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1.1. Existence of Solutions 5 solution of the initial value problem and put y(t) = e−tAx(t). Then by the product rule, ˙(t) = d dt e−tA x(t) + e−tA ˙(t) = −Ae−tAx(t) + e−tAAx(t) = e−tA(−A + A)x(t) = 0. Therefore y(t) is a constant. Setting t = 0 shows y(t) = x0, and uniqueness follows. The claims on the solution space follow by noting that for every t ∈ R the map x0 −→ x(t, x0) : Rd → Rd is a linear isomorphism. Hence also the map x0 −→ x(·,x0) : Rd → sol(A) is a linear isomorphism. Let e1 = (1, 0,..., 0) , . . . , ed = (0, 0,..., 1) be the standard basis of Rd. Then x(·, e1), . . . , x(·, ed) is a basis of the solution space sol(A) and the corresponding fundamental solution is eAt. Note that the solutions of ˙ = Ax are even real analytic, i.e., given by a convergent power series in t. As an easy consequence of the solution formula one obtains the following continuity properties. Corollary 1.1.2. (i) The solution map (t, x0) → x(t, x0) = eAtx0 : R × Rd → Rd is continuous. (ii) For every M 0 the map A → x(t, x0) = eAtx0 : {A ∈ gl(d, R) | A ≤ M} → Rd is uniformly continuous for x0 ∈ Rd with x0≤ M and t in a compact time interval [a, b],a b. Proof. (i) This follows, since for x0,y0 ∈ Rd and s, t ∈ R, eAsx0 − eAty0 ≤ eAs − eAt x0 + eAt x0 − y0 . (ii) Let A , B≤ M. One has eAtx0 − eBtx0 ≤ eAt − eBt x0 = ∞ n=0 (An − Bn) tn n! x0 . For ε 0 there is N ∈ N such that for all t ∈ [a, b], ∞ n=N+1 (An − Bn) tn n! ≤ ∞ n=N+1 ( A n + B n ) |t|n n! ≤ ∞ n=N+1 2 |Mt|n n! ε.