Let $T\in\mathbb{R}_{> 0}$ and $f\in L^{\infty}((0,T);W^{1,\infty}(\mathbb{R}))$ be given and consider for $(t_{0},x_{0})\in [0,T]\times\mathbb{R}$ the initial value problem for $t\in[0,T]$ \begin{equation} z'(t)=f(t,z(t)),\quad \quad z(t_{0})=x_{0}.\tag{IVP} \label{IVP} \end{equation} Then, $f$ satisfies the Caratheodory conditions Wikipedia: Caratheodory's existence Theorem and we obtain a global solution of the \ref{IVP}. Let us denote this solution by $z_{(t_{0},x_{0})}$ for given initial datum $(t_{0},x_{0})$. Then, the solution satisfies (this is the definition of a Caratheodory solution) the integral equality $$ z_{(t_{0},x_{0})}(t)=x_{0}+\int_{t_{0}}^{t}f(s,z_{(t_{0},x_{0})}(s))\;\mathrm{d} s. \label{IE}\tag{IE} $$
Do we have then some stability results regarding the initial datum? It seems like we have a Lipschitz-continuity, i.e. for $x_{0},x_{1}\in\mathbb{R}$ we have -- using \ref{IE} $$ z_{(t_{0},x_{0})}(t)-z_{(t_{0},x_{1})}(t)=x_{0}-x_{1}+\int_{t_{0}}^{t}f(s,z_{(t_{0},x_{0})}(s))-f(s,z_{(t_{0},x_{1})}(s))\;\mathrm{d} s. $$ By the stated regularity assumption of $f$ in the second component, it looks like we can estimate this to obtain $$ \left|z_{(t_{0},x_{0})}(t)-z_{(t_{0},x_{1})}(t)\right|\leq |x_{0}-x_{1}| +\|\partial_{2}f\|_{L^{\infty}(L^{\infty})}\int_{t_{0}}^{t} \left|z_{(t_{0},x_{0})}(s)-z_{(t_{0},x_{1})}(s)\right|\mathrm{d}s. $$ Applying Gronwall's Lemma yields the Lipschitz-continuity w.r.t the spatial initial datum. In particular, we have $\tfrac{\partial}{\partial x_{0}} z_{(t_{0},x_{0})}(t)$ exists and is essentially bounded (Rademacher).
Now my question: For $f\in C^{1}$ this derivative also satisfies a solution formula -- dependent on the solution $z_{(t_{0},x_{0})}$. Can we have the same in this weaker setting? Since we now know that the derivative exists a.e., can we perform the following computation: Differentiating both sides in \ref{IE}, we have $$ \tfrac{\partial}{\partial x_{0}} z_{(t_{0},x_{0})}(t)=1+\int_{t_{0}}^{t}\partial_{2}f(s,z_{(t_{0},x_{0})}(s))\tfrac{\partial}{\partial x_{0}} z_{(t_{0},x_{0})}(s)\,\mathrm{d} s. $$ Clearly, this is a linear integral equality and we can solve it, but what puzzles me is why the term on the right hand side should even be well-defined. $\partial_{2}f$ is only $L^{\infty}$, thus the composition of $\partial_{2}f$ with $\tfrac{\partial}{\partial x_{0}} z_{(t_{0},x_{0})}$ does not seem to be well defined (measurability).
Is this formula for $\tfrac{\partial}{\partial x_{0}} z_{(t_{0},x_{0})}$ not existent or am I missing a crucial point about measurability and integration?
Thanks in advance to everyone for reading this long question! Alex
I believe that the standard theory gives answers to your questions. In particular, look at Remark 18.4.16 (et cetera) of Kurzweil, Ordinary Differential Equations Introduction to the Theory of Ordinary Differential Equations in the Real Domain.