Ordinary Differential Equations II

Transcription

1 Ordinary Differential Equations II February Linearization of an autonomous system We consider the system (1) x = f(x) near a fixed point x. As usual f C 1. Without loss of generality we assume x = (otherwise set y = x x and consider the equation y = g(y), where g(y) = f(x + y)). Since f C 1, it is differentiable at, meaning that where is the Jacobi matrix and f(x) = Ax + R(x), 1 f 1 () n f 1 () A = f () = f n () n f n () (2) lim x R(x) x =. Hence, solutions of (1) should behave like solutions of the linear system x = Ax, when x. This is the linearization of (1) at. Theorem. (1) If all eigenvalues of f () have negative real part, then is asymptotically stable. (2) If there is an eigenvalue with positive real part, then is unstable. 1

2 Remark: In contrast to the linear case, we can t say anything in general if all eigenvalues have real part and eigenvalue λ with Re λ =. Example. We find all fixed points of { x = x + y 2 y = x + 2y 3 and determine their stability properties. The fixed points are given by the equations { x + y 2 = x + 2y 3 =. Solutions: x = 1, y = 1 and x = 9, y = 3. Jacobi matrix: ( ) 1 2y. 1 2 (x, y) = ( 1, 1): ( ) Eigenvalues: Re λ ± > unstable. λ ± = 3 2 ± i 7 2. (x, y) = ( 9, 3): ( ) Eigenvalues: λ 1 = 1 and λ 2 = 4. Re λ 2 > unstable. 2

3 Proof of stability We begin by proving the first part of the theorem. Lemma 1. If is a norm on R n, constants c 1, c 2 > s.t. where is the Euclidean norm. c 1 x x c 2 x x R n, Proof. 2nd ineq.: Write x = n k=1 x ke k, e k standard basis. Then ( n n ) x x k e k e k x. k=1 1st ineq.: x x is continuous on R n, since But then Hence, k=1 } {{ } :=c 2 x y x y c 2 x y. c 1 := inf x = min x >. x =1 x =1 x x c 1, x x c 1 x. Lemma 2. Let A R n n with Re λ < for all eigenvalues λ. inner product, on R n s.t. Ax, x σ x, x, x R n for some σ >. Proof. We show that has the required properties: x, y = e ta x e ta y dt e ta x is exp. decaying for all x R n integral converges. The formula defines a symmetric, bilinear form. x, x = e ta x 2 dt. Equality e ta x = t x =. Ax, x = Ae ta x e ta 1 d x dt = 2 dt eta x 2 dt [ ] 1 = 2 eta x 2 = 1 2 x 2 σ x, x for some σ > by Lemma 1. 3

4 Theorem 1. If all eigenvalues of f () have negative real part, then is as. stable. Proof. Let L(x) = x, x, with, as in Lemma 2. Then L(x), with equality iff x =, and L f (x) = 2 Ax, x + 2 R(x), x, where Let x = x, x. Lemma 1 lim x R(x) x =. Cauchy-Schwarz for, Ax, x σ x, x. R(x), x R(x) x = R(x) x, x, x so if x is suff. small. Hence, R(x), x σ x, x, 2 L f (x) σ x, x when x B ε (), ε > suff. small. Thus, L is a strict Liap. function in B ε (). Proof of instability To prove the second part of the theorem, we first introduce an analogue of Lyapunov functions for instability (sometimes called Chetaev functions ). Theorem 2. Assume that a C 1 function G: U R, U R n open neighbourhood of x and an open subset V of U s.t.: (1) x V, (2) G(x) > and Ġf(x) > when x V, (3) G(x) = when x V U. Then x is an unstable fixed point. U V x G = 4

5 Proof. Choose ε > so that B ε (x ) U. It suffices to prove that every trajectory starting in B ε (x ) V leaves B ε (x ). Assume that y B ε (x ) V and choose ϱ so that < ϱ < G(y). Let = {x B ε (x ) V : G(x) ϱ}. This is a compact set and Ġf(x) > on, so k := min Ġ f (x) >. Moreover, dg(φ(t, y)) k when Φ(t, y). dt Hence, G(Φ(t, y)) kt + G(y) as long as Φ(t, y). It follows that Φ(t, y) must leave. This must happen where G > ϱ the trajectory leaves B ε (x ). U B " y V x Using this theorem, we can prove the instability of a fixed point, provided that the linearization has an eigenvalue with positive real part. We assume again that x =. Theorem 3. If f () has an eigenvalue with positive real part, then is unstable. Proof. Recall that C n = ker(a λ 1 I) a 1 ker(a λ k I) a k, λ 1,..., λ k the distinct eig. values of A, ker(a λ j I) a j generalized eigenspace corresponding to λ j. Write R n = V 1 V 2, where V 1 is the direct sum of all gen. eigenspaces corresponding to eigenvalues with Re λ, and V 2 the sum of the gen. eigenspaces for eigenvalues with Re λ > (note that complex eigenvalues and eigenvectors come in complex conjugate pairs). By assumption, V 2 {}. The restriction of A to V 2 only has eigenvalues with negative real part. Lemma 2 inner prod., 2 on V 2 such that Ax, x 2 σ 2 x, x 2, for some σ 2 >. Choose < σ 1 < σ 2. If V 1 {} (otherwise easy modification), the restriction of A σ 1 I to V 1 only has eigenvalues with negative real part, and hence inner prod., 1 on V 1 s.t. Ax, x 1 σ 1 x, x 1. 5

6 Define G(x) = x 2, x 2 2 x 1, x 1 1, where x = x 1 + x 2 with x j V j. Write with R j (x) V j. Then R(x) = R 1 (x) + R 2 (x), Ġ f (x) = 2 Ax 2, x Ax 1, x R 2 (x), x R 1 (x), x 1 1 2σ 2 x 2, x 2 2 2σ 1 x 1, x ϱ(x) = (σ 1 + σ 2 )( x 2, x 2 2 x 1, x 1 1 ) + (σ 2 σ 1 )( x 1, x x 2, x 2 2 ) + ϱ(x), where ϱ(x) = 2 R 2 (x), x R 1 (x), x 1 1 and lim x ϱ(x) x 1, x x 2, x 2 2 =. ( x 1, x x 2, x 2 2 is a norm on R n ). Hence Ġ f (x) (σ 1 + σ 2 )( x 2, x 2 2 x 1, x 1 1 ) = (σ 1 + σ 2 )G(x), if x is suff. small. Taking U = B ε () and V = {x U : G(x) > } we have V, G(x) > and Ġf(x) > when x V G(x) = when x V U. Thus is unstable by the previous theorem. 6