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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 $[-1,1]$.

  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?
  2. 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......

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  • $\begingroup$ You can show that to the maximal eigenvalue there exists an eigenvector with representant a continuous positive function. I'm not sure if the eigenspace of this eigenvalue is necessarily one-dimensional. $\endgroup$ – s.harp Mar 26 '18 at 23:49
  • $\begingroup$ @s.harp I agree. I might be wrong but I don't think the paper I am reading (which link can be found above) used the simplicity of $\lambda_1$ to establish a positive answer to Q2. Quoting the author:" Since the kernel of $K$ is positive, it follows from the characterisation of the fundamental eigenvalue as the maximum of a certain quadratic functional that the fundamental eigenfunction does not change sign." $\endgroup$ – Chee Han Mar 27 '18 at 0:47
  • $\begingroup$ What do you mean with the "order" of the integral kernel? I do not have access to the paper right now. Let $f=\min\{1/2-|x-1/2|,0\}$ and $g=\min\{1/2-|x+1/2|,0\}$ so that $k(x,y)=f(x)f(y)+g(x)g(y)$. This is a continuous positive kernel, however $f-g$ is a non-positive eigenvector to the maximal eigenvalue $1$. $\endgroup$ – s.harp Mar 27 '18 at 18:36
  • $\begingroup$ The kernel in this case is the fundamental solution of the 2D Laplacian, i.e. $K(x,y) = \ln|x-y|$. @s.harp $\endgroup$ – Chee Han Mar 28 '18 at 0:35
  • $\begingroup$ This kernel is not a continuous function, it is not clear to me why $K$ should be a compact operator in this case. $\endgroup$ – s.harp Mar 28 '18 at 22:53

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