Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 67 (Perturbation and Stability Methods)

For fixed $\mu \ne 0$, a large positive root must lie close to a pole of the tangent function: away from its poles, $\tan x$ cannot balance the unbounded right side. Label the adjacent poles by $a_n=(n+1/2)\pi$ and write $x_n=a_n+{\delta }_n$. The Taylor series of $-\cot \delta$ givesSeek the asymptotic expansion $\delta =A/a_n+B/a_n^3+\cdots$. The coefficients of $a_n$ and $a_n^{-1}$ giveThus the large roots near tangent poles satisfyThe first correction already supplies the requested dependence on $\mu$. For $\mu >0$ the root approaches the pole from below; for $\mu <0$ it approaches from above. This labels roots by their nearby poles, avoiding an irrelevant finite shift in the enumeration of positive roots.

No positivity restriction on $\mu$ is required. The expansion requires $\mu \ne 0$ fixed and $|\mu |a_n\gg 1$, so the displacement is small. It is not uniform as $\mu \to 0$. At $\mu =0$ the roots are exactly $n\pi$, a different leading sequence; these cannot be recovered by setting $\mu =0$ in the pole expansion. If $\mu$ varies with $n$, its size must be checked against the small-displacement and successive-term conditions rather than using the fixed-parameter remainder blindly.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore

The method of multiple scales introduces independent variables for the fast oscillation and its slow modulation, for example $t_0=t$, $T=ϵt$, and, when needed, $T_2={ϵ}^{2}t$. The derivative becomes $d/dt={\partial }_{t_0}+ϵ{\partial }_T+{ϵ}^2{\partial }_{T_2}+\cdots$. One expands the solution in $ϵ$ while allowing its leading amplitudes and phases to depend on the slow times.

A solvability condition in the method of multiple scales removes resonant forcing of each fast homogeneous mode. Otherwise a correction contains a secular term, such as $ϵt\sin t$, and an ordinary asymptotic expansion fails on long times. The resulting slow amplitude equations incorporate that accumulated effect in the leading solution. One must still check that the amplitudes remain within the assumed weak-coupling regime and that omitted slower effects have not accumulated.

Assume $0<ϵ\ll 1$ and $X,Y,\alpha =O(1)$. For $n=2$, use the method of multiple scales with $T=ϵt$ and complex amplitudesAt first order the forcing equations areThe product has frequencies $\pm 1,\pm 3$. Its $e^{it}$ coefficient is $\overline{A}B$, while it has no $e^{2it}$ coefficient. Removing the secular terms givesThe leading initial conditions give $A(0)=X/2$ and $B=Y/2$. Put $b=Y/2$, $A=p+iq$. The primary parametric resonance of a two-to-one oscillator pair is governed byFor $b^2>{\alpha }^2$, choose $\lambda >0$. Solving with $p(0)=X/2,q(0)=0$ yieldsIf $b^2<{\alpha }^2$, let $\nu =\sqrt{{\alpha }^2-b^2}/2$; replace $\cosh (\lambda T)$ by $\cos (\nu T)$ and $\sinh (\lambda T)/\lambda$ by $\sin (\nu T)/\nu$. On the boundary $b^2={\alpha }^2$, take the continuous limit:Thus exponential growth occurs when $|Y|>2|\alpha |$ and $X\ne 0$, on the scale $t=O({ϵ}^{-1})$. Equality is a marginal case that can produce linear slow growth: for $b=\alpha \ne 0$ the displayed initial phase excites it, whereas for $b=-\alpha$ it does not. If $X=0$, $x$ stays identically zero and the other oscillator is exactly uncoupled. The leading $y$ amplitude is constant; its backreaction is a higher-order effect on this time scale.

These are leading approximations for bounded slow time $T$, hence $t=O({ϵ}^{-1})$, while the amplitudes stay $O(1)$. First-order corrections can enforce the initial velocities exactly; the leading formula alone has an $O(ϵ)$ initial-velocity mismatch. In the unstable case the expansion must not be extrapolated through arbitrarily large amplification, where weak coupling and omitted terms cease to be uniform.

