# How to tell which of several possible asymptotic forms a numerical solution to an ODE is converging to

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 $$x \to 0$$?

• 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. May 23 at 20:18
• @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? May 23 at 20:29
• It’s not obvious to me how that is defined for $x<1$ but one would be a better linear fit May 23 at 20:35
• @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 $\arg\left(F_1\right) = -3\mathrm{\pi}/4$. May 23 at 20:43
• @Daniel Hatton : What is the fifth-order ODE from which the fourth order ODE is comming from ? 14 hours ago