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A-stability removes the scalar linear decay restriction on the step size; it does not replace accuracy, nonlinear solvability or a suitable system norm. Its importance is clearest for a stiff differential equation, where rapidly decaying modes coexist with slowly varying quantities of interest. On the Dahlquist test equation $y^{\prime }=\lambda y$, the exact solution decays for $\mathrm{Re}\lambda <0$. A stable time integrator should not turn that decay into numerical growth.

For a one-step method write $y_{n+1}=R(z)y_n$, where $z=h\lambda$ and $R$ is its stability function. Its linear stability domain consists of parameters with $|R(z)|\le 1$ and a well-defined update. The method is A-stable when this domain contains the closed left half-plane. A consistent order-$p$ one-step method has $R(z)=e^z+O(z^{p+1})$ near zero, but that local agreement alone says nothing about stability at large negative $z$.

For a Runge-Kutta method with stage matrix $A$, weights $b$ and $1=(1,\dots ,1{)}^T$, eliminating the stages on the test equation givesAn explicit finite-stage method has a polynomial $R$. A nonconstant polynomial is unbounded along the negative real axis, so no consistent explicit Runge-Kutta method is A-stable. The explicit Euler method, with $R(z)=1+z$, is stable in the disk $|1+z|\le 1$ and on the negative real axis only for $-2\le z\le 0$. A stiff mode with a large negative eigenvalue therefore forces an unnecessarily small step even after its exact contribution has almost disappeared.

The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z{|}^2=1-2\mathrm{Re}z+|z{|}^2\ge 1$, so it is A-stable. The trapezoidal rule hasand is also A-stable. The implicit midpoint method has the same scalar stability function, although its nonlinear update differs from the trapezoidal rule. The first method has order one and the latter two order two. Their implicit solves trade greater work per step for freedom from the scalar stiff stability restriction.

Damping of very stiff modes separates A-stability from L-stability. For a one-step method, the latter additionally requires $R(z)\to 0$ as $z$ tends to infinity in the left half-plane. Backward Euler method satisfies this. The trapezoidal and midpoint rules instead have $R(z)\to -1$ along the negative real axis: a very stiff mode can alternate in sign with little damping even though its amplitude remains bounded. L-stability is often preferable for diffusion and rapid relaxation, while retaining oscillatory modes can be desirable in conservative dynamics.

For a linear multistep method, writeIts test-equation amplification roots satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. Absolute stability requires every root to lie in the unit disk or on its boundary with all unit-modulus roots simple; all parasitic modes matter. At $z=0$ this is the zero-stability root condition. Consistency requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$, and the Dahlquist equivalence theorem combines consistency and zero-stability to characterize convergence. Thus a claim of stability obtained by canceling a common factor is not a proof about the original recurrence and its arbitrary consistent starting data.

Explicit linear multistep methods cannot be A-stable either. If ${\beta }_k=0$ and ${\alpha }_k\ne 0$, some lower coefficient of $\rho -z\sigma$ becomes unbounded as $z\to -\infty$, while the leading coefficient stays fixed. If all roots remained in the unit disk, every coefficient-to-leading-coefficient ratio, an elementary symmetric function of those roots, would remain bounded. This contradiction forces an unstable root.

An important implicit example is BDF2 method, whose characteristic equation is $(3-2z){\zeta }^2-4\zeta +1=0$. A unit root $\zeta =e^{i\theta }$ givesThe leading coefficient vanishes only at $z=3/2$, outside the left half-plane. Near a small negative $z$, both roots are inside the disk; no unit-circle crossing can occur in the connected open left half-plane. On the imaginary axis the boundary equality can occur only at $z=0$, where the roots are $1,1/3$. Hence BDF2 is A-stable and its two roots tend to zero for very large negative $z$, giving strong stiff-mode damping. This is a complete boundary-locus argument, not just a boundary plot.

