One-dimensional flows

Fixed points, stability and bifurcations on the line and the circle.

In this chapter 3

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Flows on the line and circle

Flows on the line and circle

Method

Flows on the line

Adapted

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Shows how to read the long-time behaviour of an autonomous equation $\dot x = f(x)$ off the graph of $f$ — fixed points, the sign of $f'$ at each one as a decay or growth rate, and the theorem that every solution is monotone, so that nothing on the line can oscillate.

Prerequisites: Existence and uniqueness of solutions, Taylor's theorem, Limits of sequences and completeness

Note(What a one-dimensional flow can and cannot do)Original

An autonomous equation in one unknown,

\dot x = f(x) ,

is the smallest nonlinear problem there is, and almost none of them can be solved in closed form. They do not need to be. The whole qualitative story — which states the system settles into, from which starting values, how fast, and which solutions run away to infinity — can be read off the graph of $f$ without integrating anything, and this unit says how.

The reason the reading is so complete is a restriction that will not survive into two dimensions: a solution of $\dot x = f(x)$ is a monotone function of $t$ . It cannot overshoot, it cannot ring, and it cannot repeat. Everything either approaches a zero of $f$ or leaves every bounded set. Oscillation needs either a second dimension, which is the subject of Classification of linear planar systems and what follows it, or a phase space that closes on itself, which is Flows on the circle.

Definition(The phase line)AdaptedAdapted from Strogatz (1994), §2.1, pp. 16-17, and §2.2, pp. 18-19; Hirsch, Smale & Devaney (2013), §1.1, pp. 2-3

Let $I \subseteq \mathbb{R}$ be an open interval and let $f : I \to \mathbb{R}$ be continuously differentiable. The equation

\dot x = f(x) , \qquad x(0) = x_0 \in I ,

is called a one-dimensional or first-order system. It is autonomous: the right-hand side does not contain $t$ , so the rule of motion at a point of $I$ is the same at every instant. By Existence and uniqueness of solutions the initial-value problem has exactly one solution on some time interval around $t = 0$ , and we write $x(t)$ for it and $(t_-, t_+)$ for the largest interval on which it is defined.

The geometric reading is to regard $I$ not as an axis along which graphs of solutions are drawn, but as the space in which the motion happens — the phase line. A number $x \in I$ is a state; $f(x)$ is the velocity assigned to that state; the assignment $x \mapsto f(x)$ is a vector field on the line. A solution $x(t)$ is then the position at time $t$ of a point carried by that field, and its image in $I$ is its trajectory. The picture of $I$ with the trajectories marked, arrows pointing right where $f > 0$ and left where $f < 0$ , is the phase portrait. Strogatz's way of holding the picture in mind is to imagine a fluid filling the line and moving with local velocity $f$ , a solution being a speck carried along by it.

A fixed point (equivalently, an equilibrium point) is a zero of $f$ : a number $x^\ast \in I$ with $f(x^\ast) = 0$ . The constant function $x(t) \equiv x^\ast$ is then a solution, and in the fluid picture $x^\ast$ is a place where the fluid is at rest. We adopt Hirsch, Smale and Devaney's definitions of what it means for such a point to be stable. The fixed point $x^\ast$ is

stable if every neighbourhood $O$ of $x^\ast$ contains a neighbourhood $O_1$ of $x^\ast$ such that every solution starting in $O_1$ is defined for all $t > 0$ and stays in $O$ for all $t > 0$ ;
asymptotically stable (a sink, or an attractor) if in addition $O_1$ can be chosen so that every solution starting in it satisfies $x(t) \to x^\ast$ as $t \to \infty$ ;
unstable (a source, or a repeller, when the flow leaves on both sides) if it is not stable.

In phase portraits a sink is drawn as a filled circle and a source as an open one; a fixed point attracting on one side and repelling on the other is drawn half filled and called half-stable.

Method(Linear stability: the sign of

f'

at a fixed point)AdaptedAdapted from Strogatz (1994), §2.4, pp. 24-25; Hirsch, Smale & Devaney (2013), §1.2, pp. 6-7, and §8.4, p. 175

Let $f$ be continuously differentiable on an open interval $I$ and let $f(x^\ast) = 0$ for some $x^\ast \in I$ . Write $\lambda = f'(x^\ast)$ .

If $\lambda < 0$ , then $x^\ast$ is asymptotically stable. Quantitatively: for every $\varepsilon \in (0, -\lambda)$ there is a $\delta > 0$ such that every solution with $\lvert x_0 - x^\ast\rvert \le \delta$ exists for all $t \ge 0$ and obeys $\bigl\lvert x(t) - x^\ast \bigr\rvert \ \le \ \bigl\lvert x_0 - x^\ast \bigr\rvert \, e^{(\lambda + \varepsilon) t} \qquad (t \ge 0) .$
If $\lambda > 0$ , then $x^\ast$ is unstable. Quantitatively: for every $\varepsilon \in (0, \lambda)$ there is a $\delta > 0$ such that any solution starting with $0 < \lvert x_0 - x^\ast\rvert \le \delta$ obeys $\lvert x(t) - x^\ast\rvert \ge \lvert x_0 - x^\ast\rvert e^{(\lambda - \varepsilon)t}$ for as long as it stays within $\delta$ of $x^\ast$ , and therefore leaves that neighbourhood in finite time.
If $\lambda = 0$ , no conclusion follows. All four possibilities occur: $x^\ast = 0$ is asymptotically stable for $\dot x = -x^3$ , unstable for $\dot x = x^3$ , half-stable for $\dot x = x^2$ , and stable but not asymptotically stable for $\dot x = 0$ .

The number $\lvert \lambda\rvert$ is a rate and its reciprocal $1/\lvert\lambda\rvert$ is the characteristic time of the fixed point: the time in which a small deviation from $x^\ast$ changes by a factor $e$ .

Strogatz states the criterion in the form that the perturbation $\eta = x - x^\ast$ obeys $\dot\eta \approx \lambda\eta$ and so grows or decays exponentially at rate $\lambda$ , and identifies $1/\lvert\lambda\rvert$ as the time scale, without the $\varepsilon$ that parts 1 and 2 carry here (pp. 24-25); the $\varepsilon$ is the price of turning the approximation into an inequality valid for the nonlinear equation, and the proof below pays it. Part 1 is also the one-dimensional case of Hirsch, Smale and Devaney's theorem that an equilibrium whose linearisation has every eigenvalue of negative real part is asymptotically stable (p. 175), the single eigenvalue here being $\lambda$ .

Idea of the proof(Why the sign of

f'

decides)Original

Near $x^\ast$ the graph of $f$ is a line of slope $\lambda$ through the zero, bent by an amount that is small compared with the deviation itself. A line of negative slope through $x^\ast$ points towards $x^\ast$ on both sides, and a line of positive slope points away. The only work is to show that the bending cannot overturn the verdict, and the trick is not to track the position but its square: $\frac{d}{dt}(x-x^\ast)^2$ is $(x - x^\ast)^2$ times a coefficient that the mean value theorem exhibits as an average of $f'$ over the segment from $x^\ast$ to $x$ , and continuity of $f'$ pins that average within $\varepsilon$ of $\lambda$ . The differential inequality then integrates to the stated exponential bound, with no linearisation and no neglected terms.

Proof(Linear stability)Original

Write $u(t) = x(t) - x^\ast$ , so $u(0) = u_0 = x_0 - x^\ast$ and $\dot u = f(x^\ast + u)$ .

Step 1: an exact factorisation of the right-hand side. Since $f$ is continuously differentiable and $f(x^\ast) = 0$ , the fundamental theorem of calculus applied to $s \mapsto f(x^\ast + su)$ on $[0,1]$ gives

f(x^\ast + u) \;=\; f(x^\ast + u) - f(x^\ast) \;=\; u \int_0^1 f'(x^\ast + su)\,ds \;=:\; u\, c(u) ,

valid for every $u$ with $x^\ast + u \in I$ . This is exact, not an approximation; $c$ is continuous and $c(0) = f'(x^\ast) = \lambda$ . (It is the integral form of the first-order Taylor expansion of Taylor's theorem, arranged so that the remainder is absorbed into the coefficient rather than left over.)

Step 2: choosing the neighbourhood. Let $\varepsilon > 0$ . Because $f'$ is continuous at $x^\ast$ , there is $\delta > 0$ with $[x^\ast-\delta, x^\ast+\delta] \subset I$ and

\lvert f'(y) - \lambda\rvert < \varepsilon \qquad \text{whenever } \lvert y - x^\ast\rvert \le \delta .

If $\lvert u \rvert \le \delta$ then every point $x^\ast + su$ with $s \in [0,1]$ also lies within $\delta$ of $x^\ast$ , so averaging the displayed bound over $s$ gives

\lambda - \varepsilon \ < \ c(u) \ < \ \lambda + \varepsilon \qquad \text{for } \lvert u\rvert \le \delta . \tag{$\ast$}

Step 3: the differential inequality. Put $w(t) = u(t)^2$ . Wherever the solution satisfies $\lvert u(t)\rvert \le \delta$ ,

\dot w = 2u\dot u = 2u^2 c(u) = 2\,c(u)\,w ,

so by $(\ast)$ , $2(\lambda - \varepsilon) w \le \dot w \le 2(\lambda+\varepsilon)w$ . Multiplying the right-hand inequality by the positive factor $e^{-2(\lambda+\varepsilon)t}$ turns it into $\frac{d}{dt}\bigl[e^{-2(\lambda+\varepsilon)t} w(t)\bigr] \le 0$ , so $e^{-2(\lambda+\varepsilon)t}w(t)$ is non-increasing and

w(t) \ \le \ w(0)\, e^{2(\lambda+\varepsilon)t} , \qquad\text{that is}\qquad \lvert u(t)\rvert \ \le \ \lvert u_0\rvert\, e^{(\lambda+\varepsilon)t} . \tag{$\dagger$}

The same manipulation with the left-hand inequality gives $\lvert u(t)\rvert \ge \lvert u_0\rvert e^{(\lambda-\varepsilon)t}$ .

Step 4: part 1. Suppose $\lambda < 0$ and let $0 < \varepsilon < -\lambda$ , so $\lambda + \varepsilon < 0$ ; take $\delta$ from step 2 and $\lvert u_0\rvert \le \delta$ . As long as the solution stays in $[x^\ast-\delta, x^\ast+\delta]$ , bound $(\dagger)$ holds and gives $\lvert u(t)\rvert \le \lvert u_0\rvert \le \delta$ , so the solution cannot reach the boundary of that interval; since it also stays in a compact subset of $I$ , it does not blow up and is defined for all $t \ge 0$ . Hence $(\dagger)$ holds for all $t \ge 0$ , and as $\lambda + \varepsilon < 0$ it forces $x(t) \to x^\ast$ . Both parts of the definition of asymptotic stability are met, with $O_1 = (x^\ast-\delta, x^\ast+\delta)$ serving for any given $O$ once $\delta$ is shrunk to fit inside $O$ .

Step 5: part 2. Suppose $\lambda > 0$ and let $0 < \varepsilon < \lambda$ , so $\lambda - \varepsilon > 0$ ; take $\delta$ from step 2 and $0 < \lvert u_0\rvert \le \delta$ . While the solution remains in $[x^\ast-\delta, x^\ast+\delta]$ we have $\lvert u(t)\rvert \ge \lvert u_0\rvert e^{(\lambda-\varepsilon)t}$ , whose right-hand side exceeds $\delta$ once $t > (\lambda-\varepsilon)^{-1}\log(\delta/\lvert u_0\rvert)$ . So the solution leaves the interval by that time. Since this happens for starting values arbitrarily close to $x^\ast$ , no neighbourhood $O_1$ of the kind required for stability exists when $O = (x^\ast-\delta, x^\ast+\delta)$ .

