I have a PDE in the domain $z\in (0,\infty) $:

$$\frac{\partial^2 W(x,z)}{\partial x^2} + \frac{\partial^2 W(x,z)}{\partial z^2} + \frac{1}{l^2}W(x,z)=0$$

Now, through the variable separable method I introduce $W(x,z)=Q(x)G(z)$ and substitute in the above equation. Thereby,

$$G(z)\frac{\partial^2 Q}{\partial x^2} + Q(x)\frac{\partial^2 G}{\partial z^2} + \frac{1}{l^2}Q(x)G(z)=0$$

Diving by $Q(x)G(z)$ on both sides and rearranging the terms we get

$$\frac{l^2}{Q(x)}\frac{\partial^2 Q}{\partial x^2} + \frac{l^2}{G(z)}\frac{\partial^2 G}{\partial z^2} + 1=0$$

Now, I introduce the separation constant $\lambda$ and then I get the two ODE's in the following form:

$$\frac{\partial^2 Q}{\partial x^2} +\frac{\lambda Q(x)}{l^2} = 0$$ $$\frac{\partial^2 G}{\partial z^2} +\frac{\lambda G(z)}{l^2} + 1 = 0$$

But, the solution is in the form of $\exp(ik_xx + ik_yx + wt)$, where $r$ is the positve solution of a quadratic equation. Can someone suggest how to proceed further? and Where I have done wrong? Thanks for the attention. Here, the initial condition is $W=0$ for $x=z=0$.

  • $\begingroup$ I think after rearranging you skipped the 1 at some point. It must be included in one of the equations, for $G$ or for $Q$. $\endgroup$ – denklo Nov 7 '18 at 13:12
  • $\begingroup$ Where is $t$ coming from in your solution? $\endgroup$ – Andrei Nov 7 '18 at 15:39
  • $\begingroup$ it basically is the arbitary phase $\phi$ which I take it as $\omega t$. $\endgroup$ – newstudent Nov 7 '18 at 15:52
  • $\begingroup$ $W(x,z)=0$ is solution of the PDE and agrees with the condition $W(0,0)=0$. So, is is a trivial answer to your question. Are you sure that the wording of your question is exactly what is asked for in the original wording of the problem ? $\endgroup$ – JJacquelin Nov 8 '18 at 8:25

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