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# 5.1: Sturm-Liouville problems

• • Contributed by Jiří Lebl
• Assistant Professor (Mathematics) at Oklahoma State University

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### 5.1.1 Boundary value problems

We have encountered several diﬀerent eigenvalue problems such as:

$X''(x)+ \lambda X(x)=0$

with diﬀerent boundary conditions

$$X(0)=0~~~~~X(L)=0~~~~~{\rm{(Dirichlet)~or,}}$$

$$X'(0)=0~~~~~X'(L)=0~~~~~{\rm{(Neumann)~or,}}$$

$$X'(0)=0~~~~~X(L)=0~~~~~{\rm{(Mixed)~or,}}$$

$$X(0)=0~~~~~X'(L)=0~~~~~{\rm{(Mixed),...}}$$

For example for the insulated wire, Dirichlet conditions correspond to applying a zero temperature at the ends, Neumann means insulating the ends, etc…. Other types of endpoint conditions also arise naturally, such as the Robin boundary conditions

$hX(0)-X'(0)=0~~~~~ hX(L)+X'(L)=0,$

for some constant $$h$$. These conditions come up when the ends are immersed in some medium.

Boundary problems came up in the study of the heat equation $$u_t=ku_{xx}$$ when we were trying to solve the equation by the method of separation of variables. In the computation we encountered a certain eigenvalue problem and found the eigenfunctions $$X_n(x)$$. We then found the eigenfunction decomposition of the initial temperature $$f(x)=u(x,0)$$ in terms of the eigenfunctions

$f(x)= \sum_{n=1}^{\infty}c_nX_n(x).$

Once we had this decomposition and found suitable $$T_n(t)$$ such that $$T_n(0)=1$$ and $$T_n(t)X(x)$$ were solutions, the solution to the original problem including the initial condition could be written as

$u(x,t)= \sum_{n=1}^{\infty}c_nT_n(t)X_n(x).$

We will try to solve more general problems using this method. First, we will study second order linear equations of the form

$\frac{d}{dx}\left( p(x)\frac{dy}{dx} \right)-q(x)y+\lambda r(x)y=0.$

Essentially any second order linear equation of the form $$a(x)y''+b(x)y'+c(x)y+\lambda d(x)y=0$$ can be written as (5.1.5) after multiplying by a proper factor.

Example $$\PageIndex{1}$$: Sturm-Liouville Problem

Put the following equation into the form (5.1.5):

$x^2y''+xy'+(\lambda x^2-n^2)y=0.$

Multiply both sides by $$\frac{1}{x}$$ to obtain

$\frac{1}{x}(x^2y''+xy'+(\lambda x^2-n^2)y)=xy''+y'+ \left( \lambda x -\frac{n^2}{x}\right)y= \frac{d}{dx}\left( x \frac{dy}{dx} \right)-\frac{n^2}{x}y+\lambda xy=0.$

The so-called Sturm-Liouville problem1 is to seek nontrivial solutions to

$\frac{d}{dx}\left( p(x)\frac{dy}{dx} \right)-q(x)y+\lambda r(x)y=0,~~~~~a<x<b, \\ \alpha_1y(a)-\alpha_2y'(a)=0, \\ \beta_1y(b)+\beta_2y'(b)=0.$

In particular, we seek $$\lambda$$s that allow for nontrivial solutions. The $$\lambda$$s that admit nontrivial solutions are called the eigenvalues and the corresponding nontrivial solutions are called eigenfunctions. The constants $$\alpha_1$$ and $$\alpha_2$$ should not be both zero, same for $$\beta_1$$ and $$\beta_2$$.

Theorem 5.1.1. Suppose $$p(x), p'(x),q(x)$$ and $$r(x)$$ are continuous on $$[a,b]$$ and suppose $$p(x)>0$$ and $$r(x)>0$$ for all $$x$$ in $$[a,b]$$. Then the Sturm-Liouville problem (5.1.8) has an increasing sequence of eigenvalues

$\lambda_1<\lambda_2<\lambda_3< \cdots$

such that

$\lim_{n \rightarrow \infty} \lambda_n= +\infty$

and such that to each $$\lambda_n$$ there is (up to a constant multiple) a single eigenfunction $$y_n(x)$$.

