Now the open intervals A(t)with t E [to, t, cover the whole closed bounded interval to, ti, and taking a finite number of tk, k=1, . m such that [to, ti is completely covered by A(tk) yields the desired partition subdivision of to, til exist cod, o t h th due to the coutinuous differentiablity of g, for every u> o therd Jf(xo0(t)+6-,3o(t)+y,1)-f(o(t),3(1,t)≤C(|62|+|) and lg(xo(1)+6x,y0(t)+6y,t)-A(t)by≤C|bx+列 for all t∈R,bz∈Rn,by∈ R satisfying t∈{to,t1],|1x-x0()≤r,1,-3(t)≤r Fort∈△klet lk=(2P5)/2 Then. for 6(t)=x(t)-x0(t),y(t)=y(t)-o(t), we have x|≤C1(1x|+|6y|k) e|6k≤-ql6yk+C1|6+C1 (10.2) as long as 8, du are sufficiently small, where C1, g are positive constants which do not depend on k. Combining these two derivative bounds yield 18=|+(EC1/q)15, D)<C218-+EC2 for some constant C2 independent of k. Hence 162(7k-1+r)|≤er(62(7k-1)+(eC1/q)(7k-1))+C3 for T E0, Tk-Tk-1. With the aid of this bound for the growth of 8=I, inequality(10. 2) yields a bound for 8,li 6(7k-1+7)≤exp(-qr/e)|b(xk-1)+C4(|62(k-1)+(eC1/q)b(k-1))+C46, which in turn yields the result of Theorem 10.13 Now the open intervals �(t) with t → [t0, t1] cover the whole closed bounded interval [t0, t1], and taking a finite number of t ¯k, k = 1, . . . , m such that [t0, t1] is completely covered by �(t ¯k) yields the desired partition subdivision of [t0, t1]. Second, note that, due to the continuous differentiability of f, g, for every µ > 0 there exist C, r > 0 such that ¯ ¯ ¯ ¯ |f(x0(t) + �x, y0(t) + �y, t) − f(x0(t), y0(t), t)| ≈ C(|�x| + |�y|) and ¯ ¯ ¯ ¯ ¯ |g(x0(t) + �x, y0(t) + �y, t) − A(t)�y| ≈ C|�x| + µ|�y| ¯ for all t → R, � ¯ x → Rn, �y → Rm satisfying ¯ ¯ t → [t0, t1], |�x − x0(t)| ≈ r, |�y − y0(t)| ≈ r. For t → �k let |�y|k = (�y � Pk�y) 1/2 . Then, for �x(t) = x(t) − x0(t), �y(t) = y(t) − y0(t), we have |� ˙ x| ≈ C1(|�x| + |�y|k), �|� ˙ y|k ≈ −q|�y|k + C1|�x| + �C1 (10.2) as long as �x, �y are sufficiently small, where C1, q are positive constants which do not depend on k. Combining these two derivative bounds yields d (|�x| + (�C1/q)|�y|) ≈ C2|�x| + �C2 dt for some constant C2 independent of k. Hence |�x(�k−1 + � )| ≈ eC3� (|�x(�k−1)| + (�C1/q)|�y(�k−1)|) + C3� for � → [0, �k − �k−1]. With the aid of this bound for the growth of |�x|, inequality (10.2) yields a bound for |�y|k: |�y(�k−1 + � )| ≈ exp(−q�/�)|�y(�k−1)| + C4(|�x(�k−1)| + (�C1/q)|�y(�k−1)|) + C4�, which in turn yields the result of Theorem 10.1