Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 65 / 1

Substitution into the Newell–Whitehead–Segel equation gives $0=(\mu -q^2-R_0^2)R_0$. Thus a nonzero roll has $R_0^2=\mu -q^2>0$; at equality the branch meets the zero solution. The two signs of $R_0$ differ by a constant phase modulation.

The sign of the transverse zigzag instability must be calculated from the differential operator. Put $D={\partial }_X+i{\partial }_Y^2/2$ and $A=R_0{e}^{iqX+i\varphi (Y)}$. The chain rule givesConsequently the change in the printed functional density isFor a Lyapunov functional one integrates this density over the transverse coordinate too, or uses a transverse spatial average, with periodic or decaying boundary conditions. A fast-$x$ average alone leaves a density depending on $Y$ and is not itself a global monotonicity statement. Indeed $D^{†}=-D$ under these boundary conditions, and the functional derivative of the resulting spatial integral is $\delta V/\delta \overline{A}=\mu A+D^{2}A-|A{|}^{2}A=A_T$. Hence $dV/dT=2⟨|A_T{|}^2⟩\ge 0$: stable rolls locally maximize this functional.

With the plus sign actually printed in $D$, $q<0$ makes both quadratic contributions to $\Delta V$ negative. These rolls are stable to long-wave transverse phase modulation. For $q>0$ a sufficiently long-wave displacement increases $V$ and is unstable. At $q=0$ the second-derivative stiffness vanishes, but the fourth-derivative and nonlinear tilt terms still decrease $V$. Linearization makes the conclusion explicit:Thus for $q>0$ the unstable band is $0<k^2<4q$, with maximum growth at $k^2=2q$; in a finite periodic cell the band must contain an allowed Fourier mode. Replacing $D$ by ${\partial }_X-i{\partial }_Y^2/2$ reverses the sign of the detuning associated with the zigzag instability. It would therefore be incorrect to import that opposite sign convention into the printed equation. Transverse stability does not exclude the distinct longitudinal Eckhaus instability.

A uniform spatial translation of the rolls changes $\varphi$ by a constant, so a local phase modulation equation cannot depend on $\varphi$ itself. Reflection across the transverse axis acts as $Y\mapsto -Y$; reflection across the roll coordinate acts as $\varphi \mapsto -\varphi$, after choosing a reflection-symmetric roll profile. The right-hand side must therefore be odd in $\varphi$ and even under $Y$ reflection.

At the lowest orders, ${\varphi }_{YY}$ and ${\varphi }_{YYYY}$ have these symmetries. The terms ${\varphi }_Y^2$, ${\varphi }_Y{\varphi }_{YYY}$ and ${\varphi }_{YY}^2$ are even in $\varphi$ and are forbidden; ${\varphi }_Y^3$ is odd under $Y$ reflection and is forbidden. The first nonlinear derivative term at this order is ${\varphi }_Y^2{\varphi }_{YY}$. Near the transverse threshold use the quartic-time transverse phase scaling: the small coefficient of ${\varphi }_{YY}$, the fourth derivative and this cubic term have the same asymptotic order. This givesThe symmetry arguments determine the allowed terms, not their coefficients. A well-posed short-wave regularization requires $\gamma >0$. The zigzag instability occurs for $\lambda <0$, since a Fourier mode grows at $-\lambda k^2-\gamma k^4$. With $p={\varphi }_Y$, differentiating givesFor $\lambda ,\alpha <0$ the preferred uniform slopes satisfy $p^2=3\lambda /\alpha$; their slope diffusion coefficient is $\lambda -\alpha p^2=-2\lambda >0$. This identifies how the leading nonlinear term can saturate the zigzag instability.