Moreover, if $$q(x) \geq 0$$ and $$\alpha_1, \alpha_2,\beta_1, \beta_2 \geq 0$$, then $$\lambda_n \geq 0$$ for all $$n$$.

Problems satisfying the hypothesis of the theorem are called regular Sturm-Liouville problems and we will only consider such problems here. That is, a regular problem is one where $$p(x), p'(x),q(x)$$ and $$r(x)$$ are continuous, $$p(x)>0$$, $$r(x)>0$$, $$q(x) \geq 0$$, and $$\alpha_1, \alpha_2,\beta_1, \beta_2 \geq 0$$. Note: Be careful about the signs. Also be careful about the inequalities for $$r$$ and $$p$$, they must be strict for all $$x$$!

When zero is an eigenvalue, we usually start labeling the eigenvalues at $$0$$ rather than at $$1$$ for convenience.

Example $$\PageIndex{2}$$:

The problem $$y''+ \lambda y, 0<x<L,y(0)=0$$, and $$y(L)=0$$ is a regular Sturm-Liouville problem. $$p(x)=1,q(x)=0,r(x)=1$$, and we have $$p(x)1>0$$ and $$r(x)1>0$$. The eigenvalues are $$\lambda_n=\frac{n^2 \pi^2}{L^2}$$ and eigenfunctions are $$y_n(x)=\sin(\frac{n \pi}{L}x)$$. All eigenvalues are nonnegative as predicted by the theorem.

Exercise $$\PageIndex{1}$$:

Find eigenvalues and eigenfunctions for

$y''+\lambda y=0,~~~~~y'(0)=0,~~~~~y'(1)=0.$

Identify the $$p,q,r,\alpha_j,\beta_j$$. Can you use the theorem to make the search for eigenvalues easier? (Hint: Consider the condition $$-y'(0)=0$$)

Example $$\PageIndex{3}$$:

Find eigenvalues and eigenfunctions of the problem

$y''+\lambda y=0,~~~~~0<x<1, \\ hy(0)-y'(0)=0,~~~~~y'(1)=0,~~~~~h>0.$

These equations give a regular Sturm-Liouville problem.

Exercise $$\PageIndex{2}$$:

Identify $$p,q,r,\alpha_j,\beta_j$$ in the example above.

First note that $$\lambda \geq 0$$ by Theorem 5.1.1. Therefore, the general solution (without boundary conditions) is

$y(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x)~~~~~~\rm{if~}\lambda >0,$

$y(x)=Ax + B~~~~~~~~~~~~~~~\rm{if~}\lambda =0.$

Let us see if $$\lambda=0$$ is an eigenvalue: We must satisfy $$0=hB-A$$ and $$A=0$$, hence $$B=0$$ (as $$h>0$$), therefore, $$0$$ is not an eigenvalue (no nonzero solution, so no eigenfunction).

Now let us try $$h>0$$. We plug in the boundary conditions.

$0=hA- \sqrt{\lambda}B, \\ 0=-A\sqrt{\lambda}\sin(\sqrt{\lambda})+B\sqrt{\lambda}\cos(\sqrt{\lambda}).$

If $$A=0$$, then $$B=0$$ and vice-versa, hence both are nonzero. So $$B=\frac{hA}{\sqrt{\lambda}}$$, and $$0=-A \sqrt{\lambda}\sin(\sqrt{\lambda})+\frac{hA}{\sqrt{\lambda}}\sqrt{\lambda}\cos(\sqrt{\lambda})$$. As $$A \neq 0$$ we get

$0=- \sqrt{\lambda}\sin(\sqrt{\lambda})+h\cos(\sqrt{\lambda}),$

or

$\frac{h}{\sqrt{\lambda}}= \tan \sqrt{\lambda}.$

Now use a computer to ﬁnd $$\lambda_n$$. There are tables available, though using a computer or a graphing calculator is far more convenient nowadays. Easiest method is to plot the functions $$\frac{h}{x}$$ and $$\tan(x)$$ and see for which they intersect. There is an inﬁnite number of intersections. Denote by $$\sqrt{\lambda_1}$$ the ﬁrst intersection, by $$\sqrt{\lambda_2}$$ the second intersection, etc…. For example, when $$h=1$$, we get that $$\sqrt{\lambda_1}\approx 0.86, \sqrt{\lambda_2}\approx 3.43, ...$$. That is $$y_1 \approx 0.74,y_2 \approx 11.73,...$$, …. A plot for $$h=1$$ is given in Figure 5.1. The appropriate eigenfunction (let $$A=1$$ for convenience, then $$B= \frac{h}{\sqrt{\lambda}}$$) is

