Linearization and Local Stability, differential equations lesson

Close to an equilibrium, a curved vector field looks almost like the field of a linear system, and the behavior of a linear system is read from its eigenvalues. Can the stability of a nonlinear equilibrium be read off from that linear approximation? This lesson builds the approximation from the Jacobian matrix, shows that it decides local stability whenever no eigenvalue has zero real part, and marks the two places where it says nothing, borderline eigenvalues and states far from the equilibrium.

Key terms

Jacobian matrix

Definition Let $t$ be the independent variable, with primes denoting derivatives with respect to $t$ , and consider an autonomous system $X' = F(X)$ with state $X = (x_1, \dots, x_n)^{T}$ and field $F = (F_1, \dots, F_n)^{T}$ continuously differentiable on an open set $D \subseteq \mathbb{R}^n$ . The Jacobian matrix of $F$ at a state $X$ is the $n \times n$ matrix of first partial derivatives

J(X) = \left[\frac{\partial F_i}{\partial x_j}(X)\right].

Its entry in row $i$ and column $j$ measures how the $i$ th rate changes when the $j$ th state component changes. For a planar system $x' = f(x, y)$ , $y' = g(x, y)$ it is

J = \begin{pmatrix} f_x & f_y \\ g_x & g_y \end{pmatrix}.

Evaluated at one particular state, it is a constant matrix.

In plain words Rows belong to the rates and columns to the state variables. Differentiate each rate with respect to each variable, then substitute the state of interest.

Example For the predator-prey model $x' = x(2 - y)$ , $y' = y(x - 1)$ of the previous lesson, the rates are $f = 2x - xy$ and $g = xy - y$ , so

J(x, y) = \begin{pmatrix} 2 - y & -x \\ y & x - 1 \end{pmatrix}.

At the origin this is $\operatorname{diag}(2, -1)$ , and at the coexistence state

J(1, 2) = \begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}.

Linearization

Definition Let $X_*$ be an equilibrium, so $F(X_*) = 0$ , and write a nearby state as $X = X_* + h$ with displacement $h$ . Because $F$ is differentiable at $X_*$ , with $J = J(X_*)$ ,

F(X_* + h) = Jh + R(h),

where the remainder satisfies $\lVert R(h) \rVert / \lVert h \rVert \to 0$ as $h \to 0$ , and $\lVert \cdot \rVert$ is the Euclidean length. Since $h' = X'$ , the displacement obeys $h' = Jh + R(h)$ exactly. Dropping the remainder gives the linearization of the system at $X_*$ ,

h' = J(X_*)\,h,

a linear system with constant coefficients, which the eigenvalue methods for linear systems solve completely. It approximates the nonlinear system only for small $h$ . Far from $X_*$ it is a different equation.

In plain words Move the origin to the equilibrium and keep only the part of the field that is proportional to the displacement. Near the equilibrium the discarded part is small compared with the size of the displacement, so the linear system is a good local guide.

Example The system $x' = x(1 - x)$ , $y' = -2y$ couples logistic growth with simple decay. Its equilibria are $(0, 0)$ and $(1, 0)$ , and its Jacobian is

J(x, y) = \begin{pmatrix} 1 - 2x & 0 \\ 0 & -2 \end{pmatrix}.

At $(1, 0)$ write $x = 1 + u$ and $y = v$ , so $h = (u, v)^{T}$ . Then $x(1 - x) = -u - u^2$ exactly, and the displacement obeys

\begin{gathered} u' = -u - u^2, \\[0.4em] v' = -2v. \end{gathered}

The linearization drops $-u^2$ and keeps $u' = -u$ , $v' = -2v$ , which is $h' = J(1, 0)h$ with $J(1, 0) = \operatorname{diag}(-1, -2)$ . At the origin the same steps give $u' = u - u^2$ and the linearization $h' = \operatorname{diag}(1, -2)\,h$ .

