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If we have

  1. An equation $E$ for which the conditions for the existence of a solution are satisfied but we can't prove the uniqueness of the solutions.
  2. A perturbed equation $E_p$ of $E$ which the existence and the uniqueness of solution are satisfied.
  3. The solutions $S_p$ of $E_p$ converge strongly to the solutions $S$ of $E$.

Question: Do (2) and (3) implies the uniqueness of solution in (1)?

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    Dear Hafid: additional information should be posted as a comment or edited into the question statement.2011-01-30

1 Answers 1

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I don't know the exact context, but if I understood the problem correctly I'd say that this is clearly false. As a counterexample consider the equation $E$:

$x = x$

and the perturbed equation $E_p$ for $p > 0$:

$(1+p)x = x+p$

$E_p$ of course has a unique solution that for $p \rightarrow 0$ converges to 1; actually it's constantly 1 for every $p \ne 0$, so no matter what is your definition of "strong" this convergence is strong:

$(1+p)x-x = p$

$x(1+p-1) = p$

$x=p/p=1$

Now $x=1$ is of course also a solution of $E$... but I think $E$ has multiple solutions :-)

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    Ex$a$mple Let u'+Au =f .................. (E) u'+Au+pBu=f ............... (Ep) (E) $a$nd (Ep) are two PDE or ODE, A and B are two differential operators, p is a parameter. When p goes to zero, we get E and the solution of (Ep) converges to the solution of (E).2011-01-30