2
$\begingroup$

Is there a way to get Mathematica to find the asymptotic solution, i.e. $r\rightarrow \infty$, of the following equation? It is unable to find the full solution. (a is a real number.)

 DSolve[-f''[r] - 1/r f'[r] + (Log[r]+a) f[r] == 0, f[r], r]

Just to be clear, the goal is to find analytic solutions, not numerical ones.

$\endgroup$
  • $\begingroup$ Could you clarify what you mean by asymptotic solution? Do you mean $r\to\infty$? $\endgroup$ – Chris K Jul 11 '18 at 14:20
  • $\begingroup$ Yes, that's right. $\endgroup$ – 121 Jul 11 '18 at 14:27
4
$\begingroup$

You can use AsymptoticDSolveValue to find the asymptotic approximation of f centered at a:

AsymptoticDSolveValue[-f''[r]-1/r f'[r]+(Log[r]+a) f[r]==0,f[r],{r,a,2}]

(-a + r - (-a + r)^2/(2 a)) C[2] + C[1] (1 - 1/2 (-a + r)^2 (-a - Log[a]))

$\endgroup$
  • $\begingroup$ Could you explain what you mean by an "asymptotic approximation"? Instead of a, could you have found the solution around infinity, analogous to how Taylor expansions can be performed around infinity? $\endgroup$ – 121 Jul 11 '18 at 14:56
  • $\begingroup$ @121 Yes, you can just replace a with Infinity, but AsymptoticDSolveValue is unable to return a result when a Log is included. If you replace Log[r] with r or 1/r then it will work. $\endgroup$ – Carl Woll Jul 11 '18 at 17:21
  • $\begingroup$ You have hit the nail in the head as to why I was asking the question in the first place. With either $r$ or $1/r$, the full solutions are known. $\endgroup$ – 121 Jul 11 '18 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.