Diagonalizing the Frozen-Coefficient Amplitude System and Finding Its Growing Mode

Hello. We have now isolated the exact local wave mechanism inside an affine Boussinesq background:

$$\dot{\Theta }=a(t)\Omega ,\dot{\Omega }=b(t)\Theta ,$$

where

$$a(t)=-\frac{J\zeta (t)\cdot G(t)}{\lambda |\zeta (t){|}^2},b(t)=\lambda {\zeta }_1(t).$$

The next move is deliberately local: temporarily freeze the background geometry and regard $a$ and $b$ as constants. This turns the feedback loop into a two-dimensional constant-coefficient ODE. Diagonalizing it reveals exactly when the loop is amplifying, at what exponential rate, and which precise relation between temperature and vorticity amplitudes realizes pure growth.

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The frozen amplitude system

Write the amplitude vector as

$$X(t)=\left(\begin{array}{c}\Theta (t)\\ \Omega (t)\end{array}\right).$$

With $a$ and $b$ frozen, the system is

$$\frac{dX}{dt}=MX,M=\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right).$$

The two off-diagonal entries represent the two physical stages already identified:

• $a\Omega$ is the velocity induced by vorticity displacing the old temperature gradient;
• $b\Theta$ is the horizontal derivative of temperature producing vorticity.

Neither amplitude directly creates more of itself. Growth, if it occurs, is a two-step feedback loop.

A quick multiplication makes the structure unusually transparent:

$$M^2={\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right)}^2=\left(\begin{array}{cc}ab& 0\\ 0& ab\end{array}\right)=abI.$$

Equivalently, differentiating either equation once gives

$$\ddot{\Theta }=ab\Theta ,\ddot{\Omega }=ab\Omega .$$

Thus the sign of the single scalar $ab$ decides whether the feedback is hyperbolic and amplifying, or oscillatory.

Substituting the Boussinesq coefficients gives

$$ab=-\frac{(J\zeta \cdot G){\zeta }_1}{|\zeta {|}^2}.$$

Notice again that $\lambda$ has disappeared. Frequency changes the relative scale of $\Theta$ and $\Omega$, but the leading-order amplification rate is set by geometry: the direction of the wavevector relative to the old temperature gradient and the distinguished horizontal buoyancy direction.

There are three cases:

Sign of $ab$	Dynamics	Meaning
$ab>0$	Two real eigenvalues of opposite sign	One exponentially growing and one exponentially decaying mode
$ab<0$	Purely imaginary eigenvalues	Oscillation, not exponential amplification
$ab=0$	Degenerate system	The feedback loop is broken; amplitudes are constant or at most linear in time

The layer construction must therefore arrange $ab>0$ during each short growth interval. It is not enough for the temperature-to-vorticity coupling or the vorticity-to-temperature coupling to be nonzero separately: their signs must cooperate.

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Diagonalization in general form

Suppose

$$ab>0,\mu =\sqrt{ab}>0.$$

The characteristic polynomial of $M$ is

$$\det (M-\rho I)={\rho }^2-ab,$$

so its eigenvalues are

$${\rho }_{+}=\mu ,{\rho }_{-}=-\mu .$$

Assuming $a\ne 0$, the associated eigenvectors can be chosen as

$$v_{+}=\left(\begin{array}{c}1\\ \mu /a\end{array}\right),v_{-}=\left(\begin{array}{c}1\\ -\mu /a\end{array}\right).$$

The growing eigenline is therefore

$$\boxed{\Omega =\frac{\mu }{a}\Theta =\frac{b}{\mu }\Theta .}$$

If the initial amplitude vector lies exactly on this line, then it stays on it and is multiplied by $e^{\mu t}$:

$$\left(\begin{array}{c}\Theta (t)\\ \Omega (t)\end{array}\right)=e^{\mu t}\left(\begin{array}{c}\Theta (0)\\ \Omega (0)\end{array}\right).$$

The opposite eigenline,

$$\Omega =-\frac{\mu }{a}\Theta ,$$

decays as $e^{-\mu t}$.

More generally, any initial state decomposes into growing and decaying components:

$$X(0)=c_{+}{v}_{+}+c_{-}{v}_{-},$$

and hence

$$X(t)=c_{+}{e}^{\mu t}{v}_{+}+c_{-}{e}^{-\mu t}{v}_{-}.$$

The construction chooses its initial amplitudes close to, or exactly on, the growing eigenline. This is not merely a claim that the system is “unstable”: it is a controlled selection of the unstable direction.