Step 6: part 3. Each of the four equations has $f(0) = f'(0) = 0$ . For $\dot x = -x^3$ separation of variables gives $x(t)^{-2} = x_0^{-2} + 2t$ for $x_0 \neq 0$ , so $\lvert x(t)\rvert = \lvert x_0\rvert (1 + 2x_0^2 t)^{-1/2} \to 0$ : asymptotically stable, but decaying like $t^{-1/2}$ rather than exponentially. For $\dot x = x^3$ the same computation gives $x(t)^{-2} = x_0^{-2} - 2t$ , so $\lvert x(t)\rvert \to \infty$ as $t \uparrow 1/(2x_0^2)$ : unstable, indeed blowing up in finite time. For $\dot x = x^2$ the field points right on both sides, so the origin attracts from the left and repels on the right: half-stable, hence not stable. For $\dot x = 0$ every point is a fixed point, so solutions neither approach nor leave: stable, not asymptotically stable. $\square$

Method(Monotonicity, and the impossibility of oscillation)AdaptedAdapted from Strogatz (1994), §2.6, pp. 28-29

Let $f$ be continuously differentiable on an open interval $I$ and let $x(t)$ solve $\dot x = f(x)$ on its maximal interval $(t_-, t_+)$ .

No crossing. If $x(t_1) = x^\ast$ for some $t_1$ and some fixed point $x^\ast$ , then $x(t) \equiv x^\ast$ . Equivalently, a non-constant solution never attains the value of a fixed point, at any finite time.
Monotonicity. $x$ is either strictly increasing on all of $(t_-,t_+)$ , or strictly decreasing on all of it, or constant, according to the sign of $f(x_0)$ .
No periodic solutions. The only periodic solutions of $\dot x = f(x)$ are the constant ones. In particular there are no oscillations: no overshoot, no ringing, no sustained cycle.
The only two fates. As $t \uparrow t_+$ , either $x(t)$ tends to a fixed point of $f$ , or it leaves every compact subinterval of $I$ . If $I = \mathbb{R}$ and $x$ is bounded above (for increasing $x$ ) or below (for decreasing $x$ ), then $t_+ = \infty$ and $x(t)$ converges to a fixed point.

Strogatz states 2, 3 and (for $I = \mathbb{R}$ ) 4 and attributes them to the topology of the line: a point moving monotonically along a line never returns to where it started (pp. 28-29).

Idea of the proof(Why nothing on a line can oscillate)Original

To turn around, a solution must pass through an instant at which its velocity is zero — and its velocity is $f$ evaluated at its own position, so at that instant it is sitting on a fixed point. But a fixed point already carries a solution, the constant one, and uniqueness allows only one solution through a given state. So a solution that touches a fixed point is that constant solution and never moved at all. A moving solution therefore never changes the sign of its velocity: it is monotone, and a monotone function that is not constant cannot return to an earlier value.

The last part is the monotone convergence theorem plus one observation: a monotone solution that converges must be slowing to a stop, and the only places where the velocity vanishes are the fixed points.

Proof(Monotonicity and its consequences)Original

Part 1. Suppose $f(x^\ast) = 0$ and $x(t_1) = x^\ast$ . The constant function $y(t) \equiv x^\ast$ solves $\dot y = f(y)$ with $y(t_1) = x^\ast$ . Since $f$ is continuously differentiable it is locally Lipschitz, so by Existence and uniqueness of solutions the initial-value problem posed at time $t_1$ with value $x^\ast$ has only one solution, and two solutions that agree at one time agree on their whole common interval. Hence $x \equiv x^\ast$ .

Part 2. Set $g(t) = \dot x(t) = f(x(t))$ , a continuous function of $t$ . Suppose $g$ took both a strictly positive and a strictly negative value, say $g(t_a) > 0 > g(t_b)$ with $t_a < t_b$ . By the intermediate value theorem there is $t_c \in (t_a,t_b)$ with $g(t_c) = 0$ , that is, $f(x(t_c)) = 0$ , so $x(t_c)$ is a fixed point. By part 1, $x$ is then constant and $g \equiv 0$ , contradicting $g(t_a) > 0$ . So $g$ does not change sign.

If $g$ is nowhere zero it has one strict sign throughout, and $x$ is strictly monotone. If $g(t_c) = 0$ for some $t_c$ , part 1 makes $x$ constant. In either case the sign is that of $g(0) = f(x_0)$ .

Part 3. Let $x$ be periodic with period $T > 0$ , so $x(t+T) = x(t)$ for all $t$ . Then $x$ is defined for all $t$ and $x(T) = x(0)$ , which is incompatible with strict monotonicity. By part 2, $x$ is constant.

Part 4. Suppose $x$ is increasing (the decreasing case is the same with the inequalities reversed, or follows by applying this one to $-x$ , which solves $\dot y = -f(-y)$ ). Let $J \subset I$ be a compact subinterval, and suppose $x(t) \in J$ for all $t \in [0, t_+)$ . Then $x$ is increasing and bounded above by $\max J$ , so by the monotone convergence theorem $L = \lim_{t \uparrow t_+} x(t)$ exists and lies in $J \subset I$ . If $t_+$ were finite the solution would extend past $t_+$ from the state $L \in I$ , contradicting maximality; so $t_+ = \infty$ .

It remains to show $f(L) = 0$ . Suppose instead $f(L) = m \neq 0$ . Since $x$ increases to $L$ we have $m \ge 0$ , for if $m < 0$ then $f < 0$ on a neighbourhood of $L$ by continuity, and $x$ entering that neighbourhood would have to decrease. So $m > 0$ . By continuity of $f$ there is $\sigma > 0$ with $f(y) > m/2$ for $\lvert y - L\rvert \le \sigma$ , and there is $t_0$ with $L - \sigma < x(t) < L$ for $t \ge t_0$ . Then $\dot x(t) > m/2$ for all $t \ge t_0$ , so $x(t) > x(t_0) + (m/2)(t - t_0) \to \infty$ , contradicting $x < L$ . Hence $f(L) = 0$ and $x(t) \to L$ , a fixed point.

If no compact subinterval contains the trajectory for all time, then by definition $x$ leaves every compact subinterval of $I$ ; when $I = \mathbb{R}$ and $x$ is increasing this means $x(t) \to \infty$ (possibly in finite time, as for $\dot x = 1 + x^2$ ). $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Strogatz (1994), §2.4, Example 2.4.3, pp. 25-26, and §2.5, Example 2.5.1, p. 27

Autonomy — that $f$ does not depend on $t$ — is used everywhere, and is the whole reason a one-dimensional phase space is available. It is needed in part 1 of the monotonicity proof (a constant function is a solution only because $f(x^\ast) = 0$ holds at all times), and it is what makes the velocity a function of position alone, so that the graph of $f$ is a complete description. The non-autonomous equation $\dot x = -x\cos t$ has the oscillating solutions $x_0 e^{-\sin t}$ , so part 3 fails at once without it.
Uniqueness, which continuous differentiability of $f$ supplies, is used exactly once, in part 1 of the monotonicity proof, and everything after it rests on that. Dropping it costs the no-crossing statement: for $\dot x = 3x^{2/3}$ , which has a continuous but non-Lipschitz right-hand side at the origin, the function equal to $0$ for $t \le 0$ and to $t^3$ for $t > 0$ is a solution that sits on the fixed point $x^\ast = 0$ for all negative time and then leaves it. (Monotonicity happens to survive in that particular example because $f \ge 0$ there, but the argument above no longer delivers it.) The related non-uniqueness example $\dot x = x^{1/3}$ is in Existence and uniqueness of solutions.
$f(x^\ast) = 0$ in the stability theorem is what makes the factorisation of step 1 possible; without it the right-hand side has a non-zero constant term and there is no fixed point to be stable.
$f'$ continuous at $x^\ast$ , not merely the existence of $f'(x^\ast)$ , is used in step 2 to make $(\ast)$ hold on a whole interval rather than at the single point $x^\ast$ . The exponential bound is a statement about an interval and needs an interval's worth of control.
$\lambda \neq 0$ is essential and not a technicality: part 3 of the statement exhibits four equations with the same linearisation $\dot\eta = 0$ and four different answers. When $\lambda = 0$ , fall back on the sign of $f$ on each side of $x^\ast$ , which always settles the question; the four equations listed there realise the four possible sign patterns, namely $(+,-)$ for $-x^3$ , $(-,+)$ for $x^3$ , $(+,+)$ for $x^2$ , and $f$ vanishing identically. The degenerate case is not a curiosity: it is precisely where bifurcations happen, which is the subject of Bifurcations in one dimension.
Boundedness in part 4 cannot be dropped. $\dot x = 1 + x^2$ with $x_0 = 0$ has the solution $\tan t$ , which reaches infinity at $t = \pi/2$ : the solution is monotone and has no fixed point to converge to, and $t_+$ is finite.

Method(Reading a one-dimensional flow off the graph of

f

)AdaptedAdapted from Strogatz (1994), §2.2, pp. 18-19, and §2.4, pp. 24-25; Hirsch, Smale & Devaney (2013), §1.2, pp. 6-7

Given $\dot x = f(x)$ with $f$ continuously differentiable on an interval $I$ :

Find the fixed points. Solve $f(x) = 0$ on $I$ . If the equation cannot be solved in closed form, split $f$ into two pieces whose graphs you know, $f = g - h$ , and locate the intersections of $y = g(x)$ and $y = h(x)$ instead; the fixed points are the $x$ -coordinates of the intersections.
Mark the sign of $f$ on each interval between consecutive fixed points. One test value per interval is enough, because $f$ is continuous and vanishes only at the fixed points. Draw a rightward arrow where $f > 0$ and a leftward arrow where $f < 0$ . This is the phase portrait, and it already determines the fate of every initial condition: the flow carries $x_0$ towards the first fixed point in the direction of its arrow, or out of $I$ if there is none.
Classify each fixed point by the arrows on its two sides: in-in is a sink, out-out a source, and in-out or out-in is half-stable.
Get the rates. Compute $\lambda = f'(x^\ast)$ at each fixed point. Negative $\lambda$ confirms a sink, positive a source, and the magnitude gives the characteristic time $1/\lvert\lambda\rvert$ . If $\lambda = 0$ the arrows of step 2 still give the classification; only the rate is lost, and the approach is then slower than any exponential.
Sketch $x$ against $t$ if required. The curve is monotone, with slope $f(x)$ ; it is concave up where the phase point is speeding up (that is, where $f f' > 0$ ) and concave down where it is slowing down, and it approaches a horizontal asymptote at each sink.
Get exact times when you need them. Because the equation is separable, the time to travel from $a$ to $b$ with no fixed point in between is $t(a \to b) = \int_a^b \frac{dx}{f(x)} ,$ which is finite precisely when $f$ does not vanish on the closed interval between the two. Its divergence at a fixed point is the analytic form of part 1 of the theorem above.

Example(A quadratic flow: classification, rates and an exact transit time)Original

Take

\dot x = f(x) = -x^2 + 6x - 8 = -(x-2)(x-4) , \qquad x \in \mathbb{R} .

