Perturbation: compute an approximation to the solution of the equation $y+\epsilon\sin y=x^2$

Question

Compute approximation to the solution of the equation $y+\epsilon \sin y=x^2$ using perturbation method. Assume that terms involving powers of $\epsilon$ of order 3 or more can be ignored.

So far I only have: $$(y_0 +\epsilon y_1 +\epsilon ^2 y_2 +...) + \epsilon \sin(y_0 +\epsilon y_1 +\epsilon ^2 y_2 +...)$$

Any help appreciated!

Answer 1

Use a series expansion for the $\sin$ term. You can truncate it and collect the terms according to the power of $\epsilon$ and then solve.

Answer 2

Slugger's answer is perfectly correct. A slightly different way of thinking about the problem has been useful to me in past.

Let us restate the problem as follows. Let $\theta_0 \in \mathbb{R}$ be fixed and consider the problem of computing $y = y(t)$, such that $$ \label{main}\tag{1} y(t) + t \sin y(t) = \theta_0, $$ Here $t$ plays the role of $\epsilon$ and we are prepared to restrain $t$ to a small interval around $0$, say, $$t \in I_\delta = (-\delta,\delta).$$ Now, iqnoring, questions of differentiablity for the moment, we assume that the function $y : I_\delta \rightarrow \mathbb{R}$ exists and is smooth and start to collect information about the Taylor series of $y$ at $t=0$. We have $$y(0) = \theta_0.$$ Moreover, by differentiating \eqref{main} with respect to $t$ we have $$ y'(t) + \sin(y(t)) + t\cos(y(t)) y'(t) = 0 \quad \Rightarrow \quad y'(0) = - \sin(y(0)) = -\sin(\theta_0).$$ Repeated differentiation with respect to $t$ will allow you to recursively identify the value of higher order derivatives, but the expressions become ever more horrific. We have $$ y''(t) + \cos(y(t))y'(t) + \cos(y(t))y'(t) - t\sin(y(t))y'(t)^2 + t\cos(y(t))y''(t) = 0,$$ from which it follows that $$ y''(0) = - 2\cos(y(0))y'(0) = 2 \cos(\theta_0) \sin(\theta_0) = \sin(2\theta_0).$$

I will leave the problem of obtaining $y'''(0)$ open and instead return to the question of the existence and differentiability of $y$. I expect that this is outside the intended scope of the answer to the original problem, but there is no harm in stating these things explicitly.

To that end the inverse function theorem is the critical tool. Consider the related function $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $$ \phi(z,t) = (z+t\sin(z),t)^T. $$ It is clear that $\phi$ is differentiable and the Jacobian is $$ D\phi(z,t) = \begin{bmatrix} 1 + t\cos(z) & \sin(z) \\ 0 & 1 \end{bmatrix}.$$ In particular, it is clear that $D\phi(z,t)$ is nonsingular for all $(z,t) \in \Omega = \mathbb{R} \times (-1,1)$. It follows by the inverse function theorem that there exists an open neighborhood $U \subset \Omega$ of $(\theta_0,0)$ such that $\phi$ maps $U$ one to one and onto $\phi(U)$ and has a smooth inverse $\psi : \phi(U) \rightarrow U$. Let $\psi = (\psi_1,\psi_2)^T$ where obviously $\psi_2(\theta,t) = t$. Since $\phi \circ \psi = \text{Id}$ we have $$ \theta = \phi_1(\psi(\theta,t)) = \psi_1(\theta,t) + t \sin(\psi_1(\theta,t)).$$ It follows that $$y(t) = \psi_1(\theta,t)$$ is precisely the function that we have been investigating above. In particular, the question of existence and differentiability are settled by the inverse function theorem.