Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 72 (Perturbation and Stability Methods)

For $ϵ>0$, dominant balance for algebraic roots gives two roots on the scale $\sqrt{ϵ}$ and a third on the scale ${ϵ}^{2/3}$. On the first scale put $t=\sqrt{ϵ}s$. The equation becomes $s^5-s^3+\sqrt{ϵ}=0$. The derivative of $s^5-s^3$ at either $s=1$ or $s=-1$ is two, so the first correction is $-\sqrt{ϵ}/2$. On the second scale put $t={ϵ}^{2/3}v$; then $1-v^3+{ϵ}^{1/3}{v}^5=0$. Writing $v=1+a{ϵ}^{1/3}+\cdots$ gives $a=1/3$. Thus all three real-root expansions for positive $ϵ$ areThese are fractional-power root expansions. To check that none is missing, the derivative of the polynomial is $t^2(5t^2-3ϵ)$. Its nonzero critical points are $\pm \sqrt{3ϵ/5}$; for sufficiently small positive $ϵ$ the polynomial is positive at the negative one and negative at the positive one. Its three monotone ranges therefore contain exactly these three real roots. The remaining two small roots correspond to the nonreal cube roots of unity in the second balance.

The printed limit does not specify the sign of $ϵ$. If a two-sided real limit is intended, also put $ϵ=-\eta$ with $\eta >0$. The polynomial $t^5+\eta t^3-{\eta }^3$ is strictly increasing and has just one real root. The same dominant balance for algebraic roots gives the negative-parameter branch

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

The stationary points of the exponent are $t=1/3$ and $t=1$, because ${\varphi }^{\prime }(t)=(3t-1)(t-1)$. The first is a local maximum with $\varphi (1/3)=4/27$ and ${\varphi }^{\prime \prime }(1/3)=-2$; the second is a local minimum. Alsoso the endpoint overtakes the interior maximum only when $b>4/3$.

For fixed $0<b<1/3$, the upper endpoint is the unique maximum. Linearizing the exponent there gives the Laplace endpoint estimate $e^{\lambda \varphi (b)}/(\lambda {\varphi }^{\prime }(b))$. At $b=1/3$, only half of the local Gaussian integral is included. For $1/3<b\le 4/3$, the whole Gaussian near $1/3$ contributes. At $b=4/3$ an endpoint has the same exponential height, but its prefactor is ${\lambda }^{-1}$, smaller than the interior ${\lambda }^{-1/2}$ prefactor. For $b>4/3$, the endpoint has strictly greater height and dominates.

Consequently all fixed-$b$ regimes areThe estimate follows from Laplace's method; the tail toward negative infinity is exponentially harmless since the cubic tends to negative infinity.

Introduce the slow time $T=ϵt$ and use the method of multiple scales with $x_0=A(T)\cos \psi$, $\psi =t+\theta (T)$. The initial data give $A(0)=\sqrt{2}$ and $\theta (0)=-\pi /4$. At first order the equation for the correction isThe averaged amplitude for position-dependent damping follows from the solvability condition in the method of multiple scales, which removes the resonant sine and cosine terms. With brackets denoting a full-period average, this givesThe wave phase average vanishes because $f(A\cos \psi )\sin \psi \cos \psi$ changes sign under $\psi \mapsto 2\pi -\psi$.

For the quadratic position-dependent damping, $⟨{\cos }^2\psi {\sin }^2\psi ⟩=1/8$ and $⟨{\sin }^2\psi ⟩=1/2$, soSolving this logistic equation with $A^2(0)=2$ gives $A(T)=2/\sqrt{1+e^{-T}}$. Thus the uniformly valid leading approximation isHere $T_{\ast }$ is any fixed finite positive constant. Bounded nonresonant corrections and the averaging estimate make the error uniform on this slow-time interval. The leading expression meets the position datum exactly and the velocity datum to leading order; the first-order correction supplies its $O(ϵ)$ adjustment.

Use the same method of multiple scales and its solvability condition in the method of multiple scales. The position-dependent damping function is odd, so the lemma that odd position-dependent damping has zero first-order amplitude drift applies. In its wave amplitude average, $\psi \mapsto \psi +\pi$ changes $\sin (A\cos \psi )$ to its negative while preserving ${\sin }^2\psi$, so the average is zero. Hence $A^{\prime }=0$; the wave phase average again gives ${\theta }^{\prime }=0$.

Equivalently, expand the position-dependent damping in its uniformly convergent power series for bounded $A$. Each term ${\cos }^{2m+1}\psi \sin \psi$ has only even sine Fourier harmonics, so none resonates with the unit-frequency oscillator. With the initial wave amplitude and wave phase unchanged, the leading uniform approximation isThere is no first-order slow drift. Effects at the next order accumulate only an $O(ϵ)$ correction on this time scale.

Let $q_0$ be the stationary point and let $H=D^2\beta (q_0)$ be its real symmetric Hessian matrix. First suppose the stationary point is interior and nondegenerate, the oscillatory integral amplitude and oscillatory integral phase are sufficiently smooth, and boundary contributions are absent or smaller than the stationary contribution. Diagonalize $H$ by an orthogonal change of coordinates, with eigenvalues $h_1,h_2$. Locally,Scaling each principal coordinate by ${\lambda }^{-1/2}$ and evaluating its Fresnel integral gives the two-dimensional stationary-phase formula:Here $\mathrm{sgn}H$ is the number of positive eigenvalues minus the number of negative ones: $2$, $0$ or $-2$. The formula gives the leading nonzero term when $\alpha (q_0)\ne 0$. A partition of unity isolates this neighbourhood; integration by parts away from stationary points makes the remaining interior contribution smaller. For a smooth compactly supported oscillatory integral amplitude the local error is $O({\lambda }^{-2})$.

The printed assumption of a single stationary point alone is insufficient for an unconditional answer. A degenerate oscillatory integral phase $\beta =x^4+y^2$ with a smooth oscillatory integral amplitude supported near the origin has only one stationary point, but scales as ${\lambda }^{-3/4}$. Boundary terms can also contribute at order ${\lambda }^{-1}$: on the unit disk with oscillatory integral amplitude one and $\beta =x^2+y^2$,whose stationary contribution is $i\pi /\lambda$ and whose boundary contribution is $-i\pi e^{i\lambda }/\lambda$. Thus the boxed formula is the standard nondegenerate interior stationary contribution, with the stated hypotheses; the data as printed do not determine all possible leading behaviours.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.