How to derive the substituted Partial Differential Equation I have a second order partial differential equation. 
$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial z^2} + \frac{2}{l}\frac{\partial U}{\partial z}=0$.
I need to introduce a perturbation $\zeta(x,t)$ (i.e. introduction of a perturbation) into the above equation in order to shift the coordinates. Boundary conditions $U(0)=0$
Here, $z = \zeta(x,t)=\hat{\zeta}exp(ik\cdot x+\omega t)$, where, $k$ is a two-dimensional wave vector and $\omega$ the wave number. 
After substitution I get the following equation : 
$-k^2\hat{\zeta}exp(ik\cdot x + \omega t)\frac{\partial^2U}{\partial \zeta^2} + \frac{2}{l}\frac{\partial U}{\partial\zeta}=0$. As, $\frac{\partial^2 U}{\partial x^2} = \frac{\partial^2 U}{\partial \zeta^2}\frac{\partial^2 \zeta}{\partial x^2}$. 
But, when I look at the solution for this problem it is given as,
$U = exp(-2z/l) - 1 + \hat{\zeta}exp(ik\cdot x+\omega t - qz)$, where $q$ is the solution of a quadratic equation. 
So I'm stuck at the substituted equation, as it will not result in the given solution. 
Can anyone point out where I'm going wrong in the substitution? 
 A: Actually, this equation is screaming for the substitution
$$
U(x,z)=e^{-\alpha z}W(x,z)+c.
$$
where $c$ is an arbitrary constant that will not be essential for the following argument.
Then,
$$
\frac{\partial U}{\partial z}=-\alpha e^{-\alpha z}W(x,z)+e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}
$$
and
$$
\frac{\partial^2 U}{\partial z^2}=\alpha^2e^{-\alpha z}W(x,z)-2\alpha e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}+e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial z^2}.
$$
Putting all this in the original equation one gets
$$
e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial x^2}+
\alpha^2e^{-\alpha z}W(x,z)-2\alpha e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}+e^{-\alpha z}\frac{\partial^2 W(x,z)}{\partial z^2}
-\frac{2}{l}\alpha e^{-\alpha z}W(x,z)+\frac{2}{l}e^{-\alpha z}\frac{\partial W(x,z)}{\partial z}=0.
$$
I can remove the first order derivative by the choice $\alpha=l^{-1}$ and I can also remove the exponential. This will yield
$$
\frac{\partial^2 W(x,z)}{\partial x^2}+\frac{\partial^2 W(x,z)}{\partial z^2}
-\frac{1}{l^2}W(x,z)=0
$$
where I have substituted the $\alpha$ value obtained above. This equation is very well-known and can be solved by variable separation. As already stated in the comments, you will need to specify the boundary conditions. Anyway, to recover your kind of solution, you will need to look for solutions in the form
$$
W(x,z)=\hat\zeta\exp(ik_xx+ik_zz+i\phi),
$$
being $\phi$ an arbitrary phase that you have taken to be $\omega t$. You will get the dispersion relation
$$
k_x^2+k_z^2+\frac{1}{l^2}=0.
$$
A given domain and the boundary conditions will fix $k_x$ and $k_z$ properly.