Step 1: fixed points. $f(x) = 0$ at $x^\ast = 2$ and $x^\ast = 4$ .

Step 2: signs. $f(0) = -8 < 0$ , $f(3) = 1 > 0$ , $f(5) = -3 < 0$ . So the flow is leftward on $(-\infty,2)$ , rightward on $(2,4)$ and leftward on $(4,\infty)$ .

Step 3: classification. The arrows point away from $2$ on both sides and towards $4$ on both sides: $x^\ast = 2$ is a source, $x^\ast = 4$ is a sink.

Step 4: rates. $f'(x) = -2x + 6$ , so

\lambda_2 = f'(2) = 2 > 0 , \qquad \lambda_4 = f'(4) = -2 < 0 ,

confirming step 3 and giving both characteristic times as $1/\lvert\lambda\rvert = 0.5$ . Every solution starting with $x_0 > 2$ converges to $4$ ; every solution with $x_0 < 2$ decreases, and since $f < 0$ and $\lvert f\rvert$ grows quadratically there, it reaches $-\infty$ in finite time. The solution from $x_0 = 2$ is the constant $2$ .

Step 5: an exact transit time. Start at $x_0 = 3$ and ask how long the phase point takes to reach $x = 3.9$ . There is no fixed point in $[3, 3.9]$ , so step 6 of the method applies. Partial fractions give

\frac{1}{(x-2)(4-x)} = \frac{1}{2}\left(\frac{1}{x-2} + \frac{1}{4-x}\right) ,

because $(x-2) + (4-x) = 2$ . Hence

t = \int_3^{3.9} \frac{dx}{(x-2)(4-x)} = \frac{1}{2}\Bigl[\log(x-2) - \log(4-x)\Bigr]_3^{3.9} = \frac{1}{2}\log\frac{x-2}{4-x}\bigg|_3^{3.9} .

At $x = 3$ the bracket is $\tfrac12\log 1 = 0$ ; at $x = 3.9$ it is $\tfrac12\log(1.9/0.1) = \tfrac12\log 19$ . So

\boxed{\;t \;=\; \tfrac{1}{2}\log 19 \;=\; 1.4722194895832202\ldots\;}

Step 6: check it two ways. Numerically integrating $\dot x = -(x-2)(x-4)$ from $x(0) = 3$ with an adaptive solver gives $x(1.47221948958) = 3.8999999999994$ , agreeing to twelve figures. And the linear estimate near the sink predicts a deviation decaying as $e^{-2t}$ , so shrinking the deviation from $1$ (at $x=3$ ) to $0.1$ (at $x=3.9$ ) should take about $\tfrac12\log 10 = 1.1513$ . The true time is $28\%$ longer, because the deviation starts at $1$ , which is not small compared with the distance $2$ between the fixed points, and the flow there is slower than the linearisation at $x^\ast = 4$ suggests. The linear rate is a statement about the final approach, not about the whole journey.

Application(Overdamped motion, gradient flow and the potential)AdaptedAdapted from Strogatz (1994), §2.6, pp. 29-30, and §2.7, pp. 30-31

The physical setting in which one-dimensional flows arise exactly, rather than as a caricature, is overdamped motion: a particle of mass $m$ in a force field $F(x)$ , moving through a medium that exerts a drag $-b\dot x$ , obeys

m\ddot x = F(x) - b\dot x .

When the drag dominates the inertia the $m\ddot x$ term may be dropped and what is left is first order,

\dot x = f(x) , \qquad f = F/b .

Strogatz's picture is a bead in a vat of honey. (The neglect is not uniformly valid: it fails during a brief initial transient in which the inertial and damping terms are comparable, and only after that does the first-order description take over — a singular limit, treated in Boundary layers and matched asymptotics.)

In this reading, part 3 of the monotonicity theorem is a physical statement with no mystery in it: a mass whose inertia has been thrown away cannot overshoot, so it cannot oscillate. Damped oscillation needs the inertia, and undamped oscillation needs the inertia and no damping; both live in the two-dimensional phase plane of The damped and driven harmonic oscillator, not on a line.

The same content is packaged as a potential. Define $V$ by

f(x) = -\frac{dV}{dx} , \qquad \text{so} \qquad V(x) = -\int^x f(s)\,ds

up to an additive constant; every one-dimensional flow has one, because every continuous function has an antiderivative. Along any solution the chain rule gives

\frac{d}{dt} V\bigl(x(t)\bigr) = V'(x)\,\dot x = -f(x)\cdot f(x) = -\bigl[f(x)\bigr]^2 \ \le\ 0 ,

with equality exactly at the fixed points. So $V$ never increases along a trajectory: the system always runs downhill, and the fixed points are the critical points of $V$ , with local minima the sinks and local maxima the sources. This is the statement that every flow on the line is a gradient flow, and it gives a second reason why nothing can cycle — a cycle would have to return $V$ to its starting value, which a strictly decreasing quantity forbids. It is also the point of contact with equilibrium thermodynamics and Landau theory, where $V$ is a free energy and the stable states of the system are its minima; the bifurcations of Bifurcations in one dimension are then the changes in the shape of that free-energy landscape as a control parameter is turned. For the quadratic example above, $V(x) = \tfrac13 x^3 - 3x^2 + 8x$ , with a local maximum at $x = 2$ (value $V = 20/3$ ) and a local minimum at $x = 4$ (value $V = 16/3$ ), exactly matching the source and the sink found there.

The defining relation $f = -dV/dx$ no longer has a single-valued solution once the phase space is a circle and $f$ has no zeros, which is how Flows on the circle escapes this unit's conclusions and manages to oscillate.

Theorem

Bifurcations in one dimension

Adapted

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Proves that a fixed point of $\dot x = f_a(x)$ can only be created, destroyed or change stability where $f_a'$ vanishes, gives the saddle-node theorem with the four derivative conditions that produce it, lists the normal forms of the saddle-node, transcritical and pitchfork bifurcations, and works through an imperfect pitchfork whose hysteresis loop has exactly computable corners.

Prerequisites: Flows on the line, Implicit and inverse function theorems, Taylor expansion and critical points in several variables

Note(What changes when a parameter is turned)Original

On the line the dynamics is poor — by Flows on the line, every solution runs monotonically to a fixed point or off to infinity. What makes one-dimensional systems worth a chapter is what happens to that picture when the equation itself is moved: a family $\dot x = f_a(x)$ , with $a$ a load, a growth rate, a pump strength, a harvesting quota. Almost always a small change of $a$ moves the fixed points slightly and changes nothing qualitative. At isolated values of $a$ it does something else: fixed points appear out of nowhere, collide and annihilate, or swap their stability, and the system's long-term state changes discontinuously. Those values are the bifurcation points, and the short list of ways it can happen in one dimension is the content of this unit.

The list is short for a reason that the first theorem below makes precise. Nothing can happen where $f_a'(x^\ast) \neq 0$ , because the implicit function theorem pins the fixed point down and Flows on the line fixes its type from the sign of $f_a'$ . So every bifurcation lives at a degenerate fixed point, and the possibilities are classified by which derivative is the first not to vanish.

Definition(Families, bifurcations and their diagrams)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, p. 175, and §1.1, p. 4; Strogatz (1994), §3.0, p. 44, and §3.6, pp. 70-71

Let $f : I \times J \to \mathbb{R}$ be smooth, where $I \subseteq \mathbb{R}$ is an interval of states and $J \subseteq \mathbb{R}$ an interval of parameter values. Writing $f_a(x) = f(x,a)$ , the one-parameter family

\dot x = f_a(x) , \qquad a \in J ,

is a differential equation for each fixed $a$ . Throughout, a prime denotes $d/dx$ , so $f_a'(x) = \partial f/\partial x\,(x,a)$ , and derivatives with respect to the parameter are written $\partial f/\partial a$ . A fixed point $x^\ast$ of $f_a$ is called hyperbolic if $f_a'(x^\ast) \neq 0$ and degenerate (or non-hyperbolic) otherwise.

A bifurcation occurs at $a = a_0$ if the qualitative structure of the flow changes there: in every neighbourhood of $a_0$ there are parameter values whose phase portraits are not equivalent, the differences in practice being the number of fixed points, their stability types, or both. The value $a_0$ is a bifurcation point. Strauss's one-line description is that a bifurcation is a fork in a road of solutions that depends on a parameter.

Two pictures carry the information.

The bifurcation diagram plots the fixed points $x^\ast$ vertically against the parameter $a$ horizontally, drawing the locus $\{(a,x) : f_a(x) = 0\}$ with solid curves where the fixed point is stable and broken curves where it is unstable. A vertical slice at $a$ is the phase line of $f_a$ .
When there are two parameters, the stability diagram partitions the parameter plane into the regions on which the number of fixed points is constant, the boundaries between regions being the parameter values at which bifurcations happen.

Theorem(Hyperbolic fixed points persist)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, p. 176

Let $f$ be $C^1$ on $I \times J$ , let $f_{a_0}(x_0) = 0$ and suppose

f_{a_0}'(x_0) \ \neq\ 0 .

Then there are open intervals $I_0 \ni x_0$ and $J_0 \ni a_0$ and a $C^1$ function $p : J_0 \to I_0$ with $p(a_0) = x_0$ such that, for each $a \in J_0$ , the only zero of $f_a$ in $I_0$ is $p(a)$ . Moreover $f_a'(p(a))$ has the same sign as $f_{a_0}'(x_0)$ throughout $J_0$ , so the single fixed point in $I_0$ keeps its type: a sink stays a sink and a source stays a source.

Consequently a bifurcation of $\dot x = f_a(x)$ at $a = a_0$ requires a degenerate fixed point: it can only occur at a point $x_0$ with $f_{a_0}(x_0) = 0$ and $f_{a_0}'(x_0) = 0$ .

Idea of the proof(Persistence)Original

Fixed points are the zeros of $f_a$ , and a zero at which the graph crosses the axis transversally cannot be removed by a small deformation of the graph: it can only slide. The implicit function theorem is exactly the statement that it slides smoothly, and continuity of $f'$ then says the crossing stays a crossing of the same sense.

Proof(Persistence)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, p. 176

Apply the implicit function theorem to $F(x,a) = f(x,a)$ at $(x_0,a_0)$ . We have $F(x_0,a_0) = 0$ and $\partial F/\partial x\,(x_0,a_0) = f_{a_0}'(x_0) \neq 0$ , which is the hypothesis the theorem needs to solve for $x$ in terms of $a$ . It therefore supplies open intervals $I_0 \ni x_0$ , $J_0 \ni a_0$ and a $C^1$ map $p : J_0 \to I_0$ with $p(a_0) = x_0$ , such that for $(x,a) \in I_0 \times J_0$ the equation $F(x,a) = 0$ holds exactly when $x = p(a)$ . That is the uniqueness assertion.

For the sign, note that $(a) \mapsto f_a'(p(a))$ is continuous on $J_0$ , being a composition of the continuous $\partial F/\partial x$ with the continuous $a \mapsto (p(a),a)$ , and it is non-zero at $a_0$ . Shrinking $J_0$ so that it stays non-zero — possible by continuity — it keeps the sign it has at $a_0$ , since a continuous real function that changes sign on an interval must vanish somewhere by the intermediate value theorem. By the linear stability criterion of Flows on the line the type of $p(a)$ is read off that sign, so the type does not change.

