TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS

TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS ALEX HUMMELS Abstract. This paper proves a theorem that gives conditions for the topological equivalence of linear ordinary differential equations. Contents 1. Introductory Concepts 1 1.1. Phase Flows, Diffeomorphisms, and Vector Fields 1 1.2. Linear ODEs 2 1.3. Complexification and Decomplexification 3 1.4. Equivalence Relations between Phase Flows 3 2. Main Theorem 3 3. Lemma 2.4 5 3.1. The Lyapunov Function 5 3.2. The Homeomorphism h 8 Closing Remarks 9 Acknowledgments 9 References 9 1. Introductory Concepts 1.1. Phase Flows, Diffeomorphisms, and Vector Fields. We begin with the notion of a phase flow, a concept essential to ordinary differential equations, or ODEs. A phase flow describes the motion of a point in a deterministic system, or one whose future is uniquely determined by its present state. Definition 1.1. A set of mappings {g t g t : M M, t R} is a one-parameter group of transformations of the set M if s, t R g t+s = g t g s. Then the pair (M, {g t }) is a phase flow. In this definition we assume that g 0 is the identity transformation. Definition 1.2. A differentiable, bijective function f : U V, for U, V open subsets of R n, is a diffeomorphism if f 1 : V U is differentiable. Now if ODEs are to be linked to phase flows, the transformations of the oneparameter group of a phase flow must be differentiable. For this reason we make the following definition. 1

2 ALEX HUMMELS Definition 1.3. Let M be a manifold. The set {g t } is a one-parameter group of diffeomorphisms if 1) {g t } is a one-parameter group of transformations of M 2) The mapping g : R M M defined by g(t, x) = g t (x) is differentiable Remark 1.4. Condition 1) implies that for every s, t R that g t+s = g t g s. So g 0 = g t+( t) = g t g t. Thus as g 0 is the identity transformation, (g t ) 1 = g t for all t. Then by condition 2) g t, (g t ) 1 are differentiable so for every t we have g t is a diffeomorphism. Now we can look at a specific vector field linked to a one-parameter group of diffeomorphisms. We begin with choosing the flow (M, {g t }) where {g t } is a oneparameter group of diffeomorphisms of some subset M R n. Definition 1.5. The phase velocity, v(x), of this flow at some point x M is v(x) = d g t x dt t=0 i.e. the vector describing the motion of the phase flow at point x. For all phase flows described by one-parameter groups of diffeomorphisms in Euclidean space, the phase velocity is a vector and is defined at each point. Thus, for any such phase flow we can create a vector field by assigning the phase velocity to each point in the phase space. This is called the phase velocity vector field. When the function g t is unknown, the right side of the above equation is commonly written as ẋ. For the remainder of this paper we shall use the ẋ notation. From here we arrive at the concept of a linear ODE. 1.2. Linear ODEs. Definition 1.6. Let A : R n R n be a linear transformation. A linear ordinary differential equation is one with phase space R n and phase velocity v(x) = Ax, i.e. ẋ = Ax. The proof of our theorem will also require a useful theorem about linear ODEs Theorem 1.7. Consider the linear ODE ẋ = Ax for x R n. Then there exists a unique solution ϕ : R R n with initial condition ϕ(0) = x 0. This is a direct consequence of the existence and uniqueness theorem for firstorder ordinary differential equations. In particular, the solution is defined by ϕ(t) = e ta x 0.

TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS 3 1.3. Complexification and Decomplexification. The concepts of complexification and decomplexification will become useful to us in proving a later lemma in this paper. Any real vector space V with basis {v j } determines a complex vector space C V = C V with basis {v j }. We call C V the complexification of V. Each vector of C V is a complex linear combination v = c j v j of the v j, and V C V is identified with the subspace of those vectors whose coordinates c j are real. Similarly, any complex vector space W with basis {w j } = {a j +ib j } can be viewed as a real vector space R W with basis vectors {a j, b j }. Thus, 2dim(W ) = dim( R W ). We call R W the decomplexification of W. Remark 1.8. R ( C V ) V and C ( R W ) W for all real vector spaces V and complex vector spaces W. Before moving on, we must also consider complexification and decomplexification in terms of operators. We define a complexified operator as follows Definition 1.9. If A : U V is linear then C A : C U C V is a complex linear map defined by the equation C A(x + iy) = Ax + iay for x, y U. We can also consider the eigenvalues of C A. Definition 1.10. A scalar λ C is an eigenvalue of a real operator A : U U if there exists a nonzero vector v C U, called an eigenvector, such that C Av = λv. This is important as real operators can have complex eigenvalues, which can only be accounted for by complexified operators. 1.4. Equivalence Relations between Phase Flows. We now switch gears to introduce another important topic - equivalence of phase flows. Definition 1.11. Two phase flows (R n, {f t }) and (R n, {g t }) are equivalent if there exists a bijection h : R n R n such that for every t R that h f t = g t h. Now within this broader definition, we can more precisely specify certain types of equivalences between phase flows based on the properties of our mapping h. 1) If h, h 1 are linear, then {f t } and {g t } are linearly equivalent 2) If h, h 1 are differentiable then {f t } and {g t } are differentiably equivalent 3) If h, h 1 are continuous, then {f t } and {g t } are topologically equivalent Remark 1.12. From here we observe that 1) 2) 3). Now we have all the tools necessary to state the main theorem of this paper and prove it. 2. Main Theorem Theorem 2.1. Let A, B : R n R n be linear operators with no purely imaginary eigenvalues. Then the linear ODEs ẋ = Ax and ẋ = Bx are topologically equivalent if and only if A and B have the same number of eigenvalues with positive real parts. Proving this theorem will require three different lemmas, each more difficult to prove than the last.

4 ALEX HUMMELS Lemma 2.2. Let A 1, B 1 : R n R n and A 2, B 2 : R m R m be linear. If the pairs of systems ẋ = A 1 x and ẋ = B 1 x as well as ẋ = A 2 x and ẋ = B 2 x are topologically equivalent, then the system defined by x 1 = A 1 x 1, x 2 = A 2 x 2 is topologically equivalent to the system defined by x 1 = B 1 x 1, x 2 = B 2 x 2. Proof of Lemma 2.2. This is an immediate result of the fact that the product of homeomorphisms is a homeomorphism. Lemma 2.3. If the operator A : R n R n has no purely imaginary eigenvalues then R n can be written as the direct sum of A-invariant subspaces, R m and R m+, such that for all eigenvalues λ of A R m, Re(λ) < 0 and for all eigenvalues µ of A R m +, Re(µ) > 0. Proof of Lemma 2.3. Consider C A in Jordan normal form and consider a single Jordan block B: λ 1 0... 0. 0................... 0....... 1 0...... 0 λ Let e 1,..., e m be the basis for this block. If λ R then the e j are real vectors and we re done with this block. If λ / R then none of the e j are real vectors. In this case we see, from the definition of Jordan normal form, that the subspace spanned by the vectors e 1,..., e n corresponds to a Jordan block B. λ 1 0... 0. 0................... 0....... 1 0...... 0 λ As λ λ we find the bases of these subspaces are linearly independent over C. We can then define a real subspace of R n C R n by span{e j + e j, (e j e j )i} If we complexify this space we get the span of the {e j } and {e j }. Thus as {e j } {e j } is linearly independent over C, this real subspace has dimension 2m. Also, the eigenvalues of A restricted to this subspace have strictly positive or negative real part by assumption. Thus, continuing this process with each block we can split up R n into a direct sum as required. Lemma 2.4. Let A : R n R n be linear with no purely imaginary eigenvalues. If Re(λ) > 0 for all eigenvalues λ, then the systems ẋ = Ax and ẋ = x are topologically equivalent. If Re(λ) < 0 for all eigenvalues λ, then the systems ẋ = Ax and ẋ = x are topologically equivalent. This lemma requires much more care than the other two and shall be dealt with after the main theorem is proven. For now, we will assume the lemma and proceed to the proof of the main theorem.

TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS 5 Proof of Theorem 2.1. Suppose A and B have the same number of eigenvalues with positive real parts (m + ) and those with negative real parts (m ). By lemma 2, both the operators A and B can be split into invariant subspaces A +, B + R m+ and A, B R m by the real part of their eigenvalues. By lemma 2.4, the systems ẋ = A + x and ẋ = B + x are both topologically equivalent to the standard system ẋ = x for x R m+. Similarly, it follows that ẋ = A x and ẋ = B x are topologically equivalent to ẋ = x for x R m. Then by lemma 2.2 we get both ẋ = Ax and ẋ = Bx are topologically equivalent to x 1 = x 1, x 2 = x 2 for x 1 R m+, x 2 R m. Thus, we must have that ẋ = Ax and ẋ = Bx are topologically equivalent to each other. Suppose ẋ = Ax and ẋ = Bx are topologically equivalent to each other and A has m 1 positive real part eigenvalues whereas B has m 2 positive real part eigenvalues. Using lemma 2.3, we can split A and B up into invariant subspaces according to the real part of their eigenvalues as before: A + R m1, A R n m1, B + R m2, B R n m2. Now any point in A cannot be mapped to a point not in B. Any point not in B will tend to infinity as t tends to 0, so any map between that point and any point in A must be discontinuous at 0. A similar argument holds for A + and B +. So ẋ = A + x and ẋ = B + x must be topologically equivalent, as well as ẋ = A x and ẋ = B x. However, homeomorphisms only exist between real vector spaces of the same dimension (a result which will not be proven in this paper). Thus, m 1 = m 2 as required. 3. Lemma 2.4 3.1. The Lyapunov Function. This lemma hinges on the construction of a specific quadratic form called a Lyapunov function. The theorem guaranteeing the existence of the Lyapunov function is as follows: Theorem 3.1. Let A : R n R n be linear and let all eigenvalues λ of A be such that Re(λ) > 0. Then there exists a positive definite quadratic form q in R n such that the derivative of q along the vector field Ax is positive definite. In our proof we will consider the operator B = C A : C R n C R n and a positive definite quadratic form q : R C n R whose derivative along the vector field of R Bz is positive definite. This approach allows us to handle the case of complex eigenvalues, as mentioned in section 1.3. With this route, we can restrict z to the canonical copy of R n in C n to get the actual theorem. Now, we begin to construct the function guaranteed by the theorem above. We begin by choosing any basis and letting q(z) = z, z = n i=1 z iz i where the z i are the coordinates of z in our basis. This q will be our Lyapunov function. Now

6 ALEX HUMMELS we compute the directional derivative. R Bzq(z) = lim h 0 q(z + hbz) q(z) h z + hbz, z + hbz z, z = lim h 0 h h z, Bz + h Bz, z + h 2 Bz, Bz = lim h 0 h = z, Bz + Bz, z = Bz, z + Bz, z = 2Re Bz, z So we have shown that the derivative of our positive definite quadratic form q along the vector field R Bz is a quadratic form. Now we only need to prove that this is in fact positive definite. In order to do so, we must find a special basis. If we have an eigenbasis for B, then we get the following ( n ) 2Re Bz, z = 2Re λ i z i z i = 2 i=1 n Re(λ i z i 2 ) By assumption, we know Re(λ i ) > 0 for every i so R Bzq(z) is positive definite. However, B doesn t necessarily have an eigenbasis, so we have to find another option. The intuitive route would be to find a basis in which B is almost diagonalizable and would thus almost have an eigenbasis. We formalize this notion of such a basis in the sublemma below. Sublemma 3.2. Let B : C n C n be linear and ε > 0. Then there exists a basis in C n in which A is upper triangular with entries above the main diagonal less than ε. Proof of Sublemma 3.2. Again, we use the existence of a Jordan normal form. Consider B in Jordan normal form, where it can be decomposed into Jordan blocks of the form λ 1 0... 0. 0................... 0....... 1 0...... 0 λ So we see Be i = λe i + e i 1 for i > 1 where e 1,..., e n is the basis of the specific block. Now letting e i = ( ε 2) i 1ei, we get a new basis e 1, e 2,..., e n. So for i > 1 Be i = ( ) i 1 ε Be i = 2 ( ) i 1 ε (λe i + e i 1) = λe i + ε 2 2 i=1 [ ( ) i 2 ε e i 1] = λe i + ε 2 2 e i 1.

TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS 7 Thus we get each Jordan block in a form with all entries above the diagonal less than ε. ε λ 2 0... 0. 0................... 0....... ε 2 0...... 0 λ Repeating this for each block we get the sublemma as required. Now that we have this basis, we need to prove that the directional derivative of q along R Bz using this basis is positive definite. Here we use yet another sublemma before getting into the actual proof. Sublemma 3.3. If the quadratic form a(x) = n i,j=1 a ijx i x j is positive definite then there exists ε > 0 such that if b(x) = n i,j=1 b ijx i x j and every b ij < ε then every quadratic form a(x) + b(x) is also positive definite. Proof of Sublemma 3.3. We begin by considering a(x) on the unit sphere. As a(x) is continuous and the unit sphere is compact, i.e. a closed and bounded set, a(x) achieves both a supremum and infimum on the sphere. Then since a(x) > 0 for all x 0 the infimum, α, is such that a(x) α > 0. If we have b ij < ε then we get the following inequality for b(x) on the sphere n n b(x) = b ij x i x j b ij < n 2 ε i,j=1 i,j=1 As we want a(x) + b(x) > 0 for all x 0 we must have a(x) > b(x). It suffices to choose ε > 0 such that α > n 2 ε. Thus we find that if 0 < ε < α/n 2 then a(x)+b(x) is positive on the unit sphere, and is therefore positive definite. Remark 3.4. By the argument above, we get that any positive definite quadratic form a(x) can be bounded above and below like so: α x 2 a(x) β x 2, 0 < α β After all this work, we are finally ready to prove Theorem 3.1. Proof of Theorem 3.1. Let ε > 0 and consider q(z) in the basis guaranteed by sublemma 3.2. We have already shown that q and R Bzq(z) are quadratic forms regardless of basis, so we only need to show that R Bzq(z) > 0 for all z. Now we split up R Bzq(z) as follows n R Bzq(z) = 2Re Bz, z = 2Re b ij z j z i + 2Re b ij z j z i i,j=1 b ij z j z i = 2Re i=j i=j 1 We can do this because in our basis b ij = 0 for i > j and i < j 1. We also know b ii = λ i, i.e. the diagonal entries are the eigenvalues of B. From this we get that 2Re i=j b ijz j z i = 2Re n i=1 λ i z i 2 is positive definite as Re(λ i ) > 0 for all i. In our basis we know that b ij < ε for any ε > 0. Therefore by sublemma 3.2 we get 2Re b ij z j z i + 2Re b ij z j z i = R Bzq(z) > 0 z i=j i=j 1

