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For real base-flow velocity and positive Reynolds number, multiply the homogeneous Squire equation by the conjugate mode and integrate with zero endpoint values. Taking real parts gives the displayed negative temporal growth rate. This proves decay of homogeneous Squire modes; the coupled shear-flow system can still be unstable or show transient growth because wall-normal velocity can force the Squire variable.

A nonlinear term $f(|x|)Nx$ with real skew-symmetric matrix $N$ contributes zero to the derivative of the quadratic energy $|x{|}^2/2$. It changes orientation rather than magnitude directly. If a separate linear generator is a non-normal matrix, that reorientation can still alter later transient growth, so energy cancellation alone does not make the nonlinear evolution equivalent to the linear one.

The spectral radius of the matrix exponential gives the lower bound through the spectral abscissa $s(A)$. Differentiating the squared Euclidean norm along $q^{\prime }=Aq$ and bounding its Hermitian quadratic form gives the upper bound through the Euclidean logarithmic norm ${\mu }_2(A)$. A non-normal matrix can have ${\mu }_2(A)>0$ despite $s(A)<0$, allowing transient growth before its ultimate decay.

For $x^{\prime }=Lx$ and $E=\Vert x{\Vert }_2^2$, the Hermitian matrix $H=(L+L^{\ast })/2$ gives $\dot{E}=2x^{\ast }Hx$. Thus some initial condition grows immediately iff the largest eigenvalue of $H$ is positive. All trajectories have nonincreasing energy iff $H$ is a negative semidefinite matrix. This differs from requiring negative real parts for the eigenvalues of $L$: a non-normal matrix can satisfy modal decay while allowing transient growth.

A smooth inviscid stratified parallel flow with gradient Richardson number at least $1/4$ everywhere has no exponentially growing two-dimensional normal modes. Put $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and take its imaginary part:The integral is positive for a nonzero mode when the numerator is nonnegative, forcing $c_i=0$. This is a modal stability theorem, not a prohibition on transient growth. Maslowe's review discusses the theorem and the role of critical layers.

Multiply the homogeneous Squire equation by $\bar{\eta }$ and integrate over the channel. The zero boundary values remove the integration by parts boundary term, givingFor real $U$ and $k$, the middle term is purely imaginary. Taking real parts yields the dissipation identity for a homogeneous Squire mode:The strict inequality holds for a nonzero mode with zero endpoint values and positive Reynolds number. Hence all homogeneous Squire modes are temporally stable. In the complete shear-flow perturbation system, the Squire variable can also be forced by the wall-normal velocity; this homogeneous-mode result does not rule out instability of that coupled system or its transient growth.

For $k>0$, put $q=\sqrt{k^2+l^2}$ and define the Squire transformationThe two-dimensional Orr-Sommerfeld equation then has the same transverse operator because $k^{\prime 2}=q^2$, and the same viscous coefficient because $k^{\prime }{R}^{\prime }=kR$. Its boundary conditions on $v$ and $v^{\prime }$ are unchanged. Thus it has the same eigenfunction and the same phase-speed eigenvalue.

If the original disturbance grows, $\mathrm{Im}c>0$, so the transformed disturbance also grows. Since $R^{\prime }\le R$, every unstable three-dimensional mode maps to an unstable two-dimensional mode at a no larger Reynolds number. It follows that the minimum critical Reynolds number is attained among two-dimensional disturbances. This is the Squire theorem for temporal normal-mode onset, not a statement that three-dimensional transient amplification is impossible. The case $k=0$ is treated separately without dividing by $k$; its streamwise-uniform viscous normal modes decay, although coupling can produce transient growth.

In the linear problem, ${\dot{x}}_2=-x_2/(4R)$, so $x_2(t)=x_{20}{e}^{-t/(4R)}$. Solve ${\dot{x}}_1+x_1/R=x_2(t)$ with an integrating factor to obtain the complete solution:Equivalently the matrix exponential isBoth eigenvalues of $L$ are negative, so each solution eventually decays, despite the possible transient growth generated by $b$.

Choose $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and let $Q=N^2-(U^{\prime }{)}^2/4$. The result isBecause $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part givesIf $4N^2>(U^{\prime }{)}^2$ everywhere, the integral is strictly positive for every nonzero eigenfunction, so a mode with $c_i\ne 0$ is impossible. Thus the flow is neutrally stable to these inviscid normal modes. This is the Miles–Howard theorem, expressed as a lower bound $\mathrm{Ri}>1/4$ on the gradient Richardson number. It excludes exponential modal growth; it does not by itself exclude transient growth.

A matrix is non-normal when it does not commute with its adjoint matrix:Nonorthogonal decaying eigenmodes can interfere constructively and produce transient growth. For example,has two negative eigenvalues but is non-normal. Starting from $x(0)=(0,1{)}^T$ givesAt $t=\log 2$, its squared Euclidean norm is $2.5^2+0.25^2>1=|x(0){|}^2$. The energy grows transiently even though both eigenmodes eventually decay.

For temporal normal-mode stability of a viscous parallel shear flow, any unstable three-dimensional mode maps by the Squire transformation to an unstable two-dimensional mode at a no larger Reynolds number. Hence the first temporal modal instability occurs among two-dimensional disturbances. The theorem does not exclude three-dimensional transient growth or forced responses.