0
$\begingroup$

I want to use the perturbation method to find a solution to this nonlinear ODE:

$y'' + \epsilon y' + sin(y)=0$

I have this so far:

attempt $u(t) = u_0 + \epsilon u_1 + \epsilon^2 u_2 + ...$ as a solution.

After substituting, using Taylor expansion for $sin(u)$, and sorting, I have the following:

$\epsilon^0$ term: $u_0'' + sin(u_0)=0$.

I can see 1 solution to this ODE which is $u_0(t)=sin(t)+C$. Is there a systematic way to solve this kind of equation?

$\epsilon^1$ term: $u_1'' +u_1 + u_0' - \frac{1}{6} \pmatrix{3\\2} u_0^2 u_1 + ... =0$.

I am not sure if this is correct. Please help.

$\endgroup$
1
  • $\begingroup$ Your solution to the zeroth order ODE is incorrect. The technique you might be looking for here is called "two timing." $\endgroup$ Apr 22, 2022 at 5:03

1 Answer 1

1
$\begingroup$

Your zeroth order equation can be solved by multiplying by $y'$ to get: $1/2dy'^2/dt=d/dt(\cos(y(t))$, by integrating you get: $1/2y'^2=\cos(y(t))+E$ and then integrating once more to get:

$y=\int \sqrt{2\cos(y(t))+2E}dt$. Now obviously you can plug instead of $\cos(y(t))$ its Taylor series, and then expand with a Taylor series of $\sqrt{1+x}$...

Ho ho, the horror... reminds me my thesis... :-D

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .