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Perturbation Theory & Stability Analysis T. Weinhart, A. Singh, A.R. Thornton May 17, 2010

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Page 1: perturbation theory

Perturbation Theory &Stability Analysis

T. Weinhart, A. Singh, A.R. ThorntonMay 17, 2010

Page 2: perturbation theory

1/21

1 Perturbation Theory

2 Algebraic equationsRegular PerturbationsSingular Perturbations

3 Ordinary differential equationsRegular PerturbationsSingular PerturbationsSingular in the domain

4 The non-linear springNon-uniform solutionUniform solution using Linstead’s MethodPhase-space diagram

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 3: perturbation theory

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Introduction

Perturbation TheoryConsiders the effect of small disturbances in the equation to thesolution of the equation.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 4: perturbation theory

3/21

A first exampleLet x− 2 = 0. Then x = 2. What about

x− 2 = ε cosh(x)?

Assume ε 1 and substitute x = x0 + εx1 + ε2x2 + . . .

(x0 + εx1 + . . . )− 2 = ε cosh(x0 + εx1 + . . . )

and solve separately for each order of ε:

. . . ⇒ x ≈ 2 + ε cosh(2)

ConclusionsThe effect of the small parameter ε is small; therefore theperturbation is said to be regular. Otherwise we call it singular.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 5: perturbation theory

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Order symbol notation

Let f(x), g(x) be two functions defined around x = x0.We say, f = o(g) as x→ x0 iff lim

x→0f/g = 0.

We say, f = O(g) (of order) as x→ x0 iff limx→0

f/g = const 6= 0.

We say, f ∼ g (goes like) as x→ x0 iff limx→0

f/g = 1.

Examples:1 x2 = o(x) as x→ 0, b/c x2/x→ 0.2 3x2 = O(5x2) as x→ 0, b/c (3x2)/(5x2)→ 3/5.3 sin(x) ∼ x as x→ 0, b/c sin(x)/x→ 1.4 sin(x) = x+O(x3) as x→ 0, b/c (sin(x)− x)/x3 → 1/6.

5 sin(x) ∼ x− x3

3! as x→ 0, b/c sin(x)/(x− x3/3!)→ 1.So the solution to x− 2 = ε cosh(x) is x ∼ 2 + ε cosh(2) as ε→ 0.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 6: perturbation theory

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The Fundamental Theorem of Perturbation TheoryIf

A0 +A1ε+ · · ·+Anεn +O(εn+1) = 0

for ε→ 0 and A0, A1, . . . independent of ε, then

A0 = A1 = · · · = An = 0.

That is why we could solve separately for each order of ε:

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 7: perturbation theory

Regular Perturbations 6/21

Regular Perturbations

x2 − 3x+ 2 + ε = 0, ε 1.

Assume the roots have expansion x = x0 + εx1 + ε2x2 + . . . :

(x0 + εx1 + . . . )2 − 3(x0 + εx1 + . . . ) + 2 + ε = 0,

. . . ⇒ x = 1 + ε+ ε2 +O(ε3), 2− ε− ε2 +O(ε3).

Compare with exact solution,

x =3±√

1− 4ε2

=3± (1− 2ε− 2ε2 + . . . )

2

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 8: perturbation theory

Singular Perturbations 7/21

Singular Perturbations

εx2 − 2x+ 1 = 0, ε 1.

Assume the roots have expansion x = x0 + εx1 + ε2x2 + . . . :

ε(x0 + εx1 + . . . )2 − 2(x0 + εx1 + . . . ) + 1 = 0.

. . . ⇒ x = 1/2 + 1/8ε+ 1/16ε2 +O(ε3)

But where is the second solution? Compare with exact solution,

x =1±√

1− εε

=1± (1− ε/2− ε2/8 + . . . )

ε

. . . ⇒ x = 1/2 + 1/8ε+O(ε2), 2/ε− 1/2 +O(ε).

We can find the second solution by solving for w = εx, so

w2 − 2w + ε = 0.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 9: perturbation theory

8/21

Ordinary differential equations

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 10: perturbation theory

Regular Perturbations 9/21

Regular Perturbations

x+ x = εx2, x(0) = 1.

Assume the roots have expansion x = x0 + εx1 + ε2x2 + . . . :

(x0 + εx1 + . . . ) + (x0 + εx1 + . . . ) = ε(x0 + εx1 + . . . )2,x0(0) + εx1(0) + · · · = 1.

. . . ⇒ x = e−t + ε(e−t − e−2t) +O(ε2).

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 11: perturbation theory

Singular Perturbations 10/21

Singular PerturbationsHeat equation: Temperature x for a stone in water

εx = 1− x, x(0) = 0.

