I want to find a solution $u(x,y)$ to the laplace equation $u_{xx}+u_{yy}=0$ in the region $[0,\pi]^2$ under the following conditions: $$\frac{\partial u}{\partial x}(0,y)=\frac{\partial u}{\partial x}(\pi,y)=0$$ $$u(x,0)=\sin^2(x)=f(x) \quad y\in[0,\pi] \\ u(x,\pi)=\sin^2(2x)=g(x) \quad x\in [0,\pi] $$
so i started by dividing the solution to 4 parts, i solved for $u_0$ and $u_2$, but when i wanted to find the solution for the $\frac{\partial u_4}{\partial x}(0,y)=0$ condition i got that $u_3=0$.
here is what I tried
I tried to change the B.C to be $\frac{\partial u}{\partial x}(0,y)=f(y)$ for some arbitrary odd periodic $2\pi$ function $f\in C^1$. and just to make things simple:$$u(\pi,y)=u(x,0)=u(x,\pi)=0$$
so by setting some $x$ we can find the fourier coefficients of the function to be
$$\hat{u}_{xx}(x,n)-n^2\hat{u}(x,n)=0 \\
\hat{u}_x(0,n)=\hat{f}(n)
$$
if $n=0$: then $\hat{u}_{xx}(x,0) = 0$ therefore $\hat{u}(x,n)=A_0x+B_0$
from the first B.C we see that $B_0=-A_0\pi$ and from the second $A_0=\hat{f}(0)=0$ since $f$ is odd, so $\hat{u}(x,0)=0$
if $n\neq0$: then $\hat{u}(x,n)=A_ne^{nx}+B_ne^{-nx}$
, from the first B.C. we get $A_ne^{n\pi}+B_ne^{-n\pi}=0 \Rightarrow B_n=-A_ne^{2\pi n}$
from the second B.C.
$\hat{u}_x(x,n)=n(A_ne^{nx}-B_ne^{-nx}) \Rightarrow \hat{u}_x(0,n)=\hat{f}(n)=n(A_n-B_n)$ so $$A_n=\frac{\hat{f}(n)}{n(1+e^{2\pi n})}$$ but if I'll go back to my original problem ,setting $f(x)=0$ then $A_n,B_n=0$ and $\hat{u}(x,n)=0$ and so does $u(x,y)$.
so is there any other approach to this question?