Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 76 (Perturbation and Stability Methods)

The dominant balance for algebraic roots separates one small root of a polynomial from three large ones. For the small root of a polynomial, write $z=\epsilon +a{\epsilon }^3+O({\epsilon }^5)$ in $z=\epsilon (1+z^2{)}^2$. Comparing coefficients gives $a=2$, soThis is a real analytic function of real $\epsilon$ near zero, since the derivative with respect to $z$ of $z-\epsilon (1+z^2{)}^2$ is one at $(0,0)$.

For the large roots of a polynomial, use the real cube root $s=\mathrm{sgn}(\epsilon )|\epsilon {|}^{1/3}$ and set $z=w/s$. The transformed equation isIts three nonzero leading roots of a polynomial are $w=\omega$, with ${\omega }^3=1$. Writing $w=\omega +a{s}^2+O(s^4)$ gives $a=4a+2{\omega }^2$, hence $a=-2{\omega }^2/3$. ThereforeThe choice $\omega =1$ is the second real root of a polynomial; the other two form a complex conjugate pair for either sign of $\epsilon$. The nonzero transformed roots of a polynomial are simple, so their Taylor series in $s^2$ exist locally and exhaust the three large roots of a polynomial. Together with the small one these give all four roots of a polynomial for sufficiently small nonzero $\epsilon$. At $\epsilon =0$ itself the rational equation has only the finite root zero; the other three escape to infinity. Multiplying by $(1+z^2{)}^2$ has not introduced roots at $z=\pm i$, since those points do not solve the multiplied equation.

Use the unscaled Finite Fresnel integrals of this paper. Termwise integration of the absolutely convergent Taylor series for sine and cosine gives the full small-argument expansionsIn particular,These Taylor series converge at every finite argument, although cancellation makes them unattractive for large arguments.

For the large-argument asymptotic expansions, combine the two Finite Fresnel integrals into $C+iS$. The standard Fresnel integral givesFor the tail $J(\lambda )={\int }_{\lambda }^{\infty }{e}^{i{t}^2}dt$, integration by parts using $(e^{i{t}^2}{)}^{\prime }=2it{e}^{i{t}^2}$ givesFor example $a_1=1/4$, $a_2=-3i/8$, and $a_3=-15/16$. Repeated integration by parts bounds the remainder after any fixed number of terms by the order of the next term. Equivalently the Fresnel tail expansion isHere $(-1)!!=1$. The first three terms of the requested cosine Fresnel integral, counting its constant term, areFor comparison, retaining one further correction in both Finite Fresnel integrals givesUnlike the small-argument Taylor series, these large-argument asymptotic expansions must be truncated; adding indefinitely many terms at a fixed argument does not improve the approximation.

A calculator needs elementary arithmetic and its built-in sine and cosine, together with stored approximation coefficients. A convenient global construction uses compactified approximation from endpoint asymptotics. Define the exact real tail functions byThen $P(0)=Q(0)=A$, while $P\sim 1/(2\lambda )$ and $Q\sim 1/(4{\lambda }^3)$ at infinity. Consequentlyextend continuously to the compact interval $0\le u\le 1$, with $p(1)=1/2$ and $q(1)=1/4$. Choose stored polynomial approximations $p_N,q_N$ on that interval, constrained by the small-argument Taylor series and large-argument asymptotic expansions, and calibrated across the intervening interval. The resulting elementary formula isUniform errors ${\delta }_p,{\delta }_q$ in the stored polynomial approximations give $|C_{\mathrm{calc}}-C|\le {\delta }_p+{\delta }_q$ for every positive argument. Existence of arbitrarily accurate such approximations follows from the Weierstrass approximation theorem. A Padé approximant with a denominator having no positive zero is an alternative implementation. Matching the two end expansions alone is not an error guarantee: the middle interval must also be controlled. For very small arguments, switching to the short Taylor series avoids cancellation. Ordinary finite-precision limitations in evaluating ${\lambda }^2$ and reducing its trigonometric phase remain those of the calculator itself.

