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Consider a domain $\mathcal{D}$ which is a right circular cylinder in $\mathbb{R}^3$, with radius 1 and height 1, say. The boundary of $\mathcal{D}$, which I denote by $\partial\mathcal{D}$ consists of two parts:

(a) The unit disk on the plane $z=0$, which I denote by $\mathcal{F}$.

(b) The unit disk on the plane $z=-1$, together with the vertical sides $x^2+y^2=1, z\in(-1,0)$, which I denote by $\mathcal{B}$.

My problem involves establishing existence of minimisers of the Dirichlet energy $J(\Phi)=\int_\mathcal{D} |\nabla\Phi|^2\, dV$ over $H^1(\mathcal{D};\mathbb{R})$, with the constraint $\int_\mathcal{F} \Phi\, dA=0$. The plan is to approach it using the Direct Method of Calculus of Variations, but I have few questions at the moment:

(1) Is the domain $\mathcal{D}$ a Lipschitz domain? I don't really have a good geometric interpretaton of Lipschitz domain, but I suspect the answer is no, since a Lipschitz domain must have zero Lebesgue measure on its boundary but the Lebesgue measure of $\mathcal{F}$ is nonzero.

(2) Is there any variant of Poincare inequality for both Lipschitz/non-Lipschitz domain that also taking into account that the mean of $\Phi$ over part of the boundary, in this case is $\mathcal{F}$ (instead of the usual full domain $\mathcal{D}$) is 0? The reason is that one wants to show that $J(\Phi)$ is coercive over $H^1(\mathcal{D};\mathbb{R})$, and for Dirichlet energy the common way is to estimate it using Poincare inequality; but here I am not working with $H_0^1(\mathcal{D};\mathbb{R})$, and $\Phi$ does not have zero mean over $\mathcal{D}$.

(3) How do I deal with the integral constraint $\int_\mathcal{F} \Phi\, dA=0$. The single way I know is to show that the integral constraint is weakly continuous with respect to the $H^1(\mathcal{D};\mathbb{R})$ norm, but here my integral constraint is defined over $\mathcal{F}$. One sufficient condition (I think) is to require the trace embedding to be compact, but I suspect that claim depends strongly on the domain (whether or not it's Lipschitz). Also, how can we define a trace embedding onto parts of the boundary which has positive measure? I don't even know if this is doable, I have always thought the definition of trace only makes sense because of zero Lebesgue measure of the boundary.

(4) If there exists such trace embedding onto $L^2(\mathcal{F})$, then I think the space $Y$, which consists of functions in $H^1(\mathcal{D};\mathbb{R})$ satisfying the given integral constraint, is a Hilbert space. It is well known that the Dirichlet energy is weakly lower semicontinuous over $H^1(\mathcal{D};\mathbb{R})$, but does this result carries through over the space $Y$?

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  • $\begingroup$ Why all this fancy footwork for a problem that has a trivial explicit solution? $\endgroup$ – Michael Renardy Feb 23 '17 at 13:18
  • $\begingroup$ @MichaelRenardy My intention is to learn how to dealt with the integral side constraint that is only defined on parts of the boundary. I agree that (1) and (4) are both redundant questions. The Dirichlet energy functional is simply a toy problem, my actual problem consists of (quadratic) functional of two variables. As I mentioned, the only way I know how to deal with integral constraint is by showing that they are weakly continuous, but I'd be happy to hear your thoughts. $\endgroup$ – Chee Han Feb 23 '17 at 16:08
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Your domain is a Lipschitz (graph) domain. The 3-dimensional (!) Lebesgue measure of $\mathcal{F}$ is indeed zero, whereas $\mathcal{F}$ has positive measure for the boundary measure $\sigma$ (which coincides with the 2-dimensional Hausdorff measure). This should help you to solve your other questions. The second and third are answered in the paper "Coercivity for elliptic operators and positivity of solutions on Lipschitz domains" by Haller-Dintelmann and Rehberg (there even for "weak" Lipschitz domains).

Compactness of the trace operator as you want is not directly stated in the paper, but follows from one of the cited sources: You need (as one way to do it) Corollary 2 in Maz'yas "Sobolev spaces", Ch. 1.4.7, which essentially says that the real interpolation space $(L^2(\Omega),H^1(\Omega))_{\tau,1}$ is continously embedded into $L^2(\overline\Omega,d\sigma) \stackrel{.}{=} L^2(\partial\Omega)$. Interpolation theory then tells you that in fact $H^1(\Omega) \hookrightarrow (L^2(\Omega),H^1(\Omega))_{\tau,1}$ is a compact embedding, because $H^1(\Omega) \hookrightarrow L^2(\Omega)$ is so, making $H^1(\Omega) \hookrightarrow L^2(\partial\Omega)$ compact as well. I think that you should find compactness properties of the trace operator in the common literature about Sobolev spaces, though (as long as Lipschitz domains are considered), so you do probably not need to mess around with interpolation here.

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  • $\begingroup$ I will check out the paper later today, thanks for the reference! Does a similar result holds for $\mathcal{F}$ instead of $\partial\mathcal{D}$ thou? Actually, now that I know the embedding is compact, any weakly convergent sequence in $H^1(\mathcal{D})$ maps to a strongly convergent sequence in $L^2(\partial\mathcal{D})$; does that implies that the sequence is also strongly convergent in $L^2(\mathcal{F})$, where $\mathcal{D}, \mathcal{F}$ are defined as above in my question? $\endgroup$ – Chee Han Sep 26 '16 at 15:50
  • $\begingroup$ Now that I think of it, my subsequent question sounds really dumb and obvious. $\endgroup$ – Chee Han Sep 26 '16 at 19:15

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