OP wishes to add a new constraint $2\int_{-1}^1 f(x)\,dx=1$. This can be accommodated by choosing $$f(x)=\begin{cases} \epsilon&{\rm for}\;\; |x|>\epsilon^2,\\ \frac{1-4\epsilon+4\epsilon^2}{8\epsilon^2}&{\rm for}\;\;|x|<\epsilon^2. \end{cases} $$ So we are adding a long and narrow stub of width $\epsilon^2$ and length $ \simeq 1/\epsilon^2$ perpendicular to the rectangular domain $D$ of width $\epsilon$. As $\epsilon\downarrow 0$ the function $u(x,y)$ will not leak out into the stub, because the wave length $\lambda^{-1/2}\simeq\epsilon$ is much larger than the width $\epsilon^2$ of the stub, so the normalization of $u$ and the integral $I_G$ will not be affected in the small-$\epsilon$ limit. Hence we can still make $I_G$ grow arbitrarily large $\propto 1/\sqrt\epsilon$.
_{Notice that $u$ decays exponentially in the stub, because of the imaginary wave vector when the width is less than the wave length, so indeed we can make the stub arbitrarily long, to assure a total area of unity, without affecting the integrals of $u^2$ and $I_G$.}

Illustration of the domain $D$, a rectangle of width $2\epsilon$ and a stub of width $2\epsilon^2$.