The consequence follows by contraposition: if $f_{a_0}'(x_0) \neq 0$ at every zero $x_0$ of $f_{a_0}$ , then near each of them nothing changes as $a$ varies near $a_0$ , so no bifurcation occurs there. $\square$

Theorem(The saddle-node bifurcation)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, Theorem (Saddle-Node Bifurcation) and the three-part description preceding it, pp. 176-177

Definition (Hirsch, Smale and Devaney, p. 176). The family $\dot x = f_a(x)$ undergoes a saddle-node bifurcation at $a = a_0$ when a window $I \times J$ can be chosen — $I$ an interval of states containing the point where the collision happens, $J$ an interval of parameter values containing $a_0$ — in which the count of fixed points is

\#\{x \in I : f_a(x) = 0\} \;=\; \begin{cases} 2, & a \in J \text{ on one side of } a_0, \\ 1, & a = a_0, \\ 0, & a \in J \text{ on the other side.}\end{cases}

(Hirsch, Smale and Devaney tabulate the three counts with two equilibria for $a < a_0$ and none for $a > a_0$ , and then add that the two sides may be exchanged; since the theorem's hypotheses do not say which orientation occurs, the definition above allows either.)

Theorem. Suppose $\dot x = f_a(x)$ is a first-order differential equation, with $f$ smooth in $x$ and $a$ jointly, for which

\text{(i) } f_{a_0}(x_0) = 0, \quad \text{(ii) } f_{a_0}'(x_0) = 0, \quad \text{(iii) } f_{a_0}''(x_0) \neq 0, \quad \text{(iv) } \frac{\partial f_a}{\partial a}(x_0) \neq 0 \text{ at } a = a_0 .

Then the equation undergoes a saddle-node bifurcation at $a = a_0$ .

The two equilibria that exist on one side of $a_0$ are one sink and one source: they are the two branches of a curve with a vertical tangent at $(a_0,x_0)$ in the bifurcation diagram, and they collide and annihilate as $a$ passes through $a_0$ . The name is borrowed from the planar version, in which the colliding points are a saddle and a node; on a line there is nothing deserving the word saddle, as Strogatz notes (p. 47), and the bifurcation also goes by fold, turning-point and blue sky.

Idea of the proof(Saddle-node)Original

Hypothesis (ii) is the obstruction to solving $f_a(x) = 0$ for $x$ in terms of $a$ , which is why the picture is interesting at all. Hypothesis (iv) says we may solve the other way round, for $a$ in terms of $x$ : the set of fixed points near $(x_0,a_0)$ is the graph of a function $a = a(x)$ , drawn sideways in the bifurcation diagram. Differentiating that relation twice turns the remaining two hypotheses into statements about the shape of this graph — (ii) makes it horizontal at $x_0$ , (iii) makes it curve — so it is a parabola-like arc with a strict extremum at $x_0$ . A horizontal line $a = \text{const}$ meets such an arc twice on the curved side, once at the extremum, and not at all beyond it. Counting intersections is counting fixed points, and that is the three-case conclusion.

Proof(Saddle-node)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, pp. 177-178

Write $G(x,a) = f(x,a)$ , so that fixed points of $f_a$ are the solutions of $G = 0$ .

Step 1: invert the roles of $x$ and $a$ . By (i), $G(x_0,a_0) = 0$ , and by (iv),

\frac{\partial G}{\partial a}(x_0,a_0) = \frac{\partial f_a}{\partial a}(x_0)\bigg|_{a=a_0} \neq 0 .

This is the hypothesis the implicit function theorem needs in order to solve for the second variable. It gives open intervals $I \ni x_0$ and $J \ni a_0$ and a smooth $a : I \to J$ with $a(x_0) = a_0$ such that, for $(x,\alpha) \in I \times J$ ,

G(x,\alpha) = 0 \iff \alpha = a(x) . \tag{1}

So inside the window $I \times J$ the fixed points of $f_\alpha$ are exactly the preimages $a^{-1}(\alpha)$ , and the theorem is now a question about the shape of the single function $a$ .

Step 2: the curve is horizontal at $x_0$ . Differentiate the identity $G(x,a(x)) \equiv 0$ once with respect to $x$ :

\frac{\partial G}{\partial x} + \frac{\partial G}{\partial a}\,a'(x) = 0 , \tag{2}

all partials evaluated at $(x,a(x))$ . At $x = x_0$ the first term is $f_{a_0}'(x_0) = 0$ by (ii), and $\partial G/\partial a \neq 0$ by step 1, so

a'(x_0) = 0 .

Step 3: the curve is strictly curved at $x_0$ . Differentiate (2) again, remembering that both partials are evaluated along the curve, so each contributes two terms by the chain rule:

\frac{\partial^2 G}{\partial x^2} + 2\,\frac{\partial^2 G}{\partial x \,\partial a}\,a'(x) + \frac{\partial^2 G}{\partial a^2}\,\bigl(a'(x)\bigr)^2 + \frac{\partial G}{\partial a}\,a''(x) = 0 .

Evaluate at $x = x_0$ and use $a'(x_0) = 0$ from step 2: the middle two terms drop out, leaving

a''(x_0) = -\,\frac{\partial^2 G/\partial x^2\,(x_0,a_0)}{\partial G/\partial a\,(x_0,a_0)} = -\,\frac{f_{a_0}''(x_0)}{\partial f_a/\partial a\,(x_0)} \ \neq\ 0

by (iii) and (iv). (Hirsch, Smale and Devaney reach the same value at $x_0$ by differentiating the quotient $a'(x) = -(\partial G/\partial x)/(\partial G/\partial a)$ obtained from (2); p. 177.)

Step 4: count the intersections. By steps 2 and 3, $a$ has a critical point at $x_0$ with non-zero second derivative, so by the second-derivative test it is a strict local extremum — a strict local minimum if $a''(x_0) > 0$ , a strict local maximum if $a''(x_0) < 0$ . Take the minimum case; reversing the sign of the parameter axis converts the maximum case into it, and that exchange of the two sides is the second orientation allowed by the definition. Shrink $I$ to $(x_0 - \rho, x_0 + \rho)$ on which $a'$ is negative for $x < x_0$ and positive for $x > x_0$ — possible because $a''$ is continuous and positive at $x_0$ , so $a'$ increases strictly near $x_0$ and vanishes only there. Then $a$ decreases strictly on $(x_0-\rho, x_0]$ and increases strictly on $[x_0, x_0+\rho)$ . Put $\mu = \min\{\lim_{x \downarrow x_0-\rho} a(x), \lim_{x \uparrow x_0+\rho} a(x)\}$ and let $\alpha$ range over $(a_0, \mu)$ . For such $\alpha$ the equation $a(x) = \alpha$ is solved exactly once on each of the two monotone pieces, hence twice in $I$ ; at $\alpha = a_0$ only $x = x_0$ solves it; and for $\alpha < a_0$ nothing does, since $a \ge a_0$ throughout $I$ . By (1) these solutions are the fixed points of $f_\alpha$ lying in $I$ , so the counts are $2$ , $1$ , $0$ as $\alpha$ moves up through $a_0$ — the definition's pattern, with its two sides exchanged.

Step 5: the two fixed points are a sink and a source. For $\alpha$ slightly above $a_0$ let $x_- < x_0 < x_+$ be the two fixed points. By (2), $f_\alpha'(x_\pm) = \partial G/\partial x = -a'(x_\pm)\,\partial G/\partial a$ , and $a'(x_-) < 0 < a'(x_+)$ by the monotonicity in step 4 while $\partial G/\partial a$ keeps one sign near $(x_0,a_0)$ . So $f_\alpha'(x_-)$ and $f_\alpha'(x_+)$ have opposite signs, and the stability criterion of Flows on the line makes one of them a sink and the other a source. $\square$

Theorem(The three normal forms)AdaptedAdapted from Strogatz (1994), §3.1, pp. 45-49, §3.2, pp. 50-51, and §3.4, pp. 55-59

The following four families are the prototypes, or normal forms, of the bifurcations named. In each, $r$ is the parameter and the bifurcation is at $r = 0$ .

Saddle-node, $\dot x = r + x^2$ . For $r < 0$ there are two fixed points $x^\ast = \pm\sqrt{-r}$ , with $-\sqrt{-r}$ a sink and $+\sqrt{-r}$ a source; at $r = 0$ they merge into the half-stable point $x^\ast = 0$ ; for $r > 0$ there are none and every solution runs to $+\infty$ . (Strogatz's other prototype $\dot x = r - x^2$ is the same picture reflected, with $+\sqrt{r}$ a sink and $-\sqrt{r}$ a source for $r > 0$ .)
Transcritical, $\dot x = rx - x^2$ . There are fixed points $x^\ast = 0$ and $x^\ast = r$ for every $r$ , and $f'(0) = r$ while $f'(r) = -r$ . So for $r < 0$ the origin is a sink and $x^\ast = r$ a source; for $r > 0$ the origin is a source and $x^\ast = r$ a sink. The two points do not disappear at $r = 0$ ; they pass through each other and swap type, which is called an exchange of stabilities.
Supercritical pitchfork, $\dot x = rx - x^3$ . The origin is a fixed point for all $r$ , with $f'(0) = r$ : a sink for $r < 0$ and a source for $r > 0$ . For $r > 0$ two further fixed points $x^\ast = \pm\sqrt{r}$ exist, both sinks since $f'(\pm\sqrt r) = -2r < 0$ . One branch becomes three, hence the name. The equation is unchanged by $x \mapsto -x$ , and the pair of new fixed points is symmetric under that reflection while each one individually is not: a broken symmetry.
Subcritical pitchfork, $\dot x = rx + x^3$ . Again the origin is a sink for $r < 0$ and a source for $r > 0$ , but now the extra fixed points $x^\ast = \pm\sqrt{-r}$ exist for $r < 0$ and are sources, since $f'(\pm\sqrt{-r}) = -2r > 0$ . For $r > 0$ nothing bounds the motion and $x(t)$ reaches infinity in finite time from every $x_0 \neq 0$ .

Why "normal form". Let $f(x,r)$ be smooth and suppose a saddle-node bifurcation occurs at $(x^\ast, r_c)$ , so $f(x^\ast,r_c) = 0$ and $\partial f/\partial x\,(x^\ast,r_c) = 0$ . Expanding by Taylor's theorem in several variables about that point and discarding the terms of order $(r-r_c)^2$ and $(x - x^\ast)^3$ and higher,

\dot x \;=\; \underbrace{f(x^\ast,r_c)}_{=\,0} \;+\; (x - x^\ast)\underbrace{\frac{\partial f}{\partial x}}_{=\,0} \;+\; (r-r_c)\frac{\partial f}{\partial r} \;+\; \tfrac12 (x-x^\ast)^2 \frac{\partial^2 f}{\partial x^2} \;+\;\cdots \;=\; \alpha\,(r - r_c) + \beta\,(x-x^\ast)^2 + \cdots ,

all derivatives evaluated at $(x^\ast,r_c)$ , with $\alpha = \partial f/\partial r$ and $\beta = \tfrac12\,\partial^2 f/\partial x^2$ . Those are exactly the quantities that hypotheses (iv) and (iii) of the saddle-node theorem require to be non-zero, and rescaling $x$ and $r$ then turns the displayed equation into $\dot x = \pm 1 \cdot r + x^2$ . Strogatz presents this expansion as the reason the prototypes deserve the name (pp. 49-50), and notes that a complete theory of normal forms does more than this, producing a change of variable that makes the reduction exact rather than approximate (p. 52).

