If we have a system of PDE of the form:

$$\begin{cases} \dfrac{\partial y}{\partial t}(t,x)=D\Delta y+F(t,x,f(x),y) ,\ (t,x)\in (0,T)\times\Omega\\ \dfrac{\partial y}{\partial \nu}(t,x)=0,\ (t,x)\in (0,T)\times\partial\Omega \\ y(0,x)=y_{0}(x),\ x\in\Omega\end{cases}$$

with a unique solution $y=y^f$(mild solution). All derivatives are considered in the distributional sense (weak derivatives).Here $\Omega\subset\mathbb{R}^N$ is an open and bounded set with $C^1$ boundary, $y=(y_1,y_2,...,y_m)$, $F=(F_1,F_2,\dots,F_m)$, $D=diag(d_1,d_2,\dots, d_m),\ d_i\in\mathbb{R},\ \forall\ i=1,2,...,m$ and $f:\Omega\to\mathbb{R}$ a function in $L^1(\Omega)$. We consider the following approximation of the system:

$$\begin{cases} \dfrac{\partial y}{\partial t}(t,x)=D\Delta y+F(t,x,f_n(x),y) ,\ (t,x)\in (0,T)\times\Omega\\ \dfrac{\partial y}{\partial \nu}(t,x)=0,\ (t,x)\in (0,T)\times\partial\Omega \\ y(0,x)=y_{0}(x),\ x\in\Omega\end{cases}$$,

with the unique solution $y^{f_n}$.

How can we prove that if $f_n\to f$ in $L^1(\Omega)$, then $y_i^{f_n}\to y_i^f$ in $L^p((0,T)\times\Omega),\ \forall\ i=1,2,\dots,n$, for some $p\in [1,\infty)$? Are there some results in that sense? Maybe with a more general setting...?

Are there some estimates?

$$||y^{f_n}-y^f||\leq C||f_n-f||_{L^1(\Omega)}$$.

  • $\begingroup$ What are conditions on $F$? $\endgroup$ – Andrew Jan 30 at 18:26
  • $\begingroup$ We can assume any type of conditions (Lipschitz-type probably). I only want a method or some references about that type of parabolic problems. $\endgroup$ – Bogdan Jan 30 at 18:47
  • $\begingroup$ Do you have the same question for an equation rather for a system? $\endgroup$ – Giorgio Metafune Jan 31 at 10:26
  • $\begingroup$ It will be welcomed. $\endgroup$ – Bogdan Jan 31 at 10:51

This is only a sketch of an argument that can be used. Assume that $F$ is Lipscthitz and let $y_1,y_2$ be the solutions corresponding to $f_1, f_2$. If $v=y_2-y_1$, then $$|v_t-\Delta v|=|F(f_2,y_2)-F(f_2,y_1)+F(f_2, y_1)-F(f_1,y_1)| \le L(|v|+|f_2-f_1|)$$ with zero bc and initial value. If $T(t)$ is the semigroup generated by $\Delta$ with the appropriate bc, then $$|v(t)| \le L\int_0^t T(t-s)|v(s)|ds +L\int_0^t T(t-s)(|f_2-f_1|)ds.$$ Next use the fact that $T(t)$ is contractive in $L^p$ and maps $L^1$ to $L^p$ with norm less that $Ct^{-N/p'}$. If $t \le \delta$, taking $L^p$ norms (in space) we get $$\|v(t)\|_p \le L\delta \sup_{0 \le t \le \delta}\|v(t)\|_p+CL\|f_2-f_1\|_1\int_0^\delta t^{-N/p'}dt.$$ Now take $p<N/(N-1)$ so that $t^{-N/p'}$ is integrable near $0$ and take the supremum of the left hand side over $[0, \delta]$. If $L\delta \le 1/2$ this gives $$ \sup_{0 \le t \le \delta} \|v(t)\|\le C_1 \|f_2-f_1\|_1.$$ Now the argument can be iterated to larger intervals.

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  • $\begingroup$ Isn't that a common question on PDE's? Continuity of the solution for an approximated problem? I didn't find anything on the literature. Do you know some references? Thank you very much for the answer! $\endgroup$ – Bogdan Feb 2 at 22:58
  • 1
    $\begingroup$ Yes it is, and the use of semigroups, though not strictly necessary makes it clean. Note that in argument I wrote you can use directly Gronwall lemma instead of taking a small delta. If you work in the same Banach space then continuous depends on data is exatly what a semigroup gives. In your situation some more effort is needed to go from 1 to p. Concerning references, I think that a book dealing with applications of semigroups to nonlinear equations gives all the ideas (e.g. Haroux-Cazenave). $\endgroup$ – Giorgio Metafune Feb 3 at 8:06

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