"Proof" that the perturbative method for DE is sensible.

Question

Whenever I try to learn more about the perturbative method for solving DEs, they all contain the same prescription: Just expand the function into a perutrbative series $$ f(x) = x_0 + \varepsilon x_1 + \varepsilon^2x_2 + \cdots $$ And the reader is magically supposed to understand why the different orders, (the zeroth order, first order, etc) are different orders. Whatever that means. This series does not bear the resemblence of anything I have ever seen. Perhaps unless $ f$ is analytical, $\varepsilon = x$, $x_i$ are the derivatives of $ f$ evaluated at the point $x=0$ divided by $i!$, in which case this is a Taylor series. My guess is that this is not the case at all. I reckon some explaination is needed:

(1) It is not at all clear to me what the terms $\varepsilon$ are, are they functions of $x$? Are the $x_i$s functions of $x$? Are neither functions of $x$?

(2) Why can we expect such a series to exist? Some kind of justification is needed. What are sufficient requirements to rigorously prove that such a series exist?

(3) What does it mean that something is of order $j$? How can I determine if something is not order $j + \frac{1}{2}$ or something else? How can we make this precise?

(4) After this has been put into place, can you use the above to "show" that the perturbative method is sensible in solving an ordinary differential equation? The best would perhaps be to prove a bound on the error we are making?

Answer 1

Let's look at a typical example:

$$ \dfrac{dy}{dx} = F(x , y, \epsilon) = y + \epsilon x y^2,\ y(0) = 1 $$

The basic theorem on analytic solutions of ODE's (see e.g. Birkhoff and Rota, "Ordinary Differential Equations", section V.10) says that since $F$ is jointly analytic as a function of the variables $x, y, \epsilon$, this problem has a unique solution $y = Y(x, \epsilon)$ which is jointly analytic in the variables $x, \epsilon$ in some neighbourhood of $[0,0]$. In particular, that says there is a series expansion of the solution

$$Y(x,\epsilon) = \sum_{n=0}^\infty Y_n(x) \epsilon^n$$

where $Y_n(x)$ are analytic functions of $x$ in some fixed interval $[-a,a]$ around $0$, and the series converges uniformly and absolutely for $x$ in this interval and $\epsilon$ in some interval $[-b,b]$ around $0$. Moreover, the basic properties of analytic functions tell us that the series for $\dfrac{\partial Y}{\partial x}$ obtained by differentiating term-by-term $$ \dfrac{\partial Y}{\partial x} = \sum_{n=0}^\infty Y'_n(x) \epsilon^n$$ converges uniformly and absolutely in $[-a,a] \times [-b,b]$ as well, again to an analytic function, and so does the series for $Y(x,\epsilon)^2$ obtained by expanding out the square $$ Y(x,\epsilon)^2 = \sum_{n=0}^\infty \left(\sum_{k=0}^n Y_k(x) Y_{n-k}(x)\right) \epsilon^n$$ Thus we can plug these series in to the differential equation, and because series coefficients of analytic functions are uniquely determined we can equate corresponding coefficients to get a sequence of equations that recursively determines the $Y_n$.

$$ \eqalign{Y_0' &= Y_0\ \text{with}\ Y_0(0)=1\cr Y_1' &= Y_1 + x Y_0^2\ \text{with}\ Y_1(0) = 0\cr Y_2' &= Y_2 + 2 x Y_0 Y_1\ \text{with}\ Y_2(0) = 0\cr}$$ etc.