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The integrated travelling wave equation is $v^{\prime }=Q(v)/\epsilon$. A finite limiting value $a$ at either end must satisfy $Q(a)=0$. Otherwise continuity of $Q$ makes $v^{\prime }$ eventually have a fixed sign and an absolute value bounded below, which is incompatible with convergence to $a$. ThereforeSubtracting yields the Rankine-Hugoniot conditionFor distinct end states this givesDistinctness is needed for the printed quotient. If $u_l=u_r$, the identity is $0=0$. A nonconstant global profile is strictly monotone by the scalar ordinary differential equation, so cannot have equal finite end states. The equal-state profiles here are constant and their representation allows any $\sigma$.

Insert the travelling wave into the viscous scalar conservation law. With $s=x-\sigma t$ the equation becomesand one integration givesFor a nonconstant profile, $Q(v(s))$ never vanishes. Indeed, the autonomous ordinary differential equation $v^{\prime }=Q(v)/\epsilon$ has unique local solutions since $Q$ is $C^1$; reaching an equilibrium would force the whole solution to be constant. Separation and $c=v(0)$ therefore giveA different choice of reference point gives $s-s_0$ on the left, expressing the translation freedom of the travelling wave.

The printed formula needs a nonconstant-profile qualification. Constant profiles also solve the PDE, but their denominator vanishes at their constant value, so the separated integral is not defined. They must be included separately as equilibrium solutions of the integrated ordinary differential equation.

Take $z$ positive upward, and write $b=\tilde{\rho }g$. For a straight steady viscous buoyant conduit, the inner and outer axial equations areRegularity gives $w_i^{\prime }(0)=0$, the wall has $w_o(R)=0$, and velocity and tangential traction are continuous at $r=a_0$: $w_i=w_o$, $\lambda \mu w_i^{\prime }=\mu w_o^{\prime }$. With no interfacial tension specified, normal traction is continuous; for this straight steady flow that sets $P_i=P_o$ at the interface. Thus the modified gradients are common in this case.

When their value is zero, integration givesConsequentlyThe large center-to-interface speed ratio explains why the inner parabolic component dominates at small viscosity ratio.

For a slowly varying radius, set $B=b-P_{i,z}$. The local inner solution has $w_i=w_a+B(a^2-r^2)/(4\lambda \mu )$, soThe shear-driven outer interface speed has scale $a^{2}B\ln (R/a)/\mu$. The pressure-driven return flow needed for zero total flux adds a velocity of order $q/R^2$ and a shear-driven contribution of order $a^{2}B/\mu$. Relative to the inner parabolic velocity these terms are $O(\lambda \ln (R/a))$ and $O((a/R{)}^2)$, which are small under the stated assumptions. Axial viscous derivatives are also smaller because the plume is slender. ThereforeFor clarity, the outer pressure-gradient estimate must include both return flux and transmitted interface shear. Their flux scales are $R^4{P}_{o,z}/\mu$, $a^2{R}^{2}B/\mu$ and $q$. Zero total flux hence impliesThis justifies the negligible outer pressure gradient. We subsequently choose its negligible pressure reference as zero, as in the reduced model.

Near the moving interface the outer radial source flow is $u_r=a{a}_t/r$: the interface's axial advection is smaller because its axial velocity is small compared with the inner core speed. Its radial strain is ${\partial }_r{u}_r=-a{a}_t/r^2$. Continuity of normal traction at $r=a$, neglecting the much smaller inner normal viscous stress, givesWith $A=a^2/a_0^2$, this is $P_i=\mu A_t/A$ after dropping $P_o$. Conservation of plume volume is $\pi (a^2{)}_t+q_z=0$.

Let the axial and velocity scales beSince $\mu U_0/(b{\ell }^2)=1$, substitution gives the conduit equationThe buoyant flux and the viscous normal-stress pressure are both retained at this scaling.