Apply the method of multiple scales with independent fast time $t$ and slow time $T=\epsilon t$, and write $x=x_0(t,T)+\epsilon x_1(t,T)+\cdots$. The chain rule gives $d/dt={\partial }_t+\epsilon {\partial }_T$. At leading order,At first order,A forcing component at the natural frequency produces a secular term in $x_1$. Thus the solvability condition in the method of multiple scales is orthogonality to both $\sin \theta$ and $\cos \theta$. Define the fast-phase average, holding $R,\varphi ,T$ fixed, bySince $⟨{\sin }^2\theta ⟩=⟨{\cos }^2\theta ⟩=1/2$ and $⟨\sin \theta \cos \theta ⟩=0$, the amplitude-phase equations areThey are leading averaged equations, not exact equations for a purely sinusoidal solution of a general nonlinear oscillator.

For the cubic conservative force, $f=-R^3{\cos }^3\theta$, odd symmetry gives $⟨{\cos }^3\theta \sin \theta ⟩=0$, while $⟨{\cos }^4\theta ⟩=3/8$. HenceIf $x(0)=x_{\ast }$ and $x^{\prime }(0)=v_{\ast }$, leading initial matching givesand thereforeon bounded slow time intervals for fixed bounded initial data. The positive frequency shift agrees with the increased restoring force of $x^{\prime \prime }+x+\epsilon x^3=0$. The zero initial state is the exact zero solution, with irrelevant phase. If a first waveform correction is desired, the remaining first-order forcing is $-R_0^3\cos 3\theta /4$, so $x_1=R_0^3\cos 3\theta /32$ plus a homogeneous correction chosen to match the initial data at the next order. This exhibits explicitly why the leading harmonic ansatz is approximate.

For the oscillator coupled to slow damping feedback, expand $y=Y(T)+\epsilon y_1(t,T)+\cdots$, alongside the method of multiple scales oscillator expansion. The leading equation ${\partial }_{t}Y=0$ makes $Y$ independent of fast time. The oscillator force evaluated on $x_0=R\cos \theta$ is $2YR\sin \theta$. The amplitude-phase equations therefore giveThe next equation for the damping variable isA bounded periodic $y_1$ exists only when the mean of its right side vanishes. Thus the full leading averaged system isSet $R_0=\sqrt{x_{\ast }^2+v_{\ast }^2}$, ${\varphi }_0=atan2(-v_{\ast },x_{\ast })$, and $Y_0=y_{\ast }$. These specify the arbitrary leading initial conditions.

For $g(x)=x$, the phase average is zero. ConsequentlyThus $x(t)=R_0{e}^{-Y_0\epsilon t}\cos (t+{\varphi }_0)+O(\epsilon )$ and $y(t)=Y_0+O(\epsilon )$. Positive $Y_0$ damps the oscillator, negative $Y_0$ amplifies it, and zero $Y_0$ leaves the leading amplitude unchanged. The fast correction $y_1=R\sin \theta$ plus an initial-matching constant confirms that $y$ can oscillate slightly even though $Y$ is constant. For $R_0=0$, the exact solution is $x=0$, $y=y_{\ast }$.

For the logarithmic feedback, assume $R_0>0$. Its phase singularities are integrable, and the supplied integral givesWriting $L=\log (R/2)$ turns the nonlinear amplitude-phase equations into the linear pairso $L_{TT}+L=0$. With $L_0=\log (R_0/2)$,The first integral $L^2+Y^2=L_0^2+Y_0^2$ provides a direct check, and the amplitude remains positive on every finite slow time interval. At the retained averaged order, the physical approximation is $x\simeq R(\epsilon t)\cos (t+{\varphi }_0)$, $y\simeq Y(\epsilon t)$.

The logarithmic case needs an interpretation at oscillator zeros. The original right side is undefined at $x=0$, so a globally classical solution through each crossing, or a solution from $x_{\ast }=v_{\ast }=0$, cannot be asserted. At a transverse crossing, $\log |x|\sim \log |t-t_{\ast }|$ is locally integrable; $y$ can be interpreted as absolutely continuous, with its equation holding away from the crossings and almost everywhere. In that integrable interpretation the displayed averaged formulas apply formally for nonzero leading amplitude. The standard smooth-forcing averaging theorem cannot be invoked without addressing this singularity. The phrase arbitrary initial conditions must exclude the identically zero state for this logarithmic problem.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

With the streamfunction convention $(u,v)=({\psi }_y,-{\psi }_x)$, disturbance vorticity is $\omega =-\Delta \psi$. The basic Couette flow has constant vorticity $-1$. Linearizing the two-dimensional Euler equations for an inviscid fluid in vorticity form givesThe second term vanishes, soThis derivation uses incompressibility and an inviscid homogeneous fluid, without introducing a viscous decay term.

