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Derivative of Banach-space valued functions, and weak solutions

  1. Mar 26, 2012 #1
    Hello everyone,

    I'm new to this forum and this is my first post. I hope this is the right place to ask the following.

    I've been reading up a little on PDE theory lately, and the following point has been causing me a bit of confusion.

    In the definition of the weak solution of a parabolic PDE, one considers Banach-space valued functions

    [itex]u\in L^2\left(0,T; H_0^1(\Omega)\right)[/itex]​

    such that

    [itex]u' \in L^2\left(0,T; H^{-1}(\Omega)\right)[/itex]​

    Here [itex]T>0[/itex] and [itex]\Omega\subset\mathbb{R}^n[/itex] ([itex]n\ge 3[/itex]) is bounded open.

    Does [itex]u'[/itex] here refer to the weak derivative of [itex]u[/itex], i.e. the function defined by

    [itex]\int_0^T u'(t)\phi(t) \;\mathrm{d}t := -\int_0^T u(t) \phi'(t) \;\mathrm{d}t \quad \mbox{for all $\phi\in C^\infty_c(0,T)$} [/itex]​

    (Here the integrals are Bochner integrals.)

    If so, how does one make sense of the above expression? Specifically, my gripe is that the left-hand side technically belongs to [itex]H^{-1}(\Omega)[/itex] while the right-hand side belongs to [itex]H_0^1(\Omega)[/itex]. Is one supposed to identify both sides via the self-duality of [itex]H_0^1(\Omega)[/itex] as a Hilbert space?

    In which case, what is the rationale for distinguishing between [itex]L^2\left(0,T; H_0^1(\Omega)\right)[/itex] and [itex]L^2\left(0,T; H^{-1}(\Omega)\right)[/itex] (for example, in the definition of the weak solution to a parabolic PDE) in the first place?
  2. jcsd
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