$y_n(x)=\cos(\sqrt{\lambda_n}x)+\frac{h}{\sqrt{\lambda_n}}\sin(\sqrt{\lambda_n}x).$

When $$h=1$$ we get (approximately)

$y_1(x) \approx \cos(0.86x)+ \frac{1}{0.86} \sin(0.86x),~~~~~y_2(x) \approx \cos(3.43x)+ \frac{1}{3.43} \sin(3.43x),~~~~~....$ Figure 5.1: Plot of $$\frac{1}{x}$$ and $$\tan x$$.

### 5.1.2 Orthogonality

We have seen the notion of orthogonality before. For example, we have shown that $$\sin(nx)$$ are orthogonal for distinct $$n$$ on $$[0, \pi]$$. For general Sturm-Liouville problems we will need a more general setup. Let $$r(x)$$ be a weight function(any function, though generally we will assume it is positive) on $$[a, b]$$. Two functions $$f(x)$$, $$g(x)$$ are said to be orthogonal with respect to the weight function $$r(x)$$ when

$\int_a^bf(x)g(x)r(x)dx=0.$

In this setting, we deﬁne the inner product as

$\langle f,g \rangle \overset{\rm{def}}= \int_a^bf(x)g(x)r(x)dx,$

and then say $$f$$ and $$g$$ are orthogonal whenever $$\langle f,g \rangle=0$$. The results and concepts are again analogous to ﬁnite dimensional linear algebra.

The idea of the given inner product is that those $$x$$ where $$r(x)$$ is greater have more weight. Nontrivial (nonconstant) $$r(x)$$ arise naturally, for example from a change of variables. Hence, you could think of a change of variables such that $$d \xi =r(x)dx$$.

We have the following orthogonality property of eigenfunctions of a regular Sturm-Liouville problem.

Theorem 5.1.2. Suppose we have a regular Sturm-Liouville problem

$\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=0, \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0.$

Let $$y_j$$ and $$y_k$$ be two distinct eigenfunctions for two distinct eigenvalues $$\lambda_j$$ and $$\lambda_k$$. Then

$\int_a^by_j(x)y_k(x)r(x)dx=0,$

that is, $$y_j$$ and $$y_k$$ are orthogonal with respect to the weight function $$r$$.

Proof is very similar to the analogous theorem from § 4.1. It can also be found in many books including, for example, Edwards and Penney [EP].

### 5.1.3 Fredholm alternative

We also have the Fredholm alternative theorem we talked about before for all regular Sturm-Liouville problems. We state it here for completeness.

Theorem 5.1.3 (Fredholm alternative). Suppose that we have a regular Sturm-Liouville problem. Then either

$\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=0, \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0,$

has a nonzero solution, or

$\frac{d}{dx} \left( p(x) \frac{dy}{dx} \right) -q(x)y+\lambda r(x)y=f(x), \\ \alpha_1y(a)- \alpha_2y'(a)=0, \\ \beta_1y(b)+ \beta_2y'(b)=0,$

has a unique solution for any $$f(x)$$ continuous on $$[a,b]$$.

This theorem is used in much the same way as we did before in § 4.4. It is used when solving more general nonhomogeneous boundary value problems. The theorem does not help us solve the problem, but it tells us when a unique solution exists, so that we know when to spend time looking for it. To solve the problem we decompose $$f(x)$$ and $$y(x)$$ in terms of the eigenfunctions of the homogeneous problem, and then solve for the coeﬃcients of the series for $$y(x)$$.

