Matrix-valued functions

Transcription

1 Matrix-valued functions Aim lecture: We solve some first order linear homogeneous differential equations using exponentials of matrices. Recall as in MATH2, the any function R M mn (C) : t A(t) can be thought of as a matrix of functions A(t) = (a ij (t)) ij with each a ij : R C. We thus have Defn The derivative of the matrix-valued function A(t) is Similarly, we may integrate t d A(t + t) A(t) (A(t)) = lim = t 0 t t 0 A(t) = ( t t 0 ) a ij (t). ij ( ) daij. ij Rem In the special case that A(t) = y(t) is vector-valued, differentiation is also given by d... d so is linear. Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

2 First order matrix DE We will be interested in solving DEs in the vector-valued function y(t) of the following form. First-order DE with constant co-efficients Let A M nn (C) & v(t) be a C n -valued fn. dy = Ay(t) + v(t) is a first order differential equation with constant co-efficients. It is homogeneous if v(t) = 0. We may also specify an initial condition such as y(0) = w for some given w C n. We keep this as standard notation throughout. Rem Let V be the space of infinitely differentiable functions from R C n & suppose v(t) V. Then the DE above is linear in the following sense. Note that T = d A : V V is linear & the eqn can be re-written as T y = v. Hence homogenous solns correspond to ker T. Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

3 Product rule Below we let A(t) be a fn with values in M lm (C) & B(t) be a fn with values in M mn (C). Prop d da (A(t)B(t)) = B(t) + A(t)dB. In particular, for a constant matrix B M mn (C) we have d da (A(t)B) = B. Proof is just a computation e.g. if l = m = 2, n = then given fns of t, a, a 2, a 2, a 22, b, b 2 we have [( ) ( )] d a a 2 b = d ( ) a b + a 2 b 2 a 2 a 22 b 2 a 2 b + a 22 b 2 ( a = b + a 2 b 2 + a b + a ) 2b 2 a 2 b + a 22 b 2 + a 2 b + a 22b 2 ) ( ) b + b 2 = ( a a 2 a 2 a 22 ( a a 2 a 2 a 22 ) ( b b 2 ) Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

4 Properties of the exponential function Prop Let A, B M nn (C) be commuting matrices. exp(0) = I 2 exp(a + B) = exp(a) exp(b) 3 exp(a) = exp( A). Proof. For ) note exp(0) = Also 2) & ) = 3) so we prove 2) only. We assume the standard (Cauchy) product formula for power series. exp(a + B) = k 0 = k 0 k! (A + B)k = k 0 i+j=k, i,j 0 k! i!j! Ai B j = i 0 i+j=k, i,j 0 i! Ai ( ) k A i B j i j 0 j! Bj = exp(a) exp(b) Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

5 Derivative of the exponential function Theorem Let A M nn (C). d (exp(ta)) = A exp(ta). 2 For any w C n, y(t) = exp(ta)w is a soln to dy condition y(0) = w. = Ay(t) with initial Proof. For ) all the entries e ij (t) of exp(ta) are analytic fns of t so we may differentiate term by term d (exp(ta)) = d k! Ak t k = (k )! Ak t k k 0 k = A k (k )! Ak t k = A exp(ta) For 2) note y(0) = exp(0)w = I w = w & the product rule gives dy = d (exp(ta)w) = d (exp(ta)) w = A exp(ta)w = Ay(t). Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

6 Uniqueness of soln Theorem y(t) = exp(ta)w gives the unique soln to the initial value problem, dy = Ay(t), y(0) = w Proof. Saw last slide, that y(t) = exp(ta)w solves the IVP. We show uniqueness by giving another method of solving the IVP. We first triangularise A with some change of co-ords matrix C GL n (C) so that C AC = U for some upper triangular U. We change co-ords to z(t) = C y(t). The new initial condn is z(0) = C w = x & the new DE is d (Cz(t)) = ACz(t). Left multiplying by C & using the product rule this simplifies to dz = C d (Cz(t)) = C ACz(t) = Uz(t). Suffice show new IVP in z(t) has a unique soln. Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

7 Proof by cascaded system of DEs We proceed by back-subtitution. Write U = (u ij ) ij, z(t) = (z (t),..., z n (t)) T, x = (x,..., x n ) T & look first at the last row of dz = Uz(t). We solve first uniquely for z n (t) which is possible as it is the soln to the first order linear ODE dz n = u nnz n (t), z n (0) = x n. We now look at the second last row of dz = Uz(t) to solve uniquely for z n (t). Indeed, this has a unique soln as it is the soln to dz n = u n,n z n (t) + u n,n z n (t), z n (0) = x n. which can be re-written as the linear st order ODE dz n u n,n z n (t) = u n,n z n (t), & z n (t) has been determined uniquely. z n (0) = x n Continuing in this fashion, we see that the IVP must have a unique soln. This completes the proof of the thm. Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

8 Example E.g Solve the IVP dy = Ay(t), y(0) = (2, 0,, )T if A = C(J (5) J 3 (2))C & 0 0 C = Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

9 Example cont d Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0

10 Space of homogeneous solns Let V be the space of infinitely differentiable functions from R C n. Prop Let A M nn (C) and Y be the set of solns y(t) to the DE dy = Ay(t). Then Y is n-dimensional with basis the columns of exp(ta), that is, if for j =,..., n we write y (j) (t) = (exp(ta) j,..., exp(ta) nj ) T, then {y () (t),..., y (n) (t)} is a basis for Y. Proof. Any y(t) = (y (t),..., y n (t)) T Y is a linear combn of columns of exp(ta) since y(t) = exp(ta)y(0) = y (0)y () (t) y n (0)y (n) (t). Hence the columns span Y. Lin indep of the columns follows easily from the fact that exp(ta) is invertible so exp(ta) : C n Y is an isomorphism. Daniel Chan (UNSW) Lecture 27: Homogeneous differential equations Semester / 0