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For a self-adjoint operator $L$, solving $Lu=f$ requires $f$ to be orthogonal to every $v\in \ker L$, because $⟨v,Lu⟩=⟨Lv,u⟩=0$. When the range is closed, this condition is also sufficient: $(\mathrm{ran}L{)}^{\perp }=\ker L$ and closedness gives $\mathrm{ran}L=(\ker L{)}^{\perp }$. In finite dimensions the range is automatically closed. This Fredholm solvability condition determines amplitude equations by projecting perturbative forcing onto critical eigenfunctions.

For zero initial value and slope, the causal Green function is $G_0(x,s)=H(x-s)\sin [k(x-s)]/k$, with $k\ne 0$ and $H$ the Heaviside step function. For zero values at endpoints $a<b$, put $\kappa =|k|$. The Dirichlet Green function isBoth are continuous at the source and have a first-derivative jump of one, giving $L_{x}G=\delta (x-s)$. The Dirichlet inverse exists when $\sin [\kappa (b-a)]\ne 0$; a zero denominator signals a nontrivial homogeneous mode and a Fredholm solvability condition. The endpoint kernel is the zero-conjugation case of the Dirichlet Green function for a conjugated harmonic oscillator. Boundary conditions determine which kernel must be integrated against a forcing term.

Use $\xi =x-\hat{X}(T)$, $T=\epsilon t$. At order $\epsilon$, the moving-profile expansion givesThe slow time derivative of $\epsilon R$ is of order ${\epsilon }^2$. Multiplication by $A_0^{\prime }$ and integration eliminate $\mathcal{L}R$ by the Fredholm solvability condition. Thus the leading velocity isFor $\nu (x,T)=K\delta (x+cT)$ the Dirac delta samples $\xi =-\hat{X}-cT$. From the first-order profile equation,ConsequentlyThe prefactor $q$ in the printed profile is important for the position shift in $Z$. This is a leading-order velocity in slow time; the physical velocity is $d\hat{X}/dt=\epsilon {\hat{X}}_T+O({\epsilon }^2)$.

The relative position obeys the autonomous moving-defect locking of a cubic-quintic front equation $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. For $K\ne 0$, a finite constant-velocity front must have constant $Z$, and hence must travel with the defect at ${\hat{X}}_T=-c$. Such locked fronts exist exactly whenFor the strict inequality, their positions areLinearizing the relative-position equation gives ${\eta }_T=K\beta {\mathrm{sech}}^2(\beta Z_{\ast }/2)\tanh (\beta Z_{\ast }/2)\eta$. Therefore the root with $K{Z}_{\ast }<0$ is stable and the other root unstable. At $c=K$, they coalesce at $Z=0$ in a saddle-node bifurcation. Since $Z_T=K{\beta }^2{Z}^2/4+O(Z^4)$ there, the double root attracts from only one side and is not two-sided asymptotically stable. Locking within its existence range still requires an initial relative position in the stable root's basin; beyond the unstable root the front escapes instead.

These positional stability conclusions also describe the leading weakly perturbed front. Indeed $v=A_0^{\prime }>0$ has no zeros, and $\mathcal{L}=({\partial }_{\xi }+v^{\prime }/v)({\partial }_{\xi }-v^{\prime }/v)$ is nonpositive in the $L^2$ inner product. The limiting coefficients $f^{\prime }(0)=-3{\alpha }^2/16$ and $f^{\prime }(\sqrt{q})=-3{\alpha }^2/4$ are negative. Thus the unperturbed shape modes are damped, with the neutral translation mode singled out by the solvability condition; the defect determines its stability at first order.

If $c\ne 0$ and the locking inequality fails, $Z_T$ has no finite zero and has the sign of $c$. The separation grows without bound, the localized forcing of the front decays exponentially, and ${\hat{X}}_T\to 0$: the defect passes the front, which approaches a stationary position in laboratory coordinates. The same escaping behavior occurs outside the locking basin when roots exist. If $c=0$ and $K\ne 0$, the stationary defect drives the front away from it with ever decreasing speed; $Z$ grows in magnitude logarithmically rather than tending to a finite locked position. Finally, when $K=0$, the front remains stationary for every $c$; when also $c=0$, every front position is a neutral equilibrium. These degenerate cases must not be included by dividing by $K$.

