# Nonlinear equation analysis withe epsilon value [closed]

Consider the nonlinear equation
$$\frac{d^2x}{dt^2}+\epsilon\sin(x)=0,~~\epsilon \ll 1\\ x(0)=0,~~\dot{x}(0)=1$$

and find...
A. The value of $$x_0$$ as $$\epsilon$$ goes to $$0$$
B. The first order term $$x_1$$ and the second order term $$x_2$$
C. Plot $$x(t)$$ for $$x_0$$, $$x_1$$, and $$x_2$$
D. Assemble the solution $$x=x_0(t)+\epsilon x_1(t)+\epsilon^2 x_2(t)$$. Plot $$x(t)=x_0(t)$$, $$x=x_0(t)+\epsilon x_1(t)$$ and $$x+x_0(t)+\epsilon x_1(t)+\epsilon^2 x_2(t)$$ on the same graph.

So the thought is that you can expand the sine function around $$x=x_0$$ to obtain the equations you need. Then, using Laplacian transforms, $$x_1 and x_2$$ are solvable.

## closed as off-topic by MisterRiemann, the_candyman, Davide Giraudo, Namaste, KReiserDec 8 '18 at 0:22

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• What is $x_0, x_1$ and $x_2$? Please, provide more details. Most importantly, add also your efforts. Otherwise, I feel that this question will be closed soon and you won't get any answer. – the_candyman Dec 7 '18 at 18:14
• Where did you find this, or what prompted you to conceive it? It it essential that ε be small? And is there some slick answer? I would say it's not massively different from SHM, as when x gets to the domain in which sine levels-off, it will still be follwing a downward-curving parabola. – AmbretteOrrisey Dec 7 '18 at 18:38
• Whoops, did not mean to post. I simply wanted to save for later. Regardless, I updated with more information and direction that I am headed. – Pascal Dec 7 '18 at 20:12
• @Pascal -- I've done that! It's easy, isn't it, when you're using the fields here as your personal equation-editor ... but who would be so delinquent as to do a thing like that!!? ¶ Whhoops! I've just said ... I've done it!! – AmbretteOrrisey Dec 7 '18 at 20:48

You have to solve for the terms in $$x=x_0+ϵx_1+ϵ^2x_2+...$$, so that comparing the powers of $$ϵ$$ you get $$\ddot x_0=0,~~x_0(0)=0,~\dot x_0(0)=1\\ \ddot x_1=-\sin x_0,~~x_1(0)=0,~\dot x_1(0)=0\\ \ddot x_2=-\cos(x_0)\,x_1,~~x_2(0)=0,~\dot x_2(0)=0\\$$

In another approach, multiply with $$2\dot x$$ and integrate to find $$1=\dot x^2+2ϵ(1-\cos x)=\dot x^2+4ϵ\sin^2\frac x2$$ This is a circle equation that can be parametrized via $$\dot x=\cos\phi(t)$$, $$2\sqrtϵ\sin\frac x2=\sin\phi(t)$$ or $$x=2\arcsin\frac{\sin\phi(t)}{2\sqrtϵ}$$. Compare the expressions for the first derivative $$\cosϕ(t)=\dot x=\frac{\cosϕ(t)\,\dot ϕ(t)}{\sqrt{ϵ-\frac14\sin^2ϕ(t)}}$$ Now you can expand $$1=\dot ϕ(t)\left(ϵ-\frac14\sin^2ϕ(t)\right)^{-1/2}$$ in powers of $$ϵ$$ and integrate.