Let \(v_1,v_2:\mathbb R\to\mathbb R\) be continuous and satisfy \[ v_1(x)<v_2(x) \] for every \(x\). Let \(\varphi_1,\varphi_2\) solve \[ x'=v_1(x) \qquad\text{and}\qquad x'=v_2(x), \] respectively, on an interval \((a,b)\). If \[ \varphi_1(t_0)=\varphi_2(t_0) \] for some \(t_0\in(a,b)\), prove that \[ \varphi_1(t)\le\varphi_2(t) \] for every \(t\in(t_0,b)\).
Solution
Define \[ h(t)\coloneqq\varphi_2(t)-\varphi_1(t). \]
1. At every \(t\in(a,b)\) for which \(h(t)=0\), one has\[ h'(t)>0. \]
If \(h(t)=0\), then \[ \varphi_1(t)=\varphi_2(t)=x \] for some \(x\in\RR\). Using the two differential equations, \[ \begin{aligned} h'(t) &= \varphi_2'(t)-\varphi_1'(t)\\ &= v_2(x)-v_1(x) >0. \end{aligned} \]
2. There is \(\delta>0\) such that\[ h(t)>0 \]for every \(t\in(t_0,t_0+\delta)\).
The hypothesis gives \(h(t_0)=0\), and step 1 gives \(h'(t_0)>0\). Hence \[ \lim_{t\downarrow t_0} \frac{h(t)-h(t_0)}{t-t_0} = h'(t_0) >0. \] For all sufficiently small \(t-t_0>0\), the quotient is positive, and therefore \(h(t)>0\).
3. In fact,\[ h(t)>0 \]for every \(t\in(t_0,b)\).
Suppose otherwise. Then there is \(s\in(t_0,b)\) with \(h(s)\leq0\). Choose \(\delta>0\) as in step 2 with \(t_0+\delta<s\). By continuity of \(h\), there is at least one zero of \(h\) in \([t_0+\delta,s]\). Let \[ \tau \coloneqq \min\{t\in[t_0+\delta,s]:h(t)=0\}. \] Then \[ h(t)>0 \qquad (t_0+\delta\leq t<\tau), \] while \(h(\tau)=0\). Since \(h\) is differentiable at \(\tau\), \[ h'(\tau) = \lim_{t\uparrow\tau} \frac{h(\tau)-h(t)}{\tau-t} \leq0. \] But step 1 applies at the contact point \(\tau\) and gives \(h'(\tau)>0\), a contradiction.
4. Therefore, for every \(t\in(t_0,b)\),\[ \boxed{\varphi_1(t)<\varphi_2(t)}. \]
By the definition of \(h\), step 3 says \[ \varphi_2(t)-\varphi_1(t)>0. \] This is stronger than the required inequality \(\varphi_1(t)\leq\varphi_2(t)\).
5. Q.E.D.
Step 4 proves the desired comparison.