In solving a PDE problem I managed to reduce it to following equation for the unknown functions $\phi(x,y)$ and ${\bf A}(x,y)$:
$\nabla \phi(x,y) = \nabla \times {\bf A}(x,y)$,
defined on a 2D square, with boundary conditions on $\phi(x,y)$ and/or on its derivative normal to the surface. What I am interested in determining is the function $\nabla \phi(x,y)$.
My question is: can I claim that the equation I find implies that
$\nabla \phi(x,y) = 0$ and $\nabla \times {\bf A}(x,y) = 0$?
[I know it implies that $\nabla^2 \phi(x,y) = 0$, which can in principle be solved given be B.C.s, but this seem not to be of much avail. Such equation would need to be solved numerically, with numerically defined and quite noisy boundary conditions.]
Edit: in the equation above only $A_z(x,y)$ matters. Hence, it can be rewritten as
$\nabla \phi(x,y) = - ({\hat {\bf z}} \times \nabla) A_z(x,y)$,
Does this help for anything?
Since the vector field $\textbf{A}$ has only one component, I'll skip the subscript for ease of notation and say $\textbf{A} = A \hat{\textbf{z}}$.
Then
$$ \nabla \times \textbf{A} = \frac{\partial A}{\partial y}\hat{\textbf{x}} - \frac{\partial A}{\partial x} \hat{\textbf{y}} $$
which follows that
\begin{align} \frac{\partial \phi}{\partial x} &= \frac{\partial A}{\partial y} \\ \frac{\partial \phi}{\partial y} &= -\frac{\partial A}{\partial x} \end{align}
Therefore, $\phi$ and $A$ are harmonic conjugates, i.e. there exists a holomorphic, complex-valued function $f: \Bbb C \to \Bbb C $ such that
$$ f(z) = f(x+iy) = \phi(x,y) + iA(x,y) $$
Then, a simple counter-example to your claim would be $f(z) = e^z$, or $\phi(x,y) = e^x\sin y$ and $A(x,y) = e^x\cos y$