L ECTURE 4: D YNAMICAL S YSTEMS 3 I NSTRUCTOR : G IANNI A. D I C ARO - - PowerPoint PPT Presentation

LECTURE 4: DYNAMICAL SYSTEMS 3

INSTRUCTOR: GIANNI A. DI CARO

15-382 COLLECTIVE INTELLIGENCE – S18

2

EQUILIBRIUM

§ A state 𝒚" is said an equilibrium state of a dynamical system 𝒚̇ = 𝒈(𝒚), if and only if 𝒚" = 𝒚 𝑢; 𝒚";𝒗 𝑢 = 0 , ∀ 𝑢 ≥ 0 § If a trajectory reaches an equilibrium state (and if no input is applied) the trajectory will stay at the equilibrium state forever: internal system’s dynamics doesn’t move the system away from the equilibrium point, velocity is null: 𝒈 𝒚" = 0

3

IS THE EQUILIBRIUM STABLE?

Stable equilibrium Unstable equilibrium Neutral equilibrium When a displacement (a force) is applied to an equilibrium condition: Metastable equilibrium § Why are equilibrium properties so important? § For the same definition of an abstract model

• f a (complex) real-world scenario

4

SANDPILES, SNOW AVALANCHES AND META-STABILITY

Abelian sandpile model (starting with

• ne billion grains pile in the center)

5

LYAPUNOUV VS. STRUCTURAL EQUILIBRIUM

𝑞 𝑞 𝑞

§ Lyapunouv equilibrium: stability of an equilibrium with respect to a small deviation from the equilibrium point

§ Structural equilibrium: is the equilibrium persistent to (small) variations in the structure of the systems? à Sensitivity to the value of the parameters of the vector field 𝒈

6

IS THE EQUILIBRIUM (LYAPUNOUV) STABLE?

§ An equilibrium state 𝒚" is said to be Lyapunouv stable if and only if for any ε > 0, there exists a positive number 𝜀 𝜁 such that the inequality 𝒚 0 − 𝒚" ≤ 𝜀 implies that 𝒚 𝑢; 𝒚 0 ,𝒗 𝑢 = 0 − 𝒚" ≤ ε ∀ 𝑢 ≥ 0 § An equilibrium state 𝒚" is stable (in the Lyapunouv sense) if the response following after starting at any initial state 𝒚 0 that is sufficiently near 𝒚" will not move the state far away from 𝒚" 𝑢

7

IS THE EQUILIBRIUM (LYAPUNOUV) STABLE?

What is the difference between a stable and an asymptotically stable equilibrium?

8

IS THE EQUILIBRIUM ASYMPTOTICALLY STABLE?

§ If an equilibrium state 𝒚" is Lyapunouv stable and every motion starting sufficiently near to 𝒚" converges (goes back) to 𝒚" as 𝑢 → ∞ , the equilibrium is said asymptotically stable 𝑢 𝜁,𝜀 𝜁 →0 as 𝑢 → ∞

9

SOLUTION OF LINEAR ODES

§ The general form for a linear ODE: 𝒚̇ = 𝐵𝒚, 𝒚 ∈ ℝ<, 𝐵 an 𝑜×𝑜 coefficient matrix § A solution is a differentiable function 𝒀 𝑢 that satisfies the vector field § Theorem: Linear combination of solutions of a linear ODE If the vector functions 𝒚(@) and 𝒚(A) are solutions of the linear system 𝒚̇ = 𝒈(𝒚), then the linear combination 𝑑@𝒚(@) + 𝑑A𝒚(A) is also a solution for any real constants 𝑑@ and 𝑑A § Corollary: Any linear combination of solutions is a solution By repeatedly applying the result of the theorem, it can be seen that every finite linear combination 𝒚 𝑢 = 𝑑@𝒚 @ (𝑢) + 𝑑A𝒚 A (𝑢) + …𝑑E𝒚 E (𝑢)

• f solutions 𝒚 @ , 𝒚 A ,…,𝒚 E is itself a solution to 𝒚̇ = 𝒈(𝒚)

