8
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ Cross-posted: math.stackexchange.com/questions/1824841 $\endgroup$
    – Watson
    Jun 14, 2016 at 11:54
  • 3
    $\begingroup$ The reason this fails is really that Neumann eigenfunctions can have a lot of variation along the boundary, so you'll be charged a steep prize $\int_{M_2\setminus M_1} |\nabla u|^2$ when you extend them to the larger region. $\endgroup$ Jun 15, 2016 at 4:23

3 Answers 3

7
$\begingroup$

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: $$ \lambda_2(M_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda_2(M_2) $$

$\endgroup$
1
  • 3
    $\begingroup$ By smoothing the rectangles in this example, one can also produce smooth convex domains which do not have monotonicity, and relaxing convexity, there are also counterexamples among annuli. See Ni-Wang, "On the First Positive Neumann Eigenvalue." $\endgroup$
    – Neal
    Jun 15, 2016 at 16:29
5
$\begingroup$

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.

$\endgroup$
1
0
$\begingroup$

@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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.