For reference, the full propagator is

$$e^{Mt}=\cosh (\mu t)I+\frac{\sinh (\mu t)}{\mu }M.$$

That identity follows immediately from $M^2={\mu }^{2}I$. It is a compact alternative to diagonalizing explicitly.

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The paper’s normalized growth geometry

The paper chooses a particularly revealing frozen configuration:

$$D=0,G=-A{e}_2,A>0.$$

Here $e_2=(0,1)$, so the old temperature gradient points vertically downward. Since $D=0$, the phase vector $\zeta$ does not evolve during this idealized growth episode.

Parameterize the wavevector as

$$\zeta =re(\varphi ),e(\varphi )=(\sin \varphi ,\cos \varphi ),r=|\zeta |.$$

The angle $\varphi$ is measured from the positive $x_2$-axis towards the positive $x_1$-axis. Thus

$${\zeta }_1=r\sin \varphi .$$

Also,

$$J\zeta =r(-\cos \varphi ,\sin \varphi ).$$

Taking the dot product with $G=-A{e}_2$,

$$J\zeta \cdot G=-Ar\sin \varphi .$$

Therefore

$$a=-\frac{J\zeta \cdot G}{\lambda r^2}=\frac{A\sin \varphi }{\lambda r},$$

while

$$b=\lambda {\zeta }_1=\lambda r\sin \varphi .$$

The system becomes

$$\frac{d}{dt}\left(\begin{array}{c}\Theta \\ \Omega \end{array}\right)=\left(\begin{array}{cc}0& \frac{A\sin \varphi }{\lambda r}\\ \lambda r\sin \varphi & 0\end{array}\right)\left(\begin{array}{c}\Theta \\ \Omega \end{array}\right).$$

For

$$0<\varphi <\pi ,$$

we have $\sin \varphi >0$. Both off-diagonal couplings are positive, and

$$ab=A{\sin }^2\varphi .$$

Hence the eigenvalues are

$$\boxed{{\mu }_{\pm }=\pm \sqrt{A}\sin \varphi .}$$

Be careful with the placement of the square root: the growth rate is

$$\sqrt{A}\sin \varphi ,$$

not $\sqrt{A\sin \varphi }$.

The ideal growth rate is largest when

$$\varphi =\frac{\pi }{2},$$

meaning that the wavevector is horizontal:

$$\zeta =r{e}_1.$$

Then the temperature has maximal horizontal variation, so it sources vorticity efficiently; simultaneously, the induced velocity aligns optimally with the old vertical temperature gradient.

At the limiting orientations $\varphi =0$ and $\varphi =\pi$,

$${\zeta }_1=0,$$

so the wave has no horizontal temperature derivative. Both couplings vanish in this normalized example, and the feedback loop stops.

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Rescaling the vorticity amplitude

The factors $\lambda r$ in the matrix obscure a simpler symmetric structure. Define a rescaled vorticity amplitude

$$Y=\frac{\Omega }{\lambda r}.$$

Since $\lambda$ and $r$ are frozen during this episode, the system becomes

$$\dot{\Theta }=A\sin \varphi Y,\dot{Y}=\sin \varphi \Theta .$$

Equivalently,

$$\frac{d}{dt}\left(\begin{array}{c}\Theta \\ Y\end{array}\right)=\sin \varphi \left(\begin{array}{cc}0& A\\ 1& 0\end{array}\right)\left(\begin{array}{c}\Theta \\ Y\end{array}\right).$$

Now define the two eigen-coordinates

$$Z_{+}=\Theta +\sqrt{A}Y,Z_{-}=\Theta -\sqrt{A}Y.$$

Differentiate them:

$${\dot{Z}}_{+}=\sqrt{A}\sin \varphi Z_{+},$$ $${\dot{Z}}_{-}=-\sqrt{A}\sin \varphi Z_{-}.$$

Thus diagonalization has converted the coupled physical amplitudes into two independent scalar laws:

$$Z_{+}(t)=Z_{+}(0)e^{\sqrt{A}\sin \varphi t},$$ $$Z_{-}(t)=Z_{-}(0)e^{-\sqrt{A}\sin \varphi t}.$$

