Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 336 (Perturbation Methods)

Choose the square root with $-\pi /2<\arg z<3\pi /2$, so the branch cut is the negative imaginary axis. The change of variable $u=e^{-i\pi /4}\sqrt{z}$ maps this sheet onto $\mathrm{Re}u>0$ and gives $z=i{u}^2$. Writing the exponential function as $e^{k\varphi (z)}$, its phase becomes exactly quadratic:The saddle point is therefore $u=1$, or $z_s=i$. On the descending line $u=1+iv$, $v\in \mathbb{R}$, the phase is $-1-v^2$. Its image is the parabolaThe contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A contour deformation in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no branch point or pole is crossed. Since $dz=-2(1+iv)dv$, the Gaussian integral and the odd function $v{e}^{-k{v}^2}$ giveHere the method of steepest descent actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.

For a fixed pole away from the saddle point, the smooth amplitude there is $(i-z_0{)}^{-1}$. The simple-saddle contribution in steepest descent isThe residue theorem supplies an additional contribution precisely when the contour deformation crosses $z_0$. The region swept above the original contour has positive orientation: the original contour from left to right followed by the reversed saddle contour encloses it counterclockwise. Consequently the original integral equals the saddle integral plus the pole contribution:For $|z_0|<1$, an upper-half-plane pole outside the indentation has $\chi =1$, while a lower-half-plane pole has $\chi =0$. The entire upper unit half-disk lies below the saddle parabola, so there is no further case inside that half-disk. With a fixed indentation radius $\delta$, a pole inside the upper semicircle, $|z_0|<\delta$, is also below the original contour and has $\chi =0$; real points in the indentation gap are excluded as well. Taking $\delta <|z_0|$ for a fixed nonzero pole gives the usual upper/lower classification. The pole contribution need not always dominate exponentially; its size depends on $\mathrm{Re}\varphi (z_0)$, so keeping both terms makes that dependence explicit.

A pole on the original contour requires a stated Cauchy principal value or an indentation prescription. A pole parameter on the negative imaginary branch cut still defines the integral: the contour stays on its fixed sheet, and only the denominator uses $z_0$. Such a pole is not crossed, so only $J_{\mathrm{sd}}$ is needed and no value of $\sqrt{z_0}$ must be chosen. The residue exponential above is evaluated only when $\chi =1$. At $z_0=0$ the original indentation excludes the singular point, so no additional residue is crossed. These conventions matter before taking any limiting pole position.

The ordinary saddle-point approximation with a nearby pole loses uniformity when the pole lies within the width of the saddle's Gaussian function profile. The distinguished limit isUse the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and defineOn the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful partial fraction decompositionThe first term contains the nearby pole; the second is regular at the saddle and contributes $O(e^{-k}{k}^{-1/2})$. Approaching from below the descending contour means $\mathrm{Im}{s}_d<0$. The residue theorem then gives the Gaussian saddle-pole transitionThe exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\varphi (z_0)=-k-s_d^2/4$ recovers the pole exponential in the requested expression.

The Gaussian pole integral expresses the bracket as $i\pi w(s_d/2)$, where $w$ is the Faddeeva function. Thus a convenient uniform leading answer isThe error estimate applies for bounded scaled pole position on the indicated side. In the overlap $1\ll |s_d|\ll \sqrt{k}$ below the real axis, the integral contributes $-2\sqrt{\pi }/s_d$ to leading order while the residue remains $2\pi i{e}^{-s_d^2/4}$. Since $s_d\sim \sqrt{k}(z_0-i)$, this reproduces the near-saddle limit of the separate saddle and residue terms. Boundary values on the descending contour are limits of this combined expression; the isolated real-axis pole integral itself needs a Cauchy principal value prescription.

The solution and its first derivative are continuous at $x=1$: a jump would create a Dirac delta distribution or its derivative, absent from the forcing. Put $r=x-1$, $A=a+1$ and $B=b$. The outer expansions, fixed by the respective endpoint boundary conditions, areBalancing diffusion and advection near $r=0$ gives the interior layer at a simple zero of advection with inner variable $z=r/\epsilon$. The leading inner expansion solves $Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$, so matching to $A$ and $B$ givesFor the next term define the Gaussian drift primitivesWriting $H$ for the Heaviside step function, let $F(z)=zH(-z)+(\sqrt{\pi }/4)\mathrm{erf}|z|$. Its value and first derivative match at zero and $(D^2+2zD)F=2zH(-z)$. Thuswhere the logarithmic overlaps fixHere $C_1,C_2$ are the constants in the supplied large-positive-$z$ limits of $E_0,E_1$. The terms $\epsilon \log \epsilon$ are essential switchback terms; discarding them would not give accuracy through $O(\epsilon )$.

Subtracting the common overlap from the inner expansion and outer expansions gives the additive composite expansionThe standard overlap subtraction leaves $\epsilon rH(-r)$. Replacing $H(-r)$ by $(1-\mathrm{erf}z)/2$ changes the value only by $O({\epsilon }^2)$ uniformly and gives a composite with continuous first derivative. The last term restores the endpoint values through the retained order. When $b=a+1$, $d=0$ and the order-one error-function jump disappears. An $O(\epsilon )$ interior adjustment remains to accommodate the leading outer derivative mismatch, together with logarithmic matching when $A\ne 0$.

