Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 336 (Perturbation Methods)

Factor out the elementary exponential and writeUse a matched asymptotic expansion with $0<\lambda \ll \delta \ll 1$. In the outer part,The given identity for the Euler--Mascheroni constant, integrated by parts, impliesHenceapart from pure higher matching powers in $\delta$.

In the inner part, use $t={\lambda }^2/s$:Since $s={\lambda }^2/t\le {\delta }^2$, expand the last exponential. At the required order,With $a={\lambda }^2/{\delta }^2$, this givesThe overlap terms ${\delta }^2$ and $\log \delta$ cancel, leaving the logarithmic switchback termFinally,so

The exponential phase isIts saddle points satisfyFor positive real $\lambda$, the original positive-real contour passes through $s=\lambda$; the singularity at $s=0$ separates it from the other saddle. At the contributing saddle,Set $s=\lambda +{\lambda }^{1/2}t$. ThenThe method of steepest descent therefore gives

WriteThe saddle equation is unchanged, soThe contour connected to the original positive-real integration path passes through $s_{+}$. Since $\Phi (s_{+})=0$, its steepest-descent branch hasFor $s=r{e}^{i\psi }$ and ${\lambda }^2=i{L}^2$,Thus the contour equation isIt approaches the origin along the positive imaginary axis, passes through $r=L$, $\psi =\pi /4$, and tends to infinity along the positive real axis. Along both ends the real part of the phase tends to $-\infty$, so the contour rises from the origin nearly vertically, bends through the first-quadrant saddle, and leaves nearly horizontally. The other saddle lies on a different constant-imaginary-phase contour and does not connect to this deformation.

Introduce independent fast and slow timesand seek $u=u_0+ϵu_1+\cdots$. ThenAt leading order,so writeAt order $ϵ$,The solvability condition in the method of multiple scales removes the resonant sine and cosine components. Averaging over one fast period gives the amplitude-phase equations for a weakly perturbed oscillatorTherefore $u\sim R(T)\cos [t+\varphi (T)]$ without the secular growth that a single-time expansion would produce.

At leading order,soThe required period averages areHenceThe initial data give $R(0)=1$ and $\varphi (0)=0$, soThus the perturbation produces a slow frequency shift but no leading amplitude decay:

NowThereforewhereas $R{\varphi }^{\prime }=0$ because $⟨{\cos }^3\theta \sin \theta ⟩=0$. Integrating with $R(0)=1$ givesand henceThe leading uniformly slow-decaying oscillation is

Set $X=ϵx$. The equation becomesSince $k$ is positive for every $X\ge 0$, there is no classical turning point. The leading WKB approximation for a slowly varying oscillator isThe condition $y(0)=0$ selects a sine, and $y_x(0)=1$ fixes its coefficient. Sincethe result isThe WKB validity measureis uniformly $O(ϵ)$ and tends to zero at both ends of $X\ge 0$. With no turning point, the approximation is therefore uniformly valid on the entire stated half-line.

Use the outer expansionsThe reduced leading equation isFor the symmetric fundamental mode, $y_0=a\cos ({\omega }_{0}x)$. The reduced second-order problem can impose the displacement conditions $y_0(\pm 1)=0$ but not both clamped-slope conditions. The smallest positive root is

Near the right endpoint introduce the stretched coordinateand near the left endpoint use $X_{-}=(1+x)/ϵ$. The leading outer solution is $O(ϵ)$ in either inner region, so put $y=ϵY_0+\cdots$. The dominant inner equation isAt either endpoint, clamping gives $Y_0(0)=Y_0^{\prime }(0)=0$, bounded matching excludes the growing exponential, and the outer behavior requiresThus the two leading inner solutions have the same form:They show explicitly that each clamped endpoint has a boundary layer of width $O(ϵ)$.

At order $ϵ$, the outer equation isThe symmetric solution iswhich is the stated resonant particular solution plus a homogeneous normalization term.

As $x=1-ϵX_{+}$ approaches the right endpoint from the outer region,The large-$X_{+}$ inner expansion isMatching the constant terms givesThe left endpoint gives the same condition by symmetry. Therefore

Away from the thin layer, set $ϵ=0$. The outer equation isand the boundary condition selected by this first-order problem is the right-end condition. Thuswhich satisfies $y_{\mathrm{out}}(1)=1$ but has $y_{\mathrm{out}}(0)=e^{3/8}$.

The missing left condition is supplied by an outflow boundary layer. Put $X=x/ϵ$ and $y=Y(X)$. After multiplication by $ϵ$, the equation isAt leading order,The conditions $Y_0(0)=1$ and $Y_0\to y_{\mathrm{out}}(0)=e^{3/8}$ as $X\to \infty$ give

The common overlap is $e^{3/8}$. The additive composite expansion is thereforeIt satisfies both endpoint values up to exponentially small or higher-order errors and is uniformly valid to $O(1)$ on $0\le x\le 1$.