Linear stability and the phase plane

Theorem

Linear stability and the phase plane

T-067Home MU-205Threads change

Statement

Let \( f:U\to\mathbb{R}^2 \) be a \( C^1 \) vector field on an open set \( U\subseteq\mathbb{R}^2 \), let \( x_0\in U \) satisfy \( f(x_0)=0 \) (an equilibrium of \( \dot x=f(x) \)), and let \( A=Df(x_0) \) be the Jacobian at \( x_0 \), with eigenvalues \( \lambda_1,\lambda_2 \). Suppose \( x_0 \) is hyperbolic: \( \operatorname{Re}(\lambda_i)\neq 0 \) for \( i=1,2 \). Then (a) [Lyapunov linearised stability] if \( \operatorname{Re}(\lambda_1),\operatorname{Re}(\lambda_2) \lt 0 \) then \( x_0 \) is asymptotically stable for the nonlinear flow, while if \( \operatorname{Re}(\lambda_i) \gt 0 \) for some \( i \) then \( x_0 \) is unstable; and (b) [Hartman–Grobman] there is a neighbourhood of \( x_0 \) on which the nonlinear flow of \( \dot x=f(x) \) is topologically conjugate to the linear flow \( \dot u=Au \), so the qualitative phase portrait near \( x_0 \) (node, saddle, or spiral) is exactly that of the linearisation, determined by \( \tau=\operatorname{tr}A \), \( \Delta=\det A \) via the sign pattern of \( \Delta \) and of the discriminant \( \tau^2-4\Delta \).

Why it matters

Nonlinear systems almost never admit closed-form solutions, yet the local behaviour near an equilibrium is exactly the piece of the phase portrait one can compute by hand: differentiate once, find the eigenvalues of a \( 2\times 2 \) matrix, and read off stable node, saddle, spiral, and so on. This theorem is the rigorous justification for that shortcut – it says the sketch you draw from the linearisation is not merely suggestive but, away from the borderline (non-hyperbolic) cases, an accurate local picture of the true nonlinear flow.

It is also the hinge on which the entire phase-plane methodology of MU-205 turns: nullclines and direction fields tell you the global shape, but it is linear stability analysis that tells you what happens at the fixed points those curves meet, converting a qualitative sketch into a classified one.

Hypotheses

\( f \) is \( C^1 \) near \( x_0 \). Without differentiability there is no Jacobian to linearise with. Take \( f(x,y)=(\sqrt{|y|}\,\operatorname{sgn}y,\,-x) \) near the origin: \( f \) is continuous but not differentiable at points with \( y=0 \), so \( A=Df(0,0) \) simply does not exist and the theorem has no linear object to invoke.

\( x_0 \) is an equilibrium: \( f(x_0)=0 \). The Taylor expansion \( f(x_0+u)=Au+R(u) \) used throughout the proof relies on the constant term vanishing; if \( f(x_0)\neq 0 \) the point is not even a fixed point of the flow, so "stability of \( x_0 \)" is not a meaningful question.

Hyperbolicity: no eigenvalue of \( A \) has zero real part. Drop this and the linearisation is silent. \( \dot x=y,\ \dot y=-x-\varepsilon x^2y \) has \( A=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \) with eigenvalues \( \pm i \) (a linear centre) at every \( \varepsilon \); yet for \( \varepsilon \gt 0 \) the true equilibrium is an asymptotically stable spiral and for \( \varepsilon \lt 0 \) it is unstable – the sign of the cubic term, invisible to \( A \), decides the outcome.

Proof

We prove part (a) in full (this is Lyapunov's linearised-stability theorem, valid in \( \mathbb{R}^n \)); part (b) is the Hartman–Grobman theorem, which we state precisely and cite, since a full proof (construction of the conjugacy via a contraction-mapping argument on the space of bounded homeomorphisms) is a substantial independent result beyond a single-theorem treatment. We prove the stable case of (a) in detail and indicate the unstable case, which is structurally dual.

f(x_0+u) = Au + R(u), \qquad \frac{R(u)}{|u|}\to 0 \text{ as } u\to 0

Taylor's theorem with the Lagrange/Peano remainder for a \( C^1 \) function, using \( f(x_0)=0 \) and \( Df(x_0)=A \); this is the definition of differentiability of \( f \) at \( x_0 \), not an approximation we merely hope holds. A

u=x-x_0 \implies \dot u = Au+R(u), \qquad R\in C^0,\ R(0)=0,\ R(u)=o(|u|)