Reversing the diffusion sign changes the leading inner equation to $-Y_0^{\prime \prime }+2z{Y}_0^{\prime }=0$. Its nonconstant solution grows like the integral of $e^{z^2}$, so bounded matching forces the same leading value $C$ on both sides. The bulk is $C+r$ on the left and $C$ on the right; both endpoint conditions are instead supplied by decaying endpoint layers for reversed diffusion, of width ${\epsilon }^2$. A weaker width-$\epsilon$ interior adjustment matches derivatives and selects $C$. Indeed, its first-order equation has a non-growing solution only ifThe leading structure is consequentlywith the smaller interior correction understood. Unlike the positive-diffusion case, the endpoint values are not transported into two distinct order-one inner limits.

An integrating factor explains the second hint and the first-order inner calculation. Formultiplication by $e^{z^2}$ gives $(e^{z^2}{Y}^{\prime }{)}^{\prime }=e^{z^2}(2a_{1}z+a_2+a_3\mathrm{erf}z)$. The term $a_{1}z$ already produces $2a_{1}z$, and integrating the remaining terms twice gives $a_2{E}_0+a_3{E}_1$, in terms of the Gaussian drift primitives. The homogeneous solutions are a constant and the error function. HenceFor the piecewise forcing in the inner problem, $F$ replaces $a_{1}z$ and provides the continuously matched particular solution for $2zH(-z)$. The coefficients $2c$ and $2d$ then account for the $2Y_0$ forcing.

The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

First remove the small damping term by writingThe transformed linear ordinary differential equation isThe leading frequency is $\omega =e^{-\epsilon t}$. Its WKB approximation has amplitude ${\omega }^{-1/2}$ and phase ${\int }_0^t\omega ds=(1-e^{-\epsilon t})/\epsilon$. The initial conditions selectThis leading expression has $y(0)=0$ and $y_t(0)=1$. Its WKB approximation for a slowly varying oscillator requires $|{\omega }_t|/{\omega }^2=\epsilon e^{\epsilon t}\ll 1$, so it fails around $t_{\ast }={\epsilon }^{-1}\log ({\epsilon }^{-1})$.

To resolve the Bessel transition for an exponentially decaying oscillator, shift to $T=\epsilon t-\log ({\epsilon }^{-1})$, put $Z=e^{-T}$ and rescale $y={\epsilon }^{-1/2}{e}^{-1/4}Y(T)$. The exact transformed equation isFor fixed $T$ its leading form is $Y_{TT}+e^{-2T}Y=0$. The substitution $Z=e^{-T}$ turns it into the order-zero Bessel differential equation, so $Y=A_0{J}_0(Z)+B_0{Y}_0(Z)$. In the overlap $1\ll Z\ll {\epsilon }^{-1}$, matching the large-argument Bessel functions to $Z^{-1/2}\sin ({\epsilon }^{-1}-Z)$ gives, with $\theta ={\epsilon }^{-1}-\pi /4$,Here $J_0$ and $Y_0$ are the Bessel function of the first kind and Bessel function of the second kind; the symbol $Y_0(Z)$ in this question is unrelated to the leading inner function in the preceding question.

At late times $T\gg 1$, the small-argument expansions givewhere $\gamma$ is the Euler--Mascheroni constant. The solution becomes asymptotically linear rather than maintaining the exponentially growing WKB envelope. Its leading late-time slope is ${\epsilon }^{1/2}{e}^{-1/4}\sqrt{2/\pi }\cos \theta$. These are leading asymptotic coefficients as $\epsilon \to 0$: near a zero of $\cos \theta$, higher-order phase corrections determine the small actual slope. The formula is not an absolute-error estimate uniform to arbitrarily late times. The exact late slope of an exponentially damped oscillator, obtained from a Kummer function and a Wronskian, provides a separate check even near those exceptional phases.

The transition equation reduces to the Bessel differential equation because $d/dT=-Zd/dZ$ and $d^2/d{T}^2=Z^2{d}^2/d{Z}^2+Zd/dZ$. Thus $Y_{TT}+e^{-2T}Y=0$ becomes $Z^2{Y}_{ZZ}+Z{Y}_Z+Z^{2}Y=0$.

The Bessel function of the first kind has $J_0(Z)=1+O(Z^2)$ at zero, while the Bessel function of the second kind has $Y_0(Z)=(2/\pi )(\log (Z/2)+\gamma )+O(Z^2|\log Z|)$. Since $\log Z=-T$, these yield a constant and a linear function of $T$, explaining the late-time behavior. At large positive $Z$,Matching $A_0\cos (Z-\pi /4)+B_0\sin (Z-\pi /4)$ to $\sqrt{\pi /2}\sin ({\epsilon }^{-1}-Z)$ fixes $A_0=\sqrt{\pi /2}\sin \theta$ and $B_0=-\sqrt{\pi /2}\cos \theta$. The large- and small-argument asymptotic expansions thereby connect the oscillatory WKB approximation to the nonoscillatory late-time solution through one Bessel transition for an exponentially decaying oscillator.