Existence, not uniqueness of variational problem

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Let $(H,langle cdot , cdot rangle)$ a Hilbert space, $a:Htimes H rightarrow mathbb R$ a bounded coercive bilineal form and $F:Hrightarrow mathbb R$ linear and bounded. It is well-known that the Lax-Milgram theorem assures that there exists an unique $u in H$ such that
$$
a(u,v) = F(v) quad forall v in H.
$$
I was wondering, can we drop any hypothesis of the Lax-Milgram theorem just to guarantee that there exists a solution but it is not necessarily unique?







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    Let $(H,langle cdot , cdot rangle)$ a Hilbert space, $a:Htimes H rightarrow mathbb R$ a bounded coercive bilineal form and $F:Hrightarrow mathbb R$ linear and bounded. It is well-known that the Lax-Milgram theorem assures that there exists an unique $u in H$ such that
    $$
    a(u,v) = F(v) quad forall v in H.
    $$
    I was wondering, can we drop any hypothesis of the Lax-Milgram theorem just to guarantee that there exists a solution but it is not necessarily unique?







    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $(H,langle cdot , cdot rangle)$ a Hilbert space, $a:Htimes H rightarrow mathbb R$ a bounded coercive bilineal form and $F:Hrightarrow mathbb R$ linear and bounded. It is well-known that the Lax-Milgram theorem assures that there exists an unique $u in H$ such that
      $$
      a(u,v) = F(v) quad forall v in H.
      $$
      I was wondering, can we drop any hypothesis of the Lax-Milgram theorem just to guarantee that there exists a solution but it is not necessarily unique?







      share|cite|improve this question











      Let $(H,langle cdot , cdot rangle)$ a Hilbert space, $a:Htimes H rightarrow mathbb R$ a bounded coercive bilineal form and $F:Hrightarrow mathbb R$ linear and bounded. It is well-known that the Lax-Milgram theorem assures that there exists an unique $u in H$ such that
      $$
      a(u,v) = F(v) quad forall v in H.
      $$
      I was wondering, can we drop any hypothesis of the Lax-Milgram theorem just to guarantee that there exists a solution but it is not necessarily unique?









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      share|cite|improve this question




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      asked Jul 20 at 18:37









      Gonzalo Benavides

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