Proper boundary conditions for potential flow around cylinder

Question

I am computing the stationary, incompressible, inviscid and irrotational flow around a circular cylinder using a discretization in general coordinates.

I derived a PDE and proper boundary conditions in the physical domain (left image) given by $G := \{(x_1,x_2 : R_1^2 \leq x_1^2 + x_2^2 \leq R_2^2, x_2 \geq 0\}$. This PDE is $\Delta \phi = 0$, where $\phi$ is the perturbation potential.

Then the assignment asks to simplify the Laplacian in general coordinates. For this is calculate the covariant/contravariant basis vectors and the Jacobian. Next I define the mapping $\mathbf{x}(\xi)$.

But I am stuck on deriving proper boundary conditions in the logical domain (right image). These have to be defined in terms of the covariant/contravariant basis vectors and the metric tensor. As a hint, it is given: Do not use the specific mapping at this stage. Use the fact that the boundaries of the domain are of the form $\xi^\alpha(\mathbf{x}) = $ constant, and therefore the unit normal vector on the boundary is given by $\pm \mathbf{a}^{(\alpha)}/|\mathbf{a}^{(\alpha)}|$

I don't see how to use the covariant/contravariant basis vectors. We have a new basis for the physical domain G for a polar coordinate system, so what do they have to do with the computational domain?

The boundary conditions in invariant form:

1. Along the cylinder(flow is inviscid) $0 = \mathbf{u}\cdot\mathbf{n} = -U_\infty n_1$, where $\mathbf{u}$ is the velocity field given by $u_\alpha = \frac{\partial \Phi}{\partial x_\alpha}$ and $\mathbf{\Phi}$ the total velocity potential function given by $\Phi = U_\infty x_1 + \phi$. Now $U_\infty$ is the free stream velocity in the $x_1$ direction and $\phi$ is the perturbation potential.
2. On the horizontal boundary of the physical domain we have $\frac{\partial u}{\partial x_2} = 0$, so $\frac{\partial^2 \phi}{\partial x_1x_2} = 0$ and $\frac{\partial^2 \phi}{\partial x_2^2} = 0$.
3. Along the top of the domain we have a homogeneous Dirichlet boundary condition for the perturbation potential. $\phi(\mathbf{x}) = 0$ for $\mathbf{x}$ on this boundary.

I left out a lot of specifics, because I'm not sure if these are needed.

Answer 1

Be careful with the Neumann BC. $\partial u/\partial x=0$ does no imply that $\partial^2 u/\partial x^2=0$, e.g. $u=x^2$.

You have the BC defined in the physical domain... and you have an expression that relates physical coordinates $x$ with computational coordinates $ξ$ in the computational domain $x=x(ξ)$, or vice-versa. Substitute these in your BC and apply the chain rule. If the map is well defined, you will not have any problem computing its inverse.