I know that $u_1 = ln|x^2+y^2|$ is harmonic. Knowing that $u_2 = 0$ is harmonic, I can see that boundary values for $u_1$ on unit circle coincide with boundary values of $u_2$.
My question is does this contradict the uniqueness of solutions to boundary value problems? I am not quite sure if 0 could be counted as a solution nor am I sure about whether absolute values have any role. If you could let me know, I would greatly appreciate it. Thanks.
I think that by "unit circle" you mean $D=\{(x,y): x^2+y^2 \le 1\}$. But your function $u_1$ is only defined for $(x,y) \ne (0,0)$.
Let $D_0:= D \setminus \{(0,0)\}$. Then $u_1$ and $u_2$ are harmonic on $D_0$ but $u_1 \ne u_2$ on $ \partial D_0$, since $(0,0) \in \partial D_0$, and $u_1$ is not defined in this point.