Change of variables translating \( x_0 \) to the origin; stability of \( x_0 \) for \( \dot x=f(x) \) is by definition equivalent to stability of \( u=0 \) for this shifted system. A

\operatorname{Re}(\lambda_i)\le -2\alpha \lt 0 \ (i=1,2) \implies \exists\, M\ge 1 : \ \|e^{At}\|\le Me^{-\alpha t}, \quad t\ge 0

Standard spectral estimate for matrix exponentials: put \( A \) in (real) Jordan form \( A=PJP^{-1} \); each block of \( e^{Jt} \) is a polynomial in \( t \) times \( e^{\lambda_i t} \), and polynomial growth is dominated by any strictly smaller exponential rate, so \( \|e^{At}\|\le M e^{(\max_i\operatorname{Re}\lambda_i+\alpha)t}\le Me^{-\alpha t} \) for a suitable constant \( M \) depending on \( P \). B

A^{\mathsf T}P+PA=-I \ \text{has a unique solution } P=P^{\mathsf T}\succ 0

Lyapunov equation lemma: because every eigenvalue of \( A \) has negative real part, \( P=\int_0^\infty e^{A^{\mathsf T}t}e^{At}\,dt \) converges (by Step 3, the integrand is bounded by \( M^2e^{-2\alpha t} \)), is symmetric, and is positive definite since \( u^{\mathsf T}Pu=\int_0^\infty |e^{At}u|^2dt\ge 0 \) with equality only at \( u=0 \) (as \( e^{At} \) is invertible); differentiating under the integral sign gives \( A^{\mathsf T}P+PA=\int_0^\infty \frac{d}{dt}\big(e^{A^{\mathsf T}t}e^{At}\big)dt=\big[e^{A^{\mathsf T}t}e^{At}\big]_0^\infty=-I \). C

V(u)=u^{\mathsf T}Pu \implies c_1|u|^2\le V(u)\le c_2|u|^2 \ \text{for some } 0\lt c_1\le c_2

\( P \) is symmetric positive definite (Step 4), so its eigenvalues \( \mu_{\min},\mu_{\max}\gt 0 \) give \( \mu_{\min}|u|^2\le u^{\mathsf T}Pu\le \mu_{\max}|u|^2 \) by the Rayleigh quotient characterisation of eigenvalues of a symmetric matrix. A

\dot V(u) = u^{\mathsf T}(A^{\mathsf T}P+PA)u + 2u^{\mathsf T}PR(u) = -|u|^2 + 2u^{\mathsf T}PR(u)

Chain rule applied to \( V(u(t)) \) along a solution of Step 2, then substitution of the Lyapunov equation from Step 4. B

\forall \varepsilon\gt 0\ \exists\,\delta\gt 0: |u|\lt\delta \implies |R(u)|\le \varepsilon|u| \implies \dot V(u)\le\big(-1+2\|P\|\varepsilon\big)|u|^2

Definition of \( R(u)=o(|u|) \) from Step 1 (choose \( \varepsilon \)), Cauchy–Schwarz on \( 2u^{\mathsf T}PR(u)\le 2\|P\|\,|u|\,|R(u)| \). Fixing \( \varepsilon=\frac{1}{4\|P\|} \) gives \( \delta \) such that \( \dot V(u)\le -\tfrac12|u|^2 \) on \( \{0\lt |u|\lt\delta\} \). B

u=0 \text{ is asymptotically stable for } \dot u=Au+R(u)

Lyapunov's direct (second) method: \( V \) is positive definite and \( \dot V \) is negative definite on a neighbourhood of \( 0 \) (Steps 5, 7), which is exactly the hypothesis of the Lyapunov asymptotic stability theorem, cited here rather than reproved. Combined with Step 2 this yields asymptotic stability of \( x_0 \) for the original system, proving the stable half of (a). B

\operatorname{Re}(\lambda_i)\gt 0 \text{ for some } i \implies x_0 \text{ unstable}

