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

Transcription

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

2 1/21 1 Perturbation Theory 2 Algebraic equations Regular Perturbations Singular Perturbations 3 Ordinary differential equations Regular Perturbations Singular Perturbations Singular in the domain 4 The non-linear spring Non-uniform solution Uniform solution using Linstead s Method Phase-space diagram

3 2/21 Introduction Perturbation Theory Considers the effect of small disturbances in the equation to the solution of the equation.

4 3/21 A first example Let x 2 = 0. Then x = 2. What about x 2 = ɛ cosh(x)? Assume ɛ 1 and substitute x = x 0 + ɛx 1 + ɛ 2 x (x 0 + ɛx ) 2 = ɛ cosh(x 0 + ɛx ) and solve separately for each order of ɛ:... x 2 + ɛ cosh(2) Conclusions The effect of the small parameter ɛ is small; therefore the perturbation is said to be regular. Otherwise we call it singular.

5 4/21 Order symbol notation Let f(x), g(x) be two functions defined around x = x 0. We say, f = o(g) as x x 0 iff lim x 0 f/g = 0. We say, f = O(g) (of order) as x x 0 iff lim x 0 f/g = const 0. We say, f g (goes like) as x x 0 iff lim x 0 f/g = 1. Examples: 1 x 2 = o(x) as x 0, b/c x 2 /x x 2 = O(5x 2 ) as x 0, b/c (3x 2 )/(5x 2 ) 3/5. 3 sin(x) x as x 0, b/c sin(x)/x 1. 4 sin(x) = x + O(x 3 ) as x 0, b/c (sin(x) x)/x 3 1/6. 5 sin(x) x x3 3! as x 0, b/c sin(x)/(x x 3 /3!) 1. So the solution to x 2 = ɛ cosh(x) is x 2 + ɛ cosh(2) as ɛ 0.

6 5/21 The Fundamental Theorem of Perturbation Theory If A 0 + A 1 ɛ + + A n ɛ n + O(ɛ n+1 ) = 0 for ɛ 0 and A 0, A 1,... independent of ɛ, then A 0 = A 1 = = A n = 0. That is why we could solve separately for each order of ɛ:

7 Regular Perturbations 6/21 Regular Perturbations x 2 3x ɛ = 0, ɛ 1. Assume the roots have expansion x = x 0 + ɛx 1 + ɛ 2 x : (x 0 + ɛx ) 2 3(x 0 + ɛx ) ɛ = 0,... x = 1 + ɛ + ɛ 2 + O(ɛ 3 ), 2 ɛ ɛ 2 + O(ɛ 3 ). Compare with exact solution, x = 3 ± 1 4ɛ 2 = 3 ± (1 2ɛ 2ɛ ) 2

8 Singular Perturbations 7/21 Singular Perturbations ɛx 2 2x + 1 = 0, ɛ 1. Assume the roots have expansion x = x 0 + ɛx 1 + ɛ 2 x : ɛ(x 0 + ɛx ) 2 2(x 0 + ɛx ) + 1 = x = 1/2 + 1/8ɛ + 1/16ɛ 2 + O(ɛ 3 ) But where is the second solution? Compare with exact solution, x = 1 ± 1 ɛ ɛ = 1 ± (1 ɛ/2 ɛ2 / ) ɛ... x = 1/2 + 1/8ɛ + O(ɛ 2 ), 2/ɛ 1/2 + O(ɛ). We can find the second solution by solving for w = ɛx, so w 2 2w + ɛ = 0.

9 8/21 Ordinary differential equations

10 Regular Perturbations 9/21 Regular Perturbations ẋ + x = ɛx 2, x(0) = 1. Assume the roots have expansion x = x 0 + ɛx 1 + ɛ 2 x : (ẋ 0 + ɛẋ ) + (x 0 + ɛx ) = ɛ(x 0 + ɛx ) 2, x 0 (0) + ɛx 1 (0) + = x = e t + ɛ(e t e 2t ) + O(ɛ 2 ).

