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Differentiate the material derivative of the passive scalar component by component. With ${\sigma }_{ij}={\partial }_j{u}_i$,The chain rule along a Lagrangian trajectory $\dot{X}=u(X,t)$ therefore gives the covector evolution of a material scalar gradientThe transpose matters: a gradient is transported as a covector, rather than as a material separation vector. When $u_3=0$, the third row of $\sigma$ vanishes, andLet $k_h=(k_1,k_2{)}^T$ and setFor $c>0$, the matrix exponential gives the exact horizontal solutionThe two projections are onto the eigenspaces with eigenvalues $-c$ and $+c$. If $p_{-}\ne 0$, write $p_{-}=k_0(\cos \varphi ,\sin \varphi {)}^T$ with $k_0=|p_{-}|>0$. ThenThe stated growing asymptotic requires this nonzero projection. An initial gradient entirely in the other eigenspace decays, and when $c=0$ the horizontal gradient is constant.

For constant vertical shear, write $h=({\sigma }_{13},{\sigma }_{23}{)}^T$. Integrating the third component using the exact horizontal solution yieldsConsequently, if $D=h\cdot (\cos \varphi ,\sin \varphi )$ is nonzero, $k_3\sim -k_{0}D{e}^{ct}/c$, andIf $D=0$, exponential growth of $k_3$ is absent, but the ratio still tends to zero on the growing horizontal branch. Thus exponential vertical growth is generic, not unconditional. The scalar-gradient inclination in a steady planar strain is the tangent of the angle between the scalar's normal and the horizontal plane. Equivalently, it is the cotangent of the inclination of an isoconcentration surface to that plane. Large $\alpha$ corresponds to nearly horizontal scalar surfaces; small $\alpha$ corresponds to nearly vertical ones.

For the probability calculation, take $\Gamma ,\Lambda >0$. The independent random variables $a,b$ with normal distributions of variance ${\Gamma }^2$ make $c$ distributed according to the Rayleigh distribution, withThe advertised Gaussian-over-Rayleigh inclination distribution also needs $D$ to be independent of $(a,b)$; its Gaussian marginal alone does not imply this. Under that intended assumption, for $A>0$,The cumulative distribution function is zero for $A\le 0$. Differentiating givesand the probability density function is zero for negative $A$. Its integral is one because the cumulative distribution function tends to one at infinity. Independence is a real hypothesis: $D=(\Lambda /\Gamma )a$ has exactly the stated Gaussian marginal but makes $\alpha \le \Lambda /\Gamma$ almost surely, contradicting the unbounded support of this density. A sufficient physical model is isotropic Gaussian vertical shear, independent of $a,b$, whose projection in any horizontal direction has variance ${\Lambda }^2$.

If the strain varies slowly, the horizontal gradient must align with the current growing direction, and the vertical-to-horizontal ratio must adjust before that strain changes. Both adjustment times are of order $c^{-1}$. Thus the local condition is $c{\tau }_{\sigma }\gg 1$, and the typical condition for the specified statistics isThis is an adiabatic approximation, not a uniform statement about arbitrarily weak instantaneous strains: samples with $c{\tau }_{\sigma }\lesssim 1$ retain memory of earlier strain. The instantaneous eigenvector must likewise turn slowly compared with $c$.

For rapidly fluctuating vertical shear, insert the large-time horizontal solution into its stochastic differential equation:The linear combination $W=\cos \varphi W^{(1)}+\sin \varphi W^{(2)}$ is a Brownian motion: it has independent Gaussian increments and increment variance $({\cos }^2\varphi +{\sin }^2\varphi )dt=dt$. For the signed inclination $\beta =k_3{e}^{-ct}/k_0$, the Itô product rule introduces no extra cross term because the exponential factor is deterministic. It follows thatThis is an Ornstein-Uhlenbeck process, explaining why white-noise vertical shear gives an Ornstein-Uhlenbeck inclination. Its Fokker-Planck equation isIn a stationary state the Fokker-Planck probability current $j=-cBp-(g^2/2)p^{\prime }$ is constant. A nonzero constant current cannot give an integrable, nonnegative density at both ends of the real line, so $j=0$. Integrating $p^{\prime }/p=-2cB/g^2$ and normalizing with the Gaussian integral givesThe stationary variance is $g^2/(2c)$. More explicitly the explicit Ornstein-Uhlenbeck solution has mean $\beta (0)e^{-ct}$ and variance $g^2(1-e^{-2ct})/(2c)$, confirming convergence to this density. Here $c>0$ and $g\ne 0$ are needed for the stated smooth stationary density; at $c=0$ there is no confinement, while $g=0$ gives a point mass at zero.