Exercise 11, § 1

a) On the box P : $|t - t_0| < a,\ |x - x_0| < b$ let g and h be two continuous real functions such that $g(t, x) < h(t, x)$ on P. Let u (resp. v) be an integral of $x' = g(t, x)$ (resp. $x' = h(t, x)$) such that $u(t_0) = x_0$ (resp. $v(t_0) = x_0$) defined on an interval $[t_0, t_0 + c[$; show that for $t_0 < t < t_0 + c$ one has $u(t) < v(t)$ (consider the supremum $\tau$ of the t for which this inequality holds).

b) Let u be the maximal integral of $x' = g(t, x)$ corresponding to the point $(t_0, x_0)$ (exerc. 10), defined on an interval $[t_0, t_0 + c[$, with values in $|x - x_0| < b$. Show that in every compact interval $[t_0, t_0 + d]$ contained in $[t_0, t_0 + c[$ the minimal and maximal integrals of the equation $x' = g(t, x) + \varepsilon$ corresponding to the point $(t_0, x_0)$ are defined once $\varepsilon > 0$ is small enough, and converge uniformly to u when $\varepsilon$ tends to 0 through values > 0.

c) Let g and h be two real continuous functions defined on P and such that $g(t, x) \leq h(t, x)$ on P. Let $[t_0, t_0 + c[$ be an interval on which are defined an integral u of $x' = g(t, x)$ such that $u(t_0) = x_0$, and the maximal integral v of $x' = h(t, x)$ corresponding to the point $(t_0, x_0)$. Show that $u(t) \leq v(t)$ on this interval (reduce to case a) using b)).