0
$\begingroup$

I am trying to solve the following S-L problem:

$\Phi_{xx}-2\Phi_x+\lambda\Phi=0$

$\Phi_x(-1)=\Phi_x(1)=0$

where $\Phi_{xx}$ is the second derivative of $\Phi$. Professor's solution is available here (in Spanish, I'm afraid).

We first do: $m^2-2m+\lambda=0$ to get $m=1\pm \sqrt{1-\lambda}$. So I have to try out the different values that make m a complex, real or 0.

My question arises when trying to solve the case in which $1-\lambda<0$. Then $\lambda>1$ and I assign, for simplicity, $m=1\pm jb$, where $b=\sqrt{1-\lambda}$.

The solutions must be of the form $\Phi(x)=C_1 e^{jbx}+C_2e^{-jbx}=e^x(A \cos(bx)+B \sin(bx))$.

Applying the boundary conditions, we get to a 2 unknowns, 2 equations linear system. Skipping the necesary algebra, I get that, in order for this to have a non-trivial solution, the following must be true:

$\sin(2b)(1+b^2)=0$

Fortunately, this matches up with the solution provided. Then, the professor only considers the case in which $\sin(2b)=0$. I agree to that. But my question is, why shouldn't I also try to get an eigenvalue from $(1+b^2)=0$?

I mean: $(1+b^2)=0$ so $b=\sqrt{ -1}$. And solving for $\lambda$, $b^2=1-\lambda=-1$ and therefore $\lambda=2$.

Why isn't this $\lambda$ a valid eigenvalue?

Thank you very much for your attention $:)$

  • 0
    @Tunococ Hi, thank you very much. As you say,$b$is actually $\sqrt{\lambda-1}$. Now I see that it also has to be a real number. Thank you!2012-08-23

0 Answers 0