NOTES ON LINEAR ODES

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1 NOTES ON LINEAR ODES JONATHAN LUK We can now use all the discussions we had on linear algebra to study linear ODEs Most of this material appears in the textbook in 21, 22, 23, 26 As always, this is a preliminary version: if you have any comments or corrections, please send them to me Given a linear ODE 1 General solutions to linear systems { u(0) = u 0, we know that the unique solution is defined for all t R is given by u(t) = e At u 0 Given our discussions on taking exponentials of matrices, we immediately have Theorem 11 Let A be an (n n) complex matrix The l-th component of any solution to takes the following form u l (t) = p l,j (t)e λjt, λ j eigenvalue of A where for each l, j, p l,j (t) is a polynomial Proof Recall that every matrix is similar to a matrix in the Jordan canonical form, ie, there exists S such that S 1 AS = J, where is in the Jordan canonical form Moreover, as we discussed earlier, every non-zero entry of e Jt λit tk takes the form e k! (for some eigenvalue λ i some k N) This then implies the theorem Remark 12 Notice that not all functions of the above form are solutions Remark 13 Suppose the characteristic polynomial is given by χ A (λ) = Π m (λ λ i) νi, where λ i are distinct Then the polynomials p l,j (t) have degree strictly smaller than ν j (Exercise: Why?) If A is real, we want to find an expression for general real solutions These are in particular the solutions one obtain when given real initial data Before we proceed, let us just make a simple observation that if A is an (n n) real matrix λ C \ R is an eigenvalue, then its complex conjugate λ is also an eigenvalue (This is because if Av = λv for v C n \ {0}, then A v = λ v) With this observation, we have the following corollary: Corollary 14 Let A be an (n n) real matrix Suppose µ 1,, µ m are all of its real eigenvalues α 1 ± iβ 1,, α s ± iβ s are all of its eigenvalues in C \ R Then the l-th component of any real solution to takes the following form u l (t) = m p l,i (t)e µit + s e αkt (q l,k (t) sin(β k t) + r l,k (t) cos(β k t)), k=1 where p l,i (t), q l,k (t) r l,k (t) are real polynomials 1

2 2 JONATHAN LUK Proof Suppose u(t) is a real solution By Theorem 11, there exists a complex polynomial p l,j such that u l (t) = p l,j (t)e λjt λ j eigenvalue of A =Re( p l,j (t)e λjt ) λ j eigenvalue of A m s = p l,i (t)e µit + e αkt (q l,k (t) sin(β k t) + r l,k (t) cos(β k t)) k=1 for some real polynomials p l,i (t), q l,k (t) r l,k (t) 2 Large time behavior of solutions to linear ODEs Proposition 21 Let A be an (n n) complex matrix such that all eigenvalues λ j satisfy Re(λ j ) < α < 0 Then, for any solution u(t) to, we have as t e αt u(t) 0 Proof By Theorem 11, each component u l (t) takes the form u l (t) = p l,j (t)e λjt λ j eigenvalue of A Since Re(λ j ) < α < 0, e αt (RHS) 0 for each l as t The proposition hence follows Although Proposition 21 above shows that all solutions converge to 0 as t, it does not actually show that 0 is a stable equilibrium (This is because Proposition 21 is in principle consistent with solutions becoming very large before converging to 0) Nevertheless, we have the following more quantitative proposition 1 : Proposition 22 Let A be an (n n) complex matrix such that all eigenvalues λ j satisfy Re(λ j ) < α < 0 Then there exists Λ > 0 such that for every solution u(t) to, for all t 0 u(t) Λe αt u(0) Proof Since u(t) = e At u(0), it suffices to prove that e At op Λe αt By our discussions on linear algebra, there exists an invertible matrix S such that S 1 AS = J, where J is in the Jordan canonical form Hence, e At = Se Jt S 1 Recall that every non-zero entry of e Jt λit tk takes the form e k! (for some eigenvalue λ i some k N) By Problem 1 in HW3, we know that e Jt op ( ij (ejt ) ij 2 ) 1 2 (where (e Jt ) ij denotes the ij-th entry of e Jt ) Therefore, there exists a constant C > 0 such that e αt e Jt op C e (α+re(λj))t (1 + t + + t n 1 ), λ j eignevalue of A which is bounded for all t 0 since Re(λ j ) < α Finally, since we obtain the conclusion of the proposition e At op S op e Jt op S 1 op, Next, we consider the case where some eigenvalues can have positive real parts 1 Note that strictly speaking this proposition only give the boundedness of e αt u(t) instead of its decay Nevertheless, since the inequality Re(λ i ) < α < 0 is strict, there exists β > 0 such that Re(λ i ) < β < α < 0 Hence, by applying Proposition 22 with β in place of α, we see that its conclusion indeed implies Proposition 21

