Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 77 / Section II

For the real-coefficient cubic-quintic amplitude equation, the trivial pattern $A=0$ exists for every $\mu$. A nonzero spatially uniform pattern has arbitrary constant phase and intensity $q=|A{|}^2>0$ satisfying $\mu +\alpha q-q^2=0$. ThusNonzero patterns exist for $\mu \ge -{\alpha }^2/4$. The upper branch $q_{+}$ exists throughout that range; a distinct positive lower branch $q_{-}$ exists only for $-{\alpha }^2/4<\mu <0$ and reaches zero at $\mu =0$.

For a homogeneous perturbation, write $A=R{e}^{i\varphi }$. Then $\dot{R}=R(\mu +\alpha R^2-R^4)$ and $\dot{\varphi }=0$. The radial eigenvalue at a nonzero pattern iswhile the phase eigenvalue is zero. Therefore the upper branch is stable modulo its arbitrary phase for $\mu >-{\alpha }^2/4$; the lower branch is unstable. The phase-neutral direction prevents asymptotic attraction to one particular complex amplitude, but permits orbital stability of the circle of equivalent patterns.

Conduction is asymptotically stable for $\mu <0$ and unstable for $\mu >0$. At $\mu =0$, the positive cubic term destabilizes arbitrarily small nonzero amplitudes, even though its linear eigenvalue vanishes. At $\mu =-{\alpha }^2/4$, the merged nonzero branch is semistable from above only: $\dot{R}=-R(R^2-\alpha /2{)}^2$, so amplitudes just below it eventually decrease toward zero. Thus the bistable interval is $-{\alpha }^2/4<\mu <0$, with the endpoints treated separately rather than declared stable merely because an eigenvalue is zero.

For the intended nontrivial interface, take the limiting amplitude at the patterned end to be $R_{-}>0$; the identically zero solution is otherwise a trivial exception. Splitting the stationary real-coefficient cubic-quintic amplitude equation into real and imaginary parts givesThe imaginary equation implies $R^2{\varphi }^{\prime }=J$, a constant. The radial first integral isA regular solution has finite $E$. As $R\to 0$, the nonnegative $J^2/(2R^2)$ term would diverge unless $J=0$. Hence ${\varphi }^{\prime }=0$ wherever the nontrivial front has positive amplitude. This conclusion does not assume in advance that the patterned end has no phase gradient.

For a heteroclinic orbit connecting two constant states, $R^{\prime }$ tends to zero at both ends. Thus $E=V(0)=0$, while the nonzero end must obey both $\mu +\alpha q-q^2=0$ and $V(\sqrt{q})=0$. Eliminating $\mu$ givesThis is the stationary front of a cubic-quintic amplitude equation: a unique equal-potential parameter in the bistable interval.

Existence, not just a necessary condition, follows from the first integral. At ${\mu }_M$, $V(R)=-R^2(R^2-q{)}^2/6$. The descending solution obeys $R^{\prime }=-R(q-R^2)/\sqrt{3}$, which integrates toIt connects the stable upper pattern at $x\to -\infty$ to conduction at $x\to +\infty$, with arbitrary translation $x_0$ and constant phase ${\varphi }_0$.

Assume $\alpha >0$, so a finite nonzero critical wavenumber exists. The linear growth rate of a Fourier mode is $\lambda (k)=-\alpha +\mu k^2-k^4$. Maximizing over $k$ and setting that maximum to zero givesSet $K=k_c^2=\sqrt{\alpha }$. The resonant three-wave triad has wavevectors satisfy ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$ and ${\mathbf{k}}_i\cdot {\mathbf{k}}_j=-K/2$ for $i\ne j$. With the slow time and amplitude scaling, every growth, quadratic and cubic term first enters at order ${ϵ}^3$. Projection onto each critical mode removes the nonresonant correction ${\Theta }_3$ by the method of multiple scales solvability condition.

The parameter displacement contributes ${\mu }_{2}KA$ to the first mode. For the quadratic term, use $\nabla \cdot ({\Theta }_1\nabla {\Theta }_1)=\frac{1}{2}{\nabla }^2({\Theta }_1^2)$. The resonant coefficient of ${\Theta }_1^2$ at ${\mathbf{k}}_1$ is $2\bar{B}\bar{C}$, so its contribution is $+\gamma K\bar{B}\bar{C}$.

For the cubic gradient term, an ordered triple of modes with wavevectors $\mathbf{p},\mathbf{q},\mathbf{r}$ summing to ${\mathbf{k}}_1$ contributes $(\mathbf{p}\cdot \mathbf{q})({\mathbf{k}}_1\cdot \mathbf{r})$ times its amplitude product. The three self-interaction permutations yield $-3K^2|A{|}^{2}A$. The six permutations of $({\mathbf{k}}_1,{\mathbf{k}}_2,-{\mathbf{k}}_2)$ yield $-2K^2-4({\mathbf{k}}_1\cdot {\mathbf{k}}_2{)}^2=-3K^2$, and likewise for the third mode. Therefore the hexagonal convection amplitude equations areFor a nonzero convection roll, $B=C=0$ and $|A{|}^2={\mu }_2/(3K)$, requiring ${\mu }_2>0$. Choose its phase so $A=a>0$. The radial perturbation decays, but a perturbation in the two other orientations givesThe eigenvalues are $\pm |\gamma |Ka$. ThusThis is roll instability from broken up-down symmetry. A growing homogeneous amplitude perturbation already proves instability, without a sideband calculation. The source's slightly asymmetric setting means nonzero quadratic coupling. If $\gamma =0$ is allowed literally, the displayed leading equations instead have a continuous sphere of constant-total-intensity states and neutral orientation perturbations. They then establish marginality, not strict growing-mode instability; higher-order effects would decide further selection. Likewise $\alpha =0$ would invalidate the assumed finite-$k_c$ triad scaling.