Hopf Bifurcation and Oscillation Onset, differential equations lesson

Compare a planar system at a range of parameter values. Below some value every nearby solution spirals into a steady state, and just above it nearby solutions settle onto a small steady oscillation whose size grows from zero as the parameter moves on. How does the oscillation appear, and what decides whether it attracts the solutions around it? This lesson studies the Hopf bifurcation, in which a pair of complex eigenvalues of the Jacobian crosses the imaginary axis. The eigenvalues locate the threshold, and a nonlinear coefficient decides whether the new oscillation attracts or repels. Throughout, $t$ is time, a dot means a derivative with respect to $t$ , and the state is $X = (x, y)^{T}$ .

Key terms

Hopf bifurcation

Definition Consider a planar family $\dot X = F(X, \mu)$ with a real parameter $\mu$ , where $F$ has continuous partial derivatives up to order three. Suppose that at $\mu = \mu_0$ the point $X_*(\mu_0)$ is an equilibrium whose Jacobian has the eigenvalues $\pm i\omega_0$ with $\omega_0 > 0$ . Since $0$ is not an eigenvalue, the equilibrium continues as a smooth branch $X_*(\mu)$ for $\mu$ near $\mu_0$ , and the Jacobian along it has eigenvalues $\alpha(\mu) \pm i\omega(\mu)$ . A Hopf bifurcation is the birth of a small periodic orbit from the equilibrium, or its collapse into it, as $\mu$ passes $\mu_0$ . The Hopf bifurcation theorem gives conditions for it. Suppose

\begin{gathered} \alpha(\mu_0) = 0, \quad \omega_0 > 0, \\[0.4em] \alpha'(\mu_0) \ne 0, \quad \ell_1 \ne 0, \end{gathered}

where $\ell_1$ is the first Lyapunov coefficient at $\mu_0$ . Then there are a neighborhood $N$ of $X_*(\mu_0)$ and an interval $I$ around $\mu_0$ with the following property. For each $\mu$ in $I$ on the side where $\alpha'(\mu_0)(\mu - \mu_0)$ and $\ell_1$ have opposite signs, the neighborhood $N$ contains exactly one periodic orbit. It is a limit cycle around the equilibrium, its size is proportional to $\sqrt{\lvert \mu - \mu_0 \rvert}$ to leading order, and its period is close to $2\pi/\omega_0$ . For the other values of $\mu$ in $I$ , including $\mu_0$ itself, there is no periodic orbit in $N$ . The cycle is attracting when $\ell_1 < 0$ and repelling when $\ell_1 > 0$ . The theorem is local in both the state and the parameter, so it says nothing about large oscillations elsewhere.

In plain words At the threshold the spiral around the equilibrium turns from inward to outward or back. On one side of the threshold a small loop surrounds the equilibrium, and it shrinks onto the equilibrium as the parameter approaches the threshold, with a size proportional to the square root of the parameter’s distance from the threshold. The sign of one nonlinear coefficient decides whether the loop attracts or repels.

Example For the lesson’s system

\begin{aligned} \dot x &= \mu x - y - x(x^2 + y^2), \\[0.4em] \dot y &= x + \mu y - y(x^2 + y^2), \end{aligned}

the origin is an equilibrium for every $\mu$ , with Jacobian and eigenvalues

\begin{gathered} A(\mu) = \begin{bmatrix} \mu & -1 \\ 1 & \mu \end{bmatrix}, \\[0.4em] \lambda = \mu \pm i. \end{gathered}

So $\mu_0 = 0$ , $\alpha(\mu) = \mu$ , $\omega = 1$ and $\alpha'(0) = 1$ , and the cards below show that $\ell_1 = -1$ . The theorem predicts an attracting cycle for small $\mu > 0$ , and the polar form finds it exactly. It is the circle of radius $\sqrt{\mu}$ , traversed with period $2\pi$ .

Eigenvalue crossing

Definition Along the branch $X_*(\mu)$ let $A(\mu)$ be the Jacobian of $F$ at $X_*(\mu)$ . Its eigenvalues form a complex pair $\alpha(\mu) \pm i\omega(\mu)$ with $\omega(\mu) > 0$ exactly when $\det A > \frac14(\operatorname{tr} A)^2$ , and then

\begin{gathered} \alpha = \tfrac12 \operatorname{tr} A, \\[0.4em] \omega = \sqrt{\det A - \tfrac14 (\operatorname{tr} A)^2}. \end{gathered}

The pair crosses the imaginary axis at $\mu_0$ if $\alpha(\mu_0) = 0$ and $\alpha$ changes sign at $\mu_0$ . The transversality condition $\alpha'(\mu_0) \ne 0$ asks for a crossing with nonzero speed, and it implies the sign change. If $\alpha'(\mu_0) > 0$ , the equilibrium is hyperbolic for $\mu \ne \mu_0$ near $\mu_0$ , so linearization shows that it is asymptotically stable for $\mu$ slightly below $\mu_0$ and unstable for $\mu$ slightly above. At $\mu_0$ itself linearization decides nothing. The nonzero imaginary part $\omega_0$ gives the oscillation its frequency at the threshold. A real eigenvalue passing through $0$ is a different situation, which typically changes the number of equilibria, as in the one-dimensional bifurcations.

