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.
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.