Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 68 (Perturbation and Stability Methods)

For real $x$, the function $F(x)=x^2{e}^{-x}$ has derivative $F^{\prime }(x)=x(2-x)e^{-x}$. It decreases from infinity to zero on the negative half-line, increases from zero to $4e^{-2}$ on $(0,2)$, and then decreases to zero. Hence sufficiently small positive $ϵ$ gives exactly three real roots: two near zero and one tending to positive infinity.

Set $\delta =\sqrt{ϵ}$ and write either small root as $x=s\delta +A{\delta }^2+O({\delta }^3)$, with $s=\pm 1$. Expanding $x^2{e}^{-x}$ gives ${\delta }^2+s(2A-1){\delta }^3+O({\delta }^4)$. Matching its value to ${\delta }^2$ fixes $A=1/2$. Thus the two asymptotic expansions areFor the large root, put $L=\log (1/ϵ)$. Taking logarithms gives $x-2\log x=L$, so $x/L\to 1$. Writing $x=L+r$ then gives $r=2\log L+2\log (1+r/L)$. Iteration yieldsIts two asymptotic expansion terms are of different logarithmic orders; a regular power series in $ϵ$ cannot describe it.

As an exact check, the real branches of Lambert W give $x_{-}=-2W_0(\sqrt{ϵ}/2)$, $x_{+}=-2W_0(-\sqrt{ϵ}/2)$ and $x_{\mathrm{large}}=-2W_{-1}(-\sqrt{ϵ}/2)$. If complex roots are intended instead, there are infinitely many, given by $-2W_j(\pm \sqrt{ϵ}/2)$. For each fixed nonprincipal branch their two logarithmic terms follow from $W_j(z)=L_j-\log L_j+O(\log |L_j|/|L_j|)$, where $L_j=\log z+2\pi ij$ and logarithms are continued on that branch. The three real roots are the ones selected above.

One convenient version of Watson's lemma is the following. Suppose $G(t)$ is integrable away from zero and has, as $t\downarrow 0$, an asymptotic expansion $G(t)\sim {\sum }_{j\ge 0}{c}_j{t}^{\beta +j-1}$, where $\beta >0$. On a finite interval, or on an infinite interval with suitable exponential growth control,To prove a finite truncation, split the integral at a fixed small $\delta >0$. The part away from zero is exponentially small. On $(0,\delta )$ subtract the first $N$ local terms and bound the remainder by $C{t}^{\beta +N-1}$; its integral is $O({\lambda }^{-\beta -N})$. Extending each polynomial integral to infinity changes it only exponentially, and the substitution $u=\lambda t$ produces the Gamma integral. A little-o local remainder gives the corresponding little-o asymptotic remainder. This is a proof by localization and remainder bounds, rather than merely formal integration of a series.

For the monotone-phase integral, assume the derivatives needed below exist and are bounded near $a$, and the rest of the integral is controlled. Put $t=\varphi (a)-\varphi (x)$. Since ${\varphi }^{\prime }(a)<0$, this is a regular endpoint coordinate andDifferentiating with $dx/dt=-1/{\varphi }^{\prime }$ gives $G(0)=-g(a)/{\varphi }^{\prime }(a)$ and $G^{\prime }(0)=g^{\prime }(a)/{\varphi }^{\prime }(a{)}^2-g(a){\varphi }^{\prime \prime }(a)/{\varphi }^{\prime }(a{)}^3$. Watson's lemma therefore gives the two-term monotone-endpoint Laplace expansionAn $O(e^{\lambda \varphi (a)}{\lambda }^{-3})$ remainder follows, for example, from a sufficiently smooth transformed amplitude with bounded second derivative near zero. The displayed second coefficient can vanish for special amplitudes without changing the expansion.

