Inequality related to differential equation For $f\in C^1(\mathbb{R},\mathbb{R}), u,v\in C^1([0,\infty),\mathbb{R})$ such that
$$u'(t)\leq f(u(t)),$$
$$v'(t)\geq f(v(t)), \text{ and}$$
$$u(0)\leq v(0).$$ 
Show that $u(t)\leq v(t)$ for all $t\geq 0$
 A: Define $h(t)=v(t)-u(t)$, this is then in $\mathcal{C}^1([0,\infty),\mathbb{R})$. We want to show that $h$ is always non-negative. However observe that for all $t$, $$h^\prime(t)\geq f(v(t))-f(u(t)) .$$
In particular, $h^\prime(t)\geq 0$ when $u(t)=v(t).$ i.e, when $h(t)=0.$ Intuitively, we would then expect $h$ to always stay above the $x$-axis since it is continuously differentiable (note $h(0)\geq 0$). This idea can be made rigorous with calculus.
A: The following is a fairly standard result. (Sometimes the comparison is with
a solution, but the basic technique is the same.)
Pick $T>0$, then $f$ is uniformly Lipschitz with constant $L$ for $t \in [0,T]$.
Suppose $u(t_2) > v(t_2) $ for some $t_2 > 0$. Let $t_1 = \sup \{ t \in  [0,t_2] | u(t) \le v(t) \}$. Then $u(t) > v(t)$ for $t \in (t_1,t_2]$.
Let $\delta = u-v$, then $\delta(t_1) = 0$ and $\delta'(t) \le f(u(t))-f(v(t))\le L \delta(t)$ for
$t \in [t_1,t_2]$. This gives ${d \over dt} [e^{-L(t-t_1)} \delta(t)] \le 0$
and hence $\delta(t) \le 0$ for $t \in [t_1,t_2]$, which is a contradiction.
Hence $u(t) \le v(t)$ for all $t \in [0,T]$. Since $T$ was arbitrary, we are finished.