3 NOTES ON LINEAR ODES 3 Theorem 23 Suppose A is a (n n) complex matrix Assume that there are no eigenvalues on the imaginary axis that every eigenvalue satisfies Re(λ i ) > α for some α > 0 Then C n decomposes as so that C n = P + P, e αt e At u 0 0, as t, for u 0 P (21) e αt e At u 0, as t, for u 0 P + \ {0} (22) Proof Definitions of P + P Recall that C n can be decomposed as C n = ker((λ 1 I A) ν1 ) ker((λ k I A) ν k ) ker((λ k+1 I A) ν k+1 ) ker((λ m I A) νm ) Without loss of generality (by otherwise relabelling), assume that Re(λ 1 ),, Re(λ k ) > α > 0 Re(λ k+1 ), Re(λ n ) < α < 0 Define P + = ker((λ 1 I A) ν1 ) ker((λ k I A) ν k ) P = ker((λ k+1 I A) ν k+1 ) ker((λ m I A) νm ) Clearly, C n = P + P Property of P Let us first consider the case where v i ker((λ i I A) νi ) for some i = k + 1,, m Recall that A = B + N where B is diagonalizable, N n = 0 BN = NB Moreover, we showed that Bv i = λ i v i Hence, for some C > 0, we have e αt e At v i = e αt e Bt e Nt v i = e αt e λit e Nt v i = e (α+re(λi))t e Nt v i Ce (α+re(λi))t (1 + t + + t n 1 ) 0 Since any u 0 P can be written as a sum of such v i s, the desired estimate for e αt e At u 0 follows from the above discussion the triangle inequality Property of P +, Step 1: A reduction We claim that to estimate e At u 0 for u 0 P +, it suffices to consider the case where u 0 ker((λ i I A) νi ) \ {0} for some i = 1,, k To see this, first notice that if u 0 P +, then e At u 0 P + for all t 0 Moreover, we can define a norm (Exercise: Check that it is a norm) on P + by k v = v i, where v = v v k, v i ker((λ i I A) νi ) Since P + is finite dimensional, must be equivalent to on P +, ie, there exists C > 0 such that for every v P +, C 1 v v C v Now suppose we can show that e αt e At v i whenever v i ker((λ i I A) νi ) \ {0} for some i = 1,, k Given u 0 P + \ {0}, u 0 = v v k, v i ker((λ i I A) νi ) At least one of these v i s is nonzero Hence, e αt e At u 0 C 1 e αt e At u 0 C 1 sup e αt e At v i i Property of P +, Step 2 Now let u 0 ker((λ i I A) νi ) \ {0} for some i = 1,, k We have e At u 0 = e Bt e Nt u 0 = e λit e Nt u 0 = e Re(λi)t e Nt u 0 (23) We claim that e Nt u 0 is bounded below for large t To see this, note that since N n = 0, e Nt = n 1 i=0 ti N i i! Hence, there exists a largest i 0 0 such that N i0 u 0 0 Moreover, since all the other non-zero terms have smaller powers of t, for t sufficiently large, e Nt u ti 0 N i 0 i 0! u 0 =: γ > 0 Hence, by (23), we have that if u 0 ker((λ i I A) νi ) \ {0} for some i = 1,, k, Together with part 1, this concludes the proof e αt e At u 0 = e αt+re(λi)t e Nt u 0 γe αt+re(λi)t

