Infer boundedness from differential inequality $\frac{dx}{dt} \leq x(t)^2 + y(t)$? The original problem comes from a paper saying that
$$\frac{d}{dt}\|u\|_{H^s} \leq C\|u\|_{H^s}^2 + \|f(t)\|_{H^s},$$
where $\|\cdot\|_{H^s}$ denotes the Sobolev norm of order $s$, and $f(t) \in L^1(0,T;H^s)$. Then the author concludes that $x \in L^\infty(0,T; H^s)$. But I think it is fine to simplify it into a real-valued differential inequality for positive function $x(t)$:
$$\frac{dx}{dt} \leq x(t)^2 + y(t),$$
where $y(t)$ is some integrable function. How do we see $x(t)$ is also bounded in some interval, say $[0,T]$? If the power of $x(t)$ in the RHS is $1$, then we could use Gronwall inequality to conclude the result. But it is not the case here. Could anyone help with it? I appreciate any hint and suggestion!
Edit: On another paper I read that:
$x(t) \leq g(t)$, where $g(t)$ solves
$$
\frac{dg}{dt} = g(t)^2 + y(t), g(0) = x(0).
$$
And $T$ is chosen as the any number such that $\sup_{t\in[0,T]} g(t) < \infty$.
Questions:

*

*Does the ODE $$
\frac{dg}{dt} = g(t)^2 + y(t), g(0) = x(0)
$$ has a solution?

*Why $x(t) \leq g(t)$ on $[0,T]$?

 A: This is not a complete answer as it doesn't consider the case that your function $y$ is merely integrable, but it illustrates the idea. Let's suppose that $y$ is sufficiently nice such that the IVP
\begin{equation}
\dot{g}(t)=g(t)^2+y(t),\quad g(0)=x_0
\end{equation}
has a solution $g\in C^1([0,T]).$ Furthermore, consider a function $x\in C^1([0,T])$ that satisfies
\begin{equation}
\dot{x}(t)< x(t)^2+y(t),\quad x(0)=x_0
\end{equation}
for all $t\in [0,T]$.
Claim: The function $h(t):=g(t)-x(t)> 0$ for all $t\in(0,T]$.
Proof: Since $\dot{h}(0)>0$, there exists a small interval where $h$ is strictly positive. Consider the minimal $\tilde{t}\in(0,T]$ that satisfies $h(\tilde{t})=0$. Since $h(t)>0$ for $t\in(0,\tilde{t})$ and $h(\tilde{t})=0$, we conclude that $\dot{h}(\tilde{t})\leq 0$. On the other hand, we have
\begin{equation}
\dot{h}(\tilde{t})=\dot{g}(\tilde{t})-\dot{x}(\tilde{t})>g(\tilde{t})^2-x(\tilde{t})^2=0,
\end{equation}
which is obviously a contradiction. Therefore no such point $\tilde{t}$ exists and $h(t)>0$ for $t\in(0,T]$.
If you don't have a strict inequality, then the argument is similar but a bit more tedious to write down.
A: After discussing with my advisor, I think we need some more assumption on $u$, that is, we assume that $u \in L^1(0,T;H^s)$. We first note that the ODE
$$
\frac{dg}{dt} = g(t)^2 + y(t), g(0) = x(0)
$$
has a local solution because the right-hand-side is locally Lipschitz and integrable in time. Denote its maximal existence time $T_{max}$, and let $T < T_{max}$.
Then, notice that
$$
\frac{d(\|u\|_{s} - g)}{dt} \leq (\|u\|_s+g)(\|u\|_s-g),
$$
then a Gronwall type argument applies, as we assume that $u \in L^1(0,T;H^s)$, and $g$ remains bounded on $[0,T]$. It follows that $\|u\|_s \leq g$ on $[0,T]$.