Dual construction: split \( \mathbb{R}^2=E^s\oplus E^u \) into the (possibly trivial) stable and unstable generalised eigenspaces of \( A \); on \( E^u \) an analogous Lyapunov equation \( A^{\mathsf T}Q+QA=I \) (solved on the unstable subspace, where \( \|e^{-At}\|\) decays) produces a function that is positive and strictly increasing along trajectories starting arbitrarily close to \( 0 \) with nonzero \( E^u \)-component, forcing them to leave any fixed neighbourhood; this is Chetaev's instability theorem applied to the linearisation, cited rather than reproved in full. C

\text{[Hartman–Grobman]} \quad \exists \text{ homeomorphism } h \text{ on a neighbourhood of } x_0 \text{ with } h\circ\phi_t = \psi_t\circ h

Cited theorem (Hartman 1960, Grobman 1959): here \( \phi_t \) is the local nonlinear flow of \( \dot x=f(x) \) and \( \psi_t(u)=e^{At}u \) the linear flow, and hyperbolicity of \( A \) is exactly the hypothesis under which the conjugacy \( h \) exists; the construction solves a fixed-point equation for \( h-\mathrm{id} \) in the Banach space of bounded continuous maps, using the exponential dichotomy of Step 3 applied on both \( E^s \) and \( E^u \). Because \( h \) is a homeomorphism (not merely continuous), it carries closed orbits, spiralling, and separatrices of the linear picture to topologically identical structures in the nonlinear phase portrait, which gives part (b): the local phase portrait type is read off from \( \tau=\operatorname{tr}A,\ \Delta=\det A \) exactly as for the linear system \( \dot u=Au \). C

Result

\operatorname{Re}(\lambda_i)\neq 0 \ \forall i \implies \Big[\max_i\operatorname{Re}\lambda_i\lt 0 \Rightarrow x_0 \text{ asympt. stable}\Big],\ \Big[\exists i:\operatorname{Re}\lambda_i\gt 0\Rightarrow x_0 \text{ unstable}\Big],\ \text{phase portrait} \cong \dot u = Au

Reading. At a hyperbolic equilibrium, throw away every term of the vector field past first order, keep only the Jacobian \( A \); the eigenvalues of \( A \) alone tell you whether trajectories converge to or flee the equilibrium, and (up to a continuous, invertible warping of the picture) what shape they trace while doing so – node, saddle, or spiral, exactly as for the linear system.

Scope. Local statement only: it describes behaviour in some neighbourhood of \( x_0 \), whose size is not given by the theorem. It says nothing about non-hyperbolic equilibria (purely imaginary or zero eigenvalues), and Hartman–Grobman's conjugacy is topological, not differentiable – it preserves the qualitative type of orbit but not, in general, quantities like curvature or the precise rate of approach along a spiral.

Corollaries & converses

If \( \tau=\operatorname{tr}A\lt 0 \) and \( \Delta=\det A\gt 0 \) (equivalently both eigenvalues have negative real part, by Vieta's formulas \( \lambda_1+\lambda_2=\tau,\ \lambda_1\lambda_2=\Delta \)), then \( x_0 \) is asymptotically stable, and it is a node if \( \tau^2-4\Delta\ge 0 \), a spiral if \( \tau^2-4\Delta\lt 0 \).
If \( \Delta\lt 0 \) then the eigenvalues are real with opposite sign, so \( x_0 \) is a saddle and always unstable, regardless of \( \tau \).
A hyperbolic equilibrium of a planar system is isolated: since \( \det A\neq 0 \) (as \( 0 \) is not an eigenvalue), the implicit function theorem shows no other zero of \( f \) accumulates at \( x_0 \).
Converse fails. Asymptotic stability of the nonlinear equilibrium does not force hyperbolicity or negative-real-part eigenvalues: \( \dot x=-x^3 \) has \( x=0 \) asymptotically stable (via \( V=x^2 \)) yet \( A=Df(0)=0 \), a non-hyperbolic (indeed zero) linearisation. So the theorem gives a sufficient, not necessary, criterion.
The instability half's converse also fails in the same way: \( \dot x = x^3 \) is unstable at \( 0 \) with the same zero linearisation, so the linearisation alone cannot distinguish this from the stable example above – higher-order (Lyapunov function / centre-manifold) analysis is required whenever \( A \) is not hyperbolic.