Hyperbolic equilibrium

Definition An equilibrium $X_*$ is hyperbolic if no eigenvalue of $J(X_*)$ has real part zero. The linearization theorem, for $F$ continuously differentiable near $X_*$ , has two parts. If every eigenvalue of $J(X_*)$ has negative real part, then $X_*$ is locally asymptotically stable. If at least one eigenvalue has positive real part, then $X_*$ is unstable. At a hyperbolic equilibrium one of the two parts always applies, so the linearization decides stability. More is true there. By the Hartman-Grobman theorem, near a hyperbolic equilibrium a continuous change of coordinates carries the trajectories of the nonlinear system onto those of its linearization, so, for example, a saddle of the linearization is a saddle of the nonlinear system. For a planar system with $\tau = \operatorname{tr} J(X_*)$ and $\Delta = \det J(X_*)$ , both eigenvalues have negative real part exactly when $\tau < 0$ and $\Delta > 0$ , and $X_*$ fails to be hyperbolic exactly when $\Delta = 0$ , or when $\tau = 0$ and $\Delta > 0$ .

In plain words When every eigenvalue has a definite inward or outward rate, the small nonlinear terms are too weak near the equilibrium to change the verdict. Only an eigenvalue on the imaginary axis leaves the question open.

Example For $x' = x(1 - x)$ , $y' = -2y$ the matrix $J(1, 0) = \operatorname{diag}(-1, -2)$ has eigenvalues $-1$ and $-2$ , so $(1, 0)$ is a hyperbolic sink and locally asymptotically stable. The matrix $J(0, 0) = \operatorname{diag}(1, -2)$ has eigenvalues $1$ and $-2$ , so the origin is a hyperbolic saddle and unstable. States with small $x > 0$ are pushed away from it, while states on the $y$ -axis decay toward it. By contrast, the coexistence state $(1, 2)$ of the predator-prey model has $\tau = 0$ and $\Delta = 2$ , with eigenvalues $\pm i\sqrt{2}$ , so it is not hyperbolic.

Local asymptotic stability

Definition An equilibrium $X_*$ of $X' = F(X)$ is stable, also called Lyapunov stable, if for every $\varepsilon > 0$ there is a $\delta > 0$ such that every solution with $\lVert X(0) - X_* \rVert < \delta$ exists for all $t \ge 0$ and satisfies $\lVert X(t) - X_* \rVert < \varepsilon$ for all $t \ge 0$ . It is locally asymptotically stable if it is stable and there is an $\eta > 0$ such that every solution with $\lVert X(0) - X_* \rVert < \eta$ satisfies $X(t) \to X_*$ as $t \to \infty$ . An equilibrium that is not stable is unstable. The word local means that both promises are made only for starts in some neighborhood of $X_*$ , whose size the definition leaves open. When every eigenvalue of $J(X_*)$ has negative real part the approach is even exponential. There are constants $K$ , $\alpha > 0$ and $\eta > 0$ such that

\begin{aligned} &\lVert X(t) - X_* \rVert \\ &\quad \le Ke^{-\alpha t}\lVert X(0) - X_* \rVert \end{aligned}

for all $t \ge 0$ whenever $\lVert X(0) - X_* \rVert < \eta$ .

In plain words Stability means that small disturbances stay small. Asymptotic stability adds that they also die out. Local means that this is promised only for disturbances that are small enough, with no claim about large ones.

Example For $x' = x(1 - x)$ , $y' = -2y$ , start at $x(0) = \tfrac{1}{4}$ and $y(0) = 2$ . Separating the logistic equation gives

\begin{gathered} x(t) = \frac{1}{1 + 3e^{-t}}, \\[0.4em] y(t) = 2e^{-2t}. \end{gathered}

To check, the derivative is $x'(t) = 3e^{-t}/(1 + 3e^{-t})^2$ , and $x(1 - x)$ is the product of $1/(1 + 3e^{-t})$ and $1 - x = 3e^{-t}/(1 + 3e^{-t})$ , which is the same fraction, while $y' = -4e^{-2t} = -2y$ . As $t \to \infty$ the state tends to the sink $(1, 0)$ , and for large $t$ the gaps $1 - x(t) \approx 3e^{-t}$ and $y(t) = 2e^{-2t}$ shrink at the rates $-1$ and $-2$ given by the eigenvalues. This start is not especially close to $(1, 0)$ , so its convergence is more than the local theorem promises.

