Figure 5.13: Bifurcation diagram for the system (5.10). For µ < 0, the origin is a stable equilibrium. For each µ > 0, the origin is an unstable equilibrium but there is a stable limit cycle with amplitude √
µ.
of the long-term behavior of solutions as a function of the parameter µ. To do so, we must create a three-dimensional plot illustrating how the value of µ affects the behavior of x and y as t → ∞. The bifurcation diagram appears in Figure 5.13.
Notice that the family of limit cycle solutions forms a surface (a paraboloid, to be specific).
Like other types of bifurcations, Andronov-Hopf bifurcations can be either super-critical or subsuper-critical. The above is an example of the supersuper-critical case, in which stable limit cycles are spawned as µ increases beyond a critical value. It is also possible to have subcritical Andronov-Hopf bifurcations in which a stable equilib-rium loses stability as µ increases, and a family of unstable limit cycle solutions is destroyed in the process.
Qualitatively, all Andronov-Hopf bifurcations exhibit the sort of behavior exhib-ited by Equations (5.10). Let β 6= 0 and consider the following linear,
constant-coefficient system with a pair of complex conjugate eigenvalues µ ± βi:
Now let us introduce nonlinearity be adding functions with “quadratic and higher-order” terms to the right hand sides of these two equations; i.e.,
x0 = µx − βy + p(x, y)
y0 = βx + µy + q(x, y). (5.11)
Here, we assume that the functions p(x, y) and q(x, y) are analytic functions of two variables. That is, p(x, y) and q(x, y) can be represented in terms of convergent power series expansions
Notice that we have only included quadratic and higher-order terms, and the sub-scripts used on the coefficients correspond to the exponents on x and y. Roughly speaking, p and q are nothing more than generalizations of polynomials in two vari-ables. By design, the system (5.11) has exactly one equilibrium (the origin), and the linearization about the origin is given by
· x0 If µ = 0, then the origin is a center for the linearized system.
In preparation for stating a criterion for the occurrence of an Andronov-Hopf bifurcation in (5.11), we will need the following definition.
Definition 5.3.2. For the system (5.11), the Lyapunov number is defined as σ = 3π
For an explanation of where this mysterious quantity comes from, see Section 3.4 of the text of Perko [8]. Specifically, read the discussion of Poincar´e maps (a test for the stability of periodic solutions).
The Lyapunov number provides a quick way of testing whether the system (5.11) actually experiences an Andronov-Hopf bifurcation at µ = 0.
Theorem 5.3.3. (Hopf Bifurcation Theorem). If σ 6= 0, then the system (5.11) experiences an Andronov-Hopf bifurcation at µ = 0. If σ < 0, a unique, stable limit cycle bifurcates from the origin as µ increases from negative to positive. For small, positive µ, the amplitude of the limit cycle solution is approximately proportional to
√µ. Similarly, if σ > 0, then a unique unstable limit cycle bifurcates from the origin as µ decreases from positive to negative.
Example. In our previous example, we found that the system (5.10) x0 = −y + x(µ − x2− y2)
y0 = x + y(µ − x2− y2)
suffers an Andronov-Hopf bifurcation at µ = 0. In our above notation, β = 1, p(x, y) = −x3− xy2 and q(x, y) = −x2y − y3. Notice that there are no “quadratic”
terms—i.e., terms in which the exponents of x and y sum to 2. This implies that a20, a11, a02, b20, b11, and b02 are all 0. For the “cubic” terms, we have
a30 = −1, a12 = −1, a21 = 0, a03 = 0, and
b30 = 0, b12 = 0, b21 = −1, b03 = −1.
The Lyapunov number is σ = 3π
2 [3(−1 − 1) + (−1 − 1)] = −12π < 0,
and it follows that a supercritical Andronov-Hopf bifurcation occurs at µ = 0.
It is possible to state a stronger version of the Hopf Bifurcation Theorem which allows us to estimate the period and amplitude of the limit cycles that are born via an Andronov-Hopf bifurcation. Suppose that a system x = f (x; µ) experiences an Andronov-Hopf bifurcation at some critical parameter value µ = µc. Let λ(µ) = α(µ)+ iω(µ) denote the eigenvalues of a complex conjugate pair of eigenvalues whose real parts change sign at µ = µc. At the bifurcation point, ω(µc) 6= 0 and α(µc) = 0. Moreover, the eigenvalues satisfy a transversality condition α0(µc) 6= 0, which
basically just states that the eigenvalues really are moving from negative to positive real part (or vice-versa) as µ passes µc. Let us consider the supercritical case in which α0(µc) > 0, so that the real parts of the pair of eigenvalues are changing from negative to positive. The Hopf Bifurcation Theorem states that for µ slightly larger than µc, there exists a family of stable limit cycle solutions whose amplitudes are roughly proportional to√
µ − µc. Moreover, if µ − µc is reasonably small, the period of the limit cycle solutions is approximately 2π/ω(µc). For details, refer to the text of Hassard et al. [4].