The Second Dahlquist barrier says that a convergent A-stable ordinary linear multistep method has order at most two. This theorem does not limit implicit Runge-Kutta methods or multiderivative multistep methods. In particular the A-stable third-order two-step multiderivative method in question 1 uses $y^{\prime \prime }$ and satisfies different order conditions. Among collocation Runge-Kutta methods, the standard family results give Gauss collocation methods order $2s$ with diagonal $[s/s]$ Padé approximants to $e^z$ as stability functions, and Radau IIA methods order $2s-1$ with subdiagonal $[s-1/s]$ approximants. Both families are A-stable; the Gauss functions have a nonzero limit at infinity, whereas Radau IIA is L-stable. The corrected two-stage right-Radau method, for example, haswhich matches the exponential through cubic order and tends to zero at infinity. Its poles $2\pm i\sqrt{2}$ lie in the right half-plane. For $z=-r+iy$, $r\ge 0$, direct subtraction givesproving its A-stability directly.

Sectorial A-alpha stability relaxes the requirement to a sector around the negative real axis, useful when the spectrum of a stiff problem remains in that sector; higher-order backward differentiation formulas illustrate the usefulness of this weaker requirement. For a normal linear system $y^{\prime }=Ly$, unitary diagonalization reduces the numerical update to its scalar factors $R(h{\lambda }_j)$. For a nonnormal system, a scalar eigenvalue test alone does not give a uniform bound in the chosen norm. Nonlinear stability is different again: B-stability asks for contractivity when $\mathrm{Re}⟨f(u)-f(v),u-v⟩\le 0$, and algebraic stability of a Runge-Kutta method is a standard sufficient criterion. Thus A-stability addresses a precise and valuable linear test, while step selection still has to meet accuracy requirements and the actual implicit equations must be solved on the relevant branch.

Eigenvalue analysis proves mesh-uniform stability when it is accompanied by control of eigenvectors or an energy norm; the signs of eigenvalues alone are insufficient. For an evolution equation, the method of lines first produces $u_h^{\prime }=L_h{u}_h$. A time integrator then gives an amplification matrix $S_{h,\delta }$, for example $S_{h,\delta }=R(\delta L_h)$ for a Runge-Kutta method. The useful stability requirement on a fixed physical time interval iswhere $C_T$ is independent of the refining spatial and temporal meshes. The semidiscrete analogue is $\Vert e^{t{L}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. These are bounds on propagation of initial errors and residuals, not merely statements that each finite matrix has a bounded solution over its own finite time interval.

If $L_h$ is a normal matrix in the chosen discrete inner product, it has a unitary eigenbasis. ThenFor dissipative problems it is sufficient that every $\delta {\lambda }_j$ lie in the integrator's linear stability domain. A diagonalizable family also suffices if its eigenvector matrices have uniformly bounded condition numbers: the same maximum is multiplied by $\Vert V_h\Vert \Vert V_h^{-1}\Vert$. Without that uniform bound the argument does not establish mesh stability. For finite-time estimates, an amplification modulus $1+O(\delta )$ can also be acceptable; strict contraction at every step is a stronger property.

For a constant-coefficient stencil on the full lattice or a periodic grid, von Neumann stability analysis provides the eigenvectors explicitly. A Fourier mode diagonalizes each translation-invariant spatial difference, giving a scalar symbol ${\lambda }_h(\theta )$ or a small matrix symbol for a system. The discrete Fourier transform and Parseval identity turn a uniform modal bound into an actual discrete L2 norm bound. The two-dimensional stencil in question 3 has $\mathrm{Re}{\lambda }_h=-(1-\cos \eta {)}^2/h\le 0$, so its continuous-time amplification is contractive; a later time integrator must still be checked. With boundaries one must analyze the boundary rows as well: a whole-line Fourier symbol cannot detect an unstable boundary closure.

For the heat equation $u_t=\kappa u_{xx}$ on a unit interval with zero Dirichlet data, the centred second difference has a sine eigenbasis and eigenvaluesExplicit Euler method multiplies mode $j$ by $1+\delta {\lambda }_j$. Requiring all factors to have modulus at most one yields $\kappa \delta /h^2\le 1/[2{\sin }^2((N-1)\pi /(2N))]$ on a given grid, and the sharp grid-independent sufficient limit isThese parabolic time-step restrictions express the growth of the largest-magnitude diffusion eigenvalue like $h^{-2}$. Backward Euler method instead has factor $(1-\delta {\lambda }_j{)}^{-1}$ and the Crank-Nicolson method factor $(1+\delta {\lambda }_j/2)/(1-\delta {\lambda }_j/2)$, both bounded by one for every $\delta \ge 0$. Their unconditional stability follows from A-stability, with much stronger damping of the shortest scales for backward Euler.