In plain words Follow the two eigenvalues as the parameter moves. They are mirror images across the real axis, and at the threshold they pass through the imaginary axis together, from the stable half-plane to the unstable one or back. Transversality says that they really move across, instead of touching the axis and turning back or crossing it with zero speed.

Example For the lesson’s system $\operatorname{tr} A = 2\mu$ and $\det A = \mu^2 + 1$ , so $\alpha(\mu) = \mu$ , $\omega = 1$ and $\alpha'(0) = 1$ . If the linear terms were $\mu^2 x - y$ and $x + \mu^2 y$ instead, the eigenvalues $\mu^2 \pm i$ would touch the imaginary axis at $\mu = 0$ and turn back. The polar form would then be $\dot r = \mu^2 r - r^3$ , with an attracting cycle of radius $\lvert \mu \rvert$ for every $\mu \ne 0$ , on both sides of the threshold, so the one-sided picture of the theorem is lost.

Polar normal form

Definition The polar normal form of a Hopf bifurcation is the family

\begin{gathered} \dot r = \mu r + a r^3, \\[0.4em] \dot\theta = \omega + b r^2 \end{gathered}

in polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ , with constants $\omega > 0$ , $a \ne 0$ and $b$ . In the coordinates $x$ and $y$ it is the polynomial system

\begin{aligned} \dot x &= \mu x - \omega y \\[0.4em] &\quad + (ax - by)(x^2 + y^2), \\[0.4em] \dot y &= \omega x + \mu y \\[0.4em] &\quad + (bx + ay)(x^2 + y^2), \end{aligned}

whose Jacobian at the origin has the eigenvalues $\mu \pm i\omega$ . Besides $r = 0$ , the radial equation has the root $r_*$ with $r_*^2 = -\mu/a$ , which exists when $\mu$ and $a$ have opposite signs. For small $r_*$ the circle $r = r_*$ is a periodic orbit with period $2\pi/(\omega + b r_*^2)$ , and the radial derivative there is

\begin{aligned} \frac{\partial \dot r}{\partial r}\bigg\vert_{r = r_*} &= \mu + 3a r_*^2 \\[0.4em] &= -2\mu. \end{aligned}

Near a Hopf point that satisfies the theorem’s conditions, a smooth change of coordinates and a shift and rescaling of the parameter bring the radial equation to $\dot r = \mu r + a r^3$ plus higher-order terms, with an angular speed close to $\omega_0$ , and the coefficient $a$ is then the first Lyapunov coefficient.

In plain words The normal form is the simplest system with a Hopf bifurcation. Its radius and angle separate, so the size of the oscillation obeys a one-dimensional equation of its own, and the cycle is an equilibrium of that equation.

Example The lesson’s system is the normal form with $\omega = 1$ , $a = -1$ and $b = 0$ . Its polar equations $\dot r = \mu r - r^3$ and $\dot\theta = 1$ follow from $x\dot x + y\dot y = \mu r^2 - r^4$ and $x\dot y - y\dot x = r^2$ . For $\mu > 0$ the cycle has radius $\sqrt{\mu}$ and radial derivative $-2\mu < 0$ . Take $\mu = \frac14$ , so the cycle has radius $\frac12$ . The solution

(x(t), y(t)) = \tfrac12(\cos t, \sin t)

checks directly. Since $x^2 + y^2 = \frac14$ , the term $\mu x$ cancels $-x(x^2 + y^2)$ and the term $\mu y$ cancels $-y(x^2 + y^2)$ . What remains is $\dot x = -y$ and $\dot y = x$ , which this pair satisfies. The radial derivative there is $-\frac12$ , so a small error in the radius decays, while a shift in phase stays as it is.

