NONLINEAR SYSTEMS:
CONSERVED QUANTITIES AND LYPANUOV FUNCTIONS
J. WONG (FALL 2019)
Topics covered • Conservative systems
◦ Using conserved quantities to plot the phase plane ◦ Finding conserved quantities
◦ What qualitative behavior is allowed? ◦ Example: kinetic/potential energy ◦ Theorem on (nonlinear) centers • More examples
◦ Lotka-Volterra example
◦ Deducing solution behavior from the conserved quantity • Lyapunov functions
◦ Energy arguments (without a conserved quantity) ◦ The simpler case (stability)
◦ The improved result (asymptotic stability) ◦ Example: damped oscillator (loss of energy)
1. Conservative systems
A conservative system has a (non-trivial) quantity that is constant along each solution. That is, there is a function E(x, y) (an ’energy’, often literally an energy) such that
d
dtE(x(t), y(t)) = 0 (1.1)
for any solution curve (x(t), y(t)). In this section, we will adopt the physics notation of ˙x = dx
dt.
For the most part, t-derivatives are taken ‘along solutions’, i.e. the expression is evaluated at x(t), y(t) (so there are chain rule terms). For instance,
˙
E means d
dtE(x(t), y(t)).
The condition (1.1) implies that solutions lie on level sets of the energy E, i.e. on the curves {E(x, y) = C}.
Note that solutions curves may only be a subset of the level set. The fact that solutions are contained in the level sets of E (which will fill the entire plane) means that by analyzing E we can more or less deduce all the solution behavior.
Important: ‘Non-constant’ is needed! Every constant function is trivially conserved: E(x, y) = const. =⇒ ∂E/∂t = 0
We require E to be non-constant everywhere. Precisely, this means
E is not constant on any disk {|x − x0| < r}. (1.2)
Connection: exact equations Exact ODEs are similar, but as first order equations they form only branches of the 2d system. For instance,
x = ˙y, y = ˙−x =⇒ E(x, y) = x2+ y2 = C and solutions (x(t), y(t)) are circles. The exact equation for y(x),
2x + 2yy0 = 0
has the same potential, but solutions y(x) are semi-circles. Note that the method for obtain-ing the conserved quantity is the same as for the potential of exact equations (see below).
1.1. Finding conserved quantities. One useful trick is to try to write dy
dx = ˙y/ ˙x =
g(x, y) f (x, y).
If a conserved quantity exists, this should be an exact equation, so it can be solved by that procedure to find the potential. Note that while we have technically eliminated t to find y(x), the method will yield a correct E(x, y) for the original system.
Practical note: This is easy to check (since we have a condition for exactness), so it is good to try if a conserved quantity is suspected before moving onto more complicated techniques. Example 1: The simple harmonic oscillator (e.g. small swings of a pendulum)
¨
x + ω2x = 0 is written in system form as
˙x = y, ˙y = −ω2x. Divide the second equation by the first:
dy dx = ˙ y ˙x = − ω2x y =⇒ y dy dx = −ω 2x.
This is exact (with N = y and M = ω2x) and also separable. Integrate to get E(x, y) := 1 2y 2 + ω 2 2 x 2 = const. which is the sum of the kinetic (y = velocity) and potential energy.
Example 2: Not all conserved quantities are actual energies. Consider the equations ˙x = x(1 − y), y = ay(x − 1)˙
(we’ll do the analysis of this example later). Assume x, y > 0 and compute dy dx = y(x − 1) ax(1 − y) =⇒ a 1 − y y dy dx = x − 1 x . Now integrate to get a conserved quantity E:
a(log y − y) = x − log x + C =⇒ E(x, y) = a(log y − y) − x + log x.
The ‘energy’ E (not really an energy) is conserved for solutions that start in the quadrant {x, y > 0} (note that this is an invariant set, so E never becomes undefined).
2. Example: kinetic/potential energy A ball of mass m = 1 rolls down a hill with a shape
F (x) = 1 4x
4− 1 2x
2.
