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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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What kind of equation? Algebraic? Differential? Why is there a tag of [probability]? – Willie Wong Jan 30 '11 at 11:03
Exemple Let u'+Au =f .................. (E) u'+Au+pBu=f ............... (Ep) (E) and (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). – user6407 Jan 30 '11 at 20:12
Dear Hafid: additional information should be posted as a comment or edited into the question statement. – Willie Wong Jan 30 '11 at 20:13

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:

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

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


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

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Nice and simple example. In physics, one would say that the perturbed system might break the symmetry of the unperturbed system, thus leading to a (possibly) unique solution, while the unperturbed system has many more solutions. I wonder however if finding the solution of the perturbed system, then taking the limit to zero perturbation and applying the symmetry group of the original unperturbed equation would not give us all the solutions of the unperturbed problem. It works with your example: the original equation is invariant for scaling the variable. – Raskolnikov Jan 30 '11 at 14:35
@Raskolnikov: I don't think so. Consider $0 < a_1 < a_2 < \ldots < a_k$. Let $P(x) = (x-a_1)\cdot (x-a_2)^2(x-a_3)^2\cdots(x-a_k)^2$. It has $k$ distinct roots. But the perturbed equation $P(x) + p^2 = 0$ only has s unique solution. The solution $s_p$ converges to $a_1$. Your statement is of course true in certain more restricted setting, where the degeneracy can be attributed to symmetries. But in general I think there's way too much freedom for symmetries to solve everything. – Willie Wong Jan 30 '11 at 15:52
Yeah, I think you're right. What do you mean by "s" solutions though? – Raskolnikov Jan 30 '11 at 16:20
@Raskolnikov: $s_p$ is the unique root to $P(x) + p^2 = 0$ above. (s for solution, p is the parameter...) Sorry for the confusion. Or do you mean in the sentence "$P(x) + p^2 = 0$ only has s unique solution"? That's a typo, I hit the 's' key instead of 'a' key. :p – Willie Wong Jan 30 '11 at 16:51
It's the latter one I meant. Thanks for clearing that up. – Raskolnikov Jan 30 '11 at 17:21

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