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Uniqueness for ode coming from parabolic pde

  1. Apr 20, 2009 #1
    Hey all,
    I was working a little on parabolic pde, and came across this (comes up in regularity theory). Consider a Hilbert triple [itex]V\subset H\subset V^*[/itex] (continuous embeddings) and a linear operator [itex]A(t)[/itex] from V to V*, where t ranges in some interval [0,T]. Now let [itex]w\in H^1(0,T;V^*)\cap L^2(0,T;V)[/itex] solve

    [tex]w'=A(t)w-\int_0^t A'(\tau)w(\tau) d\tau, \quad w(0)=0 [/tex].

    I want to show that this implies w=0. How could I do that?

    I tried multiplying by w and integrating by parts, which results in

    [tex] 1/2 (w,w) +\int_0^t a(s,w(s),w(s))ds= -\int_0^t (\int_0^s A(\tau)w(\tau d\tau,w(s))ds,[/tex]

    where a(s,w(s),w(s)) is the induced quadratic form satisfying [itex] a(s,w(s),w(s))\geq \alpha ¦w(s)¦_V-\beta ¦w(s)¦_H[/itex] for constants >0, uniformly in t. How does this help me?

    Best regards...cliowa
     
  2. jcsd
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