For $u_t=-a{u}_x$, $a>0$, explicit time stepping with a backward spatial difference givesThus the upwind finite difference scheme is contractive exactly for $0\le \nu \le 1$, a Courant–Friedrichs–Lewy condition matching the direction and speed of information propagation. A centred spatial difference has purely imaginary eigenvalues, and forward Euler gives $|S{|}^2=1+{\nu }^2{\sin }^2\theta >1$. Under a fixed positive hyperbolic Courant number, high-frequency data grow by a fixed factor in each of $O(1/h)$ steps, proving instability as the mesh is refined. This is not a claim that no exceptionally small coupled step can ever give a finite-time bound: if $\delta =O(h^2)$, the excess per step is only $O(\delta )$, but the usual hyperbolic scaling is unstable. Higher-order explicit integrators can possess nontrivial imaginary-axis stability intervals and are then suitable for skew-adjoint advection or the energy-form first-order wave equation.

Variable coefficients usually prevent Fourier diagonalization, but may preserve a self-adjoint spatial operator. The positive-face diffusion discretization in question 5 is an example. Its discrete energy identity makes $A+B$ symmetric negative definite, so a unitary eigenbasis still exists even without a closed eigenvalue formula. More generally, a dissipative operator $L_h$ in a discrete energy inner product obeyswhich directly proves a uniform semigroup estimate. The dissipative Cayley-transform contraction also follows without normality: for $\alpha \ge 0$,Taking $v=(I-\alpha L_h{)}^{-1}u$ proves contraction of the Crank–Nicolson factor. The contractivity of split Crank-Nicolson diffusion then follows from multiplying two contraction bounds; simultaneous eigenvectors of the two directional matrices are unnecessary.

The negative spectra do not imply uniform semidiscrete stability example shows what can go wrong. LetBoth eigenvalues are $-1$ for every mesh, but applying the propagator to the second unit vector at $t=1$ gives norm at least $1/(eh)$. The evolution is therefore not mesh-uniformly stable. For a discrete matrix, a unit-modulus eigenvalue with a nontrivial Jordan block gives powers growing polynomially in the step number. Even strictly interior eigenvalues can allow large transient growth when eigenvectors are badly conditioned. A pseudospectrum or numerical range of an operator, and especially a direct energy estimate, can reveal behavior missed by eigenvalues.

A resolvent estimate provides another way to see the needed uniformity. If $\Vert S_h^n\Vert \le C$ for every $n$, then for $|z|>1$,This shows directly that a uniform power estimate imposes more than the location of the spectrum. Conversely, resolvent-to-power estimates in varying dimension must themselves be uniform; a bound with constants depending on the number of grid points is not enough.

For a linear multistep method, each spatial eigenvalue gives an amplification polynomial, and all roots must satisfy the root condition for a multistep method, including simplicity of any unit-modulus root. Merely following the root approximating $e^{\delta \lambda }$ misses parasitic modes. The temporal and spatial analyses must therefore be combined with the appropriate zero-stability and uniform modal bounds.

Finally stability connects to convergence. If the exact grid samples have normalized residual ${\tau }^n$ and the error obeys $e^{n+1}=S_h{e}^n+\delta {\tau }^n$, iteration givesA residual tending uniformly to zero then gives convergence on $[0,T]$. For a well-posed linear initial-value problem and a consistent linear finite-difference approximation, this is the forward implication of the Lax equivalence theorem. The practical conclusion is to analyze the full operator including boundaries, select a time method whose stability region contains the relevant scaled spectrum, and verify that eigenvectors or an energy estimate make the resulting bounds uniform as the grid is refined.

For a nonconstant invasion travelling wave, the relevant end states are $w(-\infty )=k$ and $w(+\infty )=0$. These fronts do not exist at every nonnegative speed: the phase plane calculation gives the threshold $2\sqrt{k}$. The constant solutions $w\equiv 0$ and $w\equiv k$ are, of course, nonnegative travelling waves for every speed, including zero.

