6.1: Introduction

In physics many problems arise in the form of boundary value problems involving second order ordinary differential equations. For example, we might want to solve the equation

\[a_2(x)y'' + a_1(x)y' + a_0(x)y = f(x) \label{6.1} \]

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

\[L = a_2(x) \dfrac{d^2}{dx^2} + a_1(x) \dfrac{d}{dx} + a_0(x). \nonumber \]

Then, Equation \ref{6.1} takes the form

\[Ly = f. \nonumber \]

As we saw in the general boundary value problem (4.20) in Section 4.3.2, we can solve some equations using eigenvalue expansions. Namely, we seek solutions to the eigenvalue problem

\[L \phi = \lambda \phi \nonumber \]

with homogeneous boundary conditions and then seek a solution as an expansion of the eigenfunctions. Formally, we let

\[y=\sum_{n=1}^{\infty} c_{n} \phi_{n}. \nonumber \]

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 as was done in Section 4.3.2. [Otherwise, we would have to solve a infinite coupled system of algebraic equations instead of an uncoupled and diagonal system.

It turns out that any linear second order 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 Sturm-Liouville operator. We will highlight some of the properties of such operators and prove a few key theorems, though this will not be an extensive review of Sturm-Liouville theory. The interested reader can review the literature and more advanced texts for a more in depth analysis.

We define the Sturm-Liouville operator as

\[\mathcal{L}=\dfrac{d}{d x} p(x) \dfrac{d}{d x}+q(x) \label{6.2} \]

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

\[\mathcal{L} u=-\lambda \sigma(x) u \nonumber \]

or

\[\dfrac{d}{d x}\left(p(x) \dfrac{d u}{d x}\right)+q(x) u+\lambda \sigma(x) u=0 \label{6.3} \]

for \(x \in(a, b)\). 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 regular. Otherwise, it is called singular.

We also need to impose the set of homogeneous boundary conditions

\[\begin{gathered}
\alpha_{1} u(a)+\beta_{1} u^{\prime}(a)=0 \\
\alpha_{2} u(b)+\beta_{2} u^{\prime}(b)=0
\end{gathered} \label{6.4} \]

The \(\alpha\)'s and \(\beta\)'s 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, \(u(a)=0\) and \(u(b)=0\). For \(\alpha_{i}=0\), we have Neumann boundary conditions. In this case, \(u^{\prime}(a)=0\) and \(u^{\prime}(b)=0\). 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.

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 \(u(a) = u(b)\) and \(u′(a) = u′(b)\). 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 (Equation \ref{6.1}) occur often. We now show that Equation (\ref{6.1}) can be turned into a differential equation of Sturm-Liouville form:

\[\dfrac{d}{d x}\left(p(x) \dfrac{d y}{d x}\right)+q(x) y=F(x) . \label{6.5} \]

Another way to phrase this is provided in the theorem:

Theorem 6.1

Any second order linear operator can be put into the form of the Sturm-Liouville operator (6.2).

The proof of this is straight forward, as we shall soon show. Consider the equation (6.1). If \(a_{1}(x)=a_{2}^{\prime}(x)\), then we can write the equation in the form

\[\begin{aligned}
f(x) &=a_{2}(x) y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{0}(x) y \\
&=\left(a_{2}(x) y^{\prime}\right)^{\prime}+a_{0}(x) y
\end{aligned} \label{6.6} \]

This is in the correct form. We just identify \(p(x)=a_{2}(x)\) and \(q(x)=a_{0}(x)\).

However, 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 \(a_{2}^{\prime}(x)=2 x \neq a_{1}(x)\). The linear differential operator in this equation is not of Sturm-Liouville type. But, we can change it to a Sturm Liouville operator.

In the Sturm Liouville operator the derivative terms are gathered together into one perfect derivative. This is similar to what we saw in the first chapter 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 (6.1) so that it can be written in Sturm-Liouville form. We first divide out the \(a_{2}(x)\), giving

\[y^{\prime \prime}+\dfrac{a_{1}(x)}{a_{2}(x)} y^{\prime}+\dfrac{a_{0}(x)}{a_{2}(x)} y=\dfrac{f(x)}{a_{2}(x)}. \nonumber \]

Now, we multiply the differential equation by \(\mu\):

\[\mu(x) y^{\prime \prime}+\mu(x) \dfrac{a_{1}(x)}{a_{2}(x)} y^{\prime}+\mu(x) \dfrac{a_{0}(x)}{a_{2}(x)} y=\mu(x) \dfrac{f(x)}{a_{2}(x)}. \nonumber \]

The first two terms can now be combined into an exact derivative \(\left(\mu y^{\prime}\right)^{\prime}\) if \(\mu(x)\) satisfies

\[\dfrac{d \mu}{d x}=\mu(x) \dfrac{a_{1}(x)}{a_{2}(x)}. \nonumber \]

This is formally solved to give

\[\mu(x)=e^{\int \dfrac{a_{1}(x)}{a_{2}(x)} d x} . \nonumber \]

Thus, the original equation can be multiplied by factor

\[\dfrac{\mu(x)}{a_{2}(x)}=\dfrac{1}{a_{2}(x)} e^{\int \dfrac{a_{1}(x)}{a_{2}(x)} d x} \nonumber \]

to turn it into Sturm-Liouville form.

\[a_{2}(x) y^{\prime \prime}+a_{1}(x) y^{\prime}+a_{0}(x) y=f(x), \label{6.7} \]

In summary, Equation \ref{6.1} can be put into the Sturm-Liouville form

\[\dfrac{d}{d x}\left(p(x) \dfrac{d y}{d x}\right)+q(x) y=F(x) \label{6.8} \]

where

\[\begin{gathered}
p(x)=e^{\int \dfrac{a_{1}(x)}{a_{2}(x)} d x}, \\
q(x)=p(x) \dfrac{a_{0}(x)}{a_{2}(x)}, \\
F(x)=p(x) \dfrac{f(x)}{a_{2}(x)}
\end{gathered} \label{6.9} \]

Example \(\PageIndex{2}\)

For the example above,

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

Solution

We need only multiply this equation by

\[\dfrac{1}{x^{2}} e^{\int \dfrac{d x}{x}}=\dfrac{1}{x}, \nonumber \]

to put the equation in Sturm-Liouville form:

\[\begin{aligned}
0 &=x y^{\prime \prime}+y^{\prime}+\dfrac{2}{x} y \\
&=\left(x y^{\prime}\right)^{\prime}+\dfrac{2}{x} y .
\end{aligned} \label{6.10} \]