1
$\begingroup$

Consider $P$ a linear partial differential operator in $\Bbb R ^n$. Consider some boundary condition given in the generic form $C(u) = 0$, that guarantees a unique solution (if any) of $Pu = 0$.

Let $(u_i)_{i \in I}$ be a net of distributions such that $\lim \limits _{i \in I} P u_i = 0$ and $\lim \limits _{i \in I} C(u_i) = 0$. Is it possible, then, to conclude that there exist a distribution $u$ such that $\lim \limits _{i \in I} u_i = u$, $Pu = 0$ and $C(u) = 0$?

Playing with words, if I have a net of approximate solutions of a PDE, are its elements approximations of a solution of that PDE? That is, can I use them to construct a solution?

(I have chosen to work with distributions since they seem the most convenient, feel free to modify this context suiting your needs. If stronger hypotheses are needed, please say so in comments below and I shall edit my question to reflect this.)

Improve this question
$\endgroup$
11
  • 1
    $\begingroup$ How is this different from asking whether $f(x_i)$ converges to $f(x)$ when $x_i$ converges to $x$ on a general topological space? Obviously they agree when $f$ is continuous. $\endgroup$
    – Igor Khavkine
    Aug 22, 2015 at 10:58
  • $\begingroup$ I can't imagine a situation in which one naturally has a net of a solution of a PDE which is not a sequence. But maybe that's only my limited imagination… $\endgroup$
    – Dirk
    Aug 22, 2015 at 11:00
  • $\begingroup$ @IgorKhavkine I guess the difference is that the OP does not assume that $x_i$ converges to $x$ but only that $f(x)$ converges to zero and wants to conclude the convergence of the $x_i$. $\endgroup$
    – Dirk
    Aug 22, 2015 at 11:02
  • 2
    $\begingroup$ @Dirk, if that is the case, then there are obvious counter examples like $P=d/dx$ and $u_i = i$ (constant functions) with $i \in \mathbb{N}$. $\endgroup$
    – Igor Khavkine
    Aug 22, 2015 at 11:11
  • 1
    $\begingroup$ @IgorKhavkine This counterexample does not work with unique solutions (as requested by the OP or the edited question with explicit boundary conditions). $\endgroup$
    – Dirk
    Aug 22, 2015 at 12:30

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.