The unheaded transition request concerns a cubic endpoint-to-saddle transition. Near $x=0$ its exponent has the expansionAt the cubic endpoint $x$ has size ${\lambda }^{-1/3}$. Requiring the linear term to contribute on the same scale gives $1-\alpha =O({\lambda }^{-2/3})$. Thus the distinguished limit and its uniform leading integral areFor fixed $\nu$, the substitution $x={\lambda }^{-1/3}t$ makes the omitted exponent terms $O({\lambda }^{-2/3})$ on bounded $t$ intervals. Cubic decay controls the remaining tail, while the original interval away from zero is exponentially negligible. The estimate is uniform for $\nu$ in compact sets. In the printed range $\alpha \le 1$, $\nu \ge 0$; the same transition integral also makes sense for negative fixed $\nu$.

At $\nu =0$, the Gamma integral gives $h(0)=6^{1/3}\Gamma (1/3)/3$, recovering the cubic endpoint regime. For $\nu \to +\infty$, scale $t=u/\nu$ and expand $e^{-u^3/(6{\nu }^3)}$ to obtain $h(\nu )={\nu }^{-1}-{\nu }^{-4}+O({\nu }^{-7})$. Its leading term gives ${\lambda }^{-1/3}/\nu =1/[\lambda (1-\alpha )]$, recovering the ordinary endpoint regime in their overlap. Both matching limits are therefore checked explicitly.

For fixed $0\le \alpha <1$, the phase derivative is $-1+\alpha \cos x\le -(1-\alpha )<0$ on the integration interval. Its maximum is at $x=0$, with phase zero and derivative $-(1-\alpha )$. The Laplace endpoint estimate, or the substitution $x=t/[\lambda (1-\alpha )]$, givesThis fixed-parameter formula is not uniform as $\alpha \uparrow 1$; its diverging coefficient signals the distinguished limit derived in the parent solution.

At $\alpha =1$, the phase is $-x+\sin x=-x^3/6+O(x^5)$, so the ordinary nonstationary endpoint formula fails. In the degenerate endpoint in Laplace's method, take $x={\lambda }^{-1/3}t$ and extend the localized integral to infinity:The last equality follows by $u=t^3/6$. It is the $\nu =0$ value of the cubic Laplace transition integral after its normalization is rescaled to the coefficient $1/6$.

Let $X=ϵx$ and suppose $k^2(X)$ stays positive and sufficiently smooth in the region considered. Write $y=\exp [{ϵ}^{-1}{S}_0(X)+S_1(X)+\cdots ]$ in ${ϵ}^2{y}_{XX}+k^2(X)y=0$. The leading and first transport equations of the WKB approximation areWith $K=\sqrt{k^2}>0$, they give $S_0^{\prime }=\pm iK$ and $S_1=-\frac{1}{2}\log K+\text{constant}$. Thus the leading WKB approximation for a slowly varying oscillator isThe amplitude is fixed by the transport equation, not just by replacing the constant wave number in a sinusoid. The local validity condition is that variation of $K$ over a wavelength is small, for example $|dK/dx|/K^2\ll 1$.

If $k^2=-{\kappa }^2<0$, the same calculation gives the evanescent formAt a classical turning point where $k^2$ vanishes, these amplitudes diverge and the separated WKB descriptions cease to be uniform. For a simple zero, an Airy turning-point connection formula replaces them in an $O({ϵ}^{2/3})$ interval in $X$; it connects oscillation to exponential behavior with the usual quarter-phase shift. A higher-order zero needs a different local canonical problem, so the Airy statement assumes a simple turning point.

For a smooth positive weight $r$ on the closed interval, let $T={\int }_0^{\pi }\sqrt{r(x)}dx$. The local wave number in the Sturm-Liouville problem is $\sqrt{\lambda }\sqrt{r}$, and the condition at zero selects a sine. Imposing the condition at $\pi$ gives the WKB quantization conditionFor weighted normalization ${\int }_0^{\pi }r|y_n{|}^{2}dx=1$, rapidly oscillating averaging gives $C_n\sim \sqrt{2/T}$. There is no turning-point phase correction here because $r$ stays positive at both endpoints with Dirichlet boundary conditions.

