L ECTURE 5: D YNAMICAL S YSTEMS 4 I NSTRUCTOR : G IANNI A. D I C ARO - - PowerPoint PPT Presentation

LECTURE 5: DYNAMICAL SYSTEMS 4

INSTRUCTOR: GIANNI A. DI CARO

15-382 COLLECTIVE INTELLIGENCE – S18

2

LINEAR MULTI-DIMENSIONAL MODELS

§ A two-dimensional example: 𝑦̇$ = −4𝑦$ − 3𝑦) 𝑦̇) = 2𝑦$ + 3𝑦) 𝒚(0) = (1,1) 𝒚 = 𝑦$ 𝑦) 𝐵 = −4 −3 2 3 § Eigenvalues and Eigenvectors of 𝐵: 𝜇$ = 2, 𝒗$ = 1 −2 𝜇) = −3, 𝒗) = 3 −1 § For the case of linear (one dimensional) growth model, 𝑦̇ = 𝑏𝑦, solutions were in the form: 𝑦 𝑢 = 𝑦4𝑓67 § 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 = 𝒚4𝑓𝑩7 hold? What about operator 𝑩? (real, positive) (real, negative)

3

SOLUTION (EIGENVALUES, EIGENVECTORS)

§ The eigenvector equation: 𝐵𝒗 = 𝜇𝒗 § Let’s set the solution to be 𝒚 𝑢 = 𝑓97𝒗 and lets’ verify that it satisfies the relation 𝒚̇ 𝑢 = 𝐵𝒚 § Multiplying by 𝐵: 𝐵𝒚(𝑢) = 𝑓97𝐵𝒗 , but since 𝒗 is an eigenvector: 𝐵𝒚 𝑢 = 𝑓97𝐵𝒗 = 𝑓97(𝜇𝒗) § 𝒗 is a fixed vector, that doesn’t depend on 𝑢 → if we take 𝒚 𝑢 = 𝑓97𝒗 and differentiate it: 𝒚̇ 𝑢 = 𝜇𝑓97𝒗, 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 = 𝑓97𝒗

𝒚 𝑢 = 𝑑$𝑓9>7𝒗$ + 𝑑)𝑓9?7𝒗)

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

4

SOLUTION (EIGENVALUES, EIGENVECTORS)

𝒚 𝑢 = 𝑑$𝑓9>7𝒗$ + 𝑑)𝑓9?7𝒗)

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

𝒚 𝑢 = −4/5𝑓)7𝒗$ + 3/5𝑓FG7𝒗)

𝑦$ 𝑢 = − 4 5 𝑓)7+ 9 5 𝑓FG7 𝑦) 𝑢 = 8 5 𝑓)7− 3 5 𝑓FG7

𝑦) 𝑦$ 𝒗$ 𝒗𝟑

(1,1)

§ Except for two solutions that approach the origin along the direction of the eigenvector 𝒗) =(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)

5

TWO REAL EIGENVALUES, OPPOSITE SIGNS

𝑦) 𝑦$ 𝒗$ 𝒗𝟑

(1,1)

§ The straight lines corresponding to 𝒗$ and 𝒗𝟑 are the trajectories corresponding to all multiples of individual eigenvector solutions 𝐷𝑓97𝒗: 𝒗$: 𝑦$ 𝑢 𝑦) 𝑢 = 𝑑$ 𝑓)7 1 −2 𝒗): 𝑦$ 𝑢 𝑦) 𝑢 = 𝑑) 𝑓FG7 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$,𝑦)) à It’s a line in the phase space (e.g., for 𝒗$:

N? N> 𝑢 = O>? O>> = −2)

6

TWO REAL EIGENVALUES, SAME SIGN

𝒚 𝑢 = 𝑑$𝑓9>7𝒗$ + 𝑑)𝑓9?7𝒗)

Node

𝑦̇$ = −2𝑦$ 𝑦̇) = 𝑦$ − 4𝑦) § Asymptotically Stable or unstable behavior depending on the sign of 𝜇$) § Trajectories either moving away from the equilibrium to infinite-distant away (when 𝜇 > 0), or moving directly toward, and converge to equilibrium (when 𝜇 < 0). § The trajectories that are the eigenvectors move in straight lines.

7

ONE REAL, REPEATED EIGENVALUE

§ Case with two linearly independent eigenvectors Focus Proper node (star point)

𝒚 𝑢 = 𝑓9>7(𝑑$𝒗$ + 𝑑)𝒗))

§ Every nonzero solution traces a straight-line trajectory: constant slope, direction given by the linear combination of the eigenvectors It is unstable if the eigenvalue is positive; asymptotically stable if the eigenvalue is negative.

