4 1. Autonomous Linear Differential and Difference Equations A linear differential equation (with constant coeﬃcients) is given by a matrix A ∈ gl(d, R) via ˙(t) = Ax(t), where ˙ denotes differentiation with respect to t. Any differentiable function x : R −→ Rd such that ˙(t) = Ax(t) for all t ∈ R, is called a solution of ˙ = Ax. An initial value problem for a linear differential equation ˙ = Ax consists in finding for a given point x0 ∈ Rd a solution x(·,x0) that satisfies the initial condition x(0,x0) = x0. The differential equation ˙ = Ax is also called a time invariant or autonomous linear differential equation. The parameter t ∈ R is interpreted as time. A description of the solutions is based on the matrix exponential: For a matrix A ∈ gl(d, R) the exponential eA ∈ Gl(d, R) is defined by eA = ∞ n=0 1 n! An ∈ Gl(d, R). This series converges absolutely, i.e., ∞ n=0 1 n! An≤ ∞ n=0 1 n! A n = e A ∞, and the convergence is uniform for bounded matrices A, i.e., for every M 0 and all ε0 there is δ0 such that for all A, B ∈gl(d, R) with A , B≤M A − B δ implies eA − eB ε. This follows since the power series of the scalar exponential function is uni- formly convergent for bounded arguments. Furthermore, the inverse of eA is e−A and the matrices A and eA commute, i.e., AeA = eAA. Note also that for an invertible matrix S ∈ Gl(d, R), eSAS−1 = SeAS−1. The same properties hold for matrices with complex entries. The solutions of ˙ = Ax satisfy the following properties. Theorem 1.1.1. (i) For each A ∈ gl(d, R) the solutions of ˙ = Ax form a d-dimensional vector space sol(A) ⊂ C∞(R, Rd) over R, where C∞(R, Rd) = {f : R −→ Rd | f is infinitely often differentiable}. (ii) For each initial value problem the solution x(·,x0) is unique and given by x(t, x0) = eAtx0,t ∈ R. (iii) For a basis v1,...,vd of Rd, the functions x(·,v1),...,x(·,vd) form a basis of the solution space sol(A). The matrix function X(·) := [x(·,v1),...,x(·,vd)] is called a fundamental solution of ˙ = Ax and ˙ (t) = AX(t),t ∈ R. Proof. Using the series expression etA = ∑ ∞ n=0 An n! tn, one finds that the matrix etA satisfies (in Rd·d) d dt etA = AetA. Hence eAtx0 is a solution of the initial value problem. To see that the solution is unique, let x(t) be any

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2014 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.