Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 60 / 3

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.

Let the uniform nonzero amplitude be $R$, so $\mu =R^2-\alpha R$ and $B=0$. Write $r=R(2R-\alpha )$. A perturbation with Fourier wavenumber $k=n\pi /L$ has stability matrixThe constant $B$ mode is disallowed because the conserved mean is prescribed. At $k=0$ the only allowed mode has eigenvalue $-r$, so homogeneous stability requires $r>0$. Then every nonzero-mode trace is negative, and its determinant is smallest in sign at the smallest allowed $k^2$. The exact periodic stability criterion for a conserved-field uniform amplitude isA zero in either inequality is a marginal case requiring nonlinear analysis. The zero-amplitude branch, separately, has amplitude growth rates $\mu -k^2$ and nonzero conserved-field growth rates $-\sigma k^2$, and is strictly stable when $\mu <0$.

For the long-domain interpretation one may take $\alpha >0$ by reversing the sign of $A$, and first consider the positive upper branch $R>\alpha /2$. It is homogeneously stable even though sufficiently long spatial modulations grow wheneverFor $0<\delta <1$, this givesAt a fixed finite $L$ the actual unstable part is smaller: it must also satisfy $(\pi /L{)}^2<\alpha R-2(1-\delta )R^2$. For $\delta \ge 1$, the right side is positive for every positive upper-branch $R$, so every such state is unstable to sufficiently long waves. The physical mechanism is feedback from the conserved field: an increase in $A^2$ reduces local $B$, and the term $-AB$ reinforces that increase.

There are two important limits on the wording of the requested stability conclusion. First,never vanishes at a finite regular point of this branch. At its fold the derivative is infinite. Thus a condition of zero $dR/d\mu$ cannot literally occur here. The meaningful contrast is that a positive-slope, homogeneously stable branch can still have a growing spatial mode.

Second, a long-wave conclusion cannot be asserted for every finite periodic domain. For example, $\alpha =1$, $\delta =2$, $\sigma =1$, $R=1$, $\mu =0$, $L=1$ give $r=1$ and $k^2+r-2\delta R^2=n^2{\pi }^2-3>0$ for every nonzero integer $n$. This state is strictly stable even though $\delta >1$. Nor does the unrestricted statement include every sign of $R$: $\alpha =1$, $\delta =2$, $R=-1/10$, $\mu =11/100$ give $r=3/25$ and $r-2\delta R^2=2/25>0$, hence stability for every period and every $\sigma >0$. These examples distinguish the exact stated real-variable periodic problem from its intended positive-amplitude, long-layer interpretation.

At a periodic steady state, two spatial integrations of $0=(B+\delta A^2{)}_{XX}$ make $B+\delta A^2$ constant. The zero-mean constraint fixes that constant to $\delta ⟨A^2⟩$. Define $\beta =1-\delta >0$ and $\nu =\mu -\delta ⟨A^2⟩$; the stationary amplitude equation isFor a broad localized plateau, the two transition layers occupy $O(\ell )$ length while the plateau and exterior occupy $2hL$ and $2(1-h)L$. Provided both lengths are large compared with the front width $\ell$, spatial averaging gives $⟨A^2⟩=h{R}^2+O(R^2\ell /L)$.

Multiplying the stationary equation by $A_X$ gives the first integralIn the well-separated-front approximation, $A\to 0$ with $A_X\to 0$ outside, so $E=0$. On the plateau $A\to R\ne 0$, giving both $\nu +\alpha R-\beta R^2=0$ and $W(R)=0$. Substitution of the first into the second gives $R^3(\beta R/4-\alpha /6)=0$, henceThe equal-potential condition is therefore $2{\alpha }^2=-9\nu \beta$. It is a front balance, not an integration of the stationary equation without its $A_X$ multiplier.

These plateau values also produce actual isolated fronts. At this balance $W(A)=-\beta A^2(A-R{)}^2/4$, so for $\alpha >0$ the increasing front satisfies $A_X=\sqrt{\beta /2}A(R-A)$. It isFor $\alpha <0$ the sign-reversed construction gives the negative plateau. Two oppositely oriented well-separated fronts give the leading periodic mesa solution approximation; their weak interaction and tails supply finite-$L$ corrections.

Finally $\nu =\mu -\delta h{R}^2$ gives the coexistence fraction of a conserved-field amplitude mesa:The leading existence range is the interval obtained from $0<h<1$:This assumes $\alpha \ne 0$ and $L\gg \ell$, with $hL,(1-h)L\gg \ell$. At the two displayed ends a transition pair can no longer be treated as well separated, so they are asymptotic bounds rather than exact finite-period bifurcation values. For $\alpha =0$ this particular nonzero plateau balance collapses to $R=0$ and does not yield the mesa solution family.