Linearizing about unit area gives $a_T+2a_Z-a_{TZZ}=0$. ThusWave crests propagate upward. The group velocity is $2(1-k^2)/(1+k^2{)}^2$: it is downward for $|k|>1$, so the phase direction is not the direction of every wave packet.

For a travelling wave $f(\zeta )$, integrating once and using its uniform far field givesDivide by $f^2$, multiply by $f^{\prime }/f$ and integrate again. A convenient potential, defined up to an additive constant, isAt the positive crest $f=\alpha f_0$, the derivative is zero. Subtracting the far-field potential yieldsThis is the solitary-wave amplitude-speed relation for the conduit equation. A nontrivial elevation wave has $\alpha >1$ and $c>2f_0$; as $\alpha \downarrow 1$, the speed tends to the long-wave speed $2f_0$. Positive area is required throughout the derivation.

Let the actual dimensional thickness be $d=ϵah$, and let $\bar{z}$ increase downward. The leading curvature of the outer free surface isThe Young–Laplace equation gives $p_{\bar{z}}=-\gamma (d_{\bar{z}}/a^2+d_{\bar{z}\bar{z}\bar{z}})$. In the lubrication approximation, the downward velocity at distance $y$ from the wall is $(\rho g-p_{\bar{z}})(dy-y^2/2)/\mu$, from no slip and zero free-surface shear. Integrating this profile givesCircumferential corrections are higher order in $d/a$, so volume conservation is $d_{\bar{t}}+q_{\bar{z}}=0$ at leading order. DefineThen the thin liquid film on a vertical cylinder satisfiesThe circumferential-curvature term $h_z$ is retained along with axial curvature.

The perturbation calculation givesLong waves with $0<|k|<1$ grow, while $|k|>1$ decay; the fastest growth is $1/12$ at $|k|=1/\sqrt{2}$. Disturbances drift downward at speed $G$. The dimensional unstable wavelengths exceed $2\pi a$, with most unstable wavelength $2\pi \sqrt{2}a$. This is the thin-film version of the Rayleigh–Plateau instability driven by circumferential curvature and opposed by short-wave axial curvature.

For a travelling wave, one integration fixed by $H\to 1$ givesor $H^{\prime \prime \prime }+H^{\prime }=3c(H-1)/H^3-G(1-H^{-3})$. In the large positive-speed core, substitute the expansion using the supplied assumptions. The left side at order $c^{2/3}$ is $H_2^{\prime \prime \prime }+H_2^{\prime }$, while the gravity term is order one and $3c/H^2$ is only order $c^{-1/3}$. Thus $H_2^{\prime \prime \prime }+H_2^{\prime }=0$. Its general solution is a constant plus a sine and cosine. The vanishing leading height and slope at both endpoints selectThe core length is $2\pi$ in the axial coordinate; its amplitude is fixed by transition matching.

At either edge set $X=(3c{)}^{1/3}(x-x_e)$, with $x_e=0$ at the trailing edge and $x_e=2\pi$ at the leading edge. Keeping the height $H$ of order one, the exact wave equation becomesThus the transition equation is $H^3{H}_{XXX}=H-1$. Its linearization about $H=1$ has exponents $1$ and $-1/2\pm i\sqrt{3}/2$. Behind the pulse, approaching the uniform film as $X\to -\infty$ permits only the positive real exponent. The positive-growing, nontrivial branch has one amplitude that is removed by translation, giving the relevant unique $H_{-}$. In front, decay as $X\to +\infty$ permits the two real oscillatory amplitudes; translation removes one parameter and leaves a one-parameter family $H_{+}$. These are the unstable manifold and stable manifold dimensions for the transition system. The literal boundary condition also admits the constant solution and an opposite departing branch, so uniqueness here is specifically for the branch matching a raised positive pulse.

