Real Analysis – Non Convex Optimization Problem in W_0^{1,2}

calculus-of-variationsreal-analysisreference-requestsobolev-spaces

Let $0< \alpha \ll 1$. I'm trying to minimize $\int_0^\pi |f'|^2 dx$ over the functions $f \in W_0^{1,2}([0,\pi])$ (or at least find "good" lower bound in terms of $\alpha$) such that satisfy the following constraints:
$$
\begin{cases}
\int_0^\pi f^2 dx = 1, \\
\int_{\pi/3}^{2\pi/3} f^2 dx \leq \alpha.
\end{cases}
$$

In other words, I'm trying to find a function that vanishes at $0$ and $\pi$, has most of its mass near the end points, and minimizes its Dirichlet energy.

I do have a trivial lower bound given by Poincare (Wirtinger) inequality which is not in terms of $\alpha$.
I don't know where to start studying this problem.

I have tried taking Fourier transform of $f$ but the second constraint gets complicated. To be honest, I started with the formulation in the Fourier space which is the following:
$$
\begin{cases}
\mbox{find sequence $(a_n)_n$ that minimizes} &\sum_n na_n^2, \\
\mbox{subject to:} &\sum_n a_n^2 = 1, \\
& \int_{\pi/3}^{2\pi/3} (\sum_n a_n sin(n x))^2 dx \leq \alpha.
\end{cases}
$$

Does anyone know where to start or any related reference?

Best Answer

You can treat this as a problem with two Lagrange multipliers. Then by standard methods, a minimizer $f$ has to exist (by convexity in $f'$) and has to be a weak solution to $$-f'' + \lambda f + \mu f \chi_{[\pi/3,2\pi/3]} = 0$$ with $\lambda, \mu \in \mathbb{R}$ and $\mu \leq 0$ because the constraint is one-sided.

Solving this equation on its sub-intervals and using the boundary condition gives you $$ f(x) = \begin{cases} a_1 \sin(\lambda x) &\text{ for } x \in [0,\pi/3] \\ b_1 \sin((\lambda + \mu)x) +b_2 \cos((\lambda + \mu)x) &\text{ for } x \in [\pi/3,2\pi/3] \\ a_2 \sin(\lambda (x-\pi)) &\text{ for } x \in [2\pi/3,\pi]. \end{cases} $$

Using a bit of regularity theory on $f'' = - \lambda f - \mu f \chi_{[\pi/3,2\pi/3]} \in L^2([0,\pi])$ gives you $f \in W^{2,2}([0,\pi])$, so since we are in 1d, $f'$ is absolutely continuous. This gives you two equalities at $\pi/3$ and $2\pi/3$ each.

You also can explicitly calculate $\int_0^\pi f^2 = 1$ and have $\int_{\pi/3}^{2\pi/3} f^2 = \alpha$ or $\mu = 0$.

This gives you a system of 6 (independent, if I am not mistaken) equations on 6 variables. I will not try to solve it here, but my guess would be that it is a tedious but doable task. You will likely end up with a countable family of solutions, but selecting the smallest should be easy.

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