Comparison theorem for ODE

Denote $h=f-g$. Assume there is $b>a$ such that $h(b)>0$. Since $h$ is continuous and $h(a)\leq0$ thus there exists $c\in[a,b)$ such that $h(x)>0$ for $x\in(c,b]$ and $h(c)=0$ ($c=\inf\{d\in[a,b]: f|_{[d,b]}>0\}$). We thus get that for $x\in[c,b]$ \begin{align*} h'\leq\Phi(f(x),x)-\Phi(g(x),x)\leq L|f(x)-g(x)|=Lh, \end{align*} from which $h(b)\leq h(c)e^{L(b-c)}=0$ which is a contradiction.

Without Lipschitz continuity of $\Phi$, it is false. For a counterexample take $\Phi(y,x)=(3/2)y^{1/3}$ and let $g(x)=0$, $f(x)=x^{3/2}$, and $a=0$.

(This is the standard example for nonuniqueness of solutions with a non-Lipshitz right hand side.)