Math 211 Math 211 Lecture #36 Forced Harmonic Motion Nonlinear - PowerPoint PPT Presentation

1 Math 211 Math 211 Lecture #36 Forced Harmonic Motion Nonlinear Systems November 21, 2001

2 Forced, Damped Harmonic Motion Forced, Damped Harmonic Motion x ′′ + 2 cx ′ + ω 2 0 x = A cos ωt • Ch. polynomial: P ( λ ) = λ 2 + 2 cλ + ω 2 0 • General Solution x ( t ) = G ( ω ) A cos( ωt − φ ) + x h ( t ) . � Transient term x h ( t ) dies out exponentially. � Steady-state solution x p ( t ) = G ( ω ) A cos( ωt − φ ) . 0 − ω 2 ) 2 + 4 c 2 ω 2 . � ◮ Gain: G ( ω ) = 1 / ( ω 2 ◮ Phase: φ = arccot � ( ω 2 0 − ω 2 ) / 2 cω � . Return

3 Steady-State Solution Steady-State Solution x p ( t ) = G ( ω ) A cos( ωt − φ ) . • The forcing function is A cos ωt . • The steady-state response is oscillatory. � The amplitude is G ( ω ) times the amplitude of the forcing term. � The steady-state oscillation is at the forcing frequency. � There is a phase shift of φ/ω . Return

4 Interacting Species Interacting Species • Two species with populations x 1 & x 2 . • Interaction between the species can be helpful or detrimental. • Basic model x ′ 1 = r 1 x 1 x ′ 2 = r 2 x 2 • r 1 & r 2 are the reproductive rates. Return

5 Reproductive Rates Reproductive Rates • If x 2 = 0 the reproductive rate for x 1 is r 1 = a 1 − b 1 x 1 . � a 1 > 0 ⇒ natural growth. � a 1 < 0 ⇒ natural decline. � b 1 = 0 Malthusian growth. � b 1 > 0 logistic growth. Return

6 • If x 2 > 0 the reproductive rate for x 1 is r 1 = a 1 − b 1 x 1 + c 1 x 2 . � c 1 > 0 ⇒ interaction is helpful to x 1 . � c 1 < 0 ⇒ interaction is detrimental to x 1 . � The reproductive rate for x 2 is r 2 = a 2 − b 2 x 2 + c 2 x 1 . • The model for interacting species is x ′ 1 = ( a 1 − b 1 x 1 + c 1 x 2 ) x 1 x ′ 2 = ( a 2 − b 2 x 2 + c 2 x 1 ) x 2 Return

7 Predator Prey Model Predator Prey Model Rabbits & foxes, fish & sharks, and cottony cushion scale insect & ladybird beetle. • F = fish & S = sharks. F ′ = ( a − bS ) F S ′ = ( − c + dF ) S or F ′ = ( a − eF − bS ) F S ′ = ( − c + dF ) S a = 3 , b = 3 , c = 1 , d = 3 , e = 3 . Return Interacting species Reproductive rates

8 Competing Species Competing Species Cattle and sheep. • x 1 and x 2 competing for resources. x ′ 1 = ( a 1 − b 1 x 1 + c 1 x 2 ) x 1 x ′ 2 = ( a 2 − b 2 x 2 + c 2 x 1 ) x 2 � a i > 0 , b i > 0 , & c i < 0 • Example: x ′ = (5 − 2 x − y ) x y ′ = (7 − 2 x − 3 y ) y Return

9 Linearization Linearization The principal idea of differential calculus: • Approximate nonlinear mathematical objects by linear ones. • Example: Approximate the function f ( y ) near y 0 by a linear function. f ( y 0 + h ) = f ( y 0 ) + f ′ ( y 0 ) h + R ( h ) R ( h ) where lim = 0 . h h → 0 � The linear function is L ( h ) = f ( y 0 ) + f ′ ( y 0 ) h. Return

10 Linearization of an ODE Linearization of an ODE y ′ = f ( y ) • Assume f ( y 0 ) = 0 and f ′ ( y 0 ) � = 0 . • Set y = y 0 + u . Get u ′ = f ( y 0 + u ) = f ′ ( y 0 ) u + R ( u ) • Approximate by the linear differential equation u ′ = f ′ ( y 0 )˜ ˜ u Return Taylor’s theorem

11 • If f ′ ( y 0 ) � = 0 the equilibrium point of the linearization at 0 has the same stability properties as that of the nonlinear equation at y 0 . � f ′ ( y 0 ) > 0 ⇒ y 0 is unstable. � f ′ ( y 0 ) < 0 ⇒ y 0 is asymptotically stable. • We can solve the linearization explicitly. Return

12 Linearization of a Planar System Linearization of a Planar System x ′ = f ( x, y ) y ′ = g ( x, y ) • Asume ( x 0 , y 0 ) is an equilibrium point, so f ( x 0 , y 0 ) = g ( x 0 , y 0 ) = 0 Return

13 We have by Taylor’s theorem f ( x 0 + u, y 0 + v ) = ∂f ∂x ( x 0 , y 0 ) u + ∂f ∂y ( x 0 , y 0 ) v + R f ( u, v ) g ( x 0 + u, y 0 + v ) = ∂g ∂x ( x 0 , y 0 ) u + ∂g ∂y ( x 0 , y 0 ) v + R g ( u, v ) where R f ( u, v ) u 2 + v 2 → 0 and R g ( u, v ) √ √ u 2 + v 2 → 0 Return System

14 • Set x = x 0 + u and y = y 0 + v . The system becomes u ′ = ∂f ∂x ( x 0 , y 0 ) u + ∂f ∂y ( x 0 , y 0 ) v + R f ( u, v ) v ′ = ∂g ∂x ( x 0 , y 0 ) u + ∂g ∂y ( x 0 , y 0 ) v + R g ( u, v ) Return Taylor’s theorem

15 Linearization at ( x 0 , y 0 ) Linearization at ( x 0 , y 0 ) u ′ = ∂f u + ∂f ˜ ∂x ( x 0 , y 0 )˜ ∂y ( x 0 , y 0 )˜ v v ′ = ∂g u + ∂g ˜ ∂x ( x 0 , y 0 )˜ ∂y ( x 0 , y 0 )˜ v • This is a linear system. � We can solve it explicitly. � Does it give information about the original nonlinear system? Return Original system Nonlinear system Matrix form

16 Matrix Form of the Linearization Matrix Form of the Linearization v ) T and introduce the Jacobian matrix Set u = (˜ u, ˜ ∂f ∂f   ∂x ( x 0 , y 0 ) ∂y ( x 0 , y 0 ) J =   ∂g ∂g   ∂x ( x 0 , y 0 ) ∂y ( x 0 , y 0 ) • The linearization becomes u ′ = J u . Return Linear system Original system

17 Theorem: Consider the planar system x ′ = f ( x, y ) y ′ = g ( x, y ) where f and g are continuously differentiable. Suppose that ( x 0 , y 0 ) is an equilibrium point. If the linearization at ( x 0 , y 0 ) has a generic equilibrium point at the origin, then the equilibrium point at ( x 0 , y 0 ) is of the same type. Return Matrix form Generic

18 Generic Equilibrium Points Generic Equilibrium Points • Saddle, nodal source, nodal sink, spiral source, and spiral sink. � All occupy large open subsets of the trace-determinant plane. • Nongeneric types � Center and others. Occupy pieces of the boundaries. Return Theorem

19 Examples Examples • Predator prey • Competing species • Center x ′ = y + αx ( x 2 + y 2 ) y ′ = − x + αy ( x 2 + y 2 ) Return Theorem Generic