Use the translation-covariant phase expansion with $\xi =x+\varphi (Y,T)$, $Y=\epsilon y$ and $T={\epsilon }^{4}t$. Write all fast profiles at $\xi$, including the corrections. LetThe Swift–Hohenberg equation has steady equation $r{w}_0-B^2{w}_0-w_0^3=0$, and differentiation gives $L{w}_0^{\prime }=0$. On a fast periodic cell $L$ is a self-adjoint operator. We assume the translation mode is its only zero mode; otherwise one must impose the corresponding additional Fredholm solvability conditions and retain additional amplitudes.

Since ${\nabla }^2={\partial }_{\xi }^2+{\epsilon }^2{\mathcal{D}}_Y^2$, its squared operator is $B^2+2{\epsilon }^{2}B{\mathcal{D}}_Y^2+{\epsilon }^4{\mathcal{D}}_Y^4$. The time derivative first contributes at order ${\epsilon }^4$. At order ${\epsilon }^2$,The Fredholm solvability condition is orthogonality of this forcing to $w_0^{\prime }$. Periodic integration by parts givesThus the condition is $2J{\varphi }_{YY}=0$. General slowly curved rolls require $J=0$ at leading order. If $J=O({\epsilon }^2)$, this order-${\epsilon }^2$ residual first enters the order-${\epsilon }^4$ equation, so the leading order-${\epsilon }^2$ problem is solvable. In particular,is precisely the near-threshold condition. Defining $N=⟨(w_0^{\prime }{)}^2⟩>0$ and $J=-{\epsilon }^2\lambda N/2$ keeps the residual as the phase-diffusion term at the next order.

There is an important chain rule qualification in this part. The printed value of $\gamma$ follows from the expansion. Its printed value of $\alpha$ follows only if transverse differentiation of the translated correction profiles is omitted. For a consistent translation-covariant phase expansion, that omission is generally nonzero.

At the leading transverse threshold letwhere the last conditions fix the phase gauge. Put $p={\varphi }_Y$, $r_1={\varphi }_{YY}$, $s_1={\varphi }_{YYY}$ and $t_1={\varphi }_{YYYY}$, and use $w_2=C_1(\xi )r_1+C_2(\xi )p^2$. At order ${\epsilon }^4$ the Fredholm solvability condition givesThe first two contributions follow directly by differentiating:Choose the reflection-symmetric roll with $w_0$ even. Then $C_1$ is odd and $C_2$ even. Projection onto the odd translation mode eliminates $p{s}_1$, $r_1^2$ and $p^4$, leaving exactly the symmetry-allowed phase modulation equation. Its coefficients areandIf one freezes $C_j$ at the unshifted coordinate when taking transverse derivatives, ${\partial }_Y^2{w}_2=C_1{t}_1+2C_2(r_1^2+p{s}_1)$, and its projection contributes only the $\gamma$ term. The other two projected terms then give the printed expression ${\alpha }_{\mathrm{print}}=6⟨w_0^{\prime }(w_0^{\prime \prime \prime }+w_0{C}_1{C}_2)⟩/N$. But freezing these profiles is inconsistent with the translated $w_0$ in $L$ for a general finite phase displacement. The printed expression needs the extra assumption $⟨w_0^{\prime }B(C_1^{\prime \prime }+5C_2^{\prime })⟩=0$, which is not an identity.

A small-amplitude calculation demonstrates the difference. For a critical roll of leading amplitude $\eta$ and fast wavenumber $\kappa$, the steady equation and $J=0$ giveThe third harmonic coefficient in $w_0$ follows by dividing its cubic forcing $-{\eta }^3\cos (3\kappa \xi )/4$ by the fast eigenvalue $-64$. Substituting in $L{C}_j$ gives the displayed correction profiles; the near-zero first-harmonic eigenvalue in the even problem must be retained to obtain the first harmonic of $C_2$. Periodic orthogonality then givesThus the supplied nonlinear coefficient is not generally an exact consequence of the finite-amplitude translated-roll expansion. The corrected coefficient above, the printed calculation under its additional freezing assumption, and the genuine near-threshold phase modulation equation are all distinguished explicitly.