How to find other solutions to this vectorproblem? Suppose I have a vector field $\mathbf{A}(x,y,z)$, of which I know:
$$ \mathbf{A}(x,y,0)=(1+\alpha x)\hat{z}$$
Thus, I know the value of $\mathbf{A}$ in the $xy$-plane. Say, within $|x|,|y|\leq\frac{1}{2}$.
Furthermore, I have the following requirements for $\mathbf{A}$.
$$\nabla\cdot\mathbf{A}=0, \\ \nabla\times\mathbf{A}=0,$$
which have to be satisfied in $|x|,|y|,|z|\leq{\frac{1}{2}}$.
I want to find the vector field $\mathbf{A}$ that satisfy all of the above conditions, at least for the given boundaries, but for larger (infinite?), domains as possible.
I did find the following solution, but, with some rather crude assumptions, so I wonder if there are any other approaches to solve the problem.
Assumption 1: There is no $y$-dependency.
Assumption 2: $\displaystyle\frac{d\mathbf{A}_x}{dx}=0$.
Under these assumptions, one can easily obtain from the curl-requirement, that
$$\mathbf{A}=\alpha z\hat{x}+(1+\alpha x)\hat{z}$$
But, is this the only one? I am especially interested in other solutions which do no show $y$-dependency, and, even more interested if there is a solution $\displaystyle\lim_{z\to\infty}\mathbf{A}_x<\infty$. A proof that the solution that I obtained straightforwardly is the only one obviously also counts as an answer.
 A: You solution is unique.
Suppose there existed two vector fields $A=\nabla u$ and $B=\nabla v$ satisfying your conditions. Then $A-B = \nabla(u-v)$ where $w=u-v$


*

*is harmonic;

*is constant on the $xy$ plane;

*satisfies $\partial_z w = 0$ on the $xy$ plane.


Pick a point on the $xy$ plane, e.g. the origin. At that point,
$$\partial_x^a \partial_y^b w = 0$$
for any $a,b$ since $w$ is constant on the plane. Moreover
$$\partial_x^a \partial_y^b \partial_z w = 0$$
by property (3). Since $w$ is harmonic, 
$$\partial_x^a\partial_y^b\partial_z^{c+2} w= -\partial_x^{a+2}\partial_y^b\partial_z^c w - \partial_x^a\partial_y^{b+2}\partial_z^c w$$
and it follows that all partial derivatives of $w$ at the point vanish, and so since harmonic functions are analytic, $w$ is constant.
Therefore $A-B=0$ and $A$ is unique.
A: Note that $\vec{A} = \vec{\nabla} \phi$, where $\phi$ is a harmonic function. In addition, we want $$\vec{A} \cdot \vec{e}_z = (1+ax) + \sum_{k=1}^{\infty} f_k(x,y) z^k \text{ i.e., }
\phi(x,y,z) = f(x,y) + (1+ax)z + \sum_{k=2}^{\infty} g_k(x,y) z^k$$
Here is one possibility that gives rise to a large class of candidates:
Let $$\vec{A} = \vec{\nabla}\left(\psi(x,y) + (1+ax)z \right)$$ where $\psi(x,y)$ is any $2$D harmonic functions.