Fails without

Drop hyperbolicity. \( \dot r=0,\ \dot\theta=1 \) in polar form around the origin (a pure rotation, \( A=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \), eigenvalues \( \pm i \)) has every orbit a closed circle: neither stable nor unstable in the strict Lyapunov sense used here, and no conjugacy to a linear saddle/node/spiral exists because none of those linear types is orbit-equivalent to a continuum of closed curves through every neighbourhood of the origin.
Drop \( C^1 \) smoothness. \( \dot x = -x + 2|x| \) on \( \mathbb{R} \) (one-dimensional analogue for clarity) is continuous but only piecewise linear, non-differentiable at \( 0 \); the two one-sided slopes are \( -1 \) and \( 1 \), so no single Jacobian value exists and the classification via a single eigenvalue is simply undefined, even though the qualitative behaviour (attracting from the left, repelling to the right) is perfectly well-posed and computable by direct case analysis.
Confuse "no positive-real-part eigenvalue" with stability. If one eigenvalue is exactly zero and the other negative (e.g. \( A=\operatorname{diag}(0,-1) \)), the theorem's hypotheses are not met (not hyperbolic) and stability genuinely depends on the nonlinear terms along the zero-eigenvalue (centre) direction, via centre-manifold reduction, which lies outside this theorem entirely.

Common errors

Classifying an equilibrium from \( \tau,\Delta \) when an eigenvalue has zero real part (e.g. \( \Delta=0 \), or \( \tau=0,\Delta\gt 0 \) giving a linear centre): the theorem simply does not apply there, yet students routinely label such points "centre" or "stable" for the nonlinear system without further justification.
Forgetting to translate coordinates: linearising \( f \) itself instead of \( f(x_0+u) \), i.e. evaluating the Jacobian at the origin instead of at the equilibrium \( x_0\neq 0 \) under study.
Treating the Hartman–Grobman conjugacy as differentiable and quoting linearised eigenvalues as the actual (nonlinear) rate of decay; the conjugacy is only a homeomorphism, so exponents can be quantitatively wrong even where the qualitative type (e.g. "spiral") is correctly identified.
Sign errors in the discriminant condition: using \( \tau^2-4\Delta \) to decide node vs. spiral while forgetting that \( \Delta\gt 0 \) is also required for the equilibrium to be a stable/unstable node or spiral rather than a saddle.
Asserting the converse: inferring "nonlinear system is asymptotically stable, therefore the Jacobian has negative-real-part eigenvalues" – false, as the corollaries section shows with \( \dot x=-x^3 \).

Discussion

The theorem packages two logically separate results that are almost always used together. Lyapunov's 1892 linearised-stability theorem is a statement purely about stability (does \( |u(t)|\to 0 \)?) and needs no planarity; it holds verbatim in \( \mathbb{R}^n \). Hartman–Grobman, proved independently by Philip Hartman and David Grobman around 1959–60, is a statement about the shape of the orbit structure, and is what licenses the pictorial vocabulary – "node," "saddle," "spiral" – carrying over unchanged from the linear phase portrait to the nonlinear one. It is only in the plane that this classification collapses to the simple \( (\tau,\Delta) \) diagram taught in MU-205; in \( \mathbb{R}^n \) hyperbolic equilibria are classified by the dimensions of the stable and unstable manifolds, a strictly richer invariant.

The Lyapunov-equation proof given above is worth dwelling on because it is a template used throughout control theory and nonlinear dynamics: whenever the linear part of a system is "nice" (here, Hurwitz), one manufactures a quadratic Lyapunov function from that linear part alone, then treats the nonlinear remainder as a perturbation too small to disturb the sign of \( \dot V \) near the equilibrium. The same \( P \) solving \( A^{\mathsf T}P+PA=-I \) reappears verbatim in linear-quadratic control and in numerical stiffness analysis.

A genuinely subtle point is what hyperbolicity buys beyond stability: it buys structural stability of the local picture under \( C^1 \)-small perturbations of \( f \) itself, not just under the specific nonlinear remainder \( R \) present in a given problem. This is why hyperbolic equilibria are the generic, "robust" case in applications, while non-hyperbolic ones (centres, saddle-nodes) are exactly the loci where small changes in a model's parameters can qualitatively change the dynamics – the subject matter of bifurcation theory, which begins precisely where this theorem's hypotheses fail.