### 5.1.4 Eigenfunction series

What we want to do with the eigenfunctions once we have them is to compute the eigenfunction decomposition of an arbitrary function $$f(x)$$. That is, we wish to write

$f(x)= \sum_{n=1}^{\infty}c_ny_n(x),$

where $$y_n(x)$$ the eigenfunctions. We wish to ﬁnd out if we can represent any function $$f(x)$$ in this way, and if so, we wish to calculate (and of course we would want to know if the sum converges). OK, so imagine we could write $$f(x)$$ as (5.1.24). We will assume convergence and the ability to integrate the series term by term. Because of orthogonality we have

$\langle f,y_m \rangle = \int_a^bf(x)y_m(x)r(x)dx \\ = \sum_{n=1}^{\infty}c_n \int_a^by_n(x)y_m(x)r(x)dx \\ =c_m \int_a^by_m(x)y_m(x)r(x)dx= c_m \langle y_m,y_m \rangle .$

Hence,

$c_m= \frac{\langle f,y_m \rangle}{\langle y_m,y_m \rangle}= \frac{\int_a^bf(x)y_m(x)r(x)dx}{\int_a^b(y_m(x))^2r(x)dx}.$

Note that $$y_m$$ are known up to a constant multiple, so we could have picked a scalar multiple of an eigenfunction such that $$\langle y_m,y_m \rangle=1$$ (if we had an arbitrary eigenfunction $$\tilde{y}_m$$, divide it by $$\sqrt{\langle \tilde{y}_m,\tilde{y}_m \rangle}$$). When $$\langle y_m,y_m \rangle=1$$ we have the simpler form $$c_m=\langle f,y_m \rangle$$ as we did for the Fourier series. The following theorem holds more generally, but the statement given is enough for our purposes.

Theorem 5.1.4. Suppose $$f$$ is a piecewise smooth continuous function on . If $$y_1,y_2, \ldots$$ are the eigenfunctions of a regular Sturm-Liouville problem, then there exist real constants $$c_1,c_2, \ldots$$ given by (5.1.26) such that (5.1.24) converges and holds for $$a<x<b$$.

Example $$\PageIndex{4}$$:

Take the simple Sturm-Liouville problem

$y''+ \lambda y=0, ~~~~~0<x<\frac{\pi}{2}, \\ y(0)=0,~~~~~ y' \left( \frac{\pi}{2}\right)=0.$

The above is a regular problem and furthermore we know by Theorem 5.1.1 that $$\lambda \geq 0$$.

Suppose $$\lambda = 0$$, then the general solution is $$y(x)Ax+B$$, we plug in the initial conditions to get $$0=y(0)=B$$, and $$0=y'(\pi/2)=A$$, hence $$\lambda=0$$ is not an eigenvalue. The general solution, therefore, is

$y(x)=A\cos(\sqrt{\lambda}x)+B\sin(\sqrt{\lambda}x).$

Plugging in the boundary conditions we get $$0=y(0)=A$$ and $$0=y'(\pi/2)=\sqrt{\lambda}B\cos(\sqrt{\lambda}\frac{\pi}{2})$$. $$B$$ cannot be zero and hence $$\cos(\sqrt{\lambda}\frac{\pi}{2}=0)$$. This means that $$\sqrt{\lambda}\frac{\pi}{2}$$ must be an odd integral multiple of $$\frac{\pi}{2}$$, i.e. $$(2n-1)\frac{\pi}{2}=\sqrt{\lambda_n}\frac{\pi}{2}$$. Hence

$\lambda_n=(2n-1)^2.$

We can take $$B=1$$. Hence our eigenfunctions are

$y_n(x)= \sin((2n-1)x).$

Finally we compute

$\int_0^{\frac{\pi}{2}}(\sin((2n-1)x))^2dx=\frac{\pi}{4}.$

So any piecewise smooth function on $$[0, \pi/2]$$ can be written as

$f(x)=\sum_{n=1}^{\infty}c_n\sin((2n-1)x),$

where

$c_n= \frac{\langle f,y_n \rangle}{\langle y_n,y_n \rangle}= \frac{\int_0^{\frac{\pi}{2}}\sin((2n-1)x)dx}{\int_0^{\frac{\pi}{2}}(\sin((2n-1)x))^2dx}= \frac{4}{\pi} \int f(x)\sin((2n-1)x)dx.$

Note that the series converges to an odd $$2\pi$$-periodic (not $$\pi$$ -periodic!) extension of $$f(x)$$.

Exercise $$\PageIndex{3}$$

In the above example, the function is deﬁned on $$0<x< \pi/2$$, yet the series converges to an odd $$2\pi$$-periodic extension of $$f(x)$$. Find out how is the extension deﬁned for $$\pi/2 <x< \pi$$.

1Named after the French mathematicians Jacques Charles François Sturm (1803–1855) and Joseph Liouville (1809–1882).