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After some thinking I've come to the following conclusion.

Consider the initial value problem $$\text{(P)}\begin{cases}x'(t)=f(t,x(t)),\quad t\geq t_0\\x(t_0)=x_0 \end{cases}$$ where $f:D\subset\mathbb{R}^2\to\mathbb{R}$, $(t_0,x_0)\in D$. Suppose there are $t_1>t_0$ and continuous functions $L ,U:[t_0,t_1]\to\mathbb{R}$, such that $$(t_0,x_0)\in V:=\{(t,x)\in\mathbb{R}^2:t\in[t_0,t_1],\,L(t)\leq x-x_0\leq U(t)\}\subset D.$$ Suppose also there exist continuous $g,h:[t_0,t_1]\to \mathbb{R}$ such that for every $(t,x)\in V$ $$g(t)\leq f(t,x)\leq h(t)$$ and for every $t\in[t_0,t_1]$ $$L(t)\leq \int_{t_0}^{t}g(s)\,ds=,\quad \int_{t_0}^{t}h(s)\,ds\leq U(t).$$ Suppose also $f$ is Lipschitz in $x$ in $D$. Then there's a unique function $x$ defined on the interval $[t_0,t_1]$ that satisfies (P). Moreover it holds that $L\leq x-x_0\leq U$.

I'd like to know where this result was first stated (or a stronger version of it).

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  • $\begingroup$ (i) There are two questions in this post. According to MathOverflow guidelines, there should only one question in one post. (ii) Concerning the uniqueness of the solution, clearly there should be some condition on $f$, such that $f(t,x)$ being Lipschitz in $x$. $\endgroup$
    – Iosif Pinelis
    Commented Feb 18 at 0:47
  • $\begingroup$ I forgot to mention the Lipschitz part. $\endgroup$
    – aleph2
    Commented Feb 18 at 7:07
  • $\begingroup$ You still have two questions, and they are not really well posed. Yes, it is easy to see that the result is correct. However, given that this is easy to see, why do you think the result is stated as some proposition or theorem in the literature? $\endgroup$
    – Iosif Pinelis
    Commented Feb 18 at 15:33
  • $\begingroup$ Because it seems such a basic result, I wanted to know who first gave a stronger proposition. I updated the question(s). Thanks for your feedback. $\endgroup$
    – aleph2
    Commented Feb 18 at 17:24
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    $\begingroup$ Note that it is true with weak inequalities, $L\le G$ and $H\le U$. The set $\{u\in C([t_0,t_1]): (t,u(t))\in V\}$ is closed and invariant for the operator $T:u\mapsto x_0+\int_{t_0}^\cdot f(s,u(s))ds$ (which is a contraction wrto an equivalent norm) $\endgroup$
    – Pietro Majer
    Commented Feb 18 at 22:47

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