Matrix Exponentials, Hyperbolic Fixed Points, Dulac's Negative Criterion and Lyapunov Functions

Question 1

What type of equations does the matrix exponential method solve?

Answer 1

-linear differential systems with constant coefficients

Question 2

Taylor Series of an Exponential Function in 1D

Answer 2

e^(at) = 1 + at + 1/2! at² + 1/3! at³ + …

Question 3

Taylor Series of an Exponential Function of a Matrix

Answer 3

exp(At) = I + At + 1/2! A²t² + 1/3! A³t³ + …

• where A is an nxn matrix and I is the identity matrix
• this series always converges and can be computed in a closed form

Question 4

Proposition:

Let A be a square matrix with constant entries…

Answer 4

…then:

i) the exponential matrix X(t)=exp(At) satisfies the differential equation X’=AX
ii) any solution X(t) of the differential equation X’=AX satisfying the initial condition X(0)=I is given by X(t)=exp(At)
iii) solution of the initial value problem |x’=A|x, |x(0)=|xo is given by |x(t) = exp(At)*|xo

Question 5

Properties of Matrix Exponentials

Answer 5

1) exp(0A) = I
2) if AB=BA, then:
exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
3) if A = UBU^(-1) then, exp(UBU^(-1)) = Uexp(B)U^(-1)
4) if A is 2x2 matrix with first row α, 0 and second row 0,β, then exp(A) is a 2x2 matrix with entries first row: e^α, 0 and second row: 0, e^β

Question 6

What are the two cases for the matrix exponential method?

Answer 6

Case a -> λ1≠λ2

Case b -> λ1=λ2

Question 7

Case B: λ1=λ2

Solution

Answer 7

-let (2x22) matrix A have eigenvalues λ1=λ2, then:

exp(At) = e^(λ1t) * (I + t(A-λ1I))

Question 8

Proposition

λ1=λ2

Answer 8

-let A be a 2x2 matrix with such that λ1=λ2 (i.e. eigenvalues coincide)
-then:
(A-λ1I)² = |0
-this is also true for nxn matrices with coninciding eigenvalues:
(A-λ1I)^n = 0

Question 9

Nilpotent

Definition

Answer 9

-a matrix B such that B^n = |0 for some n∈ℕ is called a nilpotent matrix

Question 10

Case B) λ1=λ2

Proof

Answer 10

A = A - λ1I + λ1I
-let B = A - λ1I
A = B + λ1I
-since I commutes with any matrix, [B,λ1I]=0, B and λ1I commute
exp(At) = exp(Bt + λ1It) = exp(Bt)exp(λ1It)
-expand the exponentials using the Taylor series:
exp(Bt) = I + Bt + B²t²/2! + …., by B²=0 by the λ1=λ2 proposition
exp(Bt) = I + Bt + 0 + 0 + … = I + Bt
exp(λ1It) = exp(2x2) with entries tλ1, 0, 0, tλ1
= exp(2x2) with entries e^(tλ1), 0, 0, e^(tλ1) = e^(tλ1)I
-sub back in:
exp(At) = (I + Bt) (e^(tλ1)I)
= e^(tλ1)(I + Bt)
= e^(tλ1)(I + t(A-λ1I))

Question 11

Case A) λ1≠λ2

Solution

Answer 11

exp(At) = U(2x2)U^(-1)

-where the 2x2 has entries e^(tλ1), 0, 0, e^(tλ2)

Question 12

Case A) λ1≠λ2

Proof

Answer 12

-consider 2x2 matrix A with eigenvalues λ1, λ2 such that λ1≠λ2
-has eigenvalues |v1, |v2 which will be linearly independent since λ1≠λ2
-let U=(|v1, |v2)
AU = A(|v1,|v2) = (A|v1, A|v2) = (λ1|v1,λ2|v2) = (|v1,|v2)(2x2)
-where 2x2 has entries, λ1, 0, 0, λ2, so:
AU = U(2x2)
AUU^(-1) = A = U(2x2)U^(-1)
-take exponential:
exp(At) = exp(U(2x2)U^(-1)) where 2x2 now has entries tλ1, 0, 0, tλ2
exp(At) = exp(U(2x2)U^(-1)) = Uexp(2x2))U^(-1)
exp(At) = U(2x2)U^(-1)
-where (2x2) now has entries e^(tλ1), 0, 0, e^(tλ2)

Question 13

Behaviour at Hyperbolic vs Non-Hyperbolic Fixed Points

Answer 13

• at a non-hyperbolic point a small peterbation of the system can completely change the nature stability of the fixed poin
• at a hyperbolic fixed point this doesn’t happen

Question 14

Hyperbolic Fixed Point

Linear System Definition

Answer 14

• a fixed point at the origin of the linear system |x’=A|x is hyperbolic if each eigenvalue of A has non-zero real part
• otherwise it is called a non-hyperbolic fixed point

Question 15

Hyperbolic Fixed Point

Non-Linear System Definition

Answer 15

-a fixed point |x* of a non-linear system is hyperbolic if it is hyperbolic for the corresponding linearised system

Question 16

Hartman-Grobman Theorem

Answer 16

• the local phase portrait near a hyperbolic fixed point is ‘topologically equivalent’ to the phase portrait of the linearised system
• here ‘topological equivalence’ means there is a homomorphism that maps one phase portrait onto the other, i.e. trajectories map onto trajectories and the orientation is preserved

Question 17

Lotka-Volterra System

Answer 17

-Lotka-Volterra systems are used to model populations:
x’ = x(a-bx-cy)
y’ = y(d-ey-fx)
-where x and y are the populations of two different species
-a and d are the growth of the population from reproduction
-b and e are control of population size from amount of food, disease etc.
-c and f are competition terms from interaction between the two populations
-looking at the phase portrait, each point can be treated as an initial population and following the trajectory shows how the that population would evolve with time

Question 18

Dulac’s Negative Criterion

Theorem

Answer 18

let x’=f(x,y) and y’=g(x,y) be a continuously differentiable vector field in a simply connected region R⊆ℝ
-if there exists a continuously differentiable (in R) function h(x,y) such that:
∂h(x,y)f(x,y)/∂x + ∂h(x,y)g(x,y)/∂y
-is positive (or negative) throughout R, then there are no periodic orbits lying entirely within R

Question 19

Dulac’s Negative Criterion

Proof

Answer 19

-proof by contradiction

Question 20

Lyapunov Function

Definition

Answer 20

• suppose that x*ϵℝ^n is a fixed point of a dynamical system |x’=|F(|x)
• let R be an open neighbourhood of |x*, and R^ be the closure of R (i.e. R including the boundary)
• let V(x): R^->ℝ be a continuously differentiable function
• a function V(x) satisfying the following conditions
  i) V(|x) = 0
  ii) V(x)>0 for all xϵR, x≠x, (V is positive definite)
  iii) V’<0 for all xϵR, x≠x*, (V’ is negative definite)
• is called a Lyapunov function

Question 21

Weak Lyapunov Function

Definition

Answer 21

• a function V(x) satisfying the following conditions:
  i) V(|x)=0
  ii) V(|x)>0 for all xϵR, x≠x, (V is positive definite)
  iii) V’≤0 for all xϵR, x≠x*, (V is negative semi-definite)
• is called a weak Lyapunov function

Question 22

Implications of the Existence of a Lyapunov Function

Answer 22

1) there are no other fixed points except |x* in R^
2) for Lyapunov or weak Lyapunov, there is a region B⊆R of |x* such that all trajectories starting in B will stay in B, for Lyapunov (not weak) existence also implies that trajectories starting in B tend towards the fixed point |x*
3) there are no periodic orbits in R^
4) |x* is an asymptotically stable point or if V is a weak Lyapunov function, |x* is a stable point

Question 23

Stable Fixed Point of a Dynamical System

Informal Definition

Answer 23

-a fixed point |xof a dynamical system |x’=|F(|x) is called stable if solutions with initial conditions sufficiently close to |x exist for all t>0 and stay close to |x* at any t>0

Question 24

Stable Fixed Point of a Dynamical System

Formal Definition

Answer 24

• a fixed point |x* of a dynamical system |x’=|F(|x) is called stable if:
• > for any ε>0 there exists 𝛿>0 such that all solutions |x(t) with initial conditions |x(0)=|xo, ||x0-|x|0 and ||x(t)-|x|

Question 25

Asymptotically Stable Fixed Point of a Dynamical System

Definition

Answer 25

-a fixed point is asymptotically stable if it is stable AND |x(t)->|x as t->∞

Question 26

Lyapunov Functions and Stability of Fixed Points

Answer 26

• the existence of a weak Lyapunov function guaranties the stability of the fixed point
• the existence of a Lyapunov function guaranties the asymptotic stability of the point

Question 27

Global Lyapunov Functions

Answer 27

• if a Lyapunov function V(x) exists for all |xϵℝ^n then it is a global Lyapunov function
• this means that the system has only one fixed point which is a global attractor
• i.e. every trajectory tends to this fixed point as t->∞
• and the existence of a global Lyapunov function rules out any periodic orbits

Question 28

How to find a Lyapunov function

Answer 28

• there is no set formula
• good candidates for the fixed point |x=0 are:
  i) V(x) = Σanxn²
  ii) V(x) = Σ an*xn^(2mn)
• all n are subscripts, sums taken between n=1 and n=N, mnϵℕ
• for some choices of positive constants an and positive integers an