Set $f(A)={\mu }_{0}A+\alpha A^3-A^5$ and $V(A)={\mu }_0{A}^2/2+\alpha A^4/4-A^6/6$. Multiplication of the stationary reaction-diffusion equation by $A_x$ gives the first integral $A_x^2/2+V(A)=E$. A regular front with finite endpoint limits has $A_x\to 0$ there, and its nonzero endpoint must satisfy $f(C)=0$. Since $V(0)=0$, the first integral gives $E=0$ and $V(C)=0$. Write $q=C^2>0$. The two endpoint conditions areEliminating ${\mu }_0$ gives the Maxwell balance for a scalar reaction-diffusion front:This is a necessary condition for a stationary front, because the two spatially homogeneous states must have equal potential. It is also sufficient here. With these values, $-2V(A)=A^2(A^2-q{)}^2/3$. For the increasing positive front chooseIntegrating this logistic equation and writing $\beta =2q/\sqrt{3}=\sqrt{3}\alpha /2$ givesThe positive constant multiplying the exponential is arbitrary and can be absorbed into $X_0$; choosing it to be $q$ gives the printed normalization exactly. The profile tends to zero and $q$ at the two ends and its first-order equation implies $A_0^{\prime \prime }+f(A_0)=0$. Its negative is the front to $-\sqrt{q}$, with the same squared profile.

Differentiation of the stationary equation produces the translation modeThe operator $\mathcal{L}$ is self-adjoint. Twice applying integration by parts gives, for $R$ with vanishing boundary Wronskian,Bounded $R,R^{\prime }$ suffice, because the derivatives of the front decay exponentially. This proves the identity as a Fredholm solvability condition associated with translation invariance.

For a stationary finite-wavenumber instability, the relevant Landau–Ginzburg model is the real Ginzburg–Landau equation: its coefficients are real although its pattern amplitude is complex. The complex amplitude records both the strength and the position of the underlying pattern. Assume a simple critical pair of Fourier modes $\pm k_c$, a quadratic nondegenerate maximum of the linear growth rate, and no additional conserved or marginal mode. The last assumption is essential: the conserved field in the previous problem requires an additional evolution equation.

To derive the model, write a real evolution equation schematically as $u_t=L(r)u+N_2(u,u)+N_3(u,u,u)+\cdots$, with symmetric multilinear nonlinearities. Put $r={\epsilon }^2{r}_2$, $X=\epsilon x$, $T={\epsilon }^{2}t$ and use the method of multiple scales expansionHere $L(k_c;0)v=0$. At the next order the noncritical mean and second harmonic are slaved to the critical mode:where $L(0;0)w_0=-2N_2(v,\overline{v})$ and $L(2k_c;0)w_2=-N_2(v,v)$. These inverses exist under the no-other-critical-mode hypothesis. Let $v^{\ast }$ be an adjoint null vector normalized by $⟨v^{\ast },v⟩=1$. At order ${\epsilon }^3$, the Fredholm solvability condition requires the resonant forcing to be orthogonal to $v^{\ast }$. The cubic resonant contribution isThis shows why the cubic coefficient includes feedback from slaved harmonics as well as any original cubic nonlinearity. The resulting amplitude equation, after naming its linear control coefficient $\mu$, isHere $\lambda (k)$ is the critical linear dispersion relation for growth. The condition $g>0$ gives a supercritical bifurcation; $g<0$ calls for higher-order saturation, for example the quintic real Ginzburg-Landau equation.

A concrete check is the cubic Swift–Hohenberg equation $u_t={\epsilon }^2\mu u-(1+{\partial }_x^2{)}^{2}u-g_0{u}^3$. With $u=\epsilon (A{e}^{ix}+\overline{A}{e}^{-ix})+\cdots$, its linear operator contributes $4A_{XX}$ at resonant order, while the cubic resonance is $3|A{|}^{2}A$. Hence $A_T=\mu A+4A_{XX}-3g_0|A{|}^{2}A$. This directly realizes the perturbation-theory coefficients above.

Symmetry independently constrains the form. A fast spatial translation rotates $A$ by a constant phase, so every nondifferential monomial must transform like $A$. The leading allowed nonlinear term is $|A{|}^{2}A$, and a quadratic term is forbidden. A spatial reflection maps $A(X)$ to $\overline{A}(-X)$; it makes the leading coefficients real and rules out a drift term in a reflection-symmetric rest frame. The vanishing first derivative of the linear growth rate at $k_c$ is consistent with the absence of a leading first spatial derivative. These arguments constrain terms but do not calculate $\xi ,g$ or replace the solvability calculation. Breaking reflection or studying an oscillatory instability generally leads to the complex Ginzburg–Landau equation with genuinely complex coefficients.

