Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 68 / 4

Let $U$ be real and twice continuously differentiable, take a nonzero real wave number, and write $c=c_r+i{c}_i$. Exponential instability requires $k{c}_i>0$, in particular $c_i\ne 0$, so $U-c$ has no zero on the channel. Divide the Rayleigh equation for inviscid shear flow by $U-c$, multiply by ${\psi }^{\ast }$ and integrate. The wall conditions remove the boundary term, givingIts imaginary part isThe weight is nonnegative and positive wherever the nontrivial eigenfunction does not vanish. A nontrivial solution of this regular second-order equation cannot vanish on an interval. If $U^{\prime \prime }$ had one strict sign without changing sign, the integral could not vanish. If $U^{\prime \prime }$ were identically zero, the real identity would instead give the contradiction $I=0$. Thus $U^{\prime \prime }$ must take both signs, and continuity gives an interior zero. This proves Rayleigh's inflection-point theorem: an unstable profile must have an inflection point, in fact a sign change of $U^{\prime \prime }$ rather than merely an isolated zero without sign change.

Let $U_s=U(y_s)$ at an inflection point supplied by the preceding identity. Its imaginary part, divided by nonzero $c_i$, gives $\int U^{\prime \prime }|\psi {|}^2/|U-c{|}^{2}dy=0$. Its real part isAdding $(c_r-U_s)$ times the zero integral replaces $U-c_r$ by $U-U_s$:Therefore the integrand's profile factor must be strictly negative somewhere in the open channel:This is the Fjørtoft theorem. The positive weight and the strictly negative integral establish the strict inequality, not merely a nonpositive value. It is a stronger necessary condition than Rayleigh's inflection-point theorem, and neither condition alone proves instability.

Use the real diffusive interpretation $\gamma >0$, $U,{\mu }_0\in \mathbb{R}$, with $\nu >0$. Without a positive real diffusion coefficient the printed ordering inequality and real oscillator coordinate would need different hypotheses. Remove the drift by $\eta =e^{Ux/(2\gamma )}v$. Direct differentiation reduces the linear complex Ginzburg-Landau equation toSet $v=b(\xi )e^{-i({\omega }_0+ϵ{\omega }_1)t}$ and $\xi =\sqrt{ϵ}fx$. The resulting equation isCanceling its order-one coefficient and normalizing the quadratic term givesThe supplied decay quantization has $\lambda =2j+1$, $j=0,1,2,\dots$. Equivalently the Hermite functions $b_j=e^{-{\xi }^2/2}{H}_j(\xi )$ solve this harmonic-oscillator problem, where $H_j$ is the physicists' Hermite polynomial. Thus ${\omega }_1=-i(2j+1)\sqrt{\nu \gamma }$ and the exact temporal growth rates areThe global modes of a quadratically confined Ginzburg-Landau equation areFor every fixed positive $ϵ$, the negative quadratic spatial exponent dominates the linear drift exponent and the polynomial at both infinities; all these normal modes are spatially bounded. The largest growth rate is $s_0$, so the least-node normal mode is temporally non-growing, and all normal modes are non-growing, whenEquality is a neutral ground-state normal mode; strict inequality gives decay. This proves the printed sufficient condition for a bounded stable global normal mode and identifies its stronger meaning: stability of the leading global normal mode. Existence of merely some stable higher-$j$ normal mode is not equivalent to the inequality, since sufficiently large $j$ gives $s_j<0$ even above the leading-mode threshold. In the quadratic-coefficient model the reduction and eigenvalues are exact, not just a first-order approximation to a more general profile.