Nonhyperbolic equilibrium

Definition An equilibrium $X_*$ is nonhyperbolic if at least one eigenvalue of $J(X_*)$ has real part zero, such as a zero eigenvalue or a pair $\pm i\omega$ with $\omega > 0$ . If another eigenvalue has positive real part, $X_*$ is still unstable by the linearization theorem. Otherwise the linearization does not decide stability. The neglected nonlinear terms decide it, and the same linearization can belong to an asymptotically stable equilibrium, to one that is stable but not asymptotically stable, or to an unstable one. Another argument is needed, such as an exact solution, a conserved quantity or a Lyapunov function.

In plain words An eigenvalue with zero real part means that the linear part has no pull toward the equilibrium and no push away from it in some direction, so the small terms that linearization throws away have the final word.

Example The scalar equations $x' = -x^3$ and $x' = x^3$ both have the equilibrium $0$ with Jacobian $0$ , so they share the linearization $h' = 0$ . For $x(0) = a \ne 0$ , separation gives

x(t) = \frac{a}{\sqrt{1 \pm 2a^2t}},

where the plus sign solves $x' = -x^3$ for all $t \ge 0$ and the minus sign solves $x' = x^3$ for $t < 1/(2a^2)$ . The first solution tends to $0$ , but only at the rate of $1/\sqrt{2t}$ , more slowly than any decaying exponential, so $0$ is asymptotically stable for $x' = -x^3$ without being exponentially stable. The second becomes infinite as $t$ increases to $1/(2a^2)$ , so $0$ is unstable for $x' = x^3$ . For $x' = 0$ , with the same linearization, every solution is constant, and $0$ is stable but not asymptotically stable. The predator-prey coexistence state is a planar case. Its linearization has eigenvalues $\pm i\sqrt{2}$ and closed elliptical orbits, and the conserved quantity of the previous lesson is what shows that the nonlinear trajectories close as well.

Local versus global behavior

Definition A statement about an equilibrium $X_*$ is local when it is made only for starts in some neighborhood of $X_*$ , and global when it is made for every start in the state space. The basin of attraction of an asymptotically stable equilibrium is the set of all initial states whose solutions exist for all $t \ge 0$ and tend to $X_*$ . The equilibrium is globally asymptotically stable if it is stable and its basin is the whole state space. Linearization gives only local conclusions, with no estimate of how large the neighborhood is, and a system with more than one equilibrium cannot have a globally asymptotically stable one.

In plain words The Jacobian sees only a small window around the equilibrium. Whether distant states also return is a separate question, and only the full nonlinear equations answer it.

Example For $x' = x(1 - x)$ , $y' = -2y$ the sink $(1, 0)$ attracts every start with $x(0) > 0$ , because then $x(t) \to 1$ and $y(t) \to 0$ . Starts with $x(0) = 0$ stay on the $y$ -axis and tend to the saddle $(0, 0)$ instead. The start $x(0) = -1$ , outside the population interpretation but allowed by the equations, gives

x(t) = \frac{1}{1 - 2e^{-t}},

which decreases to $-\infty$ as $t$ increases to $\ln 2$ , so this solution does not even exist for all $t \ge 0$ . The basin of $(1, 0)$ is the half plane $x > 0$ , not the whole plane, even though the local verdict at $(1, 0)$ is as strong as a linearization can give.

Check yourself

Find the equilibria of $x' = y$ , $y' = x - x^3 - y$ and classify each one by linearization.

Show the answer

An equilibrium needs $y = 0$ and $x - x^3 = x(1 - x)(1 + x) = 0$ , so the equilibria are $(0, 0)$ , $(1, 0)$ and $(-1, 0)$ . The Jacobian is

J(x, y) = \begin{pmatrix} 0 & 1 \\ 1 - 3x^2 & -1 \end{pmatrix}.

