Find a harmonic function in the first quadrant, Find a harmonic function $\phi$(x,y) in the first quadrant with the boundary values $\phi$(x,0) = -1 for x>0, and $\phi$(0,y) = 1 for y>0.  Is this function unique?
My attempt was this: 
Consider the principal branch of log(z), cutting away the negative real axis.  Then Log(z) is analytic away from the branch cut, and its real and imaginary parts are harmonic conjugates.
Look at the harmonic function Arg(x,y) in the first quadrant.
Consider the piecewise-defined function:
$$\phi =  Arg(x,0) - 1 $$
$$\phi =  Arg(0,y)\frac{2}{\pi} $$  
My gut feeling is that I haven't done it correctly, and that I may need to first construct some linear fractional transformation or other conformal mapping, and then compose the Arg function with it - and scaling Arg(f(z)) suitably to achieve -1, +1 on the boundary.
As for uniqueness, going back to my paragraph above, if indeed I have to construct some conformal first, then $\phi$ is not unique, since there can be many different compositions of conformal mappings that lead to the same outcome, e.g., there are many ways to compose a mapping that maps the UHP to the unit disk.  Then, composing these different mappings with the harmonic function, Arg(z), produces a different harmonic function.
Any hints or solutions are welcome.
Thanks,
 A: The function $f(x,y)=xy$ is harmonic and it vanishes on the coordinate lines $x=0$ and $y=0$. You can add this function to any solution you have and get a new solution. So you won't get uniqueness unless you restrict to some class of solutions, such as (perhaps) bounded functions.
You have the right idea for the solution you want: the arg function is constant on the positive x and y axes, and you can offset by a constant and then multiply by a constant to get what you want.
A: If we start with first principles, then we have the following PDE.
$$\nabla ^2 \phi(r,\theta) = 0 \,\,\text{for}\,\,0\le r, 0\le \theta\le \pi/2 \tag 1$$
with $\phi(r,\theta=0)=-1$ and $\phi(r,\theta =\pi/2)=1$.  The problem is not well-posed since there is no condition imposed on the boundary $r=R\to \infty$.
We can write a general solution to $(1)$ as
$$\phi(r,\theta)=A+B\theta+\sum_{n=1}^{\infty}r^n(a_n\cos n\theta+b_n\sin n\theta)$$
Applying the boundary conditions reveals that
$$A+\sum_{n=1}^{\infty}a_nr^n=-1\implies a_n=0, A=-1$$
$$B\pi/2-1+\sum_{n=1}^{\infty}b_n\sin(n\pi/2)r^n=1\implies b_{2n-1}=0, B=\frac{4}{\pi}$$
Therefore, we can write
$$\bbox[5px,border:2px solid #C0A000]{\phi(r,\theta)=\frac{4}{\pi}\theta-1+\sum_{n=1}^{\infty}c_nr^{2n}\sin(2n\theta)}$$
for any suitable coefficients $c_n$.
A: Hint: Consider $\text {Arg}\  z - \pi/4.$
