Besides periodic solutions and equilibria, can there be non-constant solutions of x0 = f (x) which remain bounded? The answer is yes, as we illustrate via the following examples.
A heteroclinic orbit. Consider the second-order ODE
u00+ cu0 − u(u − β)(u − 1) = 0, (4.2) where c and β are positive constants and 0 < β < 12. Introducing w = u0, we write the equation as a system of two first-order equations
u0 = w
w0 = u(u − β)(u − 1) − cw = u3 − (1 + β)u2 + βu − cw. (4.3) Note that there are three equilibria for this system. Setting u0 = 0 forces w = 0, and then setting w0 = 0 forces u = 0, u = β, or u = 1. The Jacobian matrix for this system is
Jf (u, w) =
· 0 1
3u2− 2(1 + β)u + β −c
¸ .
At the equilibrium (u, w) = (0, 0), the Jacobian is
which has characteristic equation λ2 + cλ − β = 0. Since the determinant of Jf (0, 0) is −β < 0, we immediately conclude that (0, 0) is a hyperbolic equilib-rium and that there is a saddle at the origin. The eigenvalues of Jf (0, 0) are λ± = 12
h
−c ±p
c2+ 4β i
, the larger of which is positive and the smaller of which is negative. To calculate an eigenvector for the positive eigenvalue λ+, we form the matrix convenient to treat v1 as our free variable so that
· v1
is an eigenvector for λ+, and a similar calculation reveals that
· 1 λ−
¸
is an eigenvector for the negative eigenvalue λ−. The spans of these two eigenvectors form the unstable and stable subspaces, and since λ− < 0 < λ+ the orientation of these vectors is as sketched in Figure 4.5. The same sorts of calculations show that the equilibrium (u, w) = (1, 0) is also a saddle and has a similar orientation as the saddle at the equilibrium (Figure 4.5).
The behavior at the equilibrium (u, w) = (β, 0) is a bit different. The Jacobian matrix is
and we note that the determinant β − β2 is positive because 0 < β < 1. Hence, this
E (0,0)
sE (0,0)
uE (1,0)
sE (1,0)
u00 11
00 11
00
11
u
w
β 1
Figure 4.5: Two saddle equilibria for Equations (4.3).
equilibrium is not a saddle like the other two. In fact, the eigenvalues
−c ±p
c2− 4(β − β2) 2
both have negative real part. It follows that this equilibrium is either a stable node or a stable focus, depending upon the relative sizes of the constants β and c.
Interesting fact: If the constant c is chosen appropriately, we can force the stable manifold Ws(0, 0) at the origin to coincide with the unstable manifold Wu(1, 0) of the equilibrium (1, 0). This forms a special trajectory in the phase plane which connects the two equilibria, and there is a name for this type of solution.
Definition 4.2.1. Suppose that x∗ and x∗∗are two distinct equilibria of the system x0 = f (x). A solution x(t) with the property that
t→−∞lim x(t) = x∗ and lim
t→∞x(t) = x∗∗
is called a heteroclinic orbit.
In our example above, it is actually possible to use a special trick to find the heteroclinic orbit analytically. Indeed, consider the simpler ODE given by u0 = au(u − 1), where a > 0 is a constant. Unlike our original system, this ODE has only two equilibria: u = 0 and u = 1. It is easy to check that 0 is stable and 1 is unstable. We claim that for special choices of the constants a and c, the solutions of this simpler differential equation are also solutions of the original second-order equation (4.2). To see this, we will substitute u0 = au(u − 1) into Equation (4.2).
Since u0 = a(u2 − u), we calculate that
00 0 0 0 2
00 11
00 11
00
11 u
w
β
1 t
u
0 1
Figure 4.6: A heteroclinic orbit for Equation (4.2). The left panel shows the uw-phase plane, and the heteroclinic orbit appears in bold. The right panel shows a graph of u versus t. Note that u → 0 as t → ∞ and u → 1 as t → −∞.
Equation (4.2) becomes
a2u(u − 1)(2u − 1) + cau(u − 1) − u(u − β)(u − 1) = 0 which, after factoring out u(u − 1), can be rewritten as
u(u − 1)£
(2a2− 1)u + (β + ca − a2)¤
= 0.
The only way the left hand side, a function of u, could be identically equal to 0 is if both
2a2− 1 = 0 and β + ca − a2 = 0.
The solution of this system of two equations is a = 1
√2 and c = √ 2
µ1 2− β
¶ .
Note that c > 0 since 0 < β < 12. In summary, for these special choices of c and a, the solution of the equation u0 = au(u − 1) is also a solution of the original equation (4.2). It is straightforward to solve this equation by separation of variables.
