Classification of Planar Fixed Points

Technique: direct solution of the linear ODE via eigendecomposition (real case) and real canonical form (complex case), followed by qualitative analysis of the solution formulas.

Step 1 (relations between $T,D,\lambda_i$). Write $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, so $T=a+d$, $D=ad-bc$. The characteristic polynomial is

By Vieta's formulas for this monic quadratic, its roots $\lambda_1,\lambda_2$ satisfy $\lambda_1+\lambda_2=T$ and $\lambda_1\lambda_2=D$. This proves (1) and is the algebraic backbone of every subsequent case, since it converts statements about $(T,D)$ into statements about $\lambda_1,\lambda_2$ and vice versa.

Step 2 (Case $D<0$: saddle). Since $D=\lambda_1\lambda_2<0$, the two roots of the real quadratic $\lambda^2-T\lambda+D=0$ are real (its discriminant $\Delta=T^2-4D \ge -4D>0$) and have opposite signs (their product is negative). Say $\lambda_1<0<\lambda_2$. Because $D\ne0$, $A$ is invertible, so $0$ is not an eigenvalue and $A$ has two linearly independent real eigenvectors $\mathbf v_1,\mathbf v_2$ (distinct real eigenvalues of a real matrix always admit linearly independent eigenvectors, since eigenvectors for distinct eigenvalues are linearly independent by a standard linear-algebra argument: if $c_1\mathbf v_1+c_2\mathbf v_2=0$, applying $A$ and $\lambda_1$-multiples and subtracting forces $c_2=0$, then $c_1=0$). In the eigenbasis, the general solution is

If $c_2\ne0$, then since $\lambda_2>0$, $\|\mathbf x(t)\|\to\infty$ as $t\to\infty$; if $c_2=0,c_1\ne0$, then since $\lambda_1<0$, $\mathbf x(t)\to0$ as $t\to\infty$ but $\|\mathbf x(t)\|\to\infty$ as $t\to-\infty$. Thus the line $\operatorname{span}(\mathbf v_1)$ is invariant and attracting in forward time (the stable manifold), the line $\operatorname{span}(\mathbf v_2)$ is invariant and attracting in backward time (the unstable manifold), and every other solution approaches $\operatorname{span}(\mathbf v_2)$ asymptotically as $t\to+\infty$ and $\operatorname{span}(\mathbf v_1)$ as $t\to-\infty$. This is precisely the saddle pattern; the origin is unstable because the $c_2\ne0$ solutions leave every neighborhood.

Step 3 (Case $D>0,\Delta>0$: node). Here $\Delta=T^2-4D>0$ gives two distinct real roots $\lambda_1\ne\lambda_2$, and $D=\lambda_1\lambda_2>0$ forces them to have the same sign; that common sign is the sign of $T=\lambda_1+\lambda_2$ (a sum of two same-signed reals has that sign). As in Step 2, eigenvectors $\mathbf v_1,\mathbf v_2$ for the distinct eigenvalues are linearly independent, and

Suppose $T<0$, so $\lambda_1,\lambda_2<0$; say $\lambda_2<\lambda_1<0$ (order them so $|\lambda_1|<|\lambda_2|$). Then every term decays, so $\mathbf x(t)\to0$ as $t\to\infty$ for every initial condition: the origin is asymptotically stable, a stable node. Moreover for $c_1\ne0$,

and since $\lambda_2-\lambda_1<0$, the second term vanishes relative to the first as $t\to\infty$, so the direction of $\mathbf x(t)$ tends to $\operatorname{span}(\mathbf v_1)$: trajectories become tangent to the eigenline of the eigenvalue closer to zero (the "slow" eigenvalue) as $t\to\infty$. If $T>0$ both eigenvalues are positive; running $t\to-t$ reduces to the previous case, so all solutions blow up as $t\to+\infty$ and the origin is an unstable node, unstable by the same divergence argument as in Step 2 applied with both exponents positive.

Step 4 (Case $D>0,\Delta=0$: degenerate node). Here $\lambda_1=\lambda_2=\lambda=T/2$, and since $D=\lambda^2>0$ we have $\lambda\ne0$. Two subcases by the Jordan structure of $A$:

• If $A=\lambda I$, every nonzero vector is an eigenvector, $\mathbf x(t)=e^{\lambda t}\mathbf x(0)$, and orbits are rays from the origin (a star node).
• Otherwise $A$ is not diagonalizable and has a single Jordan block; choosing a generalized eigenvector $\mathbf v_2$ with $(A-\lambda I)\mathbf v_2 = \mathbf v_1\ne0$ and $(A-\lambda I)\mathbf v_1=0$, the general solution is

Since $e^{\lambda t}\to0$ dominates any polynomial factor $t$ as $t\to\infty$ when $\lambda<0$ (because $|t|e^{\lambda t}\to0$ by L'Hôpital or by comparing $e^{\lambda t}$ to $1/|t|$ for large $t$), we get $\mathbf x(t)\to0$ if $\lambda<0$ ($T<0$, stable) and blow-up if $\lambda>0$ ($T>0$, unstable). This proves (4).

