09 - Lyapunov’s method, first integrals
Introduction
Example. Consider the following nonlinear system
x' y x3 f x, y 1 y' x y3 f x, y 2
The equilibrium point/points of the system can be found by solving the algebraic equation system x' 0, y' 0 y x3 0, x y3 0 y y9 y 1 y8 0 x 0, y 0.
The only equilibrium point is the origin: 0, 0 .
f f 1 1 x y 3 x2 1 The Jacobian matrix is f ' x, y f f 1 3 y2 2 2 x y
0 1 1 The coefficient matrix of the linearized system is f ' 0, 0 det Λ 2 1 0 1 0 1 Λ Λ the eigenvalues are i. Since and are not hyperbolic (that is, Re 0), the stability of the Λ12 Λ1 Λ2 Λ12 origin cannot be decided based on the eigenvalues. The phase portrait of the system is the following:
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1.0 0.010
0.5 0.005
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0.010 1.0 0.010 0.005 0.000 0.005 0.010 1.0 0.5 0.0 0.5 1.0 The trajectories go around the origin but it cannot be seen if they converge to the origin or diverge from it. 2 diffeq-09.nb
Lyapunov’s method: We consider the function V x, y x2 y2, where x, y 2 and investigate whether the value of V increases or decreases along the trajectories. The nullclines of V are circles centered at the origin.
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increases diverge from Observation: If the value of V along the trajectories then the trajectories decreases converge to unstable the origin, so the origin is . asymptotically stable
Let x t , y t , t be an arbitrary solution and let V t V x t , y t , t .
V x, y x2 y2
2 2 Then the derivative of V t x t y t with respect to the time t is the following:
d V t 2 x t x ' t 2 y t y' t 2 x x ' 2 y y' dt V x, y f x, y V x, y f x, y x 1 y 2 2 x y x3 2 y x y3 2 x4 y4 0 for all x, y 0, 0
Since the derivative of V is negative, then the value of V decreases along the trajectories. So the trajectories converge towards the origin and thus the origin is asymptotically stable. We obtained the phase portrait not only in a neighbourhood of the origin but on the whole phase plane.
This idea can be formulated (for two-dimensional systems) as follows.
Lyapunov’s method diffeq-09.nb 3
Definition. Let M 2 be a region and let f : M 2 and V : M be continuously differentiable functions. Then the derivative of V with respect to the system x' f x, y , y' f x, y (or the Lie- 1 2 derivative of V) is the following scalar product:
L V x, y V ' x, y f x, y V x, y f x, y V x, y f x, y f x 1 y 2
where f x, y f x, y , f x, y . 1 2
Remark. As we have seen in the example, if x t , y t is a solution of the system d x' f1 x, y , y' f2 x, y then for the function V t V x t , y t , V t Lf V x t , y t . It means dt that the sign of the function L V x, y indicates whether the value of V increases or decreases along f the trajectories. The main point of Lyapunov’s method is to find a function V that is monotonic along the trajectories and the monotonicity of V can be decided without calculating the solutions. It is an important special case when the value of V is constant along the trajectories.
Definition. The function V is a first integral of the system x' f x, y , y' f x, y if L V x, y 0. 1 2 f
Theorem (Lyapunov’s stability theorem). Let M 2 be a region, let f : M 2 be a continuously differentiable function and let p M be an equilibrium point of the autonomous system x ' t f x t , that is, f p 0. Suppose there exists an open neighbourhood U M of p and a continuously differen- tiable function V : U such that V p V q for all q U\ p . Then
if L V q 0 for all q U\ p then p is a stable equilibrium point; f if L V q 0 for all q U\ p then p is an asymptotically stable equilibrium point; f if L V q 0 for all q U\ p then p is an unstable equilibrium point.s f Exercises
