The corresponding minimization problem with the area constraint replaced by a perimeter constraint (whatever notion of perimeter one would consider) is in general not well-posed. Indeed, for every $L>0$, one can construct a sequence of smooth connected and closed sets $K_n\subset\overline\Omega$ of perimeter $L$ approaching a subset of $\partial \Omega$ so that $\lambda_1(\Omega\setminus K_n)\downarrow \lambda_1(\Omega)$ as $n\to\infty$ (notice that by regularity there is no doubt on the notion of perimeter of $K_n$).
Therefore, as in Problem 2, we restrict the class of admissible obstacles to convex sets.
For a fixed $L\in (0,P(\overline{\Omega}))$, consider the {minimization} problem
\begin{equation}\label{prob3}
\min \{ \lambda_1(\Omega\setminus K) : \; \text{$K\subset \overline{\Omega}$, $K$ closed and convex, $P(K)=L$}\}.
\end{equation}
The existence of a minimizer is a consequence of the compactness of the class of convex sets and of the continuity of the perimeter w.r.t Hausdorff convergence of convex sets.
Notice that, for particular domains $\Omega$ and small values $L$, it is still possible to have trivial solutions.
For example, if the boundary $\partial\Omega$ contains a segment and if $L$ is smaller
than twice the length of such a segment, then every segment contained in $\partial \Omega$ of perimeter $L$ is a minimizer.
On the other hand, if $L$ is large enough,
every minimizer has positive Lebesgue measure, since minimizing sequences will not be able to degenerate to a segment. In any case, one expects that every minimizer touches the boundary $\partial \Omega$. ...
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