Exercise 14, § 1

a) Let $\omega(t, z)$ be a continuous real function, $\geq 0$, defined for $0 < t \leq a$ and $z \geq 0$. Suppose that $z = 0$ is the only integral of $z' = \omega(t, z)$ defined for $0 < t \leq a$ and such that $\lim_{t \to 0} z(t) = 0$ and $\lim_{t \to 0} z(t)/t = 0$. Let I = $[t_0, t_0 + a[$, let S be a ball with centre $x_0$ in E, and f a map from $I \times S$ into E, such that for any $t \in I$, $x_1 \in S$ and $x_2 \in S$ one has $\|f(t, x_1) - f(t, x_2)\| \leq \omega(|t - t_0|, \|x_1 - x_2\|)$. Show that, on an interval with left-hand endpoint $t_0$ contained in I, the equation $x' = f(t, x)$ can have only one solution u such that $u(t_0) = x_0$. (Argue by contradiction: if v is a second integral of $x' = f(t, x)$, bound $\|u(t) - v(t)\|$ below on an interval with left-hand endpoint $t_0$, with the help of exerc. 13 b), and thus obtain a contradiction.)

Apply to the case where $\omega(t, z) = k(z/t)$ with $0 \leq k < 1$ (cf. IV, p. 200, exerc. 8).

b) The result of a) applies for $\omega(t, z) = z/t$; but show by an example in this case that if $\mathbf{u}, \mathbf{v}$ are two integrals to within $\varepsilon$ of $x' = f(t, x)$, equal to $x_0$ at the point $t_0$, it is not possible to bound $\| \mathbf{u}(t) - \mathbf{v}(t) \|$ above by a number depending only on $t$ (and not on the function $f$). (Take for $f$ a continuous map of $\mathbf{R}_+ \times \mathbf{R}$ into $\mathbf{R}$, equal to $x/t$ for $t \geq \alpha$ and for $0 < t < \alpha$ and $|x| \leq t^2/(\alpha - t)$, and independent of $x$ for the other points $(t, x)$.)

c) Let $\theta$ be a continuous real function, $\geq 0$ on the interval $[0, a]$. Show that if the integral $\int_0^a \frac{\theta(t)}{t} \, dt$ converges then the result of a) applies for $\omega(t, z) = \frac{1 + \theta(t)}{t} z$; on the other hand, if $\int_0^a \frac{\theta(t)}{t} \, dt$ is infinite, give an example of a continuous function $f$ such that $\| f(t, x_1) - f(t, x_2) \| \leq \frac{1 + \theta(|t - t_0|)}{|t - t_0|} \| x_1 - x_2 \|$, but such that the equation $x' = f(t, x)$ has infinitely many integrals equal to $x_0$ at the point $t_0$ (method similar to that of $b$) ).