0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.