A “Proof” of the Sturm-Liouville Theorem/Problem

IMAGE CREDIT: NASA/JPL: This shows Jupiter’s Great Red Spot; a storm that has been occurring for over 300 years now. Quite recently, however, observations show that the Spot appears to be shrinking in size. 

About a week ago, I was looking through my notebooks and came across an unfinished problem posed by one of my professors. Unfortunately, I was not able to solve the problem during the semester. However, I thought it might be something interesting to consider. I did a quick search and found that the problem he gave us was to prove the Sturm-Liouville Theorem.

The main brute-force method to analytically solving a given partial differential equation is the separation of variables. This method is heavily used by physicists and in doing so transforms the initial boundary value problem (IBVP) into a Sturm-Liouville problem in which we have an ordinary differential equation and linear homogeneous boundary conditions.

A quick statement of the problem:

Consider the following Sturm-Liouville Problem:

\displaystyle \frac{d}{dx}\bigg\{p(x)\frac{d\phi(x)}{dx}\bigg\}-q(x)\phi(x)+\lambda^{2}w(x)\phi(x)=0 (1.1)

\displaystyle \alpha_{1}\phi(L)+\alpha_{2}\frac{d\phi(L)}{dx}=0 (1.2)

\displaystyle \beta_{1}\phi(R)+\beta_{2}\frac{d\phi(R)}{dx}=0 (1.3)

In the context of the problem, let \displaystyle p(x)=1, q(x)=0, w(x)=1. About the \displaystyle \lambda^{2}, I am aware that the original differential equation had a single \displaystyle \lambda. However, I am not sure of the implications to the problem by removing one of the \displaystyle \lambda, so I am including the problem as stated by my professor.  So, Eq.(1.1) becomes

\displaystyle \frac{d^{2}\phi(x)}{dx^{2}}+\lambda^{2}\phi(x)=0 (1.4)

So the problem is to prove that if there exists two unique non-trivial eigenfunctions, \displaystyle \phi_{n}(x),\phi_{m}(x) such that \lambda_{n}\neq \lambda_{m} and that for n\neq m, then on the interval [L,R]

\displaystyle \int_{L}^{R}\phi_{n}(x)\phi_{m}(x)dx = 0. (1.5)

The problem also has the properties

  • An infinite sequence of eigenvalues: \displaystyle \lambda_{1} < \lambda_{2} < ... < \lambda_{n} < ... < \infty.
  • For each eigenvalue \lambda_{n} there exists a unique, nontrivial (i.e. non-zero) solution \displaystyle \phi_{n}(x), not including multiples of said solution.

Much of the background information (e.g. the properties and general form of the problem) was obtained from the following sources:


Farlow, S.J., Partial Differential Equations for Scientists and Engineers. Dover. 1993.

“Proof”*.  Recall from above the Eqs. (1.4), (1.2), and (1.3). Let the following equations be indexed for n:

\displaystyle \frac{d^{2}\phi_{n}(x)}{dx^{2}}+\lambda_{n}^{2}\phi_{n}(x)=0  (i.)

\displaystyle \alpha_{1}\phi_{n}(L)+\alpha_{2}\frac{d\phi_{n}(L)}{dx}=0  (ii.)

\displaystyle \beta_{1}\phi_{n}(R)+\beta_{2}\frac{d\phi_{n}(R)}{dx}=0  (iii.)

Similarly,  let the following equations be indexed for m:

\displaystyle \frac{d^{2}\phi_{m}(x)}{dx^{2}}+\lambda_{m}^{2}\phi_{m}(x)=0  (iv.)

\displaystyle \alpha_{1}\phi_{m}(L)+\alpha_{2}\frac{d\phi_{n}(L)}{dx}=0  (v.)

\displaystyle \beta_{1}\phi_{m}(R)+\beta_{2}\frac{d\phi_{m}(R)}{dx}=0  (vi.)

