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Apologies for the vague title, it was getting rather long so I decided to just explain more in the body of the text.

I am curious about the state of understanding for existence and uniqueness of solutions to elliptic PDE on regions with multiple boundary components. In particular, is it always sufficient to impose Dirichlet boundary conditions on just one of the boundary components?

As an illustrative example, take the situation of the Laplacian operator on an annulus centered at the origin in the complex plane. In this case, it is sufficient to impose a Dirichlet boundary condition on the outside of the annulus. Convolution with the Poisson kernel will solve this problem on the disk ("filling in" the annulus), and then appealing to some unique continuation theorem will tell you that this is the unique solution on the annulus as well.

In fact, it is clear from the above discussion that imposing boundary conditions on both components of the boundary will yield a PDE with no solution for almost every choice of boundary conditions.

Is there any known generalization of the observation given in the example above? I would imagine it gets more complicated since not every PDE has something like the Poisson kernel, but I don't really have a sense for how "special" the case of the Laplacian is.

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  • $\begingroup$ Be careful. Your claim about the annulus would imply that $u\equiv 0$ is the only harmonic function in $B_1 \setminus B_{1/2}$ with $u=0$ on $\partial B_1$, but consider $u(z) = \log(|z|)$. $\endgroup$
    – user126920
    Commented Aug 30, 2018 at 4:37
  • $\begingroup$ Ah, my mistake. A unique continuation argument will not hold because the derivatives of two solutions with the same prescribed boundary values will not necessarily agree to infinite order at the boundary - in the case of $u = 0$ and $u = \log(|z|)$, their derivatives with respect to the normal coordinate are different. Is this correct? $\endgroup$ Commented Aug 30, 2018 at 13:19
  • $\begingroup$ For unique continuation, you need the functions to agree on an open set, and the boundary of $B_1$ isn't open. $\endgroup$
    – user126920
    Commented Aug 30, 2018 at 16:32

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