§ The general form for an ODE: 𝒚̇ = 𝒈(𝒚), where 𝒈 is a 𝑜 -dim vector field

10

FUNDAMENTAL AND GENERAL SOLUTION OF LINEAR ODES

§ Theorem: Linearly independent solutions If the vector functions 𝒚 @ , 𝒚 A ,…, 𝒚 < are linearly independent solutions of the 𝑜-dim linear system 𝒚̇ = 𝒈(𝒚), then, each solution 𝒚(𝑢) can be expressed uniquely in the form: 𝒚 𝑢 = 𝑑@𝒚 @ (𝑢) + 𝑑A𝒚 A (𝑢) + …𝑑<𝒚 < (𝑢) § Corollary: Fundamental and general solution of a linear system If solutions 𝒚 @ , 𝒚 A ,…, 𝒚 < are linearly independent (for each point in the time domain), they are fundamental solutions on the domain, and the general solution to a linear 𝒚̇ = 𝒈(𝒚), is given by: 𝒚 𝑢 = 𝑑@𝒚 @ (𝑢) + 𝑑A𝒚 A (𝑢) + …𝑑<𝒚 < (𝑢)

11

GENERAL SOLUTIONS FOR LINEAR ODES

§ Corollary: Non-null Wronskian as condition for linear independence The proof of the theorem uses the fact that if 𝒚 @ , 𝒚 A ,…, 𝒚 < are linearly independent (on the domain), then det 𝒀 𝑢 ≠ 0 𝒀(𝑢) = 𝑦@@(𝑢) ⋯ 𝑦@<(𝑢) ⋮ ⋱ ⋮ 𝑦<@(𝑢) ⋯ 𝑦<<(𝑢) Therefore, 𝒚 @ , 𝒚 A ,…, 𝒚 < are linearly independent if and only if W[𝒚 @ , 𝒚 A ,…, 𝒚 < ](𝑢) ≠ 0

Wronskian

§ Theorem: Use of the Wronskian to check fundamental solutions If 𝒚 @ , 𝒚 A ,…, 𝒚 < are solutions, then the Wroskian is either identically to zero or else is never zero for all 𝑢 § Corollary: To determine whether a given set of solutions are fundamental solutions it suffices to evaluate W[𝒚 @ , 𝒚 A ,…,𝒚 < ](𝑢) at any point 𝑢

12

STABILITY OF LINEAR MODELS

§ Let’s start by studying stability in linear dynamical systems … § The general form for a linear ODE: 𝒚̇ = 𝐵𝒚, 𝒚 ∈ ℝ<, 𝐵 an 𝑜×𝑜 coefficient matrix § Equilibrium points are the points of the Null space / Kernel of matrix 𝐵 𝐵𝒚 = 𝟏, 𝑜×𝑜 homogeneous system § Invertible Matrix Theorem, equivalent facts: § 𝐵 is invertible ⟷ det 𝐵 ≠ 0 § The only solution is the trivial solution, 𝒚 = 𝟏 § Matrix 𝐵 has full rank § det 𝐵 = ∏ 𝜇U

< UV@

, all eigenvalues are non null § … § In a linear dynamical system, solutions and stability of the origin depends on the eigenvalues (and eigenvectors) of the matrix 𝐵

13

RECAP ON EIGENVECTORS AND EIGENVALUES

Geometry: § Eigenvectors: Directions 𝒚 that the linear transformation 𝐵 doesn’t change. § The eigenvalue 𝜇 is the scaling factor of the transformation along 𝒚 (the direction that stretches the most)

Algebra: § Roots of the characteristic equation § 𝑄 𝜇 = 𝜇𝑱 − 𝐵 𝒚 = 0 → det 𝜇𝑱 − 𝐵 = 0 § For 2×2 matrices: det 𝜇𝑱 − 𝐵 = 𝜇A − 𝜇 tr 𝐵 + det 𝐵 § Algebraic multiplicity 𝒐: each eigenvalue can be repeated 𝑜 ≥ 1 times (e.g., (𝜇 − 3)A, 𝑜 = 2) § Geometric multiplicity 𝒏: Each eigenvalue has at least one or 𝑛 ≥ 1 eigenvectors, and only 1 ≤ 𝑟 ≤ 𝑛 can be linearly independent § An eigenvalue can be 0, as well as can be a real or a complex number

