Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 (Perturbation and Stability Methods)

A regular perturbation admits an asymptotic expansion about the limiting problem on the relevant scale without losing its solutions or conditions. A singular perturbation changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a polynomial may lose its degree.

The two finite roots approach $s=\pm \sqrt{2}$. Set $x=s+ϵa+O({ϵ}^2)$. Substitution into the polynomial gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a regular perturbation of a simple algebraic root, withThe limiting quadratic has only two roots, whereas the original equation has three. The missing root is a divergent algebraic root in a singular perturbation. The dominant balance for algebraic roots sets $x={ϵ}^{-1}X$, giving $X^3+X^2=2{ϵ}^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a{ϵ}^2+O({ϵ}^4)$; the left side is $a{ϵ}^2+O({ϵ}^4)$, so $a=2$. Hence the second nonzero term occurs at order $ϵ$, not order one:For small nonzero real $ϵ$ these three branches are real and distinct; the divergent branch changes sign with $ϵ$. The two bounded branches are regular even though the full root problem is singular.

The integrand is an entire function, so the defining contour is immaterial. A rectangular contour, closed far to the right where the Gaussian function decays, gives the exact horizontal-tail identityThe constant is the half-line Gaussian integral. Expand the endpoint amplitude $e^{-s^2}=1-s^2+\cdots$ in this Watson lemma. Alternatively, repeated integration by parts gives the same expansion, including purely imaginary $z$, because every derivative of $e^{-s^2}$ is integrable on the positive real axis. Thus, on each fixed ray in the stated quadrant,The remainder is measured on the exponential endpoint scale. When that scale grows, this formula does not resolve terms exponentially small relative to it.

Write $z=r{e}^{i\theta }$. The endpoint term has modulus $e^{-r^2\cos 2\theta }/(2r)$. For $0\le \theta <\pi /4$ it is exponentially small, and at $\theta =\pi /4$ it is oscillatory of order $r^{-1}$. For $\pi /4<\theta \le \pi /2$ it dominates the constant. The quadrant asymptotics of the error function therefore give the requested leading asymptotic expansionsThe boundary ray $\theta =\pi /4$ belongs to the constant-dominated case because of the extra factor $1/r$. On the imaginary axis this gives $f(ir)\sim i{e}^{r^2}/(2r)$, consistent with $f(ir)=i{\int }_0^r{e}^{s^2}ds$. The constant/exponential dominance exchange is an anti-Stokes line phenomenon; it should not be confused with dropping the constant everywhere in the quadrant.

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the Euler--Mascheroni constant. The primary normalization is given by dlmf.nist.gov/6.6.E6 .

The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The dominated convergence theorem therefore gives its limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn particular, the leading term is $(2/\pi )\log x$.

For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint stationary phase method and the Fresnel integral yieldA smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the Bessel function of the second kind asymptoticThe additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.

Use primes for differentiation with respect to $X$. Since $X=ϵx$, substituting the amplitude-phase representation into the differential equation and separating real and imaginary parts givesThe second equation is exactly $(\sigma A^2{)}^{\prime }=0$. Thus, with a real constant $C$ and $A\ne 0$,This exact amplitude-phase equation is the Ermakov-Pinney equation for the oscillator. Conversely, any real nonvanishing $A$ satisfying it and $\sigma =C/A^2$ gives an exact complex solution. Existence is not an asymptotic assumption: choose independent real solutions $u,v$ of the original equation, form $y=u+iv$, and set $A=(u^2+v^2{)}^{1/2}$. Their nonzero constant Wronskian prevents simultaneous zeros. The phase derivative satisfies $\sigma A^2=u{v}_x-u_{x}v$, so it is constant. Complex conjugation gives a second branch, and linear combinations recover real solutions.

