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.


Does "Fourier method" mean using Fourier series? What holomorphic function can I use to transfer square to a half plane?

  • $\begingroup$ Here I think the Fourier Method is separation of variables and Fourier Series; thought I agree the term is a bit ambiguous. $\endgroup$ – User8128 Jun 25 '17 at 19:27
  • $\begingroup$ A circle to a half plane you can do a Möbius transformation. But a square to a circle... hmm. Maybe you can use that level sets of $\lim_{k\to \infty} x^k + y^k$ become more and more square-like the larger $k$ is. $\endgroup$ – mathreadler Jun 25 '17 at 20:15
  • $\begingroup$ @mathreadler Thanks I've already come across that after I asked the question. I thought square to a circle could be done by polar coordinates $\endgroup$ – Puzzled student Jun 25 '17 at 20:39
  • $\begingroup$ I am sorry this problem should probably be solved by separation of variables. Similar problem here $\endgroup$ – Puzzled student Jun 25 '17 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.