I've previously mentioned the ordinary differential equation

$$12P\left(f\left(x\right)\right)^3f''''\left(x\right)+12\left(3P-1\right)\left(f\left(x\right)\right)^2f'\left(x\right)f'''\left(x\right)+4\left(3P+2\right)\left(f\left(x\right)\right)^2\left(f''\left(x\right)\right)^2+12\left(P-1\right)f\left(x\right)\left(f'\left(x\right)\right)^2f''\left(x\right)-\frac{9P^2}{Q^4}x = 0$$

where $P$ and $Q$ are control parameters - $P$ is real and positive and $Q$ is real and almost certainly negative. I have boundary conditions

$$f\left(1\right) = 1$$ $$f'\left(1\right) = 0$$ $$f''\left(1\right) = 0$$

(and also $f\left(0\right) = 0$, but we'll ignore that one for now). The differential equation has a movable singular point where $f\left(x\right) = 0$, which one might anticipate will make things hard,

I've applied the first steps of Frobenius' method, and from the results I expect the asymptotic form of the solution as $x \to 0$ to be one of the following:

  • $f\left(x\right) \simeq F_0$
  • $f\left(x\right) \simeq F_1x\left(\ln\left(x\right)\right)^{3/4}$
  • $f\left(x\right) \simeq F_2x$
  • $f\left(x\right) \simeq F_3x^{9P/\left(9P-2\right)}$ or
  • $f\left(x\right) \simeq \frac{2^{5/4}I\sqrt{3P}}{5^{1/4}\left(9P-10\right)^{1/4}Q}x^{5/4}$

where the $F_i$ are arbitrary constants and $I \in \left\lbrace1,-1,\mathrm{i},-\mathrm{i}\right\rbrace$.

Let's say I pick a value of $f'''\left(1\right)$ and launch a finite-difference integration of the differential equation starting from $x = 1$, towards $x = 0$. How can I detect, from the finite-difference results, which of the five possible asymptotic forms the numerical solution is converging on as $x \to 0$?

  • 1
    $\begingroup$ Do a log-log regression between $x$ and $f$ near $x=0$ and the slope tells you the dominant power of $x$ in that regime. $\endgroup$
    – whpowell96
    May 23 at 20:18
  • $\begingroup$ @whpowell96 Good thinking, but the $F_1x\left(\ln\left(x\right)\right)^{3/4}$ and $F_2x$ forms both have the same limiting slope, no? $\endgroup$ May 23 at 20:29
  • 1
    $\begingroup$ It’s not obvious to me how that is defined for $x<1$ but one would be a better linear fit $\endgroup$
    – whpowell96
    May 23 at 20:35
  • $\begingroup$ @whpowell96 $f\left(x\right)$ has to be real to make physical sense, and I'd be astonished if $f\left(x\right)$ weren't positive throughout $0 < x \leq 1$, so presumably $\arg\left(F_1\right) = -3\mathrm{\pi}/4$. $\endgroup$ May 23 at 20:43
  • 1
    $\begingroup$ @Daniel Hatton : What is the fifth-order ODE from which the fourth order ODE is comming from ? $\endgroup$
    – JJacquelin
    14 hours ago


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