Consider the pendulum equation $\theta''=a-\sin\theta,\theta(0,a)=\theta'(0,a)=0$. I am trying to find $\frac{d}{da}\theta(t,a)$ at $a=0$.
Taking the derivative of the equation with respect to a gives $\frac{d}{da}\theta''(t,a)=1-\cos\theta(t,a)\frac{d}{da}\theta(t,a)$. Evaluating this at $a=0$ and let $u(t)=\frac{d}{da}\theta(t,a)|_{a=0}$, then $u''(t)=1-u(t)\cos\theta(t,0)$. In addition, $u(0)=\frac{d}{da}\theta(0,0)=\frac{d}{da}\theta(0,a)|_{a=0}=0$ and $u'(0)=\frac{d}{da}\theta'(0,0)=\frac{d}{da}\theta'(0,a)|_{a=0}=0$.
If I could find a way to simplify $\cos\theta(t,0)$ then I think I would be almost done, but I'm not sure how. $\theta(t,0)=\theta(t,a)|_{a=0}$ and so$ \theta'(t,0)=\theta'(t,a)|_{a=0}$, but is there any way to relate this to $u(t)$? Am I missing something obvious?
Edit:
Indeed I missed the obvious uniqueness theorem, because $\theta(t,0)=\theta(t,a)|_{a=0}$ and then $\theta''(t,0)=-\sin\theta(t,0)$ from the original equation. But thate quaton clearly has the trivial solution, which by uniqueness is the only one. Thus, $\cos\theta(t,0)=1$ and the equation for $u(t)$ becomes $u''(t)=1-u(t)$.