Homogeneous systems of linear ordinary differential equation - Lecture 8 ∗ September 27, 2012 Transition matrix ẋ(t) = A(t)x(t), x(t0 ) = x0 where A(t) is a m × n matrix whose entries are continuous functions of time t. The solution is x(t) = Φ(t, t0 )x0 , where Φ(t, t0 ) is the transition matrix given in terms of the Peano-Baker series: Z tZ s Z t A(s)A(σ) dσds + . . . A(s) ds + Φ(t, t0 ) = I + t0 t0 t0 Special case A(t) ≡ A is a constant n × n matrix: Z tZ sZ Z tZ s Z t A2 dσds + A ds + Φ(t, t0 ) = I + t0 t0 Z tZ t0 s t0 Z σ A3 dτ dσds = t0 t0 Z tZ Z t0 t t0 A3 dτ dσds + . . . t0 s A3 (σ − t0 ) dσds t0 (s − t0 )2 ds 2 t0 1 1 = A3 (t − t0 )3 = A3 (t − t0 )3 6 3! = The k-th term Z t Z σk−1 t0 t0 σ ··· t0 Z A3 σ0 Ak dσ0 dσ1 . . . dσk−1 = t0 1 k A (t − t0 )k k! The Peano-Baker series can be written as: ∞ Φ(t, t0 ) = I+A·(t−t0 )+ X (A · (t − t0 ))k 1 1 2 A (t−t0 )2 + A3 (t−t0 )3 +. . . = = eA·(t−t0 ) 2! 3! k! k=0 ∗ This work is being done by various members of the class of 2012 1 2 Dynamic Systems Theory Scalar systems with constant coefficients ẋ = ax, x(0) = x0 ⇔ x(t) = eat · x0 Solve this by more elementary means dx dx = ax ⇒ dx = axdt ⇒ = adt dt x Z t a dτ = at + C ⇒ x = eat · eC ⇒ ln(x) = ẋ = ax ⇒ 0 and x(0) = x0 ⇒ eC = x0 For scalar systems with time dependent coefficients ẋ = a(t)x(t), x(0) = x0 we can write the solution x(t) = e Rt 0 a(σ) dσ x0 For a n × n matrix whose entries are functions of t Z tZ s Z t A(s)A(σ) dσds + . . . A(s) ds + Φ(t, t0 ) = I + t0 t0 6= e Rt t0 A(σ) dσ t0 (in general) There is a very special case in which the matrix case has a simplication that is similar to the scalar case: ẋ = f (t)Ax(t), x(0) = x0 where f (t) is a scalar cotinuous function and A is a constant n × n matrix. Z t Z tZ σ1 f (σ)A dσ + f (σ1 )f (σ2 )A2 dσ2 dσ1 + . . . 0 0 0 Z t Z σ1 Z t f (σ) dσ · A + =I+ f (σ1 )f (σ2 ) dσ2 dσ1 A2 + . . . Φ(t, t0 ) = I + 0 Let γ(t) = Rt 0 0 0 f (s) ds, the above series can be rewritten as Φ(t, t0 ) = I + γ(t)A + 1 1 [γ(t)]2 A2 + [γ(t)]3 A3 + . . . 2! 3! (Exercise: evaluate the k-th term in the series to show that this is correct). Rt Seeing this pattern, we write Φ(t, t0 ) = eγ(t)A = e( 0 f (s) ds)A . 