Common misconception. Many students treat "linearisation" as an approximation whose accuracy simply needs to be checked numerically for a given problem; the theorem instead gives an exact qualitative guarantee (same topological type, definite stability verdict) valid in some neighbourhood, with no approximation error in the classification itself once hyperbolicity holds – only the size of that neighbourhood, and any quantitative rates, are left unspecified.

Worked examples

\ddot\theta + b\dot\theta + \sin\theta = 0,\quad 0\lt b\lt 2 \ \implies\ \dot x=y,\ \dot y=-\sin x - by

Standard reduction of the damped pendulum to a first-order planar system with \( x=\theta \) (angle), \( y=\dot\theta \) (angular velocity); equilibria are \( (n\pi,0),\ n\in\mathbb{Z} \), and by \( 2\pi \)-periodicity it suffices to examine \( (0,0) \) and \( (\pi,0) \). A

Df(x,y)=\begin{pmatrix}0 & 1\\ -\cos x & -b\end{pmatrix} \implies A_{(0,0)}=\begin{pmatrix}0&1\\-1&-b\end{pmatrix},\ \ A_{(\pi,0)}=\begin{pmatrix}0&1\\1&-b\end{pmatrix}

Direct differentiation of \( f(x,y)=(y,-\sin x-by) \), evaluated at each equilibrium; \( f\in C^\infty \) so the theorem's smoothness hypothesis holds everywhere. A

\text{At } (0,0):\ \tau=-b,\ \Delta=1 \implies \tau^2-4\Delta = b^2-4\lt 0 \text{ (since } 0\lt b\lt 2)

\( \tau=\operatorname{tr}A,\ \Delta=\det A \) computed from the matrix in Step 2; hyperbolic since \( \Delta=1\neq 0 \) rules out a zero eigenvalue and \( \tau=-b\neq 0 \) rules out purely imaginary ones for \( b\gt 0 \). B

\Delta\gt 0,\ \tau\lt 0,\ \tau^2\lt 4\Delta \implies (0,0) \text{ is a stable spiral (asymptotically stable focus)}

By the Corollaries: \( \Delta\gt 0,\tau\lt 0 \) gives asymptotic stability (part (a) of the theorem), and negative discriminant \( \tau^2-4\Delta\lt 0 \) gives complex conjugate eigenvalues, hence a spiral rather than a node, and by part (b) the true nonlinear phase portrait near the hanging-down equilibrium is a damped oscillation into rest, matching physical intuition. B

\text{At } (\pi,0):\ \tau=-b,\ \Delta=\det\begin{pmatrix}0&1\\1&-b\end{pmatrix}=-1\lt 0

Same computation at the inverted-pendulum equilibrium; \( \Delta\lt 0 \) means the eigenvalues are real with opposite signs (product \( \lambda_1\lambda_2=\Delta\lt 0 \)), regardless of the value of \( \tau \). A

(0,0): \text{stable spiral (all } 0\lt b\lt 2\text{)} \qquad (\pi,0): \text{saddle, always unstable}

Reading. The hanging-down equilibrium of the damped pendulum is a genuine local attractor for every positive damping \( b\lt 2 \) (underdamped regime), while the inverted equilibrium is a saddle and hence unstable for every value of \( b \), including \( b=0 \) – no amount of linear damping stabilises an inverted pendulum, matching the well-known fact that active control is needed there.

\dot x = -x+y+x^3, \qquad \dot y = -2y - x^2y

A nonlinear planar system with equilibrium at the origin ( \( f(0,0)=(0,0) \) ); \( f\in C^\infty \), so hypotheses of smoothness and \( f(x_0)=0 \) hold. A

A = Df(0,0) = \begin{pmatrix}-1 & 1\\ 0 & -2\end{pmatrix}, \qquad \lambda_1=-1,\ \lambda_2=-2

\( A \) is upper triangular so its eigenvalues are its diagonal entries; both real and negative, so the origin is hyperbolic and, by the theorem, an asymptotically stable node. A

A^{\mathsf T}P+PA=-I \implies P=\begin{pmatrix}\tfrac12 & \tfrac16\\ \tfrac16 & \tfrac13\end{pmatrix}

