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Let $C=\Omega \times (0,\infty)$. We want to find a solution $v \in H^1(C)$ such that given $u \in H^{\frac 12}(\Omega)$, $$\int_0^\infty\int_\Omega \nabla v \nabla \varphi + v_y\varphi_y = 0\quad\forall \eta \in H^1(C), \quad\eta(x,0) \equiv 0$$ $$v(x,0) = u(x)$$ where eg. the $n(x,0) = 0$ means in the sense of trace, so $T\eta = 0$ for $T:H^1(C) \to H^{\frac 12}(\Omega)$. This is a weak solution of the problem: $-\Delta_C v = 0$, $Tv=u$, $\partial_\nu u = 0 $ on $\partial\Omega \times (0,\infty)$ (zero Neumann BC).

Now, if $u \equiv 1$, then $v \equiv 1$ would be solution however $1 \notin L^2(C)$. If instead we ask for $v \in X(C)$, where $X$ is a space that involves only the derivatives and the trace onto $\Omega$ at $y=0$, then $1$ is a solution in that space, and is unique. The standard way to get solutions in $H^1(C)$ is to impose a mean value zero condition on the data $u$.

My problem is I seem to have a proof that there is a solution $v \in H^1(C)$ with $Tv=u$ for $u\equiv 1$. Here is my "proof":


Let $U \in H^1(C)$ be such that $TU = 1$. Then the difference $d=v-U$ solves $$\int_0^\infty\int_\Omega \nabla d \nabla \varphi + d_y\varphi_y = -\int_0^\infty\int_\Omega \nabla U \nabla \varphi + U_y\varphi_y \quad\forall \eta \in H^1(C), \quad\eta(x,0) \equiv 0\tag{1}$$ $$d(x,0) = 0.$$ There is a unique $d \in H^1(C)$ solving this because we can define a functional $J:\{d \in H^1(C) \mid Td = 0 \text{ and } \int_\Omega d(x,y)\;\mathrm{d}x = 0 \text{ a.e. $y$}\} \to \mathbb{R}$ such that $$J(d) = \frac{1}{2}\int\int|\nabla d|^2 + d_y^2 + \int\int \nabla U \nabla d + u_y d_y.$$ This is convex, coercive (by Poincare, due to the mean value zero part of the domain), proper etc. It is then easy to show (1) holds (the minimiser solves the related variational problem for all test functions in the domain of $J$, and then we can remove the mean value condition required on the test function). Now it remains to set $v= d+ U \in H^1(C)$, and this solves the original problem.


So where is the fault in my argument??

Note that the domain of $J$ does not require a $d$ with $\int_\Omega d(x,0) = 0$; which would rule out the initial data.

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closed as off-topic by Michael Renardy, Stefan Kohl, Denis Serre, András Bátkai, Ryan Budney Feb 2 '15 at 6:23

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  • $\begingroup$ It looks like you are missing a boundary condition on $\partial\Omega$. $\endgroup$ – Michael Renardy Jan 31 '15 at 19:12
  • $\begingroup$ Yes, it is zero Neumann BCs. $\endgroup$ – jamesC Jan 31 '15 at 19:37
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    $\begingroup$ Hint 1: You are getting an apparent contradiction even in the degenerate case when $\Omega$ is a zero-dimensional point and $C$ is a half-line (note that there is clearly no $H^1$ solution to $\Delta v = 0, Tv = 1$ in this case). Hint 2: check all of your claims about $J$ carefully. $\endgroup$ – Terry Tao Feb 1 '15 at 2:03
  • $\begingroup$ @TerryTao Thanks for the comment. (On the non-degenerate case): my feeling is the domain of $J$ may not be well-defined. The function $y \mapsto \int_\Omega d(x,y) =: (Md)(y)$ is in fact such that $Md \in H^1(0,\infty)$ for $d \in H^1(C)$, hence $Md \in C^0([0,\infty))$. So if we take the continuous representative then we must have $Md(y) = 0$ for all $y$ including $y=0$. Could you please give another hint to what you meant? $\endgroup$ – jamesC Feb 1 '15 at 18:14
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    $\begingroup$ In the degenerate case, the mean zero condition means that the domain of J consists solely of the zero function. This suggests that one take a closer look at your claim that "one can remove the mean value condition required on the test function". $\endgroup$ – Terry Tao Feb 1 '15 at 20:48