Problem
Find harmonic function $u(x,y)$ on domain $\Omega=(0,1/2)\times(0,1/2)$ such that:
$u(x,0)=x$ ... $x \in (0,1/2)$
$u(1/2,y)=1/2-2y^2$ ... $y\in (0,1/2) $
$u(x,1/2)=0$ ... $x \in (0,1/2)$
$u(0,y)=0$ ... $y \in (0,1/2)$
Hint: First find harmonic polynomial $F(x,y)$ such that harmonic function $v(x,y)=u(x,y)-F(x,y)$ is zero in the corners of the square. Then use Fourier method.
Solution atempt:
I did the easy part and found $F(x,y)=x-2xy$ but now I have
$\Delta v(x,y)=0$
$v(x,0)=0$ ... $x \in (0,1/2)$
$v(1/2,y)=-2y^2+y$ ... $y\in (0,1/2) $
$v(x,1/2)=0$ ... $x \in (0,1/2)$
$v(0,y)=0$ ... $y \in (0,1/2)$
I guess I have to solve it and then just write: $u(x,y)=v(x,y)+F(x,y)$
Hypothetical solution
What can be meant by Fourier method?< In other examples it meant using Fourier transform on whole equation, solving algebraic equation and using inverse Fourier transform.
The method with Fourier transform can be used to find "fundamental solution" meaning solution of equation $\Delta e(x,y)=0, y>0, e(x,0)=\delta(x)$ the fundamental solution is: $$e(x,y)=\frac{y}{\pi}\frac{1}{x^2+y^2}$$ Other solutions on the same domain are obtained by convolution with boundary condition. If the domain is different we use function which transforms the domain. I don't see any Fourier transform here. Maybe the Fourier method means using Fourier series of boundary condition which of course only works for periodic functions. I should probably somehow transfer the square $\Omega=(0,1/2)\times(0,1/2)$ to half plane ${\mathbb R}\times(0,\infty)$ and make the boundary condition periodic. Then make Fourier series and it might work.
Question
Does "Fourier method" mean using Fourier series? What holomorphic function can I use to transfer square to a half plane?