At the origin it has $\tau = -1$ and $\Delta = -1$ , so its eigenvalues are the roots of $\lambda^2 + \lambda - 1 = 0$ , namely $(-1 \pm \sqrt{5})/2 \approx 0.618$ and $-1.618$ . The origin is a hyperbolic saddle and unstable. At $(\pm 1, 0)$ the Jacobian has $\tau = -1$ and $\Delta = 2$ , the eigenvalues are the roots of $\lambda^2 + \lambda + 2 = 0$ , namely $(-1 \pm i\sqrt{7})/2$ , and their real part $-\tfrac{1}{2}$ is negative. Both of these equilibria are hyperbolic sinks, so they are locally asymptotically stable.

The system $x' = -y - x(x^2 + y^2)$ , $y' = x - y(x^2 + y^2)$ has a linearization at the origin with eigenvalues $\pm i$ . A classmate concludes that the origin is a center surrounded by closed orbits. Use $r^2 = x^2 + y^2$ to decide what really happens.

Show the answer

The cubic terms have zero first derivatives at the origin, so the Jacobian there is the rotation matrix with rows $(0, -1)$ and $(1, 0)$ , whose eigenvalues are $\pm i$ . The origin is nonhyperbolic, and the linearization decides nothing. Along a solution,

\begin{aligned} xx' + yy' &= -xy - x^2r^2 \\ &\quad + xy - y^2r^2 \\[0.4em] &= -r^4, \end{aligned}

so $(r^2)' = 2(xx' + yy') = -2(r^2)^2$ . This separable equation gives

r(t)^2 = \frac{r_0^2}{1 + 2r_0^2t},

where $r_0$ is the starting distance, so every solution tends to the origin, and the angle obeys $\theta' = (xy' - yx')/r^2 = 1$ . Trajectories spiral into the origin, which is asymptotically stable, and there are no closed orbits. The classmate trusted the linearization at a nonhyperbolic equilibrium. With the signs of the cubic terms reversed the linearization is the same, but $(r^2)' = 2r^4$ and the origin is unstable.

For the system $x' = -x + x^3$ , $y' = -y$ , show that the origin is locally asymptotically stable. Does the solution with $x(0) = 2$ and $y(0) = 0$ tend to the origin, and what is the basin of attraction of the origin?

Show the answer

The Jacobian is $\operatorname{diag}(-1 + 3x^2, -1)$ , which is $\operatorname{diag}(-1, -1)$ at the origin, a hyperbolic sink. So the origin is locally asymptotically stable. For the start $(2, 0)$ the second equation keeps $y = 0$ , and the first is solved by

x(t) = \frac{2}{\sqrt{4 - 3e^{2t}}}.

It has $x(0) = 2$ , and both $x'$ and $-x + x^3$ equal $6e^{2t}(4 - 3e^{2t})^{-3/2}$ . This solution becomes infinite as $t$ increases to $\tfrac{1}{2}\ln\tfrac{4}{3} \approx 0.144$ , so it never approaches the origin. More generally, $y(t) \to 0$ for every start, while the sign of $x(x^2 - 1)$ shows that $x(t) \to 0$ when $\lvert x(0) \rvert < 1$ , that $x$ stays at $\pm 1$ when $\lvert x(0) \rvert = 1$ , and that $\lvert x \rvert$ grows without bound when $\lvert x(0) \rvert > 1$ . The basin is the strip $-1 < x < 1$ . Its edges are the lines $x = \pm 1$ , which contain the saddles $(\pm 1, 0)$ with Jacobian $\operatorname{diag}(2, -1)$ .

Linearization and Local Stability

Key terms

Jacobian matrix

Linearization

Hyperbolic equilibrium

Local asymptotic stability

Nonhyperbolic equilibrium

Local versus global behavior

Check yourself

How this lesson connects

Builds on

Used later in