Advanced Calculus fi..

654 Advanced Calculus, Fifth EditionWe can proceed as before, taking $,'O'(t) = 2, for a 5 t 5 b and defining $y'(t) by(9.65). These functions are then well defined and continuous for a 5 t 5 b, withgraphs in R. We let M be chosen so thatand have againWe proceed as before to obtain (9.66), choosing h so that b - i 5 h, i - a 5 h. Therest of the argument is the same, and we obtain the desired solution &(t). It shouldbe remarked that now (9.67) is valid for a 5 t I b, and hence (9.68) follows, withthe derivatives at t = a, b understood as derivatives to the right and left, respectively.111. (Uniqueness). We return to the general case of I. .Let us suppose that wehave found a second solution xi = +i(t) (i = 1 ,..., n), It - il < h with +;(i) = Xi.Since this is a solution of (9.54),(+((t)( = I Fi(t, +l(t), .. . ) I 5 M as long as thesolution remains in the rectangular region R. Since k; 5 Mh by (9.63), it followsthat the solution cannot leave R for (t- il < h (Problem 8 following Section 9.9)and must therefore have a graph in R.From (9.54) we can writeand hence from (9.65),Alsosince IF, (t , X I, . . . , x,)l 5 M in R. By induction as before, using (9.61), we concludethat for It - il < h,%'MN converges, so that MN +- 0 as N +- m. Therefore the sequenceThe series$IN'(t) converges uniformly to +, (t) for It - il < h. But we already know that~$,'~'(t) converges to &(t) on this interval. Therefore $,(t) = +,(t) for Jt - tJ < h,as asserted.