I want to study the large $x$ solution to a Riccati equation. After listening to the lectures on Mathematical Physics by Carl Bender, I have fallen in love with asymptotic analysis. But, by no means do I have a full understanding of it.
I transformed the Riccati equation into the following second order differential equation:
$$u''(x)-Q(x) u(x)=0\qquad Q(x)=\frac{n(n+2)}{4x^2}+\frac{a^2 b^2\,e^{-2x}}{x^{2n+1}},\enspace\text{with}\,\, a,b>0, \,\,n=\{0,1,2,\ldots\}$$
defined on $0<x<\infty$, subject to the initial condition: $u(1)=1$, $u'(1)=1$. The solution to this differential equation gives the solution y(x) to the Riccati equation: $$ y_\text{sol}(x)=-\frac{(n+2)}{2a}x^{n+1} + \frac{x^{n+2}}{a}\frac{d}{dx}\ln u(x) $$
For concreteness take $n=0$ and $a=100$, $b=1$.
My Attempt
My numerical investigations revealed that the large $x$ behavior of $u$ is very insensitive to the initial conditions. (The hunch I got from this is that the dependence on the initial condition would appear in the negligible part of the solution). And $y_\text{sol}$ at large $x$ tends to a non-zero constant which I am interested in.
I apply WKB. But, before doing so I recognized this problem as arising from a 3-dimensional QM problem where $x$ is a radial coordinate. I recalled some obscure claim by Langer (1937) that in order to get an 'accurate WKB result' we must rescale $x$ by putting $x=e^z$ to make the domain of the problem over the entire real numbers $-\infty<z<\infty$. [I have no understanding of why this must be done from an asymptotic analysis point of view; can someone help me with this?]. Then if I put $u(x)=e^{z/2}w(z)$, I get a new second order differential equation
$$w''(z) - P(z)w(z)=0\qquad P(z)=\frac{(n+1)^2}{4}+e^{2z}\frac{a^2b^2\,e^{-2(e^{z})}}{(e^{z})^{2n+1}}$$
for which I can apply WKB. The end result (after transforming back to $u$ in $x$-space) is:
$$ u(x)\sim A \frac{1}{Q_L(x)^{1/4}}\Big(e^{\int_1^x dx' \sqrt{Q_L(x')}}+B e^{-\int_1^x dx' \sqrt{Q_L(x')}}\Big) $$ Here, $Q_L = \displaystyle\frac{(n+1)^2}{4x^2}+\frac{a^2 b^2\,e^{-2x}}{x^{2n+1}}$ is slightly different from $Q$ due to the brief transformation to $z$-space, and $A$ and $B$ are constants fixed by boundary conditions.
Problems
When I plot this result against the numerically integrated solution for $u(x)$ and also for $y(x)$, I find that WKB does a fantastic job approximating the small $x$ behavior of the exact solution, but doesn't get quite get the large $x$ behavior correctly. WKB does get leading the asymptotic behavior: $u'(x)/u(x) \sim \frac{n+2}{2}\frac{1}{x}$ as $x\rightarrow\infty$. But I need the next-to-leading asymptotic form which I expect to be $\sim 1/x^{n+2}$. But WKB approximation tells me that it is $\sim e^{-2x}$, which is very subdominant compared to what I am expecting. I am puzzled by this because I thought WKB was supposed to uniformly get the right answer for all $x$! I tried even tried taking the WKB approximation to next to leading order in hopes of getting the expected next-to-leading order asymptotic behavior, but I still keep getting highly subdominant terms... what's going on??
Perhaps I should follow a different strategy to determine the asymptotic behavior?
Plot Here is a plot of $y(x)$ satisfying $y(1)=1$ for $n=0$ and $a=100$, $b=1$:
Magenta curve: numerically integrated. Blue curve: WKB result.