In the specified example $r(x)=(x+1{)}^{-2}$, so $T=\log (1+\pi )=L$ and ${\lambda }_n^{\mathrm{WKB}}=(n\pi /L{)}^2$. To solve exactly, set $t=\log (x+1)$ and $y=\sqrt{x+1}w(t)$. Direct differentiation reduces the Euler-Cauchy equation to $w^{\prime \prime }+(\lambda -1/4)w=0$ on $0<t<L$, with zero endpoint values. Hence the inverse-square weighted Dirichlet spectrum isValues $\lambda \le 1/4$ give no nonzero solution satisfying both endpoint conditions. The WKB eigenfunction shape after leading quantization is actually identical to this exact shape; the leading eigenvalue omits $1/4$.

Using the exact eigenvalue as the reference, its relative error isThus $n\ge 3$ is sufficient and necessary among positive integers. Referencing the approximate value instead gives $n>5L/\pi$, and the same integer answer.

Use normal modes $e^{ikx+st}$, so $s=-i\omega$ in the convention $e^{ikx-i\omega t}$. For a translation-invariant, well-posed linear initial value problem on the whole line, localized data are propagated by a causal Green function. The Fourier transform in space and Laplace transform in time give an inverse integral whose denominator is the dispersion relation $D(k,s)$. The initial temporal contour is a Bromwich line to the right of every real-wave-number temporal singularity.

The Briggs-Bers technique lowers that temporal contour while deforming the spatial contour to avoid its poles. A spatial pinch point occurs when two spatial branches originating on opposite sides of the original Fourier contour collide and prevent further deformation. Algebraically a simple candidate satisfies $D=D_k=0$, with $D_s\ne 0$ and $D_{kk}\ne 0$; these equations alone do not establish the pinch topology. A same-side collision is a false spatial saddle in quartic dispersion. The opposite-side requirement is also explicit in the primary numerical study of pinch-point detection.

The rightmost contributing pinch determines the fixed-position exponential growth of the impulse response, ordinarily with a $t^{-1/2}$ saddle point prefactor. A positive rate is absolute wave-packet instability. If real-wave-number normal modes grow but the impulse response decays at each fixed position, it is convective wave-packet instability; growth can then occur along moving rays. If no temporal normal mode grows, the system is stable, with equality thresholds treated as marginal rather than exponential decay. Along a ray $x=vt$, apply the same argument to $s(k)+ikv$, whose saddle point condition is $s^{\prime }(k)+iv=0$. Thus the laboratory frame is part of the absolute/convective definition.

For the quartic problem the temporal dispersion relation isWrite $a=\mathrm{Re}{a}_1$, $b=\mathrm{Re}{a}_2$, $c=\mathrm{Re}{a}_3$. For $b>0$ the causal Fourier kernel is absolutely convergent for positive time and the finite maximum temporal growth rate isThe usual initial-value framework fails if $b<0$, or if $b=0$ and $a>0$, because arbitrarily large real wave numbers grow arbitrarily fast: no finite Bromwich line can start to their right. This is a high-frequency ill-posedness obstruction, not an absolute instability criterion. If $b=0$ and $a\le 0$, the growth bound is finite, but the dispersive limiting case needs oscillatory Fourier inversion or a limiting-absorption construction when there is no quadratic damping. It must not be justified by the absolutely convergent quartic-diffusion argument. The following explicit calculation first assumes $b>0$ and $a_2\ne 0$.

The double-root candidates areRather than assume all are physical, compute the causal resolvent at the impulse location. Put $S=s+a_3$, and for large $\mathrm{Re}s$ factorThen ${\kappa }_1{\kappa }_2=\sqrt{S/a_2}$ and $({\kappa }_1+{\kappa }_2{)}^2=-a_1/a_2+2\sqrt{S/a_2}$, with sheets selected from that large-$s$ region. Partial fraction decomposition and ${\int }_{\mathbb{R}}(k^2+{\kappa }^2{)}^{-1}dk=\pi /\kappa$ give the quartic impulse-response Laplace resolventAll roots in this expression are continued from the causal region, not selected afresh at a candidate collision. The two possible singularities are now transparent. $S=0$ is a square-root singularity when $a_1\ne 0$. The extra singularity requires $\sqrt{S}=a_1/(2\sqrt{a_2})$. Taking $\sqrt{a_2}$ with positive real part and putting $z=a_1/\sqrt{a_2}$, this lies on the physical $\sqrt{S}$ sheet precisely when $\mathrm{Re}z>0$. If that real part is zero the candidate is on a cut boundary and has nonpositive real $S$, so it cannot lie to the right of $S=0$. For $\mathrm{Re}z<0$, the apparently growing candidate may be on the wrong sheet altogether.