Remark(Where the hypotheses are used, and what the degenerate cases are for)AdaptedAdapted from Hirsch, Smale & Devaney (2013), §8.5, pp. 177-178; Strogatz (1994), §3.2, p. 50, and §3.4, pp. 55-56

(i) $f_{a_0}(x_0) = 0$ is what puts a fixed point at $(x_0,a_0)$ for anything to happen to.
(ii) $f_{a_0}'(x_0) = 0$ is used in step 2, and it is also the content of the persistence theorem: without it there is no bifurcation at all, just a fixed point sliding smoothly.
(iii) $f_{a_0}''(x_0) \neq 0$ is used only in step 3, to make the curve of fixed points genuinely curved. Drop it and the conclusion fails: for $\dot x = a + x^3$ the fixed point $x^\ast = -a^{1/3}$ exists and is unique for every $a$ , so there is never more than one and no saddle-node occurs, although (i), (ii) and (iv) all hold at $(0,0)$ .
(iv) $\partial f_a/\partial a\,(x_0) \neq 0$ is used in step 1 and in step 5, and it is what the transcritical and pitchfork families violate. For $\dot x = rx - x^2$ and $\dot x = rx - x^3$ at $(x_0,r_c) = (0,0)$ we have $\partial f/\partial r = x$ , which vanishes at $x_0 = 0$ . So those two bifurcations are consistent with the theorem; they simply are not covered by it.
Why the saddle-node is nevertheless the typical case. Hirsch, Smale and Devaney's genericity argument (pp. 177-178): a bifurcation forces the two equalities (i) and (ii), which is already two conditions; having imposed them, the next derivatives $f''$ and $\partial f/\partial a$ will in general be non-zero, and those are exactly (iii) and (iv). Vanishing of either is an extra coincidence, and a coincidence needs a reason.
The reasons, in practice, are two. A constraint — a fixed point that must exist for every parameter value, like the zero-population state of a growth model or the zero-amplitude state of an oscillator — forces $f(x_0,a) = 0$ for all $a$ , hence $\partial f/\partial a\,(x_0) = 0$ , and the generic bifurcation of such a family is transcritical (Strogatz, p. 50). A symmetry $f_a(-x) = -f_a(x)$ , as in a system with a left-right symmetry, forces all even-order $x$ -derivatives at $x_0 = 0$ to vanish, killing $f''$ as well, and the generic bifurcation is then a pitchfork (Strogatz, pp. 55-56). The three normal forms are therefore not three unrelated accidents: they are the generic case, the generic case with one fixed point pinned, and the generic case with a reflection symmetry.
What breaks a pitchfork. Because the pitchfork needs an exact symmetry, it is destroyed by any imperfection that breaks it. The example below does this explicitly, and what is left is a pair of saddle-node bifurcations — the generic case reasserting itself.
Critical slowing down. At the bifurcation point the linearisation vanishes, $f_r'(x^\ast) = 0$ , so the stability criterion of Flows on the line gives no rate and the approach to the fixed point is not exponential. For $\dot x = -x^3$ , which is the supercritical pitchfork at $r = 0$ , the decay is like $t^{-1/2}$ . Physicists call the resulting divergence of the relaxation time near a bifurcation critical slowing down (Strogatz, p. 56).

Method(Locating and identifying a bifurcation)AdaptedAdapted from Strogatz (1994), §3.1, Example 3.1.2, pp. 47-48, and §3.4, Example 3.4.1, pp. 57-58; Hirsch, Smale & Devaney (2013), §8.5, pp. 176-178

Given $\dot x = f(x,a)$ :

Find the candidate points. Solve the pair of equations $f(x,a) = 0 , \qquad \frac{\partial f}{\partial x}(x,a) = 0$ simultaneously. By the persistence theorem these are the only places a bifurcation can occur. Geometrically, the second equation says the graph of $f_a$ is tangent to the $x$ -axis.
If the fixed points cannot be found in closed form, split and intersect. Write $f(x,a) = g(x,a) - h(x)$ with the two graphs known, and read fixed points off their intersections and the flow direction off which graph is higher; the bifurcation is where the graphs become tangent, which is the pair of conditions $g = h$ , $\partial g/\partial x = h'$ .
If the parameter enters more simply than the state does, solve for the parameter. The locus $f(x,a) = 0$ is often the graph of a function $a = a(x)$ even where it is not the graph of $x = x(a)$ ; tabulating $(x, a(x))$ plots the whole bifurcation diagram with no root-finding at all. This is both the computational shortcut and the proof route of the saddle-node theorem.
Test the four conditions (i)-(iv) at each candidate $(x_0,a_0)$ . All four satisfied: saddle-node, and the bifurcation diagram has a vertical tangent there.
If (iv) fails, look for the reason. Check whether $x_0$ is a fixed point for all $a$ (expect transcritical: confirm by checking $\partial^2 f/\partial x^2 \neq 0$ and $\partial^2 f/\partial x\partial a \neq 0$ at the point) or whether $f_a$ is odd in $x - x_0$ (expect pitchfork: the sign of $\partial^3 f/\partial x^3$ then distinguishes supercritical, negative, from subcritical, positive).
Reduce to a normal form by translating the bifurcation point to the origin, $u = x - x_0$ and $s = a - a_0$ , and expanding. The leading terms identify the type; the coefficients give the scale of the new fixed points, for instance $u^\ast \approx \pm\sqrt{-\alpha s/\beta}$ for the saddle-node form $\dot u = \alpha s + \beta u^2$ .
Draw the diagram and mark each branch stable or unstable using the sign of $f_a'$ along it. With two parameters, draw the stability diagram in the parameter plane as well, its boundaries being the curves found in step 1 with $x$ eliminated.

Example(An imperfect pitchfork and its hysteresis loop)Original

Take the two-parameter family

\dot x = f(x,h) = h + rx - x^3

with $r$ held fixed at $r = 3$ and the imperfection parameter $h$ as the control:

\dot x = h + 3x - x^3 .

At $h = 0$ this is the supercritical pitchfork at $r = 3 > 0$ , with a source at $x^\ast = 0$ ( $f' = 3$ ) and sinks at $x^\ast = \pm\sqrt3$ ( $f' = 3 - 9 = -6$ ). Any $h \neq 0$ destroys the symmetry $x \mapsto -x$ and so, by the last remark, must destroy the pitchfork. Here is what replaces it.

Step 1: the candidate points. Step 1 of the method asks for the simultaneous solutions of

h + 3x - x^3 = 0 , \qquad \frac{\partial f}{\partial x} = 3 - 3x^2 = 0 .

The second gives $x_0 = \pm 1$ , and substituting into the first gives $h_0 = x_0^3 - 3x_0 = \mp 2$ . So there are exactly two candidates,

(x_0, h_0) = (-1, \,2) \qquad\text{and}\qquad (x_0,h_0) = (1,\,-2) .

(These agree with Strogatz's general formula $h_c(r) = \tfrac{2r}{3}\sqrt{r/3}$ for the critical imperfection, which at $r = 3$ gives $h_c = 2\cdot 1 = 2$ .)

Step 2: test the four conditions at $(-1,2)$ . With $f(x,h) = h + 3x - x^3$ :

f(-1,2) = 2 - 3 + 1 = 0 ,\quad f'(-1) = 3 - 3 = 0 ,\quad f''(-1) = -6x\big|_{x=-1} = 6 \neq 0 ,\quad \frac{\partial f}{\partial h} = 1 \neq 0 .

All four hold, so the saddle-node theorem applies: a saddle-node bifurcation at $h = 2$ . The same four hold at $(1,-2)$ , with $f''(1) = -6$ .

Step 3: the normal form. Put $u = x + 1$ and $s = h - 2$ and expand exactly:

\dot u = (2+s) + 3(-1+u) - (-1+u)^3 = (2+s) + (-3+3u) - (-1 + 3u - 3u^2 + u^3) = s + 3u^2 - u^3 .

To leading order $\dot u = s + 3u^2$ , the saddle-node normal form with $\alpha = 1$ and $\beta = 3$ . These match the formulas of the preceding statement: $\alpha = \partial f/\partial h = 1$ and $\beta = \tfrac12 f''(-1) = \tfrac12\cdot 6 = 3$ . The two fixed points for $s < 0$ sit at $u^\ast \approx \pm\sqrt{-s/3}$ .

Step 4: the fixed points at the two critical values, exactly. At $h = 2$ ,

2 + 3x - x^3 = -\bigl(x^3 - 3x - 2\bigr) = -(x+1)^2 (x - 2) ,

so the fixed points are the double root $x = -1$ and the simple root $x = 2$ . At $h = -2$ , replacing $x$ by $-x$ gives $-2 + 3x - x^3 = -(x-1)^2(x+2)$ : a double root at $x = 1$ and a simple root at $x = -2$ . These are exact, not numerical.

Step 5: the hysteresis loop. Sweep $h$ slowly upwards from $h = -3$ , with the state always at whichever sink it currently occupies.

At $h = -3$ the cubic $x^3 - 3x + 3$ has discriminant $-4(-3)^3 - 27\cdot 3^2 = 108 - 243 = -135 < 0$ , so there is a single real root, $x^\ast = -2.10380340\ldots$ , and it is the only fixed point. The state sits there.
At $h = -2$ a second, half-stable, fixed point is born at $x = 1$ ; just above $h = -2$ it splits into a source and a sink, but the state stays on the lower sink, which it has been tracking continuously.
Through the whole range $-2 < h < 2$ there are three fixed points and the state remains on the lower branch, which at $h = 0$ is $x = -\sqrt3$ .
At $h = 2$ the lower sink collides with the source at $x = -1$ and both vanish. The state, left with no local equilibrium, is carried by the flow to the only fixed point remaining, $x = 2$ . It jumps from $x = -1$ to $x = 2$ , a discontinuous change of amplitude $3$ in response to an arbitrarily small change in $h$ .
Now sweep $h$ back down. The state stays on the upper branch past $h = 2$ , past $h = 0$ (where the upper branch is $x = +\sqrt3$ ), and all the way to $h = -2$ , where that branch collides with the source at $x = 1$ and disappears. The state then jumps down from $x = 1$ to $x = -2$ , again by $3$ .

The sweep up and the sweep down therefore follow different paths: the response depends on the history, not only on the present value of $h$ . The loop's two corners are at

\boxed{\;(h,x) = (2,-1) \to (2,2) \quad\text{and}\quad (h,x) = (-2,1) \to (-2,-2) ,\;}

so the hysteresis loop has width $\Delta h = 4$ in the control parameter and jump height $\Delta x = 3$ at each end.

Step 6: check. Each jump must start at a double root and end at the remaining simple root of the same cubic, and the factorisations in step 4 confirm both: $-(x+1)^2(x-2)$ at $h = 2$ and $-(x-1)^2(x+2)$ at $h = -2$ . The two cubics are related by $x \mapsto -x$ , $h \mapsto -h$ , which is the residual symmetry of the family and explains why the two jumps have the same height. The normal-form prediction of step 3 can also be checked against the exact roots: at $s = -0.03$ it gives $u^\ast = \pm\sqrt{0.03/3} = \pm 0.1000$ , so $x^\ast \approx -1.1000$ and $-0.9000$ , while the exact roots of $1.97 + 3x - x^3 = 0$ near $-1$ are $-1.09840$ and $-0.89826$ — right to three figures, with the error coming from the discarded $-u^3$ . (The third root, $1.99666$ , is the far branch that the two colliding points will jump to.)