Step 5 (Case $D>0,\Delta<0,T\ne0$: spiral). Now the quadratic has complex conjugate roots $\lambda_{1,2}=\alpha\pm i\beta$ with $\alpha=T/2\ne0$, $\beta=\sqrt{-\Delta}/2\ne0$ (real since $\Delta<0$). A real matrix with complex eigenvalue $\alpha+i\beta$ ($\beta\ne0$) has a complex eigenvector $\mathbf w=\mathbf p+i\mathbf q$ with $\mathbf p,\mathbf q\in\mathbb R^2$ linearly independent over $\mathbb R$ (if they were dependent, $\mathbf w$ would be a real scalar multiple of a real vector, forcing $\beta=0$, a contradiction). In the real basis $(\mathbf p,\mathbf q)$, $A$ takes the real canonical form

The system $\dot{\mathbf y} = \begin{pmatrix}\alpha&-\beta\\\beta&\alpha\end{pmatrix}\mathbf y$ has solution, in polar coordinates $r=\|\mathbf y\|$, $\theta=\arg(y_1+iy_2)$,

(obtained by writing $y_1+iy_2 = re^{i\theta}$ and differentiating: $\dot y_1+i\dot y_2 = (\alpha+i\beta)(y_1+iy_2)$ gives $\dot r + ir\dot\theta = (\alpha+i\beta)r$ after dividing by $e^{i\theta}$, and comparing real/imaginary parts). Hence $r(t) = r(0)e^{\alpha t}$ and $\theta(t)=\theta(0)+\beta t$: an exponential spiral, traversed with constant nonzero angular speed $\beta\ne0$ so that trajectories genuinely wind around the origin (not merely oscillate along a line). Since $P$ is a fixed linear (hence bounded, invertible) change of coordinates, $\|\mathbf x(t)\|\to0 \iff \|\mathbf y(t)\|\to 0 \iff \alpha<0$, and likewise for divergence. Thus if $T<0$ ($\alpha<0$) the origin is a stable spiral ($\mathbf x(t)\to0$ for every solution, with infinitely many rotations as $t\to\infty$), and if $T>0$ ($\alpha>0$) it is an unstable spiral (rotating outward, $\|\mathbf x(t)\|\to\infty$).

Step 6 (Case $D>0,T=0$: center). Now $\alpha=T/2=0$ and $\beta=\sqrt D\ne0$. By Step 5's computation with $\alpha=0$: $\dot r=0$, so $r(t)\equiv r(0)$ is constant, and $\theta(t)=\theta(0)+\beta t$ increases linearly. In the coordinates $\mathbf y$, every nonzero orbit is therefore a circle of radius $r(0)$ traversed at constant angular speed $\beta$, hence a closed periodic orbit with period $2\pi/|\beta|$. Applying the fixed invertible linear map $P$ transforms these circles into ellipses (linear images of circles), so every nonzero orbit of the original system $\dot{\mathbf x}=A\mathbf x$ is a closed ellipse traversed periodically. In particular no solution tends to $0$ or to $\infty$: the origin is stable (orbits starting near $0$ stay on a nearby closed ellipse for all time) but not asymptotically stable. This is the center classification.

Step 7 (hyperbolicity and structural stability, sketch/summary). By definition the origin is hyperbolic exactly when $\operatorname{Re}\lambda_1,\operatorname{Re}\lambda_2\ne0$; from Step 1's Vieta relations this holds precisely in the saddle case ($D<0$), the node cases ($D>0,T\ne0$, whether $\Delta\gtrless0$), and the spiral case ($D>0,\Delta<0,T\ne0$) — i.e., every case except the center ($T=0$) and, as a boundary degeneracy, any case with $D=0$ (not covered above, corresponding to a zero eigenvalue and a non-isolated or line of equilibria). A small perturbation of the entries of $A$ moves $(T,D)$ continuously; if $(T,D)$ lies strictly off the axis $T=0$ (for $D>0$) and strictly off $D=0$, a sufficiently small perturbation keeps it in the same open region of the $(T,D)$-plane (saddle region $D<0$; node regions; or spiral regions), preserving the qualitative type. If $T=0$ exactly (center) or $D=0$, an arbitrarily small perturbation can push $(T,D)$ off the axis into a spiral, node, or saddle region — so these non-hyperbolic cases are not structurally stable, establishing the final claim of the theorem. $\blacksquare$