4 4 JONATHAN LUK Remark 24 Notice that (21) defines P uniquely, but (22) does not define P + uniquely We will nevertheless from now on use P ± to mean the subspaces constructed in Theorem 23 Remark 25 P is called the stable subspace (or stable manifold, or incoming manifold) P + is called the unstable subspace (or unstable manifold, or outgoing manifold) Remark 26 As long as P + {0}, 0 is an unstable equilibrium Moreover, for u 0 = u + + u with u ± P ±, e At u 0 as long as u + 0 In particular, there exists an open dense set U C n such that u 0 U implies e At u 0 Remark 27 The discussion above assumes that A is a complex matrix In the case where A is real, in fact one can also decompose R n = R P + R P, where R P ± = P ± R n The above discussion does not cover the case where at least one eigenvalue lies on the imaginary axis That case is in general more complicated we will not formulate a general theorem In the following two examples, we will see that even if 0 is the only eigenvalue, there can be rather different long time behavior In particular, in Example 28, 0 is a stable equilibrium while in Exapmle 29, 0 is not a stable equilibrium Example 28 Let A be the (2 2) zero matrix In this case, any solution to = 0 is constant in time In particular, it neither grows nor decays as t Example 29 In the second example, consider A = [ The eigenvalues are also 0 as in the previous example Nevertheless, [ ] [ ] [ e At 1 t (u0 ) u 0 = 1 (u0 ) = 1 + t(u 0 ) (u 0 ) 2 (u 0 ) 2 Hence, the solution grows as t if (u 0 ) 2 0 Finally, let us note that in both Proposition 21 Example 28, 0 is a stable equilibrium, but they have somewhat different long time behavior in that u(t) 0 in Proposition 21 but u(t) 0 in general in Example 28 It is therefore useful to make the following definition: Definition 210 (Asymptotic stability) We say that a solution ū(t) to an ODE with initial data ū(0) = ū 0, which is defined for all t 0, is asymptotically stable if both of the following hold: ū is stable, there exists r > 0 such that for any data u 0 B(ū 0, r), the corresponding solution u(t) satisfies as t ] u(t) ū(t) 0 Remark 211 Let us note that the assumption that ū is stable is necessary (Exercise: Why?) 3 Equivalence of linear systems Definition 31 Let A B be two (n n) real matrices Two linear ODEs linearly are said to be topologically/continuously smoothly/diffeomorphically f : R n R n such that u (t) = Bu(t) ] equivalent as real ODEs if there exists an invertible function

5 linear both f f 1 are continuous smooth, as maps : R n R n for all t R NOTES ON LINEAR ODES 5 f e At = e Bt f Remark 32 It is easy to check (Exercise) that these are indeed equivalence relations Remark 33 It is clear from the definition that linear equivalence = smooth equivalence = topological equivalence Remark 34 One can also entertain the various notions of equivalence for complex ODEs We will however not discuss that in these notes We now study each of these notions of equivalences First, we start with linear equivalence: Proposition 35 Let A B be two real (n n) matrices The ODEs u (t) = Bu(t) are linearly equivalent as real ODEs if only if A B are similar Proof It is clear that if there is an invertible matrix 2 C such that C 1 AC = B, then Ce At = e Bt C Conversely, if there is an invertible matrix C such that Ce At = e Bt C, by differenting evaluating at t = 0, we obtain AC = BC We then turn to smooth equivalence It turns out that it is the same as linear equivalence Theorem 36 Two linear real ODEs are linearly equivalent (as real ODEs) if only if they are smoothly equivalent Proof As we noted in Remark 33, linear equivalence = smooth equivalence is easy We now prove the other implication Let A B be two real (n n) matrices Suppose there is a smooth map f : R n R n with f 1 well-defined smooth such that f e At = e Bt f Step 1 We first note that since e At 0 = 0, we must have that e Bt f(0) = f(0) Define f(x) = f(x) f(0) Clearly f is smooth, invertible, f 1 is smooth We check that f(e At x) = f(e At x) f(0) = e Bt f(x) f(0) = e Bt (f(x) f(0)) = e Bt f(x) (31) Step 2 Since f is differentiable f(0) = 0, we have f(x) = (D f 0 )x + G(x), where (D f 0 ) is a (constant) (n n) invertible matrix, G satisfies lim x 0 G(x) x = 0 Now, using (31), we have (D f 0 )e At x e Bt (D f 0 )x = G(e At x) + e Bt G(x) In particular, given fixed t R, x R n \ {0}, for every ɛ > 0, there exists δ > 0 sufficiently small such that for some C > 0 This implies Since ɛ is arbitrary, we have The conclusion follows δ (D f 0 )e At x e Bt (D f 0 )x G(δe At x) + e Bt G(δx) Cδɛ x 2 Notice that all linear maps are given by matrices (D f 0 )e At x e Bt (D f 0 )x Cɛ x (D f 0 )e At = e Bt (D f 0 )

