4.1: Sturm-Liouville Operators

In physics many problems arise in the form of boundary value problems involving second order ordinary differential equations. For example, we will explore the wave equation and the heat equation in three dimensions. Separating out the time dependence leads to a three dimensional boundary value problem in both cases. Further separation of variables leads to a set of boundary value problems involving second order ordinary differential equations.

In general, we might obtain equations of the form

$$a_{2}(x) y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{0}(x) y=f(x)\label{eq:1}$$

subject to boundary conditions. We can write such an equation in operator form by defining the differential operator

$$L=a_{2}(x) D^{2}+a_{1}(x) D+a_{0}(x),\nonumber$$

where $D=d / d x$. Then, Equation $\eqref{eq:1}$ takes the form

$$L y=f \text {. }\nonumber$$

Recall that we had solved such nonhomogeneous differential equations in Chapter 2. In this section we will show that these equations can be solved using eigenfunction expansions. Namely, we seek solutions to the eigenvalue problem

$$L \phi=\lambda \phi\nonumber$$

with homogeneous boundary conditions on $\phi$ and then seek a solution of the nonhomogeneous problem, $L y=f$, as an expansion over these eigenfunctions. Formally, we let

$$y(x)=\sum_{n=1}^{\infty} c_{n} \phi_{n}(x) .\nonumber$$

However, we are not guaranteed a nice set of eigenfunctions. We need an appropriate set to form a basis in the function space. Also, it would be nice to have orthogonality so that we can easily solve for the expansion coefficients.

It turns out that any linear second order differential operator can be turned into an operator that possesses just the right properties (self-adjointedness) to carry out this procedure. The resulting operator is referred to as a SturmLiouville operator. We will highlight some of the properties of these operators and see how they are used in applications.

We define the Sturm-Liouville operator as

$$\mathcal{L}=\frac{d}{d x} p(x) \frac{d}{d x}+q(x) .\label{eq:2}$$

The Sturm-Liouville eigenvalue problem is given by the differential equation

$$\mathcal{L}=-\lambda\sigma (x)y,\nonumber$$

or

$$\frac{d}{d x}\left(p(x) \frac{d y}{d x}\right)+q(x) y+\lambda \sigma(x) y=0,\label{eq:3}$$

for $x \in(a, b), y=y(x)$, plus boundary conditions. The functions $p(x), p^{\prime}(x)$, $q(x)$ and $\sigma(x)$ are assumed to be continuous on $(a, b)$ and $p(x)>0, \sigma(x)>0$ on $[a, b]$. If the interval is finite and these assumptions on the coefficients are true on $[a, b]$, then the problem is said to be a regular Sturm-Liouville problem. Otherwise, it is called a singular Sturm-Liouville problem.

We also need to impose the set of homogeneous boundary conditions

$$\begin{align} &\alpha_{1} y(a)+\beta_{1} y^{\prime}(a)=0, \nonumber \\ &\alpha_{2} y(b)+\beta_{2} y^{\prime}(b)=0 .\label{eq:4} \end{align}$$

The $\alpha^{\prime}$ s and $\beta^{\prime}$ are constants. For different values, one has special types of boundary conditions. For $\beta_{i}=0$, we have what are called Dirichlet boundary conditions. Namely, $y(a)=0$ and $y(b)=0$. For $\alpha_{i}=0$, we have Neumann boundary conditions. In this case, $y^{\prime}(a)=0$ and $y^{\prime}(b)=0$. In terms of the heat equation example, Dirichlet conditions correspond to maintaining a fixed temperature at the ends of the rod. The Neumann boundary conditions would correspond to no heat flow across the ends, or insulating conditions, as there would be no temperature gradient at those points. The more general boundary conditions allow for partially insulated boundaries.

Note

• Dirichlet boundary conditions - the solution takes fixed values on the boundary. These are named after Gustav Lejeune Dirichlet ( $1805-1859$ ).
• Neumann boundary conditions - the derivative of the solution takes fixed values on the boundary. These are named after Carl Neumann (1832-1925).

Another type of boundary condition that is often encountered is the periodic boundary condition. Consider the heated rod that has been bent to form a circle. Then the two end points are physically the same. So, we would expect that the temperature and the temperature gradient should agree at those points. For this case we write $y(a)=y(b)$ and $y^{\prime}(a)=y^{\prime}(b)$. Boundary value problems using these conditions have to be handled differently than the above homogeneous conditions. These conditions leads to different types of eigenfunctions and eigenvalues.

