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Let

  • $\Omega_i\subseteq\mathbb{R}^n$ be a domain
  • $\lambda_i$ be the first weak eigenvalue of $-\Delta$ in $\Omega_i$

It's easy to verify that $\Omega_1\subseteq\Omega_2$ implies $\lambda_1\ge \lambda_2$, by considering the corresponding Rayleigh quotients.

However, while I'm quite sure that we don't need to have strict inequality, i.e. $\lambda_1>\lambda_2$, I failed to find an elegant counter example.

0xbadf00d
  • 13,422

1 Answers1

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It does has strict inequality: Let $f_1$ satisfies $-\Delta f_1 = \lambda f_1$ and $\|f_1\|_{L^2(\Omega_1)} = 1$ on $\Omega_1$. If $\Omega_1 \subset \Omega_2$, then the function

$$g(x) = \begin{cases} f_1(x) & \text{ if } x\in \Omega_1 \\ 0 & \text{ if not.}\end{cases}$$

is a $W^{1, 2}_0(\Omega)$ function, $\|g\|_{L^2(\Omega_2)} = \|f_1\|_{L^2(\Omega_1)}=1$ and

$$\lambda_1= \frac{\int_{\Omega_2} \|\nabla g\|^2}{\|g\|_{L^2(\Omega_2)}}=\int_{\Omega_2} \|\nabla g\|^2 = \int_{\Omega_1} \|\nabla f_1\|^2.$$

If $\lambda_2 = \lambda_1$, then the Rayleigh quotients is attained by $g$, which implies that $g$ is also the first eigenfunction. But it is impossible as the first eigenfunction must be strictly positive (Note $g$ is not strictly positive).

  • Are you sure that we can apply the (strong) maximum principle for weak eigenfunctions, too? Moreover, your $f_1$ is a (classical) eigenfunction. Does that make any difference? – 0xbadf00d Jun 28 '15 at 19:21
  • @0xbadf00d : I am implicitly using that weak eigenfunctions are all smooth. –  Jun 28 '15 at 21:25