Consider the following pde: $$-\Delta u =0$$ in $(0,\pi) \times (0,4)$ with boundary conditions:
$$u(0,y) = u(\pi,y) \ \text{for} \ y\in (0,4)$$
and $$u(x,0) = \cos x \sin x \ \text{for} \ x\in (0,\pi)$$ and $$u(x,4) = \cos x \sin x +4 \ \text{for} \ x\in (0,\pi)$$
I tried seperation of variables $u(x,y)=f(x)g(y)$
Inserting this yields two odes: $$g''(y)-cg(y)=0$$ and $$f''(x) + cf(x)=0$$ for some constant$c$
Consider the first ode. The characterstic polynomial is given by $\lambda^2=c$ Therefore there are 3 cases to consider:
$c=0: g(y)= Ay+b$
$c>0: g(y)= Ae^{\sqrt{c}y} + Be^{-\sqrt{c}y}$
$c<0: g(y)= A\cos(\sqrt{-c} y) + B\sin(\sqrt{-c} y)$
How can I now use my boundary conditions. I only know $f(x)g(0) = f(x) \sin x \cos x$ I do not know how I can insert them?
The periodic boundary condition tells you that the solution is $\pi$-periodic in $x$-direction, or at least can be interpreted as such if suitably extended outside the rectangle. This means that in that direction you get a Fourier series expansion $$ u(x,y)=\frac{a_0(y)}2+\sum a_k(y)\cos(2kx)+b_k(y)\sin(2kx). $$ This gives the same result as the separation approach, but starting from a different direction.
Comparing the conditions on the upper and lower boundaries gives $b_1(0)=b_1(4)=\frac12$, $a_0(4)=8$, all other coefficients are zero at $y=0$ and $y=4$. Further, $$ -Δu(x,y)=-\frac{a_0''(y)}2+\sum_k[4k^2a_k(y)-a_k''(y)]\cos(2kx)+[4k^2b_k(y)-b_k''(y)]\sin(2kx). $$ Now comparing coefficients and solving the simple DE gives $$ a_0(y)=2y,\\ b_1(y)=\frac{\cosh(2(y-2))}{2\cosh(4)}, $$ all other coefficient functions being zero.