Saddle-node bifurcations. First, we discuss one common mechanism for the birth or destruction of an equilibrium solution. The canonical example to keep in mind is given by the ODE
dx
dt = µ + x2. (5.1)
Suppose µ < 0. The equilibria satisfy x2 + µ = 0, which implies that x = ±√
−µ are the two equilibria of the system. To determine whether they are stable, we compute that the “Jacobian” of f (x) = x2+ µ is simply f0(x) = 2x. By calculating f0(√
−µ) = 2√
−µ > 0, we find that √
−µ is an unstable equilibrium. Similarly, since f0(−√
−µ) = −2√
−µ < 0, it follows that −√
−µ is a stable equilibrium. Next, suppose that µ = 0. There is only one equilibrium, namely x = 0. Although it is non-hyperbolic, it is easy to check (via separation of variables) that this equilibrium is unstable. Finally, suppose µ > 0. Then x2+ µ > 0 for all x, implying that there are no equilibria. In summary, as µ increases from negative to positive, two equilibria (one stable and one unstable) merge and annihilate each other, leaving no equilibria if µ > 0. Clearly µ = 0 is a bifurcation point, and this type of bifurcation is called a node bifurcation. A minor adjustment of Equation (5.1) reveals that saddle-node bifurcations can also create new equilibria as the parameter µ increases. The equation x0 = −µ + x2 experiences a saddle-node bifurcation at µ = 0, creating two equilibria for µ > 0.
Bifurcation diagrams. One very common and useful way to visualize bifurcations of a system x0 = f (x; µ) is to sketch a graph of the equilibrium values of x as a func-tion of the bifurcafunc-tion parameter µ. As an illustrafunc-tion, the bifurcafunc-tion diagram for Equation (5.1) appears in Figure 5.1. By convention, unstable equilibria are plotted as dashed curves and stable equilibria are plotted as solid curves. In Figure 5.1, the dashed curve corresponds to the unstable equilibrium x =√
−µ and the solid curve corresponds to the stable equilibrium x = −√
−µ. The bifurcation diagram allows us to visualize the qualitative behavior of solutions for various choices of initial con-ditions and the parameter µ. For example, if we start at the initial condition x0 in Figure 5.1, the unstable equilibrium will repel us and x will decrease towards the stable equilibrium (as indicated by an arrow in the Figure). The bifurcation diagram for x0 = −µ + x2 can be obtained by reflecting the parabola in Figure 5.1 across the vertical axis.
Transcritical bifurcations. The saddle-node bifurcation is a very common way for equilibria to be created or destroyed as a parameter µ is varied. We now describe a mechanism by which two equilibria can exchange their stability. The canonical
xo
0000 00 1111 11
unstable
stable
x
µ no
equilibria
Figure 5.1: Saddle-node bifurcation in Equation (5.1). For µ < 0, there are two equilibria: x = ±√
−µ. The larger of these is unstable (dashed curve) and the smaller is stable (solid curve). For µ > 0, there are no equilibria.
example of a transcritical bifurcation is given by the equation dx
dt = µx − x2. (5.2)
Setting dx/dt = 0, we find that there are two equilibria: x = 0 and x = µ. Note that the former is an equilibrium independent of the choice of µ. To check the stability of these equilibria, we compute that the “Jacobian” of f (x) = µx − x2 is simply f0(x) = µ − 2x. Since f0(0) = µ, we conclude that x = 0 is stable if µ < 0 and unstable if µ > 0. Similarly, since f0(µ) = −µ, we conclude that x = µ is stable if µ > 0 and unstable if µ < 0. In this case, µ = 0 is the bifurcation point, and the two equilibria exchange their stability there. The corresponding bifurcation diagram appears as Figure 5.2, and the arrows in the figure indicate how x would change starting from various choices of x(0) and µ. For example, if µ > 0 and we choose any initial condition x0 ∈ (0, µ), then x(t) → µ as t → ∞.
