# Asymptotic behavior of the solution of $x^4 \frac{d^2y}{dx^2}+ \frac{1}{4}y=0$ near $0$

Can you help me find the leading asymptotic behaviors about the irregular singular point $x=0$ of $$x^4 \frac{d^2y}{dx^2}+ \frac{1}{4}y=0$$

So far I have got $y(x) = c_{1}8\exp(2/x)+c_{2}8\exp(-2/x)$, is this on the right track for the answer?

-
What do you mean by asymptotic behavior? – Davide Giraudo Feb 23 '12 at 21:01
I it means is the equation analytic. Im not sure tho. – steven Feb 23 '12 at 21:10
Mathematica gives the following: inputting "DSolve[x^4 f''[x] + 1/4 f[x] == 0, f[x], x]" gives a general solution of the form "E^(I/(2 x)) x C[1] - I E^(-(I/(2 x))) x C[2]", for some constants C[1] and C[2]. Thus, it seems likely that some variable substitution of your original equation leads a differential equation with constant coefficients. Note that your equation is irregular singular at $x=0$, which explains why the solution has an essential singularity around this point. – A Walker Feb 23 '12 at 22:20
I have got $y(x) ~ c_{1}8\exp{2/x}+c_{2}8\exp{-2/x}$, is this on the right track for the answer? – steven Feb 23 '12 at 23:17

In my answer I switch $x$ to $t$, please don't get confused.

We just prove $t=0$ is an irregular singular point of that equation. First, recall the definition & the theorems with

$$y''+p(t)y'+q(t)y=0$$

step 1. If $\lim\limits_{t \to t_0} p(t)$ & $\lim\limits_{t \to t_0} q(t)$ are finite, the $t=t_0$ is called an ordinary point, otherwise, it's called a singular point.

So, you can transform the equation above to

$$y''+\frac{1}{4t^4}y=0$$

and find $q(t)=\frac{1}{4t^4}$. Check $\lim\limits_{t \to 0} q(t)$ are infinite, because $t_0=0$ in your question that is the point you want to prove the irregular singular point. In this step, you prove the point is not ordinary point but singular point.

Next step is needed for that singular point is irregular singular point, not regular singular point.

step 2. If $t=t_0$ is singular point, but both $\lim\limits_{t \to t_0} (t-t_0)p(t)$ & $\lim\limits_{t \to t_0} (t-t_0)^2q(t)$ are finite, the $t_0$ is called a regular singular point. Otherwise, it is an irregular singular point.

Above, we find $q(t)=\frac{1}{4t^4}$. Check $\lim\limits_{t \to 0} (t-0)^2q(t)$ are infinite. Finally, we prove that $t_0=0$ is irregular singular point. And that equation shows asymptotic behavior as $t \to 0$.

If you want to solve the case that $y$ behaves severely bad, then Carlini-Liouville-Green method will help you!

-