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At a positive spatially homogeneous equilibrium point, $v=u^2$ and $2+u-v=0$. Hence $(u,v)=(2,4)$. The reaction Jacobian matrix there isThe two-dimensional Routh-Hurwitz stability criterion gives strict linear asymptotic stability for $a>2$, and instability for $0<a<2$.

For a spatial mode with $s=k^2$, the reaction-diffusion linearization is $J_s=J-\mathrm{diag}(s,ds)$. When $a>2$, its trace is negative for every $s\ge 0$, whileA growing mode exists precisely when this quadratic is negative at some $s>0$. Its minimum must occur at positive $s$, and must be strictly negative, givingWith $z=d/a$, the latter inequality becomes $4z^2-28z+1>0$. Selecting the root compatible with $2d>a$ yields the Turing instability thresholdThe unstable band isOn a bounded domain a boundary condition must permit a mode in this band. Equality at the threshold gives a neutral mode, not positive growth.

At $a=2$, the homogeneous linearization has purely imaginary eigenvalues, so it is not strictly linearly asymptotically stable. If “stable” is read as nonlinear stability, the boundary deserves attention. Put $x=u-2$, $y=(v-4-x)/\sqrt{3}$, and $x=r\cos \theta$, $y=r\sin \theta$. The exact equations give $\dot{r}=r^{2}F(\theta )$, $\dot{\theta }=\omega +rG(\theta )$, where $\omega =2\sqrt{3}$,Expanding $dr/d\theta$ through cubic order, ${\int }_0^{2\pi }F=0$ cancels the quadratic displacement. The correction from the quadratic change in $r$ also integrates to zero, because it is proportional to $(\int F{)}^2$. Since ${\int }_0^{2\pi }FG=\pi /(2\sqrt{3})$, the Poincaré return map isThis cubic return-map stability of a weak focus shows that $a=2$ is a weak attracting focus. The strict linear range used in the Turing instability calculation is $a>2$; the positive homogeneous equilibrium is also nonlinearly asymptotically stable at the boundary $a=2$. For that boundary, $\mathrm{tr}{J}_s=-(1+d)s<0$ for every nonzero spatial mode. The same strict Turing instability threshold and unstable band therefore also apply there, with the homogeneous mode understood through nonlinear stability.

Assume first $\beta >0$ and $d\ge 0$. The nonnegative equilibrium points areand, when both coordinates are nonnegative,The Jacobian matrix at $O$ has eigenvalues $1,1-d$, so $O$ cannot attract an initially positive uninfected population. At $U$ the eigenvalues are $-1,\beta -d$. At $I$, for $d<1$, they are $d(1+\beta )-\beta$ and $-(1-d)$. At an interior $C$, the equilibrium equations reduce the Jacobian matrix toThus the coexistence equilibrium is asymptotically stable precisely when $\beta /(1+\beta )<d<\beta$.

For positive initial populations these local results identify the global outcome. The relation $\dot{N}=N(1-N)-dy$ bounds all trajectories. With the Dulac function $1/(xy)$, the divergence of the rescaled vector field is $-1/y-1/x<0$, excluding interior periodic orbits. Boundary saddles cannot attract an interior trajectory except at the merging thresholds. The resulting long-term outcomes areAt $d=\beta /(1+\beta )$, $C$ merges into $I$; at $d=\beta$, it merges into $U$. The merged equilibria still attract interior trajectories, but convergence along the zero-eigenvalue direction is algebraic. For example at $d=\beta$, put $x=1+u$, $y=v$. The local centre manifold has $u=-(1+\beta )v+O(v^2)$, and $\dot{v}=-{\beta }^2{v}^2+O(v^3)$. At the other threshold, using $y=(1-d)+v$, $x=u$, gives $v=(\beta -1)u+O(u^2)$ and $\dot{u}=-{\beta }^2{u}^2+O(u^3)$.

The invariant axes also matter: $y=0$ leads to $U$ whenever $x(0)>0$; $x=0$ leads to $I$ for $d<1$ and to extinction for $d\ge 1$. At $d=1$, $I$ coincides with $O$. If $\beta =0$, disease declines for $d>0$ and any initially positive $x$ tends to $U$. If also $d=0$, every point $x+y=1$ is an equilibrium and the initial population proportions are preserved.

