Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 48 (Perturbation Methods)

Put $L=\log \lambda$ and integrate by parts using $x^{-2}=-d(x^{-1})/dx$. The exponential cutoff has derivative $\lambda e^{-x}{e}^{-\lambda e^{-x}}$, so the endpoint at infinity vanishes and the lower endpoint givesWith $u=\lambda e^{-x}$ this becomesThe mass of the transformed integral lies at $u=O(1)$, whereTo control this asymptotic expansion, split at $u=e^{-L/2}$ and $u=e^{L/2}$. On the central interval the geometric-series remainder is bounded by $2(\log u{)}^2/L^3$, whose integral against $e^{-u}$ is finite. The lower tail has length $e^{-L/2}$ and denominator at least $L$; the upper tail is exponentially small in $e^{L/2}$, with the original denominator at least one. The omitted tails of the logarithmic moments are likewise negligible. Thus extending the moments to $(0,\infty )$ is justified.

The Gamma function gives ${\int }_0^{\infty }{e}^{-u}du=1$, and the first logarithmic moment is $-\gamma$, with $\gamma$ Euler's constant. ThereforeThis is the moving Gumbel cutoff in an inverse-power tail: replacing the original cutoff by a sharp step at $x=\log \lambda$ gives the first term, but its smooth transition supplies the Euler-constant correction.

The phase $f(x)=x(1-\cos x)$ is nonnegative on the interval and vanishes only at its two endpoints. An interval away from both endpoints contributes exponentially little, so both endpoint contributions must be expanded before their orders are ranked.

At the left endpoint,The Laplace's method scale is $x={\sigma }^{-1/3}z$. Hence its leading contribution isThe substitution $v=z^3/2$ evaluates the integral by the Gamma function. Since there is no quartic term in the phase, the next left-end correction is relatively $O({\sigma }^{-2/3})$, and thus absolutely $O({\sigma }^{-1})$.

At the right endpoint put $r=2\pi -x$. Thereso $r={\sigma }^{-1/2}z$ givesThe leading right-end integral is a Gaussian integral. Its first correction is also absolutely $O({\sigma }^{-1})$. The mixed cubic and quadratic Laplace endpoints therefore produce the first two global termsKeeping only the dominant cubic endpoint would miss the second term, which is larger than that endpoint's own first correction.

For real $s$, the multiplier $e^{ias}$ has modulus one andThe triangle inequality applied to the integral therefore givesThe hypotheses $s+\sigma >0$ and $t>0$ ensure endpoint convergence. Differentiation in $s$ is justified on compact parameter intervals by integrable bounds with an additional factor $1+|\log x|$. Differentiating both the exponential multiplier and the power of $x$ makes the logarithmic denominator cancel:Here the scaled Gamma function integral is obtained by $u=tx$. Integrating this derivative from zero to $s$ yieldsThe same modulus bound also gives the useful parameter-shift remainder bound for a logarithmic Laplace integralbecause the denominator magnitude is at least $a$.

Fix a small $s>0$, for example $s<\sigma /2$, independently of $t$. Taking imaginary parts of the identity in part (a) givesThe parameter-shift remainder bound for a logarithmic Laplace integral makes the last term $O(t^{-\sigma -s})$, smaller than $t^{-\sigma }{L}^{-k}$ for every fixed $k$.

The endpoint function $h$ is analytic near zero. For any fixed $N$, its Taylor expansion isIntegrating this expansion against $e^{-Lr}$ useswhile the error is $O(L^{-N-2})$ and the extension from $s$ to infinity contributes an exponentially small tail. This is Watson's lemma, and givesEquivalently the coefficients have the stated exponential generating function $h(r)={\sum }_{n\ge 0}{\beta }_n{r}^n/n!$. In particular,Thus the apparent $t^{-\sigma }(\log t{)}^{-1}$ term vanishes: the first nonzero terms are $t^{-\sigma }[\Gamma (\sigma )(\log t{)}^{-2}+2{\Gamma }^{\prime }(\sigma )(\log t{)}^{-3}+\cdots ]$. The derivatives of the gamma function supply the successive logarithmic coefficients.

The relevant boundary values are $a=\pi$ and $\sigma =0$, but the original complex integral defining $g(0)$ requires $\sigma >0$. Substituting zero into it is invalid: its real part has a nonintegrable $dx/(x\log x)$ singularity near zero. Its imaginary part is different, since for positive $\sigma$This has a finite limit as $\sigma \downarrow 0$. On $(0,1)$, $x^{\sigma -1}\le x^{-1}$, whose displayed logarithmic denominator makes the integral finite; at infinity, a fixed upper bound on $\sigma$ and exponential damping provide an integrable dominator. Thus dominated convergence gives $N(t)={\lim }_{\sigma \downarrow 0}F(t,\pi ,\sigma )$.

For the asymptotics, take the imaginary-part parameter identity before passing to that limit, with a fixed $0<s<1/2$. Its endpoint factor converges toby the Gamma reflection formula. Near zero, $\sin (\pi r)\Gamma (\sigma +r)$ is bounded uniformly for small $\sigma \ge 0$, because the sine zero compensates the gamma pole. At $s>0$, the remainder integral itself remains absolutely convergent at $\sigma =0$ and is $O(t^{-s})$. ConsequentlyThe endpoint function is analytic with value one, so Watson's lemma proves the logarithmic expansion of Ramanujan's functionIn particular, with $\gamma$ Euler's constant,This use of imaginary-part regularization at a zero Mellin exponent matters: the positive-$\sigma$ coefficient ${\beta }_0$ in part (b) is always zero, whereas the boundary endpoint function has $h_0(0)=1$. The fixed-positive-parameter expansion is not uniform as $\sigma \downarrow 0$, so its coefficients cannot simply be sent to that boundary term by term.

Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Here the zero-parameter solution is $y_0=1$, but stopping there erases a weakly excited growing mode. Expand $y=1+\epsilon y_1+\cdots$ at fixed $x$. The first correction satisfiesand thereforeThe $\epsilon x^2/2$ term eventually dominates the constant, so this is another singular perturbation on the unbounded domain.

Using the exact logarithmic-variable system from part (a), the initial values are now $z(0)=\epsilon$ and $w(0)=0$. Its linear growing/stable modes have matching coefficientsThus $w\sim Cx$ is order one only at $x=O({\epsilon }^{-2})$, rather than $O({\epsilon }^{-1})$. This is the nonlinear crossover delayed by a small unstable-mode coefficient. The first change in the size of $y$ occurs much earlier, at $x=O({\epsilon }^{-1/2})$, when its weak growing mode becomes comparable to the initial constant; that is not yet the derivative-nonlinearity crossover.

At $X=Cx=O(1)$ the same logistic crossover in a weakly nonlinear Euler equation gives $w=X/(1+X)$ and $H(X)=X-\log (1+X)$. The nonlinear-region solution is now $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-3}H({\epsilon }^{2}x)$ at the lowest order. A composite asymptotic expansion covering all these regions isIt gives $y(1)=1$ exactly, and its initial derivative is $O({\epsilon }^3)$ rather than zero, within leading accuracy. At fixed $x$ its expansion is precisely $1+\epsilon [(x^2-1)/2-\log x]+O({\epsilon }^2)$; for $Cx\ll 1$ it retains the weak $A{x}^2$ growth, and for $Cx\gg 1$ it gives $y\sim x/\epsilon$. The positive equilibrium and boundedness argument from part (a) applies after the initially zero derivative immediately becomes positive. Thus the same eventual slope saturation occurs, but much farther downstream. Both the seeding calculation and the changed scale are essential to a leading approximation on the entire half-line.

Write $T=\epsilon t$ and $q(T)=2e^{-T}$. The forcing acts on the fast initial-time scale, so first remove a localized particular solution. Expanding the coefficient there givesA particular choice is $p_1=(t+1/2)e^{-2t}$, henceIts initial value is $1+\epsilon /2$ and its initial derivative is $-2+O({\epsilon }^2)$. The localized forcing determines WKB phase constants; omitting it before imposing the initial conditions would miss the permanent free oscillation.

For the homogeneous part, the slow-time equation is ${\epsilon }^2{u}_{TT}+q^{2}u=0$. Write its WKB approximation as $A(T)[C\cos \Phi +B\sin \Phi ]$, normalizing $A(0)=1$. The leading amplitude transport equation gives $A=e^{T/2}$ and leading phase ${\epsilon }^{-1}{\int }_0^{T}q(s)ds=2(1-e^{-T})/\epsilon$.

To include every $O(\epsilon )$ contribution on bounded slow-time intervals, the first phase correction is needed. If the local phase wavenumber is $q+{\epsilon }^2{q}_2$, the real part of the amplitude-phase equation gives $q_2=A_{TT}/(2qA)=e^T/16$. The WKB phase correction for a slowly varying oscillator, anchored to zero at the initial time, is thereforeThere is no first-order bulk amplitude correction in this normalization. At $t=0$, the homogeneous initial value must be $-\epsilon /2+O({\epsilon }^2)$ and its initial derivative $1+O({\epsilon }^2)$. Since $A_t^{\prime }(0)=\epsilon /2$ and ${\Phi }_t^{\prime }(0)=2+O({\epsilon }^2)$, these give $C=-\epsilon /2+O({\epsilon }^2)$ and $B=1/2+O({\epsilon }^2)$. The requested forced approximation isfor $0\le \epsilon t\le T_0$ with fixed $T_0$. The approximate initial value is exactly one through this order and its derivative is $-1+O({\epsilon }^2)$. The particular-solution residual is uniformly localized and $O({\epsilon }^2)$; the corrected homogeneous residual is $O({\epsilon }^3)$ on the long interval, producing $O({\epsilon }^2)$ solution error after integration over times of order ${\epsilon }^{-1}$.

The WKB approximation requires the relative frequency-change parameterIt therefore fails when $\epsilon e^{\epsilon t}=O(1)$, even though the coefficient has no finite-time zero. This is the Bessel transition for an exponentially decaying oscillator. Introduce the shifted slow time and amplitudeThe exact rescaled equation isFor fixed $\tau$ its right side is smaller than every algebraic power of $\epsilon$. The leading transition problem is consequentlyWith $Z=2e^{-\tau }$ this becomes $Z^2{U}_{0,ZZ}+Z{U}_{0,Z}+Z^2{U}_0=0$, the order-zero Bessel differential equation. This identifies the governing equation without solving it.

The amplitude-phase series also supplies overlap data while $\epsilon e^T\ll 1$, although its bounded-$T$ absolute error estimate is not uniform up to the transition. For matching, combine the earlier sine and cosine terms into $\frac{1}{2}\sin (\Phi -\epsilon )$ through first order. In the overlap $\epsilon \ll e^{\tau }\ll 1$, their phase isto the retained order. Thus amplitude and phase matching as $\tau \to -\infty$ requireThese are oscillatory amplitude-phase conditions, rather than pointwise ratio statements at zeros. The smaller $e^{\tau }/16$ phase term records the first overlap correction if required. The rapidly varying matching constant ${\alpha }_{\epsilon }$ must be retained as data; there is no single epsilon-independent limiting phase. The two matching conditions determine the leading transition solution, which is deliberately left unsolved.