lecture_35 (dragged) 2 - d dt = A , (0) = has the unique...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MA 36600 LECTURE NOTES: MONDAY, APRIL 20 3 To see why this suffices, frst recall the Product Rule: d dt x = d dt ° Ψx 0 ± = ² d dt Ψ ³ x 0 =[ ] x 0 = Ax . As For the initial condition: x (0) = Ψ (0) x 0 = Ix 0 = x 0 . We now show that Ψ ( t ) satisfes the initial value problem above. We have the expansion Ψ ( t )= ´ k =0 A k t k k ! = I + A t + A 2 t 2 2 + A 3 t 3 6 + ··· = Ψ (0) = I . This Function has the derivative d dt Ψ = 0 + A + A 2 t + A 3 t 2 2 + ··· = ´ k =1 A k t k 1 ( k 1)! = A · ´ k =1 A k 1 t k 1 ( k 1)! = . Using these ideas, it is easy to see that in general the initial value problem
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d dt = A , (0) = has the unique solution ( t ) = exp ( A t ) . Keep in mind that the order oF the matrices here does matter! The matrix exp ( A t ) is not a solution to the initial value problem....
View Full Document

This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online