0

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?

Incognito
  • 555

1 Answers1

1

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.

Davis Yoshida
  • 1,560
  • 8
  • 16
  • 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? – Incognito Apr 09 '14 at 21:43
  • 1
    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. – Davis Yoshida Apr 09 '14 at 23:11
  • 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 $\lambda$. – Incognito Apr 10 '14 at 16:48