First integrals
1. Find a first integral for the following systems and sketch the phase portrait in the x, y plane.
a) x' 2 y, y' x b) x' 1, y' sin x c) x' y, y' 4 x d) x' 1, y' x e) x' y, y' 2 f) x' 2, y' 3
x ' f y Remark. In the special case when the planar system has the form then by multiplying the y' g x equations we obtain g x x' f y y'. Integrating both sides with respect to t and then using the dx substitution formula (x' dt dt dx), it follows that g x x f y y. Thus, if F ' f and G' g dt then V x, y F y G x is a first integral. 2. Consider the equation of motion of the harmonic oscillator: m x '' D x 0 where m 0, D 0 are 4 diffeq-09.nb
constants. Transform the equation into a linear system and find a first integral for the system. Find the energy conservation for the system and sketch the phase portrait in the x, x' plane. 3. Decide whether the function V x, y is a first integral for the following system: x' x y, y' x2 y2
if a) V x, y x ln y x2 y x 0, y 0 y2 b) V x, y 2 ln x x 0, y 0 x2 4. Decide whether the function V x, y 3 x ln x 3 y 2 ln y x 0, y 0 is a first integral for the following system: x' 2 x 3 x y, y' 3 x y y Solution
1. a) x' 2 y, y' x
The product of the equations: x x' 2 y y' Integrating both sides with respect to the time: x x ' t 2 y y ' t d x x2 Using the substitution formula (x' t d t d x): x x 2 y y y2 c d t 2 x2 The function V x, y y2 is a first integral for the system 2 d 1 Verification: V x t , y t 2 x x ' 2 y y' x 2 y 2 y x 0 dt 2 The trajectories go along the nullclines of V. x2 The equation of the trajectories: y2 C (ellipses centered at the origin) 2 Direction of the trajectories: for example at 1, 1 : x' 2 0 x increases y' 1 0 y decreases diffeq-09.nb 5
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3 2 1 0 1 2 3 1. b) x' 1, y' sin x
The product of the equations: sin x x' 1 y' Integrating both sides with respect to the time: sin x x ' t 1 y ' t d x Using the substitution formula (x' t d t d x): sin x x 1 y cos x y c d t The function V x, y cos x y is a first integral for the system d Verification: V x t , y t sin x x ' 1 y' sin x 1 1 sin x 0 dt The trajectories go along the nullclines of V. The equation of the trajectories: cos x y C or y cos x C Direction of the trajectories: x' 1 0 x increases the trajectories go to the right Remark: y' sin x 0 if 2 k x 2 k k is an integer y increases Π Π Π y' sin x 0 if 2 k x 2 2 k k is an integer y decreases Π Π Π Π 6 diffeq-09.nb
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10 5 0 5 10 1. f) x' 2, y' 3
The product of the equations: 3 x' 2 y' Integrating both sides with respect to the time: 3 x ' t 2 y ' t d x Using the substitution formula (x' t d t d x): 3 x 2 y 3 x 2 y c d t The function V x, y 3 x 2 y is a first integral for the system d Verification: V x t , y t 3 x' 2 y' 3 2 2 3 0 dt Remark: V x, y 2 y 3 x or V x, y 5 3 x 2 y 7 is also a first integral. The trajectories go along the nullclines of V. The equation of the trajectories: 3 x 2 y C (straight lines) Direction of the trajectories: x' 2 0 x increases y' 3 0 y increases diffeq-09.nb 7
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10 5 0 5 10 2. Consider the equation of motion of the harmonic oscillator: m x '' D x 0 where m 0, D 0 are constants. Transform the equation into a linear system and find a first integral for the system. Find the energy conservation for the system and sketch the phase portrait in the x, x' plane. Let x x x ' x ' x x ' x or x ' y 1 1 2 1 2 x2 x ' D D D D x2 ' x '' x x1 x2 ' x1 y' x m m m m D The product of the equations: x x ' y y' m D D x2 y2 Integrating both sides with respect to the time: x x y y c m m 2 2 1 1 The function V x, y D x2 m y2 is a first integral for the system 2 2 1 1 The meaning of the terms: D x2 is the potential energy (x is the displacement) and m y2 2 2 is the kinetic energy (y x ' is the velocity) V is total mechanical energy of the system 1 1 The equation of the trajectories: D x2 m y2 C (ellipses centered at the origin) 2 2 Example: 8 diffeq-09.nb
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Lyapunov functions
5. Investigate the stability of the equilibrium points of the following system
x' x3 y, y' x y3
using the Lyapunov function V x, y x2 y2. 6. Investigate the stability of the origin in the following system
x ' 3 y x5, y ' 2 x y5
using the Lyapunov function V x, y 2 x2 3 y2. 7. Investigate the stability of the origin in the following systems using the Lyapunov func- tion
V x, y a x2 b y2 where a 0, b 0.