For the WKB approximation, work on an interval where $K=|k|>0$ is smooth and bounded away from zero. Choose the positive phase branch and an $ϵ$-independent positive $C$. The equations naturally admit even-power expansionsThere is no order-$ϵ$ phase correction in this convention. Leading order gives ${\sigma }_0=K$ and $A_0=(C/K{)}^{1/2}$. Comparing order ${ϵ}^2$ in the real equation yieldsConsequentlyThe constants absorb the normalization. The opposite phase branch changes the sign of $\sigma$ and its correction. A zero of $k$ is a classical turning point where this particular expansion fails; the exact amplitude-phase equation itself remains valid for a suitable complex solution.

For the Sturm-Liouville problem, take $ϵ={\lambda }^{-1/2}$ and use the interval coordinate as $X$, so $K(X)=X+1$ on $[0,1]$. Combining the conjugate branches to satisfy the left Dirichlet boundary condition produces a sine of the phase integral from zero. The right Dirichlet boundary condition requires an integer multiple of $\pi$. No turning-point phase shift is present. We haveThus the two-term Dirichlet WKB quantization isSolving first for the square root and then squaring givesAs an independent sign check, the Liouville transformation of a second-order equation with $t=X+X^2/2$ and $v=\sqrt{X+1}y$ gives $-v_{tt}-3v/[4(2t+1{)}^2]=\lambda v$ on an interval of length $3/2$. The mean transformed potential is $-3/16$, consistent with the constant spectral correction.

Assume $ϵ\downarrow 0$ and fixed $A>0$. Away from the degenerate endpoint $x=0$, the reduced equation is $A{y}_0^{\prime }+y_0=0$. The positive drift makes the rapidly varying homogeneous mode decay as one moves away from the left boundary, so the outer solution is selected by the right boundary condition:It does not satisfy the left value: its limit there is $L=e^{1/A}$.

If the layer width is $\delta$, diffusion is $O(ϵ/{\delta }^2)$, while $Ax{y}^{\prime }$ is $O(A)$ because the drift vanishes linearly. Their dominant balance gives the square-root boundary layer at a vanishing driftWriting $y=Y(\xi )$ gives $Y^{\prime \prime }+\xi Y^{\prime }+[\sqrt{ϵ}/A^{3/2}]\xi Y=0$. The first nonzero inner expansion satisfies $Y_0^{\prime \prime }+\xi Y_0^{\prime }=0$, hence $Y_0^{\prime }=B{e}^{-{\xi }^2/2}$. Imposing $Y_0(0)=1$ and matching $Y_0(\infty )=L$ givesThe overlap region is $\sqrt{ϵ/A}\ll x\ll 1$ for fixed $A$. The additive composite expansioncombines the two leading approximations and satisfies the left value exactly and the right value up to an exponentially small term. Only the first nonzero terms were requested; higher matching terms need not be added.

For fixed $A<0$, the decaying fast mode is at the other endpoint. The outer solution is now fixed by $y(0)=1$, giving $e^{-x/A}$. At $x=1$ the drift does not vanish, so the right layer has width $O(ϵ/|A|)$, with leading transient proportional to $e^{-|A|(1-x)/ϵ}$. In particular its leading inner profile is $e^{-1/A}+(1-e^{-1/A})e^{-\eta }$, where $\eta =|A|(1-x)/ϵ$. The vanishing drift at the left endpoint does not require an order-one mismatch layer for these data.

For $A=0$, the reduced equation is $x{y}_0=0$, which cannot supply the nonzero endpoint values. The equation becomes an Airy ordinary differential equation: $x={ϵ}^{1/3}\xi$ gives $Y^{\prime \prime }+\xi Y=0$. The exact solutions are combinations of $\mathrm{Ai}(-x/{ϵ}^{1/3})$ and $\mathrm{Bi}(-x/{ϵ}^{1/3})$. For $x>0$ they oscillate, with local wavelength of order $\sqrt{ϵ/x}$, rather than forming a decaying exponential layer matched to a smooth nonzero outer profile. The $O({ϵ}^{1/3})$ region resolves the turning endpoint, but oscillations extend throughout the interval. Boundary-value resonances can occur at isolated parameter values. Thus setting $A=0$ is not a continuous substitution into either fixed-sign layer approximation.

