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Consider a compact domain $\Omega \subset \mathbb{R}^n$ with smooth boundary for simplicity. Consider the Laplacian operator with zero Dirichlet boundary conditions on $\Omega$. It is well-known that $-\Delta$ has discrete spectrum $0 = \lambda_0 < ... \leq \lambda_k \leq ...$ going to infinity. I am curious in understanding how the "spectrum" of the Laplacian changes when one alters the boundary condition, in the following sense:

Are there constants $\mu$ for which we can solve $-\Delta u = \mu u$ such that $u$ has constant non-zero value on the boundary? (We are not fixing the constant apriori, but that does not matter in any case).

Can such $\mu$ be negative? How many such $\mu$ could be there?

Is there a way of understanding if and how the new "spectrum" $\mu_k$ is related to $\lambda_k$? For starters, is there a relation between $\mu_1$ and $\lambda_1$? I would be highly grateful for any suggestions. My main interest is in dimension $n = 2$, though it is unclear to me if that matters.

Edit: I have put "spectrum" in quotes because, as pointed out by Denis Serre and Nate Eldredge, these are not eigenvalues per se.

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    $\begingroup$ Imposing a non-zero boundary condition is a mathematical non-sense. An eigenvalue problem is always homogeneous. You could change the boundary operator $B$, but you should keep an homogeneous condition $Bu=0$. $\endgroup$
    – Denis Serre
    Commented Mar 2, 2018 at 3:12
  • $\begingroup$ @DenisSerre Please forgive my confusion. But why can I not expect, for example, a solution to $\Delta u = \lambda_1 u$ with $u \equiv 5$ on the boundary? Of course, such a $u$ might not exist, but why is it nonsensical to ask this? $\endgroup$
    – SMS
    Commented Mar 2, 2018 at 3:22
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    $\begingroup$ Then this $\lambda_1$ will not be an eigenvalue. $\endgroup$
    – Denis Serre
    Commented Mar 2, 2018 at 3:32
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    $\begingroup$ The notion of "eigenvalue" makes sense for linear operators on a vector space. A set of functions with boundary condition $1$ is not a vector space. $\endgroup$
    – Nate Eldredge
    Commented Mar 2, 2018 at 4:24
  • $\begingroup$ The problem with inhomogeneous boundary conditions has a unique solution for every $\mu$ that is NOT an eigenvalue of the homogeneous problem. Voting to close and downvoting the question (why did anyone upvote it?). $\endgroup$
    – Michael Renardy
    Commented Mar 2, 2018 at 13:42

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