Solving $u_{xx} + u_{yy} = 0$ subject to $u(x, 0) = u(0, y) = 0$ $ u(x, 1) = \sin(x)$, $u(1, y) = y^2$

Question

I tried to proceed as expected: set $u = X(x)Y(y)$, then you get $$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda$$. Assume $\lambda>0$, and $\lambda = z^2$ then you get $$X = C_1e^{-zx}+C_2e^{zx}$$, $$Y = C_3\cos(zy)+C_4\sin(zy)$$. So then we get the from the first two conditions $X(0) = C_1+C_2 = 0$, and $Y(0) = C_3 = 0$. Then we get $$X(x) = C_2\sinh(zx)C_4\sinh(z) = \sin(x)$$ and $$C_2\sinh(z)C_4\sin(zy) = y^2$$ from the second two conditions. I have no idea where to go from there.

Answer 1

The boundary conditions at point $(1,1)$ introduce a singularity $$\begin{cases} u(x,1)=\sin(x) \quad\to\quad u(1,1)=\sin(1)\\ u(1,y)=y^2 \quad\to\quad u(1,1)=1 \end{cases}$$

This can be overcome with the solution in term of Fourier series: