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Let $Q = \Omega \times (0,C)$ where $\Omega$ is a bounded domain, write $(x,y) \in Q$ for $x \in \Omega$ and $y \in (0,C)$. Is the problem $$\Delta_{(x,y)}v = 0\quad\text{in $Q$}$$ $$\frac{\partial v(x,y)}{\partial \nu} = 0 \quad \text{on $\partial\Omega \times (0,C)$}$$ $$v(x,0) = u\quad \text{given in $H^{\frac 12}(\Omega)$}$$ $$v(x,C) = 0$$ well-posed is the sense that it has a weak solution $v \in H^1(\Omega \times (0,C))$ (or some homogenous space) with $v(x,0)=u$ and: $$\int_0^C \int_\Omega \nabla_{(x,y)}v\nabla_{(x,y)}\varphi = 0$$ for each $\varphi \in H^1(\Omega \times (0,C))$ such that $\varphi(x,0)=\varphi(x,C)=0$?

I am only interested in weak solutions, not classical ones. I was hoping to apply a Lax-Milgram argument but I don't see what to do with condition $v(x,0) = u$. It cannot be "removed" by using a translation argument due to the imposition $v(x,C) = 0$. We can suppose that $u$ has average integral zero if it helps. Any advice appreciated...

I wanted a nice weak form so I decided to have test functions vanishing at $y=C$. Every book I see regarding elliptic equations with disjoint boundaries seems to have one Dirichlet conditions on one part of the boundary and one Neumann conditions on the other part, but never two different Dirichlet conditions. So I have no idea how to approach this problem.

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  • $\begingroup$ Are you using $\nu$ for two different things? $\endgroup$
    – Deane Yang
    Apr 5, 2015 at 16:22
  • $\begingroup$ hmm, not that I can see. The solution is a $v$ and the unit normal is the $\nu$ @DeaneYang. $\endgroup$ Apr 5, 2015 at 20:36
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    $\begingroup$ Sorry. My eyes are bad. $\endgroup$
    – Deane Yang
    Apr 5, 2015 at 20:59

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This particular boundary value problem for the Laplace operator is called the Zaremba problem. There are several possibilities to solve it. The probably easiest one is to minimize the functional $$ I(f) = \int_Q |\nabla f|^2\,dxdy $$ over the set of functions $f\in H^1(C)$ satisfying $f(\cdot,0)=u$ and $f(\cdot,C)=0$. The (unique) minimum (use the Poincaré inequality) is the (unique) solution $v$.

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