Consider the following integral eigenvalue equation $u = \lambda Ku$, where $K\colon L^2(F)\longrightarrow L^2(F)$ is a symmetric, compact, self-adjoint operator with positive, continuous kernel of order 0. Here, $F$ is the closed interval $[-1,1]$.
- Viewing the eigenvalue equation as $Ku = u/\lambda = \mu u$, with $\mu = 1/\lambda$, spectral theorem asserts that there exists a sequence $\mu_n$ of positive real "eigenvalues" converging to $0$, which means that $\mu_n$ is a bounded sequence. With this in mind, the term "fundamental eigenvalue" of $u = \lambda Ku$ refers to the smallest $\lambda$, say $\lambda_1$. Is this correct?
- Assuming the interpretation of fundamental eigenvalue above makes sense, is it true that the sign of the corresponding eigenfunction $u_1$ does not change?
I am hoping for a proof/explanation to Q2 from the variational characterisation point of view. I am not entirely sure but I believe $\lambda$ has the following variational characterisation: $$ \lambda_1 = \max_{u\in L^2(F), u\neq 0} \frac{\langle Ku,u\rangle}{\langle u,u\rangle} $$ where $\langle\cdot,\cdot\rangle$ is the standard $L^2$ inner product. If it helps, this problem arises when one is looking for time-harmonic solutions of the linear water waves problem in the half-space $\mathbb{R}_-^2$; in the literature this is known as the ice-fishing problem. I doubt this is relevant, but in any cases $Ku$ is actually the single-layer potential restricted to $F$.
EDIT: I thought more about this yesterday and found out that this problem is very similar to the Laplace-Dirichlet eigenvalue problem. Evans' PDE textbook, Chapter 6.5 Theorem 2 (ii) [page 358], asserts that the eigenfunction corresponding to the first nontrivial eigenvalue is positive within the domain. But again I am unsure about this......