8

ONE REAL, REPEATED EIGENVALUE

§ Case with two linearly independent eigenvectors

𝒚 𝑢 = 𝑓9>7(𝑑$𝒗$ + 𝑑)𝒗))

9

ONE REAL, REPEATED EIGENVALUE

𝑦$ 𝑦)

All solutions except for the equilibrium diverge to infinity Repulsive focus (Improper node) Negative eigenvalue § Case with one linearly independent eigenvectors

𝒚 𝑢 = 𝑑$𝒗$𝑓9>7 + 𝑑)(𝒗$𝑢𝑓9>7 + 𝒗𝑓9>7)

Positive eigenvalue § One eigenvalue solution: 𝑓9>7𝒗$ § Need to find another solution, linearly independent § From solutions of the form: 𝑢𝑓9>7𝒗$ + 𝒗 𝑓9>7 § 𝒗 is a generalized eigenvector that can be determined from 𝐵 − 𝜇$𝐽 𝒗$ = 𝒗 ~ (Node + Spiral)

10

IMAGINARY EIGENVALUES

§ Roots of the characteristic equation are complex numbers: § If Λ is a complex eigenvalue, then its conjugate Λ S is also an eigenvalue. § If 𝒗 is a complex eigenvector of Λ, then 𝒗 S, the complex conjugate of its entries, is an eigenvector associated to Λ S § The solutions for the two conjugate eigenvalues: § 𝑓(9TUV)7𝒗 = 𝑓97𝑓UV7𝒗 = 𝑓97(cos𝜈𝑢 + 𝑗 sin𝜈𝑢)𝒗 § 𝑓(9FUV)7 𝒗 S = 𝑓97𝑓FUV7 𝒗 S = 𝑓97(cos 𝜈𝑢 − 𝑗 sin𝜈𝑢) 𝒗 S Λ± = 𝜇 ± 𝑗𝜈 § Let 𝒗 =

𝒗𝟐 𝒗𝟑 , 𝒗

S =

𝒗 S𝟐 𝒗 S𝟑 , 𝑣$ = 𝛽$ + 𝑗𝛾$, 𝑣) = 𝛽) + 𝑗𝛾)

𝑓(9TUV)7𝑣$ = 𝑓97(cos𝜈𝑢 + 𝑗 sin𝜈𝑢) (𝛽$ + 𝑗𝛾$) 𝑓(9TUV)7𝑣) = 𝑓97(cos 𝜈𝑢 + 𝑗 sin𝜈𝑢) (𝛽) + 𝑗𝛾)) 𝑓(9FUV)7𝑣 b$ = 𝑓97(cos𝜈𝑢 − 𝑗 sin𝜈𝑢) (𝛽$ − 𝑗𝛾$) 𝑓(9FUV)7𝑣 b) = 𝑓97(cos 𝜈𝑢 − 𝑗 sin𝜈𝑢) (𝛽) − 𝑗𝛾))

𝑓(9TUV)7𝒗 = 𝑓(9FUV)7 𝒗 S =

11

IMAGINARY EIGENVALUES

𝑓97[(𝛽$cos𝜈𝑢 − 𝛾$ sin𝜈𝑢) + 𝑗(𝛾$ cos 𝜈𝑢 + 𝛽$ sin𝜈𝑢)]

𝑓(9TUV)7𝒗 = 𝑓(9FUV)7 𝒗 S =v

𝑓97[(𝛽)cos𝜈𝑢 − 𝛾) sin𝜈𝑢) + 𝑗(𝛾) cos𝜈𝑢 + 𝛽) sin𝜈𝑢)] 𝑓97[(𝛽$cos𝜈𝑢 − 𝛾$ sin𝜈𝑢) − 𝑗(𝛾$ cos 𝜈𝑢 + 𝛽$ sin𝜈𝑢)] 𝑓97[(𝛽)cos𝜈𝑢 − 𝛾) sin𝜈𝑢) − 𝑗(𝛾) cos𝜈𝑢 + 𝛽) sin𝜈𝑢)] 𝒁$ = 𝑓97 𝛽$ cos𝜈𝑢 − 𝛾$ sin𝜈𝑢 𝛽) cos𝜈𝑢 − 𝛾) sin𝜈𝑢 𝒁) = 𝑓97 𝛾$ cos 𝜈𝑢 + 𝛽$ sin𝜈𝑢 𝛾) cos 𝜈𝑢 + 𝛽) sin𝜈𝑢

𝑓(9TUV)7𝒗 = 𝒁$ + 𝑗𝒁) 𝑓(9FUV)7 𝒗 S = 𝒁$ − 𝑗𝒁)