Put $z=x-ct$ and $p=w^{\prime }$. The travelling-wave reduction of a reaction-diffusion system givesThe equilibrium points in the phase plane are $(k,0)$ and $(0,0)$. At $(k,0)$ the characteristic equation is $r^2+cr-k=0$, so there is one positive and one negative eigenvalue. Thus it is a saddle equilibrium. Its descending unstable manifold has $w<k$, $p<0$ and locallyAt $(0,0)$ the eigenvalues areFor $c>2\sqrt{k}$ it is a stable node; for $0<c<2\sqrt{k}$ it is a stable spiral. The repeated-root case $c=2\sqrt{k}$ must be included in the existence argument rather than inferred from distinct roots.

Suppose $c\ge 2\sqrt{k}$. Choose $a>0$ with $a(c-a)\ge k$; for example $a=(c-\sqrt{c^2-4k})/2$. Consider the closed triangular regionOn the horizontal side $p=0$, the vector field has $p^{\prime }=-w(k-w)\le 0$, pointing into $\mathcal{T}$. On the vertical side $w=k$, it has $w^{\prime }=p\le 0$, again pointing inward. For the sloping side use $h=p+aw$, so that on $h=0$,Thus trajectories cannot cross into $h<0$. The remaining boundary point $(0,0)$ is an equilibrium point and cannot be crossed in finite time by a nonconstant trajectory, by uniqueness theorem for ordinary differential equations. This proves forward invariance of $\mathcal{T}$. The argument works unchanged at $c=2\sqrt{k}$ with $a=\sqrt{k}$.

The descending unstable manifold of $(k,0)$ initially enters $\mathcal{T}$. Its forward trajectory remains in this compact region, so it exists for all positive $z$. Within the interior, $p<0$: a first contact with $p=0$ at $0<w<k$ would have $p^{\prime }<0$, contrary to arrival from below. Hence $w$ decreases strictly and has a limit $\ell \in [0,k)$. To identify the limit, its omega-limit set lies on the line $w=\ell$ and is invariant. Any trajectory remaining on this line has $p=w^{\prime }=0$; its second equation then forces $\ell (k-\ell )=0$. Since $\ell <k$, it follows that $\ell =0$, and the only possible omega-limit set is $(0,0)$. Backward in $z$, the unstable manifold tends to $(k,0)$ by its construction. We have therefore obtained a positive decreasing heteroclinic orbit and hence a nonnegative travelling wave for every $c\ge 2\sqrt{k}$.

For completeness, the threshold is necessary for such a front. When $0<c<2\sqrt{k}$, put $\nu =\sqrt{k-c^2/4}$, $X=w$, and $Y=(p+cw/2)/\nu$. Near the zero equilibrium point the system becomesFor a nonzero trajectory its polar angle satisfiesA trajectory approaching zero therefore makes infinitely many rotations and must cross $X=w=0$. It cannot be a nonnegative front. At $c=0$, the first integralis conserved. It has values $k^3/6$ and $0$ at the two desired end states, so no heteroclinic orbit connects them. ConsequentlyThis is the phase-plane construction of a Fisher travelling front, including the critical speed. The statement for all $c\ge 0$ must distinguish these nonconstant fronts from the two constant travelling waves.

The natural expectation is orbital stability, since a spatial translation of a travelling wave is another travelling wave. Thus convergence for ordinary perturbations, when it holds, is to a suitably translated front. It also matters which perturbations are allowed in the leading edge: the zero state ahead of this front is linearly unstable, and a sufficiently shallow positive tail can select a different speed. One should not assert asymptotic stability for arbitrary disturbances without a norm and a class of initial data. The requested essential spectrum calculation is a necessary part of such an analysis, not by itself a complete nonlinear stability proof.

In the moving coordinate $z=x-ct$, write $u(x,t)=w_c(z)+v(z,t)$. Expansion of the reaction term givesThe first equation to study is its linearization,and the associated eigenvalue equation $\mathcal{L}\varphi =\lambda \varphi$. Differentiating the front equation gives $\mathcal{L}{w}_c^{\prime }=0$, the neutral translation mode whenever it belongs to the chosen function space. Ahead of the front the limiting coefficient is $k$, so in unweighted L2 space the limiting dispersion relation is $\lambda =-{\xi }^2+ic\xi +k$: its rightmost real part is positive. An exponential weight penalizes precisely these disturbances ahead of the front.

