The Cauchy problem for Laplace equation in unit cube Here is the question:
We are given a laplace equation $\Delta u=0$ in $Q:=(0,1)\times(0,1)$. 
Q1: What are some conditions you can put on this to get uniqueness/existence?
Q2:What if you wanted to prescribe $u(x,0)=f(x)$ and $u_y(x,0)=g(x)$ on $[0,1]\times\{0\}$? What about the existence and uniqueness?
For Q1 I simply put $u(x,y)=f(x,y)$ on the boundary of $Q$, and we require that $f$ is continuous on $\partial Q$. Hence by Perron method, we have both existence and uniqueness of $u$.
But I stuck on Q2. This one looks to me is a Cauchy problem, or part of it... But anyway, here is what I tried so far.
For simplicity, I assume $f$ and $g$ are all $C^\infty$ and I define a new function $\bar{g}$ such that $g=\bar g$ on $[0,1]\times \{0\}$ and 
$$ \int_{\partial Q}\bar g \,d\mathcal H^1=0 $$ 
so that at least I match the requirement for $u$ being the solution of $\Delta u=0$ in $Q$.
My next job is to define $\bar f$ on $\partial Q$ such that $f=\bar f$ on $[0,1]\times \{0\}$ and hence we could use, maybe Perron again, to solve this equation. However, once $\bar f$ is fixed, then $u$ will be determent completely by $\bar f$ and hence I can not control what is $u_y(x,0)$, it will be determent by $f$ but not $g$...
I guess may be I should build some special $\bar f$ so that $\bar f$ is somehow related to $g$? I am not sure... Please help me here!
Also, what about uniqueness? In my opinion, if we have a solution, then this solution has to be unique. This is the fact that if $u=\partial_\nu u =0$ on the smooth part of boundary then $u\equiv0$. Actually, this is an exercise one can find in Gilbert and Trudinger.
 A: Uniqueness
To show uniqueness holds in Q2, it suffices to prove that only the zero function matches the boundary condition $u=u_y=0$. A harmonic function can be reflected across a line segment where it vanishes, by $u(x,-y)=-u(x,y)$. Indeed, this preserves the local mean value property both on the boundary and away from it. Now we have a harmonic function in $(0,1)\times (-1,1)$ such that both $u$ and $u_y$ vanish on $(0,1)\times \{1\}$; hence $u_x=0$ too. A quick way to finish the proof is to invoke complex analysis: $u_x-iu_y$ is holomorphic and vanishes on a line segment, hence is identically zero.
(There ought to be a proof that works in higher dimensions, but I can't think of it now.)
Existence
To show existence fails in Q2, you can also use the reflection argument from above. If there is a solution with $f\equiv 0$, then $g$ is a restriction of a harmonic function to a line segment inside of its domain, hence $g$ must be real-analytic. So, existence fails for any function space that allows non-analytic functions.
Another approach is to consider a smaller rectangle $[0,1]\times [0,1/2]$ and solve the Dirichlet problem there with non-smooth data on the top boundary. Then take $f(x)=u(x,0)$ and $g(x)=u_y(x,0)$.  If there was a solution $v$ with this data, then by uniqueness $u\equiv v$. But then $u$ has a smooth extension beyond $[0,1]\times [0,1/2]$, a contradiction.