Step 7: the stability diagram. Letting $r$ vary again, the same calculation gives tangency at $x_0 = \pm\sqrt{r/3}$ and hence $h = \pm h_c(r)$ with $h_c(r) = \tfrac{2r}{3}\sqrt{r/3}$ , defined for $r > 0$ . The two curves bound a wedge in the $(r,h)$ plane inside which there are three fixed points and outside which there is one; they meet tangentially at the origin, which is Strogatz's cusp point (p. 70), and the surface of fixed points above the $(r,h)$ plane is the cusp catastrophe surface, whose fold is what the state falls off when it jumps.

Application(Thresholds, buckling and the catastrophe picture)AdaptedAdapted from Strogatz (1994), §3.3, pp. 53-55, §3.4, p. 55, and §3.6, pp. 72-73; Strauss (2008), §14.4, Example 1, pp. 402-403

A transcritical bifurcation: the laser threshold. Strogatz's simplified laser model (after Haken) tracks the number $n(t)$ of photons in the cavity. Stimulated emission creates photons through chance encounters between a photon and an excited atom, so the gain is proportional to the product $nN$ , where $N$ counts the excited atoms; loss through the end mirrors is proportional to $n$ alone:

\dot n = G n N - k n .

Emission de-excites the atoms, so if the pump holds $N_0$ atoms excited in the absence of lasing, the actual number is $N = N_0 - \alpha n$ . Substituting,

\dot n = (G N_0 - k)\,n - \alpha G\, n^2 ,

which is the transcritical normal form with $r = GN_0 - k$ (and the quadratic coefficient $\alpha G$ , removable by rescaling $n$ ). The state $n^\ast = 0$ exists for every pump strength — no photons is always an equilibrium, the constraint that forces the transcritical case — and it is a sink while $N_0 < k/G$ : the device is a lamp. At $N_0 = k/G$ the origin loses stability to the branch

n^\ast = \frac{G N_0 - k}{\alpha G} ,

which is positive and stable beyond the threshold: the device lases. The threshold is not a property of any one atom but of the collective equation, and it is an exchange of stabilities between two branches that cross.

A pitchfork: buckling. Load a vertical column with a weight $\lambda$ . For small $\lambda$ the straight configuration is stable; past a critical load it goes unstable and the column buckles, left or right with equal probability. The governing equation has the left-right symmetry that forces a pitchfork, and the two buckled states are the symmetry-broken pair. Strauss works the corresponding boundary-value problem, $u'' + \lambda \sin u = 0$ on $[0,l]$ with $u'(0) = u'(l) = 0$ for the bending angle $u(x)$ , and reports that linearising $\sin u \approx u$ gives the eigenvalues $\lambda_n = n^2\pi^2/l^2$ , that the nonlinear problem bends each of these vertical lines into a branch leaning to the right, and that the first branch, at $\lambda = \pi^2/l^2$ , is a pitchfork whose two buckled states are the stable ones while the trivial state and the higher branches are unstable (pp. 402-403). The dynamical variable there is a function rather than a number, so this is not literally a one-dimensional flow; the point is that the normal forms of this unit describe the bifurcation anyway, which is what makes them worth classifying.

Why hysteresis matters. The worked example above is the generic fate of a symmetric system with an imperfection, and its behaviour is the one engineers and physicists have to design around. The response to a slowly swept control is smooth and reversible as long as the state stays on one sheet of the fixed-point surface; at the fold it falls to the other sheet, discontinuously and irreversibly, and recovering the original state requires sweeping the control well past its original value. Strogatz's examples are a bead on a tilted wire and the outbreak of a forest pest; the Josephson junction of Flows on the circle shows the same loop in its current-voltage curve once its capacitance is restored. The term catastrophe for the jump, and the cusp surface that organises the whole picture, come from this geometry (pp. 72-73). In the language of Flows on the line, the potential $V$ with $f = -V'$ develops a second well as the parameter is turned; the state sits in whichever well it is in until that well disappears, and only then does it find the other.

Method

Flows on the circle

Adapted

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Shows that an autonomous flow whose phase space is a circle either has fixed points, in which case every solution locks onto one, or has none, in which case every solution is periodic with the same period $\oint d\theta/f(\theta)$ ; works out the nonuniform oscillator $\dot\theta = \omega - a\sin\theta$ completely, including the square-root divergence of the period at the saddle-node, and reads off the current-voltage curve of an overdamped Josephson junction.

Prerequisites: Flows on the line, Bifurcations in one dimension, Trigonometric integrals by residues

Note(What the circle adds)Original

The conclusion of Flows on the line was negative: a first-order autonomous equation on $\mathbb{R}$ cannot oscillate, because a monotone point on a line never comes back. Bend the line into a circle and the obstruction is gone. The motion is still monotone — the phase still never reverses — but now going forward for long enough returns the system to its starting state. That is all it takes, and it makes the circle the smallest phase space on which anything can repeat.

What results is a model of an oscillator in which there is no amplitude at all, only a phase. That sounds like a loss and is in fact the point: a great many real oscillators — a pendulum driven past its stall torque, a Josephson junction above its critical current, a neuron, a flashing firefly — settle onto an amplitude of their own and then do nothing further except advance in phase. For those systems, the one variable that matters lives on a circle, and the whole of their behaviour, including whether they lock onto a driving signal and how long a cycle takes, is in this unit.

Definition(Vector fields on the circle)AdaptedAdapted from Strogatz (1994), §4.1, pp. 94-95, and §4.2, p. 95

Write $S^1$ for the circle, whose points we label by a real number $\theta$ with the understanding that $\theta$ and $\theta + 2\pi k$ name the same point for every integer $k$ . Such a label is called a phase or an angle.

A vector field on the circle assigns to each point of $S^1$ one velocity. In practice it is given by a first-order system

\dot\theta = f(\theta)

in which $f : \mathbb{R} \to \mathbb{R}$ is continuously differentiable and $2\pi$ -periodic, $f(\theta + 2\pi) = f(\theta)$ for all $\theta$ . The periodicity is what makes the assignment well defined: the labels $\theta$ and $\theta+2\pi k$ refer to one point, and they must be given one velocity. Without it the prescription is incoherent — Strogatz's example is $\dot\theta = \theta$ , which would assign the velocity $0$ and the velocity $2\pi$ to the same point of the circle (p. 94). Restricting the label to $(-\pi,\pi]$ does not repair it; it merely moves the inconsistency to a jump in the velocity at $\theta = \pi$ .

A fixed point is a point of $S^1$ at which $f$ vanishes, and it is classified exactly as on the line, by the sign of $f'$ there; see Flows on the line. A periodic solution is a non-constant solution with $\theta(t+T) = \theta(t)$ as points of the circle for some $T > 0$ , that is, with $\theta(t+T) - \theta(t)$ a fixed non-zero multiple of $2\pi$ ; the least such $T$ is the period.

The simplest case, $f \equiv \omega$ with $\omega > 0$ constant, is the uniform oscillator: $\theta(t) = \omega t + \theta_0$ , which runs round the circle at constant angular speed and has period $T = 2\pi/\omega$ . Note there is no amplitude variable anywhere in the formulation. The nonuniform oscillator is the one-parameter deformation

\dot\theta = \omega - a\sin\theta , \qquad \omega > 0 ,\ a \ge 0 ,

in which $\omega$ is the mean of the velocity round the circle and $a$ its amplitude of variation; Strogatz lists phase-locked loops, oscillating neurons, charge-density waves, Josephson junctions and the overdamped driven pendulum among the places it arises (p. 96).

Method(The dichotomy for flows on the circle)AdaptedAdapted from Strogatz (1994), §4.0, p. 93, §4.1, p. 94, and §4.3, pp. 97-98

Let $f$ be continuously differentiable and $2\pi$ -periodic, and consider $\dot\theta = f(\theta)$ on $S^1$ . Exactly one of the following holds.

$f$ has a zero. Then there are no periodic solutions, and every solution converges to a fixed point as $t \to +\infty$ and to a (generally different) fixed point as $t \to -\infty$ . No solution is ever in danger of running off to infinity, because the circle is compact; the long-time state is always a phase-locked one, at rest on the circle.
$f$ has no zero. Then $f$ has one sign throughout, every solution is periodic, and all of them have the same period $T \;=\; \left\lvert\, \int_0^{2\pi} \frac{d\theta}{f(\theta)} \,\right\rvert ,$ which is finite. The phase advances (if $f > 0$ ) or retreats (if $f < 0$ ) by exactly $2\pi$ in each period.

The nonuniform oscillator. For $f(\theta) = \omega - a\sin\theta$ with $\omega > 0$ and $a \ge 0$ the two cases are separated by $a = \omega$ :

$a < \omega$ : no fixed points, and case 2 applies with the explicit period $T \;=\; \frac{2\pi}{\sqrt{\omega^2 - a^2}} .$ At $a = 0$ this is the uniform oscillator's $2\pi/\omega$ ; as $a$ increases the oscillation becomes more and more uneven, fastest at $\theta = -\pi/2$ and slowest at $\theta = \pi/2$ , and $T$ increases, diverging as $a \uparrow \omega$ .
$a = \omega$ : a single, half-stable fixed point at $\theta^\ast = \pi/2$ , and no periodic solutions.
$a > \omega$ : two fixed points, given by $\sin\theta^\ast = \omega/a$ , with $f'(\theta^\ast) = -a\cos\theta^\ast$ ; the one with $\cos\theta^\ast > 0$ , namely $\theta^\ast = \arcsin(\omega/a) \in (0,\pi/2)$ , is a sink, and the one with $\cos\theta^\ast < 0$ , namely $\pi - \arcsin(\omega/a)$ , is a source. Every solution but the source itself converges to the sink.

Passing $a$ through $\omega$ is therefore a saddle-node bifurcation in the sense of Bifurcations in one dimension, and it is the mechanism by which oscillation starts and stops here.

The bottleneck law. Let $f$ be as in case 2 and suppose its minimum value is small: $f(\theta_m) = r > 0$ with $r$ small, $f'(\theta_m) = 0$ and $c = f''(\theta_m) > 0$ . Then

T \;=\; \pi\sqrt{\frac{2}{r\,c}}\;\bigl(1 + o(1)\bigr) \qquad (r \to 0^+) ,

so the period diverges like $r^{-1/2}$ , that is, like the inverse square root of the distance to the saddle-node. The whole of the period is spent creeping through a narrow neighbourhood of $\theta_m$ — the bottleneck left behind by the two fixed points that have just annihilated, which Strogatz calls the saddle-node ghost (p. 99). For the nonuniform oscillator, $\theta_m = \pi/2$ , $r = \omega - a$ and $c = a$ , and the law reads $T \approx \pi\sqrt{2/(a(\omega-a))}$ , in agreement with the exact formula as $a \uparrow \omega$ .

Idea of the proof(Why the two cases, and where the period comes from)Original

Case 1 is Flows on the line transplanted. Unroll the circle onto $\mathbb{R}$ : the zeros of $f$ unroll to a set of fixed points repeating with period $2\pi$ , and the no-crossing theorem traps the unrolled solution between two consecutive ones. A trapped monotone solution converges, and it can only converge to a fixed point. Nothing ever gets round the circle, so nothing repeats.

Case 2 is the computation that case 1 forbids. With $f$ nowhere zero the equation separates and $t$ is an explicit integral of $d\theta/f(\theta)$ ; the time to advance from $\theta_0$ to $\theta_0 + 2\pi$ is the integral of the $2\pi$ -periodic integrand $1/f$ over an interval of length $2\pi$ , which does not depend on where the interval starts. That single fact is why every solution has the same period.