There is a small functional-analytic correction to the printed definition. Use the complete exponentially weighted L2 spaceThe additional requirement of unweighted L2 space membership in the printed set makes that set incomplete under its stated weighted norm. Indeed $v_N=1_{[-N,0]}$ belongs to the intersection and is a weighted Cauchy sequence, but its weighted limit $1_{(-\infty ,0]}$ does not belong to unweighted L2 space. The usual Hilbert space for this spectral argument is the completion $X_{\gamma }$. We use that completion explicitly, rather than treating the intersection as a complete space.

Multiplication $T_{\gamma }v=e^{\gamma z}v$ is a unitary map from $X_{\gamma }$ onto L2 space. Take the domain $T_{\gamma }^{-1}{H}^2(\mathbb{R})$ for $\mathcal{L}$. The transformed closed operator isThis follows from $T_{\gamma }{\partial }_z{T}_{\gamma }^{-1}={\partial }_z-\gamma$. The two limiting constant-coefficient dispersion relations, obtained by the Fourier transform, areIn particular the necessary strict leading-edge condition is ${\gamma }^2-c\gamma +k<0$. For $c>2\sqrt{k}$ this meansA particularly convenient choice isWith this choice there is no first derivative, and ${\mathcal{L}}_{\gamma }={\partial }_z^2+k-c^2/4-2w_c(z)$ is a self-adjoint differential operator on $H^2(\mathbb{R})$. This also avoids possible ambiguities between definitions of essential spectrum for nonnormal operators.

The relevant perturbation theorem is essential-spectrum invariance under relatively compact perturbations: if $A_0$ is a closed operator, $V:D(A_0)\to H$ is compact with respect to the graph norm, and $A_0+V$ is closed on the same domain, their Fredholm essential spectrum is the same. For self-adjoint operators this is the usual essential spectrum after removing isolated finite-multiplicity eigenvalues. To see why the theorem applies, regard $A_0-\lambda$ and $A_0+V-\lambda$ as bounded maps from the graph norm domain to $H$, and use compact perturbation invariance of Fredholm operators for each $\lambda$.

Define a step-potential comparison operatorBoth $A_0$ and ${\mathcal{L}}_{\gamma }$ are self-adjoint on $H^2(\mathbb{R})$. Their difference is multiplication byIt is relatively compact, not necessarily compact as a multiplication map on all of L2 space. To prove the needed compactness, take a sequence bounded in $H^2(\mathbb{R})$. On each bounded interval the Rellich-Kondrachov compactness theorem gives a subsequence converging in L2 space. A diagonal subsequence makes the products by $V$ converge locally. Outside a sufficiently large interval, $\Vert V{\Vert }_{\infty }$ is arbitrarily small and the sequence has a uniform L2 norm bound, so those tails are uniformly small. Thus the products converge globally in L2 space, proving graph-domain compactness.

Finally,Here is a direct verification. Its quadratic form is at most $(k-c^2/4)\Vert h{\Vert }_2^2$, so no spectrum lies above this endpoint. For every $\lambda \le k-c^2/4$, choose $\xi$ with $\lambda =k-c^2/4-{\xi }^2$. Normalized, increasingly long cutoffs of $e^{i\xi z}$, translated far to the right, have disjoint supports and satisfy $\Vert (A_0-\lambda )h_N{\Vert }_2\to 0$: differentiating the cutoff gives errors of orders $N^{-1}$ and $N^{-2}$. They converge weakly to zero. This is a singular sequence in the Weyl criterion, proving that every such $\lambda$ belongs to the essential spectrum. The relatively compact perturbation theorem therefore givesThis proves the weighted essential spectrum of a Fisher travelling front is strictly stable when $c>2\sqrt{k}$. In fact here the quadratic form gives more:Since $w_c\ge 0$, the full transformed spectrum lies below $-(c^2/4-k)$, and the evolution given by its linearization decays at least as $e^{-(c^2/4-k)t}$ in this weighted norm. This does not contradict translation invariance: the nonzero translation mode cannot lie in this weighted operator domain, since otherwise it would be a zero eigenfunction, contradicting the strictly negative quadratic-form bound. The weight restricts the permitted perturbations and excludes these neutral translations. Nor does a weighted L2 norm alone control the nonlinear remainder at $z\to -\infty$; a nonlinear theorem needs further boundedness or regularity assumptions.

