One-Dimensional Bifurcations, differential equations lesson

A population that grows logistically and is harvested at a constant rate settles at a stable positive level when the harvest is light, from any start above a small threshold. Raise the harvest a little past a certain value and the population collapses from every starting level, although the equation keeps its form and only one constant has changed. This lesson studies how the equilibria of an equation $x' = f(x, \mu)$ , with time $t$ as the independent variable and $x' = dx/dt$ , appear, disappear and trade stability as the parameter $\mu$ is varied. It sorts the basic changes into saddle-node, transcritical and pitchfork bifurcations and tells the supercritical pitchfork from the subcritical one.

Key terms

Bifurcation parameter

Definition Consider a scalar autonomous equation

x' = f(x, \mu),

where $f$ is continuously differentiable in $x$ and $\mu$ together, and write $f_x = \partial f/\partial x$ . The bifurcation parameter $\mu$ is a real constant. It is fixed while a solution runs, and each value of $\mu$ gives its own equation with its own phase line. A value $\mu_0$ is a bifurcation value if, arbitrarily close to $\mu_0$ , there are parameter values whose phase lines differ qualitatively from the phase line at $\mu_0$ , with a different number of equilibria or a different pattern of arrows. Near a hyperbolic equilibrium nothing of this kind happens. If $f(x_*, \mu_0) = 0$ and $f_x(x_*, \mu_0) \ne 0$ , the implicit function theorem gives, for every $\mu$ close to $\mu_0$ , exactly one equilibrium close to $x_*$ . It moves continuously with $\mu$ and keeps the sign of $f_x$ , and with it the stability. So the phase line can change its character near an equilibrium $x_*$ only at a parameter value $\mu_0$ with

\begin{gathered} f(x_*, \mu_0) = 0, \\[0.4em] f_x(x_*, \mu_0) = 0. \end{gathered}

In plain words The parameter is a dial that is set before a solution starts and stays put while it runs. A bifurcation value is a setting where the phase line changes its character, and near an equilibrium that can only happen where the derivative test of the phase line fails.

Example In the harvested logistic equation $x' = x(1 - x) - h$ the harvest rate $h \ge 0$ is the bifurcation parameter. For $h = 0.16$ the equilibria are the roots of $x^2 - x + 0.16 = 0$ , namely $0.2$ and $0.8$ . The derivative $f_x = 1 - 2x$ equals $0.6$ and $-0.6$ there, so $0.2$ is unstable and $0.8$ is asymptotically stable. For $h = \frac14$ the two have merged into $x = \frac12$ , where $f_x = 0$ . Completing the square gives

\begin{aligned} f(x) &= x(1 - x) - h \\[0.4em] &= \tfrac14 - h - \left(x - \tfrac12\right)^2, \end{aligned}

which is negative for every $x$ when $h > \frac14$ . Then there is no equilibrium, and every solution decreases at a rate of at least $h - \frac14$ . So $h = \frac14$ is the bifurcation value. A harvest rate that drifts with time while a solution runs would be a different problem, a nonautonomous equation.

Bifurcation diagram

Definition The bifurcation diagram of $x' = f(x, \mu)$ is the set of points $(\mu, x)$ with $f(x, \mu) = 0$ , drawn with the parameter $\mu$ on the horizontal axis and the equilibrium state $x$ on the vertical axis. Its curves are the equilibrium branches. A branch is drawn solid where $f_x < 0$ , so the equilibrium is asymptotically stable, and dashed where $f_x > 0$ , so it is unstable. Points with $f_x = 0$ are classified separately from the signs of $f$ . The vertical line through a fixed value of $\mu$ is the phase line of that one equation, and arrows can be drawn on it. The diagram is not a graph of a solution against time. Moving along a branch compares different equations and does not follow any solution.

In plain words Stand the phase lines for all parameter values side by side. Their equilibria line up into curves, and the picture shows at a glance which equilibria exist for each setting of the parameter and which of them attract.

Example For $x' = \mu - x^2$ the equilibria satisfy $\mu = x^2$ , a parabola opening to the right with its vertex at the origin. Since $f_x = -2x$ , the upper half $x = \sqrt{\mu}$ is solid and the lower half $x = -\sqrt{\mu}$ is dashed. Vertical lines to the left of $\mu = 0$ miss the parabola, so those equations have no equilibria. On the line $\mu = 1$ the arrows point down above $x = 1$ , up between $-1$ and $1$ , and down below $-1$ , toward the solid point and away from the dashed one.

