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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 $:)$

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There is something weird with your notation. You say $b = \sqrt{1 - \lambda}$ when $\lambda > 1$, so $b$ is purely imaginary. But you set $m = 1 \pm jb$, so $m$ would be real (if I understand correctly that $j = \sqrt{-1}$). –  Tunococ Aug 23 '12 at 0:56
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Anyway, I'll just guess you mean $b = \sqrt{\lambda - 1}$ when $\lambda > 1$. In that case, you already assume that $b$ is real, so $1 + b^2 = 0$ cannot happen. –  Tunococ Aug 23 '12 at 0:57
    
@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! –  Serge Aug 23 '12 at 15:35
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