How to apply Gronwall's lemma on a variational system I am trying to figure out how to get a bound using Gronwall's lemma while reading a paper. Assume that $u(\cdot)$ and $v(\cdot)$ are the solutions, with initial condition $u(0)=v(0)=0,$ of the variational system
$$\begin{pmatrix}\dot u(t)\\\dot v(t)\end{pmatrix}=\begin{pmatrix}0&f'(q(t))\\1&0\end{pmatrix} \begin{pmatrix} u(t)\\v(t)\end{pmatrix} + \begin{pmatrix}\frac{1}{12} \dddot p(t)\\ -\frac{1}{6} \dddot q(t)\end{pmatrix}.$$
Here, $p(\cdot)$ and $q(\cdot)$ are real valued functions with the bounds
$|\dddot p(t)|\le K(|p(t)|^2 + |f(q(t))|$), and $|\dddot q(t)|\le K|p(t)|$, where $K$ is some constant and $f$ is a $C^3$ function for which, $f', f^{(2)}, f^{(3)}$ are uniformly bounded by a constant, say $B$ in particular and $K$ depends on $B$.
For a fixed $T>0$, and an arbitrary function $g$: we define $$\Vert g(\cdot)\Vert_\infty := \sup_{0\le t \le T} |g(t)|.$$
Under the above settings, how can we apply Gronwall's lemma and the bounds for $\dddot p$ and $\dddot q$ to obtain the bound (where $K$ is some constant that may not be equal as the one above):
$$\Vert u(\cdot)\Vert_\infty + \Vert v(\cdot)\Vert_\infty \le K(\Vert p(\cdot)\Vert_\infty ^2 + \Vert f(q(\cdot ))\Vert_\infty )? $$
I am only aware of the Gronwall's lemma for a single function, and not on a variational system.
Gronwall lemma for system of linear differential inequalities this post is the closest kind of result I found but here the additional terms are constants and not functions.
I would greatly appreciate if anyone could show me how to apply Gronwall's lemma in this case and also what form the constant $K$ would be (does it depend on the time $T$ or is it just a constant independent of the rest and just some constant multiplication of the bound $K$ for $\dddot p$ and $\dddot q$?)
 A: You have
\begin{align*}
\dot{u}(t)  & =f^{\prime}(q(t))v(t)+\frac{1}{12}\dddot{p}(t),\\
\dot{v}(t)  & =u(t)-\frac{1}{16}\dddot{q}(t).
\end{align*}
Integrating
\begin{align*}
u(t)  & =u(0)+\int_{0}^{t}\left[  f^{\prime}(q(r))v(r)+\frac{1}{12}\dddot
{p}(r)\right]  dr,\\
v(t)  & =v(0)+\int_{0}^{t}\left[  u(r)-\frac{1}{16}\dddot{q}(r)\right]  dr.
\end{align*}
Define $g(t)=|u(t)|+|v(t)|$. Then,
\begin{align*}
|u(t)|  & \leq\int_{0}^{t}\left[  |f^{\prime}(q(r))||v(r)|+\frac{1}{12}
|\dddot{p}(r)|\right]  dr\leq\int_{0}^{t}\left[  |f^{\prime}(q(r))|g(r)+\frac
{1}{12}|\dddot{p}(r)|\right]  dr,\\
|v(t)|  & \leq\int_{0}^{t}\left[  |u(r)|+\frac{1}{16}|\dddot{q}(r)|\right]
dr\leq\int_{0}^{t}\left[  |g(r)|+\frac{1}{16}|\dddot{q}(r)|\right]  dr.
\end{align*}
Summing the two inequalities you get
$$
g(t)\leq\int_{0}^{t}\left[  |f^{\prime}(q(r))|+1\right]  g(r)\,dr+\int_{0}
^{t}\left[  \frac{1}{12}|\dddot{p}(r)|+\frac{1}{16}|\dddot{q}(r)|\right]  dr.
$$
You can now apply Gronwall's inequality to get
$$
g(t)\leq\int_{0}^{t}\left[  \frac{1}{12}|\dddot{p}(r)|+\frac{1}{16}|\dddot
{q}(r)|\right]  dr\exp\left(  \int_{0}^{t}\left[  |f^{\prime}(q(r))|+1\right]
\,dr\right)  .
$$
Now you can apply the bounds on $\dddot{p}$ and $\dddot{q}$ to obtain
\begin{align*}
|u(t)|+|v(t)|  & \leq\int_{0}^{t}\left[  \frac{K}{12}|p(r)|^{2}
+K|f(q(r)|+\frac{K}{16}|p(r)|\right]  dr\exp\left(  \int_{0}^{t}\left[
|f^{\prime}(q(r))|+1\right]  \,dr\right)  \\
& \leq\int_{0}^{t}\left[  \frac{K}{12}\Vert p\Vert^{2}+K\Vert f(q)\Vert
_{\infty}+\frac{K}{16}\Vert p\Vert_{\infty}\right]  dr\exp\left(  \int_{0}
^{t}\left[  \Vert f^{\prime}(q)\Vert_{\infty}+1\right]  \,dr\right)  \\
& \leq T\left[  \frac{K}{12}\Vert p\Vert^{2}+K\Vert f(q)\Vert_{\infty}
+\frac{K}{16}\Vert p\Vert_{\infty}\right]  dr\exp\left(  \left[  \Vert
f^{\prime}(q)\Vert_{\infty}+1\right]  T\right)  .
\end{align*}
You could have used also $h(t)=\sqrt{|u(t)|^{2}+|v(t)|^{2}}$.
