Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 79 / 1

Let $h=\sqrt{1-m^2}$ be the complementary modulus of the complete elliptic integral of the first kind. With $\phi =\pi /2-\theta$, write the integrand as $({\sin }^2\phi +h^2{\cos }^2\phi {)}^{-1/2}$. Its logarithmic endpoint asymptotic of the complete elliptic integral comes from $\phi =O(h)$; an expansion at fixed $\phi$ alone misses the constant accompanying the logarithm.

Choose an intermediate cutoff $h\ll d\ll 1$. In the endpoint region, the integral isOutside this region, the leading integral isThe arbitrary cutoff cancels in this matched asymptotic expansion, leavingThese are the divergent logarithm and the finite constant, the requested first two terms. If terms are instead grouped by powers of the complementary modulus, the next refinement is $K(m)=\log (4/h)+(h^2/4)[\log (4/h)-1]+O(h^4\log (1/h))$; its coefficients agree with the DLMF expansion.

There are two contributing regions: an endpoint of width $x^{-1}$ near $u=0$, and a very narrow peak near $u=1$. Both must be retained in the asymptotic expansion.

For the peak, set $a=e^{-x}$ and $u=1+av$. At fixed $v$, $\log u\sim av$ and $e^{-xu}\sim a$, so the integrand times $du$ approaches $dv/(1+v^2)$. This is the Lorentzian contribution from shrinking regularization. Its total contribution isFor example, cut off this region at $|u-1|=e^{-x/2}$: its scaled endpoints tend to infinity, while the changes in the exponential and logarithm inside the peak are exponentially small. The matched tails outside it are also exponentially small compared with the endpoint terms below.

At the endpoint, put $v=xu$ and $L=\log x$. To algebraic orders in $L^{-1}$, the regularizing denominator term is negligible there, giving the Laplace endpoint expansion with a logarithmic denominatorThis integral represents the endpoint expansion only; it is cut off before the original peak when making the argument precise. Since ${\Gamma }^{\prime }(1)=-{\gamma }_E$, with ${\gamma }_E$ the Euler--Mascheroni constant, the first two contributions to the original integral areThe endpoint correction refines to $[x{L}^2{]}^{-1}[1-2{\gamma }_E/L+3({\gamma }_E^2+{\pi }^2/6)/L^2+\cdots ]$, using derivatives of the gamma function. An application of Watson's lemma confined to $u=0$ would omit the leading peak contribution $\pi$.

The Briggs-Bers technique starts with a temporal Laplace transform and spatial Fourier transform of a localized disturbance. The temporal inversion contour must initially lie above every temporal singularity. Lower it while deforming the spatial contour away from the roots of the dispersion relation. A collision of spatial branches originating on opposite sides of that contour can create a spatial pinch point. A contributing pinch with $\mathrm{Im}{\omega }_{\ast }>0$ gives absolute wave-packet instability: exponential growth at a fixed position. Growth that is transported away while a fixed position decays is convective wave-packet instability. Solving $D=D_k=0$ finds candidate double roots, but does not by itself establish a pinch.

Substitution of the normal mode givesFor real wavenumber $k$, the temporal growth rate isA nonzero real cubic coefficient in this expression makes it unbounded above on one end of the real axis. Once that coefficient vanishes, a positive quadratic coefficient also makes it unbounded above. Hence the bounded temporal growth condition for cubic dispersion isThe maximum is then $-\mathrm{Re}\gamma$, attained at $k=0$ and, when $\mathrm{Re}\alpha =0$, at all real $k$. A finite upper bound is necessary for choosing the initial temporal inversion contour; arbitrarily fast high-wavenumber growth prevents the usual Briggs-Bers criterion from defining that causal inversion for generic localized data.

For $\beta \ne 0$, the stationary candidates satisfysoEach candidate must pass the spatial-branch pinch test before its imaginary frequency is used. In particular, under the finite-growth conditions the real-wavenumber evolution has multiplier modulus at most $e^{-t\mathrm{Re}\gamma }$. For integrable transformed initial data this bounds its fixed-position amplitude by a constant times that exponential. Thus a further necessary condition isThe false complex saddle in dissipative cubic dispersion illustrates why the second formal candidate cannot override this bound. For $\mathrm{Re}\alpha <0$, the neighborhood $k=0$ gives a fixed-position impulse response proportional to $e^{-\gamma t}/\sqrt{t}$: the cubic term is smaller on its $k=O(t^{-1/2})$ scale. It therefore supplies the usual growing pinch when $\mathrm{Re}\gamma <0$. The purely dispersive cases have stationary-phase algebraic prefactors instead. If $\beta =0$, only the quadratic stationary point remains; if both derivative coefficients vanish, the equation is local multiplication and its growth is directly $e^{-\gamma t}$.

Write $r=\sqrt{\mu /\delta }$. At first order the amplitude-phase perturbation is $A=r+\epsilon (A_1+ir\theta )+O({\epsilon }^2)$. Expanding the cubic term in the focusing nonlinear Schrodinger equation gives $A|A{|}^2=r^3+\epsilon (3r^2{A}_1+i{r}^3\theta )+O({\epsilon }^2)$. The constant terms cancel because $\delta r^2=\mu$. Separating real and imaginary parts yieldsFor the Fourier normal modes, these equations give $-i\Omega {\hat{A}}_1=-\chi r{K}^2\hat{\theta }$ and $-ir\Omega \hat{\theta }+(2\mu -\chi K^2){\hat{A}}_1=0$. Their determinant vanishes exactly whenThe focusing nonlinear Schrodinger modulation dispersion is negative for $0<K^2<2\mu /\chi$. One of the two frequencies then has positive imaginary part and grows exponentially: the constant state has modulational instability. The growth rate is $\sqrt{\chi K^2(2\mu -\chi K^2)}$, maximized at $K^2=\mu /\chi$ with value $\mu$. Modes beyond the unstable band have real frequencies. The zero-wavenumber phase mode is neutral, with possible linear phase drift from a uniform amplitude perturbation.