Interpret the Fourier transform as a distributional Fourier transform. Exponentially regularize the Heaviside step function:As $a\downarrow 0$, the first term tends to $\pi \delta (k)$ in the tempered distribution sense, since its integral is $\pi$ and it concentrates at zero. The second tends to $-i\mathrm{PV}(1/k)$. ThereforeThe principal-value convention is essential: the printed $1/(ik)$ is not an ordinary integrable function at zero. This normalization also satisfies $ik\hat{H}=1$, the Fourier transform of a derivative version of $H^{\prime }=\delta$.

The real-space expression $1/x$ means the principal-value reciprocal distribution. Apply even exponential damping. Oddness givesDifferentiate the integral with respect to $k$; its derivative is ${\int }_0^{\infty }{e}^{-ax}\cos (kx)dx=a/(a^2+k^2)$. Its value at $k=0$ is zero, so the integral equals $\arctan (k/a)$. Taking $a\downarrow 0$ givesThe limit is a tempered distribution limit: the regularized transforms are uniformly bounded and converge away from zero, allowing dominated convergence against a Schwartz function. This Fourier transform of a principal-value reciprocal argument also establishes the needed Dirichlet integral ${\int }_0^{\infty }\sin u/udu=\pi /2$ in its Abel interpretation.

The locally integrable logarithm has distributional derivative $(\log |x|{)}^{\prime }=\mathrm{PV}(1/x)$. By the Fourier transform of a derivative and the previous result,For $k\ne 0$, this gives $-\pi /|k|$. At zero the expression must be extended as a Hadamard finite-part integral, not treated as a locally integrable function. For example, define the finite-part inverse-absolute-value distribution byThen $kT=\mathrm{sgn}k$. Different reference cutoffs change $T$ only by a multiple of the Dirac delta distribution. The general solution of the transform equation is consequentlyIndeed multiplication by $k$ annihilates only a delta term among distributions supported at zero; delta derivatives would generate additional nonzero terms. This supplies the precise interpretation of the paper's formula, and the convention-dependent constant $C$ need not be found.

For a locally integrable periodic function, its Fourier series coefficients are $c_m={\int }_0^{1}f(x)e^{-2\pi imx}dx$. Window the function to one period by $w(x)=H(x)-H(x-1)$. With the stated Fourier transform convention,There is no extra factor of $2\pi$: the Fourier coefficient already integrates over a period of length one.

For a completely arbitrary periodic distribution, multiplication by a discontinuous window need not be defined, for example at a derivative of a delta supported on a period boundary. The sharp-window assertion therefore presupposes a windowing convention or enough regularity at the endpoints. The smooth-window Fourier coefficients of a periodic distribution use a smooth compactly supported partition-of-unity window $\chi$ with ${\sum }_{j\in \mathbb{Z}}\chi (x+j)=1$; then $c_m=\widehat{\chi f}(2\pi m)$ in the distribution sense. The logarithmic example below is locally integrable, so the printed sharp window is legitimate and no ambiguity affects it.

For $f(x)=\log |x-1/2|$, put $u=x-1/2$. Symmetry gives, for integer $m\ne 0$,An integration by parts has no endpoint term: $u\log u\to 0$ at zero and $\sin (\pi m)=0$ at the upper endpoint. It follows thatwhere the sine integral is $\mathrm{Si}(z)={\int }_0^z\sin t/tdt$. Its limiting value $\pi /2$ follows from the Abel-regularized calculation in part (ii); integration by parts of its tail gives $\mathrm{Si}(\pi m)=\pi /2-(-1{)}^m/(\pi m)+O(m^{-3})$ for positive integer $m$. Hence the Fourier coefficients of a periodic logarithmic singularity haveThe first term is the requested leading behaviour. Its alternating sign records the singularity at half a period, and its $m^{-1}$ decay records the logarithmic cusp. The $m^{-2}$ correction comes from the derivative mismatch across the periodically joined endpoints. Also $c_0=-1-\log 2$ and $c_{-m}=c_m$, consistent with the real symmetric profile.