Saddle-node bifurcation

Definition A saddle-node bifurcation, also called a fold, happens at $(x_0, \mu_0)$ when two equilibria near $x_0$ , one asymptotically stable and one unstable, approach each other as $\mu$ tends to $\mu_0$ from one side, merge into the single equilibrium $x_0$ at $\mu = \mu_0$ , and have no counterpart near $x_0$ for $\mu$ on the other side. If $f$ has continuous second derivatives, the conditions

\begin{gathered} f = 0, \qquad f_x = 0, \\[0.4em] f_{xx} \ne 0, \qquad f_\mu \ne 0, \end{gathered}

all evaluated at $(x_0, \mu_0)$ , guarantee it. Because $f_\mu \ne 0$ , the equilibria near $(x_0, \mu_0)$ form a single curve $\mu = m(x)$ with $m'(x_0) = 0$ and $m''(x_0) = -f_{xx}/f_\mu \ne 0$ , so the curve turns back at $x_0$ . Along it $f_x = -f_\mu\, m'(x)$ , and $m'$ changes sign at $x_0$ , so the two equilibria have opposite stability. The model case, or normal form, is $x' = \mu - x^2$ .

In plain words Two equilibria slide toward each other, meet and vanish. Before the collision one attracts and the other repels. Afterward no rest state is left in that part of the line, so solutions pass straight through it.

Example For $x' = \mu - x^2$ with $\mu > 0$ the equilibria are $\pm\sqrt{\mu}$ , and $f_x = -2x$ makes $\sqrt{\mu}$ asymptotically stable and $-\sqrt{\mu}$ unstable. For $\mu < 0$ the rate $\mu - x^2$ is negative everywhere, so there is no equilibrium and every solution decreases. The conditions hold at $(0, 0)$ with $f_{xx} = -2$ and $f_\mu = 1$ . At $\mu = 0$ the equation $x' = -x^2$ with $x(0) = x_0$ has the solution

x(t) = \frac{x_0}{1 + x_0 t}

on the interval containing $t = 0$ where $1 + x_0 t \ne 0$ , since its derivative is $-x_0^2/(1 + x_0 t)^2 = -x(t)^2$ . A start $x_0 > 0$ decreases toward $0$ as $t \to \infty$ . A start $x_0 < 0$ decreases and tends to $-\infty$ as $t$ increases to $1/\lvert x_0 \rvert$ . So the merged equilibrium $0$ is semistable. It attracts from the side $x > 0$ and repels on the side $x < 0$ , a fact the zero derivative $f_x(0, 0)$ could not reveal.

Transcritical bifurcation

Definition A transcritical bifurcation happens at $(x_0, \mu_0)$ when two branches of equilibria, each defined on both sides of $\mu_0$ , cross at $x_0$ when $\mu = \mu_0$ and exchange stability there. Exchange of stability means that the branch which is asymptotically stable for $\mu$ just below $\mu_0$ is unstable for $\mu$ just above it, while the other branch changes from unstable to asymptotically stable. It typically occurs when one equilibrium persists for every parameter value, such as $x = 0$ in an equation $x' = x\,g(x, \mu)$ . The normal form is $x' = \mu x - x^2$ .

In plain words No equilibrium disappears. Two of them pass through each other, and in passing they hand over the role of the attracting state.

Example Write $\mu x - x^2 = x(\mu - x)$ . The branches are $x = 0$ and $x = \mu$ , and $f_x = \mu - 2x$ gives

\begin{gathered} f_x(0, \mu) = \mu, \\[0.4em] f_x(\mu, \mu) = -\mu. \end{gathered}

For $\mu < 0$ the branch $x = 0$ is asymptotically stable and $x = \mu$ is unstable, and for $\mu > 0$ the roles are reversed. At $\mu = 0$ the two meet in the single semistable equilibrium of $x' = -x^2$ . Seen only as pictures, the phase lines at $\mu = -1$ and $\mu = 1$ look alike, with an unstable equilibrium below a stable one. What has changed is which branch plays which role. If $x$ is a population, only $x \ge 0$ is meaningful. For $\mu < 0$ every population then tends to $0$ , and for $\mu > 0$ every positive start tends to $\mu$ . The branch $x = \mu$ for $\mu < 0$ lies outside the model’s domain, although it is still an equilibrium of the equation.

