I have the given PDE:
\begin{equation} z_{xx}+z_{yy}=\lambda z \end{equation}
BC: $\Omega:\{z=z(x,y); 0<x<a, 0<y<b\}$
IC:
$0=z_x(0,y)=z_x(a,y), 0<y<b$ - (Neumann condition on y) $0=z(x,0)=z(x,b), 0<y<b$ - (Dirichlet condition on x)
Additional IC:
$\int\int_\Omega z^2dxdy=1$
Having the Neumann condition on y and Dircihlet condition x, we can actually just make the Ansatz immediately $z(x,y)=A_{m,n}sin\frac{n\pi}{a}xcos\frac{m\pi}{b}y$ which will give us the right solution, by using additional IC.
However, I would like to learn how separation of variables is done correctly with the necessary double-constants to include. Have a look below:
With z=XY, separation of variables gives the two ODEs:
$X_{xx}Y+Y_{yy}X=\lambda XY$ , division by XY gives:
$\frac{X_{xx}}{X}+\frac{Y_{yy}}{Y}=\lambda$
$\frac{X_{xx}}{X}=-\frac{Y_{yy}}{Y}+\lambda$
From here we form the two ODEs, with two different coefficients, as proposed by the commentators:
\begin{equation} \frac{X_{xx}}{X}=k_1, -\frac{Y_{yy}}{Y}=k_2 \end{equation}
then we get, by equaling $k_1+k_2=\lambda$
\begin{equation} \frac{X_{xx}}{X}=\lambda-k_2, -\frac{Y_{yy}}{Y}=\lambda-k_1 \end{equation}
We set $=\lambda-k_2=\mu$ and $\lambda-k_1=\gamma$ and obtain:
\begin{equation} \frac{X_{xx}}{X}=\mu, \frac{Y_{yy}}{Y}=-\gamma \end{equation}
Still, I get one trigonometric, and one hyperbolic solution, when the solution should be entirely trigonometric.
What is wrong with the separation of variable procedure here?
Give a real constant $\lambda$, you want to find a function $z(x,y)$ defined on a rectangle $R=\{ (x,y) : 0 \le x \le a,\;\; 0 \le y \le b\}$ that is a solution of the following partial differential equation on that rectangle: $$ z_{xx}+z_{yy}=\lambda z,\;\;\; z=z(x,y), $$ subject to the following conditions on the boundary of $R$: $$ z_{x}(0,y)=z_{x}(a,y)=0,\;\; 0 \le y \le b, \\ z(x,0)=z(x,b)=0,\;\; 0 \le x \le a. $$ This is a set up for separation variables, where you start by looking for solutions of the form $z(x,y)=X(x)Y(y)$, where $$ X'(0)=X'(a)=0, \\ Y(0)=Y(b)=0. $$ In order to satisfy the PDE, such a solution $z=X(x)Y(y)$ must satisfy $$ X''(x)Y(y)+X(x)Y''(y)=\lambda X(x)Y(y) \\ \frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=\lambda \\ \frac{X''(x)}{X(x)} = \lambda- \frac{Y''(y)}{Y(y)} $$ This forces the introduction of a separation constant $\mu$ such that $$ \frac{X''(x)}{X(x)}=\mu,\;\; \mu=\lambda-\frac{Y''(y)}{Y(y)}, $$ which leads to the ODEs $$ X''(x)-\mu X(x) =0, \\ Y''(y)-(\lambda-\mu)Y(y)=0. $$ It doesn't matter whether you put $\lambda$ into the $X$ equation or into the $Y$ equation. The constants are determined by the endpoint conditions for $X$, $Y$. For example, $Y(0)=Y(b)=0$ leads to non-trivial solutions only when $$ Y(y)=\sin(n\pi y/b),\;\;\; n=1,2,3,\cdots, $$ which means that only the following values of $\lambda-\mu$ can be valid: $$ \lambda-\mu = -n^2\pi^2/b^2,\;\; n=1,2,3,\cdots. $$ Similarly, the $X$ equation with endpoint conditions $X'(0)=X'(a)=0$ determines the values of $\mu$: $$ \mu = -m^2\pi^2/a^2,\;\;\; m=0,1,2,3,\cdots. $$ Here $m=0$ is valid because the constant function $1$ satisfies $X'(0)=X'(a)=0$. This is how the separation of variables solution is determined.