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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?

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  • $\begingroup$ In $J*u_1$, is the convolution in $r$? Is it $(-J* u_1) u_0$ or is it $-J * (u_1u_0)$? $\endgroup$ – Willie Wong Dec 8 '17 at 2:50
  • $\begingroup$ Also: $J$ is defined on $\mathbb{R}$, but it appears $u_1$ is only defined on $r\in [0,1]$? How do you define the convolution in this case? $\endgroup$ – Willie Wong Dec 8 '17 at 2:52
  • $\begingroup$ It is $(-J*u_1)u_2$ and * is the convolution in $r$. $\endgroup$ – user268193 Dec 8 '17 at 6:39
  • $\begingroup$ The convolution is restricted in the interval $[0,1]$. It means that $ (-J*u_1) (t, r) =\int_0^1-J(r-r') u_1(t, r')dr'$ $\endgroup$ – user268193 Dec 8 '17 at 6:44
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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.

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  • $\begingroup$ Thank you very much for your answer!! I will read carefully all the steps... About the continuity I would like to ask the following question... There is something wrong if I apply the contraction mapping theorem directly in $C_c([0,T]\times[0,1])^2$? $\endgroup$ – user268193 Dec 8 '17 at 14:48
  • $\begingroup$ I think that the map $(F_0, F_1): C_c([0,T]\times[0,1])^2\to C_c([0,T]\times[0,1])^2$ and the set $C_c([0,T]\times[0,1])^2$ is a complete metric space... Consequently, by the contraction mapping theorem, there exists a unique fixed point of $(F_0, F_1)$ in $C_c([0,T]\times[0,1])^2$ which is a local solution but should be continuous because belongs to $C_c([0,T]\times[0,1])$... There is something wrong in this argument? $\endgroup$ – user268193 Dec 8 '17 at 14:52
  • $\begingroup$ @user268193: no, nothing is wrong with your argument. I gave the version above because (a) you seem to be unsure of your proof so I gave a different version to show you that the result is true and (b) the approach I gave easily converts via difference quotients to a proof for higher regularity. Since you wrote in the final paragraph you were thinking about functions in $C^\infty_c$ I assumed you are also interested in the higher regularity case. $\endgroup$ – Willie Wong Dec 8 '17 at 16:26

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