Match the supplied trailing quadratic asymptotic to the core near zero. It gives $K_2=\kappa 3^{2/3}$, because $\frac{1}{2}{c}^{2/3}{K}_2{x}^2=\frac{1}{2}\kappa (3c{)}^{2/3}{x}^2$. The leading maximum occurs at $x=\pi$, soThe front tail has the form $H-1\sim C{e}^{-X/2}\cos (\sqrt{3}X/2+\theta )$. Its capillary waves occur ahead of the descending pulse, not behind it; the trailing tail is monotone to leading order. Converting its frequency to physical distance gives

This large solitary pulse on a cylindrical film sketch shows the leading core and the front-tail eigenmodes; the edge transitions are schematic and the ringing is enlarged for visibility. The thin-film asymptotics require both $c\gg 1$ and $ϵc^{2/3}\ll 1$, rather than taking speed to infinity at fixed film slenderness.

At the next core order, $H_0^{\prime \prime \prime }+H_0^{\prime }=-G$. Integrating and imposing zero $H_0^{\prime }$ at both endpoints givesThe constants in the two quadratic transition asymptotics match the core endpoint heights: $A_0+B_0={\beta }_{-}$ and $A_0+B_0-2\pi G={\beta }_{+}$. Hence $2\pi G={\beta }_{-}-{\beta }_{+}$, and the dimensional uniform thickness selected by matching isWith the supplied constants this fixes $G=3.75/(2\pi )$; $B_0$ is not separately fixed by these height conditions. The thickness selection and capillary wavelength follow from matched asymptotic expansion, not from the linear instability calculation.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

Let $s$ be arclength, $\mathbf{t}={\partial }_s\mathbf{r}$ the unit tangent, and $\mathbf{v}$ the local filament velocity relative to the fluid. The resistive-force theory force per unit arclength, exerted by the fluid on the filament, isThe parallel and perpendicular drag coefficients of a slender filament are proportional to dynamic viscosity. For radius $a$ and a relevant axial cutoff length $\ell \gg a$, the anisotropic slender-filament drag from Stokeslet integration gives the leading logarithmic formsThe cutoff may be set by filament length or deformation wavelength. End shape and the choice of cutoff change order-one constants inside the logarithms. Slender-body theory also includes nonlocal interactions that the local resistive-force theory approximation omits. Typically ${\xi }_{\perp }>{\xi }_{\parallel }$, with ratio tending to two in the slender limit; this drag anisotropy enables propulsion.

Write $C={\xi }_{\perp }-{\xi }_{\parallel }>0$. For a small-amplitude planar deformation, $\mathbf{t}=(1,y_x)+O(y_x^2)$ and the leading transverse velocity is $y_t$. Exact inextensibility requires the quadratic longitudinal displacement of an inextensible planar filament: if $X(s,t)=s+X_2(s,t)$ and $Y(s,t)=y(s,t)$, then $X_{2,s}=-y_s^2/2$ to this order. Its velocity is quadratic and has zero time average for a periodic material cycle without mean axial drift. Thus the fixed material abscissa in the question is leading-order notation; this correction does not change the averaged propulsive term below. An imposed mean translation would contribute its separate ordinary axial drag.

To leading nonzero order,The time-averaged deformation-induced propulsive force of a periodic planar filament and power of a periodic planar filament are consequentlyThese are the leading quadratic functionals. Their signs refer to force on the filament. For example, $y=a\sin (kx-\Omega t)$ gives negative $F_x$, opposing the direction of wave propagation.

Let $\eta (x,t)$ be an arbitrary smooth variation periodic in both variables. The first variation of the constrained functional isUsing integration by parts in $x$ and $t$, every boundary term cancels by periodic boundary conditions. ThereforeThe Euler-Lagrange equation isFor a nontrivial stroke with nonzero thrust, $\Gamma \ne 0$. Set $c=C/(\Gamma {\xi }_{\perp })$ and $v=y_t$. Then $v_t+c{v}_x=0$, whose method of characteristics solution is $v=V(x-ct)$. Integrating in time gives the travelling-wave stationarity of planar filament propulsion:Here $g$ is an arbitrary static periodic component. It contributes neither power nor mean propulsion because the time average of $y_t$ vanishes. It can be set to zero when describing the active stroke. The wave must be compatible with both specified periods: $f(x-cT)=f(x)$.