The principal steady solutions are the zero state and the Ginzburg-Landau plane wavesTheir physical carrier wavenumber is $k_c+\epsilon q$. For $q=0$ the constant phase is arbitrary, giving a circle of uniform-amplitude states. The zero state has Fourier growth rate $\mu -\xi p^2$ and is stable for $\mu <0$. For $g>0$, each existing plane wave is stable to spatially homogeneous amplitude perturbations, with radial eigenvalue $-2g{R}^2$, and neutral to a constant phase shift.

Steady modulated states are also possible. Writing $A=a(X)e^{i\varphi (X)}$ gives the conserved spatial phase current $j=a^2{\varphi }_X$ andIts first integral is $\xi a_X^2/2+\mu a^2/2-g{a}^4/4+\xi j^2/(2a^2)=E$. Boundary conditions select nonuniform periodic amplitudes, fronts or defects from this spatial dynamics. In the invariant real subspace, for example,is a steady kink joining opposite real states. It is not stable in the unrestricted complex equation: the imaginary perturbation $v=\mathrm{sech}(\sqrt{\mu /(2\xi )}X)$ has eigenvalue $\mu /2>0$ under $v_T=\xi v_{XX}+(\mu -g{A}^2)v$. This illustrates why a real-amplitude calculation can miss a phase instability. With real coefficients the Lyapunov functionalsatisfies $A_T=-\delta \mathcal{F}/\delta \overline{A}$ and $d\mathcal{F}/dT=-2\int |A_T{|}^{2}dX$ under compatible boundary conditions. It supplies a variational structure absent from a generic complex-coefficient model.

To test full linear stability of a plane wave, set $A=e^{iqX}(R+u+iv)$ with real perturbations $u,v$. Let $b=g{R}^2=\mu -\xi q^2>0$. The linear equations areFor perturbation wavenumber $p$, the sideband spectrum of a real Ginzburg-Landau plane wave isThe phase branch at small $p$ isTherefore the robust stable band on the infinite line is $q^2<\mu /(3\xi )$, narrower than the existence band $q^2<\mu /\xi$. Crossing its boundary gives the longitudinal Eckhaus instability. At the boundary the quadratic phase-diffusion coefficient vanishes and the leading nonzero term is negative of order $p^4$. More exactly, a nonzero sideband grows only when $p^2<6q^2-2\mu /\xi$; a finite periodic domain for the wave envelope may exclude that interval. Constant-amplitude stability is understood modulo its neutral global phase, and the allowed perturbation wavenumbers must always be specified.

In a second spatial dimension, an anisotropic medium that selects one roll orientation permits an equation with ${\xi }_x{A}_{XX}+{\xi }_y{A}_{YY}$. For positive coefficients its plane-wave analysis is similar, with the second-dimensional contribution adding damping. But simply replacing $A_{XX}$ by an isotropic Laplacian is not a faithful extension for rotationally invariant stationary rolls. There the critical wavevectors form a circle. Near $(k_c,0)$,This forces the anisotropic transverse scaling of an isotropic roll envelope: $X=\epsilon x$, $Y={\epsilon }^{1/2}y$, $T={\epsilon }^{2}t$. For the positive carrier convention used here the Newell–Whitehead–Segel equation isThe opposite fast-carrier convention reverses the accompanying imaginary sign and detuning convention. For a transverse phase perturbation of $A=R{e}^{iqX}$ our convention gives $\sigma =-\xi q{p}^2/k_c-\xi p^4/(4k_c^2)$. Thus $q<0$ permits the transverse zigzag instability, in addition to the longitudinal Eckhaus instability. The fourth transverse derivative and mixed derivative in the square encode the critical-circle geometry.

One complex amplitude still describes only a narrow cone of nearby roll orientations. Competing well-separated wavevectors require coupled amplitude equations; resonant triples can admit quadratic interactions and produce the hexagonal convection amplitude equations when the underlying sign symmetry allows them. Strong disorder, defect cores with large gradients, far-from-onset patterns, boundaries, mean flows and conserved fields may all lie outside the single-amplitude approximation. The method of multiple scales gives controlled leading behavior for small amplitudes and slow modulations, rather than a universal model of all two-dimensional pattern selection.

Use Schwarz conjugation of a spectral function, $F^{♯}(k)=\overline{F(\bar{k})}$. Reality of $f,q_N$ and $\lambda$ gives ${\Psi }^{♯}=\Psi$, ${\Phi }^{♯}=\Phi$ and $e^{♯}=e$. Conjugating part (c) yields the second relationwith $A^{♯}=e(ak)\Phi (k)+e(-k)\Phi (ak)+\Phi (\bar{a}k)$. The explicit $i$ changes sign, while $a$ and $\bar{a}$ are interchanged.

