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
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