 Solution

Because $⟨B⟩=0$, a periodic primitive $H$ exists. Fix its additive constant by $⟨H⟩=0$. Integrating the conserved-field equation once in space gives $H_T=\sigma (H_X+\delta A^2{)}_X$: the possible integration constant vanishes on taking the spatial mean. Treat the spatial average as the inner product defining functional derivatives. Integration by parts, with no boundary terms under periodic boundary conditions, givesThus the conserved-field real amplitude equation is a gradient flow ascending $V$, or descending $-V$, with positive diagonal mobilities. Along a classical solution,For the mixed term put $P=⟨A^4{⟩}^{1/2}$ and $Q=⟨H_X^2{⟩}^{1/2}$. The Cauchy-Schwarz inequality and a completed square giveConsequently, for $\beta =1-\delta >0$,The negative leading quartic coefficient proves the upper bound. The stronger pointwise square completionis also useful: a superlevel set $V\ge V(0)$ bounds the amplitude in $L^4$, its spatial derivative in $L^2$, and $B=H_X$ in $L^2$, because a quartic polynomial with negative leading coefficient dominates its lower-order terms.

The dynamics dissipates toward the steady-state set rather than sustaining nonconstant periodic motion. Indeed $V$ increases to a finite limit, and the time integral of $⟨A_T^2+H_T^2/(2\delta \sigma )⟩$ is finite. For global regular trajectories with the usual parabolic precompactness, every limit point is stationary, by the LaSalle invariance principle. A nonconstant periodic or recurrent orbit would force a strictly increasing $V$ to return to its original value and is impossible. Boundedness of a monotone functional alone does not prove convergence to one particular equilibrium; isolation of the limiting equilibrium or an additional convergence theorem supplies that stronger claim.