11 Singular Perturbations 10/21 Singular Perturbations Heat equation: Temperature x for a stone in water ɛẋ = 1 x, x(0) = 0. Assume the roots have expansion x = x 0 + ɛx 1 + ɛ 2 x : (ẋ 0 + ɛẋ ) = 1 (x 0 + ɛx ).... x 0 = 1 (Does not satisfy initial conditions). Again, We can find a solution by transforming, τ = ɛ 1 t, d x = 1 x, x(0) = 0. dτ... x = 1 e 2τ = 1 e 2t/ɛ.

12 Singular in the domain 11/21 Singular in the domain ẋ + ɛx 2 = 1, x(0) = 0. Assume the roots have expansion x = x 0 + ɛx 1 + ɛ 2 x : (ẋ 0 + ɛẋ ) + ɛ(x 0 + ɛx ) 2 = x t ɛ t3 2t5 ɛ This solution is non-uniform (blows up for large t). x t ɛ t3 3 is an asymptotic expansion for t < 3/ɛ.

13 12/21 The non-linear spring

14 Non-uniform solution 13/21 Duffing s equation Consider a springs with a small non-linear perturbation ẍ + ω 2 x + ɛx 3 = 0, x(0) = a, ẋ(0) = 0. with ω 2 = k/m 12 and ɛ 1. Expand: ẍ 0 + ω 2 x 0 = 0 x 0 = a cos(ωt) (1) ẍ 1 + ω 2 x 1 = a3 4 ( } 3 cos(ωt) + cos(3ωt) ). (2) {{}}{{} secular term no problem... x a cos(ωt) + ɛ a3 (cos(3ωt) cos(ωt) 12tω sin(ωt)). 32ω2 Again, this blows up for large t. How can we find a uniform solution?

15 Uniform solution using Linstead s Method 14/21 Linstead s Method 1 Assume the system has a natural frequency Ω. 2 Transform to a strained time variable τ = Ωt. 3 Expand solution x = x 0 + ɛx as well as frequency Ω = Ω 0 + ɛω Use the new variables Ω n to remove secular terms. For Duffing s equation we obtain x = a cos(τ) + ɛ a3 (cos(3τ) cos(τ)) ω2 with Ω = ω + 3a2 8ω ɛ +....

16 Phase-space diagram 15/21 Phase-space diagram Any autonomous (no explicit t-dependence) second-order system ẍ = f(x, ẋ). can be written as Then ẋ = y, dy dx = ẏ ẋ ẏ = f(x, y) f(x, y) =. y A plot of in the xy plane is called a phase-space diagram. Solution plots are called orbits.

17 For the linear spring, ẍ + ω 2 x = 0, we obtain 1 2 y2 = 1 2 ω2 x 2 + c Figure: Phase space diagram for the linear spring with ω = 1.

18 Phase-space diagram 17/21 Phase-space diagram Properties of phase-space diagrams A closed orbit corresponds to a periodic system Points where ẋ = ẏ = 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.

19 For the linear spring with dissipation, ẍ + 2γẋ + ω0 2 x = 0, we obtain x = e γt (a cos(ωt) + b sin(ωt)), ω 2 = ω0 2 γ Figure: Phase space diagram for the linear damped spring with ω 0 = 2γ = 1.

20 For the Hertian spring, ẍ + ω 2 x 3/2 = 0, we obtain 1 2 y2 = 2 5 ω2 x 5/2 + c Figure: Phase space diagram for the Hertzian spring with ω = 1.

21 For Duffings equation, ẍ + ω 2 x + ɛx 3 = 0, we obtain 1 2 y2 = 1 2 ω2 x ɛx4 + c Figure: Phase space diagram for the non-linear spring with ω = ɛ = 1.

22 Phase-space diagram 21/21 Classification of Equilibrium Points We classify equilibrium points by linearizing the system around them. We obtain the linear system ẋ = ax + by, ẏ = cx + dy. Writing this as a second order system, we get ẍ (a d)ẋ + (ad bc)x = 0 x = ae λt are solutions to this equation if λ satisfies λ 2 (a d)λ + (ad bc) = 0 Thus if solutions λ 1, λ 2 are distinct, x = a 1 e λ 1t + a 2 e λ 2t We distinguish six types of critical points: [blackboard]