partial differential equations, Laplace equation, boundary conditions

Question

For the stated problem I got the following equations after applying separation of variables, I am confused as to how to apply the mentioned boundary conditions to this problem :

\begin{align} G''(x)+(4\pi^2-c)G(x)&=0\\ 4H''(y)+cH(y)&=0 \end{align}

Answer 1

For the first part...I'm calling $c$ lambda $\lambda$

$$ G^{''}(x) + (4\pi^{2} - \lambda )G(x) = 0 \tag{1} $$

call $4\pi^{2} - \lambda = \alpha $

$$ G^{''}(x) + \alpha G(x) = 0 \tag{2} $$

$$ G(x) = c_{1} \cos(\sqrt{\alpha}x) + c_{2}\sin(\sqrt{\alpha}x) \tag{3} $$

the x boundaries are neumann

$$ \frac{\partial }{\partial x}G(x) = -\sqrt{\alpha}c_{1}\sin(\sqrt{\alpha}x) + \sqrt{\alpha}c_{2}\cos(\sqrt{\alpha}x) \tag{4} $$

put in $x=0$ and $c_{2} =0$

$$ \frac{\partial }{\partial x}G(0) = \sqrt{\alpha}c_{2} = 0\tag{5} $$

now $G(x) = c_{1}\cos(\sqrt{\alpha}x) $

$$ \frac{\partial }{\partial x}G(1) = -\sqrt{\alpha}c_{1}\sin(\sqrt{\alpha}1) = \sin(\sqrt{\alpha}) \tag{6} $$

in order to determine the eigenvalues, is this good? Can you apply the $y$ boundaries now?

Note for the next one you have

$$ H^{''}(y) + \frac{\lambda}{4}H(y) = 0 \tag{7} $$

however, the eigenvalues are going to be related now

Edit

If you simply solve the above on line $6$ and we let the boundary actually be $L$

$$\sin(\sqrt{\alpha}L) = 0 \implies \sqrt{\alpha}L = n\pi \implies \sqrt{\alpha} = \frac{n\pi}{L} \tag{8} $$

now remember what $\alpha$ is

$$ \sqrt{\alpha} = \sqrt{4\pi^{2} - \lambda} =\frac{n\pi}{L} \tag{9} $$

now $L =1$

$$4\pi^{2} - \lambda =n^{2}\pi^{2} \tag{10}$$

$$ - \lambda = n^{2}\pi^{2} - 4\pi^{2} \implies \lambda = 4\pi^{2} - n^{2}\pi^{2} \tag{11}$$