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Suppose for two given functions $f_1,f_2 \colon \mathbb{R}^2 \to \mathbb{R}$ there exist unique solutions $y_1$ and $y_2$ with the intersection of their intervals of existence $[0,\epsilon)$ to the integral equations $$y_k(x)=\int_{0}^{x} f_k(t,y_k(t)) dt$$. Moreover, suppose that $f_1(x,y)\leq f_2(x,y)$ (or strictly if that makes things easier), does it follow that $y_1 \leq y_2$ on $[0,\epsilon)$?

It's true if $f_1$ and $f_2$ are Lipschitz in $y$ and continuous in $x$ (i.e., satisfy the hypotheses of the standard uniqueness and existence theorem) or if $f_1$ and $f_2$ are decreasing in $y$ (i.e., satisfy the hypotheses of Peano's uniqueness theorem). The assumption of unique solutions is certainly necessary.

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  • $\begingroup$ The functions $f_1$ and $f_2$ are assumed continuous, right? $\endgroup$
    – Pietro Majer
    Commented Jan 24, 2011 at 16:31
  • $\begingroup$ That's one way to ensure existence of a solution, by Peano's existence theorem, but the proof of Peano's uniqueness does not require it. If both $f_k$ are continuous, then the solutions are classical solutions in that $y_k'(x)=f_k(x,y_k(x))$, and the problem becomes a matter of noticing that, $y_1(x)\leq y_2(x)$, on an interval $[0,\varepsilon)$, since $y_1'(0)\leq y_2'(0)$, and then you argue that the solutions can never cross such that $y_1(x)>y_2(x)$ after the crossing. $\endgroup$
    – AppliedSide
    Commented Jan 24, 2011 at 17:52
  • $\begingroup$ (continued) For me the problem is interesting only if at least one of the $f_k$ is not continuous, and one only has that $y_k'=f_k(x,y_k)$ a.e. $\endgroup$
    – AppliedSide
    Commented Jan 24, 2011 at 17:53
  • $\begingroup$ ah, I read your answer only after posting an answer. $\endgroup$
    – Pietro Majer
    Commented Jan 24, 2011 at 18:16

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In general the answer seems to be no: Let $$ f_1(t,y)=\begin{cases} -1, &t\leq 0,\quad y\in \mathbb R\\\ 1,& t>0,\quad y>t/2,\\\ -1,& t>0,\quad -t/2 \leq y \leq t/2,\\\ -e^{-n^2y},& t\in [1/n,1/(n-1)),\quad y< -t/2,\quad n\geq 1, \end{cases} $$ and let $f_2(t,y)=-f_1(t,-y)$ so that $f_1(t,y)< f_2(t,y)$ when $(t,y)\in \mathbb R^2$.

For $x\geq 0$ we have the solutions $y_1(x)=x$ and $y_2(x)= -x$ and in order to show that e.g. $y_1$ is unique one must show that $y_1$ cannot be such that $-x/2 \leq y_1(x)\leq x/2$ some $x>0$ and furthermore show that if $-\infty < y_1(x)< -x/2$ when $1/n \leq x \leq 1/(n-1)$ for some $n>1$ then $$ 1 > e^{n^2y(1/(n-1))}-e^{n^2y(1/n)}= \frac n{n-1} > 1. $$

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