Pitchfork bifurcation

Definition The equation $x' = f(x, \mu)$ has reflection symmetry if

f(-x, \mu) = -f(x, \mu)

for all $x$ and $\mu$ . Then $-x(t)$ is a solution whenever $x(t)$ is, because $(-x)' = -f(x, \mu) = f(-x, \mu)$ . Setting $x = 0$ gives $f(0, \mu) = 0$ , so $x = 0$ is an equilibrium for every $\mu$ . Nonzero equilibria come in pairs $\pm x_*$ , and since $f_x$ is even in $x$ , the two members of a pair have the same stability. A pitchfork bifurcation at $\mu_0$ is a change in which the branch $x = 0$ changes stability at $\mu_0$ while a symmetric pair of branches grows out of $(0, \mu_0)$ on one side of $\mu_0$ . Near the origin there is then one equilibrium on one side of $\mu_0$ and three on the other. The normal form is $x' = \mu x - x^3$ .

In plain words A symmetric equation treats $x$ and $-x$ alike, so it cannot favor either of them. When the middle equilibrium changes stability, new equilibria can only appear as a mirror-image pair, and the diagram looks like a three-pronged fork.

Example Replacing $x$ by $-x$ in $\mu x - x^3$ changes its sign, so the symmetry holds. The equilibria solve $x(\mu - x^2) = 0$ , giving $x = 0$ for every $\mu$ and $x = \pm\sqrt{\mu}$ for $\mu > 0$ . With $f_x = \mu - 3x^2$ ,

\begin{gathered} f_x(0, \mu) = \mu, \\[0.4em] f_x(\pm\sqrt{\mu}, \mu) = -2\mu. \end{gathered}

So the origin is asymptotically stable for $\mu < 0$ and unstable for $\mu > 0$ , and the pair $\pm\sqrt{\mu}$ is asymptotically stable. A constant term breaks the symmetry. In $x' = h + \mu x - x^3$ with $h \ne 0$ the value $f(0, \mu) = h$ is not zero, so the origin is no longer an equilibrium and the fork comes apart, as the last question below shows.

Supercritical and subcritical pitchforks

Definition A pitchfork bifurcation is supercritical when the symmetric pair of branches exists on the side of $\mu_0$ where $x = 0$ is unstable and the pair is asymptotically stable. It is subcritical when the pair exists on the side where $x = 0$ is asymptotically stable and the pair is unstable. The normal forms are

\begin{gathered} x' = \mu x - x^3, \\[0.4em] x' = \mu x + x^3, \end{gathered}

the first supercritical and the second subcritical. For the subcritical form the pair is $\pm\sqrt{-\mu}$ for $\mu < 0$ , where $f_x = \mu + 3x^2 = -2\mu > 0$ . At $\mu = 0$ the derivative test fails in both forms and the sign of the cubic term decides. The equilibrium $0$ of $x' = -x^3$ is asymptotically stable, and that of $x' = x^3$ is unstable.

In plain words In the supercritical case the origin hands its stability to two nearby states, so solutions settle close to where they used to. In the subcritical case the origin loses stability with no nearby stable state to take over, so solutions leave the neighborhood altogether.

Example Take $\mu = 1$ in the supercritical form and $x(0) = \frac12$ . The solution stays positive, since it cannot cross the equilibrium $0$ , so $z = x^2$ determines it. Then $z' = 2xx' = 2z(1 - z)$ is a logistic equation with $z(0) = \frac14$ , whose solution is $z = 1/(1 + 3e^{-2t})$ , and

x(t) = \frac{1}{\sqrt{1 + 3e^{-2t}}}.

Its derivative is $3e^{-2t}(1 + 3e^{-2t})^{-3/2}$ , and $x - x^3$ simplifies to the same expression, so it solves the equation for all $t$ . As $t \to \infty$ it tends to $1$ , the stable branch $\sqrt{\mu}$ . By the symmetry the start $-\frac12$ gives $-x(t)$ , which tends to $-1$ . At $\mu = 0$ the subcritical equation $x' = x^3$ has the solution $x_0/\sqrt{1 - 2x_0^2 t}$ , which is unbounded as $t$ increases to $1/(2x_0^2)$ for every $x_0 \ne 0$ , while $x' = -x^3$ has the solution $x_0/\sqrt{1 + 2x_0^2 t}$ , which tends to $0$ .