Returning to $\Theta$ and $\Omega$,

$$\Theta (t)=\frac{1}{2}\left[Z_{+}(0)e^{\sqrt{A}\sin \varphi t}+Z_{-}(0)e^{-\sqrt{A}\sin \varphi t}\right],$$ $$\Omega (t)=\frac{\lambda r}{2\sqrt{A}}\left[Z_{+}(0)e^{\sqrt{A}\sin \varphi t}-Z_{-}(0)e^{-\sqrt{A}\sin \varphi t}\right].$$

The pure growing mode is characterized by

$$Z_{-}(0)=0.$$

This is equivalent to

$$\Theta =\sqrt{A}Y,$$

or, in the original variables,

$$\boxed{\Omega =\frac{\lambda r}{\sqrt{A}}\Theta .}$$

Along this line,

$$\Theta (t)=\Theta (0)e^{\sqrt{A}\sin \varphi t},$$ $$\Omega (t)=\Omega (0)e^{\sqrt{A}\sin \varphi t}.$$

So temperature and vorticity amplitudes grow at the same exponential rate, while their ratio remains fixed.

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What actually becomes large?

The construction is ultimately concerned with derivative blowup, not necessarily blowup of the scalar temperature amplitude itself. For the sinusoidal temperature wave,

$$\vartheta (x,t)=\Theta (t)\sin (\lambda \zeta \cdot x),$$

the gradient at the origin is

$$\nabla \vartheta (0,t)=\lambda \Theta (t)\zeta .$$

Its magnitude is therefore

$$|\nabla \vartheta (0,t)|=\lambda r|\Theta (t)|.$$

A high-frequency wave can consequently have a small scalar amplitude $\Theta$ but a substantial temperature gradient. On the growing eigenline,

$$|\Omega |=\frac{\lambda r}{\sqrt{A}}|\Theta |,$$

so

$$|\nabla \vartheta (0,t)|=\sqrt{A}|\Omega (t)|.$$

The growing mode amplifies a coordinated package:

• the temperature amplitude $\Theta$;
• the vorticity amplitude $\Omega$;
• the temperature gradient $\lambda r\Theta$;
• the shear induced by the vorticity wave.

That final item is precisely why unlimited growth of a layer is not useful. The vorticity-generated shear becomes part of the affine background faced by the next layer, and its sign can obstruct the next planned amplification episode.

The paper chooses negative amplitudes on the growing eigenline:

$$\Theta <0,\Omega <0.$$

Because

$$\nabla \vartheta (0,t)=\lambda \Theta \zeta ,$$

this makes the newly accumulated temperature gradient point in the $-\zeta$ direction. The sign choice is not cosmetic: it sets the orientation and sign of the shear that must later be neutralized.

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A compact interpretation

The frozen system is a saddle, not a generic runaway process. It has:

• one one-dimensional expanding direction;
• one one-dimensional contracting direction;
• no growth when the two physical couplings have the wrong relative sign;
• a geometry-controlled rate independent of the carrier frequency $\lambda$.

The important design fact is that $\lambda$ still matters enormously for derivatives:

$$|\nabla \vartheta |\sim \lambda r|\Theta |.$$

So frequency is not used to make each single frozen episode grow faster. It is used to convert controlled amplitude gains into sharper and sharper spatial gradients, which can then be assembled across many scales.

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Takeaways

Freezing the amplitude coefficients gives

$$\frac{d}{dt}\left(\begin{array}{c}\Theta \\ \Omega \end{array}\right)=\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right)\left(\begin{array}{c}\Theta \\ \Omega \end{array}\right).$$

Because

$$M^2=abI,$$

the sign of

$$ab=-\frac{(J\zeta \cdot G){\zeta }_1}{|\zeta {|}^2}$$

determines the qualitative behavior. Exponential growth occurs exactly when $ab>0$, with rate

$$\mu =\sqrt{ab}.$$

In the paper’s normalized geometry,

$$G=-A{e}_2,\zeta =r(\sin \varphi ,\cos \varphi ),$$

the growth rate is

$$\sqrt{A}\sin \varphi$$

for $0<\varphi <\pi$, and the pure growing eigenline is

$$\boxed{\Omega =\frac{\lambda r}{\sqrt{A}}\Theta .}$$

Next, we will examine the controlled rotation that ends a growth episode by returning $\Omega$ to zero while retaining the accumulated temperature-gradient gain.