For $X\subset {\mathbb{A}}^n$ with vanishing ideal $I(X)$, the Zariski tangent space at $P$ isIt is the kernel of the Jacobian matrix of generators for the ideal, or intrinsically the dual of ${\mathfrak{m}}_P/{\mathfrak{m}}_P^2$. A point is a singular point of an algebraic variety when this tangent dimension exceeds the local dimension of the variety; otherwise it is nonsingular. For an irreducible affine variety the local dimension at a closed point equals its dimension.

There are axis equilibria $E_{\pm }=(\pm \sqrt{\mu },0)$ when $\mu \ge 0$, and $E_a=(a,a^2-\mu )$ for every $\mu$. At an axis equilibrium $(x_{\ast },0)$ the eigenvalues of the Jacobian matrix are $2x_{\ast }$ and $a-x_{\ast }$. Thus a negative axis root is a stable node when $a<x_{\ast }$ and a saddle when $a>x_{\ast }$. A positive axis root is an unstable node when $a>x_{\ast }$ and a saddle when $a<x_{\ast }$. Equalities give a zero eigenvalue.

At $E_a$,It is a saddle for $\mu <a^2$. For $\mu >a^2$, it is stable if $a<0$, unstable if $a>0$, with real node eigenvalues for $a^2<\mu \le 2a^2$ and a focus for $\mu >2a^2$. At $\mu =2a^2$ the node-focus transition is not a bifurcation of hyperbolicity when $a\ne 0$. For $a=0$, $E_a$ has imaginary eigenvalues when $\mu >0$ and the linearized system is a centre. For $a\ne 0$, the candidate local bifurcation values are $\boxed{\mu =0and\mu =a^2}$. When $a=0$ these zero-eigenvalue events coincide at $\mu =0$, while the off-axis equilibrium also has zero-real-part imaginary eigenvalues for every $\mu >0$.

The centre manifold theorem reduces local dynamics near zero-real-part eigenvalues to an invariant manifold tangent to their eigenspace; transverse directions with nonzero real parts retain their stable or unstable character. For a varying parameter, adjoin $\dot{\mu }=0$ and use an extended centre manifold for a parameter.

Take $a=1$. At $\mu =0$ near the origin, $y=0$ is an exact extended centre manifold. Its reduced equation isThis is a saddle-node bifurcation: no nearby axis equilibria for $\mu <0$, one at zero, and two for $\mu >0$. The transverse eigenvalue is positive, so the negative root is a saddle and the positive root an unstable node for small positive $\mu$.

At $\mu =1$, put $u=x-1$, $v=y$, $\nu =\mu -1$. Then $\dot{u}=2u+u^2-v-\nu$, $\dot{v}=-uv$, $\dot{\nu }=0$. Write the extended centre manifold as $u=h(v,\nu )$. Its invariance equation is $h_v(-hv)=2h+h^2-v-\nu$. Matching degrees givesHence the reduced dynamics areThe exact equilibrium branches are $v=0$ and $v=-\nu$, crossing at zero and exchanging stability in the centre direction. This is a transcritical bifurcation; the transverse eigenvalue remains positive. For $\nu <0$ the axis branch is an unstable node and the off-axis branch a saddle; for $\nu >0$ their roles reverse locally.

When $a=0$, the two parameter values merge and the origin at $\mu =0$ has a double zero eigenvalue with nonzero nilpotent linear part. The centre manifold is then two-dimensional, so the preceding one-dimensional classifications cannot simply be reused. Also the off-axis branch has imaginary eigenvalues for every $\mu >0$, with identically zero trace: this is not a generic Hopf bifurcation obtained by varying $\mu$. These degeneracies already follow without computing a new normal form.

Use the inverse change of variables $u=(x+y)/2$ and $v=(x-y)/2$. The chain rule gives the differential operatorsThe mixed partial derivatives cancel, assuming the twice continuously differentiable solution appropriate here. Consequently $f_{xy}=(g_{uu}-g_{vv})/4$.

The transformed equation isThe factors of $1/2$ belong to the inverse Jacobian matrix; using the forward transformation instead would give an incorrect factor.

