Definition of Sturm-Liouville Form
A second-order linear ODE is said to be in Sturm-Liouville Form if it has the structure of
\begin{align}
\frac{d}{dx}(p(x)\frac{dy}{dx}) + q(x)y + \lambda \rho(x) y = 0 \tag{1}
\end{align}
where \( p(x), q(x) \) are real functions of \(x\), \(\rho (x) > 0\) is a positive-definite weighting function also of \(x\), and \( \lambda \) is known as the eigenvalue of the equation that we will discuss later. Given any second-order linear ODE in the general form of
\begin{align}
\frac{d^2y}{dx^2} + R(x)\frac{dy}{dx} + Q(x)y + \lambda \rho(x) y = 0 \tag{2}
\end{align}
(divided by the leading coefficient for \( y^{\prime\prime}\) if needed) we can always multiply by an integrating factor \( r(x) = \exp(\int R(x) dx) \) (recall the integrating factor technique for first-order linear ODEs) to transform it into the form of (1):
\begin{align}
e^{\int R(x) dx}\frac{d^2y}{dx^2} + R(x)e^{\int R(x) dx}\frac{dy}{dx} + Q(x)r(x)y + \lambda \rho(x)r(x) y &= 0 \\
\frac{d}{dx}(e^{\int R(x) dx}\frac{dy}{dx})+ Q(x)r(x)y + \lambda \rho(x)r(x) y &= 0 \tag{3}
\end{align}
and we identify \(p(x) = r(x)\), \( q(x) = Q(x)r(x) \), and the weighting is now replaced by \(\rho(x)r(x)\) (note that it is still positive-definite as the exponential \( r(x) = \exp(\int R(x) dx) \) is always positive).
We can assign the linear Sturm-Liouville operator by \( \mathcal{L} = -[\frac{d}{dx}(p(x)\frac{d}{dx}) + q(x)] \) so that we can compactly write (1) as
\begin{align}
\mathcal{L}y = \lambda\rho(x) y \tag{4}
\end{align}
This is very alike the eigenvalue-eigenvector relation in Linear Algebra but now \(\mathcal{L}\) is an operator rather than a matrix. It means that for any function \( y \) that satisfies (4), the effect of \(\mathcal{L}\) is simply to multiply \( y \) by the eigenvalue factor \( \lambda \) (enlarge/shrink), plus adjusted by an additional weight \( \rho (x) \). In this case, such a function \( y \) is known to be an eigenfunction of (4), provided that it also satisfies the boundary conditions that are to be imposed.
The Sturm-Liouville operator/form is a further generalization of the canonical form for second-order ODEs (refer to this tutorial). In this case, the results about oscillation proved in the other tutorial are readily transferable with little modification. Lemma 1 is immediate by comparing with (2): \( P(x) \rightarrow r(x) \), \( Q(x) \rightarrow q(x)+\lambda\rho(x) \), and thus Theorem 2 follows. We skip Theorem 3 as an optional exercise and the updated condition of Theorem 4 is ((9) and (10) of the original): for
\begin{align}
s(x) = -\frac{p(x)y'(x)}{y(x)} \tag{5}
\end{align}
then
\begin{align}
s'(x) &= \frac{y'(x)p(x)y'(x) -y(x)(p(x)y'(x))’}{y(x)^2} \\
&= \frac{s(x)^2}{p(x)} + \frac{y(x)(q(x)y + \lambda \rho(x) y)}{y(x)^2} \\
&= \frac{s(x)^2}{p(x)} + (q(x) + \lambda\rho(x)) \tag{6}
\end{align}
by (1). If we put an extra requirement that \( p(x) > 0 \) is also positive-definite, then the argument will still work out but it now requires
\begin{align}
\int_1^{\infty} (q(x) + \lambda\rho(x)) dx = \infty \tag{7}
\end{align}
Example
The innocent case of
\begin{align}
\frac{d^2y}{dx^2} + \lambda y = 0 \tag{8}
\end{align}
is already in Sturm-Liouville Form with \( p(x) = 1 \), \( q(x) = 0 \), and a constant weight \( \rho(x) = 1 \). It is just a simple second-order constant-coefficient ODE (the simple harmonic oscillator equation) with two possible solutions \( \sin (\sqrt{\lambda} x), \cos (\sqrt{\lambda} x) \) provided that \( \lambda > 0 \). However, as the next few sections show, the Sturm-Liouville theory actually enables the implication of many important results and insights regarding the two trigonometric functions.
Exercise
Convert the following ODE into its Sturm-Liouville Form
\begin{align}
y^{\prime\prime} -2xy’ + \lambda x y = 0 \tag{9}
\end{align}
Answer
By the remark for (2) and (3), we know that the integrating factor is \( \exp(\int -2x dx) = \exp(-x^2) \). So (9) is converted as
\begin{align}
e^{-x^2}y^{\prime\prime} -2xe^{-x^2}y’ + \lambda xe^{-x^2} y &= 0 \\
(e^{-x^2}y’)’ + \lambda xe^{-x^2} y &= 0
\end{align}
the Sturm-Liouville form with \( p(x) = \exp(-x^2) \), \( q(x) = 0 \), and \( \rho(x) = xe^{-x^2} \) (this requires \(x>0\)).
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