Suppose, for simplicity, that: • T velocity is v = dx
dt
• The velocity is proportional to the slope of the curve • Physical constants are 1 (e.g. gravity acceleration g = 1)
From energy conservation, the sum of the kinetic and potential energy is constant: KE + PE = const. =⇒ E(x, y) := 1
2v
2+ F (x) = C. Differentiating and using that ˙x = v, we find that the system is
˙x = v
˙v = −F0(x) = x − x3 (2.1)
Check that the energy is conserved for the system (2.1): ˙ E = ∂ ∂t 1 2v 2 + F (x) = v ˙v + F0(x) ˙x = −vF0(x) + F0(x)v = 0 =⇒ E conserved!
Equilibria/stability: The equilibria occur where v = 0 and F0(x) = 0, so there are three: P− = (−1, 0), P0 = (0, 0), P+ = (1, 0).
Calculate the linearization (Jacobian A and its eigenvalues) for each point: • For P0, this tells us the local behavior:
A =0 1 1 0
=⇒ saddle point; eigenvectors (1, ±1)T. • For P±, we are not so lucky:
A = 0 1 −2 0
linear center.
The phase portrait (using the energy): A precise analysis of the level sets of E will determine the phase portrait. Consider the level set (contour)
{(x, v) : E(x, v) = E0}.
See plot below. This contour is made out of two branches (by solving for v directly):
v = ±p2E0− 2F (x) (2.2)
so the level set E(x, v) = E0 is made of two branches, reflected across the x axis.
Remark (symmetry): We can save half the work by exploiting symmetry. Note that the system (2.1) is unchanged under the substitutions
t → −t, v → −v
(this is a time reversal symmetry). It follows that the phase plane for v < 0 is the reflection of the phase plane for v > 0 across the x axis (v → −v), with time reversed (flip direction of arrows). Once we deduce the behavior for v > 0, this rule fills in the rest.
To find where the branches end, look for where v = 0 (where the square root is √0): v = 0 ⇐⇒ F (x) = E0.
Thus the branches (2.2) end at points (x, v) = (x−, 0) and (x, v) = (x+, 0) where x− and x+ are the closest points to x less/greater than x with F (x) = E0. This is shown graphically by drawing a horizontal line at E0 on a graph of F (x) (see below).
Now note that the two branches intersect at this point (since v = 0 for both). By uniqueness, these branches must be the same solution. It follows they are one periodic orbit that goes from x− to x+ in the upper half, then x+ back to x− on the lower half.
Symmetry: The symmetry (??) provides a more elegant argument. First, show that the solution in the upper half goes from (x−, 0) to (x+, 0), forming a curve Γ. By the symmetry, there is a solution ˜Γ on the lower half going from (x+, 0) to (x−, 0) (time reversed, reflected in v). But both intersect on the x-axis, so they are the same curve.
This is a common trick: Find half the orbit, use symmetry to ‘clone it’ and get the other half, then use uniqueness to glue them together to form one closed orbit.
Analysis: We may now describe all the solution curves by moving the value of E0 and using the analysis above to deduce the path. The nullclines and direction field give some help for the sketch (to get the shape right):
x0 = 0 ⇐⇒ v = 0 =⇒ solutions cross the x-axis vertically, y0 = 0 ⇐⇒ x = ±1, 0.
The three cases (above the middle ‘hill’, exactly on at its peak, and below).
• E0 > 0 (green): There are two points x = ±a where F (x) = E0, with a > 1.
Between −a and a we have E0 > F (x) so (2.2) is well-defined and so v > 0 for the positive branch between −a and a (and v < 0 for the negative branch.
The intersection of the branches at (±a, 0) implies they are one periodic orbit. • E0 = 0 (red): Now the branches intersect the x axis at (±
√
2, 0) and 0.