8 ALEX HUMMELS As mentioned previously, by restricting z to R n, the real theorem follows from our proof in the complex case. 3.2. The Homeomorphism h. The object of lemma 3 is finding a homeomorphism h : R n R n between the phase flows {f t } of ẋ = Ax and {g t } of ẋ = x. In fact we are only two shorts proofs away from proving lemma 3. But first, some setup. We let S be the unit ellipsoid described by S = {x R n : q(x) = 1} where q is our Lyapunov function. Now we claim that we can define h : R n R n by 1) h fixes S pointwise 2) h(f t x 0 ) = g t x 0 for every t R, x 0 S and h(0) = 0 Now all we have to do to finish lemma 3 is show that h is well-defined, h and h 1 are continuous bijections, and h f t = g t h for all t. To do so, we start with the following sublemmas. Sublemma 3.5. Let ϕ : R n R n be some nonzero solution of ẋ = Ax and define ρ : R R by ρ(t) = ln(q(ϕ(t))). Then ρ is a diffeomorphism. In particular, there exists an α, β such that 0 < α < β and α dρ dt β. Proof of Sublemma 3.5. Now by the uniqueness theorem for linear ODEs, we get that q(ϕ(t)) 0 for all t R. Else, this would violate that 0 = 0, i.e. that the point 0 does not move along the phase flow. This guarantees that ρ is well-defined and we already know q(x) is differentiable, so we get dρ. Since Ax q(x) is a positive definite quadratic form and dρ dt that 0 < α β and α dρ dt β. dt = Axq(x) q(x) is bounded, there exists an α, β such Corollary 3.6. Every nonzero point x R n can be written uniquely in the form x = f t x 0 where x 0 S, t R, and {f t } is the phase flow of ẋ = Ax. Proof of Corollary 3.6. First we consider the solution ϕ with initial condition ϕ(0) = x. Sublemma 3.5 implies that the phase flows outward from the origin at a definite rate at all times. Thus if q(x) < 1 there exists a time τ > 0 such that q(ϕ(τ)) = 1. Else if q(x) > 1 then the property holds with τ < 0. We then set x 0 = ϕ(τ) S. Then simply by flowing (backwards in the first case and forward in the second case) in time τ seconds we reach x again. Thus, x = f t x 0 where t = τ regardless of case. Then t is unique, as the phase curve passing through some x R only intersects S in a single point, x 0, by sublemma 3.5 and the monotonicity of ρ(t). Proof of Lemma 2.4. These proofs allow us to construct a one-to-one function F : R S n 1 R n \ {0} defined by F (t, x 0 ) = f t x 0. We quickly find that F is continuous, due to the continuous dependence of solutions on their initial conditions. Similarly, F 1 is one-to-one and continuous. We can actually use the same arguments with ẋ = x and {g t }, as for ẋ = x we get dρ dt = 2Re n i=1 Ax2 i n = 2. i=1 Ax2 i Thus we can create another one-to-one function G : R S n 1 R n \ {0} defined by G(t, x 0 ) = g t x 0. By the same logic G, G 1 are one-to-one, continuous functions. In fact, considering the function G F 1 : R n \ {0} R n \ {0}, we get (G F 1 )(x) = h(x) for all x 0. Thus, as h(0) = 0, we get that h is a one-to-one function. As h is continuous at 0 by sublemma 3.5, we need only prove that h f t = g t h for all t. This is obviously true at 0, and quite easy for x 0 where x = f s x 0 as (h f t )(x) = h(f t (f s x 0 )) = h(f t+s x 0 ) = g t+s x 0 = g t (g s x 0 ) = g t (h(x)) = (g t h)(x)

TOPOLOGICAL EQUIVALENCE OF LINEAR ORDINARY DIFFERENTIAL EQUATIONS 9 So we have successfully constructed a homeomorphism h to prove lemma 2.4. Closing Remarks This theorem specifically disallows purely imaginary eigenvalues, so we will take a moment to consider how their inclusion affects this argument. If we allow the eigenvalues of A to be such that Re(λ) 0 then in place of the Lyapunov function there would have to be a positive semidefinite function. This in turn invalidates our proof of the main theorem, as some representation x = f t x 0 may not be unique. The argument presented in this paper is outlined in [1]. The same source notes this theorem holds locally for ODE in general. More precisely, any nonlinear ODE ẋ = v(x) is topologically equivalent to its linear part in a small neighborhood of a fixed point, provided the linear parts of v(x) have no purely imaginary eigenvalues. Acknowledgments I would like to thank my mentor Ben Lowe for his continual guidance and encouragement on this paper. References [1] V.I. Arnold. Ordinary Differential Equations. The MIT Press. Massachusetts. 1980.