3 Dynamic Systems Theory Specific examples Recall the controlled pendulum system θ̈ + cθ̇ + g sin(θ) = u(t) l θ1∗ = 0 and θ2∗ = π are the two equilibrium points. Linearize about the stable equilibrium (θ1∗ ) – first assuming c = 0 (no friction). g ẍ + x = ū. l For today’s lecture we’re assuming ū = 0 (no forcing). Thus to put the system into the framework under discussion we first orderize as follows: p p p x2 ẋ1 = gl ẋ = glp x1 = gl x ⇒ x2 = ẋ ẋ2 = ẍ = − gl x = − gl x1 ẋ1 0 = ẋ2 −α α x1 · x2 0 p where α = gl . To solve this differential equation, we compute 0 −α α t 0 2 3 1 0 α 1 0 α 0 α 2 e =I+ t+ t + t3 + . . . −α 0 2 −α 0 3! −α 0 3 2 1 0 −α3 1 −α2 0 1 0 0 α 2 t3 t + = + t+ 0 0 α2 0 1 −α 0 2 3! α3 6 4 5 1 α4 0 1 −α6 0 1 0 α5 5 4 + t + t + t6 + . . . 0 α6 4! 0 α4 5! −α5 0 6! 1 − 12 α2 t2 + = = αt − cos(αt) − sin(αt) 1 4 4 1 6 6 4! α t − 6! α t + 1 5 5 1 3 3 3! α t + 5! α t − . . . sin(αt) cos(αt) ... 1 3 3 1 5 5 3! α t − 5! α t + . . . 1 2 2 1 4 4 1 6 6 2 α t + 4! α t − 6! α t + . . . −αt + 1− Check that this satisfies the differential equation d −α sin(αt) α cos(αt) cos(αt) sin(αt) = −α cos(αt) −α sin(αt) dt − sin(αt) cos(αt) 0 α cos(αt) sin(αt) = · −α 0 − sin(αt) cos(αt) and ! 4 Dynamic Systems Theory cos(α · 0) − sin(α · 0) sin(α · 0) cos(α · 0) = 1 0 0 =I 1 The solution to the origina problem is x1 (0) cos(αt) sin(αt) ẋ1 (t) · = x2 (0) − sin(αt) cos(αt) ẋ2 (t) s s l (cos(αt)x1 (0) + sin(αt)x2 (0)) g s l = cos(αt)x(0) + sin(αt)ẋ(0) g s r r g l g = cos t x(0) + sin t ẋ(0) l g l x(t) = l x1 (t) = g Example 2 ẋ = Ax where A is a 3 × 3 matrix 0 A = kz −ky −kz 0 kx ky −kx 0 where kx2 + ky2 + kz2 = 1. A is a skew symetric matrix (aij = −aji ). d kx(t)k = Note: The norm of the x(t) satisfying this equation is constant ( dt 0). Thus the solution “lives” on the surface of the unit sphere. 2 ky kx kx kz −kz − ky2 1 0 0 0 −kz ky ky kz t2 −kz2 − kx2 0 −kx t + ky kx eAt = 0 1 0 + kz kx kz ky kz −ky2 − kx2 0 0 1 −ky kx 0 −ky kz }| { z }| { z 0 2 2 2 2 2 2 −k (−k − k ) − k k −k k − k (k + k ) t3 + . . . z z y y z x y z y x + −kz 0 −ky ky −kx 0 1 2 2 1 3 1 2 4 1 5 = I + At + A t − At − A t + At 3! 4! 2 5! 1 3 1 5 1 1 1 2 2 = I + A t − t + t − ... + A t − t4 + t6 − . . . 3! 5! 2 4! 6! = I + A sin(t) + A2 (1 − cos(t)) 5 Dynamic Systems Theory Simplications that always work Compute e −3/2 1/2 1/2 t −3/2 . Directly 1 5/2 −3/2 1/2 0 t+ + 1/2 −3/2 1 2 −3/2 1 0 −3/2 2 t + ... 5/2 (I can’t see a pattern.) Suppose I express A w.r.t.another basis {~ p1 , p~2 } and suppose that in this λ1 0 basis A has the form . 0 λ2 A~ p1 = λ1 p~1 A~ p2 = λ2 p~2 A p~1 p~2 = λ1 p~1 λ2 p~2 = p~1 λ1 0 AP = P 0 λ2 λ1 0 −1 P AP = =Λ 0 λ2 λ1 p~2 0 0 λ2 1 = I + P ΛP −1 t + (P ΛP −1 )(P ΛP −1 )t2 + . . . 2 1 1 1 = I + P ΛP −1 t + (P Λ2 P −1 )t2 + (P Λ3 P −1 )t3 + (P Λ4 P −1 )t4 + . . . 2 3! 4! 1 2 2 1 3 3 1 4 4 = P I + Λt + Λ t + Λ t + Λ t + . . . P −1 2 3! 4! λt 1 e 0 =P P −1 0 eλ 2 t eAt = eP ΛP −1 t