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My doubt is related to the equivalence between solutions of the following Sturm-Liouville problem: \begin{equation} r^{2}f''(r) + 2rf'(r) + \{\omega^{2}r^{2} - [j(j+1)-|q|^{2})]\}f(r)=0\,,\label{SL1} \end{equation} where $r\in\mathbb{R}$ (actually $r\in [0,+\infty)$) and $j$ can take the values $j=|q|-1, |q|, |q|+1,...$(and so on). The boundary conditions (BC) are $f(r_{0})=0=f(r\to\infty)$.

It is relevant to stress that for $r<r_{0}$, where $r_{0}$ stands for a constant, this function f(r) is identically zero (by definition of my physical problem). Thus, in one hand I'm considering that the solution is given by $$f(r)=\begin{cases}0, \text{for } r<r_{0}\,,\\ f_{r_{0}}\,, \text{for } r>r_{0}\,,\end{cases}$$ where $f_{r_{0}}$ is the solution to my first eq. with the aforementioned BCs.

On the other hand, I could simply proceed with a simple change of variables such that $$\bar{r} = r - r_{0}\,,$$ which implies that my ODE turns out to be $$(\bar{r}+r_{0})^{2}f''(\bar{r}) + 2(\bar{r}+r_{0})f'(\bar{r}) + \{(\bar{r}+r_{0})^{2}\omega^{2} - [j(j+1)-|q|^{2}]f(\bar{r})\}=0\,,$$ where, now, primes denote derivative with respect to $\bar{r}$ and the BCs were shifted to $$f(\bar{r}=0)=0=f(\bar{r}\to\infty)\,.$$

My question is: Are these two solutions equivalent? Are there any problems related to my first "method"? I mean, when I find $f_{r_{0}}$ with the boundary condition applied to $r_{0}\neq 0$ I may have ignored the behavior of the function between the interval $0 \leq r < r_{0}$, right? This looks confusing since from definition I have a null function in this interval.

Thanks!

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  • $\begingroup$ It seems to me that in your second method, you'll be allowing a few more solutions than in the first. Say you find an $f_{r_0} $ that behaves linearly for small $r-r_0 $. Then, in your first formulation, your derivative operator will generate a $\delta $-function at $r=r_0 $, and you will discard this solution, since $f$ doesn't solve your equation at $r=r_0 $ (though it does everywhere else). In your second method, you disregard this consideration at $r=r_0 $ and you will accept $f_{r_0 } $ as a solution. $\endgroup$ Feb 14 '20 at 15:45
  • $\begingroup$ Dear @MichaelEngelhardt, I cannot see why there is a solution with linear behavior in this limit. Besides that, the boundary condition at $r=r_0$ has been chosen so that both side limits go to zero. That is, I was looking for some smooth solution of the ODE, which means that the first derivative acting on this radial function must not be a Dirac delta-"function". But, I may be wrong and I would be delighted if you could explain your opinion in more details. Finally, I forgot to mention that without the translation, the ODE can be put in the form of a Bessel equation. $\endgroup$
    – MLPhysics
    Feb 17 '20 at 1:01
  • $\begingroup$ I would have thought that your typical $f_{r_0 } $ would be a Bessel function with a node at $r_0 $ - so it would be proportional to $r-r_0 $ close to $r_0 $. Then the proposed solution has a discontinuous slope at $r_0 $, and the second (not the first) derivative generates a $\delta $-function, so it actually does not solve the ODE at $r_0 $ - you would presumably discard this. In your second method, you lose this information, and you'd presumably have to reintroduce it in terms of a more stringent boundary condition at $\bar{r} =0$. $\endgroup$ Feb 17 '20 at 1:12

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