a) x' y x3 b) x' 2 y x5 c) x' y x3 d) x' 3 y x y' x y3 y' x y5 y' x y3 y' 2 x y Solution
5. The system: x' x3 y, y' x y3 The Lyapunov function: V x, y x2 y2 the nullclines of V are circles centered at the origin: 2 2 diffeq-09.nb 9
x2 y2 c
The derivative of V with respect to the system is: d Lf V x, y V x t , y t V ' x, y f x, y x V x, y f1 x, y y V x, y f2 x, y dt 2 x x' 2 y y' 2 x x3 y 2 y x y3 2 x4 2 x y 2 x y 2 y4 2 x4 y4 0 if x, y 0, 0 the value of V decreases along the trajectories, so the trajectories go towards the origin (they intersect the nullclines going inwards) the origin is asymptotically stable
7. a) The system: x' y x3, y' x y3 The Lyapunov function: V x, y a x2 b y2 where a 0, b 0 the nullclines of V are ellipses centered at the origin The derivative of V with respect to the system is:
d V x t , y t 2 a x x' 2 b y y' 2 a x y x3 2 b y x y3 dt 2 a x4 2 b y4 2 a x y 2 b x y If a 1, b 1 then 2 a x y 2 b x y 2 x y 2 x y 0 and d V x t , y t 2 a x4 2 b y4 2 x4 2 y4 0 if x, y 0, 0 dt the value of V decreases along the trajectories, so the trajectories go towards the origin (they intersect the nullclines going inwards) the origin is asymptotically stable
7.b) The system: x' 2 y x5, y' x y5 The Lyapunov function: V x, y a x2 b y2 where a 0, b 0 the nullclines of V are ellipses centered at the origin The derivative of V with respect to the system is:
d V x t , y t 2 a x x' 2 b y y' 2 a x 2 y x5 2 b y x y5 dt 2 a x6 2 b y6 4 a x y 2 b x y If a 1, b 2 then 4 a x y 2 b x y 4 x y 4 x y 0 and d V x t , y t 2 a x6 2 b y6 2 x6 4 y6 0 if x, y 0, 0 dt the value of V increases along the trajectories, so the trajectories go away from the origin (they intersect the nullclines going outwards) the origin is unstable Lotka-Volterra population model 10 diffeq-09.nb
Let x t and y t respectively denote the number of preys and predators, say, rabbits and foxes at time t. We assume that the prey population is the total food supply for predators. We also assume the following:
1) Preys have unlimited food supply. Hence if there were no predators, their number would grow exponentially, that is, when y 0 we have x ' a x where a 0. 2) The prey population decreases at a rate proportional to the number of predator/prey encounters (b x y). So the differential equation for the prey population is x' a x b x y where a 0, b 0. 3) In the absence of prey, the predator population declines at a rate proportional to the current population. So when x 0 we have y' c y with c 0. 4) When there are prey in the environment, we assume that the predator population increases at a rate proportional to the predator/prey meetings, or d x y. Together we have y' c y d x y where c 0, d 0.
This simplified predator/prey system (also called the Volterra-Lotka system) is
x' a x b x y x a b y y' c y d x y y c d x
where the parameters a, b, c and d are all assumed to be positive. Since we are dealing with popula- tions, we only consider x, y 0.
The equilibrium points are the solutions of x a b y 0, y c d x 0.
c a The two solutions are 0, 0 and , . The Jacobian matrix of the system is d b a b y b x f ' x, y . d y c d x
a 0 At 0, 0 the Jacobian matrix is f ' 0, 0 . The eigenvalues are a 0 and c 0 and we 0 c Λ1 Λ2 obtain a saddle point.
