Question

4.1.3 - 4 and -1 -11 then
(a) x2, y 3 (b)x=-2. y - 3
- then​

Answer 1

Answer:

Step-by-step explanation:

Let a > 0, then for 0 < t < a., by considering (4.1.4), we have n t |x|(t) - xi (t) < |x|(0) - X:(0)| 'i J I' f,(qi(s)) - *#,G,G) ds t ... Thus, by Gronwall's inequality; M d 2“ . |x|(t) ... Also from (4.1.3) we see that {y}(t)} and so {qk(t)} are uniformly Cauchy on [b,a]. ... Q. E. D. Let x = (x1,x2,...,xn)elR", y = (y1, y2,...,ym)elf", Pec'(1R", 1R), = -v P =