# Eigenvalues of a Sturm-Liouville problem

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.

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.

• You've been busy this a.m.! +1 – Namaste Nov 24 '13 at 15:03
• 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? – CAF Nov 24 '13 at 17:48
• Are you now asking about a different part of the problem? – Amzoti Nov 24 '13 at 17:51
• 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 $\lambda > 0$. – CAF Nov 24 '13 at 17:58
• 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? – Amzoti Nov 24 '13 at 18:19