Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the Sturm-Liouville problem

$$-u''(x) + V(x)u(x) = \lambda u(x)$$

for $u : \mathbb R \rightarrow \mathbb R$, $\lambda \in \mathbb R$. I am looking for a method to find the eigenvalues $\lambda$ which produce solutions $u \in L^2(\mathbb R)$ for various choices of $V$, e.g. $V(x) = c/\cosh^2(Cx)$ or $V(x) = e^x$. Is there a general procedure for doing this? If not, can someone provide suggestions for finding the eigenvalues at least in the two cases I mentioned?

share|cite|improve this question
Yes, there is such a procedure. I recall having read something related to this in the classical book on ordinary differential equations by Coddington and Levinson. See here. – Giuseppe Negro Jan 2 '13 at 18:37
up vote 3 down vote accepted

Numerically the problem can be solved e.g. using the Numerov algorithm with the shooting method. There are only a few choices for $V$ for which there is an explicit analytic solution, some of them you can find here.

One of it is $$V(x) = \frac{c}{\cosh^2(C x)}$$ which is called the Pöschl–Teller potential. However, it only has solutions in $L^2$ if $c<0$. After proper rescaling, you will find that the eigensolutions solutions are Legendre functions.

When $V(x) =e^{x}$ I believe that the eigenvalue problem does not have any solution in $L^2$.

share|cite|improve this answer
Nice! I wasn't familiar with the Ploeschl-Teller potential. – Ron Gordon Jan 3 '13 at 0:21

The eigenvalues $\lambda$ emerge from boundary conditions that are part of the differential operator. In your case, $x \in (-\infty, \infty)$, so you may specify the behavior of $u$ as $x \rightarrow \infty$. One way to attack this is to consider the Fourier transform of $u$:

$$ \hat{u}(k) = \int_{-\infty}^{\infty} dx \: u(x) \exp{(i 2 \pi k x)} $$

Applying the integral operator to the S-L equation, and integrating by parts (and assuming that $u$ and its derivative vanish at $\infty$), we get an integral equation for $ \hat{u}$:

$$ \int_{-\infty}^{\infty} dk' \: \hat{V}(k-k') \hat{u}(k') = (\lambda + 4 \pi^2 k^2) \hat{u}(k) $$


$$ \hat{V}(k) = \int_{-\infty}^{\infty} dx \: V(x) \exp{(i 2 \pi k x)} $$

and we recover the solution from

$$ u(x) = \int_{-\infty}^{\infty} dk \: \hat{u}(k) \exp{(-i 2 \pi k x)} $$

share|cite|improve this answer
How does one solve the integral equation you get after applying the fourier transform? – user15464 Jan 2 '13 at 18:54
You can approximate the convolution by a matrix product. Be warned, however, that the functions $V$ must be $L^2 (\mathbb{R})$. – Ron Gordon Jan 2 '13 at 18:59
Can it be solved explicitly for the choices of $V$ I mentioned in the original post? – user15464 Jan 2 '13 at 19:26
The first, yes, the second, no. But with the second (or anything that is not $L^2(\mathbb{R})$), no technique will work. For those functions, you need to work over a finite interval; in such cases, other techniques may be more appropriate. – Ron Gordon Jan 2 '13 at 19:28
How would I go about solving it for $V(x) = c/\cosh^2(cx)$? – user15464 Jan 2 '13 at 19:43

There is a technique called "AIM" which stands for "asymptotic iteration method". See for instance this reference 1 and reference 2.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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