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Chapter 2 Linear Autonomous Odes

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Chapter 2 Linear Autonomous Odes Chapter 2 Linear autonomous ODEs 2.1 Linearity Linear ODEs form an important class of ODEs. They are characterized by the fact that the vector field f : Rm Rp R Rm is linear (at constant value of the parameters and time), that is for all x, y ×Rm and× a,→ b R ∈ ∈ f(ax + by,λ,t)= af(x,λ,t)+ bf(y,λ,t). This implies that the vector field may be represented by an m m matrix A(λ, t), so that × dx = A(λ, t)x. dt In this chapter we will focus on the case that the matrix A does not depend on time (so that the ODE is autonomous): dx = A(λ)x, dt It turns out that Proposition 2.1.1. The flow of an ODE is linear if and only if the corresponding vector field is linear. t1,t0 t1,t0 t1,t0 t1,t0 R Proof. The flow Φλ is linear if Φλ (ax + by) = aΦλ (x)+ bΦλ (y) for all a, b and x, y Rm. Equation (1.2.1) contains the formal definition of the flow map, from∈ which it ∈ immediately follows that linearity of Φ in x is correlated with linearity of f in x. 2.2 Explicit solution: exponential We consider the linear autonomous ODE dx = Ax, (2.2.1) dt 11 12 CHAPTER 2. LINEAR AUTONOMOUS ODES where x Rm and A gl(m, R) (an m m matrix with real entries). ∈ ∈ × In the case that m = 1, when A R, the flow map Φt of the linear ODE is equal to (multiplication by) exp(At). ∈ It turns out that we can write the flow map of a general linear autonomous ODE in this same form, if we define the exponential operator as Definition 2.2.1. Let L gl(m, R), then ∈ ∞ Lk exp(L) := k! Xk=0 You should recognize the familiar Taylor expansion of the exponential function in the case that m = 1. In the formula, Lk represents the matrix product of k consecutive matrices L. It follows that exp(L) is also a linear map (which can be represented by an m m matrix). Moreover, it is invertible, with inverse exp( L). Technically, one may worry about× the fact − whether the coefficients of the matrix exp(L) are indeed all smaller than infinity. But it is in fact not too hard to find an upper bound for any of the entries of exp(L) given that the entries of the matrix L are finite. (See exercise) Theorem 2.2.2. The linear autonomous ODE (2.2.1) with initial condition x(0) = y has the unique solution x(t) = exp(At)y. Proof. We use the definition of the derivative: d exp((t + h)A) exp(tA) exp(At) = lim − dt h 0 h → exp(hA) I = lim − exp(tA) h 0 h → 1 ∞ (hA)n = lim exp(tA) h 0 h n! → n=1 ! X hA 1 ∞ (hA)n = lim + exp(tA) h 0 h h n! → n=2 ! X ∞ Aj+2 = lim A + h hj exp(tA)= A exp(tA). h 0 (j + 2)! → j=0 ! X d Since y is constant it thus follows that dt x(t)= A exp(At)y = Ax(t). To show that the solution is unique, suppose that x(t) is a solution. We calculate (by using the chain rule) d d dx(t) (exp( tA)x(t)) = exp( tA) x(t)) + exp( tA) dt − dt − − dt = A exp( At)x(t) + exp( tA)Ax(t)=0. − − − 2.3. COMPUTATION OF THE FLOW MAP 13 Hence, x(t) = exp(tA)c where c Rm is some constant. We observe that this constant is entirely determined to be equal to∈y by the boundary condition x(0) = y. Hence the solution x(t) = exp(tA)y is unique. It is easy to extend the result to the case of initial condition x(τ) = y, yielding x(t) = exp(A(t τ))y. − Corollary 2.2.3 (Existence and uniqueness). For the linear autonomous ODE (2.2.1), through each point y in the phase space passes exactly one solution at time t = τ, namely x(t) = exp(A(t τ))y. − 2.3 Computation of the flow map 2.3.1 The planar case We consider the calculation of the flow map exp(At). Before dealing with matters in higher dimension we first focus on the case of ODEs on the plane R2. We consider the calculation of the flow map exp(At). Recall the definition of the exponential of a linear map (matrix) in Definition 