The branches starting at the leftmost endpoint terminate at 0; together they form a loop. Notice that the loop exits (0, 0) then rotates around and converges back to (0, 0). It is part of the stable and unstable manifold of (0, 0). The same occurs in the branch on the right by the same argument. 1
• −1/4 < E0 < 0 (blue): There are now two sets of periodic orbits, one set inside each of the E0 = 0 loops. When E0 ≈ −1/4, these orbits are close to the eq. points at (±1, 0) as the loops shrink in size.
Physically, the E0 = 0 loops correspond to the case where the ball has just the right amount of kinetic energy to stop exactly on the top of the middle hill (which is unstable!).
1An orbit that leaves and returns to the same equilibrium is called a homoclinic orbit. . These solutions
are often significant in dynamical systems, e.g. the height h(z) of a droplet of water on a flat surface is a homoclinic orbit in the phase plane (h, h0) starting and returning to the base height of the surface.
2.1. Stability for centers. Note that the stable equilibria are at the minima of E, and the level sets around these points show that they are indeed neutrally stable and are centers (nearby solutions orbit around the equilibrium). This result is true in general:
Theorem (Centers for conservative systems): Suppose the (planar) system x0 = F (x) has a conserved quantity E and x∗ is an isolated equilibrium. If x∗ is a minimum or maximum of E then it is a center.
The idea of the proof is that near x∗ the level sets of E are closed orbits around x∗ and E > E(x∗) (strictly) near x∗ since it is isolated (which prevents some exceptions). Note that the same is not true of a critical point in general; x∗ might be a saddle node like (0, 0) in the previous example.
Example: For the ball rolling down the hill, (±1, 0) are minima of the energy (at the value E = −1/4). To check this, either show that E ≥ −1/4 or use brute force:
H =F 00(x) 0 0 1 =⇒ H x=±1 =2 0 0 1 =⇒ H is p.d. (λ1, λ2 > 0)
Since E is conserved, the theorem says that they are both centers. Indeed, we see that (x∗, 0) is a minimum of E ⇐⇒ F00(x) > 0,
an equilibrium is a minimum of E if and only if F00(x) > 0, i.e. if and only if x is a minimum of F (the bottom of a hill). This is also why (0, 0) (at a maximum of F ) is a saddle.
The big picture: The theorem shows that centers (and periodic orbits) are typical features for conservative systems, as are saddles. Roughly,
local minimum of E =⇒ center (neutrally stable) saddle point of E =⇒ saddle node(unstable)
However, stable nodes and spirals (or unstable) are impossible for most conserved systems (see homework for the argument).
Calculus reminder: The ‘minimum can be checked either by showing that E ≥ E0 near E0(or in the opposite case, that E can be increasing or decreasing near E0), or by brute force. For brute force, recall that from calculus,
x∗ is a minimum of E ⇐⇒ the Hessian H is positive definite where the Hessian is the matrix of second partials:
Hij = ∂2E ∂xi∂xj = "∂2E ∂x2 ∂2E ∂x∂y ∂2E ∂x∂y ∂2E ∂y2 #
Positive definite here means the eigenvalues are positive. An equivalent condition is H is positive definite iff det(H) > 0 and H11, H22> 0.
3. Another conservative system Earlier, we briefly considered the equations (picking a = 2)
˙x = x(1 − y), y = 2y(x − 1)˙
for the ‘Lotka-Volterra model’ for predator-prey populations. Here we work out the details, making use of the conserved quantity
E(x, y) = 2 log y − 2y − x + log x (3.1) The modeling context is as follows:
• x and y are the prey and predator populations, respectively • In the absence of the predator (y = 0), the prey grows at a rate x
• In the absence of prey, there is no food and the predator pop. decays at a rate −2y • The predator pop. grows at a rate 2xy proportional to the population of both (more
food and more predators means more growth), while the prey population dies at a proportional rate −xy.
The goal is to determine how the populations change in time; what happens ‘in equilibrium’ when the predator and prey populations balance?
