Let the rectangle $P = {\lbrace(t,x);|t-t_0|<a, |x-x_0|<b\rbrace}\subsetℝ^2$, let $f$ and $g$ be two continuous and locally Lipschitz functions. If $g < f$ in $P$, then for $\phi$ and $\psi$ which are solutions of, respectively,
$x'=g(t,x), \hspace{0.1cm}x(t_0)=x_0$ and $x'=f(t,x),\hspace{0.1cm} x(t_0)=x_0$
defined for $0\le t\le c$, prove that $\phi(t) \le \psi(t)$ for $t_0< t\le c$
And with the same hypotheses, if $g\le f$ prove that $\phi(t) \le \psi(t)$ for $t_0\le t\le c$
What I did:
Since $0<f-g$ and $|f-g|\le M$ in $P$, with $M=sup\lbrace|(f-g)(t,x)|; (f-g)(t,x)\in P \rbrace$ , then applying Picard theorem, there exists a unique solution (call it $\sigma$) of
$x'=(f-g)(t,x), \hspace{0.1cm} x(t_0)=x_0$ in $I_\alpha = \lbrace t; |t-t_o|\le \alpha \rbrace$ with $\alpha=min\lbrace a, (b/M)\rbrace$
then, $\sigma'(t)=(f-g)(t, \sigma(t))$, $\hspace{0.1cm}$ $\sigma(t_0)=x_0$
... and this doesn't lead me to anything. Any hint on how should I proceed?