2.2.1. Example 2.3.1 (Hyperbolic). Let λ 0 A = 1 , 0 λ 2 with λ ,λ R, then 1 2 ∈ k λ1t 0 k k k λ1 t λ1t ∞ ∞ 0 λ t ∞ (At) 2 k=0 k! 0 e 0 exp(At)= = = k k = λ2t k! k! λ2 t 0 e 0 k∞=0 ! Xk=0 Xk=0 P k! Example 2.3.2 (Elliptic). Let P 0 β A = , β −0 with β R, then ∈ β2 0 A2 = − = β2I, 0 β2 − − so that for all k N ∈ 0 ( )k+1β2k+1 A2k =( )kβ2kI, A2k+1 = − . − ( )kβ2k+1 0 − so that 2 2 +1 k (βt) k k (βt) k ∞ ∞ k=0( 1) (2k)! k=0( 1) (2k+1)! cos(βt) sin(βt) exp(At)= − 2k+1 − − 2k = . k (βt) k (βt) − ∞ ( 1) ∞ ( 1) ! sin(βt) cos(βt) Pk=0 − (2k+1)! P k=0 − (2k)! P P 14 CHAPTER 2. LINEAR AUTONOMOUS ODES Example 2.3.3 (Jordan Block). Let λ 1 A = , 0 λ with λ R, then ∈ λk kλk 1 Ak = − . 0 λk so that (λt)k (λt)k λt λt ∞ ∞ k=0 k! t k=0 k! e te exp(At)= (λt)k = λt . 0 ∞ 0 e P Pk=0 k! ! P Distinct real eigenvalues dx R R2 We consider a linear autonomous ODE dt = Ax, with A gl(2, ) andx . Suppose that A has a real eigenvalue µ with orresponding eigenvector v∈with. Then if we∈ consider an initial condition on the linear subspace V that is generated by v (a line) then we find that V is an invariant subspace for the flow, that is Φt(V )= V . In fact, we may express that ODE restricted dv to V as dt = µv, and the flow map of this restricted ODE is (multiplication by) exp(µt). Suppose now that A has two distinct eigenvalues µ = µ . Then it follows that the cor- 1 6 2 responding eigenvectors v1 and v2 must be linearly independent, i.e. v1 = av2 for all a R. 2 6 ∈ In other words, v1 and v2 span the plane R . As a consequence, we may write any point as a linear combination of the two eigenvectors x = av1 + bv2, (2.3.1) and by linearity of the flow map, and knowing the flow in the directions of the eigenvectors, we obtain t t t t λ1t λ2t Φ (x) = Φ (av1 + bv2)= aΦ (v1)+ bΦ (v2)= ae v1 + be v2. (2.3.2) In order to find the matrix expression for Φt we now need to find expressions for the coefficients a and b in (2.3.1). We do this by taking the (standard matrix) product on both sides with vectors (vi⊥)⊤, where vi⊥ is perpendicular to the eigenvector vi, for i =1, 2: (v ) x (v ) x a = 2⊥ ⊤ , b = 1⊥ ⊤ . (v2⊥)⊤v1 (v1⊥)⊤v2 On a more abstract level, we may think of the vectors av1 and bv2 being obtained from the vector x by projection: 1 P1x = av1 = v1((v2⊥)⊤v1)− (v2⊥)⊤x, 1 P2x = bv2 = v2((v1⊥v2)⊤)− (v1⊥)⊤x. Note that projections P are linear maps that satisfy the property that P 2 = P . As they are linear, we can represent them by matrices: 1 1 P1 = v1((v2⊥)⊤v1)− (v2⊥)⊤, P2 = v2((v1⊥)⊤v2)− (v1⊥)⊤. 2.3. COMPUTATION OF THE FLOW MAP 15 So in terms of these expressions, we can rewrite (2.3.2) as t λ1t λ2t Φ = e P1 + e P2. (2.3.3) 1 1 Example 2.3.4. Let A = , then the eigenvalues are equal to λ = 1 and λ = 2 with 0 2 1 2 1 1 corresponding eigenvectors v = and v = . Any vector in R2 can be decomposed 1 0 2 1 into a linear combination of these eigenvectors as follows: x 1 1 =(x y) + y . y − 0 1 Hence, x 1 1 (x y)et + ye2t et e2t et x exp(At) =(x y)et + ye2t = − = − , y − 0 1 ye2t 0 e2t y so that 1 1 0 1 exp(At)= et − + e2t . 0 0 0 1 We now construct the projections using the formulas we obtained above. Choosing the perpen- 0 1 dicular vectors (for instance) as v = and v = , we obtain 1⊥ 1 2⊥ 1 − 1 1 1 1 0 1 P = v ((v⊥)⊤v )− (v⊥)⊤ = , P = v ((v⊥)⊤v )− (v⊥)⊤ = . 1 1 2 1 2 0− 0 2 2 1 2 1 0 1 The answer in Example 2.3.1 can be evaluated in the same way. We note that the above procedure can also be carried out in the higher dimensional case. In the m-dimensional case, if A has m distinct real eigenvalues λ1,...,λm then m λkt exp(At)= e Pk, Xk=1 where Pk is the projection to the one-dimensional subspace spanned by the eigenvector for eigenvalue λ with kernel equal to the (m 1)-dimensional subspace spanned by the collection k − of all other eigenvectors.
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