Skip to main content
4 events
when toggle format what by license comment
Aug 30, 2018 at 16:32 comment added user126920 For unique continuation, you need the functions to agree on an open set, and the boundary of $B_1$ isn't open.
Aug 30, 2018 at 13:19 comment added Rohil Prasad 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?
Aug 30, 2018 at 4:37 comment added user126920 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 $u(z) = \log(|z|)$.
Aug 30, 2018 at 2:39 history asked Rohil Prasad CC BY-SA 4.0