Consider Eqs.(i) and (iv.), let us multiply Eq.(i) by \phi_{m}(x) and the same with Eq.(iv.), however we multiply by \phi_{n}(x). Taking the difference yields

\displaystyle \phi_{n}^{\prime\prime}(x)\phi_{m}(x)-\phi_{m}^{\prime\prime}(x)\phi_{n}(x)+(\lambda_{n}^{2}-\lambda_{m}^{2})\phi_{n}(x)\phi_{m}(x) = 0 (2)

Let us now integrate Eq.(2) over the closed interval [L,R] to give

\displaystyle \int_{L}^{R}\phi_{n}^{\prime\prime}(x)\phi_{m}(x)-\phi_{m}^{\prime\prime}(x)\phi_{n}(x)dx+(\lambda_{n}^{2}-\lambda_{m}^{2})\int_{L}^{R}\phi_{n}(x)\phi_{m}(x)dx = 0 (3)

Let us focus on the first integral on the left-hand side.  We can separate each term to give

\displaystyle \int_{L}^{R}\phi_{n}^{\prime\prime}(x)\phi_{m}(x)dx-\int_{L}^{R}\phi_{m}^{\prime\prime}(x)\phi_{n}(x)dx (4)

We can integrate by parts for the first term

\displaystyle \int_{L}^{R}\phi_{n}^{\prime\prime}(x)\phi_{m}(x)dx= \bigg\{\phi_{n}^{\prime\prime}(x)\phi_{m}(x)\bigg\|_{L}^{R}\bigg\}-\int_{L}^{R}\phi_{m}^{\prime}(x)\phi_{n}(x)dx. (5)

and doing the same for the second term we arrive at

\displaystyle \int_{L}^{R}\phi_{m}^{\prime\prime}(x)\phi_{n}(x)dx=\bigg\{\phi_{m}^{\prime\prime}(x)\phi_{n}(x)\bigg\|_{L}^{R}\bigg\}-\int_{L}^{R}\phi_{n}^{\prime}(x)\phi_{m}(x)dx. (6)

Finding the difference, we see that the remaining integrals in (5) and (6) cancel leaving the terms in the braces. If we evaluate those terms over the interval [L,R] we get the following

\displaystyle \bigg\{\phi_{n}^{\prime}(R)\phi_{m}(R)-\phi_{n}(L)\phi_{m}(L)-\phi_{m}^{\prime}(R)\phi_{n}(R)+\phi_{n}(L)\phi_{m}^{\prime}(L)\bigg\}+(\lambda_{n}^{2}-\lambda_{m}^{2})\int_{L}^{R}\phi_{n}(x)\phi_{m}(x)dx = 0 (7).

Let us suppose now that we equate Eqs.(ii) and (v.) and Eqs.(iii) and (vi) and multiply equations (ii) and (v) by \phi_{m}(L) and \phi_{n}(L), respectively, and multiply equations (iii) and (vi) by \phi_{m}(R) and \phi_{n}(R). For Eqs.(ii) and (v) it follows that

\displaystyle \alpha_{1}(\phi_{n}(L)\phi_{m}(L)-\phi_{m}(L)\phi_{n}(L))+\alpha_{2}(\phi_{n}^{\prime}(L)\phi_{m}(L)-\phi_{m}^{\prime}(L)\phi_{n}(L))=0, (8.1)

From which we observe that if we assume that \alpha_{1},\alpha_{2}\in \mathbb{R}|\alpha_{1},\alpha_{2}\neq 0, then

\displaystyle (\phi_{n}(L)\phi_{m}(L)-\phi_{m}(L)\phi_{n}(L))+(\phi_{n}^{\prime}(L)\phi_{m}(L)-\phi_{n}(L)\phi_{m}^{\prime}(L))=0

The first term in the sum above vanishes which leads us to conclude that

\displaystyle \phi_{n}^{\prime}(L)\phi_{m}(L)-\phi_{n}(L)\phi_{m}^{\prime}(L)=0 (8.2)

By similar logic we find that for the second boundary condition

\displaystyle \phi_{n}^{\prime}(R)\phi_{m}(R)-\phi_{n}(R)\phi_{m}^{\prime}(R)=0 (9)

Therefore, returning (all the way back) to Eq.(7), we arrive at

\displaystyle (\lambda_{n}^{2}-\lambda_{m}^{2})\int_{L}^{R}\phi_{n}(x)\phi_{m}(x)dx = 0, (10)

where we assume that the eigenvalues satisfy \lambda_{n}^{2}\neq \lambda_{m}^{2}, for n\neq m. As a check, suppose that n=m, in which case the integral (Eq.(10)) becomes

\displaystyle \int_{L}^{R}\bigg\{\phi(x)\bigg\}^{2}dx > 0. (11)

Furthermore, if \lambda_{n}^{2}=\lambda_{m}^{2}, then the entire equation vanishes. Thus, this “proof”is complete. Q.E.D.

* Once again, I hesitate to call this a proof since it is not what a mathematician would call rigorous.

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