14

RECAP ON EIGENVECTORS AND EIGENVALUES

15

LINEAR MULTI-DIMENSIONAL MODELS

§ A two-dimensional example: 𝑦̇@ = −4𝑦@ − 3𝑦A 𝑦̇A = 2𝑦@ + 3𝑦A 𝒚(0) = (1,1) 𝒚 = 𝑦@ 𝑦A 𝐵 = −4 −3 2 3 § Eigenvalues and Eigenvectors of 𝐵: 𝜇@ = 2, 𝒗@ = 1 −2 𝜇A = −3, 𝒗A = 3 −1 § For the case of linear (one dimensional) growth model, 𝑦̇ = 𝑏𝑦, solutions were in the form: 𝑦 𝑢 = 𝑦c𝑓ef § The sign of a would affect stability and asymptotic behavior: x = 0 is an asymptotically stable solution if a < 0, while x = 0 is an unstable solution if a > 0, since other solutions depart from x = 0 in this case. § Does a multi-dimensional generalization of the form 𝒚 𝑢 = 𝒚c𝑓𝑩f hold? What about operator 𝑩? (real, positive) (real, negative)

16

SOLUTION (EIGENVALUES, EIGENVECTORS)

§ The eigenvector equation: 𝐵𝒗 = 𝜇𝒗 § Let’s set the solution to be 𝒚 𝑢 = 𝑓hf𝒗 and lets’ verify that it satisfies the relation 𝒚̇ 𝑢 = 𝐵𝒚 § Multiplying by 𝐵: 𝐵𝒚(𝑢) = 𝑓hf𝐵𝒗 , but since 𝒗 is an eigenvector: 𝐵𝒚 𝑢 = 𝑓hf𝐵𝒗 = 𝑓hf(𝜇𝒗) § 𝒗 is a fixed vector, that doesn’t depend on 𝑢 → if we take 𝒚 𝑢 = 𝑓hf𝒗 and differentiate it: 𝒚̇ 𝑢 = 𝜇𝑓hf𝒗, which is the same as 𝐵𝒚 𝑢 above Each eigenvalue-eigenvector pair (𝜇, 𝒗) of 𝐵 leads to a solution of 𝒚̇ 𝑢 = 𝐵𝒚 , taking the form: 𝒚 𝑢 = 𝑓hf𝒗

𝒚 𝑢 = 𝑑@𝑓hif𝒗@ + 𝑑A𝑓hjf𝒗A

§ The general solution to the linear ODE is obtained by the linear combination of the individual eigenvalue solutions (since 𝜇@ ≠ 𝜇A, 𝒗𝟐 and 𝒗𝟑 are linearly independent)

17

SOLUTION (EIGENVALUES, EIGENVECTORS)

𝒚 𝑢 = 𝑑@𝑓hif𝒗@ + 𝑑A𝑓hjf𝒗A

𝒚 0 = (1,1) 1,1 = 𝑑@(1,−2) + 𝑑A(3,−1) à 𝑑@ = −4/5 𝑑A = 3/5

𝒚 𝑢 = −4/5𝑓Af𝒗@ + 3/5𝑓opf𝒗A

𝑦@ 𝑢 = − 4 5 𝑓Af+ 9 5 𝑓opf 𝑦A 𝑢 = 8 5 𝑓Af− 3 5 𝑓opf

𝑦A 𝑦@ 𝒗@ 𝒗𝟑

(1,1)

§ Except for two solutions that approach the origin along the direction of the eigenvector 𝒗A =(3, -1), solutions diverge toward ∞, although not in finite time § Solutions approach to the origin from different direction, to after diverge from it Saddle equilibrium (unstable)

18

TWO REAL EIGENVALUES, OPPOSITE SIGNS

𝑦A 𝑦@ 𝒗@ 𝒗𝟑

(1,1)

§ The straight lines corresponding to 𝒗@ and 𝒗𝟑 are the trajectories corresponding to all multiples of individual eigenvector solutions 𝐷𝑓hf𝒗: 𝒗@: 𝑦@ 𝑢 𝑦A 𝑢 = 𝑑@ 𝑓Af 1 −2 𝒗A: 𝑦@ 𝑢 𝑦A 𝑢 = 𝑑A 𝑓opf 3 −1 § The eigenvectors corresponding to the same eigenvalue 𝜇, together with the origin (0,0) (which is part of the solution for each individual eigenvalue), form a linear subspace, called the eigenspace of λ § The two straight lines are the two eigenspaces, that, as 𝑢 → ∞, play the role of “separators” for the different behaviors of the system § The slope of a trajectory corresponding to one eigenvalue is constant in (𝑦@,𝑦A) à It’s a line in the phase space (e.g., for 𝒗@:

vj vi 𝑢 = wij wii = −2)