The corresponding trajectory in the uw-phase plane is a heteroclinic orbit. As t →
−∞, the trajectory connects to the equilibrium (u, w) = (1, 0), and as t → ∞, the trajectory connects to the equilibrium (u, w) = (0, 0). The solution exists for all time t, is bounded, and is neither an equilibrium nor a periodic orbit (see Figure 4.6).
The heteroclinic orbit forms a connection of the stable manifold Ws(0, 0) with the unstable manifold Wu(1, 0).
A homoclinic orbit. We now discuss a special type of orbit in which the stable and unstable manifolds of the same equilibrium are connected. The example we give
00
11 0011 00
11
x
t y
x
Figure 4.7: Two homoclinic orbits for Equation (4.4). The left panel shows the xy-phase plane, and the homoclinic orbits appear in bold. The right panel shows a graph of x versus t for one of the two homoclinic orbits. Note that x → 0 as t → ±∞.
is that of a double well potential. More exactly, consider the system
x0 = y and y0 = x − x3, (4.4)
which has three equilibria. Linearization indicates that (x, y) = (0, 0) is a saddle equilibrium. The equilibria (1, 0) and (−1, 0) are non-hyperbolic, and are centers for the linearized systems. It is actually possible to show that they are centers for the nonlinear system as well, by defining the energy functional E(x, y) = 12y2 −
1
2x2+14x4 and arguing that E(x, y) remains constant along phase plane trajectories.
Figure 4.7 shows a rough sketch of the phase plane for this system. Notice that the unstable manifold Wu(0, 0) at the origin happens to coincide with the stable manifold Ws(0, 0). This closed loop trajectory is not a periodic orbit—a sketch of the corresponding solution x(t) also appears in Figure 4.7. This special orbit is called a homoclinic orbit.
EXERCISES
1. This series of questions concerns the famous Lorenz system dx
dt = −σx + σy dy
dt = rx − y − xz dz
dt = xy − βz,
where σ, r, and β are positive parameters. The Lorenz system exhibits chaos if σ = 10, r = 28, and β = 8/3. Below, you will learn why chaos could never occur if
(a) Explain why, independent of the choice of initial conditions, the initial value problem for the Lorenz system is guaranteed to have a unique solution (at least locally).
(b) Show that the origin is a locally asymptotically stable equilibrium if 0 < r < 1.
(c) Show that if r > 1, then there will be a one-dimensional unstable manifold at the origin.
(d) The origin is actually globally asymptotically stable if r < 1, and the remaining parts of this exercise will guide you through a proof. To start, define the function V (x, y, z) = σ1x2+ y2+ z2. Show that
∇V • f = −2
· x −
µr + 1 2
¶ y
¸2
− 2
"
1 −
µr + 1 2
¶2#
y2− 2βz2,
where f denotes the right hand side of the Lorenz system.
(e) Show that if r < 1, then
1 −
µr + 1 2
¶2
> 0.
Then, explain why this implies that ∇V • f is strictly less than 0 (except at the origin) whenever r < 1.
(f) Finally, explain why V is a Lyapunov function and why you can conclude that the origin is a global attractor if r < 1.
2. Consider the system
dx
dt = −y − x(1 − x2− y2) dy
dt = x − y(1 − x2 − y2).
Show that this system has an unstable periodic solution and carefully sketch the phase portrait.
3. Consider the system
dx
dt = −y + x(4 − x2− y2) dy
dt = x + y(9 − x2− y2).
You may assume that the origin is the only equilibrium of this system. Classify the local stability of the origin. Then, show that this system has at least one stable, periodic solution.
4. Consider the system
dx
dt = −y + x(r2 − 6r + 8) dy
dt = x + y(r2− 6r + 8),
where r2 = x2 + y2. Use the Poincar´e-Bendixon Theorem to prove that this system has both stable and unstable periodic orbits by following these steps:
(a) Show that the origin is the only equilibrium of this system.
(b) Using the chain rule, differentiate both sides of r2 = x2+ y2 with respect to t.
Then, assuming r 6= 0 (i.e., excluding the equilibrium solution), solve for dr/dt. You should obtain an autonomous ODE for r.
(c) Using the equation for dr/dt you found in Part (b), show that dr/dt > 0 on the circle r = 1 and that dr/dt < 0 on the circle r = 3. Use the Poincar´e-Bendixon Theorem to conclude that there is at least one stable periodic orbit within the annulus 1 < r < 3.
(d) Using the equation for dr/dt you found in Part (b), show that dr/dt > 0 on the circle r = 5. Combined with the fact that dr/dt < 0 on the circle r = 3, this seems to suggest that an unstable periodic orbit exists inside the annulus 3 < r < 5. To prove this, make the substitution t 7→ −t, which “reverses” the flow of time. Then, use the Poincar´e-Bendixon Theorem to show that the resulting system has a stable periodic orbit inside the annulus 3 < r < 5. Finally, conclude that the original system (i.e., going forward in time) has an unstable periodic orbit inside that annulus.