For $n=1$, both leading oscillations have frequencies $\pm 1$, so their quadratic product contains only frequencies $0,\pm 2$. There is no first-order resonant coupling: the solvability condition in the method of multiple scales gives $A_T=i\alpha A/2$ and $B_T=0$, a frequency shift without first-order amplitude growth. The constant and second-harmonic corrections can feed back into the fundamental at second order. Consequently the first possible growth scale from this second-order resonance from quadratic oscillator coupling isThis identifies the possible slow scale, not a guaranteed instability for every parameter or initial state. In particular, the first-order $\alpha$ frequency shift may detune the second-order resonance unless appropriately tuned; a second-order amplitude analysis would decide actual growth.

Let $K(X)=|k(X)|>0$ on a smooth interval, with $X=ϵx$. For the WKB approximation for a slowly varying oscillator, writeSubstitution into ${ϵ}^2{y}_{XX}+K(X{)}^{2}y=0$ gives, at the first two orders,Thus $S_0^{\prime }=\pm iK$ and $S_1^{\prime }=-K^{\prime }/(2K)$. The leading WKB approximation isThe $K^{-1/2}$ amplitude is the transport correction accompanying the rapid wave phase. Real solutions are equivalent real sine and cosine combinations.

For validity, $K$ must be smooth, nonzero and slowly varying compared with the local wavelength. In physical $x$ derivatives, sufficient local checks are $|K_x|/K^2\ll 1$ and $|K_{xx}|/K^3\ll 1$. Indeed each displayed branch has relative residualSmooth positive $K(X)$ bounded away from zero has these properties on fixed slow intervals. At a zero of $k$ the WKB approximation fails and a classical turning point requires a different local analysis, often an Airy turning-point connection formula. Rapid variations, coefficient singularities and excessively long accumulation of wave phase error also lie outside this leading approximation.

Assume $ϵ\to 0^{+}$. Since the drift $(1+x{)}^2$ is positive, the fast mode decays to the right; the matched asymptotic expansion has a left boundary layer at $x=0$ of width $O(ϵ)$, and the outer expansion uses the condition at $x=1$.

Write $y\sim y_0+ϵy_1$. The first two equations for the outer expansion areEnforcing $y_0(1)=1$ and $y_1(1)=0$, defineThen the outer asymptotic expansion isTo derive the correction, set $y_1=Eq$. Cancellation of the homogeneous terms leaves $q^{\prime }=-2/(1+x{)}^5-1/(1+x{)}^6$; integration and $q(1)=0$ give the expression above.

In the left boundary layer, set $\xi =x/ϵ$ and $y=Y_0(\xi )+ϵY_1(\xi )+\cdots$. The equation becomes $Y^{\prime \prime }+(1+ϵ\xi {)}^2{Y}^{\prime }+ϵY=0$. Put $A=e^{1/2}$ and $B=1-A$. The leading inner equation and matching give $Y_0^{\prime \prime }+Y_0^{\prime }=0$, $Y_0(0)=1$, $Y_0(\infty )=A$, hence $Y_0=A+B{e}^{-\xi }$. At the next order,Since $q(0)=53/80$, put $C=53A/80$. The matched inner correction satisfying $Y_1(0)=0$ isFor verification, the differential operator $d^2/d{\xi }^2+d/d\xi$ maps $e^{-\xi }P(\xi )$ to $e^{-\xi }(P^{\prime \prime }-P^{\prime })$. The large-$\xi$ common part is $A+ϵ(C-A\xi )$, precisely the small-$x$ outer expansion in the overlap $ϵ\ll x\ll 1$.

Adding the inner expansions and outer expansions and subtracting their common part gives the first-order composite expansion for a positive driftThis is uniformly correct through $O(ϵ)$ on $0\le x\le 1$. It satisfies the left boundary condition exactly; at the right boundary the remaining error is exponentially small. Polynomial growth in the inner correction is harmless because it is multiplied by the decaying exponential. For the smooth nonvanishing drift, the next matched terms are uniformly $O({ϵ}^2)$.

Reversing the drift reverses the endpoint carrying the boundary layer. The outer equation is $-(1+x{)}^2{y}_0^{\prime }+y_0=0$, and now its condition comes from $x=0$:Its value at $x=1$ is $e^{1/2}$, so it cannot by itself satisfy the right boundary value. Put $\xi =(1-x)/ϵ$ there. The leading inner equation is $Y^{\prime \prime }+4Y^{\prime }=0$, with a correction proportional to $e^{-4\xi }$ that decays into the domain. Thus the outer region has size $O(1)$ and the right endpoint layer has width $O(ϵ/4)=O(ϵ)$. One would expand the coefficient near $x=1$, solve the successive equations for the inner expansion, match as $\xi \to \infty$, and form a matched asymptotic expansion by subtracting the overlap. No turning region occurs because the drift does not vanish on this interval.