Take the spatial Fourier transform of the causal Green function. The transformed equation is ${\hat{G}}_t=(\mu -iUk-\gamma k^2)\hat{G}+\delta (t)$, with zero response for $t<0$. Integration across the impulse sets $\hat{G}(k,0^{+})=1$, soInverting the Gaussian Fourier transform givesThis solves the homogeneous equation for $t>0$ and tends to the Dirac delta distribution as $t\downarrow 0$. Its spatial integral is $e^{\mu t}$, as required by the zero-wavenumber growth law.

Along a ray $x=vt$, the exponential growth rate isThe impulse grows in the moving interval of rays $U-2\sqrt{\gamma \mu }<v<U+2\sqrt{\gamma \mu }$, while its peak at $x=Ut$ grows as $t^{-1/2}{e}^{\mu t}$. At a fixed position,Thus, under the printed assumption $\mu >0$, $0<\mu <U^2/(4\gamma )$ gives convective wave-packet instability, because the growing disturbance is swept away from each fixed observation point. $\mu >U^2/(4\gamma )$ gives absolute wave-packet instability, with exponential growth at fixed position. At equality the fixed-position exponential rate is zero and the impulse decays algebraically as $t^{-1/2}$, even though its moving peak grows. The two bounding rays have the same algebraic prefactor.

These statements classify the Green function and localized initial disturbances propagated by its convolution. They do not assert that every possible initial condition decays at fixed position in the convective case: spatially uniform initial data instead produce $e^{\mu t}$. Persistent external forcing can likewise change the observed late-time response. The distinction between convective wave-packet instability and absolute wave-packet instability concerns localized disturbances in a specified frame.

Retain the factor $ϵ$ in the last temporal phase, as printed in the original PDF. Let $a=U/(2\gamma )$ and put $\eta =e^{ax}b(\xi )e^{st}$ with $s=-i{\omega }_0-iϵ{\omega }_1$. The drift-removing substitution for a Dirichlet Sturm-Liouville problem cancels the first derivative:With $\xi =\sqrt{ϵ}fx$, this becomesChoose the positive scale $f=(\nu /\gamma {)}^{1/4}$. To leave an order-one spectral parameter, cancel the numerator's order-one term. ThenThe assumed decaying solutions of the Hermite differential equation require $\lambda =2j+1$, $j=0,1,\dots$. They are proportional to $e^{-{\xi }^2/2}{H}_j(\xi )$, with physicists' Hermite polynomials. Therefore the global modes of a quadratically confined Ginzburg-Landau equation haveFor every fixed positive $ϵ$, multiplying the Hermite-Gaussian by $e^{Ux/(2\gamma )}$ still gives decay at both infinities. The strongest mode is $j=0$, so the whole-line system is globally unstable precisely whenEquality is a neutral ground mode. This spectrum is exact for the stated quadratic profile, even though the notation organizes it as an expansion in $ϵ$.

The homogeneous absolute-instability threshold in part (a) is $\mu =U^2/(4\gamma )$. The global threshold approaches that value as the profile varies increasingly slowly, but at finite $ϵ$ the quadratic confinement adds the loss $ϵ\sqrt{\gamma \nu }$. A locally absolutely unstable patch at the profile maximum is necessary but must be strong and broad enough to overcome that loss. Merely having positive local temporal growth ${\mu }_0>0$, which can be only convective wave-packet instability, is insufficient for a growing decaying global normal mode. This connects the local impulse classification to the discrete global spectrum.