6 6 JONATHAN LUK Remark 37 This does not mean if there is a smooth map f : R n R n with f 1 well-defined smooth satisfying f(e At u 0 ) = e Bt f(u 0 ), then f is linear (Exercise: Why?) Finally, we discuss topological equivalence The main result is in Theorem 317 We will first start with some preliminary discussions Proposition 38 Let A be an (n n) complex matrix Let ɛ > 0, then there exist an invertible matrix S an upper triangular matrix T such that S 1 AS = T, All the entries of T above the diagonal have size < ɛ Proof Since we know that every matrix is similar to one in the Jordan canonical form, it suffices to prove the proposition for a Jordan block λ λ 1 λi + J n := 0 λ λ Define S to be the diagonal matrix ɛ 0 S = 0 0 ɛ n ɛ n 1 Then λ ɛ λ 1 0 ɛ 0 S 1 (λi + J n )S = ɛ (n 2) 0 λ 1 0 ɛ n ɛ (n 1) 0 0 λ 0 0 ɛ n λ ɛ ɛ ɛλ ɛ 2 = ɛ (n 2) 0 ɛ n 1 λ ɛ n ɛ (n 1) 0 0 ɛ n 1 λ λ ɛ λ ɛ = 0 λ ɛ 0 0 λ

7 NOTES ON LINEAR ODES 7 Proposition 39 Let A be a complex (n n) matrix whose eigenvalues have positive real parts Then there exists a norm on C n such that d dt eat u 0 2 > 0 for all u 0 C n \ {0} Proof Let ɛ > 0 (to be chosen later) By Proposition 38, there exists S such that S 1 AS = T for some upper triangular T with off-diagonal entries < ɛ Define n z 2 := (S 1 z) i (S 1 z) i It is easy to check that is a norm We compute d dt eat z 2 = d dt SeT t S 1 z 2 = n 2Re ( (e T t S 1 ) z) i (e T t T S 1 z) i 2 min λ eigenvalue of A Re(λ) et t S 1 z 2 ) Cɛ e T t S 1 z 2, for some C > 0 depending only on n, where is the stard norm on C n given by z 2 = n z i 2 By our assumptions on the eigenvalues of A, we can choose ɛ sufficiently small so that the RHS is > 0 Theorem 310 Let A be a real (n n) matrix so that all eigenvalues have positive real parts Then are topologically equivalent as real ODEs u (t) = u(t), Proof Let be the norm as in Proposition 39 Define the homeomorphism by { e A log x x x f(x) = x 0 0 x = 0 We check that it has the desired properties: f gives the equivalence of the ODEs To see this, we check that f(e t x) = e A log et x x = e At A log x x e = e At f(x) x x f is continuous This is obvious when x 0 To see that f is continuous at 0, we check that log x x lim ea x 0 x = 0, since by Proposition 22, lim x 0 e A log x op = 0 f 1 is well-defined f 1 can be given by { f 1 e Aτ e τ y, y 0 (y) = 0 y = 0 for τ such that e Aτ y = 1 We need to show that τ exists is unique for any given y R n \ {0} The existence of τ follows from lim τ e Aτ y = +, lim τ e Aτ y = 0 the intermediate value theorem The uniqueness of τ follows from Proposition 39 f 1 is continuous We first check that f 1 is continuous away from 0 It suffices to show that the map y τ : R n \ {0} R is continuous For this, notice that by Proposition 39, for y 0, d dτ e Aτ y 2 < 0 Therefore, by the implicit function theorem 3, τ is C 1 therefore continuous at y 3 Here, let us recall the following particular case of the implicit function theorem Let g : R k R R be a C 1 function Suppose there exists (x 0, y 0 ) R k R such that g(x 0, y 0 ) = 0

