I am trying to find the leading asymptotic behaviours, using the method of dominant balance, about the irregular singular point $x=0$ for the equation $$x^4\frac{d^2y}{dx^2}+\frac{1}{4}y=0.$$

First let $y=e^{S(x)}, y'= S'e^{S(x)}, y''=[(S')^2+S'']e^{S(x)}$. Our equation then looks like $$x^4[(S')^2+S'']e^{S(x)}+\frac{1}{4}e^{S(x)}=0$$ which implies that $$x^4[(S')^2+S'']=-\frac{1}{4}.$$

We then make the assumption that $S'' << (S')^2$, allowing us to then say $$x^4(S')^2 \sim -\frac{1}{4}.$$

This then implies that $$S' \sim \pm \frac{i}{2x^2} \implies S \sim \mp \frac{i}{2x}.$$

We can see that $S'' \sim \mp \frac{i}{x^3}$ and $(S')^2=-\frac{1}{4x^4}$ which show our assumption to be true as $x \to 0$.

Now we assume that $$S(x) = \frac{i}{2x} + C(x), $$ where $C(x) = o(x^{-1})$.

Sub this into $$ S'' + (S')^2 + \frac{1}{4x^4} = 0 $$ to find $$ \frac{i}{x^3} + C'' -\frac{1}{4x^4}+\frac{i}{x^2}C'+(C')^2 +\frac{1}{4x^4} = 0. $$

After cancelling out and removing small terms we are left with $$ \frac{i}{x^3} + \frac{i}{x^2}C' \sim 0. $$

This then implies that $C \sim -\ln(x)$.

We then have that $$ S(x) = \frac{i}{2x} - \ln(x) + D(x),$$ where $D(x) = o(\ln(x)).$

Sub this into $$ S'' + (S')^2 + \frac{1}{4x^4} = 0 $$ to find $$ \frac{2}{x^2}+\frac{2i}{x^3}-\frac{2D'}{x}-\frac{1}{4x^4}+(D')^2-\frac{iD'}{x^2}+\frac{1}{4x^4}=0.$$

Again we cancel and remove very small terms to find $$\frac{2}{x^2}+\frac{2i}{x^3}-\frac{2D'}{x} \sim 0.$$

This then implies that $D' \sim \frac{1}{x} + \frac{i}{2x}.$

Hence, $D \sim \ln(x) + \frac{i}{2}\ln(x).$

Since this does not tend to $0$ as $x \to 0$ I figured I must continue with the process but the next stage gave me some very questionable results and I realised I didn't really understand what I was looking for or trying to achieve.

Any feedback on current work or clarifications on what I am supposed to be doing would be greatly appreciated.


You're doing the right thing, but remember that you're interested in the behaviour of $y(x)$ rather than $S(x)$: using your approximation, you should get that $$ y(x) = e^{S(x)} = e^{\frac{i}{2x} + \text{ln}(x)} = x\,e^{\frac{i}{2x}}, $$ for which $\lim_{x \to 0} y(x) = 0$. To check your approximations, it might be useful to know that the solutions to the original ODE are given by $$ y(x) = x\,e^{\pm \frac{i}{2x}}. $$

  • $\begingroup$ Thank you for the response, is it not neccessary to include constants in the final solution ? $\endgroup$ – Evan Feb 28 '18 at 22:11
  • $\begingroup$ Well, yes, that follows from general ODE theory: given two linearly independent solutions to your second order ODE (which are given in the last line of my answer), the general solution can be written as a linear combination of those two independent solutions -- that's where the constants come in. $\endgroup$ – Frits Veerman Mar 1 '18 at 18:11

First, I think $C \sim \log(x)$ not $-\log(x)$, because in the equation above, we have $$ \frac{i}{x^3} + C'' -\frac{1}{4x^4} \color{red}{\text{ minus }} \frac{i}{x^2}C'+(C')^2 +\frac{1}{4x^4} = 0 $$ instead of plus.

Next, let us get the equation for $D$ and as you said. The equation is $$\big(\color{blue}{S''} + C'' + \color{orange}{D''}\big) + \big(\color{red}{S'^2} + C'^2 +\color{orange}{D'^2} + \color{blue}{2S'C'} + 2 S' D' + \color{orange}{2 C' D'} \big)+ \color{red}{\frac{1}{4x^4}} = 0$$ where $S = \frac{i}{2x}$, $C = \log(x)$.

  • The $\color{red}{\text{ color red terms }}$ will cancel out because this is how we solved $S$, i.e. $$S'^2 + \frac{1}{4x^2} \sim 0 \Longrightarrow S \sim \pm\frac{i}{2x}$$ and we can easily check $$S'^2 + \frac{1}{4x^2} = 0 \Longleftarrow S = \pm\frac{i}{2x}.$$
  • The $\color{blue}{\text{ color blue terms }}$ will cancel out because this is how we solved for $C$.
  • The $\color{orange}{\text{ color orange terms }}$ are small compared to the rest of the terms.

Finally, we get

$$C'' + C'^2 + 2S'D' \sim 0$$ We get $$D' \sim 0.$$

Hope my calculations are correct...

Furthermore, if as you calcuated $C = -\log(x)$, then you should get $$D \sim -2ix.$$


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