Question:

a b Xx 1 x 0 5 1E t b t b k20 c x keZ is uniformly Cauchy on

Last updated: 9/21/2023

a b Xx 1 x 0 5 1E t b t b k20 c x keZ is uniformly Cauchy on

a b Xx 1 x 0 5 1E t b t b k20 c x keZ is uniformly Cauchy on t b t b verification of these properties is left for the reader We consider the uniform limit of the sequence x to 8 to 8B x b C which is given by x t lim x 1 1E1 b t b x 1 B x b for each te to b t b MK k 1 Thus x t lim xo 1 2 F 8 8 89 ds f s x x lim ff s x s ds x lim f s x s ds E xXx 4 ff s lim x s ds x f f s x s ds Thus the function tx 1 satisfying the integral equation Finally the validity of this integral equation has the following two implications x dx dt x 1 x ff s f s x s ds for all 1 t b t b that is X t x f s 4 d 0 f s x s ds x 0 xo j f s x s ds f t X t This now proves that the curve X 1 8 1 82 thus obtained is a solution of the initial value problem 2 3 UNIQUENESS OF A SOLUTION 20 We prove an inequality which will lead us to the uniqueness of the solutions Let Proposition 2 Gronwall s Inequality f a b 0 g a b 000 be continuous functions and A 0 a constant satisfying ft A Chapter 2 Systmes of First Order ODE g de ffs gs ds for all te ab Then ft A e Proof First we assume A 0 and put ht A S te ab Then ht 0 for all 1 a b and h t f t g t h t g t for all te a b fsgs ds for all Salt for all 16 ab Integrating this inequality over at