- #1
bolbteppa
- 309
- 41
In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.
The first issue I have is that my book began with the statement that given
$$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$
the problem [itex]L[y] \ = \ f[/itex] can be re-cast in the form [itex]L[y] \ = \ \lambda y[/itex].
Now it could be a typo on their part but I see no justification for the way you can just do that!
More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:
Given a linear second order ode
$$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$
it is an exact equation if it is derivable from a differential equation of one order lower, i.e.
$$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$
The equation is exact iff
$$a''(x) - b'(x) + c(x) = 0. $$
If [itex]F[/itex] is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.
This equation is exact iff
$$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$
If you expand this out you get the Adjoint operator
$$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$
If you expand [itex]L^*[/itex] you see that we can satisfy [itex]L \ = \ L^*[/itex] if [itex]a'(x) \ = \ b(x)[/itex] & [itex]a''(x) \ = \ b'(x)[/itex] which then turns [itex]L[y][/itex] into something of the form
$$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$
Thus we seek an integratiing factor [itex]\alpha(x)[/itex] so that we can satisfy this & the condition this will hold is that [itex]\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}[/itex]
Then we're dealing with:
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$
But again, by what my book said they magically re-cast this problem as
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$
Then calling
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$
a Sturm-Liouville problem.
My question is, how can I make sense of everything I wrote above? How can I clean it up & interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues & finding full solutions - what's going on? I want to be able to look at [itex]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/itex] & know how & why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness & integrating factors, thanks for reading!
The first issue I have is that my book began with the statement that given
$$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$
the problem [itex]L[y] \ = \ f[/itex] can be re-cast in the form [itex]L[y] \ = \ \lambda y[/itex].
Now it could be a typo on their part but I see no justification for the way you can just do that!
More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:
Given a linear second order ode
$$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$
it is an exact equation if it is derivable from a differential equation of one order lower, i.e.
$$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$
The equation is exact iff
$$a''(x) - b'(x) + c(x) = 0. $$
If [itex]F[/itex] is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.
This equation is exact iff
$$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$
If you expand this out you get the Adjoint operator
$$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$
If you expand [itex]L^*[/itex] you see that we can satisfy [itex]L \ = \ L^*[/itex] if [itex]a'(x) \ = \ b(x)[/itex] & [itex]a''(x) \ = \ b'(x)[/itex] which then turns [itex]L[y][/itex] into something of the form
$$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$
Thus we seek an integratiing factor [itex]\alpha(x)[/itex] so that we can satisfy this & the condition this will hold is that [itex]\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}[/itex]
Then we're dealing with:
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$
But again, by what my book said they magically re-cast this problem as
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$
Then calling
$$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$
a Sturm-Liouville problem.
My question is, how can I make sense of everything I wrote above? How can I clean it up & interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues & finding full solutions - what's going on? I want to be able to look at [itex]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/itex] & know how & why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness & integrating factors, thanks for reading!