At the minimal speed $c_{\ast }=2\sqrt{k}$, the best choice is $\gamma =\sqrt{k}$. Now ${\gamma }^2-c\gamma +k=(\gamma -\sqrt{k}{)}^2$, andThe essential spectrum has no positive part but touches zero; no positive exponential weight makes it strictly negative. Thus the critical front has no spectral gap and this argument gives no exponential decay rate there. Strict weighted spectral stability holds above the minimal speed; at the minimal speed it is marginal.

Take the C-star algebra $M_2(\mathbb{C})$ and chooseThen $a-1=\mathrm{diag}(0,1)$ is positive, whileis positive, since its quadratic form at $(x,y)$ is $|x+y{|}^2$. Thus $1\le a\le b$, as required. However,has determinant $-1$ and eigenvalues $5\pm \sqrt{26}$, one of which is negative. Equivalently, its quadratic form at the real vector $(5,-4)$ is $-4$. It is therefore not positive. This proves that squaring is not order preserving in a C-star algebra, even for invertible elements bounded below by the identity.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

First find the positive spatially homogeneous equilibrium. Setting the reactions to zero gives $v=u^2$ and $u/v=b$, and hence the unique positive equilibrium isThe zero pair is not an admissible positive state and is singular in the $u^2/v$ term. For homogeneous deviations, the reaction Jacobian matrix at this equilibrium isThus $\mathrm{tr}J=b-1$, $\det J=b$, and its eigenvalues areFor $0<b<1$, both real parts are negative: a real pair has positive product and negative sum, while a complex pair has real part $(b-1)/2$. The reaction equilibrium is therefore asymptotically stable, with no growing oscillatory mode. This does not exclude damped oscillations. Indeed, for $3-2\sqrt{2}<b<1$ the eigenvalues are complex; for example, $b=1/2$ gives $-1/4\pm i\sqrt{7}/4$, a stable focus with oscillatory transients. Thus a literal exclusion of every oscillatory solution is false; the intended exclusion of sustained or growing oscillations is the meaningful statement.

One can strengthen this to exclude nonlinear periodic oscillations in the positive quadrant without relying only on the local linear stability analysis. For the reactions $f=u^2/v-bu$, $g=u^2-v$, use the Dulac function $H=u^{-2}$. ThenWere there a periodic orbit wholly in the positive quadrant, the flux of $H(f,g)$ through it would vanish, since this vector is tangent to the orbit; Green's theorem would equate that flux to a strictly negative interior integral. This contradiction proves there is no sustained positive periodic oscillation when $0<b<1$, the positive-quadrant Dulac multiplier for quadratic activation argument. It leaves the damped transients entirely consistent with the conclusion.

Now introduce a perturbation $(\delta u,\delta v)=\mathbf{a}{e}^{\sigma t+ikx}$ in the quadratic activator-inhibitor model. The Fourier mode of squared wavenumber $q=k^2$ evolves underIts matrix trace and determinant areA Turing instability must start with reaction stability, requiring $0<b<1$. Then $\tau (q)<0$ for every $q\ge 0$, so a mode is unstable exactly when $D(q)<0$: its two real eigenvalues have opposite signs. This also proves the diffusion-driven onset is stationary rather than oscillatory.

The upward-opening quadratic $D(q)$ has its minimum at $q_m=(db-1)/(2d)$. To have a negative value at a physically allowed positive squared wavenumber, one needsWriting $r=db$, the second inequality is $r^2-6r+1>0$, with roots $3\pm 2\sqrt{2}$. The first inequality discards the lower branch, givingThese are the Turing threshold of the quadratic activator-inhibitor model conditions. Equivalently, in the $(b,d)$ plane the instability region lies strictly above $d=(3+2\sqrt{2})/b$ and strictly to the left of $b=1$, with $b>0$. The boundary diverges as $b\to 0^{+}$ and approaches $3+2\sqrt{2}$ as $b\to 1^{-}$.

At equality, the determinant has a double root and one growth rate is zero, while the other remains negative. Substituting $db=3+2\sqrt{2}$ into its minimum location givesBeyond onset, the unstable band isThe stated parameter region assumes continuously available wavenumbers, as on an infinite domain. With specified finite-domain boundary conditions, an allowed nonzero spatial eigenmode must additionally lie in this band; homogeneous stability alone does not ensure that a finite box admits the critical wavelength.