My question is very simple: Suppose $u,v:(a,b)\subset \mathbb{R} \to \mathbb{R}^+$ solve \begin{equation} (p(x)u'(x))'=-q(x)f(u(x)) \end{equation} \begin{equation} (p(x)v'(x))'=-r(x)g(v(x)) \end{equation} where $p>0$ and sufficiently smooth and $qf \geq rg \geq 0$ are sufficiently smooth as well. Suppose moreover that $u(a)=v(a)$ and $u'(a)=v'(a)$.
Q: Does it follow that $u \leq v$?
Note: In the classical setting for the Sturm-Picone theorem ($f(x,u)=q(x)u$, $g(x,u)=r(x)u$) one compares the zeroes of $u$ and $v$ ($u$ reaches a zero sooner than $v$) instead of claiming a full $u\leq v$ inequality. This suggests of course that the answer to my question is negative. Is there a nice counterexample?
EDIT: my original question (before this edit) had $f$ and $g$ not only depending on $u$ and $v$ respectively, but also explicitly on $x$. So \begin{equation} (p(x)u'(x))'=-f(x,u(x)) \end{equation} \begin{equation} (p(x)v'(x))'=-g(x,v(x)) \end{equation} where $f\geq g \geq 0$. In this case, we can easily construct a counterexample. Consider as domain $0 < u,v < 2$, $0 < x < \pi$. Choose $p=1$. Now draw a curve \begin{equation} v(x)=\begin{cases} &1 \text{ when } x < \frac{\pi}{2} \\ & 0.8-(x-2) \text{ when } x>2 \\ & s(x) \text{ (something smooth, smoothly connecting and with negative 2nd derivative) when } \frac{\pi}{2} \leq x \leq 2 \end{cases} \end{equation} Construct, for example by using Witney's extension theorem, a function $g(x,u)\geq 0$ with support contained in $[\frac{\pi}{2},2]\times [0.8,1.1]$ (Notice already that this domain is disjoint from the graph $(x,\cos(x))_{0 < x < \pi}$) such that $s''(x)=g(x,s(x))$. Now take $f(x,u)=u+g(x,u)$. $f\geq g \geq 0$ and $u(x)=\cos(x)$ solves $u''=-f$. Yet $v$ solves $v=-g$ while $v(3) <0< u(3)$