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I'm having trouble understanding some of the concepts related to these problems. Here's an example I'm working on:

$$y''+(\lambda+1)y=0 ; y'(0)=0,y'(1)=0$$

The characteristic equation I found was given by: $m^2 +\lambda +1=0$, which gave me $$y(x)=c_1 \cos(\sqrt{-\lambda+1}x) +c_2 \sin(\sqrt{-\lambda+1}x) $$ and

$$y'(x)= -\sqrt{-\lambda+1} c_1 \sin(\sqrt{-\lambda+1}x) + \sqrt{-\lambda+1} c_2 \cos(\sqrt{-\lambda+1}x) $$

Plugging in the boundary values I get: $y'(0)=0$ which gives $ c_2=0$ and for $y'(1)=0$, $-\sqrt{-\lambda+1} c_1 \sin(\sqrt{-\lambda+1})=0$

But I'm not sure what to do past this point. I'm not sure what it is I'm looking for. Any insight would be great.

mweiss
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Dimitri
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2 Answers2

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You're almost there. $$ y'' + (\lambda + 1) y = 0 \implies m^2 + (\lambda + 1) = 0 \implies m = \pm i\mu \implies \\ y = \left \{ \begin{array}{ll} A \cos \mu x + B \sin \mu x, && \text{if } \lambda \ne -1 \\ Dx + C, && \text{if } \lambda = -1 \end{array} \right . $$ where $\mu = \sqrt{\lambda + 1}$.

After applying BCs, second solution becomes trivial as $y = C$, so I assume $\lambda \ne -1$, then $$ y' = -\mu \left( -A\sin \mu x + B \cos \mu x\right) $$

Now, use BCs per your post, $$ y'(0) = 0 \implies B = 0,\quad \text{so}\ y' = -\mu A \sin \mu x \\ y'(1) = 0 \implies \mu A \sin \mu = 0 \implies \mu_n = \pi n $$ where $n \in (0, \infty)$. One can find that $\lambda_n = n^2 \pi^2 - 1$.

Trivial solution can be incorporated into this solution if we take $A_0 = C$, so $$ y = A_n \cos \mu_n x = A_n \cos n \pi x, \qquad n \in [0, +\infty). $$

Kaster
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    In the case $\lambda = -1$ the equation is $y''=0$ which yields the solution $y(x) = ax+b$. – Leucippus Jun 30 '14 at 23:45
  • @Leucippus I used wrong place to put it. Point is after you apply BCs you get that $a = 0$, so $y = b$. – Kaster Jun 30 '14 at 23:49
  • @Kaster : This is very true. Applying the boundary conditions did slip past me. Nice solution. – Leucippus Jun 30 '14 at 23:53
  • I messed up in the final conclusion. Principle of superposition doesn't apply here. Well, it does in general, but after using BC all coefficients except one should vanish. Will reflect that in the answer. – Kaster Jul 01 '14 at 00:07
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You're looking for such $\lambda$ that $-\sqrt{-\lambda-1} c_1 \sin(\sqrt{-\lambda-1})=0$, so you must have $\sqrt{-\lambda-1}=k\pi$ for $k \in \mathbb{Z}$, so $-\lambda=k^2\pi^2+1$

agha
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  • If that's the eigenvalue, how do you find the eigenfunction in this case? – Dimitri Jun 30 '14 at 22:18
  • Substitute $c_2=0$ and $\lambda=k^2\pi^2+1$ in your formula. – agha Jun 30 '14 at 22:20
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    It should be $\lambda = -(1+k^{2} \pi^{2})$ where $k = 0, \pm 1, \pm 2, \cdots$ – Leucippus Jun 30 '14 at 22:26
  • @Leucippus negative sign was absorbed by the $\sinh \to \sin$ and $\cosh \to \cos$ conversion. So it should be $\lambda_k = -1 + k^2 \pi^2$. Also, $k = a$ and $k = -a$ both give same eigenfunction, so only $k \ge 0$ will generate linearly independent eigenfunctions. – Kaster Jun 30 '14 at 22:30