Recall: Pendulum. Unstable Pendulum Exponential growth dominates. Equilibrium is unstable.

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Presentation transcript:

Recall: Pendulum

Unstable Pendulum Exponential growth dominates. Equilibrium is unstable.

Recall: Finding eigvals and eigvecs

Nonlinear systems: the qualitative theory Day 8: Mon Sep 20 1.How we solve it (the basic idea). 2.Why it matters. 3.How we solve it (details, examples). Systems of 1st-order, linear, homogeneous equations

Solution: the basic idea

General solution

Systems of 1st-order, linear, homogeneous equations 1.Higher order equations can be converted to 1 st order equations. 2.A nonlinear equation can be linearized. 3.Method extends to inhomogenous equations. Why important?

Conversion to 1 st order

Another example Any higher order equation can be converted to a set of 1 st order equations.

Nonlinear systems: qualitative solution e.g. Lorentz: 3 eqns  chaos Stability of equilibria is a linear problem °qualitative description of solutions phase plane diagram

2-eqns: ecosystem modeling reproduction starvation eating getting eaten

Ecosystem modeling reproduction starvation eating getting eaten OR: Reproduction rate reduced Starvation rate reduced

Equilibria

Linearizing about an equilibrium 2 nd -order (quadratic) nonlinearity

Linearizing about an equilibrium 2 nd -order (quadratic) nonlinearity small really small

The linearized system cancel

The linearized system Phase plane diagram

The “other” equilibrium Section 6 Problem 4 ?

Linear, homogeneous systems

Solution

Interpreting σ

General solution

N=2 case Recall

b. repellor (unstable)a. attractor (stable) c. saddle (unstable) d. limit cycle (neutral) e. unstable spiral f. stable spiral Interpreting two σ’s

Need N>3

b. repellora. attractor c. saddle Interpreting two σ’s both real

d. limit cycle e. unstable spiral f. stable spiral Interpreting two σ’s: complex conjugate pair

b. repellora. attractor c. saddle d. limit cycle e. unstable spiral f. stable spiral Interpreting two σ’s

The mathematics of love affairs R(t)= Romeo’s affection for Juliet J(t) = Juliet’s affection for Romeo Response to own feelings (><0) Response to other person (><0) Strogatz, S., 1988, Math. Magazine 61, 35.

The mathematics of love affairs (S. Strogatz) R(t)= Romeo’s affection for Juliet J(t) = Juliet’s affection for Romeo Response to own feelings (><0) Response to other person (><0)

Example: Out of touch with feelings

Limit cycle R J

Example: Birds of a feather

negative positive if b>a negative if b<a b<a: both negative (romance fizzles) b>a: one positive, one negative (saddle …?) both real c. saddle growth eigvec decay eigvec

Example: Birds of a feather R J

Decaying case: a>b R J

Saddle: a<b R J

R J

Homework Sec. 6, p. 89 #4: Sketch the full phase diagram: ? ? #6: Optional

Why a saddle is unstable R J No matter where you start, things eventually blow up.