Solving the \( 2\times2 \) linear system for the entries of \( P=P^{\mathsf T} \) from Step 4 of the proof: writing \( P=\begin{pmatrix}p_{11}&p_{12}\\p_{12}&p_{22}\end{pmatrix} \) and expanding \( A^{\mathsf T}P+PA \) gives \( -2p_{11}=-1 \), \( p_{11}-3p_{12}=0 \), \( 2p_{12}-4p_{22}=-1 \), solved in turn to give \( p_{11}=\tfrac12,\ p_{12}=\tfrac16,\ p_{22}=\tfrac13 \); the leading minors \( \tfrac12\gt 0 \) and \( \det P=\tfrac16-\tfrac1{36}=\tfrac{5}{36}\gt 0 \) confirm \( P\succ 0 \). C

V(x,y)=\tfrac12x^2+\tfrac13xy+\tfrac13y^2, \qquad \dot V = -x^2-y^2 + 2u^{\mathsf T}PR(u),\ \ R(u)=(x^3,-x^2y)

\( V(u)=u^{\mathsf T}Pu \) with \( P \) from Step 3; the linear part of \( \dot V \) is \( -|u|^2 \) exactly as in Step 6 of the general proof, since \( P \) solves the Lyapunov equation for this specific \( A \). B

2u^{\mathsf T}PR(u) = x^4+\tfrac13x^3y-\tfrac13x^3y-\tfrac23x^2y^2 = x^4-\tfrac23x^2y^2

Direct expansion of \( 2\big[(p_{11}x+p_{12}y)x^3+(p_{12}x+p_{22}y)(-x^2y)\big] \) using the entries of \( P \) found in Step 3. A

|x^4-\tfrac23x^2y^2|\le \tfrac53|u|^4 \implies \dot V \le -|u|^2+\tfrac53|u|^4 \lt 0 \ \text{ for } 0\lt|u|\lt\sqrt{3/5}

Bound each quartic monomial by \( |u|^4=(x^2+y^2)^2 \) (since \( |x|,|y|\le|u| \)), then the linear term \( -|u|^2 \) strictly dominates the quartic remainder on a small enough punctured ball, giving \( \dot V\lt 0 \) there – the explicit instance of the general \( \varepsilon \)-\( \delta \) domination argument of Step 7. B

V \succ 0,\ \dot V \prec 0 \text{ near } 0 \implies (0,0) \text{ is asymptotically stable}

Reading. Beyond the qualitative conclusion "stable node" read off from the eigenvalues alone, this example exhibits the actual certificate of stability: an explicit Lyapunov function, built purely from the linear part, whose decrease survives the addition of the cubic nonlinear terms on a genuine (if unspecified-radius) neighbourhood of the origin.

