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

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.

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

With the negative coefficient of $y^{\prime \prime }$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left boundary condition to the outer solution, givingThe boundary layer still has width $O(ϵ)$. Put $X=(1-x)/ϵ$ and $Y(X)=y(1-ϵX)$. The rescaled equation isIts leading inner expansion has $Y_0=A+B{e}^{-X}$. The asymptotic matching condition is $Y_0(\infty )=e^{-1/2}$, while $Y_0(0)=0$. ThusSubtracting the overlap $e^{-1/2}$ in the additive composite expansion givesIts graph starts exponentially close to one, follows the gently decreasing outer solution, and drops from near $e^{-1/2}$ to zero in a right boundary layer of width $O(ϵ)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/ϵ}$. Its right endpoint slope is $-e^{-1/2}(1+1/ϵ)$, showing the rapid final descent.