8 8 JONATHAN LUK Next, we check that f 1 is continuous at 0 Suppose there exists a sequence y n 0 In order that e Aτn y n = 1, we have τ n Hence, f 1 (y n ) e τn 0, as desired Remark 311 Since topological equivalence is an equivalence relation, the above theorem in particular shows that if A B are both real (n n) matrices so that all eigenvalues have positive real parts Then are topologically equivalent as real ODEs u (t) = Bu(t), In a completely analogous manner, we also have Proposition 312 Let A be a real (n n) matrix so that all eigenvalues have negative real part Then are topologically equivalent as real ODEs u (t) = u(t), Combining Theorem 310 Proposition 312, we have Theorem 313 Let A be a real (n n) matrix so that exactly n + eigenvalues (counting multiplicities) have positive real part, exactly n eigenvalues (counting multiplicities) have negative real part, no eigenvalues are on the imaginary axis Let I n+,n be the diagonal matrix I n+,n = diag(+1, +1,, +1, 1,, 1), where the first n + diagonal entries are +1 the last n diagonal entries are 1 Then are topologically equivalent as real ODEs u (t) = I n+,n u(t), Proof Step 1 Since R n = R P + R P, we can find S so that S 1 AS = Ã, which takes the following block diagonal form [ ] A+ 0 à = 0 A for some n + n + real matrix A +, whose eigenvalues have positive real parts, some n n real matrix A, whose eigenvalues have negative real parts Step 2 By Remark 33 Proposition 35, it suffices to show that u (t) = I m,n u(t), u (t) = Ãu(t) are topologically equivalent as real ODEs, where à is as in Step 1 By Theorem 310 Proposition 312, there exists f ± : R n± R n± which is continuous with continuous inverse such that f ± e ±It = e A±t f ± Writing x R n as x = (x +, x ), where x ± R n±, define f(x) = (f + (x + ), f (x )) It is easy to see that f has the desired properties to imply the conclusion of the Theorem In the above, we have given a sufficient condition for topological equivalence of linear ODEs It turns out that if we restrict to linear ODEs so that the corresponding matrix has no eigenvalues on the imaginary axis, then this condition is in fact necessary (see Proposition 316) To see this, we first need a fact Definition 314 We say that two metric spaces (X, d) (Y, d) are homeomorphic if there exists a continuous bijection f : X Y such that f 1 is also continuous The following is a fact that we will not prove in these notes Fact R n R m are homeomorphic if only if m = n g y (x 0, y 0 ) 0 Then, there exists an open neighborhood U of x 0 a unique C 1 function y : U R such that whenever x U g(x, y(x)) = 0

9 NOTES ON LINEAR ODES 9 Remark 315 This seemingly obvious fact is not so easy to prove Notice for instance that there exists a continuous surjection R 1 R 2 Proposition 316 Let A B be (n n) real matrices with no eigenvalues on the imaginary axis Suppose there exists a continuous f : R n R n with continuous inverse such that f e At = e Bt f Then A B have the same number of eigenvalues (counting multiplicities) with positive real parts Proof Suppose there exists an f as in the statement of the proposition Notice that R P (A) = {x R n : e At x 0 as t } R P (B) = {x R n : e Bt x 0 as t } Therefore, f must be a homeomorphism between R P (A) R P (B) By the fact above, they must have the same dimensions Combining Theorem 313 Proposition 316, we have the following Theorem 317 Let A B be (n n) matrices with no eigenvalues on the imaginary axis Then u (t) = Au(t) u (t) = Bu(t) are topologically equivalent as real ODEs if their stable subspaces have the same dimensions

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