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PDE in H^-1

  1. Jan 8, 2005 #1
    Hello,
    How can i proof the existance of a solution of a PDE on H^(-1)( Omega)?
    :mad:
     
  2. jcsd
  3. Aug 22, 2009 #2

    Reb

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    You mean (...must have meant) that, given the pde (P) [tex]\texttt{L}u=f[/tex] on an appropriate space, the right hand side belongs to [tex]H^{-1}[/tex]. The most appropriate space for the solutions is, in this case, [tex]L^{2}[/tex].

    The reason for requiring [tex]f\in H^{-1}[/tex] is that now the problem (P) can be put into variational formulation, and then the methods of functional analysis can be applied: Say, for linear problems, the Lax-Milgram Theorem. Or, for nonlinear parabolic problems, monotonicity methods.
     
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