= Solution
Because $\langle B\rangle=0$, a periodic primitive $H$ exists. Fix its additive constant by $\langle H\rangle=0$. 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>, gives
$$
\frac{\delta V}{\delta A}=\mu A+\alpha A^2-A^3+A_{XX}-AH_X=A_T,\qquad
\frac{\delta V}{\delta H}=\frac12(A^2)_X+\frac{H_{XX}}{2\delta}=\frac{H_T}{2\delta\sigma}.
$$
Thus the <conserved-field real amplitude equation> is a <gradient flow> ascending $V$, or descending $-V$, with positive diagonal mobilities. Along a classical solution,
$$
\boxed{\frac{dV}{dT}=\left\langle A_T^2+\frac{H_T^2}{2\delta\sigma}\right\rangle\geq0.}
$$
For the mixed term put $P=\langle A^4\rangle^{1/2}$ and $Q=\langle H_X^2\rangle^{1/2}$. The <Cauchy-Schwarz inequality> and a completed square give
$$
\left\langle-\frac12A^2H_X-\frac{H_X^2}{4\delta}\right\rangle
\leq\frac{PQ}{2}-\frac{Q^2}{4\delta}
=\frac{\delta P^2}{4}-\frac{(Q-\delta P)^2}{4\delta}
\leq\frac\delta4\langle A^4\rangle.
$$
Consequently, for $\beta=1-\delta>0$,
$$
V\leq\left\langle\frac\mu2A^2+\frac\alpha3A^3-\frac\beta4A^4-\frac12A_X^2\right\rangle
\leq\max_{a\in\mathbb R}\left(\frac\mu2a^2+\frac\alpha3a^3-\frac\beta4a^4\right)<\infty.
$$
The negative leading quartic coefficient proves the upper bound. The stronger pointwise square completion
$$
V=\left\langle\frac\mu2A^2+\frac\alpha3A^3-\frac\beta4A^4-\frac12A_X^2-\frac{(H_X+\delta A^2)^2}{4\delta}\right\rangle
$$
is also useful: a superlevel set $V\geq V(0)$ 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.
\b[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 $\langle A_T^2+H_T^2/(2\delta\sigma)\rangle$ 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.
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