Set $s=x+1\in [0,2]$. The drift now vanishes quadratically at the left endpoint, and the outer expansion selected from $s=2$ is proportional to $e^{1/s}$, which becomes singular as $s\downarrow 0$. The oscillatory endpoint with a quadratically vanishing drift is not an ordinary monotone layer of width ${ϵ}^{1/3}$: balancing just diffusion and drift on that scale neglects the zeroth-order $y$ term.

The formal local exponential rates obey $ϵ{\lambda }^2+s^2\lambda +1=0$. For $s\gg {ϵ}^{1/4}$ their slow and fast branches are approximately $-s^{-2}$ and $-s^2/ϵ$. They coalesce when $s^4\sim 4ϵ$. This locates the main transition a distance $O({ϵ}^{1/4})$ from the left endpoint.

A reliable procedure removes the first derivative exactly:Use WKB approximation for this transformed equation on each side of its unique positive zero $s_t=\sqrt{2}{ϵ}^{1/4}+O({ϵ}^{1/2})$. For $s>s_t$ the transformed modes are exponential; for $0<s<s_t$ they are oscillatory. Their physical amplitudes include the displayed exponential prefactor.

At $s_t$, $P^{\prime }(s_t)=O({ϵ}^{3/4})$, so the Airy turning-point connection formula usesThis layer connects the oscillatory and exponential WKB approximation branches. On the transition scale $s={ϵ}^{1/4}S$, the exact equation is ${ϵ}^{1/2}{v}_{SS}=(S^4/4-1+{ϵ}^{1/4}S)v$, displaying a small effective WKB approximation parameter ${ϵ}^{1/4}$.

Close to $s=0$, the local wavelength is $O(\sqrt{ϵ})$: setting $s=\sqrt{ϵ}\xi$ gives $Y_{\xi \xi }+Y+\sqrt{ϵ}{\xi }^2{Y}_{\xi }=0$, whose leading solutions are sine and cosine. This endpoint form imposes $y(-1)=1$. Within the oscillatory region, the prefactor changes by order one over the additional envelope scale $s=O({ϵ}^{1/3})$. The $\sqrt{ϵ}$ wavelength and ${ϵ}^{1/3}$ envelope are subscales of the same oscillatory region, not additional omitted ordinary layers.

Use the outer exponential/regular branches for $s\gg {ϵ}^{1/4}$, oscillatory WKB approximation inside the $O({ϵ}^{1/4})$ transition region, an $O({ϵ}^{5/12})$ Airy neighborhood of $s_t$, and the $O(\sqrt{ϵ})$ local endpoint form to apply the left boundary data. Match their constants and enforce the right boundary data at $s=2$; there is no separate right endpoint layer required. Matching amplitudes need not remain $O(1)$, and parameter values admitting a homogeneous Dirichlet solution require a separate solvability check rather than an assumed uniform algebraic expansion.

Here the singular perturbation has an exact conservative structure:The integrating factor for this first-order linear differential equation givesAt $x=1$ and $x=-1$ the integral terms have opposite signs while the exponential factors agree. Equal boundary data therefore force $C=0$, and normalization gives the unique exact solutionThis exposes the Gaussian interior layer of a conservative drift equation. Its central scale is $x=O(\sqrt{ϵ})$, where $x=\sqrt{ϵ}\xi$ gives $Y^{\prime \prime }+\xi Y^{\prime }+Y=0$. The central Gaussian function profile is $e^{-{\xi }^2/2}$ with amplitude $e^{1/(2ϵ)}$.

Away from the center, $|x|\gg \sqrt{ϵ}$, use exponential WKB approximation branches rather than an assumed bounded algebraic outer solution. The reduced ordinary differential equation $x{y}_0^{\prime }+y_0=0$ would give $C/x$; matching both equal positive endpoint values with that algebraic branch is impossible. The exact solution selects its zero coefficient and the exponentially large branch with a Gaussian function profile instead. Near either endpoint, a distance $O(ϵ)$ changes $y$ by order one relative to its endpoint value: with $r=(1-|x|)/ϵ$, $y=e^{r-ϵr^2/2}$. These endpoint scales describe normalization of the same global exponential solution.

