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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.

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  • $\begingroup$ Please feel free to retags/retitle, I do not know my way around here yet. $\endgroup$ Commented Dec 5, 2013 at 13:22
  • $\begingroup$ Not a full solution, but notice that if $A$ is curl-free over a simply connected region, it must be the gradient of some potential $u$, and that if furthermore $A$ is divergence-free, that $u$ is harmonic. So you are looking for harmonic functions over the ball satisfying some rather strong conditions along the $xy$ plane. $\endgroup$ Commented Dec 5, 2013 at 16:29
  • $\begingroup$ @user7530 I got that, so basically, is know at the plane, and also the derivative of $A_x$, but then? $\endgroup$ Commented Dec 5, 2013 at 20:07
  • $\begingroup$ You know that $u$ is constant (and so WLOG, zero) on the plane, and you also have Neumann conditions on $u$ along the plane. Maybe you can show uniqueness from this. $\endgroup$ Commented Dec 5, 2013 at 20:21
  • $\begingroup$ @user7530 What do you mean with WLOG? And what do you mean with constant for $u$? It can look like $1+x+z+xz$? $\endgroup$ Commented Dec 5, 2013 at 20:31

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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 $w=u-v$

  1. is harmonic;
  2. is constant on the $xy$ plane;
  3. 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.

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  • $\begingroup$ Thank you. I am not familiar with the notation $\partial_x^a$. (I do understand $\partial_x$ is the derivative wrt $x$). Can you please explain? $\endgroup$ Commented Dec 6, 2013 at 16:28
  • $\begingroup$ @Bernhard I just mean $\frac{\partial^a}{\partial x^a}$. $\endgroup$ Commented Dec 6, 2013 at 16:30
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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.

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    $\begingroup$ Thank you. If I use the harmonic function $\psi(x,y)=\exp(x)\sin(y)$, than I find in the $xy$-plane for the $y$ component of $A_y=\exp(x)\cos(y)$, which does not match my first requirement? Am I missing something? $\endgroup$ Commented Dec 6, 2013 at 8:33
  • $\begingroup$ @Bernhard What is your first requirement? As I see you only have three requirement, $\vec{A}$ is divergence free, curl free and $\vec{A}(x,y,0) = (1+ax) \hat{e}_z$. $\endgroup$ Commented Dec 6, 2013 at 22:41
  • $\begingroup$ The last statement in your comment says that $A\cdot x=0$ and $A \cdot y=0$. This was my first equation (or constraint) $\endgroup$ Commented Dec 6, 2013 at 22:42
  • $\begingroup$ @Bernhard Oh Ok. For some reason, I read it as $\vec{A}(x,y,0).\vec{e}_z = (1+ax)$. $\endgroup$ Commented Dec 6, 2013 at 23:29

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