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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 (EXPLANATION IN DETAILS PLEASE)?

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?

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 (explanation in details please)?

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 (EXPLANATION IN DETAILS PLEASE)?

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?

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 (explanation in details please)?