# perturbation method for $y' - (y-1)^2 = \epsilon \frac{y^2}{x^2}$ and $y(1) = 1$.

The question follows this exercise in Bender and Orszag.

We would like to find the asymptotic for $$y' - (y-1)^2 = \epsilon \frac{y^2}{x^2}\text{ and } y(1) = 1$$ Dropping the epsilon term, we get two solutions $$y=1 \text{ or } y = 1- \frac{1}{x-c}$$ and the answers there said we will "add perturbation terms to either $y$ or $(y-1)^{-1}$." What does this mean exactly?

And also what is a standard perturbation method for equations like $y' - (y-1)^2 = \epsilon \frac{y^2}{x^2}$?

I am only familiar with the case when $\epsilon$ only appears on the $y''$ term. Here I think we can not just plug in $y = y_0+ \epsilon y_1 + \epsilon^2 y_2+ \cdots$, since there are two possible boundary layers at $x = 0$ or $x=c$ when $y$ takes the non constant solution?

And for what type of ODEs we can simply substitute in $y = y_0+ \epsilon y_1 + \epsilon^2 y_2+ \cdots$ to get the asymptotic?

## 1 Answer

I cannot answer all of your questions, partially because I do not understand some of your questions (such as the "What does this mean exactly?" question, as this differential equation here is different from the one in the link). I do not understand why you say "Here I think we cannot just plug in [...]," since this procedure works perfectly fine in this problem. However, in this answer, I shall show you that you can still use the standard perturbation theory to deal with this differential equation.

I shall write $y=f^\epsilon$ for the solution to $$y'(x)-\big(y(x)-1\big)^2=\epsilon\,\frac{\big(y(x)\big)^2}{x^2}\text{ and }y(1)=1\,.\tag{*}$$ Note that $f^0$ is the constant function $f^0\equiv 1$. Thus, for small $\epsilon$, we expect that $f^\epsilon$ is obtained from $f^0$ by via the perturbation theory. That is, for some functions $f_0,f_1,f_2,\ldots$, we have $$f^\epsilon=f_0+\epsilon\,f_1+\epsilon^2\,f_2+\ldots\,,$$ where $f_0=f^0$. Observe that $f_k(1)=0$ for all $k=1,2,3,\ldots$.

The equation for $f_1$ is obtained by ignoring terms of oder $\epsilon^2$ or higher. From (*) and from $f_0=f^0\equiv1$, we get $$\epsilon\,f_1'(x)=\frac{\epsilon}{x^2}\text{ or }f_1'(x)=\frac{1}{x^2}\,.$$ Consequently, $$f_1(x)=\int_1^x\,\frac{1}{t^2}\,\text{d}t=1-\frac{1}{x}\,.$$

The equation for $f_2$ is obtained by ignoring terms of order $\epsilon^3$ or higher. Since $f_0(x)=1$ and $f_1(x)=1-\frac{1}{x}$, we find that, from (*), $$\left(\frac{\epsilon}{x^2}+\epsilon^2\,f'_2(x)\right)-\epsilon^2\left(1-\frac{1}{x}\right)^2=\frac{\epsilon}{x^2}\,\Biggl(1+2\,\epsilon\left(1-\frac1x\right)\Biggr)\,.$$ Therefore, $$f_2'(x)-\left(1-\frac{1}{x}\right)^2=\frac{2}{x^2}\left(1-\frac1x\right)\,,\text{ or }f_2'(x)=1-\frac{2}{x}+\frac{3}{x^2}-\frac{2}{x^3}\,,$$ yielding $$f_2(x)=\int_1^x\,\left(1-\frac{1}{t^2}\right)\,\text{d}t=(x-1)-2\,\ln(x)+3\,\left(1-\frac{1}{x}\right)-\left(1-\frac{1}{x^2}\right)\,.$$

You can continue so on and so forth, but the deeper you go, the more complicated the differential equation will be. However, for small $\epsilon$ and for $x$ near $1$, you can see that the approximation $$y(x)\approx 1+\epsilon\,\left(\frac{(x-1)}{x}\right)+\epsilon^2\,\Biggl((x-1)-2\,\ln(x)+3\,\left(1-\frac{1}{x}\right)-\left(1-\frac{1}{x^2}\right)\Biggr)$$ gives an estimate of the solution to (*) very well (check this with a numerical solver or something alike). For $x$ further away from $1$ and for larger $\epsilon$, you are going to have to find more terms.

• Thank you very much for the answer. My equation is in the comment of the answer in the link. And also when can we use this standard perturbation series for odes with $\epsilon$ term. I know when we have the singular case $\epsilon y'' + a(x) y' + b(x) y = 0$, we will need to use a WKB series instead of the stadnard perturbation series, is there a general rule to follow for which one to use? – Xiao Aug 5 '18 at 19:35
• I am not aware of criteria when the standard perturbation series should not be used, apart from the singular case. However, it is never a promise (except for few cases) that the perturbation series of any kind will produce a good approximation far away from the boundary points. Usually, you will have two choices: get more terms or do the approximations at other points. – Batominovski Aug 5 '18 at 19:45
• Thank you, and maybe one more classic example is the Duffing equation $y'' +y + \epsilon y^3 = 0$ where the standard perturbation series will diverge as $x$ goes to infinity while the exact solution stays bounded. – Xiao Aug 5 '18 at 19:50
• That is why, sometimes, I would apply perturbation theory at many $x$'s. So I don't have to deal with the divergence problem. Say, you know the value at $x=0$ but you want the value at $x=10$. I would then apply the perturbation at $x=0$ to get an approximate value at $x=0.1$. Then, using what I know at $x=0.1$, I recompute and get the value at $x=0.2$, etc, etc. So, I don't have to go far, and need only to get the perturbation series up to only a few terms. It is probably more accurate than compute the series once and use at all $x$. – Batominovski Aug 5 '18 at 20:00