For a smooth map $\mathbf{u}$, the Jacobian determinant is the determinant of its Jacobian matrix $A_{ai}={\partial }_i{u}_a$. In Einstein summation convention,Differentiate the supplied expression for $V_i$ by the product rule. The term differentiating ${\partial }_j{u}_a$ vanishes because ${\partial }_i{\partial }_j{u}_a$ is symmetric in $i,j$, while ${ϵ}_{ijk}$ is antisymmetric; the term differentiating ${\partial }_k{u}_b$ vanishes similarly. Only differentiation of $u_c$ remains:The last equality follows by the cyclic relabelling $(i,j,k)\mapsto (j,k,i)$, which has positive sign. This proves the Jacobian determinant as a divergence identity.

For a composition $\mathbf{w}=\mathbf{u}\circ \mathbf{v}$, the chain rule givesTaking determinants yieldsFor the first-order perturbation, use spatial derivative control in perturbations of the identity map: the usual smooth-family interpretation gives $D{\mathbf{u}}_t=I+tD\mathbf{F}+o(t)$ locally. Multilinearity of the determinant shows that its linear term comes from choosing one column of $tD\mathbf{F}$ and all other columns from $I$. This gives the trace:The regularity qualification matters: a pointwise $o(t)$ remainder for maps does not by itself give an $o(t)$ remainder for spatial derivatives. For instance $(x+t^2\sin (x/t^2),y,z)$ has a uniform $o(t)$ displacement and $\mathbf{F}=0$, but its Jacobian determinant at $x=0$ is $2$ for $t\ne 0$. Thus the first expansion needs the remainder in $C^1$ locally, or an equivalent smoothness assumption on the family.

Under the subsequent group law, ${\mathbf{u}}_0$ is the identity and ${\mathbf{u}}_{-t}$ is the inverse of ${\mathbf{u}}_t$. Write $j(t,\mathbf{x})=J[{\mathbf{u}}_t](\mathbf{x})$. Applying the composition formula to ${\mathbf{u}}_{t+h}={\mathbf{u}}_h\circ {\mathbf{u}}_t$ and the short-time expansion gives the Jacobian evolution of a smooth flow:HenceIf only the pointwise generator expansion is assumed initially, the group law still gives ${\partial }_t{\mathbf{u}}_t(\mathbf{x})=\mathbf{F}({\mathbf{u}}_t(\mathbf{x}))$. Uniqueness and smooth dependence for this smooth ordinary differential equation make it a smooth flow map, thereby justifying the differentiated expansion used above. Thus a divergence-free generator hasBy the change of variables formula, these maps preserve ordinary Euclidean volume and orientation. This is the Euclidean instance of a volume-preserving vector field.

Use dimensionless interaction couplings in the exponent of the Boltzmann weight. In the following formulas $K,L$ denote the physical interaction energies divided by $k_{B}T$ if that factor has not already been absorbed into the displayed Hamiltonian. This convention is required for temperature-independent numerical fixed-point coordinates.

Split the original square lattice into retained sites $E$ with even $x+y$ and eliminated sites $O$ with odd $x+y$, where coordinates are in units of $a$. Every nearest-neighbour bond joins $E$ to $O$, whereas a diagonal next-nearest-neighbour bond stays within one sublattice. Denote retained Ising spins by $s_e$ and eliminated Ising spins by ${\eta }_o$. For fixed retained Ising spins defineThe blocked Boltzmann weight in the checkerboard decimation of the square-lattice Ising model isUse the independent uniform product measure on Ising spins on $O$ to perform a cumulant expansion. Its one-spin mean vanishes, and $⟨{\eta }_o{\eta }_{o^{\prime }}{⟩}_0={\delta }_{o{o}^{\prime }}$. Consequently the first-order $K$ term, the first-order eliminated-sublattice $L$ term, and the mixed $KL$ term vanish. To the requested order,Here the truncation keeps $K^2$ and the linear term in $L$; equivalently count $L$ as order $K^2$. At higher orders the interaction family will generally not close.

Each eliminated site has four retained neighbours, soIts six distinct neighbour pairs comprise four corner pairs at distance $\sqrt{2}a$ and two opposite pairs at distance $2a$. A given retained diagonal pair shares two eliminated centres; a given opposite pair shares only one. Thus the generated pair couplings are $2K^2$ for diagonal pairs and $K^2$ for axial pairs separated by two original spacings. The original $L$ interaction directly contributes to each retained diagonal pair. No single-spin term survives spin inversion symmetry, and no three-spin or four-spin term is present in this second cumulant. This proves closure on exactly two nonconstant interactions at the retained order.

