Positive harmonic function in the first quadrant Suppose $u(x_1,x_2)$ is a positive harmonic function in $Q = (0, \infty)\times(0, \infty)$ and $u(x_1,x_2)\equiv0$ for $(x_1,x_2)\in\partial Q \setminus \{0\}$. Show that there exist two constants $C_1$ and $C_2$ such that $$u(x_1, x_2) = C_1x_1x_2+C_2\frac{x_1x_2}{|x|^4}.$$ 
I know if $u \equiv 0$ on the entire boundary $\partial Q$ one can prove $u(x_1, x_2) = Cx_1x_2$. But I don't know why $u$ is of the form $\frac{x_1x_2}{|x|^4}$ when it has singularity at the origin. Can you give me a hint or reference on this problem? Thanks a lot!
 A: Using the Riemann Mapping Theorem on a problem like this is bad, because the argument can't generalize to $\Bbb R^n$. But I give up on the "right" proof - here's an argument that must work, although I haven't worked out the details:


There exist two positive harmonic functions $u_1$ and $u_2$ in the unit disk such that if $u$ is a positive harmonic function in the disk that vanishes everywhere on the boundary except at $\pm1$ then $u=c_1u_1+c_2u_2$.


Proof: There exists a positive measure $\mu$ on the boundary such that $u=P[\mu]$, the Poisson integral of $\mu$. Now $\mu$ must be supported on $\pm1$, so $\mu=c_1\delta_1+c_2\delta_{-1}$. Let $u_1=P[\delta_1]$ and $u_2=P[\delta_{-1}]$.
Note that you can easily write down $u_1$ and $u_2$ above explicitly; I leave that to you.. Also note it's easy to give an explicit conformal mapping of $Q$ onto the unit disk, starting with the map $z\to z^2$ that takes $Q$ onto the upper half-plane.
Work out the details explicitly and you see that your $u$ is $c_1$ times something plus $c_2$ times something else; if the result is actually true those somethings must turn out to be $x_1x_2$ and $x_1x_2/|x|^4$.
