Question

(Sturm-Liouville) I'm having trouble figuring out how to put a given problem in the form of an SLDE, equation posted below shortly.

Answer 1

Let's say I have a DE of the form \[\alpha ^2 y''+\alpha y'+(\lambda^n \alpha^2-A)y=0\]
Where A is a constant and n is a natural number. How do I put something like this in the form of an SLDE? The general form of an SLDE is \[[p(x)y']'+[q(x)+\lambda r(x)]y=0\]

But I have a lambda that is not a power of one, do I need to take a square root or something like that, or can I just recognize that maybe it's just an arbitrary term (lambda squared, lambda cubed, it doesn't really matter, you can treat all of them like lambda for the purposes of fitting the form of an SLDE)-how would I deal with the lambda to a non-1 power?

@SithsAndGiggles

Answer 2

@dan815

Answer 3

You might need to recruit an integrating factor.
\[\color{red}{\alpha^2\mu(x)}y''+\color{blue}{\alpha\mu(x)}y'+(\lambda^n\alpha^2-A)\mu(x)y=0\]
To match up the general form,
\[\color{red}{p(x)}y''+\color{blue}{p'(x)}y'+q(x)y+\lambda r(x)y=0\]
we clearly need \(\alpha^2\mu(x)=p(x)\) and \(\alpha\mu(x)=p'(x)\). Differentiating the first equation, we have \(p'(x)=\alpha^2\mu'(x)\), and so we have the ODE with solution
\[\alpha \mu(x)=\alpha^2\mu'(x)~~\implies~~\mu(x)={\large e^{x/\alpha}}\]
So we can rewrite the original ODE as
\[\begin{align*}\alpha^2{\large e^{x/\alpha}}y''+\alpha{\large e^{x/\alpha}}y'+(\lambda^n\alpha^2-A){\large e^{x/\alpha}}y&=0\\
\frac{d}{dx}\left[\alpha^2{\large e^{x/\alpha}}y'\right]+(\lambda^n\alpha^2-A){\large e^{x/\alpha}}y&=0\end{align*}\]
From here, we could set \(q(x)=-A{\large e^{x/a}}\). It seems to me that the \(\lambda\) in the general form is indeed a general constant that may very well take on the form of some "new" \(\lambda\), say \(\hat{\lambda}\), with \(\hat{\lambda}=\lambda^n\). To find \(\lambda\), you would take the \(n\)-th root of \(\hat{\lambda}\). I'll have to check some literature, though, just to be sure.