Assume the roots have expansion x = x0 + εx1 + ε2x2 + . . . :

(x0 + εx1 + . . . ) = 1− (x0 + εx1 + . . . ).

. . . ⇒ x0 = 1 (Does not satisfy initial conditions).

Again, We can find a solution by transforming, τ = ε−1t,

d

dτx = 1− x, x(0) = 0.

. . . ⇒ x = 1− e−2τ = 1− e−2t/ε.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 12: perturbation theory

Singular in the domain 11/21

Singular in the domain

x+ εx2 = 1, x(0) = 0.

Assume the roots have expansion x = x0 + εx1 + ε2x2 + . . . :

(x0 + εx1 + . . . ) + ε(x0 + εx1 + . . . )2 = 1.

. . . ⇒ x ∼ t− εt3

3− ε2 2t5

15.

This solution is non-uniform (blows up for large t).x ∼ t− ε t33 is an asymptotic expansion for t <

√3/ε.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 13: perturbation theory

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The non-linear spring

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 14: perturbation theory

Non-uniform solution 13/21

Duffing’s equationConsider a springs with a small non-linear perturbation

x+ ω2x+ εx3 = 0, x(0) = a, x(0) = 0.

with ω2 = k/m12 and ε 1. Expand:

x0 + ω2x0 = 0 ⇒ x0 = a cos(ωt) (1)

x1 + ω2x1 = −a3

4( 3 cos(ωt)︸ ︷︷ ︸secular term

+ cos(3ωt)︸ ︷︷ ︸no problem

). (2)

. . . ⇒ x ∼ a cos(ωt) + εa3

32ω2(cos(3ωt)− cos(ωt)− 12tω sin(ωt)).

Again, this blows up for large t. How can we find a uniformsolution?

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 15: perturbation theory

Uniform solution using Linstead’s Method 14/21

Linstead’s Method1 Assume the system has a natural frequency Ω.2 Transform to a strained time variable τ = Ωt.3 Expand solution x = x0 + εx1 + . . . as well as frequency

Ω = Ω0 + εΩ1 + . . . .4 Use the new variables Ωn to remove secular terms.

For Duffing’s equation we obtain

x = a cos(τ) + εa3

36ω2(cos(3τ)− cos(τ)) + . . .

with Ω = ω + 3a2

8ω ε+ . . . .

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 16: perturbation theory

Phase-space diagram 15/21

Phase-space diagram

Any autonomous (no explicit t-dependence) second-ordersystem

x = f(x, x).

can be written as

x = y,

y = f(x, y)

Thendy

dx=y

x=f(x, y)y

.

A plot of in the xy plane is called a phase-space diagram.Solution plots are called orbits.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 17: perturbation theory

For the linear spring, x+ω2x = 0, we obtain 12y

2 = −12ω

2x2 + c.

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

Figure: Phase space diagram for the linear spring with ω = 1.

Page 18: perturbation theory

Phase-space diagram 17/21

Phase-space diagram

Properties of phase-space diagrams• A closed orbit corresponds to a periodic system• Points where x = y = 0 are called critical points and

represent equilibrium points of the system.• Orbits cannot intersect or cross• This plot shows only the orbit structure, all temporal

information is lost.

Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring

Page 19: perturbation theory

For the linear spring with dissipation, x+ 2γx+ ω20x = 0, we

obtain x = e−γt (a cos(ωt) + b sin(ωt)) , ω2 = ω20 − γ2.

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

Figure: Phase space diagram for the linear damped spring withω0 = 2γ = 1.

Page 20: perturbation theory

For the Hertian spring, x+ ω2|x|3/2 = 0, we obtain12y

2 = −25ω

2|x|5/2 + c.

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

Figure: Phase space diagram for the Hertzian spring with ω = 1.

Page 21: perturbation theory

For Duffings equation, x+ ω2x+ εx3 = 0, we obtain12y

2 = −12ω

2x2 − 14εx

4 + c.

-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

Figure: Phase space diagram for the non-linear spring with ω = ε = 1.

Page 22: perturbation theory

Phase-space diagram 21/21

Classification of Equilibrium PointsWe classify equilibrium points by linearizing the system aroundthem. We obtain the linear system

x = ax+ by,

y = cx+ dy.

Writing this as a second order system, we get

x− (a− d)x+ (ad− bc)x = 0

x = aeλt are solutions to this equation if λ satisfies

λ2 − (a− d)λ+ (ad− bc) = 0

Thus if solutions λ1, λ2 are distinct,

x = a1eλ1t + a2e

λ2t

We distinguish six types of critical points: [blackboard]Perturbation Theory Algebraic equations Ordinary differential equations The non-linear spring