For a pure travelling wave, its leading force and power areThus positive $x$ propulsion uses a wave travelling in the negative $x$ direction. Any smooth interior maximizer satisfying only the stated power constraint must have this travelling active deformation. If the drag were isotropic, $C=0$ and there would be no leading propulsion.

There is a qualification to the printed maximization claim: the variation proves stationarity, not the existence of an unrestricted global maximum. The bandwidth limitation in fixed-power filament optimization can be seen directly. Set $k=2\pi /\lambda$, $\Omega =2\pi /T$, and consider $y_m=a\sin (mkx+\Omega t)$. The leading expressions giveAt fixed power the formal quadratic objective increases with $m$. Eventually that sequence violates the small-slope approximation, but the approximation supplies no numerical slope or wavelength cutoff with which to define the global optimization. Even within a strict small-slope neighbourhood, a sufficiently small admixture of a higher spatial harmonic can improve a candidate interior maximum while keeping power fixed. For example, $y=a[\sqrt{1-{\eta }^2}\sin (kx+\Omega t)+\eta \sin (mkx+\Omega t)]$ retains the fundamental periods and exactly the same leading power, while its force is multiplied by $1+(m-1){\eta }^2$. For fixed $m>1$, taking $a$ and $\eta$ small keeps every slope small.

With an explicit admissible Fourier bandwidth or an additional geometric constraint, the intended conclusion can be made precise. For a finite set of nonstatic Fourier series modes proportional to $e^{i(k_{j}x-{\Omega }_{j}t)}$, with wavenumbers $k_j$ and frequencies ${\Omega }_j$,This is a weighted average of $-C{k}_j/({\xi }_{\perp }{\Omega }_j)$. A largest allowed ratio is attained by a single travelling mode, or by modes sharing the same phase speed; their superposition is still a travelling wave. Thus the Euler-Lagrange calculation yields the requested travelling-wave form, while a global maximum requires a specified admissible shape class.

Take the physical diffusion and consumption constants $D,k>0$. Use the travelling wave coordinate $z=x-vt$, with $v>0$. Then ${\partial }_t=-vd/dz$ and ${\partial }_x=d/dz$. The original PDF has the bacterial drift flux $\chi b{c}_x$: the TeX transcription's ${\chi }^b$ is erroneous. Substituting the logarithmic chemotactic sensitivity $\chi =\alpha /C$ givesPrimes here denote differentiation with respect to $z$. The second equation already shows that nutrient increases towards the front whenever the bacterial density is positive. For the positive band, $C>0$ at every finite $z$, so division by $C$ is legitimate, despite its zero limit far behind.

Integrating the first equation once giveswith constant $J$. For a localized band with no bacterial flux at infinity, $B$ and its flux terms vanish ahead, so $J=0$. Equivalently the laboratory flux $-D{B}^{\prime }+\alpha B{C}^{\prime }/C$ equals $vB$. The Keller--Segel model then has the useful first-order reductionThe far-field conditions choose this zero-flux integration constant; it should not be imposed for an arbitrary nonlocalized travelling solution. We solve its positive localized branch in the next part.

Interpret the displayed two-harmonic form as a leading term in a weakly nonlinear expansion. Write $\xi =(1+ϵ\kappa )x-t$ and ${\Psi }_0=\alpha \cos \xi +\beta \cos 2\xi$. The linear operator multiplies $\cos m\xi$ byFor $m=1,2$, this is $4ϵ\kappa +O({ϵ}^2)$ and $64ϵ\kappa +O({ϵ}^2)$, respectively. Thus detuning of a wave resonance enters at the same order as the quadratic forcing. The cosine addition formula givesProjection onto the two resonant harmonics by harmonic balance requiresThe nonzero branches are thereforeThe two signs are related by a half-period translation of $\xi$; the trivial branch $\alpha =\beta =0$ also exists.

