Solution
= Solution
Now
$$
f=-u^2u_t
=R^3\cos^2\theta\sin\theta.
$$
Therefore
$$
R'=-R^3
\langle\cos^2\theta\sin^2\theta\rangle
=-\frac{R^3}{8},
$$
whereas $R\phi'=0$ because $\langle\cos^3\theta\sin\theta\rangle=0$. Integrating with $R(0)=1$ gives
$$
\frac d{dT}R^{-2}=\frac14,
$$
and hence
$$
\boxed{
R(T)=\left(1+\frac T4\right)^{-1/2},
\qquad
\phi(T)=0}.
$$
The leading uniformly slow-decaying oscillation is
$$
\boxed{u(t)\sim
\left(1+\frac{\epsilon t}{4}\right)^{-1/2}
\cos t.}
$$