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I'm working in computational magnetohydrodynamics and I'm interested in expressing the initial conditions for the magnetic field, $\mathbf{B}$ in term of the vector potential $\mathbf{A}$ through the well know relationship \begin{equation} \mathbf{B} = \nabla \times\mathbf{A} \,. \end{equation}

For certain reasons (that are not relevant here), I wish to specify $\mathbf{A} = [A_x, A_y, A_z]^T$ in a unique way; that is, I want to remove any gauge freedom associated with the gauge transformation \begin{equation} \mathbf{A}' = \mathbf{A}+ \nabla f \,. \end{equation}

In particular, I'm interested in the axial gauge, which fixes one of the components of $\mathbf{A}$ to zero. Here let us choose the $x$ component, so that $A_x = 0$. It is well known that this gauge condition leaves some residual gauge freedom (see this answer for more details).

My question is: Does specifying a boundary condition for $\mathbf{A}$ alongside the axial gauge fully remove gauge freedom without overly restricting $\mathbf{A}$? For example, take $\mathbf{A} (-L_x,y,z) = [0,0,0]^T$, in a Cartesian domain with boundaries at $-L_x$, $-L_y$, $-L_z$ and $L_x$, $L_y$, $L_z$. If not, is it sufficient to provide further boundary conditions, say at $(x,-L_y,z)$ and/or $(x,y,-L_z)$ ?

AnthonyPadua
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Yes, you can see this by solving explicitly for the potential: $$ B_x = \partial_yA_z-\partial_yA_z \quad B_y = -\partial_xA_z \quad B_z = \partial_xA_y $$ which gives: $$ A_y(x,y,z) = A_y(x_0,y,z) +\int_{x_0}^xB_z(\xi,y,z)d\xi \quad A_z(x,y,z) = A_z(x_0,y,z)-\int_{x_0}^x B_y(\xi,y,z)d\xi $$ Therefore, to entirely specify $A$ in the axial gauge, you just need to specify $A_y,A_z$ on a slice $x=x_0$.

LPZ
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