I have read the proof that if $u:D\to\mathbb{R}$ is harmonic and $D$ is simply connected then there exists a harmonic conjugate for $u$.
I can see why simple-connectedness is required in the proof, as we have to construct a well-defined anti-derivative; but is there a counter-example that shows that, in fact, this is a necessary as well as a sufficient condition?
Edit: I see from below that it is not actually a necessary condition. So, is there a harmonic function on a non-simply-connected region that has no harmonic conjugate, is what I'm asking.
Best Answer
Take $\ln|z|$ on the annulus $1<|z|<2$. Its harmonic conjugate (locally) is $\arg z$, except one can not define it on the entire annulus continuously. This is because the $\ln z=\ln|z|+i\arg z$ function is multivalued on the annulus. You can define a continuous branch of the logarithm after cutting out any radial line segment from the annulus (what is left is then simply connected), e.g. the principal branch makes a cut along the real axis.
Instead of the annulus you can take a punctured disk $0<|z|<2$ or even the entire punctured plane $0<|z|$. The effect is the same. You can always construct harmonic conjugates locally, on disks around any point, and they are unique up to a constant, but when the domain is not simply connected it is not always possible to stitch them together coherently because you may end up with different constants when going around non-contractible loops in $\Omega$. This is exactly what happens with $\arg z$ when going around the origin.