Is this function positive?

Question

Could someone tell me if my argument is correct? Let $\rho_1:[0,1]\to [0,1]$ and $J:\mathbb R\to \mathbb R^+$, I have a system of two coupled PDE's and I proved that its solution $(u_0(t, r), u_1(t, r))$ exists unique in $C([0 ,\tilde t]\times [0,1], \mathbb R)$ and can be written implicitly as \begin{align} (0) \;u_0(t,r)=e^{-\int_0^{t}\int_{0}^1J({r-r'})u_1(s, r')dr'ds}> 0, \end{align} while \begin{align}\label{1} (1)\;u_1(t,r)=e^{-t}\rho_1(r)+\int_0^tds\;e^{-({t-s})}\int_0^1dr'J(r-r')u_1(s,r')u_0(s,r). \end{align}

I would like to prove that $u_0(t, r)>0$ and $u_1(t,r)\geq 0$ for every $(t,r)\in [0, \tilde t]\times [0,1]$. I proved it in the following way

By (0) it is obvious that $u_0({t, r})>0$ for all $({t, r})\in [0, \tilde t]\times [0,1]$. To prove that the same property holds for the function $u_1({\cdot, \cdot})$, define \begin{align} A:=\{r\in[0,1] : \rho_1(r)=0\}\quad B:=\{r\in[0,1] : \rho_1(r)>0\} \end{align} and the time \begin{align}\nonumber t^*:=\inf\{t\in (0, \tilde t]: u_1(t, r^*)\neq 0\text{ for some $r^*\in A$}\text{ or } u_1(t, r^*)=0 \text{ for some $r^*\in B$}\}, \end{align} with the convention that the infimum of the empty set is $\tilde t+1$.

If $t^*>\tilde t$ the proof follows trivially. Indeed assuming $t^*>\tilde t$ we have that for every $s\in (0, \tilde t]$ fixed, $u_1({s, r})=0$ for all $r\in A$ and $u_1({s, r})\neq 0$ for all $r\in B$.

Suppose by contradiction that there exists $\bar r\in B$ such that $u_1({s, \bar r})<0$. Since $\bar r\in B$ we have that $u_1({0, \bar r})>0$; the continuity of the function $u_1({\cdot, \cdot})$ in the first variable and the intermediate values theorem allow to conclude that there exists $s^*\in (0,s)$ such that $u_1({s^*, \bar r})=0$. It follows that $t^*\leq s^*<s\leq \tilde t$ and this contradicts the assumption $t^*> \tilde t$.

Consequently, when $t^*>\tilde t$, we can conclude that $u_1(t, r)\geq 0$ for every $(t,r)\in [0, \tilde t]\times [0,1]$.

Suppose $t^*\leq \tilde t$. We have two possibilities: \begin{align} (a)\;\exists r^*\in A: u_1(t^*, r^*)\neq 0,\qquad (b)\;\exists r^*\in B: u_1(t^*, r^*)=0. \end{align} Suppose by contradiction that (b) holds, then $u_1(t^*, r^*)=0$ and $u_1(0, r^*)>0$. By evaluating (1) in $(t^*, r^*)$ we get a contradiction.

If (a) holds we have that $u_1(t^*, r^*)\neq 0$, $\rho_1(r^*)=0$ and $u_1(t, r)\geq 0$ for every $(s, r)\in (0, t^*)\times [0,1]$.

By (1) we get that $u_1(t^*, r^*)>0$ and consequently we can conclude that $u_1({t, r})\geq 0$ for every $({t, r})\in [0,t^*]\times[0,1]$. Iterating the same procedure in the interval $[t^*,\tilde t]$ it is possible to show that $u_1(t,r)\geq 0$ for every $(t,r) \in [0,\tilde t]\times [0,1]$.

Is that correct?

Answer 1

You can, probably, do it this way, of course (I haven't checked your argument for minor errors), but since $u_1$ is continuous, you can just define $U(t)=\min(0,\min_{r\in[0,1]} u_1(t,r))\le 0$ and, ignoring pretty much everything, write $$ U(t)\ge C\int_0^t U(s)\,ds, \text{ where }C=\|u_0\|_{C([0,\widetilde t]\times[0,1])}\|J\|_{L^1([-1,1])} $$ But then the classical Gronwall lemma that you can find in any textbook finishes the story in no time. So why to reinvent the wheel without the need to negotiate a rough terrain?