Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 67 / 4

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.