b c 0 c a c a At the other equilibrium point , , f ' , d . The eigenvalues are pure imaginary, d b d b d a 0 b i a c and so we cannot conclude anything at this stage about stability of this equilibrium Λ12 point. We cannot determine the precise behavior of solutions: they could possibly spiral in toward the equilibrium point, spiral toward a limit cycle, spiral out toward “infinity” and the coordinate axes, or else lie on closed orbits. To make this determination, we search for a Lyapunov function V. Employing the trick of separation of variables, we look for a function of the form
V x, y F x G y . diffeq-09.nb 11
The derivative of V with respect to the system is
d V x t , y t x F x' y G y' x F x a b y y G y c d x . d t
d V x F x y G y We obtain 0 provided . Since x and y are independent variables, this is possible if d t d x c b y a x F x y G y and only if constant. Setting the constant equal to 1, we obtain d x c b y a
c a x F d and y G b . Integrating, we find x y F x d x c log x and G y b y a log y.
Thus the function V x, y d x c log x by a log y is constant on the solution curves of the system when x, y 0. Theorem. Every solution of the predator/prey system is a closed orbit (except the equilibrium point c a , and the coordinate axes). d b
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0 1 2 3 4 c a For any given initial populations x 0 , y 0 with x 0 0 and y 0 0, other than , , the popula- d b tions of predator and prey oscillate cyclically. No matter what the populations of prey and predator are, neither species will die out, nor will its population grow indefinitely. Cyclic variations of this kind have been observed in nature, for example, for lynx and snowshoe hare near Hudson Bay, with a cycle of about 10 years. 12 diffeq-09.nb
Bendixson’s criterion
Consider the system
x' f x, y 1 y' f x, y 2
If in a simply-connected open set E the expression div f f f has constant sign (i.e. the sign x 1 y 2 remains unchanged and the expression vanishes only at isolated points or on a curve), then the above system has no closed trajectories in the set E.
Exercises
8) Show that the following system does not have a periodic solution: x' x y2 x3 y' x y x2 y
9) Show that following system does not have a periodic solution in the domain x2 y2 4: x' x x y2 y3 y' 3 y x2 y x3
10) Show that the following system does not have a periodic solution in the domain x2 y2 2 3: x' x y x3 y' x y y3 Bendixson–Dulac theorem
Let E 2 be a simply-connected open set and let B : E be a differentiable function such that div B f B f B f has constant sign and it vanishes only at isolated points or on a curve. Then x 1 y 2 the system x' f x, y , y' f x, y has no periodic solutions lying entirely in the set E. 1 2 Exercises
11) Consider the system x' x 1 a x b y y' y 1 c x d y 1 where a, b, c, d 0. Let B x, y . Show that the system does not have a periodic solution in the set x y E x, y : x 0, y 0 .
12) Consider the system x' y y' x a y 2 x2 b y2 diffeq-09.nb 13
2 b x where a 0. Let B x, y a e . Show that the system does not have a periodic solution. Solutions
8) f x, y x y2 x3, f x, y x y x2 y 1 2 div f x, y f x, y f x, y 1 3 x2 1 x2 2 4 x2 0 for all x, y 2, so the system 1 1 2 2 does not have a periodic solution in 2. 9) f x, y x x y2 y3, f x, y 3 y x2 y x3 1 2 div f x, y f x, y f x, y 1 y2 3 x2 4 x2 y2 0, if x2 y2 4, so the system does 1 1 2 2 not have a periodic solution in the region x2 y2 4. 10) f x, y x y x3, f x, y x y y3 1 2 div f x, y f x, y f x, y 1 3 x2 1 3 y2 2 3 x2 3 y2 0, if x2 y2 2 3, so the 1 1 2 2 system does not have a periodic solution in the region x2 y2 2 3. 11) Let f x, y x 1 a x b y , f x, y y 1 c x d y . 1 2 1 1 a x B x, y f1 x, y x 1 a x b y b x y y y a 1 B x, y f1 x, y y 1 1 d y B x, y f2 x, y y 1 c x d y c x y x x d 2 B x, y f2 x, y x a d Then div B f x, y 1 B x, y f1 x, y 2 B x, y f2 x, y 0, if x 0 , y 0, so the system does y x not have a periodic solution in E. 2 b x 2 b x 2 2 12) div B f x, y a e y a e x a y 2 x b y 1 2 2 b x 2 2 b x 2 b x 2 2 b x 2 a b y a 2 a b y a 0 for all x, y 2, so the system does not have a periodic solution in 2.