5. Here is an instance of the famous FitzHugh-Nagumo nerve membrane model:
dx
dt = −x(x − 1)(x + 1) − y dy
dt = x −1 2y.
(a) Show that this system has exactly one equilibrium and that it is unstable.
0000 1111 0 0 1 1
0 0 1 1
00 11
y
x
Figure 4.8: Rectangular path to use in exercise involving FitzHugh-Nagumo model.
(b) Consider the rectangular path Γ with corners (√
3, 2√
3) (−√ 3, 2√
3) (−√
3, −2√
3) (√
3, −2√ 3)
as illustrated in Figure 4.8. By showing that the flow is directed inward on this path, use the Poincar´e-Bendixon Theorem to conclude that these equations have at least one periodic solution. Hint: You will need to parametrize each edge of the rectangle separately, and there are many possible parametrizations.
6. Consider the second-order nonlinear ODE u00+ 5
√6u0+ u(1 − u) = 0.
(a) Write the ODE as a system of two first-order ODEs by setting v = u0. Show that the resulting system has two equilibria, and that one equilibrium is a stable node while the other is a saddle.
(b) Verify that
u(t) = 1
£1 + (√
2 − 1)et/√6¤2
is a solution of the ODE and that it corresponds to a heteroclinic orbit. Plot u(t) versus t and sketch the corresponding trajectory in the uv-phase plane.
7. In this exercise, you will find a homoclinic orbit for a particular ODE. Consider the second-order, nonlinear equation
v00+ 3v2− σv = 0
where σ is a constant. This ODE could be written as a system in the usual way, by letting w = v0. We claim that there is a non-zero solution v(t) which has the property that v, v0, v00 → 0 as t → ±∞. This solution, if plotted in the vw-phase plane, would correspond to a homoclinic orbit.
(a) Multiply both sides of the above ODE by v0 and show that the resulting equation is equivalent to
d dt
(v0)2 2 = d
dt
³
−v3+ σ 2v2
´ .
(b) Now integrate both sides of the equation in Part (a), and let β denote the integration constant. Explain why we may assume that β = 0.
(c) By algebraically solving for v0, show that dv
dt = ±v√
σ − 2v.
Then, taking the minus sign for later convenience, use separation of variables to show that
t = −
Z dv
v√
σ − 2v.
(d) Recall the following definitions of hyperbolic trigonometric functions:
sinh(z) = 1
2(ez− e−z) cosh(z) = 1
2(ez+ e−z) tanh(z) = sinh(z)
cosh(z) sech(z) = 1
cosh(z)
In the integral you wrote down in Part (c), make the (hyperbolic) trigonometric substitution
v = σ
2sech2(θ) dv = −σsech2(θ) tanh(θ) dθ and note that
v√
σ − 2v = σ3/2
2 sech2(θ) tanh(θ).
Do the resulting integral to show that t = 2
√σθ + C where C is an integration constant.
(e) Use algebra to show that
v(t) = σ 2sech2
·√σ
2 (t − C)
¸ .
Then, plot the function v(t) versus t for the special case σ = 1 and C = 0 to get a sense of what the homoclinic solution would look like. Finally, set w = v0 and sketch the homoclinic trajectory in the vw-phase plane.
Chapter 5
Bifurcations
In practice, we often deal with ODEs which contain parameters (unspecified con-stants) whose values can profoundly influence the dynamical behavior of the system.
For example, suppose we model population changes for a species. The birth and death rates of the species would be examples of parameters whose values would sub-stantially impact that behavior of solutions of the underlying differential equation.
Example. Consider the ODE dx/dt = µx, where µ is a parameter. The solution of this equation is x(t) = x0eµt, where x0 = x(0). Notice that if µ > 0, the solutions exhibit exponential growth, whereas if µ < 0 we observe exponential decay. If µ = 0, solutions are constant. The critical value µ = 0 marks the “boundary” between two very different types of dynamical behavior.
Definition 5.0.2. Consider a system of ODEs of the form x0(t) = f (x; µ), where µ is a parameter. A bifurcation is any major qualitative change in the dynamical behavior of the system in response to varying the parameter µ.
In the previous example, we would say that a bifurcation occurs at µ = 0, because the equilibrium x = 0 changes from stable (µ < 0) to unstable (µ > 0).
There are many ways that the qualitative behavior of a system can be drastically altered in response to changes in parameters. Equilibria and/or periodic solutions can be created or destroyed, or they can change their stability. In what follows, we will survey several common types of bifurcations. For simplicity, we will deal primarily with ODEs with only one dependent variable. An excellent reference for this material is provided by Strogatz [11].