Choose sampling points satisfying $k_n+\lambda /k_n=2\pi in/\ell$. Put ${\epsilon }_n=(-1{)}^n$ and ${\chi }_n=\ell (\bar{a}{k}_n+\lambda /(\bar{a}{k}_n))/2$. Then $e(k_n)=e(-k_n)={\epsilon }_n$ and $e(k_n)e(\bar{a}{k}_n)e(a{k}_n)=1$. Multiply the first relation by ${\epsilon }_n$ and subtract the conjugate relation. The coefficient of $\Psi (a{k}_n)$ vanishes, leavingThus the Fourier coefficient $N_n=\Psi (k_n)={\int }_{-\ell /2}^{\ell /2}{e}^{2\pi ins/\ell }{q}_N(s)ds$ isThe endpoint terms in the derivative-free expression for $\Phi$ cancel, because the other two hyperbolic arguments sum to $-i\pi n$ and $\sinh \chi (k_n)=0$. Writing $h(k)=k-\lambda /k$, we obtain the entirely known expressionThis Dirichlet reconstruction in an equilateral triangle uses Fourier series inversion on the finite side, with convergence in the natural trace norm and pointwise away from endpoints under sufficient smoothness.

For $\lambda >0$, take $k_n=i[\pi n/\ell +\sqrt{(\pi n/\ell {)}^2+\lambda }]$. It is nonzero for every integer $n$. Moreover $\mathrm{Re}{\chi }_n=(\sqrt{3}\ell /4)[v+\lambda /v]>0$ when $k_n=iv$, so the denominator cannot vanish. For $\lambda =0$ use $k_n=2\pi in/\ell$ for $n\ne 0$; the remaining coefficient is $N_0=0$, since the divergence theorem gives $3\int q_N={\int }_D\Delta q=0$. This includes the constant harmonic solution.

For negative nonresonant parameters, use either root of the sampling equation and analytic continuation of the reconstruction. Any zero denominator at a parameter where the Dirichlet operator remains invertible is removable: the uniquely determined trace, and hence its Fourier coefficients, depend analytically on that parameter. Evaluate such values by limits rather than claiming a pole or dividing by zero.

At a true Dirichlet resonance in an equilateral triangle, the request for a uniquely determined normal trace is false. A concrete counterexample uses the triangle's affine barycentric coordinates ${\beta }_1+{\beta }_2+{\beta }_3=1$. The functionis positive inside and zero on the boundary. Since $|\nabla {\beta }_j|=2/(\sqrt{3}\ell )$, direct differentiation gives $\Delta h=-16{\pi }^{2}h/(3{\ell }^2)$. At $\lambda =-4{\pi }^2/(3{\ell }^2)$ both $q=0$ and $q=h$ have the same zero Dirichlet data, but different normal derivatives. On a side where ${\beta }_1=0$, the latter derivative is $-2\pi {\sin }^2(\pi {\beta }_2)/(\sqrt{3}\ell )$, not zero.

In general, if $-4\lambda$ is a Dirichlet eigenvalue, the Fredholm solvability condition is ${\int }_{\partial D}f{\partial }_n\varphi ds=0$ for every eigenfunction $\varphi$ at that eigenvalue. This follows from Green's second identity. When these conditions hold, the complete answer is $q=q_p+{\sum }_j{c}_j{\varphi }_j$ and $q_N^{(r)}=(q_p{)}_N^{(r)}+{\sum }_j{c}_j({\partial }_n{\varphi }_j){|}_r$, with arbitrary real coefficients $c_j$. A rotation-invariant particular solution may be used, but homogeneous additions need not be invariant. Such noninvariant eigenfunctions must exist: otherwise the complete Dirichlet eigenbasis would span only the proper rotation-invariant subspace of $L^2(D)$. Thus the source's unrestricted real parameter needs this resonance qualification in both (c) and (d).

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.

For $\lambda >0$, choose a sample $k_n$ satisfying $\chi (k_n)=i\pi n$ for each integer $n$. A convenient branch isThus $k_n+\lambda /k_n=2\pi in/l$. The undivided identity in part (d) givesRotation invariance makes all three vertex values equal, so $q(l/2)=q(-l/2)$. The exponential has the same endpoint value $(-1{)}^n$ at both ends. Integration by parts in the definition of $\Phi$ therefore givesThese are the Fourier coefficients with the positive-exponential convention. Hence the computed boundary trace isEvery quantity in this rotationally invariant Neumann reconstruction in an equilateral triangle is determined by the prescribed $f$. To check that no denominator was silently divided by zero, put $k_n=i{\kappa }_n$, ${\kappa }_n>0$. ThenNeither factor vanishes for $\lambda >0$. For sufficiently smooth compatible traces, the periodic extension of $q$ is continuous and piecewise smooth; the Fourier series recovers $q$, with the usual endpoint convention. Reality implies $Q_{-n}=\overline{Q_n}$ even though the sampling formula uses complex arguments.