Thus a central $O(\sqrt{ϵ})$ turning region, exponential outer regions on both sides, and $O(ϵ)$ endpoint normalization scales give the complete regional description. The peak is exponentially large, so no uniformly bounded regular outer expansion should be presumed. The exact expression already provides a uniform solution without further matching calculations.

Substitution of a normal mode gives the dispersion relationFor real $k$, $\sigma$ is the temporal growth rate. Since $\beta \ne 0$, its supremum is finite exactly whenFor $\beta <0$, arbitrarily short wavelengths grow arbitrarily fast. For $\beta >0$, the finite maximum temporal growth rate isThe Briggs-Bers criterion starts the inverse temporal Laplace transform above all temporal singularities and then deforms its contour downward while following the spatial roots. A finite growth bound supplies such an initial contour and a causal, high-frequency-controlled Green function. Unbounded temporal growth prevents that standard construction.

For absolute wave-packet instability, a candidate spatial pinch point must satisfy $D=0$, $D_k=0$, and $\mathrm{Im}\omega >0$. HeregivesThus candidate growing saddles require $\gamma <0$ at $k_0$, or $\gamma <{\alpha }^2/(4\beta )$ at $k_{\pm }$. Together with $\beta >0$, existence of at least one such candidate requires $\gamma <{\alpha }^2/(4\beta )$.

A growing double root is not sufficient: the roots must pinch the spatial inversion contour from opposite sides. Collisions of branches originating in the same spatial half-plane do not obstruct the relevant deformation. This distinction is part of the Briggs-Bers criterion; it is stated, for example, in the primary study doi.org/10.1017/jfm.2016.195.

An explicit false spatial saddle in quartic dispersion is $\alpha =-2$, $\beta =1$, $\gamma =1/2$. Its candidates $k=\pm i$ have $\omega =i/2$, but all real modes have $\sigma =-k^4-2k^2-1/2<0$. Each imaginary collision joins two branches in the same half-plane; neither is a relevant pinch. For $\omega =i\Omega$, the spatial roots obey $k^2=-1\pm \sqrt{1/2-\Omega }$, making those same-half-plane collisions transparent as $\Omega \downarrow 1/2$.

For this particular real, even dispersion relation one can also establish the actual threshold directly. At the origin its impulse Green function isLaplace method selects the real maximum, and gives a positive prefactor times $t^{-1/2}{e}^{{\sigma }_{\max }t}$ because $\alpha \ne 0$. Therefore the actual absolute wave-packet instability condition isThese are sufficient for this model as well as necessary. They follow after identifying relevant real saddles; the earlier algebraic double-root test alone lacks the pinch information. Equality is marginal, not exponential absolute growth.

Use the standard Fourier transform on the whole real axis. The printed lower limit is missing a minus sign. For this integrable even function, its only obstruction to smoothness is the origin, so its high-frequency algebraic terms come from its local cusp:Localize near zero with a smooth cutoff. The smooth polynomial terms, including $|x{|}^6=x^6$, contribute no algebraic cusp term; their localized transforms decay faster than any prescribed inverse power. The supplied half-line formula, interpreted away from $k=0$ in an Abel-regularized or tempered distribution sense, gives the Fourier transform of an algebraic cuspFor $p=3$ the cosine is $1$, and for $p=9$ it is $-1$. Both cusp coefficients in the local expansion are $-1$. HenceThe second coefficient is $29!=725760$. The absence of a $|k{|}^{-7}$ term follows from the smooth even power $x^6$, not from neglecting an available correction.

One can verify the signs by integration by parts on $2\mathrm{Re}{\int }_0^{\infty }{e}^{-ikx}(1+x^3{)}^{-1}dx$. Its endpoint derivatives first have nonzero relevant values $f^{(3)}(0)=-3!$ and $f^{(9)}(0)=-9!$, which contribute twice those values divided by $(ik{)}^4$ and $(ik{)}^{10}$. Higher derivatives are integrable on the half-line, justifying the displayed remainder after further integrations. Equivalently these coefficients are the Fourier decay from a derivative jump at the origin.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.