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So, I'm trying to solve the following eigenvalue problem for the eigenvalues:

$$u''(x)+\lambda^2u(x)=0$$ $$u(0)=u(1)$$ $$u'(0)=u'(1)$$

Of course, the two eigenvectors are cosine and sine, and the solution which spans the space of all solutions is as follows:

$$u(x)=A\sin(\lambda x)+B\cos(\lambda x)$$

Now, when I try to apply the BC's, I expected that I would obtained some kind of equation involving only the eigenvalue. Then, the solution of eigenvalues is usually the points at which both sides are equivalent. However, I'm finding that there are no eigenvalues which satisfy these conditions? What am I doing incorrectly here?

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  • $\begingroup$ The differential equation is precisely the equation for simple Harmonic motion $\endgroup$
    – user122283
    Apr 9, 2014 at 21:33
  • $\begingroup$ Well, yeah; of course. That's the general solution which I have above. The problem is finding the eigenvalues (or frequency). Any idea why I can't solve for them fromt eh BC's? $\endgroup$
    – Incognito
    Apr 9, 2014 at 21:35
  • $\begingroup$ Do you get the following equation: $\cos^2(\lambda) - 2\cos(\lambda) + 1 + \sin(\lambda)(\cos(\lambda) = 0$ (or something more or less like it)? $\endgroup$
    – Jared
    Apr 9, 2014 at 21:58
  • $\begingroup$ Check this related problem. $\endgroup$
    – Mhenni Benghorbal
    Apr 10, 2014 at 0:29

1 Answer 1

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Given that $u(x) = A\sin{\lambda x} + B\cos{\lambda x}$, you can simply use this definition of $u$ in conjunction with the the boundary conditions.

You have that $u(0) = u(1)$, so then $B = A\sin{\lambda} + B\cos{\lambda}$.

You also have that $u'(0) = u'(1)$, so $A\lambda = A\lambda\cos{\lambda} - B\lambda\sin{\lambda}$. This should be fairly straight forward system of equations to solve.

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  • $\begingroup$ That's what I thought, but, when i go to solve it, I find that no values for lambda solve the BC's. Also, in your first equation, the Bsin should be Bcos. Any idea what I'm doing incorrectly? $\endgroup$
    – Incognito
    Apr 9, 2014 at 21:43
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    $\begingroup$ Whoops, edited. Well, just based off of the periodcity of sine and cosine, it seems that $\lambda = 2k\pi$ for any $k \in \mathbb{Z}$ works. $\endgroup$
    – Davis Yoshida
    Apr 9, 2014 at 23:11
  • $\begingroup$ Thank you. I took a look at this again, and I'm getting that $\lambda=2n \pi$. However, what about the constants for the eigenvectors? Is there anything we can say about them? Trying to reduce them from the equations above lead us to the expression for $\lambda$. $\endgroup$
    – Incognito
    Apr 10, 2014 at 16:48

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