How to obtain the asymptotic solution for a ODE

Question

How can I obtain an asymptotic solution of $\phi$ when $x\rightarrow\infty$ for the following equation $$ \biggl[-\frac d {dx^2}+\frac{l(l+1)}{x^2}+\frac{x^2}4+\frac s x-\frac E \omega\biggr]\,\phi(x)=0 $$ I read some notes about obtainig the asymstotic solution but I didn't undrstand how to apply them to my case.

Answer 1

For large $x$ your equation reduces essentially to $$ -ϕ''(x)+ \frac{x^2}4 ϕ(x)=0, $$ all other terms are small perturbations. By the WKB approximation, the asymptotic behavior is $$ ϕ(x) \sim x^{-1/2}e^{x^2/4} $$ by the general formula that $-ϕ''(x)+ q(x) ϕ(x)=0$ has solutions asymptotic to $q(x)^{-1/4}\exp\left(\int\sqrt{q(x)}dx\right)$.

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A more careful execution of the WKB computations, as noted in the comments by @Maxim, gives a slight correction for this level of approximation. In the general WKB framework, one tries to find basis solutions in the form $ϕ(x)=e^{S(x)}$. After insertion into the ODE and removal of the exponential factor the equation $$ -S''-S'^2+\frac{l(l+1)}{x^2}+\frac{x^2}4+\frac s x-\frac E \omega=0 $$ remains. For large $x$, the derivative terms have also to be large, assume that first and second derivative are of comparable magnitudes so that the square of the first is very large against the second derivative, $S'^2\gg|S''|$. Then balance the two largest terms to get a first approximation $S_0$ with $$ S_0'^2=\frac{x^2}4\implies S_0=\pm \frac{x^2}4 $$ Now insert $S=S_0+S_1$ to get $$ S_0''+S_1''+2S_0'S_1'+S_1'^2=\frac{l(l+1)}{x^2}+\frac s x-\frac E \omega $$ and determine as the leading terms on both sides, from $S_0'\sim x$ we get $S_1'\sim \frac1x$, so that $S_1'^2,S_1''\sim \frac1{x^2}$ are small against the other terms. On the level of constants the reduced equation remains as $$ S_0''+2S_0'S_1'=-\frac E \omega,\\ \pm\frac12\pm xS_1'=-\frac E \omega,\\ \implies S_1=\left(-\frac12\mp\frac E \omega\right)\ln x. $$ One could continue further with $S=S_0+S_1+S_2+...$, but the terms quickly become unwieldy. Up to this point, the two basis solutions have the asymptotic form for $x\to\infty$ $$ S_+(x)=x^{-1/2-E/ω}\,e^{x^2/4}~~\text{ and }~~S_-(x)=x^{-1/2+E/ω}\,e^{-x^2/4} $$