Nullclines/directions: The equilibria are at
A = (0, 0), B = (1, 1) To start, we identify nullclines:
on {x = 0} : ˙x = 0 and ˙y = −2y on {y = 0} : ˙y = 0 and ˙x = x
From the above, both the x and y axis are invariant sets; solutions converge to A = (0, 0) along x = 0 and diverge to ∞ along y = 0. It follows that
{x, y > 0} is an invariant set. (3.2) The other nullclines are not invariants:
on {y = 1} : ˙x = 0 and ˙y = 2(x − 1) on {x = 1} : ˙y = 0 and ˙x = 1 − y
This splits the first quadrant into four sets S1, S2, S3, S4 as shown below, where x0, y0 each have a certain sign. This suggests solutions are closed orbits (the population cycles over time; predators overeat prey then reduce in numbers, which allows the prey population to re-grow), but we need to make an argument for this to be sure.
Linearization: For completeness, the linearization: J =1 − y −x 2y 2(x − 1) =⇒ J A = 1 0 0 −2 , J B = 0 −1 2 0 ,
It follows that A is a saddle node (with stable/unstable manifolds along the axes) and B is a linear center. Thus we need the conserved quantity to conclude that is indeed a center. Show that the center is really a center: We claim that B is a maximum of the conserved quantity E(x, y). To check this, compute the Hessian:
∂E ∂x = 1 x − 1, ∂E ∂y = 2 y − 2 =⇒ H = −1/x2 0 0 −2/y2
The eigenvalues are negative, so this is indeed a maximum. (Note that we can’t check that (0, 0) is a saddle here since E(0, 0) = −∞). From the theorem on centers for conservative systems, B is a center.
Use level sets of E to draw solutions: Here we show that orbits are closed. The following uses a mix of direct arguments and the conserved quantity; you could just plot the level sets of E but this will also illustrate the more general graphical techniques.
Claim 1: All solutions with x, y > 0 are bounded. Let x(t), y(t) be a solution with x, y > 0 with E(x, y) = E0. Write the conserved quantity as
E0 = 2f (y) − f (x), f (y) = log y − y. (3.3) Note that
f (y) → −∞ as y & 0, f (y) → −∞ as y % ∞.
Suppose y(t) diverges to ∞ (as t increases). Then f (y) → −∞, which forces f (x) → −∞ to balance it out in (3.3). Then x(t) → 0 or x(t) → ∞. But x0 is positive when y is large (from the phase plane), so it cannot be that x(t) → 0.
A similar argument shows that x(t) also cannot diverge to ∞. We can also check that x and y cannot go to 0 (the origin is unstable due to the saddle point).
Claim 2: All solutions with x, y > 0 are closed orbits. To show this, consider a solu-tion that starts on the segment I between S1 and S4 (so y = 1 and 0 < x < 1). By claim 1, the solution must loop around and re-enter S1 (it cannot diverge).2 It then must intersect the starting segment I again. But E is one-to-one on I and E is conserved for the solution, so the solution must intersect itself back at the starting point. It follows (by uniqueness), the solution is a closed orbit.3
Draw solution curves: Now we know what all the orbits starting on I look like; it is not hard to see that this covers all the possible orbits. We may now justify drawing the phase portrait as concentric closed orbits (as on the right figure).
2This is not totally rigorous, but all the technical details can be checked using E.
3The argument here is an example of a powerful tool called a Poincare map, or first return map, which
4. Energy arguments
More generally, when an ‘energy’ is not constant, but changes in a known way (e.g. always dewcreasing), it can be used to analyze solution behavior.
As a first example, recall that the ODE for the displacement of a damped spring is ¨
x + β ˙x + x = 0
where β > 0 is the damping coefficient. Set y = ˙x and write as a system: ˙x = y
˙
y = −x − βy (4.1)
If β = 0 then the conserved quantity is
E(x, y) = 1 2x
2+ 1 2y
2. (4.2)
If β > 0 then E is not conserved but it is still useful: ˙ E = x ˙x + y ˙y = xy + y(−x − by) = −by2 so it follows that ˙ E ≤ 0, E = 0 ⇐⇒ y = 0.˙ (4.3)
That is, the energy of the system is strictly decreasing except when y = 0, when ˙E = 0.
