Error bounds for non-autonomous systems with respect to input The error bounds for an ordinary differential equation:
\begin{equation}
  \dot{x}(t) = f(x(t))
\end{equation}
with respect to initial conditions $x(t_0) = x_0$, $\hat{x}(t_0)=\hat{x}_0$
\begin{equation}
  \left|\left|\hat{x}(t)-x(t)\right|\right| \leqslant e^{L\left|t-t_0\right|}\left|\left|\hat{x}(0)-x(0)\right|\right|
\end{equation}
where L is a Lipschitz constant of $f$ with respect to $x$. For a proof, see e.g. Stoer, Bulirsch.
I have read in an article [Arnold] that for a system: 
\begin{equation}
  \dot{x}(t) = f(x(t), u(t))
\end{equation}
hold a similar bounds with respect to different input signals $u(t)$ and $\tilde{u}(t)$:
\begin{equation}
  \left|\left|\tilde{x}(t)-x(t)\right|\right| \leqslant C\left(e^{L\left|t-t_0\right|}-1\right)\max_{s\in[t_0,t]}\left|\left|\tilde{u}(s)-u(s)\right|\right|
\end{equation}
where L is a Lipschitz constant of $f$ with respect to $x$. Initial conditions are assumed to be equal $\tilde{x}(0) = x(0) = x_0$. An input signal $\tilde{u}(t)$ is assumed to be a polynomial approximation of $u(t)$.
I would like to ask for a help with proving this statement. I went through the following steps:
\begin{align}
  &x(t) = x_0 + \int^t_{t_0}f(x(s), u(s)) \mathrm{ds}\\
  &\tilde{x}(t) = x_0 + \int^t_{t_0}f(\tilde{x}(s), \tilde{u}(s)) \mathrm{ds}\\
  &\tilde{x}(t) - x(t) = \int^t_{t_0} \left[f(\tilde{x}(s), \tilde{u}(s)) - f(x(s), u(s)) \right] \mathrm{ds}\\
  &\left|\left|\tilde{x}(t) - x(t)\right|\right| \leqslant \int^t_{t_0} \left|\left|f(\tilde{x}(s), \tilde{u}(s)) - f(x(s), u(s)) \right|\right| \mathrm{ds}
\end{align}
I get stuck at this step. I am not sure whether I am supposed to use Lipschitz condition:
\begin{equation}
  \left|\left|f(\tilde{x}(t), U(t)) - f(x(t), U(t)) \right|\right| \leqslant L\left|\left|\tilde{x}(t) - x(t)\right|\right| 
\end{equation}
and whether this is a correct formulation of Lipschitz condition with respect to x or whether I need to somehow use the fact about polynomial approximation. Any assistance is appreciated.
Edit:
Additional uniform Lipschitz condition of $f$ with respect to both arguments can be assumed.
 A: You can use Gronwall's inequality, we need only the following corollary,

Let $L$ and $U$ be non-negative real numbers. Suppose that$\ \ f: [t_0,t_1] \rightarrow \mathbb{R}$ is a continuous function satisfying
  $$f(t) \leq \int_{t_0}^{t} Lf(t)+M \ \,dt  $$ 
  Then 
  $$f(t) \leq \frac{L}{M} \left(\exp(t-t_0) -1\right)$$

proof. 
From Grownwall's inequality:
$$\begin{align}
f(t) & \leq  M(t-t_0) + \int_{t_0}^{t} ML(s-a)\exp\left(\int_{s}^{t}L\,dr\right) \, ds\\
& =  M(t-t_0) + \int_{a}^{t} ML(s-t_0)\exp\left(M(t-s)\right) \, ds \\
& = M(t-t_0) + \left[-M(s-t_0)\exp(L(t-s))\right]_{t_0}^{t} + \int_{t_0}^{t} M\exp(L(t-s)) \,dt \\
&= \frac{M}{L} \left(\exp(t-t_0) -1\right)\
\end{align} $$
Now we prove

Let $f \in C(\mathbb{R}^n\times \mathbb{R}^m, \mathbb{R})$ be lipschitz in the first variable with constant $L$, and in the second variable by constant $M$. If $x_i:[t_0,t_1] \rightarrow \mathbb{R}^n$ satisfy, 
  $$\begin{cases}
x_i'(t) = f(x_i(t),u_i(t)\\
x_i(t) = x_0\\
\end{cases}$$
  Then there is a constant depending only on $L$ and $M$ s.t,
  $$\|x_1(T)-x_2(T)\| \leq C\left(  e^{L(T-t_0)}-1 \right) \max_{[t_0,T]} \|u_2(t) - u_1(t) \| $$
  for all $T \in [t_0,t_1]$

proof.
Let $T \in [t_0,t_1]$. Note that for all $t \leq T$,
$$\|x_1(t) - x_2(t)\| \leq \int_{t_0}^{t} \|f(x_1(t),u_1(t)) - f(x_2(t),u_2(t))\| \,dt$$
By Lipschitz continuity:
$$\begin{align}
\|x_1(t) - x_2(t) \| & \leq \int_{t_0}^{t} L\|x_1(t) - x_2(t)\| + \|f(x_2(t),u_1(t)) - f(x_2(t),u_2(t)) \| \,dt \\
& \leq \int_{t_0}^{t} L\|x_1(t) - x_2(t)\| + M\max_{[t_0,T]} \|u_2(t) - u_1(t) \| \,dt \\
\end{align}$$
Hence by Gronwall's inequality (take $f(t) = \|x_2(t) -x_1(t) \|$ and replact $M$ in the corollary above by $M\max_{[t_0,T]} \|u_2(t) - u_1(t) \|$):
$$ \|x_1(T)-x_2(T)\| \leq \frac{M}{L} \left(  e^{L(T-t_0)}-1 \right) \max_{[t_0,T]} \|u_2(t) - u_1(t) \|  $$ 