As an illustration, suppose that the Jacobian matrix associated with a system of ODEs has eigenvalues (µ−3)±2µi. Then α(µ) = µ−3 and ω(µ) = 2µ. At µ = µc= 3, we have α(3) = 0 and ω(3) = 6 6= 0. The transversality condition α0(3) = 1 6= 0 is also satisfied, and we conclude the an Andronov-Hopf bifurcation has occurred at µ = 3. The bifurcation would be supercritical because α0(3) > 0. This means that for µ values slightly larger than 3, there exists a family of stable periodic solutions.
The periods of these limit cycles are approximately 2π/ω(3) = π/3, and their amplitudes are roughly proportional to √
µ − 3. The bifurcation diagram for such a system would look essentially identical to the one in Figure 5.13, except that the vertex of the paraboloid-like surface would be shifted from µ = 0 to µ = 3.
Here, we will not attempt to provide a comprehensive list of the various types of bifurcations that can occur. Clearly, there are many ways that the qualitative behavior of a phase portrait could be suddenly and dramatically altered as a pa-rameter µ is varied. Systems with more than one papa-rameter offer an even greater degree of flexibility. Students interested in learning more about bifurcation theory are encouraged to take an advanced course in differential equations and dynamical systems.
EXERCISES
1. For each of the following ODEs, µ denotes a real parameter. In each case, identify the equilibria of the ODE and determine which ones are stable and unstable. Then,
produce a bifurcation diagram.
(a) dx
dt = (µ − 1)x + x2 (b) dx
dt = µ − 2 + 3x2 (c) dx
dt = µ2− x2 (d) dx
dt = µx + 9x3 (e) dx
dt = 4x(µ − ex).
2. For each of the following ODEs, µ denotes a real parameter. In each case, find all bifurcation points and, if possible, classify them as one of the types we discussed in this Chapter. Then, produce a bifurcation diagram.
(a) dx
dt = (1 + µ)(x2− µx) (b) dx
dt = µ − x2+ 4x4 (c) dx
dt = µ2− x4 (d) dx
dt = (−x2+ µ4)(x2− µ2).
3. Sketch the bifurcation diagram for the equation dx
dt = (µ − x)(µ + x2) = µ2 − µx + µx2− x3,
where µ is a real parameter. If you do this correctly, you will discover two different bifurcation points: one transcritical, and one pitchfork.
4. (a) Let µ be a real parameter. Find all bifurcation points for the equation x0 = x2− µx − 2µ2, and draw a bifurcation diagram.
(b) Suppose we modify the ODE in Part (a), multiplying the right-hand side by x to obtain the equation x0 = x(x2− µx − 2µ2). Sketch the new bifurcation diagram and compare it to the bifurcation diagram from Part (a).
(c) Although the system in Part (b) has no bifurcation points, its “bifurcation dia-gram” still conveys useful information. In fact, the diagram completely characterizes how the system behaves depending upon the parameter µ and the initial condition x0. If µ 6= 0, note that there are two stable equilibria. For µ 6= 0, find the basin of at-traction of each stable equilibrium. The basin of atat-traction for an equilibrium x∗ is the set of all initial conditions x0 for which x(t) → x∗ as t → ∞.
Remarks: (i) Notice that the basins of attraction of the stable fixed points are “sep-arated” by an unstable fixed point. In this respect, unstable equilibria can be very important, as they help dictate which stable state (if any) our system will converge to as t → ∞. (ii) Suppose that f is continuously differentiable and that the first-order ODE x0 = f (x; µ) has multiple stable equilibria. Then there is always an unstable equilibrium between each pair of stable equilibria. Any solid curves in the bifurcation diagram must be separated by a dashed curve somewhere in between.
5. A “quad-furcation”: When we discussed saddle-node bifurcations, we gave an example of an ODE that has two equilibria for µ < 0 and no equilibria for µ > 0.
Create an example of an ODE with a single parameter µ which has no equilibria for µ < 0 but four equilibria for µ > 0. Identify which equilibria are stable/unstable.
6. A “five-pronged pitchfork”: When we discussed pitchfork bifurcations, we gave an example of an ODE that has one equilibrium for µ < 0 and three equilibria for µ > 0. Create an example of an ODE with a single parameter µ which has one equilibrium for µ < 0 but five equilibria for µ > 0. Identify which equilibria are stable/unstable.
7. Consider the system dx
dt = −8y + x(µ − x2− y2) dy
dt = 8x + y(µ − x2− y2),
where µ is a parameter. (a) Show that, regardless of µ, the origin is the only equi-librium. (b) Show that a Hopf bifurcation occurs when µ = 0. (c) Give a qualitative description of the periodic solutions which are created by this bifurcation. Estimate their period, assuming that µ is small. For small µ, what can you say about the amplitude of the periodic solutions? Are these solutions stable or unstable?
