A hyperbolic saddle fixed point of a map has one eigenvalue of modulus less than one and one greater than one. The corresponding stable and unstable directions persist locally as invariant sets or manifolds under the usual smoothness and invertibility assumptions. For a noninvertible map a zero stable fixed-point multiplier is possible, and the word saddle fixed point of a map describes forward contraction in one direction and expansion in the other; backward dynamics may not be unique.

A hyperbolic sink fixed point of a map has both Jacobian eigenvalues strictly inside the unit circle. Its iterates locally attract nearby initial states; complex fixed-point multipliers can make the approach spiral. Zero fixed-point multipliers are strongly contracting and can occur for noninvertible maps. A fixed-point multiplier of modulus one requires nonlinear or exact analysis.

At a differentiable fixed point, linear perturbations advance by the Jacobian matrix. A hyperbolic source fixed point of a map has both eigenvalues of modulus greater than one, so nearby perturbations expand under forward iteration of a map. Real signs can cause alternating directions and a complex pair can cause spiral expansion. Eigenvalues on the unit circle are excluded from this linear classification.

A Hopf bifurcation occurs when a simple pair of complex conjugate eigenvalues crosses the imaginary axis at nonzero frequency. A supercritical normal form of a dynamical system isWriting $z=r{e}^{i\theta }$ gives $\dot{r}=\mu r-r^3$ and $\dot{\theta }={\omega }_0$. A stable periodic orbit of radius $\sqrt{\mu }$ is born for $\mu >0$ as the equilibrium loses stability. In contrast, a subcritical Hopf has an unstable small periodic orbit on the side where the equilibrium is still stable.

The supercritical pitchfork normal form of a dynamical system isThe symmetric equilibrium $q=0$ is stable for $\mu <0$ and unstable for $\mu >0$. Two new stable equilibria $q=\pm \sqrt{\mu }$ emerge on the unstable side of the original branch. Reflection symmetry $q\mapsto -q$ preserves the pair. The new equilibria approach zero continuously as the parameter reaches the threshold.

A saddle-node bifurcation creates or destroys two equilibria when a real eigenvalue passes through zero with a nonzero quadratic normal-form term. With a parameter orientation chosen so the pair appears for $\mu >0$, useThere are no equilibria for $\mu <0$, a double nonhyperbolic equilibrium at zero, and equilibria $q=\pm \sqrt{\mu }$ for $\mu >0$. The positive branch is stable and the negative branch unstable in the active direction; additional hyperbolic directions can make them a node and a saddle equilibrium. “Supercritical” for a fold depends on this chosen parameter orientation; unlike the pitchfork and Hopf cases it is not an intrinsic distinction between the stability of two possible cubic branches.

Small $L$ makes every nonzero $k_n=n\pi /L$ large, placing them above the finite instability band, where diffusion dominates reaction. The homogeneous zero mode is already stable.

For $0<D<D_c$, the first possible band entry as $L$ increases is $n=1$ crossing $k_1=\sqrt{q_{+}}$. Hence the critical length isAt equality the mode is neutral; it is unstable just above this value, while $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus strictly positive growth has infimum $L_c$, with no attained smallest length if “unstable” means $\mathrm{Re}\lambda >0$. The conventional threshold answer is $L_c$. Larger modes cannot enter earlier because their first-entry lengths are $n{L}_c$.

If $D\ge 1/2$, both nonconstant coefficients of $\Delta (q)$ are nonnegative, so it is positive for $q\ge 0$. If $D_c<D<1/2$, its discriminant is negative, so it is again strictly positive. Therefore $\boxed{D>D_c⟹\text{no diffusion-driven instability}}$.

At $D=D_c$ the double root can only give a neutral zero eigenvalue if it coincides with an allowed spatial mode; it still gives no positive growth rate. This is the finite-wavenumber onset of a Turing instability, not an oscillatory crossing since the mode trace stays negative.

Put $q=k_n^2$. The mode matrix has trace $-1/2-(1+D)q<0$, so a growing eigenvalue occurs exactly when its determinant is negative:This upward quadratic has two positive roots only if $D<1/2$ and $D^2-5D+1/4>0$. Within $0<D<1$ this isFor such $D$, defineThe unstable modes satisfy $\boxed{\sqrt{q_{-}}<k_n<\sqrt{q_{+}}}$. These are strict inequalities: at either endpoint the leading eigenvalue is zero rather than positive.

The cosine modes satisfy both zero-flux endpoint conditions and ${\partial }_x^2\cos (k_{n}x)=-k_n^2\cos (k_{n}x)$. Linearizing the reaction terms gives $f_a\delta a+f_b\delta b$ and $g_a\delta a+g_b\delta b$. Cancelling the common mode factor yieldsThe derivatives are exactly the entries of the matrix computed in (ii). The zero mode reproduces the stable homogeneous perturbation; positive integers give the possible spatial instabilities.

The reaction Jacobian isIts trace is $-1/2$ and determinant is $1$, so both eigenvalues have negative real parts. Explicitly they are $(-1\pm i\sqrt{15})/4$. Therefore the homogeneous reaction equilibrium is a stable focus in the absence of diffusion.

At a homogeneous equilibrium, add the two reaction equations: $f+g=1-b=0$, so $b=1$. Then $f=0$ gives $a=3/4$. Thus $\boxed{(a_{\ast },b_{\ast })=(3/4,1)}$.