The primitive vectors of the retained square lattice are $a(1,1)$ and $a(1,-1)$. Rotate the axes by $45$ degrees and rescale lengths by $1/\sqrt{2}$; retained diagonal pairs become nearest neighbours, and the axial distance-$2a$ pairs become next-nearest neighbours. With $b=\sqrt{2}$, the leading checkerboard Ising decimation recursion is thereforeThe constant part of the weight is also determined: $4(K^2/2)(N/2)=N{K}^2$, so the original partition function equalsto the same order in the logarithm of the weights. That normalization changes free energy but not the retained-spin interaction operators. If physical energy couplings are kept instead, at fixed ${\beta }_T$ the relations read $K_{\mathrm{energy}}^{\prime }=L_{\mathrm{energy}}+2{\beta }_T{K}_{\mathrm{energy}}^2$ and $L_{\mathrm{energy}}^{\prime }={\beta }_T{K}_{\mathrm{energy}}^2$.

The fixed-point equations are $L_{\ast }=K_{\ast }^2$ and $K_{\ast }=3K_{\ast }^2$. Their only finite real solutions areThe origin is the high-temperature fixed point; its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, with both multipliers zero. The nontrivial candidate critical point is $(1/3,1/9)$. Linearize the discrete renormalization-group transformation there:The characteristic polynomial is ${\lambda }^2-(4/3)\lambda -2/3$, givingEigenvectors may be chosen as $(1,2/(3{\lambda }_{\pm }){)}^T$. The positive eigenvalue greater than one is the relevant direction of a fixed point; the other is irrelevant because its magnitude is less than one. Its negative sign only alternates the sign of the perturbation between steps. For a discrete map the magnitude relative to one decides relevance, not the sign rule used for continuous beta-function eigenvalues.

Let $t$ be the relevant thermal scaling coordinate, so $t^{\prime }={\lambda }_{+}t$ to first order. A correlation length measured in restored lattice units satisfies $\xi (t^{\prime })=\xi (t)/b$. If $\xi (t)\sim |t{|}^{-\nu }$, then ${\lambda }_{+}^{-\nu }=b^{-1}$. Hence the thermal exponent from a discrete renormalization map isThis is the value predicted by the stated low-order real-space renormalization group, whose nontrivial fixed point is evaluated beyond the strictly infinitesimal-coupling limit. It is an approximation, not an exact solution of the full square-lattice Ising model.

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.

The chain rule for Frechet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

For real smooth functions of compact support in a planar domain $D$, use the Dirichlet inner productChanging the positive normalization factor does not change orthogonality. For complex functions, insert complex conjugation in the second factor to obtain the corresponding Hermitian form.

Let $f:D\to \tilde{D}$ be a conformal bijection, and let $\tilde{u},\tilde{v}$ be smooth functions of compact support on $\tilde{D}$. Its real Jacobian matrix is $Df=|f^{\prime }|R$, where $R$ is a rotation. By the chain rule,The change of variables formula has Jacobian determinant $|f^{\prime }{|}^2$, so this factor cancels:This proves conformal invariance of the planar Dirichlet inner product. By completion it is also an isometry between the corresponding Dirichlet energy spaces. It asserts invariance of the energy form, not of the inhomogeneous Sobolev norm, whose $L^2$ term has a different transformation rule.

Discuss the zero-field phase boundaries, taking $h=0$ and writing $r={\mathcal{A}}_2$, $q={\mathcal{A}}_4$, $s={\mathcal{A}}_6>0$. This qualification matters because a conjugate field generally rounds the zero-field continuous transition. Nonzero stationary points satisfy $r+q{m}^2+s{m}^4=0$.

For $q>0$, the continuous phase transition occurs atOn the $r>0$ side the only stable phase is $m=0$; on the $r<0$ side the equilibrium global minima are $\pm m_0$ with $m_0^2\sim -r/q$. The magnetization therefore grows continuously from zero as $r$ changes sign.

For $q<0$, the first-order phase transition instead occurs while $r$ is positive. At phase coexistence, $m=0$ and $m=\pm m_0$ are equally deep global minimum points. With $a=m_0^2>0$, stationarity and equality of free energies giveEliminating $r$ gives $-q{a}^2/4-s{a}^3/3=0$, henceAt this coexistence value, the sextic even Landau potential factors as $f(m)=s{m}^2(m^2+3q/(4s){)}^2/6\ge 0$. This verifies that the three equal stationary values are global minima. Above this coexistence value of $r$ the disordered phase wins; below it the ordered phase wins, with a finite jump in the phase-selected magnetization. The values $r=0$ and $r=q^2/(4s)$ are the disordered and ordered spinodal points, respectively, not the coexistence boundary: the latter follows from the repeated root of $r+qa+s{a}^2=0$.