For the initial streamwise Fourier mode, write $\psi =e^{i\alpha x}a(y,t)$. Its transported Laplacian has amplitudeThe decaying Green function on the line isThis is the negative of the one-dimensional modified Helmholtz Green function for $-{\partial }_y^2+{\alpha }^2$: it is continuous and its first derivative with respect to $y$ has jump one at $y=s$. Consequently $({\partial }_y^2-{\alpha }^2)G=\delta (y-s)$, and inversion yieldsUnder the usual decay of the initial profile and its derivatives, the initial integral returns $\varphi$. There is no nonzero homogeneous solution of $a^{\prime \prime }-{\alpha }^{2}a=0$ decaying at both infinities.

For the large-time calculation, put $q=\alpha t$ and $H(s)=G(y,s)F(s)$, keeping the observation point $y$ fixed. Assume sufficient localization to justify the following integrations by parts; for example, $F$ may be a Schwartz function. The function $H$ is continuous, but its derivative in $s$ has jumpsince $G_s(y,y+)-G_s(y,y-)=1$. The Fourier decay from a Green-function derivative jump givesHere the regular second derivative is computed separately on either side of $s=y$. If it is integrable, its Fourier transform tends to zero by the Riemann-Lebesgue lemma. HenceThe cusp in the Green function, rather than a stationary point of the linear phase, supplies this leading contribution.

To obtain the streamwise velocity, differentiate the integral itself. This avoids an unjustified differentiation of a pointwise remainder. The kernel $G_y$ has a jump of $-1$ as a function of $s$ at $s=y$, so one integration by parts givesThe transverse velocity follows exactly from $v=-i\alpha \psi$, while the Laplacian equation gives vorticity without an asymptotic approximation:Thus the generic orders are $u=O(t^{-1})$, $v=O(t^{-2})$, and $\omega =O(1)$, at each fixed $y$. If $F(y)=0$, the leading coefficients vanish and faster decay can occur. For real physical disturbances, take the real part of these complex Fourier modes. The vorticity generally keeps oscillating rather than approaching a pointwise limit, even while the velocity decays. This is linear inviscid damping in unbounded Couette flow caused by phase mixing.

For arbitrary initial vorticity, the characteristics satisfy $\dot{x}=y$, $\dot{y}=0$, and $\dot{\omega }=0$. Therefore the exact solution isPutting ${\omega }_0=-F(y)e^{i\alpha x}$ recovers the preceding nondecaying mode. In general the shear is area preserving, so its transport preserves every finite Lp norm of vorticity. Local pointwise behavior for arbitrary ${\omega }_0$ depends on its spatial profile: an initially localized profile can move away from a fixed point. There is no universal pointwise nonzero limit or $t^{-2}$ decay of vorticity for arbitrary initial data.

The stated smoothness and decay of $\varphi$ alone do not guarantee the claimed algebraic rates. The intended integration by parts calculation requires control of derivatives and tails; convergence of the displayed integral alone is insufficient. Here is a counterexample to the literal unrestricted rate, with $\alpha =1$. Choose a nonnegative smooth function $r$ supported in $(-1/4,1/4)$ with $\int r=1$, and a sufficiently rapidly increasing sequence $t_n$. SetChoose $\sum e^n/t_n<\infty$. Then $F$ is smooth and integrable, $\varphi$ is smooth and tends to zero at both infinities, and every evolution integral converges absolutely. At the observation point $(x,y)=(0,0)$, all supports have $s>0$, soAt $t=t_n$, the $n$th summand is $-1/(2t_n)$. The Fourier transform of $r$ decays faster than any inverse power. Inductively choose $t_n$ so large that all earlier summands at $t_n$ together have modulus less than $1/(8t_n)$, and choose subsequent times so that ${\sum }_{m>n}1/t_m<1/(4t_n)$. The later summands then also have modulus less than $1/(8t_n)$. Thus $|\psi (0,0,t_n)|\ge 1/(4t_n)$, contradicting $O(t^{-2})$. This construction meets the printed assumptions; what it lacks is uniform derivative localization. The preceding boxed decay formulas are the correct intended result under the explicitly stated sufficient localization hypothesis.