This preview shows page 1. Sign up to view the full content.
Unformatted text preview: solve x = p ( x,t ) , x ( ) = x , solve z = p ( z,t ) , z ( ) = x + x . Then ( t ) = x + Z t p ( ( ) , ) d, ( t ) = x + x + Z t p ( ( ) , ) + f ( ) d, therefore  ( t ) ( t )  = x + Z t p ( ( ) , )p ( ( ) , )f ( ) d  x  + Z t  f ( )  d + Z t  p ( ( ) , )p ( ( ) , )  d + 1 ( t ) + Z t K ( )  ( ) ( )  d. Therefore by BellmanGronwall,  ( t ) ( t )  ( + 1 ( t )) exp R t K ( ) d ....
View Full
Document
 Fall '10
 ClaireTomlin

Click to edit the document details