Problems

Classify the equilibrium at the origin of \( \dot x = x - 2y,\ \dot y = 3x-4y \) using \( \tau \) and \( \Delta \).
Solution
\( A=\begin{pmatrix}1&-2\\3&-4\end{pmatrix} \), so \( \tau=1-4=-3 \), \( \Delta=(1)(-4)-(-2)(3)=-4+6=2 \). Since \( \Delta=2\gt0 \) and \( \tau=-3\lt0 \), both eigenvalues have negative real part: asymptotically stable. Discriminant \( \tau^2-4\Delta=9-8=1\gt0 \), so eigenvalues are real and distinct: a stable node. (Explicitly, \( \lambda^2+3\lambda+2=0\Rightarrow\lambda=-1,-2 \).) The origin is a hyperbolic, stable node.
Find and classify all equilibria of \( \dot x=x(1-x-y),\ \dot y=y(0.75-y-0.5x) \) (competing-species model).
Solution
Equilibria: \( (0,0) \), \( (1,0) \), \( (0,0.75) \), and the interior point solving \( x+y=1,\ 0.5x+y=0.75 \Rightarrow x=0.5,\ y=0.5 \). Jacobian: \( Df=\begin{pmatrix}1-2x-y & -x\\ -0.5y & 0.75-2y-0.5x\end{pmatrix} \). At \( (0,0) \): \( A=\begin{pmatrix}1&0\\0&0.75\end{pmatrix} \), eigenvalues \( 1,0.75\gt0 \): unstable node. At \( (1,0) \): \( A=\begin{pmatrix}-1&-1\\0&0.25\end{pmatrix} \), eigenvalues \( -1,0.25 \): opposite signs, saddle (unstable). At \( (0,0.75) \): \( A=\begin{pmatrix}0.25&0\\-0.375&-0.75\end{pmatrix} \), eigenvalues \( 0.25,-0.75 \): saddle (unstable). At \( (0.5,0.5) \): \( A=\begin{pmatrix}-0.5&-0.5\\-0.25&-0.5\end{pmatrix} \), \( \tau=-1,\ \Delta=0.25-0.125=0.125\gt0 \), so stable; discriminant \( 1-0.5=0.5\gt0 \): stable node. All four equilibria are hyperbolic, so linearised classification applies throughout; the interior coexistence state is the attractor.
For which values of the parameter \( k \) is the origin a hyperbolic equilibrium of \( \dot x=y,\ \dot y=-x-ky \), and classify the phase portrait as \( k \) varies over \( \mathbb{R} \).
Solution
\( A=\begin{pmatrix}0&1\\-1&-k\end{pmatrix} \), \( \tau=-k \), \( \Delta=1 \) for all \( k \). Since \( \Delta=1\neq0 \) always, the only way to lose hyperbolicity is \( \tau=0 \), i.e. \( k=0 \) (purely imaginary eigenvalues \( \pm i \), a linear centre, non-hyperbolic). For \( k\neq0 \) the origin is hyperbolic: discriminant \( \tau^2-4\Delta=k^2-4 \). If \( |k|\gt2 \): real eigenvalues, node (stable for \( k\gt2 \), unstable for \( k\lt-2 \)). If \( 0\lt|k|\lt2 \): complex eigenvalues, spiral (stable for \( 0\lt k\lt2 \), unstable for \( -2\lt k\lt0 \)). At \( k=\pm2 \): repeated real eigenvalue \( \mp1 \), a degenerate/improper node (still hyperbolic since \( \tau\neq0 \)). At \( k=0 \) the theorem does not apply; the origin is genuinely a centre for this exactly linear system, but nonlinear perturbations of it are not classified by \( A \) alone.
Show that the origin is asymptotically stable for \( \dot x=-x+xy,\ \dot y=-y-x^2 \) by (a) linearisation and (b) an explicit Lyapunov function, and compare what each method establishes.
Solution
(a) \( A=Df(0,0)=\begin{pmatrix}-1&0\\0&-1\end{pmatrix} \), eigenvalues \( -1,-1\lt0 \): hyperbolic, and by the theorem the origin is asymptotically stable, a (possibly degenerate) stable node. This only guarantees stability in some neighbourhood, of unspecified size. (b) Take \( V=\tfrac12(x^2+y^2) \). Then \( \dot V=x(-x+xy)+y(-y-x^2)=-x^2+x^2y-y^2-x^2y=-x^2-y^2=-2V \) for every \( (x,y) \), not just locally: the cubic cross-terms cancel exactly. Hence \( V(t)=V(0)e^{-2t}\to0 \) for every initial condition, so the origin is in fact globally asymptotically stable. The linearisation theorem alone could not have detected this global fact, since it is inherently local.
Let \( \dot x=-y-x^3,\ \dot y=x-y^3 \). Determine the type of the equilibrium at the origin and decide whether Theorem T-067 alone can settle stability; if not, supply an argument that does.
Solution
\( A=Df(0,0)=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \), eigenvalues \( \pm i \): purely imaginary, so the origin is not hyperbolic and Theorem T-067 does not apply – the linearisation alone cannot decide stability here. Use \( V=\tfrac12(x^2+y^2) \): \( \dot V=x(-y-x^3)+y(x-y^3)=-xy-x^4+xy-y^4=-(x^4+y^4)\le0 \), with equality only at the origin. So \( V \) is a strict Lyapunov function and, by the Lyapunov asymptotic stability theorem (the same cited result used in Step 8 of the proof, applied directly rather than via linearisation), the origin is asymptotically stable – indeed globally, by the same argument as the previous problem. This illustrates exactly the gap the theorem leaves open at non-hyperbolic equilibria, and how it is filled case-by-case with a bespoke Lyapunov function.