# Approximation of ODE [duplicate]

This question already has an answer here:

A problem I should solve is the following IVP: I need to find an approximation for small $\varepsilon$ of $$y'' + y + \varepsilon y^3 = 0, y(0) = 1, y'(0)=0$$

My approach so far was:

I assume that I already know a solution, call it $y(t)$. Then I have a look at the Taylor series at $t-\varepsilon$ in order to get the following:

$$y(t) = \sum_{k=0}^{\infty} \frac{y^{(k)}(t-\varepsilon)}{k!}(\varepsilon)^k$$

This gives me a dependency on epsilon, and can only have a look at the first two terms of the series and say that for sufficiently small epsilon the terms $\mathcal{O}(\varepsilon^2)$ can be neglected.

The next thing I do is using the IV, giving me:

$$y(0) = y(-\varepsilon)+ y'(-\varepsilon)\varepsilon = 1$$ and $$y'(0) = y'(-\varepsilon)+y''(-\varepsilon)\varepsilon = 0$$

Using both equations I get the ODE $$y(-\varepsilon) -\varepsilon^2y''(-\varepsilon)=1$$ and solve that Euler-type ODE.

I am confused about the meaningfulness of this last ODE and this approach in general. The solution will be dependent on $\varepsilon$ and I am unable to see how it should approximate the first given ODE.

Any help is greatly appreciated!

## marked as duplicate by LutzL, Community♦May 18 '16 at 16:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – LutzL May 18 '16 at 16:13
• cf. this article on the Poincaré-Lindstedt perturbation method. – Harry49 Aug 27 '18 at 7:29

## 1 Answer

That is not at all the standard approach. The standard approach is to notice that the problem is well-posed when $\varepsilon=0$, indeed the solution in this case is $\cos(x)$. Now write $y=\sum_{n=0}^\infty \varepsilon^{\alpha_n} y_n(x)$, where $\alpha_n$ is an unknown increasing sequence of numbers,$\alpha_0=0$, and $y_n$ are unknown functions. (In many situations $\alpha_n=n$.) Then substitute that into the ODE, taking $y_0$ to be the solution with $\varepsilon=0$. Then use the principle of dominant balance after that. This means that you group together all terms with a given power of $\varepsilon$ and set that aggregate equal to zero. In your case the ODE looks like

$$\sum_{n=0}^\infty \varepsilon^{\alpha_n}(y_n''+y_n) + \varepsilon \left ( \sum_{n=0}^\infty \varepsilon^{\alpha_n} y_n \right )^3 = 0.$$

The lowest order term in the expansion of $\varepsilon y^3$ is $\varepsilon y_0^3$. So select $\alpha_1=1$, then solve

$$y_1''+y_1+y_0^3 = 0,y_1(0)=0,y_1'(0)=0$$

where the initial conditions are zero because all of the initial condition information is already in $y_0$. (If the initial conditions depended on $\varepsilon$ then you might have something nonzero here.) Thus you are left to solve $y_1''+y_1+\cos(x)^3=0,y_1(0)=0,y_1'(0)=0$, which is standard albeit a bit tedious.

Then you continue up to higher powers; it becomes cumbersome eventually because it is difficult to expand the cube for a large sum, but it can be done.

This is called regular perturbation theory. When the problem with $\varepsilon=0$ is not well-posed, the techniques used are called singular perturbation theory.