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

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.