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Let $W(0,T) := \{ u \in L^2(0,T;H^{\frac 12}(\partial\Omega)) \mid u_t \in L^2(0,T;H^{-\frac{1}{2}}(\partial\Omega))\}$. Let $\gamma$ and $\xi$ denote the trace map and its right inverse.

Does there exist a solution $u \in L^2(0,T;H^1(\Omega))$ with $\gamma(u) \in W(0,T)$ such that $$\langle (\gamma u)_t, \gamma(v)\rangle + \int_\Omega \nabla u \nabla v = 0$$ for all $v \in L^2(0,T;H^1(\Omega))$?

Is there some literature about such problems? I particularly am interested in variational/energy/Galerkin methods.

The problem is that the trace mapping is not invertible, so the converting the problem to $$\int_{\partial\Omega} z_t w+ \int_\Omega \nabla \xi(z) \nabla \xi(w) = 0$$ for all $w \in L^2(0,T;H^{\frac 12}(\Omega))$ does not help.

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  • $\begingroup$ Your equation implies that u is harmonic, and with this restriction the trace map is invertible. $\endgroup$
    – Michael Renardy
    Commented Dec 20, 2014 at 2:09
  • $\begingroup$ @MichaelRenardy Right, but then I'd have to consider $\int_{\partial\Omega}z_t \gamma(v) + \int_\Omega \nabla \gamma^{-1}(z)\nabla v=0 \tag{1}$, for all $v \in L^2(0,T;H^1)$, and substiting $w=\gamma(v)$ gives a different weak formulation which is not equivalent, since the test functions are in $H^1(\Omega)$ for a.e. $t$, and the trace is not invertible on that space (inverse is not surjective). Unless some convenient density result holds, or there is some result about mixed abstract parabolic equations of form (1) I am stuck. $\endgroup$
    – DeleMax
    Commented Dec 20, 2014 at 13:11
  • $\begingroup$ Using the fact that u is harmonic, you convert $\int_\Omega \nabla u\nabla v$ to $\int_{\partial\Omega}{\partial u\over\partial n}v.$ You then end up with $z_t=Az$, where $A$ is the Dirichlet-to-Neumann map (well studied in the literature). $\endgroup$
    – Michael Renardy
    Commented Dec 20, 2014 at 16:09

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