Therefore the physical-sheet growth rate of a quartic impulse response isThese are resolvent branch singularities on the causal continuation, so the formula implements the pinch test instead of merely listing formal saddle point rates. For $a_1=0$, the zero saddle point is fourth-order and the resolvent is proportional to $S^{-3/4}$; its impulse response is proportional to $t^{-1/4}{e}^{-a_{3}t}$ and the same exponential-rate formula applies. At nondegenerate candidates the prefactor is $t^{-1/2}$.

The classification in this diffusive case is now explicit: ${\sigma }_T<0$ gives stability; ${\sigma }_A>0$ gives absolute instability; ${\sigma }_T>0$ with ${\sigma }_A<0$ gives convective instability. When ${\sigma }_T>0$ and ${\sigma }_A=0$, the nondegenerate fixed-position response still decays algebraically, at the marginal absolute/convective threshold. Other equality cases require the stated saddle point prefactors. Temporal instability supplies a growing ray: at a real maximum of $\mathrm{Re}s(k)$, choose $v=-\mathrm{Im}{s}^{\prime }(k)$ so that the moving-ray saddle point is stationary there.

Three examples show why complex coefficients and sheets matter. For $(a_1,a_2,a_3)=(2,1,1/2)$, ${\sigma }_T={\sigma }_A=1/2$, so the instability is absolute. For $(1+2i,1,1/10)$, ${\sigma }_T=3/20>0$ but ${\sigma }_A=-1/10$, so it is convective even though there is no explicit first-derivative advection term: dispersion carries packets in opposite directions. For $(-2,1,1/2)$, both physical rates are $-1/2$. The formal candidates $k=\pm i$ have $s=1/2$ but fail the physical-sheet condition, so this temporally stable example is not absolutely unstable.

On the admissible boundary $b=0$, $a\le 0$, $a_2\ne 0$, the oscillatory or damped Fourier construction has ${\sigma }_T=-c$. The surviving zero-wave-number stationary response has the same exponential rate $-c$, with a quadratic or quartic algebraic factor; damping/dispersion therefore does not generate a separate convective-growth range there. If $a_2=0$, the problem is second order and its diffusion/dispersive kernel must be used instead of the quartic formula. Application to a finite interval or nonlocalized data also changes the spatial inversion problem, so the whole-line localized-data hypothesis is essential.

Seek an outer asymptotic expansion $y=y_0+ϵy_1+\cdots$ for fixed $x>0$. The leading equation is $x{y}_0^{\prime }=0$, and the endpoint condition gives $y_0=1$. At first order, $x{y}_1^{\prime }=y_0=1$ and $y_1(1)=0$, so $y_1=\log x$. This outer correction cannot be evaluated at $x=0$.

The coefficient $x+ϵ$ selects the boundary layer coordinate $X=x/ϵ$. With $Y(X)=y(ϵX)$, the equation becomes $(1+X)Y_X=ϵY$. Its inner leading term is constant; matching it to the outer solution gives $Y_0=1$. At first order, $Y_{1X}=1/(1+X)$, hence $Y_1=\log (1+X)+C$. In the overlap $1\ll X\ll {ϵ}^{-1}$, the outer solution is $1+ϵ\log ϵ+ϵ\log X$. Matching the constants therefore givesfor fixed $X$. The logarithmic matching in a small-exponent boundary layer must retain $ϵ\log ϵ$, which is larger in magnitude than a plain order-$ϵ$ term. At the endpoint,The coefficient of a separate plain $ϵ$ term is zero.

