MAT332
Introduction to Nonlinear Dynamical Systems and Chaos
Lecture 5 & 6
Qun WANG
University of Toronto Mississauga
September 28, 2021
(UTM)
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Review
Example (Potential)
Find the phase portrait near the fixed points of the system x ̇ = f (x) with f(x)=−dV(x) forsomesmoothfunctionV.(SupposethatVhasaunique
dx
minimum)
Solution
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Learning Objective
For today’s lecture, students are expected to:
Objective
Understand the non-oscillation phenomenon
Understand the fundamental theorem for the existence and uniqueness of solutions.
Understand the dependence of solution on initial conditions. Calculate the first variation equation
Apply the stability criteria for 1D system
You do have to do some maths to follow today’s lecture—Drawing anime is not enough for today’s !
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General Strategy
Strategy (Phase Portrait near Fixed points)
1 Solve the equation f (x) = 0 to find the fixed point(s);
2 For each fixed point x0, study the value of f(x) outside x0;
3 Conclude if starting near x0, the solution will eventually leave x0 or approach x0.
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Phase Portrait near Fixed Points
Example (Too Many Fixed Point)
Find the phase portrait of the system x ̇ = f (x) near 0, where
Solution
x2sinx1, x̸=0
f (x) =
0. x=0
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Non-Oscillation Phenomena Theorem (No Periodic Orbit)
A 1-dimensional dynamical system x ̇ = f (x) has no periodic orbit. Proof.
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Loss of Uniqueness
Example
Consider the system
Can you find two different solutions?
Solution
x ̇=3×13, x(0)=0 (1) 2
1 Essentially, it is due to the fact as as x goes to 0, the derivative x 3
does not decrease accordingly and might still change dramatically.
In other words, around 0, you cannot control the change of dynamics according to the change of state in a UNIFORM way.
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Fundamental Theorem Definition
A function F(x) is said to be Lipschitz continuous in an open interval I if there exists a constant K s.t.
∀x,y∈I, |F(x)−F(y)|≤K|x−y|
(2)
Example
Let ε > 0. Consider the function f (x ) = 32 x 13 . Show that f (x ) is continuous in (−ε, ε) ;
f (x ) is NOT Lipschitz continuous in (−ε, ε)
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Fundamental Theorem
Theorem (Existence and Uniqueness)
Consider the system Suppose that F(x) is Lipschitz continuous in (a,b) with Lipschitz constant K.
1 Existence: For an initial condition a < x0 < b, there exists a solution x(t) to x ̇=f(x), x(0)=x0
defined for some time interval−τ < t < τ such that.
2 Uniqueness: Moreover, the solution is unique in the sense that if x(t) and y(t) are two such solutions with x(0) = x0 = y(0), then x(t) = y(t) on the largest interval of time about t = 0 where both solutions are defined.
Let φt(x0) = x(t) be this unique solution with φ0(x0) = x0. The term φt(x0) is also called the flow.
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Fundamental Theorem
Theorem (Continuous Dependence on Initial Condition)
The solution φt(x0) depends continuously on the initial condition x0. Moreover, let T > 0 be a time for which φt(x0) is defined for −T ≤ t ≤ T. Then for any ε > 0, there exists a δ > 0 such that if |y0 − x0| < δ, then φt(y0) is defined for −T ≤ t ≤ T and
|φt(y0)−φt(x0)|<ε, ∀−T≤t≤T. (3) We need the following famous Gronwall’s inequality to prove this
Theorem (Gronwall’s inequality)
Let v(t) be a continuous nonnegative function on a interval −τ ≤ t ≤ τ, and L, C > 0 be constant. If
then
Lv(s)ds
∀τ ≤t ≤τ
(4)
(5)
v(t)≤CeL|t|,
In particular, if C = 0, then v(t) = 0 in [−τ, τ].
v(t) ≤ C +
t 0
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Fundamental Theorem Sketch of Proof.
there exists positive numbers r and M s.t. |F(x)| ≤ M when |x − x0| ≤ r, and [x0 − r, x0 + r] ⊂ (a, b);
∀x , y ∈ (a, b), one has that |F (x ) − F (y )| ≤ |x − y | Define
τ = min{Mr , K1 }.
Let x0(t) be an arbitrary curve with image in (a,b), and construct
recursively
xi+1(t) = x0 +
Finally define C0 = maxt∈[−τ,τ] |x1(t)| − x0(t)| (Next page)
(6)
t 0
F(xi(s))ds
(7)
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Fundamental Theorem Sketch of Proof.
1 Show that
2 Show that
∀t ∈ [−τ,τ],∀i ∈ N, |xi(t)−x0| ≤ r. (8) max |xi+1(t)−xi(t)| ≤ C0λi, with λ = Kτ < 1. (9)
t∈[−τ,τ]
∀i ∈ N,
3 Show that x i (t ) converges uniformly for all t ∈ [−τ , τ ], denote the limit as
x∗(t) = limi→∞ xi (t).
4 Show that x∗(t) solves the equation in [−τ, τ] with initial condition x∗(0) = x0.
5 Show that if x (t ), y (t ) are two solutions with initial conditions x0 , y0 in an time interval [−T,T], then
t 0
K|x(s) − y(s)|ds
6 Apply Gronwall’s inequality to conclude the continuous dependence, In particular
the uniqueness of solution.
|x(t) − y(t)| ≤ |x0 − y0| +
(10)
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Fundamental Theorem
The above theorem about existence roughly tells you that if you start from a point x0 where the behaviour of your “dynamo” is not too crazy (at least it is Lipschitz continuous), then you can “drive” at least for a while (The flow exists for a while, defined on (−τ,τ) for some τ).
For a fixed F, the time interval in which the solution exists might depend on x0.
Can you give an example where the solution is not unique? (Hint: You just saw it earlier in this session...)
Can you give an example where the flow is only locally defined? (Hint: a model in population growth we have encountered last time...)
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Fundamental Theorem Tip
In practice, if the function has a continuous derivative, then it is locally Lipschitz continuous.
Proposition
Suppose that f (x) is differentiable and its derivative f ′(x) is continuous at x0 . Then there exists a r > 0 s.t. f (x ) is Lipschitz in (x0 − r , x0 + r ).
Proof.
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First Variation Equation Definition
Suppose that F(x) is a continuous function, and its derivative F′(x) is also continuous. Let φt(x0) be the flow of the equation
x ̇(t) = F(x(t)) (11) starting from x0. Then the system
v ̇(t) = F′(φt(x0))v(t) (12) is called the First Variation Equation.
F′(φt(x0)) means first calculate F′(x), then put φt(x0) into it. Don’t do it in the reverse order!
Example
What if x0 is a fixed point?
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First Variation Equation
Consider a fixed point x∗ for the differential equation x ̇ = f (x), where F and F′ are continuous.
If F′(x∗) < 0, then x∗ is an attracting fixed point; If F′(x∗) > 0, then x∗ is an repelling fixed point
If F′(x∗) = 0, then the derivative does not determine the stability type.
Example
Using the variational equation to study the fixed points of the following systems.
(1)x ̇=x3−x, (2)x ̇=x2
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