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Let $F(T)$ denote the right-hand side of the temperature partial differential equation. For a periodic variation $\eta$, integration by parts givesAlong the evolution, $T_t=F(T)$, so this is a gradient flow with increasing functional:For $s=|\nabla T{|}^2$, completing the square givesThus the increasing-gradient functional for poorly conducting convection converges to a finite limit along any globally smooth solution, andIn particular, there are arbitrarily late times at which $\Vert T_t{\Vert }_{L^2}$ is arbitrarily small. A nonstationary time-periodic solution is impossible: it would have a strictly positive increase of $V$ over a temporal period. More generally, if the orbit is precompact in a topology strong enough for the equation, its limiting states are stationary. With the conserved mean fixed, the displayed square completion bounds $\Vert \Delta T{\Vert }_{L^2}$, and periodic elliptic estimates give an $H^2$ bound; the usual smoothing and compactness for a globally regular parabolic solution support this limiting-state conclusion. The inequalities alone do not prove convergence to one uniquely specified equilibrium, and stationary families related by translation symmetry can remain.

Now take a smooth, nonconstant periodic roll $T_0(x)$. Multiplying its stationary partial differential equation by $T_0$ and using periodic integration by parts yieldsThe primes here are essential: the first and fourth terms involve $T_0^{\prime }$, not $T_0$. Write $a=T_0^{\prime }$, $b=T_0^{\prime \prime }$, $M_2=⟨a^2⟩$, $M_4=⟨a^4⟩$, and $N_2=⟨b^2⟩$. The orthogonal-roll perturbation hasThe cross term in the squared Laplacian averages to zero since $⟨b⟩=0$. Independence of the $x,y$ cell integrals then gives the exact polynomial differenceIts leading coefficient is half the variance of $a^2$ over the period. That variance is strictly positive: a nonconstant smooth periodic function has a derivative which vanishes somewhere and is nonzero somewhere, so $a^2$ is not constant. Hence arbitrarily small transverse perturbations increase $V$ above the roll value.

One can turn this energy comparison into linear instability. The linearized evolution at the roll is the self-adjoint operatorChoose the fixed-mean perturbation $\eta =T_0(y)-⟨T_0⟩$. Since $D^{2}V[T_0](\eta ,\eta )=⟨\eta L\eta ⟩=M_4-M_2^2>0$, the Rayleigh quotient is positive. The periodic self-adjoint fourth-order operator therefore has a positive eigenvalue. This proves the transverse energy instability of nonconstant temperature rolls: every nonconstant roll is unstable wherever it exists.

The existence qualification matters. Constants are also stationary and are stable on the fixed-mean subspace for $0<\mu <1$. Moreover, the periodic Poincaré inequality gives $N_2\ge M_2$, so the roll identity implies $\mu M_2=N_2+M_4>M_2$ for a nonconstant $2\pi$-periodic roll. Such rolls require $\mu >1$ on this specified cell; the conclusion for all positive $\mu$ is conditional on a nonconstant roll being available.