The two transition loci meet at the tricritical point,Two independent control parameters such as $T$ and $g$ can tune these two conditions. In a generic local phase diagram their coefficient map has a Jacobian matrix of rank two at the intersection; unspecified functions ${\mathcal{A}}_2(T,g)$ and ${\mathcal{A}}_4(T,g)$ do not by themselves determine the geometric orientation of the boundaries. The sextic term stabilizes this intersection and the negative-quartic first-order side.

The fixed-point equation for $u$ factors asThe cubic derivative is $3u^2+2u+3>0$ for every real $u$; its values at zero and one have opposite signs. Thus the possible first coordinates in the stated domain are exactly $u=0,u_{\ast },1$.

For finite $v$, the second fixed-point equation is $v[v^3-(1+u^2{)}^2]=0$. The renormalization-group fixed points are consequentlyThere are six finite points and three boundary points at infinity. Infinity is interpreted in a compactified domain, not as an ordinary real number. The reciprocal coordinate at infinite coupling $w=1/v$ transforms as $w^{\prime }=(1+u^2{)}^2{w}^4$, making $w=0$ a well-defined boundary fixed point.

At any finite fixed point $(\tilde{u},\tilde{v})$, let $\delta u=u-\tilde{u}$, $\delta v=v-\tilde{v}$. The Jacobian matrix linearization isThe only point with both coordinates strictly interior is $\tilde{u}=u_{\ast }$, $\tilde{v}=(1+u_{\ast }^2{)}^{2/3}$. Using its fixed-point relations, the matrix becomesIts eigenvalues are $\rho$ and $4$. In particular, $u_{\ast }<1/\sqrt{3}$ because the increasing cubic is already positive there, so $\rho >1$ (numerically about $1.68$). The corresponding eigenvectors can be chosen as $(1,c/(\rho -4){)}^T$ and $(0,1{)}^T$. Both discrete multipliers exceed one, henceBoth perturbations are relevant directions of a fixed point under repeated coarse-graining, rather than one stable and one unstable direction. A length-rescaling factor was not specified, so these multipliers should not be assigned numerical critical scaling exponents without additional information. At $a=0,1$, the $u$ multiplier is zero; at $v=0$ or $w=0$ the other multiplier is also zero. Thus the four corner points attract locally within the domain, the two positive finite boundary points at $a=0,1$ are saddles, and the points $(u_{\ast },0),(u_{\ast },\infty )$ have one repulsive and one attractive direction.

Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode $k$, the Stokes temperature-slaving operator maps $\theta$ to $w={\mathcal{W}}_R\theta$, where $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $w=w^{\prime \prime }=0$. The temperature evolution has linear operator ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$ and bilinear map $\mathcal{N}(\varphi ,\chi )=-{\mathbf{u}}_R[\varphi ]\cdot \nabla \chi$. Under the homogeneous thermal Dirichlet boundary conditions, ${\mathcal{L}}_{R_c}$ is self-adjoint. Normalize its critical eigenfunction as $g(z)=\sin \pi z$ and set $s=k_c^2+{\pi }^2$; the critical vertical velocity is $f=sg$.

At order $ϵ$, the critical eigenfunction equation gives ${\mathcal{L}}_{R_c}(g{e}^{i{k}_{c}x})=0$. At order ${ϵ}^2$, the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to $|A{|}^2$ and, in a general vertical-mode calculation, a second harmonic proportional to $A^2$. These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.

