1.3. Solution Formulas 11 of the linear differential equation ˙ = Dx with initial value x0 ∈ Rd is given by (1.3.1) eDtx0 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ eμ1t · · · eμdt ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ x0. More generally, suppose that A ∈ gl(d, R) is diagonalizable, i.e., there exists a transformation matrix T ∈ Gl(d, R) and a diagonal matrix D ∈ gl(d, R) with A = T −1 DT . Then the solution of the linear differential equation ˙ = Ax with initial value x0 ∈ Rd is given by x(t, x0) = T −1 eDtTx0, where eDt is given in (1.3.1). By Proposition 1.3.1 the solution determined by a Jordan block J of a matrix A determines eAtx0 in the corresponding subspace. Next we give explicit formulas for various Jordan blocks. Example 1.3.4. Let J be a Jordan block of dimension m associated with the real eigenvalue μ of a matrix A ∈ gl(d, R). Then J = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ μ 1 · · · · · · · 1 μ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ and eJt = eμt ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 t t2 2! · · tm−1 (m−1)! · · · · · · · · · · t2 2! · t 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ . In other words, for x0 = [x1,...,xm] the jth component of the solution of ˙ = Jx reads (1.3.2) xj(t, x0) = eμt m k=j tk−j (k − j)! xk. Example 1.3.5. Let J = λ −ν ν λ be a real Jordan block associated with a complex eigenvalue pair μ = λ ± ıν of the matrix A ∈ gl(d, R). Let x0 be in the corresponding real eigenspace of μ. Then the solution x(t, x0) of ˙ = Jx is given by x(t, x0) = eλtR(t)x0 with R(t) := cos νt − sin νt sin νt cos νt . This can be proved by verifying directly that this yields solutions of the differential equation. Alternatively, one may use the series expansions of sin and cos.

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.