First Lyapunov coefficient

Definition Place the equilibrium at $\mu_0$ at the origin and choose coordinates, by a linear change if necessary, in which the linear part at $\mu = \mu_0$ is a rotation with angular speed $\omega_0 > 0$ . The system at $\mu = \mu_0$ then reads

\begin{aligned} \dot x &= -\omega_0 y + f(x, y), \\[0.4em] \dot y &= \omega_0 x + g(x, y), \end{aligned}

where $f$ , $g$ and their first derivatives vanish at the origin. With subscripts for partial derivatives at the origin, a common normalization of the first Lyapunov coefficient is

\begin{aligned} 16\,\ell_1 = {} & f_{xxx} + f_{xyy} + g_{xxy} \\[0.4em] & + g_{yyy} + \frac{R}{\omega_0}, \end{aligned}

where

\begin{aligned} R = {} & f_{xy}(f_{xx} + f_{yy}) \\[0.4em] & - g_{xy}(g_{xx} + g_{yy}) \\[0.4em] & - f_{xx}\,g_{xx} + f_{yy}\,g_{yy}. \end{aligned}

With this normalization $\ell_1$ equals the cubic coefficient $a$ of the polar normal form that the system reduces to. Other books scale $\ell_1$ by a positive factor, which changes neither its sign nor whether it vanishes. The nondegeneracy condition of the Hopf theorem is $\ell_1 \ne 0$ . If $\ell_1 = 0$ , the terms up to third order cannot decide whether a small cycle exists or how stable it is, and higher-order terms take over.

In plain words At the threshold the linear part only rotates, so whether nearby solutions creep inward or outward is decided by the nonlinear terms. The first Lyapunov coefficient measures that creep. Over one turn, a solution at a small distance $r$ from the equilibrium changes its distance by about $2\pi\ell_1 r^3/\omega_0$ . A negative value means that nearby solutions slowly spiral in, and a positive value that they slowly spiral out.

Example For the lesson’s system at $\mu = 0$ the nonlinear terms are $f = -x^3 - xy^2$ and $g = -x^2 y - y^3$ . There are no quadratic terms, so $R = 0$ , and $f_{xxx} = -6$ , $f_{xyy} = -2$ , $g_{xxy} = -2$ and $g_{yyy} = -6$ give $16\,\ell_1 = -16$ . So $\ell_1 = -1$ , which is the cubic coefficient $a$ . A system that is not already in polar form is $\dot x = \mu x - y - x^3$ , $\dot y = x + \mu y$ . Here $f = -x^3$ and $g = 0$ , so $16\,\ell_1 = -6$ and $\ell_1 = -\frac38$ . Its polar form $\dot r = \mu r - r^3\cos^4\theta$ shows where the number comes from, since the average of $\cos^4\theta$ over one turn is $\frac38$ . For small $\mu > 0$ the cycle is close to the circle of radius $\sqrt{8\mu/3}$ , which is about $0.163$ for $\mu = 0.01$ .

Supercritical and subcritical Hopf bifurcations

Definition Under the conditions of the Hopf theorem, the bifurcation is supercritical when $\ell_1 < 0$ . The cycle is then attracting and exists on the side of $\mu_0$ where the equilibrium is unstable, and at $\mu_0$ the equilibrium is still asymptotically stable, although linearization cannot show it. The bifurcation is subcritical when $\ell_1 > 0$ . The cycle is then repelling, exists on the side where the equilibrium is asymptotically stable, and shrinks onto the equilibrium as $\mu$ tends to $\mu_0$ . On the other side the equilibrium is unstable and no cycle is nearby, so every solution that starts close to the equilibrium, other than the equilibrium itself, eventually leaves a small closed disk around it, and where it goes depends on the system away from the equilibrium.

In plain words In a supercritical bifurcation the oscillation switches on gently. Just past the threshold, solutions settle onto a small attracting cycle whose size grows from zero. In a subcritical bifurcation no small cycle is left to settle on once the equilibrium is unstable, so solutions move away, and where they end up depends on the rest of the system.

Example Compare $\dot r = \mu r - r^3$ with $\dot r = \mu r + r^3$ , both with $\dot\theta = 1$ and the same eigenvalues $\mu \pm i$ . In the second, the cycle $r = \sqrt{-\mu}$ exists for $\mu < 0$ , and the radial derivative there is $-2\mu > 0$ , so it repels. Starts inside it spiral into the origin, and starts outside it move away. At $\mu = 0$ the two radial equations are $\dot r = -r^3$ and $\dot r = r^3$ , with the solutions

\begin{gathered} r(t) = \frac{r_0}{\sqrt{1 + 2r_0^2 t}}, \\[0.4em] r(t) = \frac{r_0}{\sqrt{1 - 2r_0^2 t}}. \end{gathered}

The first decays to $0$ , but only like $1/\sqrt{2t}$ rather than exponentially, and the second is unbounded as $t$ increases to $1/(2r_0^2)$ . The eigenvalues are the same in both systems, so linearization alone could never tell them apart.

