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If you are looking for a physical reason, you should ask the question why the eigenfunctions are confined to M in the first place. Usually, the reasoning for this is that V is infinite outside of M. S …

This does not appear to be known. Clearly the remaining eigenfunctions are the same as those for an equilateral triangle with Dirichlet conditions on two sides and Neumann conditions on one side. It i …

If you take $C_0^\infty$ as the initial domain, then there are many self-adjoint extensions and the Dirichlet Laplacian is only one of them. The Neumann Laplacian is another one.

The answer is no. The following reference specifically discusses the case of the two-dimensional disk: http://www.staff.uni-oldenburg.de/daniel.grieser/wwwpapers/diss.pdf

In general this is not true. Let $\Omega$ be a smooth bounded domain, let $A$ be $-\Delta$ with Neumann conditions and let $B$ be $-\Delta$ with Dirichlet conditions. The $(A+iB)$ is $(-1-i)\Delta$ wi …

This is essentially equivalent to asking whether the spectrum determines the growth bound. This is well known to be false. A pioneering counterexample is due to Zabczyk.

If we take the usual trigonometric basis, it is indeed true as has been pointed out. However, there is a harder form of the question: Some of the eigenvalues are degenerate. If we allow arbritrary eig …

Set $\alpha z=w$, you get the new system $$dy/dw={1\over \alpha}\pmatrix{0&B\cos(w+\Phi_b)\cr A\cos(w+\Phi_a)&0}y.$$ Since you are interested in a case where $\alpha$ is large and $w$ is of moderate s …

Look at Kato's book on Perturbation Theory for Linear Operators.

Yes, such eigenfunctions exist. If $a>0$, $b<0$ and $\lambda<0$, the roots $k$ of $ak^4-bk^2-\lambda=0$ are complex. Let $k_1$, $k_2$ be the roots with positive real part. Assume they are different; …

Take $M_1$ to be a region consisting of two "blobs" connected by a narrow channel of length 1 and width $\epsilon$. Now choose $u_1$ and $u_2$ such that each is 1 in one of the blobs, 0 in the other, …

A counterexample is given in Section II.5.3, p. 111 of T. Kato, Perturbation Theory for Linear Operators, 2nd ed.

To get the analytic semigroup estimate, we note first of all that the operator $M$: $Mu=Lu+(v\cdot\nabla)u$, with the same boundary condition as $L$, is self-adjoint and hence generates an analytic se …