Direct integration provides the exact check:At zero its logarithm is $ϵ\log ϵ-ϵ\log (1+ϵ)=ϵ\log ϵ-{ϵ}^2+O({ϵ}^3)$. Exponentiating gives $y(0)=1+ϵ\log ϵ+O({ϵ}^2{\log }^2ϵ)$, whose remainder is indeed $o(ϵ)$. The first-order uniform composite $1+ϵ\log [(x+ϵ)/(1+ϵ)]$ also satisfies the boundary condition exactly and reproduces both expansions to the requested order.

Let $U$ be real and twice continuously differentiable, take a nonzero real wave number, and write $c=c_r+i{c}_i$. Exponential instability requires $k{c}_i>0$, in particular $c_i\ne 0$, so $U-c$ has no zero on the channel. Divide the Rayleigh equation for inviscid shear flow by $U-c$, multiply by ${\psi }^{\ast }$ and integrate. The wall conditions remove the boundary term, givingIts imaginary part isThe weight is nonnegative and positive wherever the nontrivial eigenfunction does not vanish. A nontrivial solution of this regular second-order equation cannot vanish on an interval. If $U^{\prime \prime }$ had one strict sign without changing sign, the integral could not vanish. If $U^{\prime \prime }$ were identically zero, the real identity would instead give the contradiction $I=0$. Thus $U^{\prime \prime }$ must take both signs, and continuity gives an interior zero. This proves Rayleigh's inflection-point theorem: an unstable profile must have an inflection point, in fact a sign change of $U^{\prime \prime }$ rather than merely an isolated zero without sign change.

Let $U_s=U(y_s)$ at an inflection point supplied by the preceding identity. Its imaginary part, divided by nonzero $c_i$, gives $\int U^{\prime \prime }|\psi {|}^2/|U-c{|}^{2}dy=0$. Its real part isAdding $(c_r-U_s)$ times the zero integral replaces $U-c_r$ by $U-U_s$:Therefore the integrand's profile factor must be strictly negative somewhere in the open channel:This is the Fjørtoft theorem. The positive weight and the strictly negative integral establish the strict inequality, not merely a nonpositive value. It is a stronger necessary condition than Rayleigh's inflection-point theorem, and neither condition alone proves instability.

Use the real diffusive interpretation $\gamma >0$, $U,{\mu }_0\in \mathbb{R}$, with $\nu >0$. Without a positive real diffusion coefficient the printed ordering inequality and real oscillator coordinate would need different hypotheses. Remove the drift by $\eta =e^{Ux/(2\gamma )}v$. Direct differentiation reduces the linear complex Ginzburg-Landau equation toSet $v=b(\xi )e^{-i({\omega }_0+ϵ{\omega }_1)t}$ and $\xi =\sqrt{ϵ}fx$. The resulting equation isCanceling its order-one coefficient and normalizing the quadratic term givesThe supplied decay quantization has $\lambda =2j+1$, $j=0,1,2,\dots$. Equivalently the Hermite functions $b_j=e^{-{\xi }^2/2}{H}_j(\xi )$ solve this harmonic-oscillator problem, where $H_j$ is the physicists' Hermite polynomial. Thus ${\omega }_1=-i(2j+1)\sqrt{\nu \gamma }$ and the exact temporal growth rates areThe global modes of a quadratically confined Ginzburg-Landau equation areFor every fixed positive $ϵ$, the negative quadratic spatial exponent dominates the linear drift exponent and the polynomial at both infinities; all these normal modes are spatially bounded. The largest growth rate is $s_0$, so the least-node normal mode is temporally non-growing, and all normal modes are non-growing, whenEquality is a neutral ground-state normal mode; strict inequality gives decay. This proves the printed sufficient condition for a bounded stable global normal mode and identifies its stronger meaning: stability of the leading global normal mode. Existence of merely some stable higher-$j$ normal mode is not equivalent to the inequality, since sufficiently large $j$ gives $s_j<0$ even above the leading-mode threshold. In the quadratic-coefficient model the reduction and eigenvalues are exact, not just a first-order approximation to a more general profile.