At order ${ϵ}^3$, the method of multiple scales produces the slow derivative $A_{\tau }$, the detuning term $R_2{\partial }_R{\mathcal{L}}_R$, and the two cross-advection terms involving first- and second-order fields. Project the $e^{i{k}_{c}x}$ component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by $⟨g,g{⟩}_z$. The detuning supplies $\mu A$ with $\mu \propto R_2$; interactions of horizontal wavenumbers $2k_c$ and $-k_c$ permit $\beta A^{\ast }$ with $\beta \propto \gamma$; self-interaction through the slaved mean and second harmonic supplies $-\lambda |A{|}^{2}A$. Other products have the wrong horizontal wavenumber. Reflection $x\mapsto -x$ permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes $\beta =0$ at this order. To see this, write a positive second-harmonic forcing component as $(w,\theta )=(W(z),\Theta (z))e^{2i{k}_{c}x}$, incorporating the cosine's factor $1/2$. Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor $A^{\ast }$,The integral vanishes because $g=0$ at both plates, even though $\Theta (0)$ is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times $\gamma$; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.

The same normalization makes the remaining coefficients explicit. Since ${\mathcal{W}}_{R}g=(R{k}_c^2/s^2)g$, $\mu =R_2{k}_c^2/s^2$. The quadratic second harmonic cancels for $f=sg$, while the uniform correction is ${\theta }_{20}=-s|A{|}^2\sin (2\pi z)/(2\pi )$. Projecting $-w_1{\partial }_z{\theta }_{20}$ gives $-s^2|A{|}^{2}A/2$. Thus for the literal model and this temperature normalization,A generic nonzero $\beta$ would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.

Write $A=x+iy$. Then $x_{\tau }=(\mu +\beta -\lambda (x^2+y^2))x$ and $y_{\tau }=(\mu -\beta -\lambda (x^2+y^2))y$. These are a gradient flow for $V=-\frac{1}{2}(\mu +\beta )x^2-\frac{1}{2}(\mu -\beta )y^2+\frac{\lambda }{4}(x^2+y^2{)}^2$, so local minima give stable equilibrium points. At the origin the two eigenvalues are $\mu +\beta$ and $\mu -\beta$. The origin has exponential asymptotic stability if $\mu <-|\beta |$, retains asymptotic stability with algebraic decay at $\mu =-|\beta |$, and is unstable if $\mu >-|\beta |$. At equality, ${\rho }^2=x^2+y^2$ obeys $({\rho }^2{)}_{\tau }\le -2\lambda {\rho }^4$, since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.

For $\beta >0$ the stable nonzero equilibrium points are real; for $\beta <0$ they are imaginary:The real branch has Jacobian matrix eigenvalues $-2(\mu +\beta ),-2\beta$; the imaginary branch has $-2(\mu -\beta ),2\beta$. The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when $\beta \ne 0$.

For $\beta =0$, the origin is stable for $\mu \le 0$, with algebraic decay at zero. If $\mu >0$, the circle $|A{|}^2=\mu /\lambda$ is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero-$\beta$ branches instead exhibit phase locking to one of two phases separated by $\pi$.

Consider a horizontal fluid layer under the Boussinesq approximation heated from below, rotating uniformly about its vertical axis. To make the linear formulas definite, take depth one, stress-free impermeable fixed-temperature plates, a horizontally infinite or sufficiently large periodic domain, and negligible centrifugal-buoyancy modifications. Rigid plates change the vertical eigenfunctions and thresholds and require their own boundary value problem; the explicit formulas below use the free-slip model. Let $P=\nu /{\kappa }_T$ be the Prandtl number, $\mathrm{Ta}=(2\Omega d^2/\nu {)}^2$ the Taylor number, and $R$ the Rayleigh number. Time is again measured on the thermal diffusion time.

The linearized rotating Rayleigh-Bénard convection equations about conduction areThe Coriolis force does no work on the fluid, but it couples vertical motion to vertical vorticity and thereby changes the balance of buoyancy and viscous damping. On the plates require $w=\theta =0$ and ${\partial }_z{u}_x={\partial }_z{u}_y=0$. Write $\zeta =(\nabla \times \mathbf{u}{)}_z$. Taking a curl and a double curl removes pressure and givesA normal mode with horizontal wavenumber $k>0$, vertical index $n\ge 1$ and complex growth rate $s$ has $w=W\sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$, $\theta =\Theta \sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$ and $\zeta =Z\cos (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$. Put $a=k^2+n^2{\pi }^2$. The three equations become $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, and $(s+a)\Theta =W$. Their determinant gives the rotating-convection growth-rate polynomialThis cubic includes viscous and thermal modes as well as the convective instability. It avoids excluding a root by division by $s/P+a$. Linear stability means every root has negative real part; stationary onset has $s=0$, while overstability has $s=\pm i\omega$ with nonzero angular frequency.