8. Subcritical Hopf Bifurcations. Background: Subcritical Hopf bifurcations can be dangerous in engineering applications because when an equilibrium loses stability as a parameter µ varies, solutions can suddenly jump to a far distant stable limit cycle. Consider, for example, the planar system in polar coordinates
dr
dt = µr + r3− r5 dθ dt = 1.
where µ is a bifurcation parameter.
(a) Note that r = 0 is an equilibrium. Determine the range of µ values for which this equilibrium is stable/unstable.
(b) If we write the r equation as r0 = r(µ + r2− r4), we see that other equilibria must satisfy µ = r4− r2. Carefully sketch this curve in the µr-plane. Hint: You may find it easier to first sketch µ versus r, and then flip your axes to create a plot of r versus µ.
(c) Show that the r equation has two positive equilibria if −14 < µ < 0 but only one positive equilibrium if µ ≥ 0.
(d) Without attempting to find formulas for the positive equilibria, show that if µ > 0 then the positive equilibrium is stable.
(e) Bonus: For −14 < µ < 0, show that the smaller of the two positive equilibria is unstable.
Discussion (No work required on your part!): Here is why this sort of bifurca-tion can be devastating. Suppose µ < −1/4 and our system is in equilibrium r = 0.
Now start gradually increasing µ. When we pass µ = −1/4, two limit cycles are born, corresponding to the two different positive r values which satisfy the equation µ = r4− r2. The limit cycle with larger amplitude (larger r) is stable, and the other one is unstable. Meanwhile, our equilibrium r = 0 is still stable. However, when we reach µ = 0, the unstable limit cycle is destroyed via a subcritical Hopf bifurcation, but the larger amplitude stable limit cycle still exists. As soon as the unstable limit cycle is destroyed (at µ = 0), the equilibrium r = 0 also loses its stability. Conse-quently, once µ > 0, any small amount of “noise” would throw us off the equilibrium and our solution would (rather dramatically) leap to a large-amplitude periodic solu-tion. Physically, you can imagine a very well-behaved system which suddenly begins to oscillate violently as a parameter is increased. Subcritical Hopf bifurcations have been observed in mathematical models of aeroelastic flutter (vibrations in airplane wings), and a host of other physical/biological scenarios.
This system also provides a nice example of bistability and hysteresis, which can be explained as follows. Again, suppose µ < −1/4 and we are at the stable equilibrium.
As we increase µ past 0, our equilibrium suddenly repels nearby trajectories and
suddenly forces us toward the large-amplitude stable, periodic solution. Now suppose we start to decrease µ again. This time, we have a nice, stable limit cycle until we reach µ = −1/4, when we suddenly transition back to a stable equilibrium.
Notice that the sudden transitions between two different stable states occurred at two different µ values. This phenomenon is known as hysteresis, and the fact that there are two different stable states for −1/4 < µ < 0 is known as bistability.
9. Self-oscillations in glycolysis: For particular parameter choices, a reduced form of the famous Selkov model takes the form
dx
dt = −x + 1
10y + x2y dy
dt = µ − 1
10y − x2y.
(a) Show that
(x, y) = µ
µ, 10µ 10µ2+ 1
¶
is an equilibrium.
(b) Show that this equilibrium is a stable node if µ = 0.
(c) Show that this equilibrium is an unstable focus if µ = q1
2. (d) Show that this equilibrium is a stable focus if µ = 1.
(e) Since the eigenvalues of the Jacobian matrix depend continuously on µ, the above observations suggest that at least two Andronov-Hopf bifurcations have occurred:
one between µ = 0 and µ = q
1
2 and one between µ = q
1
2 and µ = 1. Find the µ values at which the Andronov-Hopf bifurcations occur. Remark: The first bifurcation creates a family of stable, periodic solutions. The periodic solutions disappear after the second bifurcation, when the stable focus is born.
10. The purpose of this exercise is to create a bifurcation diagram for the system dx
dt = µx(1 − x) − xy dy
dt = x − y, where µ > −1 is a real parameter.
(a) The origin is an equilibrium independent of the choice of µ. Find the other equilibrium of the system.
(b) By linearizing the system at the origin, determine the ranges of µ for which the origin is stable/unstable.
(c) Let (x1, y1) denote the equilibrium you found in Part (a). Compute the Jaco-bian matrix Jf (x1, y1) and use its determinant to find the range of µ for which the equilibrium is a saddle.
(d) When the determinant of Jf (x1, y1) is positive, the origin is either a focus or a node. However, for the purposes of creating a bifurcation diagram, we need only determine whether the equilibrium is stable or unstable, and this is easily accom-plished by inspecting the trace and determinant of Jf (x1, y1). Show that the trace is negative for all µ > −1, and conclude that if the determinant is positive, then (x1, y1) is a stable node.
(e) Since the x and y coordinates of the equilibria (0, 0) and (x1, y1) happen to be equal, we need not create a three-dimensional bifurcation diagram by plotting µ versus both x and y. Because no information is lost if we neglect the y variable, sketch a bifurcation diagram of µ versus x. What are the bifurcation values of µ?
Can you classify which type(s) of bifurcations occur?