The bottleneck law is then a statement about where that integral gets its mass. When $f$ is small at just one place, $1/f$ is large there and the integral is dominated by that neighbourhood; replacing $f$ by the first two non-vanishing terms of its Taylor expansion there turns the integral into one that can be done in closed form, and extending the limits to infinity costs only a relatively small amount because the tails are where $1/f$ is bounded.

Proof(The dichotomy)Original

Case 1. Suppose $f(\theta_\ast) = 0$ for some $\theta_\ast$ , and let $\theta(t)$ be any solution, regarded as a real-valued function solving $\dot\theta = f(\theta)$ on $\mathbb{R}$ (a lift: it is the solution of the same equation on the line, read modulo $2\pi$ to get the motion on the circle). Let $Z = \{z \in \mathbb{R} : f(z) = 0\}$ , a non-empty closed set, closed because $f$ is continuous, and invariant under $z \mapsto z + 2\pi$ because $f$ is $2\pi$ -periodic. In particular every interval of length $2\pi$ meets $Z$ .

If $\theta(0) \in Z$ the solution is the constant one and there is nothing to prove, so suppose $\theta(0) \notin Z$ . Put

A = \sup\{z \in Z : z < \theta(0)\} , \qquad B = \inf\{z \in Z : z > \theta(0)\} .

Both sets are non-empty (each of $(\theta(0)-2\pi,\theta(0))$ and $(\theta(0),\theta(0)+2\pi)$ meets $Z$ , after removing the endpoint if necessary, since $Z$ is $2\pi$ -periodic and $\theta(0) \notin Z$ ), so $A$ and $B$ are finite, they belong to $Z$ because $Z$ is closed, and $A < \theta(0) < B$ with $B - A \le 2\pi$ .

By the no-crossing part of the monotonicity theorem in Flows on the line, the solution cannot attain the values $A$ or $B$ , so it stays in the compact interval $[A,B]$ for all time. Being bounded, by the last part of that same theorem it exists for all $t$ and converges as $t \to +\infty$ to a zero of $f$ , which is $A$ or $B$ ; applying the same statement to the time-reversed equation $\dot\psi = -f(\psi)$ gives convergence to the other one as $t \to -\infty$ . Finally, a periodic solution on the circle would have to advance by a non-zero multiple of $2\pi$ , which is impossible inside an interval of length at most $2\pi$ whose endpoints are never reached.

Case 2. Suppose $f$ has no zero. Being continuous on $\mathbb{R}$ , it has one sign by the intermediate value theorem; take $f > 0$ , the case $f < 0$ following by reversing time. Being continuous and $2\pi$ -periodic, $f$ attains a positive minimum $m$ on the compact interval $[0,2\pi]$ and hence on all of $\mathbb{R}$ , so $0 < 1/f \le 1/m$ and the integral in the statement is finite.

Since $\dot\theta = f(\theta) \ge m > 0$ , the solution $\theta(t)$ is strictly increasing with $\theta(t) \ge \theta_0 + mt$ , so it passes through every value above $\theta_0$ ; separating variables, the time at which it reaches the value $\vartheta > \theta_0$ is

t(\vartheta) = \int_{\theta_0}^{\vartheta} \frac{d\theta}{f(\theta)} .

Taking $\vartheta = \theta_0 + 2\pi$ gives the time to go once round:

T = \int_{\theta_0}^{\theta_0+2\pi} \frac{d\theta}{f(\theta)} = \int_0^{2\pi} \frac{d\theta}{f(\theta)} ,

the two integrals being equal because the integrand is $2\pi$ -periodic and the intervals both have length $2\pi$ . The value is independent of $\theta_0$ , and $\theta(t+T) = \theta(t) + 2\pi$ for every $t$ by applying the computation with $\theta(t)$ in place of $\theta_0$ . On the circle that says $\theta(t+T) = \theta(t)$ , so every solution is periodic with the same period $T$ .

The period of the nonuniform oscillator. Let $0 < a < \omega$ (for $a = 0$ the integral is immediately $2\pi/\omega$ ). Evaluate

T = \int_0^{2\pi} \frac{d\theta}{\omega - a\sin\theta}

by the method of Trigonometric integrals by residues: put $z = e^{i\theta}$ , so that $\sin\theta = (z - z^{-1})/(2i)$ and $d\theta = dz/(iz)$ , and the interval $0 \le \theta \le 2\pi$ becomes one anticlockwise circuit of the unit circle. Then

\omega - a\sin\theta = \omega + \frac{ia}{2}\Bigl(z - \frac1z\Bigr) ,

T = \oint_{\lvert z\rvert = 1} \frac{1}{\omega + \frac{ia}{2}(z - z^{-1})}\,\frac{dz}{iz} = \oint_{\lvert z\rvert=1} \frac{dz}{\,i\omega z - \frac{a}{2}(z^2-1)\,} = \oint_{\lvert z\rvert=1} \frac{(-2/a)\,dz}{\,z^2 - \frac{2i\omega}{a}z - 1\,} ,

the last step multiplying numerator and denominator by $-2/a$ . The denominator has the two roots

z_\pm = \frac{i\omega}{a} \pm \sqrt{1 - \frac{\omega^2}{a^2}} = \frac{i}{a}\Bigl(\omega \pm \sqrt{\omega^2-a^2}\Bigr) ,

using $\omega > a$ so that the square root is imaginary. They are distinct, so both poles are simple, and $z_+z_- = -1$ from the constant term of the monic quadratic. Now $\lvert z_-\rvert = (\omega - \sqrt{\omega^2-a^2})/a < 1$ : squaring the equivalent inequality $\omega - a < \sqrt{\omega^2-a^2}$ (both sides positive) gives $\omega^2 - 2a\omega + a^2 < \omega^2 - a^2$ , that is $a < \omega$ , which holds. Hence $\lvert z_+\rvert = 1/\lvert z_-\rvert > 1$ , and $z_-$ is the only pole inside the contour. Its residue is

\frac{-2/a}{2z_- - 2i\omega/a} = \frac{-2/a}{-\frac{2i}{a}\sqrt{\omega^2-a^2}} = \frac{1}{i\sqrt{\omega^2-a^2}} ,

and the residue theorem gives

T = 2\pi i \cdot \frac{1}{i\sqrt{\omega^2-a^2}} = \frac{2\pi}{\sqrt{\omega^2-a^2}} .

Strogatz records the same formula and offers two routes to it: residues, as here, or the half-angle substitution $u = \tan(\theta/2)$ (p. 98).

The fixed points for $a \ge \omega$ . Setting $\omega - a\sin\theta^\ast = 0$ needs $\sin\theta^\ast = \omega/a$ , which has solutions precisely when $a \ge \omega$ : the two points $\arcsin(\omega/a)$ and $\pi - \arcsin(\omega/a)$ of $[0,\pi]$ , which coincide at $\pi/2$ when $a = \omega$ . Differentiating, $f'(\theta^\ast) = -a\cos\theta^\ast = \mp a\sqrt{1 - (\omega/a)^2}$ according to the sign of $\cos\theta^\ast$ , so the point with $\cos\theta^\ast > 0$ has $f' < 0$ and is a sink, and the other has $f' > 0$ and is a source. At $a = \omega$ the two merge and $f'(\pi/2) = 0$ ; since $f(\theta) = \omega(1 - \sin\theta) \ge 0$ with equality only at $\pi/2$ , the flow approaches that point from one side and leaves on the other, making it half-stable. The transition is exactly the saddle-node picture: two fixed points collide and annihilate as $a$ decreases through $\omega$ .

The bottleneck law. Let $f(\theta_m) = r > 0$ be the minimum, $f'(\theta_m) = 0$ and $c = f''(\theta_m) > 0$ . Write $\phi = \theta - \theta_m$ . Taylor's theorem gives $f = r + \tfrac{c}{2}\phi^2 + O(\phi^3)$ near $\phi = 0$ . Fix a small $\eta > 0$ independent of $r$ and split

T = \int_{-\eta}^{\eta} \frac{d\phi}{f(\theta_m+\phi)} \;+\; \int_{\eta \le \lvert\phi\rvert,\ \lvert\phi\rvert\le\pi} \frac{d\phi}{f} .

On the second range $f$ is bounded below by a positive constant independent of $r$ (it stays away from its minimum), so that piece is $O(1)$ . On the first, replacing $f$ by $r + \tfrac c2 \phi^2$ and then extending the limits to $\pm\infty$ changes the value by $O(1)$ as well, because $\int_{\lvert\phi\rvert>\eta} d\phi/(\tfrac c2\phi^2)$ converges. What is left is elementary: substituting $\phi = \sqrt{2r/c}\,\,s$ ,

\int_{-\infty}^{\infty} \frac{d\phi}{r + \tfrac{c}{2}\phi^2} = \frac{1}{r}\sqrt{\frac{2r}{c}}\int_{-\infty}^{\infty}\frac{ds}{1+s^2} = \pi\sqrt{\frac{2}{rc}} ,

which tends to infinity as $r \to 0^+$ . So $T = \pi\sqrt{2/(rc)} + O(1)$ , which is the stated law. $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Strogatz (1994), §4.1, pp. 94-95, and §4.3, pp. 99-100

Periodicity of $f$ is used twice and both uses are essential. It is what makes $\dot\theta = f(\theta)$ an equation on the circle at all, and it is what makes $T$ independent of the starting phase in case 2, since that step consists of shifting the interval of integration. Without it there is no well-defined velocity at a point of the circle, as $\dot\theta = \theta$ shows.
Compactness of the circle replaces the boundedness hypothesis that Flows on the line needed for solutions to converge: there is no "off to infinity" option here, which is why case 1 has no escape clause. Solutions on the circle always exist for all time.
$f$ nowhere zero in case 2 is what makes $1/f$ integrable. A single zero is enough to destroy the conclusion completely, and not by a small amount: the period jumps from finite to undefined. The divergence of $\int d\theta/f$ as a zero of $f$ is approached is the analytic shadow of the no-crossing theorem — the time to reach a fixed point is infinite.
$\omega > a$ strictly in the period formula. At $a = \omega$ the formula $2\pi/\sqrt{\omega^2-a^2}$ is not merely inapplicable but correctly signals what happens, diverging exactly where the fixed point is born.
$c = f''(\theta_m) > 0$ in the bottleneck law. If the minimum of $f$ were quartically flat instead of quadratic, the exponent would change: $\int d\phi/(r + k\phi^4)$ scales like $r^{-3/4}$ , not $r^{-1/2}$ . The square-root law is the generic one because a generic minimum is quadratic — which is the same genericity argument that made the saddle-node the typical bifurcation in Bifurcations in one dimension, seen from the other side.
One dimension. Case 2 says that when there are no fixed points, all solutions are periodic with one common period. That is special to one dimension: it is a statement about an interval of initial conditions being carried round rigidly, and it is why a flow on the circle is not yet a limit cycle. A limit cycle is an isolated periodic orbit that attracts its neighbours, and it needs the planar phase space of Limit cycles and the Poincare-Bendixson theorem.

