Why $M_1 \subset M_2 \not \Rightarrow N_{M_1} (\lambda) \leq N_{M_2} (\lambda)$ for eigenvalue problem? (EDIT)

Question

We know that for a direct problem with Dirichlet Boundary Condition (with Laplacian operator) that if two domains $M_1$ and $M_2$ are such that $M_1 \subset M_2$, then $\lambda(M_2) \leq \lambda(M_1)$, and hence, $N_{M_1} (\lambda) \leq N_{M_2} (\lambda)$. Why doesn't exist a similar result for a direct problem with Neumann Boundary Condition, i.e. $M_1 \subset M_2 \not \Rightarrow N_{M_1} (\lambda) \leq N_{M_2} (\lambda)$? Is there anyone could give me a clever counterexample? I think this is related by the fact that $H^1(M_1) \not \subset H^1(M_2)$.

Precision : $N(\lambda) \equiv \text{the number of eigenvalues less than } \lambda$.

EDIT :

The Neumann eigenvalues of the rectangle with sides $a$ and $b$ are $$\nu_{k,l}=\frac{(\pi k)^2}{a^2}+\frac{(\pi l)^2}{b^2},$$ with $k,l \in \mathbb{N}_0$. So assuming that $a>b$, the first $3$ eigenvalues are $\nu_1=0$, $\nu_2=\frac{\pi^2}{a^2}$, and $\nu_3 = \frac{\pi^2}{b^2}.$ We pick $1 < a < \sqrt{2}$, and choose $b>0$ sufficiently small, so that the rectangle can be place inside the unit squre. For the unit square, the first $3$ Neumann eigenvalues are $\nu_1 ' = 0$, $\nu_2 ' = \pi^2$, and $\nu_3 ' = \pi^2$. Since $a>1$, we have $\nu_2 < \nu_2 '$, which could not happen if domain monotonicity were true.

Does this example work? If so, since the spectrum of the rectangle is the same as Dirichlet condition, why it is a counterexample for NBC but not for DBC?

Answer 1

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of side length $L$. Then $\lambda_2(M_2)=\pi^2/L^2$ (possible eigenfunction $u=\cos \pi x/L$). If we now take $M_1\subseteq M_2$ as a thin rectangle close to the diagonal, then we obtain approximately the one-dimensional Neumann eigenvalue of an interval of that length: $$ \lambda_2(M_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda_2(M_2) $$

Answer 2

Take $M_1$ to be a region consisting of two "blobs" connected by a narrow channel of length 1 and width $\epsilon$. Now choose $u_1$ and $u_2$ such that each is 1 in one of the blobs, 0 in the other, and linearly interpolated in the narrow channel. It follows that both $\int |\nabla u_1|^2$ and $\int |\nabla u_2|^2$ are order $\epsilon$. Consequently $N_{M_1}(\lambda)\ge 2$ if $\lambda>>\epsilon$. Now take $M_2$ to be a ball containing $M_1$. In $M_2$, the second eigenvalue of the Neumann problem is of order 1.

Answer 3

@Sharpie

Hi, I think your counterexample works well for NBC, but not for DBC, and such phenomenon originates from the range of indices, $k$ and $l$.

Briefly, only one of $a$ and $b$ contributes to the first positive eigenvalue in NBC, while both of $a$ and $b$ contributes to it in DBC.

For NBC, $k, l \in \mathbb{N}_0$ so that first and second eigenvalue correspond to $(k, l) = (0, 0)$ and $(k, l) = (1, 0)$, respectively.

However, it is not the case for DBC. For DBC, $k, l \in \mathbb{N}$ so that the lowest eigenvalue on small rectangle corresponds to $(k, l) = (1, 1)$, and thus $$\lambda_1 = \pi^{2}\left( {1 \over a^2} + {1 \over b^2} \right)$$. Since the lowest eigenvalue on unit square is then $2\pi^2$, it remains to compare ${1\over a^2} + {1 \over b^2}$ and $2$.

Let's do it. Since the diagonal of rectangle has endpoints on two distinct parallel sides of unit square, $a^2 + b^2 \ge 1$. Henceforth, we get $${1\over a^2} + {1 \over b^2} \ge {1\over a^2b^2}.$$ But since the area of small rectangle is maximized when it is a square, i.e. each vertex of rectangle is a midpoint of each side of unit square, $a^2b^2 \le 1/4$. So we get $${1\over a^2} + {1 \over b^2} \ge {1\over a^2b^2} \ge 4 > 2$$ which concludes the desired monotonicity.