Query on converting a differential equation to Sturm-Liouville Form

Question

I have a query on Sturm-Liouville operators written in a textbook that I am currently using for my course on Mathematical Methods in Physics.

In the book, I do agree that Sturm-Liouville equations take the form of

$p(x) \frac{d^2y}{dx^2} + r(x) \frac{dy}{dx} + q(x)y + \lambda \rho(x) y = 0$

whereby $r(x)=\frac{dp(x)}{dx}$

However, in the portion written in the book, it says that any $2^{\text{nd}}$ order differential equations in the form $p(x) \frac{d^2y}{dx^2} + r(x) \frac{dy}{dx} + q(x)y + \lambda \rho(x) y = 0$ can be converted into Sturm-Liouville form by multiplying through by a suitable integrating factor, which is given by $F(x)=\exp\int^{x}{\frac{r(u)-p'(u)}{p(u)}du}$ to give the Sturm-Liouville (S-L) form

$[F(x)p(x)y]'+F(x)q(x)y+\lambda F(x)\rho(x)y=0$ with a different but still non-negative weight function $F(x)\rho(x)$.

My question now here is, why a different weight factor? If so, what is the point of writing $\rho(x)$ in the non-(S-L) form anyway? To my knowledge, I think that when we are solving an eigenvalue equation associated with a differential operator, it is always in the form of $Ly=\lambda y$, where $L$ refers to the differential operator. So I believe that the starting equation here should not be

$p(x) \frac{d^2y}{dx^2} + r(x) \frac{dy}{dx} + q(x)y + \lambda \rho(x) y = 0$

but rather,

$p(x) \frac{d^2y}{dx^2} + r(x) \frac{dy}{dx} + q(x)y + \lambda y = 0$

that is equivalent to solving an eigenvalue equation $Ly=\lambda y$ (apart from the negative sign in $\lambda$)

Hence, the integrating factor $F(x)$ to convert a non-(S-L) differential operator $L$ in $Ly=\lambda y$ is the weight function as I feel that $F(x)Ly=\lambda F(x)y$ => $L'y=\lambda F(x)y$, where $L'$ is the new differential operator in S-L form. Indeed, if we compare with the initial form

$p(x) \frac{d^2y}{dx^2} + r(x) \frac{dy}{dx} + q(x)y + \lambda \rho(x) y = 0$

equivalent to

$Ly=\lambda \rho(x) y$,

we should deduce that $\rho(x) = F(x)$, i.e. $F(x)$ is the weight function in this case right? Or am I misinterpreting the proposed form from the book that I am referring to? Should there always be a $\rho(x)$ in a general $2^{\text{nd}}$ order ODE when we are trying to solve the eigenvalues in the initial step? Thank you.

PS: Sorry for the long question!

FYI: The book I am referring to is "Mathematical Methods for Physics and Engineering" by Riley, Hobson and Bence.

Answer 1

There are other substitutions you can use to transform the equation to a normal form. For example, the equation $$ a(x)y''(x)+b(x)y'(x)+c(x)y(x)+\lambda d(x)y(x)=0, $$ is transformed to

$$ -\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f = \lambda d(x)f(x) $$ by the substitution $y=\rho(x)f(x)$, where $$ \rho(x) = \sqrt{a(x)}\exp\left(-\int\frac{b}{2a}dx\right). $$

This substitution is far more versatile in dealing with classical equations. However, it requires $a(x)$ to be twice differentiable. The advantage is that $a(x)$ remains the coefficient of $f''$ and $d(x)$ remains the of the eigenvalue $\lambda f(x)$, which is what I believe you're after. Many classical standard forms can be derived using this substitution.

For example, the associated Legendre equation $$ (1-x^2)u''-2x(m+1)u'+[l(l+1)-m^2-m]u=0 $$ is transformed by $u=(1-x^2)^{-m/2}f$ to the classical form

$$ -\frac{d}{dx}\left((1-x^2)\frac{df}{d}\right)+\frac{m^2}{1-x^2}f=l(l+1)f. $$ This form is hard to get any other way.

Another use of this substitution is to transform an equation to potential form $$ -\frac{d^2f}{dx^2}+qf = \lambda f. $$ This requires a clever use of the above substitution, followed by a substitution of the independent variable.

ANOTHER APPROACH: Typically, the function $d$ is turned into a weight function for the inner product space when you have a form such as the one I mentioned above: $$ -\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f = \lambda d(x)f(x) $$ Then you can define the operator $$ L=\frac{1}{d}\left[ -\frac{d}{dx}\left(a(x)\frac{df}{dx}\right)+\left(\frac{(b-a')^2}{4a}+\frac{(b-a')'}{2}-c\right)f \right] $$ on the weighted $L^2$ space $L^2_{d}$ defined by $$ \langle f,g\rangle_{L^2_{d}}=\int f(x)\overline{g(x)}d(x)dx $$ In this setting the operator $L$ is symmetric with the right endpoint conditions: $$ \langle Lf,g\rangle_{L^2_d}=\langle f,Lg\rangle_{L^2_d}+\mbox{evaluation terms}, $$ and the eigenvalue problem is $Lf=\lambda f$.