# Singular Perturbation Problem? Asymptotics?

I am in the midst of solving this equation

$\epsilon \ddot{y}+\dot{y}+1-\frac{1}{(y+1)^{2}}=0$ with the boundary condition $y(0)=1$ and $\dot{y}(0)=-1$

and $\epsilon$ is small. To start off with, I asymptotically expanded $y$ to yield

$y=y_{0}+\epsilon y_{1}$

and substituted the above and solve for $\mathcal{O}(1)$ and $\mathcal{O}(\epsilon)$ for the outer solution. The thing is, how can we make sure that the asymptotic expansion chosen for $y$ is correct? What happens if I choose to expand $y$ like

$y=y_{0}+\epsilon ^2 y_{1}$ or even $y=y_{0}+\sqrt{\epsilon} y_{1}$ to solve the outer solution?

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are there any boundary values specified for your equation? –  Valentin Jul 11 '12 at 22:11
Woops!! Yes there are! –  Joe Jul 12 '12 at 10:23

Your question is, in general, concerned with selecting the correct set of gauge functions to use in your asymptotic expansion. The book "Introduction to Perturbation Methods" by Mark Holmes gives a nice description of gauge functions as well as has numerous examples and problems related to the exponent matching procedure described above.

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@ Andrew WintersWhat do you mean by Gauge Functions? Wiki wasn't very helpful when it comes to this! –  Joe Jul 12 '12 at 15:42
Essentially the gauge functions are the set of ordered, $\epsilon$ dependent functions that you use to expand some function $f$. One set of gauge functions is $\{1,\epsilon,\epsilon^2,\ldots\}$. Another set of gauge functions could be $\{1, e^{-1/\epsilon}, e^{-2/\epsilon}, \ldots\}$. They are also sometimes called scale or basis functions. –  Andrew Winters Jul 12 '12 at 16:18
I see! Thanks for the information! –  Joe Jul 12 '12 at 16:35

Assuming there is a solution with an asymptotic series $y = y_0 + \epsilon^p y_1 + \ldots$, substituting it in to the left side of the differential equation should give something whose asymptotic series is $0$ to all orders. If you used $p > 1$, there would be a problem at order $\epsilon$ (unless $\ddot{y_0}$ happened to be $0$ - that actually does occur with the solutions $y_0=0$ and $y_0 = -2$). If you used $0<p<1$, you might have a problem at order $\epsilon^p$.

Hmmm: actually you do get solutions with $p=1/2$, for example: $$y(t) = \epsilon^{1/2} e^{-2t} - \dfrac{3 }{2} \epsilon e^{-4t} + \left(\dfrac{13}{4} e^{-6t} - 4 t e^{-2t}\right) \epsilon^{3/2} + \ldots$$

EDIT: Actually, if you don't worry about initial conditions, for any $p \in (0,1)$ there will be solutions with $y(t) = y_0(t) + \epsilon^p y_1(t) + o(\epsilon^p)$ (in some neighbourhood of $t=0$). You just specify initial conditions for $y$ to differ by something like $\epsilon^p$ from those for $y_0$. But if (as you now have done) you require fixed initial conditions, things are different.

However, your current initial conditions will not work. If $y_0$ is the solution for $\epsilon = 0$, $y_0(0)=1$ would imply $\dot{y_0}(0) = -3/4$, not $-1$.

Let's try initial conditions $y(0)=1$, $\dot{y}(0) = -3/4$. If you take $y = y_0 + \epsilon^p y_1 + \ldots$ with $0 < p < 1$, the order $\epsilon^p$ term in the equation will say $\dot{y_1} + \dfrac{2 y_1}{(y_0+1)^3} = 0$, but you also need $y_1(0)=0$, and the solution to that would be $y_1 = 0$.

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I think I see what you mean. If we choose $p>1$ there is no way to balance out the $\epsilon$ terms whereas if we choose $0<p<1$, we can't balance $\epsilon ^{p}$ apart from when $p=1/2$! Is that what you are saying? –  Joe Jul 12 '12 at 11:53
@Joe: That's not what I should be saying. I'll edit... –  Robert Israel Jul 12 '12 at 16:36
Hmm, so what this means is that for $p\in(0,1)$ my boundary conditions wouldn't work and your example illustrates just that. I am tempted to try letting $p=1$ just to see if my boundary condition works. I have a strong feeling the latter will be valid though it might be wrong! Nonetheless, thanks for the helpful comment. You gave me a lot to think about! –  Joe Jul 13 '12 at 12:44
Hmm, so what this means is that for $p\in(0,1)$ my boundary conditions wouldn't work and your example illustrates just that. I am tempted to try letting $p=1$ just to see if my boundary condition works. I have a strong feeling the latter will be valid though it might be wrong! Nonetheless, thanks for the helpful comment. You gave me a lot to think about! Edit: You are absolutely right. The boundary conditions do not work, unfortunately even for $p=1$. Why do you think this occur? The equation and boundaries arose from modelling a physical phenomena. –  Joe Jul 13 '12 at 14:30
@Joe: The equation for $\epsilon=0$ says $\dot{y} + 1 - 1/(y+1)^2 = 0$. If $y(0) = 1$ that says $\dot{y}(0) = -3/4$, not $1$. So you're trying to perturb something that doesn't exist. –  Robert Israel Jul 13 '12 at 15:07
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