I would like to find the first two terms of the asymptotic approximation for $\epsilon^2 y'' = axy$ on $0\leq x < \infty$ where $y(\infty) = 0$ and $a>0$.

Here is my work so far, with the standard WKB method, we plug in $$y(x) = \exp\left[\frac{1}{\delta} \sum_{n=0}^\infty \delta^n S_n(x)\right]$$ and divide by it; our ODE simplifies to $$\epsilon^2\left[\frac{1}{\delta} S_0'' + S_1 + \cdots + \frac{1}{\delta^2} S_0'^2 + 2 \frac{1}{\delta} S_0'S_1' + \cdots \right] -ax = 0 \quad \quad (\star)$$ Our largest term $\frac{\epsilon^2}{\delta^2} S_0'^2 $must match with $-ax$, so we make $\delta = \epsilon$, and get $$S_0'^2 \sim ax$$ thus $$S_0 \sim \pm \sqrt{a} \frac{2}{3} x^{3/2}.$$ And by matching $O(\epsilon)$ terms in $(\star)$, we get $$2S_0'S_1' + S_0'' = 0$$ and we get $S_1 \sim - \frac{1}{4} \log(ax)$.

So our approximation $$y (x)\sim C_1 (ax)^{-1/4} \exp \left[\frac{1}{\epsilon} \sqrt{a} \frac{2}{3} x^{3/2}\right] + C_2 (ax)^{-1/4} \exp \left[-\frac{1}{\epsilon} \sqrt{a} \frac{2}{3} x^{3/2}\right]$$ And from our condition $y(\infty) = 0$, we get $C_1 = 0$.

But the above approximation does not make sense at $x=0$, how can we take care of this problem? I know boundary layer theory but often it has problems when the coefficient of $y'$ is constant zero. I also worked out the WKB method for a boundary layer problem here, I am not sure how would I apply to this case.

Edit: If I add a boundary layer at $x=0$, we let $\xi = x/\delta$, then the ODE in terms of delta will become $$\frac{\epsilon^2}{\delta^2} (y_0'' + \epsilon y_1'' + \cdots ) = a\xi\delta(y_0 + \epsilon y_1, + \cdots)$$ which gives $$\frac{\epsilon^2}{\delta^2} y_0'' + \frac{\epsilon^3}{\delta^2} y_1'' + \cdots - \delta a\xi y_0 + \cdots = 0$$ If I let $\delta = \epsilon$, I will get $y_0 = a\xi + b$, how would I match this with $y(x)$ from WKB method

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    $\begingroup$ This is looking like the Airy function. $\endgroup$ – Claude Leibovici Jul 18 '18 at 4:50
  • $\begingroup$ @ClaudeLeibovici, It is the Airy equation with $\epsilon^2$ before the $y''$, could you say a bit more about this please. Thanks! $\endgroup$ – Xiao Jul 19 '18 at 3:01
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    $\begingroup$ Well, without using any perturbation theory, we have the solution and we can analyze it. I must confess that I am not familiar at all with this kind of analysis using perturbation theory, $\endgroup$ – Claude Leibovici Jul 19 '18 at 3:07
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    $\begingroup$ Have a look at math.arizona.edu/~meissen/docs/asymptotics.pdf .It could be of some interest for you. $\endgroup$ – Claude Leibovici Jul 19 '18 at 3:31

Your asymptotic approximation is correct : $$y (x)\sim C_2 (ax)^{-1/4} \exp \left[-\frac{1}{\epsilon} \sqrt{a} \frac{2}{3} x^{3/2}\right] \tag 1$$ according to the boundary condition $y(\infty)=0$.

Note : A second order ODE requires at least two boundary conditions to expect having a unique solution. Since the wording of the question specifies only the condition $y(\infty)=0$, the solution remains undetermined : The coefficient $C_2$ cannot be computed.

Another key point is that an asymptotic formula is valid only for $x\to\infty$, not for $x\to 0$.So, don't expect Eq.$(1)$ makes sens at $x=0$. It is normal that the asymptotic approximation be far to be correct for small values of $x$.

If you want an approximate around $x=0$, expand $y(x)$ in power series. $$y(x)\simeq c_0+c_1x+c_2x^2+...$$ Of course, this formula is not valid for large $x$, so the boundary condition at infinity cannot be fulfilled. Moreover, two boundary conditions should be required close to $x=0$ in order to determine the coefficients $c_0$ , $c_1$, $c_2$ , …

In fact, as Claude Leibovici quite rightly wrote in comments, the exact solving involves the Airy functions. $$y(x)=c_1\text{Ai}\left(\sqrt[3]{\frac{a}{\epsilon^2}}\:x\right)+c_2\text{Bi}\left(\sqrt[3]{\frac{a}{\epsilon^2}}\:x\right)$$ Ai and Bi are the Airy functions http://mathworld.wolfram.com/AiryFunctions.html

The condition $y(\infty)=0$ is satisfied any $c_1$ and $c_2$, which shows that the condition $y(\infty)=0$ is not sufficient to fully determine a solution.

