2
$\begingroup$

Consider the problem $y'' + \lambda y = 0$ with the following boundary conditions $y'(0)=0,$ $\,\,y(1)+y'(1)=0$. Find the normalized eigenfunctions.

The normalized eigenfunctions are $\phi(n,x) = k_n \cos \sqrt{\lambda_n}\,x$, where $k_n = \left(\frac{2}{1+\sin^2 \sqrt{\lambda_n}}\right)^{1/2}$ This corresponded to the case where $\lambda > 0$. For $\lambda < 0,$ there exists only complex solutions and solutions to Sturm-Liouville problems necessarily have real eigenvalues.

However, for $\lambda = 0$, I obtain the non trivial solution $y = c_2(1-\frac{1}{2}x)$. My book says that $\lambda=0$ is not an eigenvalue, and yet I have found a non-trivial solution (i.e one where $c_1,c_2\neq0)$ Why is this?

Many thanks.

$\endgroup$
4
$\begingroup$

Given:

$$y'' + \lambda y = 0$$

with the boundary conditions $y'(0)=0,$ $\,\,y(1)+y'(1)=0$.

For $\lambda = 0$, we arrive at:

$$y(x) = c_1 + c_2 x$$

We have $y'(x) = c_2$,

Using the boundary conditions, we have:

  • $y'(0) = c_2 = 0 \rightarrow c_2 = 0$
  • $y(1) + y'(1) = c_1 + c_2 + c_2 = 0 \rightarrow c_1 + 2c_2 = 0 \rightarrow c_1 = 0$

This means that we have a trivial solution:

$$y(x) = 0$$

This does not make any contribution to the eigenfunctions.

$\endgroup$
  • $\begingroup$ You've been busy this a.m.! +1 $\endgroup$ – Namaste Nov 24 '13 at 15:03
  • $\begingroup$ Just a quick question: The relation that the eigenvalues satisfy is $\cos \beta = \beta \sin \beta$, where $\beta = \sqrt{\lambda}$. This is apparently equivalent to $\cot \beta = \beta$ in the solution, but I don't see how this is true unless $\sin \beta \neq 0$. How can we be sure of this? $\endgroup$ – CAF Nov 24 '13 at 17:48
  • $\begingroup$ Are you now asking about a different part of the problem? $\endgroup$ – Amzoti Nov 24 '13 at 17:51
  • $\begingroup$ I have the answer to the problem as noted in my opening post, however, initially I glossed over the subtlety above. Yes, sorry, it is another part of the question corresponding to the case $\lambda > 0$. $\endgroup$ – CAF Nov 24 '13 at 17:58
  • $\begingroup$ If I didn't make an error, for $\lambda > 0$, I am getting $a=0$, $b(1-\sqrt{\lambda}) \cos \sqrt{\lambda} = 0$. $b = 0$ is a trivial solution. So, we have two other choices for that relationship to equal zero, either $\cos \sqrt{\lambda} = 0$ or $1-\sqrt{\lambda} = 0$. Is that what you got? $\endgroup$ – Amzoti Nov 24 '13 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.