Example. Equation (5.2) is similar to a well-known logistic model for population growth
x0 = µx(M − x), (5.3)
where x represents population of a species, M is the maximum population that the environment can sustain, and the parameter µ measures the birth rate minus the death rate. If µ 6= 0, the equation has exactly two equilibria: x = 0 (extinction) and x = M (proliferation), neither of which depend on µ. If µ < 0, the death rate exceeds the birth rate and x = 0 is a stable equilibrium while x = M is unstable. If µ > 0, then x = M is stable and x = 0 is unstable, and we expect the population to approach the environment’s carrying capacity as t → ∞. The transcritical bifurcation at µ = 0 marks the “threshold” between the species’ extinction and proliferation. Figure 5.3
x
µ
Figure 5.2: Transcritical bifurcation in Equation (5.2). For µ < 0, there are two equilibria: x = 0 is stable (solid) and x = µ is unstable (dashed). A bifurcation oc-curs at µ = 0, and the two equilibria exchange stability. Arrows indicate how x(t) will change as t increases, starting from various choices of µ and initial conditions x(0).
Pitchfork bifurcations. One type of bifurcation that is common in physical prob-lems involving some sort of symmetry is the pitchfork bifurcation. As an example, suppose that we balance a weight on top of a vertical beam (Figure 5.4). If the weight is small, the system is in a stable equilibrium. If the weight increases beyond a certain point, the equilibrium loses its stability. Any slight deflection of the beam from a perfectly vertical position will cause the beam to buckle. If we regard this system as “two-dimensional”, then the beam will either buckle left or right. Another example is provided by a bead on a rotating hoop (Figure 5.5). For slow rotation speeds, the bead rests at the bottom of the hoop. However, this equilibrium loses its stability if the rotation speed is increased beyond a certain critical value. In these first of these examples, the bifurcation parameter µ would be the weight of the object being supported by the beam. In the latter example, the bifurcation parameter µ is the speed of rotation of the hoop.
The canonical example of a pitchfork bifurcation is provided by the equation dx
dt = µx − x3. (5.4)
The right hand side of this equation can be factored as f (x; µ) = x(µ − x2). If µ < 0, then the quadratic factor has no real roots and therefore x = 0 is the only equilibrium. On the other hand, if µ > 0, then further factorization reveals that f (x; µ) = x(√
µ − x)(√
µ + x). Thus, for positive µ there are 3 equilibria: x = 0 and x = ±√
µ. Using the same sort of stability analysis we used in our discussions of saddle-node and transcritical bifurcations, you should convince yourself that x = 0
x
M
0 extinction
proliferation
µ
Figure 5.3: Transcritical bifurcation in Equation (5.3). For µ 6= 0, there are two equilibria: x = 0 and x = M. The bifurcation at µ = 0 causes the two equilibria to exchange stability. Arrows indicate how x(t) will change as t increases, starting from various choices of µ and initial conditions x(0).
0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000 0000000000000000
1111111111111111 1111111111111111 1111111111111111 1111111111111111 1111111111111111 1111111111111111
weight
beam
Figure 5.4: A weight being supported by a vertical beam.
is stable if µ < 0 and unstable if µ > 0. Both x = ±√
µ are stable equilibria for µ > 0. The bifurcation diagram in Figure 5.6 should convince you that the name pitchfork bifurcation is indeed appropriate. As we see in the Figure, as µ increases from negative to positive, an equilibrium loses stability at the same instant that two new equilibria are born.
Subcritical versus supercritical bifurcations. One further way of classify-ing bifurcations is accordclassify-ing to whether an equilibrium gains or loses stability as the parameter µ is increased. The pitchfork bifurcation in Equation (5.4) above is considered to be a supercritical bifurcation. An example of a subcritical pitchfork bifurcation would be given by the equation x0 = µx + x3. The reader is encouraged to sketch a bifurcation diagram for this equation. You should find that x = 0 is a
0000 00 1111 11
0000 00 1111 11
0000 00 1111 11
Figure 5.5: A bead on a rotating hoop. The vertical lines indicate the axis of rotation. For slow rotation speeds (left panel), the bead rests in a stable equilibrium at the bottom of the hoop. If the rotation speed is increased, this equilibrium loses stability and the bead settles into a new stable position (right panel).
x
µ
Figure 5.6: Pitchfork bifurcation in Equation (5.4). For µ < 0, there is only one equilibrium (x = 0) and it is stable. For µ > 0, there are three equilibria: x = 0 is unstable and x = ±√
µ are stable. Arrows indicate qualitative behavior of solutions starting from various initial conditions and choices of the parameter µ.
stable equilibrium for µ < 0 and is unstable if µ > 0. For µ < 0, there are two other equilibria x = ±√
−µ, both of which are unstable. In what follows, we will rarely distinguish between subcritical and supercritical bifurcations.