For $\lambda =0$, first impose the zero-flux compatibility from part (c). For $n\ne 0$ take the nonzero root $k_n=2\pi in/l$ and use the same coefficient formula with $\lambda =0$. The coefficient $Q_0$ is an arbitrary real constant times $l$, expressing the additive constant for this problem with a Neumann boundary condition. The zero sample cannot determine it. For example, $f=0$ admits every constant solution, so no uniquely specified $q(s)$ is possible without a normalization.

For completeness, negative parameters can be handled without assuming away resonances. Let ${\varphi }_j$ be a real orthonormal eigenbasis of the Neumann Laplacian, with $-\Delta {\varphi }_j={\mu }_j{\varphi }_j$, and defineusing the corresponding oriented $f$ on each side. The weak equation and Green's first identity give the explicit coefficient relationAt a nonresonant parameter this determines the unique solution. Its common trace is the continuation of the displayed spectral Fourier coefficients from $\lambda >0$; choose either nonzero root of $k_n^2-(2\pi in/l)k_n+\lambda =0$. If an individual quotient is $0/0$ at a nonresonant parameter, use its removable limit in $\lambda$, rather than declare a new eigenvalue. For example, a double sampling root can make $k_n-\lambda /k_n=0$ without making the boundary problem singular. The weak coefficient relation proves that the solution is analytic in the parameter wherever no ${\mu }_j+4\lambda$ vanishes, so these nonresonant limits exist.

At a genuine resonance $\lambda =-{\mu }_{\ast }/4$, the Fredholm solvability condition is precisely $F_j=0$ for every ${\mu }_j={\mu }_{\ast }$. Necessity follows directly from the coefficient relation. For compatible smooth data, sufficiency follows by dividing all nonresonant coefficients and leaving the resonant ones free. The boundary functional is continuous on $H^1(D)$, so ${\sum }_j|F_j{|}^2/(1+{\mu }_j)<\infty$; away from the finite resonant eigenspace this implies ${\sum }_j(1+{\mu }_j)|F_j/({\mu }_j-{\mu }_{\ast }){|}^2<\infty$. Thus the resulting series converges in $H^1(D)$ and satisfies the weak equation, not just its individual coefficient identities. The full solution and its boundary values are thuswith convergence in the weak solution space and the associated trace sense. Rotation averaging selects the common-side component; a nonsymmetric homogeneous addition can have different traces on the three sides. Incompatible resonant data give no solution. For arbitrary real $\lambda$, compatibility and the free homogeneous components are essential; the source does not specify conditions that would select a unique trace at resonance.

Write the reaction term as $f(u;r)=u(1-u)(u-r)$ and assume $D>0$. At $r=1/2$, multiply the stationary travelling-wave reduction of a reaction-diffusion system $D{U}^{\prime \prime }+f(U;1/2)=0$ by $U^{\prime }$ and integrate once. Since $U^{\prime }=0$ at both asymptotic phases and ${\int }_0^{U}f(s;1/2)ds=-U^2(1-U{)}^2/4$, the integration constant is zero:The boundary orientation selects $U^{\prime }=-U(1-U)/\sqrt{2D}$. Separating variables gives the logistic front of a bistable cubic equation, with arbitrary center $q_0$:The arbitrary translation is expected from translation invariance.

To develop the requested perturbation theory, let $r=1/2+\epsilon a$ with signed $a=\pm 1$, introduce slow time $T=\epsilon t$, and put $z=(x-q(t))/\sqrt{D}$. Seekwhere $U_0(z)=(1+e^{z/\sqrt{2}}{)}^{-1}$. The explicit $T$ derivative of $\epsilon U_1$ is second order. Since ${\partial }_{r}f=-u(1-u)$, the first-order equation isDifferentiation of $U_0^{\prime \prime }+f(U_0;1/2)=0$ gives $L{U}_0^{\prime }=0$. The self-adjoint differential operator $L$ has this decaying translation mode, and $U_1\to 0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0^{\prime }$ and integrating by parts makes its left side zero. Therefore the translation-mode solvability for a bistable front requiresBoth integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0^{\prime }=-U(1-U)/\sqrt{2}$, soandThe resulting speed isThis derives the velocity by the Fredholm solvability condition, rather than substituting a guessed speed.