Moreover, E(x, y) has a unique minimum at (0, 0). It follows from (4.3) that solutions curves must flow from higher to lower energy. If y = 0 and x 6= 0 (not in equilibrium) then ˙6=0 so the solution will immediately move on from {y = 0} and continue to decrease in E.
Since (0, 0 is the only minimum of E (also the only equilibrium point), we can conclude that all solutions converge to (0, 0) as t → ∞. That is, (0, 0) is asymptotically stable. Moreover, it is globally asymptotically stable, meaning that solutions converge to (0, 0) for every starting point,
5. Lyapunov functions
The decreasing energy suggests a method for proving an equilibrium point is stable (which linearization cannot do for centers) and for determining the set of starting points that con-verge to it (which linearization cannot do at all).
The idea is to construct a function like an energy that is decreasing along solutions and has a (unique) minimum at the equilibrium, so they must flow to lower values of that energy and end up at the equilibrium.
Positive definite functions: Let V (x) be defined in a region Ω ⊂ Rn and let x
0 ∈ Ω. We say that V is positive definite in Ω (with minimum at x0) if
i) V (x0) = 0
ii) V (x) > 0 for all x ∈ Ω except x0
If instead V (x) ≥ 0 we say V is positive semi-definite. With the inequalities reversed, ’positive’ is replaced by ’negative’.
Note: The ‘in’ and ’with a minimum’ are often dropped for brevity when the domain or the minimum are obvious (e.g. Ω = Rn and x
0 = ~0). Examples:
• The function V (x, y) = x2 + y2
is positive definite in Ω = R2 with a minimum at (0, 0), but we usually just say V is ‘positive definite’ here.
• The function V (x, y) = x2y2 is positive semi-definite with minima where x = 0 or y = 0. It is not positive definite.
• The function V (x) = x2− x4 is positive definite in (−1, 1) but not in R.
Theorem I (stability via Lyapunov functions): Let x0 be an equilibrium point of the system ˙x = F (x). Suppose there is a function V (x) (a ‘Lyapunov function’) such that
i) V is positive definite in a region Ω containing x0
ii) ˙V is negative semi-definite in Ω (that is, ˙V = dtd( V (x(t)) ) ≤ 0 for all x(t) ∈ Ω) Then x0 is (Lyapunov) stable. Moreover, if in addition
˙
V is negative definite in Ω (i.e. ˙V = 0 only at x0) then x0 is asymptotically stable.
That is, ˙V ≤ 0 is enough to guarantee stability - neutral or asymptotic - but does not say which is true. The statements are not if and only if, just useful sufficient conditions.
Once V is found, it is easy to use. Unfortunately, there is no single procedure for ob-taining Lyapunov functions; some inspired guesswork or a physical quantity (e.g. an actual energy) is typically required. The most common ones are
V = a(x − x0)2+ b(y − y0)2 or some higher degree polynomial (V = x4+ y4 and so on).
Example 1 (nice): For instance, consider
˙x = −x, y = −y.˙ There is an equilibrium point Let
V (x, y) = x2+ y2. Then V is positive definite. We have
˙
V = 2x ˙x + 2y ˙y = −2(x2+ y2),
so ˙V is negative definite. This means that V must be decreasing along solutions, and strictly decreasing when not an the equilibrium, which suggests that solutions must converge to (0, 0). Indeed, the theorem allows us to conclude that (0, 0) is asymptotically stable.
Global convergence: Note that the theorem does not say that (0, 0) is globally attract-ing (i.e. every solution converges to (0, 0)). However, you can see that with a bit more information about V , we can sometimes improve the result. Here, we know that
˙
V = −2(x2 + y2) = −2V. Suppose we have a solution x(t) with x(0) = x0. Then
V (x(t)) = V0e−2t.
Thus V → 0 as t → ∞. But V is only 0 at (0, 0), so it must be that x → (0, 0) as t → ∞.
