Existence and uniqueness of a solution of a sort of ODE Suppose $F:\left[0,1\right] \times C^{1}\left(\left[0,1\right]\right)\rightarrow \mathbb{R}$ be a Lipschitz-Function with Lipschitz-constant $L>0$ so that 
\begin{align}
\left\vert 
F\left(t,u\right) -F\left(t,v\right)\right\vert \leq L \left\| u-v
\right\|_2    \,\,\text{for all}\,\, t\in\left[0,1\right] \,\,\text{ and }\,\,u\left(\cdot\right),v\left(\cdot\right) \in \mathcal{L}_2\left(\left[0,1\right]\right).
\end{align}
I am looking for a theorem, which ensures the existence and uniqueness of a solution of the "kind of" ODE
\begin{align}
     \beta'\left(t\right)&= F\left(t,\beta\left(\cdot\right)\right) \\
\beta\left(0\right) &= 0,
\end{align}
where $\beta\left(\cdot\right) \in C^{1}\left(\left[0,1\right]\right)$ and $\beta^{\left(1\right)}\left(\cdot\right)\in \mathcal{L}_2\left(\left[0,1\right]\right)$. 
Beware that the standard Piccard-Lindelöf theorem doesn't apply here, because $\beta$ on the righthand side of $\beta'\left(t\right)= F\left(t,\beta\left(\cdot\right)\right)$ does not depend on $t$! 
An simple example for such an $F$ is 
\begin{align}
F\left(t,\beta\left(\cdot\right)\right) = \phi\left(t\right) \int_0^1 \beta\left(s\right) f\left(s\right)  d s,
\end{align}
for given functions $f\left(\cdot\right)\in \mathcal{L}_2\left(\left[0,1\right]\right)$ and $\phi\left(\cdot\right) \in C^1 \left(\left[0,1\right]\right)$. 
 A: Here's a comment rather than a solution.
Let us consider your simple example. Let $f \in L^2[0, 1]$ and $\phi \in C[0, 1]$. If $\beta(t)$ is a nontrivial solution to
\begin{align}
\dot \beta(t) = \left(\int^1_0 \beta(s)f(s)\ ds \right) \phi(t), \ \ \beta(0) = 0.
\end{align}
Clearly, $\beta(t) \equiv 0$ is a solution. Suppose $\beta(t)$ is a nontrivial solution, then we see that
\begin{align}
\beta(t) = \left(\int^1_0 \beta(s)f(s)\ ds \right) \int^t_0\phi(s)\ ds =: \alpha(\beta, f) \int^t_0 \phi(s)\ ds 
\end{align}
or
\begin{align}
\int^1_0 \int^t_0 f(t) \phi(s)\ dsdt = 1.
\end{align}
Now, let us choose $f\equiv 1$ and $\phi \equiv 2$, then we see that the condition is met. Hence we now have
\begin{align}
\dot \beta(t) = 2\left(\int^1_0 \beta(s)\ ds \right), \ \ \beta(0) =0.
\end{align}
Again, clearly $0$ is a solution. However, $y(t) =\lambda t$  is also a solution for any $\lambda$ since
\begin{align}
2\int^1_0 y(s)\ ds=2\int^1_0\lambda s\  ds = \lambda  = \dot y(t).
\end{align}
In short, the equation imposes a necessary condition ( a consistency condition), but it's not enough to guarantee wellposedness (lack uniqueness in the above example). 
Note: I have chosen $f, \phi$ to be very nice functions. In fact, you could also consider $\phi \equiv 1$ and $f(t) = 3t$...etc. 
Additional 
Let me answer the general question about existence for the simple example. Consider
\begin{align}
\beta(t) = \lambda \int^t_0\phi(s)\ ds
\end{align}
then we see that
\begin{align}
\dot \beta = \lambda\phi(t) \ \ \text{ and } \ \ \int^1_0 f(t) \lambda\int^t_0\phi(s)\ dsdt = \lambda \cdot 1.
\end{align}
Hence 
\begin{align}
\beta(t) = \lambda \int^t_0 \phi(s)\ ds 
\end{align}
solves the simple example with initial condition $\beta(0) = 0$. 