Check yourself

For $x' = \mu + x^2$ , find the equilibria for each value of $\mu$ , decide their stability, and name the bifurcation at $\mu = 0$ .

Show the answer

Setting $\mu + x^2 = 0$ gives $x = \pm\sqrt{-\mu}$ for $\mu < 0$ , the single equilibrium $0$ for $\mu = 0$ , and no equilibrium for $\mu > 0$ . With $f_x = 2x$ , the equilibrium $-\sqrt{-\mu}$ is asymptotically stable and $\sqrt{-\mu}$ is unstable. At $\mu = 0$ the rate $x^2$ is positive on both sides of $0$ , so $0$ is semistable, attracting from the side $x < 0$ and repelling on the side $x > 0$ . Two equilibria of opposite stability merge and disappear as $\mu$ increases through $0$ , which is a saddle-node bifurcation, and the conditions hold with $f_{xx} = 2$ and $f_\mu = 1$ . Compared with $x' = \mu - x^2$ , the pair now exists for $\mu < 0$ instead of $\mu > 0$ .

For $x' = \mu - x^3$ the point $(x, \mu) = (0, 0)$ satisfies $f = 0$ and $f_x = 0$ . Is $\mu = 0$ a bifurcation value?

Show the answer

No. The rate $\mu - x^3$ is strictly decreasing in $x$ , so for each $\mu$ the only equilibrium is the real cube root $x_* = \sqrt[3]{\mu}$ . The rate is positive below $x_*$ and negative above it, so the arrows point toward $x_*$ from both sides, and $x_*$ is asymptotically stable for every $\mu$ , including $\mu = 0$ , where the equation is $x' = -x^3$ . Every phase line has exactly one attracting equilibrium, so nothing changes qualitatively. The conditions $f = 0$ and $f_x = 0$ are necessary for a change near an equilibrium, but they are not sufficient. Here the saddle-node condition $f_{xx} \ne 0$ fails, since $f_{xx} = -6x$ vanishes at the origin.

Consider $x' = 2 + \mu x - x^3$ , a pitchfork normal form with a constant added. Show that it has no reflection symmetry, show that at $\mu = 3$ it has a double equilibrium at $x = -1$ , and classify the equilibria at $\mu = 3$ . How many equilibria are there for $\mu < 3$ and for $\mu > 3$ ?

Show the answer

Replacing $x$ by $-x$ gives $2 - \mu x + x^3$ , which is not $-(2 + \mu x - x^3) = -2 - \mu x + x^3$ , so the symmetry fails. Also $f(0, \mu) = 2$ , so $0$ is never an equilibrium. At $\mu = 3$ ,

\begin{aligned} f(x) &= 2 + 3x - x^3 \\[0.4em] &= -(x + 1)^2(x - 2), \end{aligned}

so the equilibria are $-1$ , a double root, and $2$ . The factor $-(x - 2)$ is positive for $x < 2$ , so the rate is positive on both sides of $-1$ and the equilibrium $-1$ is semistable. At $2$ the rate changes from positive to negative, so $2$ is asymptotically stable. At $(x, \mu) = (-1, 3)$ we have $f = 0$ , $f_x = \mu - 3x^2 = 0$ , $f_{xx} = -6x = 6$ and $f_\mu = x = -1$ , so this is a saddle-node bifurcation.

The cubic $x^3 - \mu x - 2$ has three distinct real roots exactly when its discriminant $4\mu^3 - 108$ is positive, that is, when $\mu > 3$ . For $\mu < 3$ there is one equilibrium. It is positive, because the rate is $2$ at $x = 0$ and tends to $-\infty$ as $x \to \infty$ , and it is asymptotically stable. For $\mu > 3$ there are three, two asymptotically stable equilibria with an unstable one between them. The positive branch continues smoothly through $\mu = 3$ , and the new pair is born in a saddle-node away from the origin instead of splitting off from it.

One-Dimensional Bifurcations

Key terms

Bifurcation parameter

Bifurcation diagram

Saddle-node bifurcation

Transcritical bifurcation

Pitchfork bifurcation

Supercritical and subcritical pitchforks

Check yourself

How this lesson connects

Builds on

Used later in