Example 2 (not nice): Let’s go back to the damped spring (with b ≥ 0): ˙x = y, y = −by − x.˙
The system is LCC; there is a unique equilibrium at (0, 0) and b = 0 =⇒ neutrally stable (a center)
b > 0 =⇒ asymptotically stable (spiral or node). A Lyapunov function can be used to reach the same conclusion. To start, let
V (x, y) = 1 2(x
2 + y2),
positive definite in R2 (minimum at (0, 0)). Repeating earlier calculations, we find that ˙
V = −by2 =⇒ V is negative semi-definite.
Theorem I then implies that (0, 0) is Lyapunov stable. However, it does not distinguish between neutral and asymptotic stability. When b > 0, this result is not as strong as it should be (it should give asymptotic stability!).
5.1. Asymptotic stability (an improved theorem). The ‘improved’ version:
Theorem II (solutions with decreasing V ): Suppose V is defined in Ω and ˙V ≤ 0. Then solutions contained in Ω must approach the set where ˙V = 0 as t → ∞. Here ’approach’ means ’stays arbitrarily close to’.
Note: for this theorem to be useful, we need a separate way to show that solu-tions stay inside a region.
For the damped spring with b > −0, note that ˙
V < 0 when y 6= 0, V = 0 when y = 0.˙ (5.1) Theorem II tells us that solutions must approach {y = 0}, and the fact that there is only one equilibrium on that line implies that solutions must in fact approach (0, 0). It should be clear from the phase plane that the solution cannot end up arbitrary close to {y = 0} unless it converges to (0, 0) (the solution crosses y = 0 and continues to loop around). We won’t worry about a rigorous argument (the analysis is beyond the scope of our discussion). It follows that the origin is asymptotically stable. In fact, because V is defined on all of R2 and the result (5.1) holds everywhere, Theorem II guarantees (0, 0) is globally attract-ing (every solution converges to the origin!).
Caution: Theorem II does not always give asymptotic stability. Consider x0 = 0, y0 = −y.
Then for the same V as before (V = x2+ y2), ˙
V = −y2.
We conclude from Theorem II that all solutions approach the set {y = 0}. Since V is positive definite with a minimum at (0, 0), Theorem I tells us that (0, 0) is Lyapunov stable. It is not, however, asymptotically stable since x stays constant .
5.2. A more involved example. We have seen a few examples where the Lyapunov func-tion was an actual energy, valid over the whole domain.
For a more complicated example, consider the system
x0 = −x + y2, y0 = −xy − x2. The equilibria occur where
x = y2, x = 0 or x = −y.
Solving this we find the equilibria are at (0, 0) and (−1, 1). The Jacobian at (0, 0) is −1 + 2y 2y −y − 2x −x =−1 0 0 0 so the linearization does not predict anything about stability. Let
V = x2+ y2.
Then V is positive definite with a minimum at (0, 0). We have ˙
V = 2x(−x + y2) + 2y(−xy − x2) = −2x2− 2yx2 = −2x2(y + 1). It follows that ˙V is not negative semi-definite in all of R2, but
˙
V is negative semi-definite in {y > −1}.
Let Ω = {y > −1}. We can now use the theorems to first show solutions stay contained in Ω, then show they converge to (0, 0). Note that (0, 0) lies in Ω.
1) Stability: By Theorem I, (0, 0) is Lyapunov stable.
2) Convergence when solutions behave: Moreover, ˙V = 0 only on the set {x = 0}. But the only equilibrium point in this set is (0, 0) and it is clear from the phase plane that solutions cannot otherwise end up arbitrarily close to {x = 0} unless they converge to (0, 0). Theorem II then tells us that every solution that stays in Ω must converge to (0, 0).
3) Asymptotic stability: Now note that from (1) the origin is stable, so we know that solutions that start close must stay close. In particular this means that solutions that start close must stay in Ω. It follows from (2) that solutions that stay close must converge to (0, 0), so (0, 0) is in fact asymptotically stable.