§ Since both sum and difference of 𝒁$ and 𝒁) are solutions, also 𝒁$ and 𝒁) are solutions; moreover, it can be proved that they are linearly independent 𝒚 𝑢 = 𝑑$𝒁$ + 𝑑)𝒁) is a general solution (real, combining real solutions) 𝑦$ 𝑢 = 𝑑$𝑓97 𝛽$ cos𝜈𝑢 − 𝛾$ sin𝜈𝑢 + 𝑑)𝑓97(𝛾$ cos𝜈𝑢 + 𝛽$ sin𝜈𝑢) 𝑦) 𝑢 = 𝑑$𝑓97 𝛽) cos𝜈𝑢 − 𝛾) sin𝜈𝑢 + 𝑑)𝑓97(𝛾) cos𝜈𝑢 + 𝛽) sin𝜈𝑢)

12

(PURE) IMAGINARY EIGENVALUES

𝑦̇$ = 𝑦) 𝑦̇)= −𝑦$ 𝒚 = 𝑦$ 𝑦) 𝐵 = 1 −1 § Eigenvalues 𝐵: Λ$ = 𝑗, Λ) = −𝑗 (𝜇 = 0, 𝜈 = 1) (pure imaginary, conjugate) § General solutions: 𝑦$ 𝑢 = 𝑑$sin𝑢 − 𝑑) cos𝑢 𝑦) 𝑢 = 𝑑$cos𝑢 − 𝑑 sin𝑢 Center

𝑦$ 𝑦)

§ Periodic solutions § Some points initially move farther away, but not too far away. § The origin is stable but not attracting.

𝑦$ 𝑢 = 𝑑$𝑓97 𝛽$ cos 𝜈𝑢 − 𝛾

$ sin 𝜈𝑢 + 𝑑)𝑓97(𝛾 $ cos𝜈𝑢 + 𝛽$ sin 𝜈𝑢)

𝑦) 𝑢 = 𝑑$𝑓97 𝛽) cos𝜈𝑢 − 𝛾) sin 𝜈𝑢 + 𝑑)𝑓97(𝛾) cos𝜈𝑢 + 𝛽) sin 𝜈𝑢)

§ Eigenvectors: 𝒗 = 0 + 𝑗(−1) 1 + 𝑗(0) = −𝑗 1 𝒗 S = 𝑗 1 𝛽$ = 0, 𝛾$ = −1,𝛽) = 1,𝛾) = 0

13

THE TWO NEIGHBORHOODS OF LYAPUNOUV STABILITY

𝑦$ 𝑦)

𝜇± = ±𝑗 𝜇± = ±3𝑗 𝑦) 𝑦$ § Lyapunov stability needs two neighborhoods, 𝜁,𝜀(𝜁): In order to have solutions stay within a neighborhood 𝜻 whose radius is the larger axis of an ellipse, initial conditions must be restricted to a neighborhood 𝜀(𝜁) whose radius is no larger than the smaller axis of the solution § If the neighborhood 𝜀 𝜁 is equal to the larger axis, then an initial point could be placed on another, larger orbit, that would not satisfy the requirement to stay within the original neighborhood 𝜁

v v

𝜁 𝜀(𝜁)

14

IMAGINARY EIGENVALUES WITH REAL PARTS

§ A more general two-dimensional example: 𝑦̇$ = 𝑏𝑦$ − 𝑐𝑦) 𝑦̇)= 𝑐𝑦$ + 𝑏𝑦) 𝒚 = 𝑦$ 𝑦) 𝐵 = 𝑏 −𝑐 𝑐 𝑏 § Eigenvalues 𝐵: 𝜇± = 𝑏 ± 𝑐 (complex, conjugate) § Solutions: 𝑦$ 𝑢 = 𝑓67 (𝑑$cos𝑐𝑢 + 𝑑) sin𝑐𝑢) 𝑦) 𝑢 = 𝑓67 (−𝑑)cos𝑐𝑢 + 𝑑$ sin𝑐𝑢) 𝑦̇$ = −𝑦$ − 10𝑦) 𝑦̇)= 10𝑦$ − 𝑦) 𝑦̇$ = 𝑦$ − 10𝑦) 𝑦̇)= 10𝑦$ + 𝑦) 𝜇± = −1 ± 𝑗 𝜇± = 1 ± 𝑗

Spiral

15

DEGENERATE CASES

§ det 𝐵 = 0 → One (or more) eigenvalues are zero § If the system has zero as an eigenvalue, then there exists a line of equilibrium points (degenerate case). § Solutions are all straight-line solutions. Depending on the sign of 𝜇 the solution may tend to or get away from the line of equilibrium points parallel to the eigenvector associated to the eigenvalue

𝜇 < 0 𝜇 > 0

16

STABILITY SUMMARY

17

SUMMARY FOR TWO-DIMENSIONAL SYSTEMS

det 𝜇𝑱 − 𝐵 = 𝜇) − 𝜇 tr 𝐵 + det 𝐵

18

SUMMARY FOR TWO-DIMENSIONAL SYSTEMS

19

A ”GALAXY” OF STABILITY CASES

20

NON LINEAR SYSTEMS?

Four equilibria: 0, ±2 , (± 3, 1) Countably infinite equilibria: (𝑜𝜌,0)