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?



Why $M_1 \subset M_2 \not \Rightarrow N_{M_1} (\lambda) \leq N_{M_2} (\lambda)$ for eigenvalue problem? (EDIT)
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:
$$
\lambda\_2(M\_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda\_2(M\_2)
$$


