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Let

  • $\Omega\subseteq\mathbb{R}^n$ be a bounded domain
  • $\lambda_1$ be the first weak eigenvalue of $-\Delta$ in $\Omega$
  • $\varphi_1$ be the weak eigenfunction associated with $\lambda_1$
  • $H:=W_0^{1,2}(\Omega)$ be the Sobolev space
  • $|\;\cdot\;|_p$ be the seminorm $$|u|_p^p:=\int_\Omega|\nabla u|^p\;d\lambda^n\;\;\;\text{for }u:\Omega\to\mathbb{R}\;\text{weakly differentiable}$$
  • $\left\|\;\cdot\;\right\|_p$ be the $L^p(\Omega)$-norm and $\langle\;\cdot\;,\;\cdot\;\rangle_2$ be the scalar product in $L^2(\Omega)$

From the basic theory of the eigenvalue problem of the Laplacian, one knows that $$\lambda_1=\inf_{H\setminus\left\{0\right\}}R\;\;\;\;\;\text{with }R(u):=\frac{|u|_2^2}{\left\|u\right\|_2^2}\;\;\;\text{for }u\in H\setminus\left\{0\right\}$$ and $\varphi_1$ can be constructed such that

  • $\left\|\varphi_1\right\|_2=1$
  • $R(\varphi_1)=\lambda_1$

Since different eigenvalues correspond to $L^2$-orthogonal eigenfunctions (with $L^2$-orthogonal gradients), it's plausible to seek for the next eigenfunction $\varphi_2$ in $$U_1:=\left\{u\in H:\langle u,\varphi_1\rangle_2=0 \right\}\;.$$ We may observe, that $U_1$ is the null space of the continuous operator $\left.\langle\;\cdot\;,\varphi\rangle_2\right|_H$. Thus, $H$ is a closed subspace of $H$ and thereby a Hilbert space itself.

So, we can apply the same process used to find $(\lambda_1,\varphi_2)$ in order to find $$\lambda_2:=\inf_{U_1\setminus\left\{0\right\}}R$$ and $\varphi_2$.

Now, using the same arguments as discussed in a similar question, it's easy to see, that $\lambda_2\ge \lambda_1$ (Please don't get confused with the naming of the $\lambda_i$ in the other question).

However, is it possible that we have $\lambda_1=\lambda_2$?

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1 Answers1

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That's not possible. The first eigenspace is always one dimensional:

Let $\phi_1$ be the first eigenfunction. Then as $\nabla |\phi_1| = \pm \nabla \phi_1$ almost everywhere, $R(\phi_1) = R(|\phi_1|)$ and so $|\phi_1|$ is also an eigenfunction with eigenvalue $\lambda_1$. By maximum principle, $|\phi_1|>0$ on $\Omega_1$, which implies that $\phi_1$ is either strictly positive or negative.

Now if $\phi_1, \phi_2$ are two eigenfunctions with respect to $\lambda_1$, then we can assume that they are both strictly positive. Then fix $x_0 \in \Omega_1$ and consider

$$f(x) = \phi_2(x_0)\phi_1(x) -\phi_1(x_0)\phi_2(x)$$

Then $f$ also satisfies $-\Delta f = \lambda_1 f$ and $f(x_0) = 0$. By maximum principle again ($f$ does not change sign!), $f$ is identically zero, so

$$\phi_2(x) = \frac{\phi_2(x_0)}{\phi_1(x_0)} \phi_1(x),$$

Thus the eigenspace is one dimensional. In particular, $\lambda _1\neq \lambda_2$.

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    You could improve your answer a bit simply by invoking orthogonality (without using regularity to show that values at one point are defined) to argue that there cannot be two positive eigensolutions. –  Apr 03 '16 at 20:10