Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the leading asymptotic behaviour as $x \rightarrow \infty$ of

$$x^2y'' + (1 + 3x)y' + y = 0 $$

Can someone kindly explain me how to solve this problem? Im learning asymptotic analysis, and I want to see how the method of dominant balance works in this example. Thanks

share|improve this question
1  
Quick hint: Assume $3x >> 1$, so ignore the 1. Then find a solution of the form $x^\alpha$ for the approximation. –  p.s. Nov 26 '12 at 3:49

1 Answer 1

up vote 6 down vote accepted

The general idea of the method of dominant balance is fairly simple. Suppose you want to solve some equation $F(y)=0$ with the unknown $y$ and known $F$. In your case, $y$ is a function, $F$ is a differential operator, but the trick applies more or less regardless what we are talking about. Suppose also that $F$ is complicated and no easy explicit solution is in sight but you can split $F$ into two pieces $F=E+G$ such that the equation $E(y)=g$ is easily solvable with any right hand side and $G$ is in some sense much smaller than $E$ for all $y$ you are interested in. After solving $E(y)=0$, you get a solution $y_1$. Then you just say that you got a good approximation and all you need now is to solve $E(y)=-G(y_1)$ to compensate for the extra term. Then you get $y_2$ that solves this equation and switch to $E(y)=-G(y_2)$, etc. If you've ever seen the Picard iteration process, this should sound very familiar.

The whole art is in the splitting. In your case, you can clearly see that $1$ in $(1+3x)$ can be certainly put in $G$ but the rest may not be immediate. Unfortunately, as it turns out, you cannot chop anything off $x^2y''+3xy'+y$. Indeed, suppose you decide that the highest derivative term is in $E$ (something should be there after all). Then you get $x^2y_1''=0$. Let's consider $y_1=1$, say. Then you end up with $G(y)=(3x+1)y'+y$, so $-G(y_1)=-1$. So far, so good. Now let us find the solution to $x^2y''=-1$. We have $y'=\frac 1x+C$ and $y_2=\log x+Cx+C'$. Oops, we seem to get $y_2$ much larger than $y_1$, not a small correction as it should be! This is a clear indicator that our splitting into the dominant and the small part is totally wrong. Playing with this a bit more, we can convince ourselves that we have no choice but to use $E(y)=x^2y''+3xy'+y$. The only part that can go to $G$ is $y'$. Fortunately, solving $E(y)=0$ is not hard. You get $y=\frac 1x$ or $y=\frac{\log x}{x}$ (up to an irrelevant free constant factor). I'll do the Picard iteration game with $y_1=\frac 1x$. Since $E$ is a linear operator, it is enough to solve for the correction term rather than for the whole thing. We need $E(y)=\frac 1{x^2}$. This produces $y_2=\frac 1x+\frac c{x^2}$ where $6c-6c+c=1$, so $c=1$. Then $G(y_2)=-\frac 2{x^3}$, so $y_3=\frac 1x+\frac 1{x^2}+\frac c{x^3}$ with $12c-9c+c=2$, whence $c=\frac 12$. And so on. Just plug'n'play until bored.

This was the cookbook part of the story. The most interesting part is to justify the game. If the sequence you have obtained converges (in the appropriate space), then you can at least say that you found a solution (the limit). Let's see if here it is the case. You have, probably, realized by now that we are getting errors of the kind $c_kx^{-k}$. Each such error gives rise to the correction term $C_kx^{-k}$ with $C_k$ of size about $k^{-2}c_k$, which, in turn, produces a new error $c_{k+1}x^{-k-1}$ with $c_{k+1}$ of size about $k^{-1}c_k$. Thus we are lucky: the errors and correction terms decay quickly and uniformly near $\infty$. You may now say that the whole thing was pure misnomination and we've just done the usual power series representation. Well, that is, indeed, the case here. However, in the truly interesting situations, you get a divergent asymptotic series and the cookbook approach not followed by a careful justification part may give a totally meaningless result.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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