On physicists viewpoint, defining the properties of a boundary layer is certainly a way to introduce some boundary conditions in the range around $x=0$. Supposing that $y(x)\simeq y_0+y'_0 x$ in a layer of a given thickness $\delta$ allows to compute $c_1$ and $c_2$ as a function of $y_0$ , $y'_0$ and $\delta$, thanks to the series expansion of the Airy functions. Then the asymptotic expansion of the Airy functions will leads to the coefficient $C_2$ in the asymptotic approximation $(1)$.




I found the example from my text book. I will add the details here:

  1. First we see that in WKB approximation above $$y(x) \sim \exp\left\{\frac{1}{\epsilon}S_0 + S_1\right\} = C_2 (ax)^{1/4} \exp\left[-\frac{1}{\epsilon}\sqrt{a} \frac{2}{3} x^{3/2} \right], \quad \epsilon \rightarrow 0^+,$$ we need to have $\epsilon S_2 << 1$ because generally we have $y(x) \sim \exp\left\{\frac{1}{\epsilon}S_0 + S_1 + \epsilon S_2 + \cdots \right\}$. Solving for $S_2$ explicitly, we get $S_2 \sim C x^{-3/2}$, thus $\epsilon S_2 <<1 \Rightarrow \epsilon^{2/3} << x,$ and we have $$\color{blue}{y_{one}(x) \sim C_2 (ax)^{1/4} \exp\left[-\frac{1}{\epsilon}\sqrt{a} \frac{2}{3} x^{3/2} \right] \quad \epsilon \rightarrow 0^+ \text{ on } x>> \epsilon^{2/3}}.$$
  2. Next, when $x$ is near zero, we let $t = (a/\epsilon^2)^{1/3}x$, so our ODE $\epsilon^2 y''(x) = axy(x)$ can be rewritten as $y''(t) = ty(t)$, and this is the Airy equation. And we know the solution is $$y(t) = D_1 \text{Ai}(t) + D_2 \text{Bi}(t)$$ thus $$\color{orange}{y_{two}(x) = D_1 \text{Ai}((a/\epsilon^2)^{1/3}x) + D_2 \text{Bi}((a/\epsilon^2)^{1/3}x)}.$$
  3. Finally would like to match them on $\epsilon^{2/3} <<x<<1$. First we see $y_{one}$ and $y_{two}$ are very different, so we will approximate $y_{two}$. From a irregular singular point (ISP) analysis of the Airy equation at infinity, we will get $$\text{Ai}(t) \sim \frac{1}{2\sqrt{\pi}} t^{1/4} \exp\left[-\frac{2}{3}t^{3/2}\right], \quad t \rightarrow \infty $$ $$\text{Bi}(t) \sim \frac{1}{\sqrt{\pi}} t^{1/4} \exp\left[\frac{2}{3}t^{3/2}\right], \quad t \rightarrow \infty,$$ (the two constants $\frac{1}{2\sqrt{\pi}}$ and $\frac{1}{\sqrt{\pi}}$ are a bit more delicate and can not be obtained from the ISP analysis, but we will assume we have this fact.) So $$\color{orange}{y_{two}(x) \sim \frac{D_1}{2\sqrt{\pi}} ((a/\epsilon^2)^{1/3}x)^{1/4} \exp\left[-\frac{2}{3}((a/\epsilon^2)^{1/3}x)^{3/2}\right] + \\ \frac{D_2}{\sqrt{\pi}} ((a/\epsilon^2)^{1/3}x)^{1/4} \exp\left[\frac{2}{3}((a/\epsilon^2)^{1/3}x)^{3/2}\right] \text{ when } x>>\epsilon^{2/3} \text{ and } \epsilon \rightarrow 0^+}. $$

After matching with $y_{one}$, we get $D_2 = 0$, $D_1 = 2\sqrt{\pi}(a\epsilon)^{-1/6} C_2$.

Now getting a uniform formula is non trivial, it follows from Langer's formula we have $$y_{uniform} (x) \sim C_2 2\sqrt{\pi}(a\epsilon)^{-1/6} \text{Ai}((a/\epsilon^2)^{1/3}x).$$

  • $\begingroup$ I think the Langer's formula is the exact solution in this case. Where is the asymptotic approximation? $\endgroup$ – SimonChan Aug 11 '18 at 18:59

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