Method(Analysing a flow on the circle)AdaptedAdapted from Strogatz (1994), §4.1, p. 94, §4.3, pp. 97-100, and §4.6, pp. 110-111

Given $\dot\theta = f(\theta)$ :

Check that it is a flow on the circle: confirm $f(\theta+2\pi) = f(\theta)$ . If it is not, the problem belongs on the line.
Look for zeros of $f$ . Combine harmonics first if you can: a sum $p\cos\theta + q\sin\theta$ is $\sqrt{p^2+q^2}\,\cos(\theta - \delta)$ with $\tan\delta = q/p$ , which exposes the extreme values of $f$ at a glance and reduces many problems to the nonuniform oscillator.
If there are zeros, classify each by the sign of $f'$ , exactly as on the line. The sinks are the phase-locked states; every initial condition ends at one of them, and the characteristic time to settle is $1/\lvert f'\rvert$ at that sink. There is no oscillation.
If there are no zeros, the motion is periodic with period $T = \left\lvert \int_0^{2\pi} \frac{d\theta}{f(\theta)}\right\rvert ,$ the same for every initial condition. When $f$ is a trigonometric polynomial, evaluate the integral by the substitution $z = e^{i\theta}$ and the residue theorem, as in Trigonometric integrals by residues; the standard case is $\int_0^{2\pi} d\theta/(p + q\cos\theta) = 2\pi/\sqrt{p^2-q^2}$ for $p > \lvert q\rvert$ .
If a parameter is near the value at which the zeros appear, do not evaluate the integral: use the bottleneck law. Locate the minimum $\theta_m$ of $f$ , read off $r = f(\theta_m)$ and $c = f''(\theta_m)$ , and take $T \approx \pi\sqrt{2/(rc)}$ . This works whether or not the integral can be done, and it is the right thing to quote because it exhibits the square-root divergence.
For a driven oscillator, form the phase difference. If the system is $\dot\theta = g(\theta)$ and the drive is $\dot\Theta = \Omega$ , work with $\phi = \Theta - \theta$ , whose equation is again a flow on the circle. A sink of the $\phi$ equation is phase locking at a constant phase lag; no fixed point is phase drift, with beat period given by step 4.
Convert the period into the observable. The measured quantity is usually the mean rate $\langle\dot\theta\rangle = 2\pi/T$ in the oscillating regime and $0$ in the locked regime, which is what produces the characteristic flat-then-rising response curves of entrained systems.

Example(A flow on the circle with a non-obvious amplitude, and its bottleneck)Original

Take

\dot\theta = f(\theta) = \mu + \cos\theta - \sin\theta ,

which is $2\pi$ -periodic, so step 1 is satisfied.

Step 1: combine the harmonics. With $p = 1$ and $q = -1$ , step 2 of the method gives $\sqrt{p^2+q^2} = \sqrt2$ and

\cos\theta - \sin\theta = \sqrt2\,\cos\!\left(\theta + \frac{\pi}{4}\right) ,

as one checks by expanding the right-hand side: $\sqrt2(\cos\theta\cos\tfrac\pi4 - \sin\theta\sin\tfrac\pi4) = \cos\theta - \sin\theta$ . So

\dot\theta = \mu + \sqrt2\,\cos\!\left(\theta+\frac{\pi}{4}\right) ,

a nonuniform oscillator of mean $\mu$ and amplitude $\sqrt2$ . Its minimum is $\mu - \sqrt2$ , attained where $\cos(\theta+\pi/4) = -1$ , that is at

\theta_m = \frac{3\pi}{4} .

So there are fixed points if and only if $\mu \le \sqrt2 = 1.41421356\ldots$ , and the saddle-node is at $\mu = \sqrt2$ .

Step 2: the period at $\mu = 2$ . Here $\mu > \sqrt 2$ , so there are no fixed points and step 4 applies:

T = \int_0^{2\pi}\frac{d\theta}{2 + \sqrt2\cos(\theta+\pi/4)} = \int_0^{2\pi}\frac{d\theta}{2+\sqrt2\cos\theta} = \frac{2\pi}{\sqrt{2^2 - (\sqrt2)^2}} = \frac{2\pi}{\sqrt2} ,

the shift of the integration variable being harmless because the integrand is $2\pi$ -periodic. Hence

\boxed{\;T = \pi\sqrt2 = 4.442882938158366\ldots\;}

Numerical quadrature of $\int_0^{2\pi} d\theta/(2+\cos\theta-\sin\theta)$ returns $4.442882938158366$ , agreeing to sixteen figures.

Step 3: approaching the bifurcation. Now set $\mu = 1.45$ , just above $\sqrt2$ . The exact period is

T = \frac{2\pi}{\sqrt{\mu^2 - 2}} = \frac{2\pi}{\sqrt{1.45^2 - 2}} = \frac{2\pi}{\sqrt{0.1025}} = 19.625374\ldots ,

and direct numerical quadrature of $\int_0^{2\pi} d\theta/(1.45+\cos\theta-\sin\theta)$ confirms $19.6253737$ . The oscillator is now very jerky: it takes nearly twenty time units to go round, almost all of it spent near $\theta_m = 3\pi/4$ , where the velocity is only $\mu - \sqrt2 = 0.0357864$ , as against a velocity of $\mu+\sqrt2 = 2.8642$ at the fast point diametrically opposite.

Step 4: the bottleneck estimate, without doing the integral. Step 5 of the method needs two numbers at $\theta_m = 3\pi/4$ :

r = f(\theta_m) = \mu - \sqrt2 = 0.03578644 , \qquad c = f''(\theta_m) = -\sqrt2\cos\!\left(\theta_m + \frac\pi4\right) = \sqrt2 .

Then

T \approx \pi\sqrt{\frac{2}{rc}} = \pi\sqrt{\frac{2}{0.03578644 \times 1.41421356}} = \pi\sqrt{39.51814} = 19.749138 .

Step 5: compare. The estimate exceeds the exact value by $0.63\%$ , from an argument that used nothing but the value and the curvature of $f$ at one point. Closing on the bifurcation improves it: the two agree to leading order because $T_{\text{exact}} = 2\pi/\sqrt{(\mu-\sqrt2)(\mu+\sqrt2)}$ and $\mu + \sqrt2 \to 2\sqrt2$ , which turns the exact formula into $2\pi/\sqrt{2\sqrt2\,r} = \pi\sqrt{2/(\sqrt2 r)}$ , precisely the estimate with $c = \sqrt 2$ . The $0.63\%$ is the error in replacing $\mu+\sqrt2 = 2.8642$ by $2\sqrt2 = 2.8284$ .

Application(The current-voltage curve of an overdamped Josephson junction)AdaptedAdapted from Strogatz (1994), §4.6, pp. 108-111, and §4.5, pp. 104-106

A Josephson junction is two superconductors separated by a thin barrier. Each superconductor is described by a macroscopic wave function with its own phase, and the junction's state is the phase difference $\phi$ between them. The two Josephson relations are that the supercurrent through the barrier is

I_s = I_c \sin\phi ,

with $I_c$ the critical current, and that the voltage across it is

V = \frac{\hbar}{2e}\,\dot\phi .

Modelling the ordinary current by a resistance $R$ and the displacement current by a capacitance $C$ in parallel with the junction, and driving the whole with a constant bias current $I$ , Kirchhoff's current law gives $C\dot V + V/R + I_c\sin\phi = I$ , which after substituting the voltage-phase relation becomes

\frac{\hbar C}{2e}\,\ddot\phi + \frac{\hbar}{2eR}\,\dot\phi + I_c\sin\phi = I .

This is the equation of a damped pendulum driven by a constant torque, with $\phi$ the angle, $C$ the mass, $1/R$ the damping and $I$ the applied torque — one reason junctions have been such useful laboratory systems for nonlinear dynamics.

The overdamped limit. When the capacitance is small enough that the second-derivative term can be dropped (after a brief initial transient; the same singular limit as in Flows on the line), what is left is a flow on the circle:

\dot\phi = \frac{2eR}{\hbar}\bigl(I - I_c\sin\phi\bigr) ,

a nonuniform oscillator with $\omega = 2eRI/\hbar$ and $a = 2eRI_c/\hbar$ , so that $a$ and $\omega$ compare exactly as $I_c$ and $I$ do. The dichotomy therefore reads:

$I < I_c$ : fixed points exist, and by case 1 the phase locks at $\phi^\ast = \arcsin(I/I_c)$ . Then $\dot\phi \to 0$ , so the time-averaged voltage is $\langle V\rangle = 0$ . Current flows with no voltage at all: this is the zero-voltage supercurrent, the defining signature of the Josephson effect.
$I > I_c$ : no fixed points, so the phase runs, and by case 2 it advances by $2\pi$ in each period $T = 2\pi/\sqrt{\omega^2-a^2}$ . The mean voltage is then $\langle V\rangle = \frac{\hbar}{2e}\,\frac{1}{T}\int_0^T \dot\phi \,dt = \frac{\hbar}{2e}\cdot\frac{2\pi}{T} = \frac{\hbar}{2e}\sqrt{\omega^2 - a^2} = R\sqrt{I^2 - I_c^2} .$

The junction's current-voltage curve is therefore flat at zero up to $I_c$ and then rises with a vertical tangent, approaching the ohmic line $\langle V\rangle = IR$ for $I \gg I_c$ . The infinite slope at the threshold is the bottleneck law in disguise: it is the same $r^{-1/2}$ divergence of the period, read as a function of the bias current. Strogatz reaches the same result in his Example 4.6.1 (p. 111), and notes that restoring the capacitance makes the curve hysteretic (p. 112), a loop of the kind analysed in Bifurcations in one dimension.

A number. Take a junction with $I_c = 0.20\ \text{mA}$ and $R = 5\ \Omega$ , biased at $I = 0.25\ \text{mA}$ . Then $I > I_c$ , so

\langle V\rangle = R\sqrt{I^2 - I_c^2} = 5 \times \sqrt{(0.25)^2 - (0.20)^2}\ \text{mV} = 5\times\sqrt{0.0225}\ \text{mV} = 5 \times 0.15\ \text{mV} = 0.750\ \text{mV} .

The phase then runs at $1/T = 2e\langle V\rangle/h$ cycles per second, and with $2e/h = 4.8360\times10^{14}\ \text{Hz}\,\text{V}^{-1}$ this is

\frac{1}{T} = 4.8360\times10^{14} \times 7.50\times10^{-4}\ \text{Hz} = 3.627\times10^{11}\ \text{Hz} = 363\ \text{GHz} ,

an oscillation in the sub-millimetre band, which is what makes these devices useful as sources and detectors there. The relation $1/T = 2e\langle V\rangle /h$ is exact and involves only fundamental constants, which is why the inverse statement — a junction irradiated at a known frequency locks its phase and develops a precisely known voltage — is used to realise the volt.

Entrainment. The same equation governs an oscillator pulled by a periodic drive. Strogatz's account of the firefly Pteroptyx cribellata (after Ermentrout and Rinzel) takes a flash rhythm of natural frequency $\omega$ which speeds up or slows down in response to a stimulus of frequency $\Omega$ , $\dot\theta = \omega + A\sin(\Theta-\theta)$ ; the phase difference $\phi = \Theta-\theta$ then obeys $\dot\phi = (\Omega-\omega) - A\sin\phi$ , a nonuniform oscillator once more. Case 1 holds when $\lvert\Omega-\omega\rvert < A$ , and the firefly entrains, flashing at the drive frequency with the constant lag $\sin\phi^\ast = (\Omega - \omega)/A$ ; this is the range of entrainment $\omega - A < \Omega < \omega + A$ . Outside it, case 2 holds and the two drift apart, but non-uniformly — slowly while the firefly nearly keeps up, then quickly through the rest of the cycle — with beat period $2\pi/\sqrt{(\Omega-\omega)^2 - A^2}$ (pp. 105-106).