Harmonic and Continuous everywhere but on a curve is harmonic throughout?

Question

Suppose u is a harmonic function everywhere in a domain $\Omega$, but on a curve inside $\Omega$ , say a segment, and is continuous throughout, i.e $u\in C(\Omega)$. Can we conclude that u is harmonic throughout $\Omega$ ?

Answer 1

I am not totally sure if the following example satisfies all the conditions you require, since I am not sure about the precise meaning of " ... but on a curve inside ...", but here we go:

Take the domain $\Omega$ to be the open unit disc $\{ z : |z| < 1 \}$ in $\mathbb{C}$.

Define $u(z)$ to equal $y$ on the subdomain $\{ x+iy \in \Omega : y>0 \}$, and define $u(z)$ to be 0 on the other half of the domain, i.e. on the subdomain $\{ x+iy \in \Omega : y \le 0 \}$.

Clearly the function is harmonic in the 2 subdomains, and is continuous on the whole of $\Omega$, since the two definitions "match up" on the "curve" $y=0$.

We can see that on the whole domain $\Omega$, the function $u$ fails to be harmonic since at any point on the line $y=0$ it fails to satisfy the mean value condition for harmonic functions.

So, the question is: does this example satisfy your conditions?

Edit:

I am still unsure of the precise conditions of the question, but here is another possible counter-example.

Take $\Omega$ to be the open disc $\{ |z| \in \mathbb{C} : |z| < 2\}$, and take the closed curve inside $\Omega$ to be the unit circle $|z|=1$. Then define $u(z)$ to equal $0$ in the subdomain $\{ |z| < 1\}$ and to equal $\log|z|$ on the remainder of $\Omega$ i.e. in $\{ 1 \le |z| < 2\}$.

Again, $u$ is continuous on the whole domain and harmonic on each subdomain, but fails the mean value condition on the closed curve.