Proof - If one domain $D$ is contained in another domain $D'$, then $\lambda_n \leq \lambda_n'$

Question

I would like to understand the proof of the theorem $4$ of the book page $326$. In fact, after a good while of trying to understand that proof, I am not even sure how it works. In being clear, does someone could explain to me the intuitive idea of this proof?

Theorem $4$ : If one domain $D$ is contained in another domain $D'$, then $\lambda_n \leq \lambda_n'$.

Answer 1

This follows from the min-max principle, which is intimately related to the question you asked yesterday. Let's fix self-adjoint boundary conditions for the Laplacian $\Delta = -\sum \partial_i^2$ on a precompact domain $\Omega$. Restrict to the subspace $D$ of smooth functions satisfying those boundary conditions. Denote by $n$ the $L^2$ inner product. Then the quadratic form $q$ on $D$ defined by $q(u,v) = \int \langle \nabla u,\nabla v\rangle\ dx$ is nonnegative and has $q(u,v) = n(\Delta u, v) = n(u,\Delta v)$. (Verify this with integration by parts.)

Denote by $H(\Omega)$ the completion of $D$ with respect to the inner product $n + q$. (Verify for yourself that if the boundary conditions are Dirichlet, $H = H_0^1(\Omega)$ and if the boundary conditions are Neumann, $H = H^1(\Omega)$.) Denote by $\mathcal{R}$ the Rayleigh quotient on $H$, $\mathcal{R}(v) = q(v,v)/n(v,v)$.

Min-max principle: The $k^{th}$ eigenvalue of the Laplacian satisfies $$ \lambda_k = \min_V \max_{v\in H} \mathcal{R}(v)$$ where $V$ ranges over $k$-dimensional subspaces of $H$.

Suppose we have two domains, $\Omega'\subset\Omega$. Let us fix boundary conditions. The intuition behind the proof is this: Every subspace of $H(\Omega')$ is also a subspace of $H(\Omega)$. Therefore the minimum in the minimax characterization of $\lambda_k(\Omega)$ is taken over a "strictly larger" set of subspaces than that of $\lambda_k(\Omega')$, and so we must have $\lambda_k(\Omega)\leq\lambda_k(\Omega')$.

Of course we cannot appeal to cardinality here. To formalize this intuition, instead use the fact that $H(\Omega)$ and $H(\Omega')$ are Hilbert spaces. Show that the embedding $\Omega'\hookrightarrow\Omega$ induces an isometric embedding. Use this to finish the argument.

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This does not hold true for Neumann eigenvalues. Consider the Neumann Laplace spectra of rectangles. Can you find a counterexample to Neumann domain monotonicity? Do you see why Dirichlet monotonicity holds for rectangles, but Neumann monotonicity does not?