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

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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:

https://en.wikipedia.org/wiki/Sturm%E2%80%93Liouville_theory#Application_to_PDEs

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