Partial differential equation:

$$\left[\left(\frac{\partial\Psi}{\partial\zeta}\right)^2+2\left(\frac{\partial\Psi}{\partial\zeta}\right)+\left(\frac{2}{3}\right)\right]\frac{\partial^2\Psi}{\partial\zeta^2}=\frac{2}{3}\delta\frac{\partial^2\Psi}{\partial t^2}$$

$\delta$: constant

Boundary conditions: $\Psi(0,t)=0$, $\Psi(1,t)=0$

Answer: $\Psi(\zeta,t)=\sum_n a_n\sin(n\pi\zeta)\cos(\lambda t)$

I need to find the Fourier Series expansion of the solution of the equation satisfying the boundary conditions. Actually I have the question and answer, but I could not solve it. Could you please help me with it? I tried to solve it but then got stuck:

$$\Psi(\zeta,t)=A(\zeta)B(t)$$ $$\left[(A')^2+2A'+\frac{2}{3}\right]A''B=\frac{2}{3}\delta AB''$$ $$\frac{A''}{A}\left[(A')^2+2A'+\frac{2}{3}\right]=\frac{2}{3}\delta\frac{B''}{B}=-L$$

$$\begin{cases}A''\left[(A')^2+2A'+\frac{2}{3}\right]+LA=0 \\ B''+\frac{3L}{2\delta}B=0.\end{cases}$$

I do not know how I should continue.

  • 1
    $\begingroup$ The variables are going to frighten me. $\psi $ and all ;p $\endgroup$ – ABC Mar 24 '13 at 2:35
  • $\begingroup$ if you're able to solve it I could have written it more pretty.) $\endgroup$ – Mehmet Eren Mar 24 '13 at 15:13
  • 3
    $\begingroup$ Fourier method is really not the best for this question. Your equation is quasilinear, and not linear. The Fourier method works well with separation of variables in the linear case because there we can use the principle of superposition to add together solutions corresponding to individual Fourier modes to get a solution in general form. In this case your equation for $A$ is nonlinear. $\endgroup$ – Willie Wong Mar 25 '13 at 16:53
  • $\begingroup$ The displayed equation up top also doesn't agree with the separation of variables procedure down low. Perhaps you meant $\Psi_\zeta^2 + 2 \Psi_\zeta + 2/3$ instead of $\Psi_\zeta^2 + 2 \Psi_\zeta 2/3$? $\endgroup$ – Willie Wong Mar 25 '13 at 16:54
  • $\begingroup$ (Yes Willie Wong, you're right about the mistake, I corrected it. I guess it changed after had been edited.) And you're also right about the being nonlinear. I talked to my professor. I am going to use Galerkin Method to solve this equation. Thank you for your attention. $\endgroup$ – Mehmet Eren Mar 26 '13 at 12:54

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