As previously mentioned, equations of the form $\eqref{eq:1}$ occur often. We form. now show that any second order linear operator can be put into the form of the Sturm-Liouville operator. In particular, equation $\eqref{eq:1}$ can be put into the form

$$\frac{d}{d x}\left(p(x) \frac{d y}{d x}\right)+q(x) y=F(x) .\label{eq:5}$$

Another way to phrase this is provided in the theorem:

The proof of this is straight forward as we soon show. Let’s first consider the equation $\eqref{eq:1}$ for the case that $a_{1}(x)=a_{2}^{\prime}(x)$. Then, we can write the equation in a form in which the first two terms combine,

$$\begin{align} f(x) &=a_{2}(x) y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{0}(x) y\nonumber \\ &=\left(a_{2}(x) y^{\prime}\right)^{\prime}+a_{0}(x) y .\label{eq:6} \end{align}$$

The resulting equation is now in Sturm-Liouville form. We just identify $p(x)=a_{2}(x)$ and $q(x)=a_{0}(x)$.

Not all second order differential equations are as simple to convert. Consider the differential equation

$$x^{2} y^{\prime \prime}+x y^{\prime}+2 y=0 .\nonumber$$

In this case $a_{2}(x)=x^{2}$ and $a_{2}^{\prime}(x)=2 x \neq a_{1}(x)$. So, this does not fall into this case. However, we can change the operator in this equation, $x^{2} D+$ $x D$, to a Sturm-Liouville operator, $D p(x) D$ for a $p(x)$ that depends on the coefficients $x^{2}$ and $x . .$

In the Sturm Liouville operator the derivative terms are gathered together into one perfect derivative, $D p(x) D$. This is similar to what we saw in the Chapter 2 when we solved linear first order equations. In that case we sought an integrating factor. We can do the same thing here. We seek a multiplicative function $\mu(x)$ that we can multiply through $\eqref{eq:1}$ so that it can be written in Sturm-Liouville form.

We first divide out the $a_{2}(x)$, giving

$$y^{\prime \prime}+\frac{a_{1}(x)}{a_{2}(x)} y^{\prime}+\frac{a_{0}(x)}{a_{2}(x)} y=\frac{f(x)}{a_{2}(x)} .\nonumber$$

Next, we multiply this differential equation by $\mu$,

$$\mu(x) y^{\prime \prime}+\mu(x) \frac{a_{1}(x)}{a_{2}(x)} y^{\prime}+\mu(x) \frac{a_{0}(x)}{a_{2}(x)} y=\mu(x) \frac{f(x)}{a_{2}(x)} .\nonumber$$

The first two terms can now be combined into an exact derivative $\left(\mu y^{\prime}\right)^{\prime}$ if the second coefficient is $\mu^{\prime}(x)$. Therefore, $\mu(x)$ satisfies a first order, separable differential equation:

$$\frac{d \mu}{d x}=\mu(x) \frac{a_{1}(x)}{a_{2}(x)} .\nonumber$$

This is formally solved to give the sought integrating factor

$$\mu(x)=e^{\int \frac{a_{1}(x)}{a_{2}(x)} d x} .\nonumber$$

Thus, the original equation can be multiplied by factor

$$\frac{\mu(x)}{a_{2}(x)}=\frac{1}{a_{2}(x)} e^{\int \frac{a_{1}(x)}{a_{2}(x)} d x}\nonumber$$

to turn it into Sturm-Liouville form.

In summary,

Summary

Equation $\eqref{eq:1}$,

$$a_{2}(x) y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{0}(x) y=f(x),\label{eq:7}$$

can be put into the Sturm-Liouville form

$$\frac{d}{d x}\left(p(x) \frac{d y}{d x}\right)+q(x) y=F(x),\label{eq:8}$$

where

$$\begin{align} &p(x)=e^{\int \frac{a_{1}(x)}{a_{2}(x)} d x},\nonumber \\ &q(x)=p(x) \frac{a_{0}(x)}{a_{2}(x)},\nonumber \\ &F(x)=p(x) \frac{f(x)}{a_{2}(x)} .\label{eq:9} \end{align}$$

Example $\PageIndex{1}$

Convert $x^{2} y^{\prime \prime}+x y^{\prime}+2 y=0$ into Sturm-Liouville form.

Solution

We can multiply this equation by

$$\frac{\mu(x)}{a_{2}(x)}=\frac{1}{x^{2}} e^{\int \frac{d x}{x}}=\frac{1}{x^{\prime}}\nonumber$$

to put the equation in Sturm-Liouville form:

$$\begin{align} 0 &=x y^{\prime \prime}+y^{\prime}+\frac{2}{x} y\nonumber \\ &=\left(x y^{\prime}\right)^{\prime}+\frac{2}{x} y .\label{eq:10} \end{align}$$