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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.