Past exam of the differential equations 2 course of the University of Oxford / 2023 / 3 / a

Consider real roots. Every zero at $ϵ=0$ is $k\pi$, $k\in \mathbb{Z}$, and its derivative $\cos (k\pi )=(-1{)}^k$ is nonzero. For a regular perturbation, write $x=k\pi +\delta$. The Taylor seriesgives $\delta =(-1{)}^kϵ+O({ϵ}^3)$. Thus, through the requested order,There is no quadratic term. The exact parametrization $x_k=k\pi +(-1{)}^k\arcsin ϵ$ also verifies completeness and the remainder order for $|ϵ|<1$.

Take $ϵ\to 0^{+}$. There are two real branches, on different scales. For a root tending to zero, the Taylor series givesThe dominant balance is $x^3\sim -6ϵ$, so a fractional-power root expansion yieldsMore precisely $x_{-}=-(6ϵ{)}^{1/3}+O(ϵ)$. An ordinary power series in integer powers of $ϵ$ would miss this scale.

There is also a root escaping to positive infinity. Boundedness of the sine term makes the leading dominant balance $x-ϵx^2=0$. The nonzero branch isIndeed the equation rearranges to $ϵx_{+}^2-x_{+}=ϵ-\sin x_{+}$, which yields $x_{+}={ϵ}^{-1}+O(1)$ on this branch. The bounded oscillatory correction need not have a constant limiting value.

To check that these are all the real solutions, solve for the parameter:Positive values occur only on $(-1,0)$ and $(1,\infty )$; $x=\pm 1$ are not roots. A useful exact derivative isIt vanishes only at isolated points, so $E$ is strictly decreasing on each of these intervals. Its endpoint limits are $+\infty$ and zero on each interval. This monotone parametrization of sine perturbation roots gives exactly one root in each interval for every $ϵ>0$, proving that both asymptotic branches, and no others, have been included. If negative small parameters are allowed, the transformation $(x,ϵ)\mapsto (-x,-ϵ)$ reflects these two branches and gives all real solutions for that sign as well.