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?


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?

  • $\begingroup$ Thanks a lot for your answer. I have a problem in understanding one step... I agree with the inequality you wrote but I don't understand how to apply the Gronwell inequality... To apply the Gronwell inequality I should have the inequality you wrote in the opposite direction... Am I wrong? $\endgroup$ – user268193 Dec 7 '17 at 11:25
  • $\begingroup$ If I change sign in both sides of the inequality I get $-U(t)\leq C\int_0^t-U(s)ds$ and then $-U(t)\leq e^{Ct}$,but this shouldn't help me... Or not? $\endgroup$ – user268193 Dec 7 '17 at 11:42
  • $\begingroup$ @user268193 $-U(t)\le [-U(0)]e^{Ct}$, not just $e^{Ct}$, and $U(0)=0$. Looks like you are quite proficient in real analysis and, probably, in PDE as well, but have never used Gronwall before. I naturally wonder where they teach students like that. $\endgroup$ – fedja Dec 7 '17 at 15:27
  • $\begingroup$ Ok I am sorry, I understand my mistake now. Thank you! I have another question/curiosity, since I see that there are also versions of Gronwall's inequality that do not need the continuity of the function, could I repeat all the procedure if I had the functions $u_0(\cdot, \cdot)$ and $u_1(\cdot, \cdot)$ in $L^\infty([0, \tilde t]\times [0,1])$? In other words my question is, can I relax the hypothesis over the continuity of $u_0$ and $u_1$ and ask for them to be just bounded in the infinity norm? $\endgroup$ – user268193 Dec 7 '17 at 16:19
  • $\begingroup$ In that case, instead of the $\min_r u_1(t, r)$ I would take the $\inf_r u_1(t, r)$. $\endgroup$ – user268193 Dec 7 '17 at 16:23

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