Example. Let µ be a real parameter. Sketch the bifurcation diagram for the ODE dx
dt = fµ(x) = 5 − µ
1 + x2. (5.5)
Solution: Equilibria would satisfy the equation 5 − µ
1 + x2 = 0,
which can be manipulated to reveal that x2 = µ/5 − 1. There are no equilibria for µ < 5, but if µ > 5 then there are two equilibria:
x = ± rµ
5 − 1.
We can already conclude that this is an example of a saddle-node bifurcation. It remains to determine which equilibria are stable. Taking the derivative of fµ(x), we have
fµ0(x) = 2µx (1 + x2)2. It follows that
fµ0(p
µ/5 − 1) = 2µp
µ/5 − 1 (µ/5)2 ,
which is positive. It follows that for µ > 5, the larger of the two equilibria is unstable.
The other fixed point is stable. The bifurcation diagram appears in Figure 5.7.
Example. (Courtesy of S. Lazaryan) Let µ be a real parameter. Sketch a bifurcation diagram for the ODE
dx
dt = fµ(x) = x(x2− µ)
³
ex2 − µ
´
. (5.6)
Solution: Equilibria must satisfy the equation x(x2− µ)
³
ex2 − µ
´
= 0,
and certainly x = 0 is an equilibrium regardless of the value of µ. If µ > 0, then the
√ 2
equilibria no x
unstable
stable 5 µ
Figure 5.7: Bifurcation diagram for Equation (5.5). A saddle-node bifurcation occurs at µ = 5.
solutions, we would need µ ≥ 1. If µ > 1, algebra reveals that x = ±√
ln µ are also equilibria. In total, there are five equilibria if µ > 1.
To test the stability of these equilibria, we must compute the “Jacobian” of fµ(x), taking the derivative with respect to x. Expanding fµ(x) as
fµ(x) = x3ex2 − µxex2 − µx3+ µ2x, we use the chain and product rules to calculate
fµ0(x) = 3x2ex2+ 2x4ex2 − µex2 − 2µx2ex2 − 3µx2+ µ2.
For the equilibrium x = 0, we have fµ0(0) = −µ + µ2 = µ(µ − 1). This quantity is negative if 0 < µ < 1 and positive otherwise. We conclude that x = 0 is a stable equilibrium if 0 < µ < 1 and is unstable otherwise.
Next, assume µ > 0 and consider the pair of equilibria x = ±√
µ. Notice that x2 = µ and x4 = µ2 for both of these equilibria. Using the above formula for fµ0(x) we have
fµ0(±√
µ) = 3µeµ+ 2µ2eµ− µeµ− 2µ2eµ− 3µ2+ µ2
= 2µeµ− 2µ2 = 2µ (eµ− µ) . Since eµ > µ, we conclude that fµ0(±√
µ) > 0 for all positive µ. This implies that the two equilibria x = ±√
µ are unstable for µ > 0.
Finally, assume µ > 1 and consider the pair of equilibria x = ±√
ln µ. In this case, we have x2 = ln µ and ex2 = µ. Our formula for fµ0(x) yields
fµ0(±p
ln µ) = 3µ ln µ + 2µ(ln µ)2− µ2− 2µ2ln µ − 3µ ln µ + µ2
= 2µ ln µ (ln µ − µ) .
1 0
x
µ
Figure 5.8: Bifurcation diagram for Equation (5.6). A pitchfork bifurcation occurs at µ = 0, creating two unstable equilibria x = ±√
µ and changing x = 0 into a stable equilibrium. Another pitchfork bifurcation occurs at µ = 1, de-stabilizing x = 0 and creating two new stable equilibria, x = ±√
ln µ.
Examining the factors individually, note that (ln µ − µ) < 0 and ln µ > 0 for µ > 1.
It follows that fµ0(±√
ln µ) < 0 for all µ > 1, implying that these two equilibria are stable. A bifurcation diagram summarizing the stability of the equilibria appears in Figure 5.8.