Continuity of the solution of a Pde system Let $\rho_1:[0,1]\to [0,1]$ and $J:\mathbb R\to \mathbb R^+$ both continuous and bounded.
I have the following system of PDE's
\begin{align}
\begin{cases}
\frac{\partial}{\partial t} u_0(t,r)=- J* u_1(t,r) u_0(t,r)\\
\frac{\partial}{\partial t}u_1(t,r)=J*u_1(t,r)u_0(t,r)-u_1(t,r)\\
u_0(0,r)=1-\rho_1(r), u_1(0,r)=\rho_1(r)
\end{cases}
\end{align}
where * denotes the convolution operator.
I would like to prove that there exists unique a local solution of the previous system. I would like to prove that this solution is also continuous.
Is it correct the following argument?
I consider the maps $F_0$ and $F_1$ defined in $L_c^\infty([0, T]\times [0,1])^2$ which contains all the functions bounded by a constant $c$
\begin{align}
F_0(x(t,r),y(t,r))&=1-\rho_1(r)+\int_0^t ds\int_0^1dr'J(r-r') y(s,r') x(s,r)\\
F_1(x(t,r),y(t,r))&=\rho_1(r)+\int_0^t ds\int_0^1dr'J(r-r') y(s,r') x(s,r)-y(s,r)
\end{align}
and I can prove that, when $T$ is small enough, the map $(F_0, F_1)$ is a contraction in $L_c^\infty([0, T]\times [0,1])^2$,.
Then by the contraction mapping theorem I can conclude that there exists a unique fixed point of $(F_0, F_1)$ which is a local solution of the previous PDE's system.
Is that correct?
There is any chance to prove the continuity of my local solution?
My idea is to apply the Contraction mapping theorem in $C_c^\infty([0, T]\times [0,1])^2$ the set of all continuous function bounded by a constant $c$... Is that possible?
 A: Your formula for $F_0$ should have a negative sign in front of the nonlinearity; but that is not particularly important. 
Your argument is correct. Here's how you can get continuity: notationally consider the sequences $x_i(t,r)$ and $y_i(t,r)$ given by $x_0 = y_0 \equiv 0$ and $x_{i+1} = F_0(x_i, y_i)$ and $y_{i+1} = F_1(x_i, y_i)$. 
Since $\rho_1$ is continuous on a compact interval, you have that $\rho_1$ is uniformly continuous. Similarly $J$ is uniformly continuous on $[-1,2]$. You have the identity
$$ x_{i+1}(t,r + h) - x_{i+1}(t, r) = \rho_1(r) - \rho_1(r+h) + \iint J(r - r') y_i(s,r') \big( x_i(s, r) - x_i(s,r + h)\big) + \\
\big( J(r - r') - J(r + h - r') \big) y_i(s,r') x_i(s,r+h) ~\mathrm{d}r' ~\mathrm{d}s $$
From the convergence of $x_i, y_i$, you know that the functions are uniformly bounded by some large constant $M$. Enlarge $M$ if necessary to be greater than $J$. Then you have
$$ \big| x_{i+1}(t, r+h) - x_{i+1}(t,r)\big| \leq |\rho_1(r) - \rho_1(r+h)| + \\
t M \Big( \sup_{r'\in [0,1]} |J(r - r') - J(r + h - r')| + \sup_{s\in [0,t]} |x_i(s,r) - x_i(s, r+h)| \Big) $$ 
Now fix $T$ such that $TM < \frac12$. Given $\delta > 0$, we see that if $\epsilon > 0$ is such that


*

*$|\rho_1(r) - \rho_1(r+h)| < \delta / 100$ whenever $|h| < \epsilon$

*$|J(r) - J(r+h) | < \delta / 100$ whenever $|h| < \epsilon$

*$|x_i(s,r) - x_i(s,r+h)| < \delta$ whenever $|h| < \epsilon$


then necessarily 
$$ |x_{i+1}(t,r+h) - x_{i+1}(t,r)| < \delta$$ 
for every $|h| < \epsilon$. Using that $x_0 = y_0 = 0$ is uniformly continuous on $[0,T]\times[0,1]$ as the base case, we see that by induction (after doing essentially the same thing with also the $y$ terms) that your family $(x_i, y_i)$ has to be uniformly equicontinuous, and hence the limit is uniformly continuous. 
If $J$ and $\rho$ are smooth, then a similar argument gives you the persistence of higher (finite) regularity for short times. 