The nonresonant first correction supplies the generated mean, third harmonic and fourth harmonic. Their linear multipliers at $ϵ=0$ are $4,40,180$, giving the nonresonant partFurther resonant amplitude corrections are determined at higher order. Consequently the nonzero answer describes an asymptotic periodic travelling wave with additional harmonics. Literally retaining only the two displayed harmonics cannot be an exact nonzero solution: their square has a positive constant term, while the linear operator applied to the two cosines has no constant term. The distinction is essential to interpreting this perturbative ansatz.

With the travelling wave coordinate $\xi =x-ct$, the reaction-diffusion system becomesAt the advancing leading edge the healthy-tissue density is small. Put $f(\xi )\sim e^{-\lambda \xi }$ with $\lambda >0$. The linearization at $f=0$ givesso a nonoscillatory decaying front requires real positive roots,The pulled travelling front selects the Fisher-KPP minimum wave speed, and therefore

For the initial wound $0<x<W$, two mirror-image fronts enter from its endpoints. The left front moves right from $x=0$ and the right front moves left from $x=W$, each with speed approximately $c_{\min }$. A sequence of sketches would show the initial rectangular depression becoming two smooth monotone transition layers, the zero-density plateau shrinking, and the two layers meeting near $x=W/2$ before the profile relaxes to $S=1$. Their meeting time isThere is an additional local relaxation time of order $1/r$ if “close to one” is assigned a fixed numerical tolerance, so the displayed estimate is the dominant healing time when $W$ is large compared with the front width $\sqrt{D/r}$.

A positive travelling wave $f$ on a uniform background $f_0$ has first integral $c{f}^{\prime 2}/(2f^2)+V(f)=V(f_0)$, with $V(f)=\ln f+c/f+(f_0^2-c{f}_0)/(2f^2)$. Equating the crest and background potentials gives the displayed speed relation for crest ratio $\alpha >1$. The elevation-wave speed exceeds $2f_0$ and approaches that long-wave speed as $\alpha \to 1$. The formula requires the positive-area branch, rather than continuation through $f=0$.

For the leading periodic propulsive force of a periodic planar filament and power of a periodic planar filament functionals, the Euler-Lagrange equation for $F_x+\Gamma (\dot{W}-{\dot{W}}_0)$ gives $({\xi }_{\perp }-{\xi }_{\parallel })y_{xt}+\Gamma {\xi }_{\perp }{y}_{tt}=0$. For nonzero thrust, $y_t$ obeys an advection equation and the active travelling wave is $f(x-ct)$, with $c=({\xi }_{\perp }-{\xi }_{\parallel })/(\Gamma {\xi }_{\perp })$. A static periodic shape may be added without changing leading force or power. Stationarity is a necessary extremum condition, not by itself an existence theorem for a global maximum.

This approximation studies the limit of a viscous scalar conservation law as its positive diffusion coefficient tends to zero. Under hypotheses giving compactness and appropriate convergence, the viscous entropy dissipation selects an entropy solution of the inviscid scalar conservation law. Fixed-phase decreasing travelling waves for a strictly convex flux converge to an entropy shock. Translating each profile differently can change or destroy the limit, so phase or initial data must specify the family.

Positive $\epsilon$ adds diffusion to a one-dimensional scalar conservation law. For a sufficiently decaying smooth solution, the divergence structure gives $\frac{1}{2}\frac{d}{dt}\Vert u{\Vert }_2^2+\epsilon \Vert u_x{\Vert }_2^2=0$, so the $L^2$ energy estimate is uniform in $\epsilon$. If $|F^{\prime }|\le M$, testing against $-u_{xx}$ also yields $\Vert u_x(t){\Vert }_2^2\le e^{M^{2}t/\epsilon }\Vert u_x(0){\Vert }_2^2$. Thin travelling waves explain why a uniform gradient bound generally fails.