Check yourself

For the system $\dot x = \mu x - 2y - x(x^2 + y^2)$ , $\dot y = 2x + \mu y - y(x^2 + y^2)$ , find the eigenvalues at the origin and the polar form. For $\mu > 0$ , find the periodic orbit, its period and its stability.

Show the answer

The Jacobian at the origin has rows $(\mu, -2)$ and $(2, \mu)$ , so the eigenvalues are $\mu \pm 2i$ , with $\alpha(\mu) = \mu$ , $\omega = 2$ and $\alpha'(0) = 1$ . The rotation terms cancel in $x\dot x + y\dot y = \mu r^2 - r^4$ , and $x\dot y - y\dot x = 2r^2$ , so

\dot r = \mu r - r^3, \quad \dot\theta = 2.

For $\mu > 0$ the circle $r = \sqrt{\mu}$ is a periodic orbit. It is traversed at angular speed $2$ , so its period is $\pi$ . The radial derivative $\mu - 3r^2$ equals $-2\mu < 0$ there, so it is an attracting limit cycle, and the bifurcation at $\mu = 0$ is supercritical with $\ell_1 = -1$ . For $\mu = 0.09$ , for instance, the cycle is traced by $(0.3\cos 2t, 0.3\sin 2t)$ .

The linear system $\dot x = \mu x - y$ , $\dot y = x + \mu y$ has the eigenvalues $\mu \pm i$ , which cross the imaginary axis at $\mu = 0$ with $\alpha'(0) = 1$ . A classmate concludes that a small limit cycle is born at $\mu = 0$ . Is that right?

Show the answer

No. In polar form $\dot r = \mu r$ and $\dot\theta = 1$ , so $r(t) = r_0 e^{\mu t}$ . For $\mu < 0$ every solution spirals into the origin, and for $\mu > 0$ every solution other than the origin spirals out without bound, so neither case has a periodic orbit. At $\mu = 0$ every circle around the origin is a periodic orbit, a whole family with no isolated member, so there is no limit cycle there either. Transversality holds, but the nondegeneracy condition fails, because there are no nonlinear terms and $\ell_1 = 0$ . An eigenvalue crossing only locates a possible threshold, and the nonlinear terms are needed to produce an isolated cycle.

The system $\dot x = \mu x - y + x(r^2 - r^4)$ , $\dot y = x + \mu y + y(r^2 - r^4)$ , with $r^2 = x^2 + y^2$ , has the polar form $\dot r = \mu r + r^3 - r^5$ , $\dot\theta = 1$ . Which kind of Hopf bifurcation happens at $\mu = 0$ ? Find the periodic orbits and their stability for $-\frac14 < \mu < 0$ and for $\mu > 0$ . For $\mu = 0.1$ , where does a solution that starts close to the origin end up, and what happens when its state after a long time is used as the start for $\mu = -0.1$ ?

Show the answer

The cubic coefficient is $+1$ , so $\ell_1 = 1 > 0$ and the bifurcation at $\mu = 0$ is subcritical. A cycle has a radius $r > 0$ whose square $s = r^2$ solves $\mu + s - s^2 = 0$ , so

s_{\pm} = \frac{1 \pm \sqrt{1 + 4\mu}}{2}.

Write $\dot r = r\,G(r^2)$ with $G(s) = \mu + s - s^2$ . At a cycle the radial derivative is $2s\,G'(s) = 2s(1 - 2s)$ . For $-\frac14 < \mu < 0$ both roots are positive. The smaller one has $1 - 2s_- = \sqrt{1 + 4\mu} > 0$ , so its cycle repels, and the larger one has $1 - 2s_+ = -\sqrt{1 + 4\mu} < 0$ , so its cycle attracts. For $\mu > 0$ the root $s_-$ is negative, the origin is unstable, and only the attracting cycle with $s_+ > 1$ remains.

For $\mu = 0.1$ that cycle has $s_+ \approx 1.092$ , so $r \approx 1.045$ . A solution starting near the origin spirals out to it, so the oscillation appears at a size of about $1$ instead of growing from zero. For $\mu = -0.1$ the cycles are at $r \approx 0.336$ , repelling, and $r \approx 0.942$ , attracting. A start at $r \approx 1.045$ lies outside both, where $G < 0$ , so it moves in to the cycle at $r \approx 0.942$ and keeps oscillating, although the origin is stable again. Only for $\mu < -\frac14$ , after the two cycles have merged at $r = 1/\sqrt{2}$ and disappeared, do all solutions return to the origin. This dependence on the history of the parameter is called hysteresis.

Hopf Bifurcation and Oscillation Onset

Key terms

Hopf bifurcation

Eigenvalue crossing

Polar normal form

First Lyapunov coefficient

Supercritical and subcritical Hopf bifurcations

Check yourself

How this lesson connects

Builds on

Used later in