2
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $(1)$.

http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/06/

http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/06/

$\endgroup$
0
$\begingroup$

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 $$y_{uniform} (x) \sim C_2 2\sqrt{\pi}(a\epsilon)^{-1/6} \text{Ai}((a/\epsilon^2)^{1/3}x).$$

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.