# Second derivative in coordinate invariant form

To solve stationary, incompressible, inviscid and irrotational flow around a circular cylinder, I am using general coordinates. Since the flow is symmetrical, we only consider the upper half of the plane.

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.

In general coordinates, the expression for $\Delta \varphi$ can be expressed as (using the Einstein summation convention) \begin{align*} \Delta \phi = \frac{1}{\sqrt{g}}\frac{\partial}{\partial \xi^\alpha}\left(\sqrt{g}g^{\alpha\beta}\frac{\partial\varphi}{\partial\xi^\beta}\right), \end{align*} where $\xi(\mathbf{x})$ is the (inverse) coordinate mapping and $g$ is the (contravariant) metric tensor.

On the cylinder itself, the BC is given by $$\nabla\phi\cdot\mathbf{n} = -U_\infty n_1,$$ where $U_\infty$ is is the free stream velocity in the $x_1$ direction and $\mathbf{n}$ is the normal on the surface of the cylinder. In general coordinates, this BC becomes $$\frac{\partial \varphi}{\partial \xi^2} = -U_\infty (\mathbf{a}^{(2)})_1,$$ where $\mathbf{a^{(\alpha)}}$ is the contravariant basis vector.

The other boundary condition (horizontal left and right of the cylinder) are given by (see derivation below) $$\frac{\partial^2 \varphi}{\partial x_2^2} = 0 \ \text{ and } \ \frac{\partial^2 \varphi}{\partial x_1\partial x_2} = 0$$ and I would like to also write this is in coordinate invariant form (in terms of the covariant/contravariant basis vectors and the metric tensor), but I have no idea where to start. Any suggestions would be helpful.

EDIT The derivation of the symmetric boundary condition is as follows. It is given that the velocity field $\mathbf{u}$ can be expressed as $u_i = \partial_i{\Phi}$, where $\Phi$ is the velocity potential function given by $$\Phi = U_\infty x_1 + \phi.$$

I thought that on the horizontal boundary

$$\frac{\partial\mathbf{u}}{\partial x_2} = \mathbf{0}$$

Leading to the two boundary conditions given above.

• The horizontal boundaries are symmetric boundaries, so should it not be $\frac{\partial\phi}{\partial x_2} = 0$ on these two boundaries ? Feb 21, 2018 at 6:34
• @PraveenChandrashekar That would make things a lot simpler. I added the initial derivation to my post. Feb 21, 2018 at 8:13
• If $u=(u_1,u_2)$ is the velocity, then on the symmetry boundary $u_2 = 0$. This gives $\frac{\partial \phi}{\partial x_2} = 0$ which is enough to solve your problem. What you are trying to specify is too much information which cannot be used as a boundary condition for Laplace equation. Once you solve the problem, you can a posteriori verify if those conditions are satisfied upto some numerical errors. Feb 21, 2018 at 9:05

IMPORTANT

To keep things clear... the Laplacian operator $\triangle$ must come with ONLY one BC in each point of the boundary, this is, you can not specify 2 BC as you propose on the boundary. To se this.. I ask you sth. How many BC you need in the boundary to solve this unidimensional BVP? $$\frac{d^2 u}{d x^2}=0,\quad x\in[x_1,x_2]$$

I think the BC you are looking for this boundary is the following: $$\vec{n}^T\sqrt{g^{\alpha\beta}}\left[\vec{grad}(\phi)\right]=0$$ Where $g^{\alpha\beta}$ is the contravariant metric tensor, $\vec{n}$ is the exterior normal vector to upper and lower boundaries and the gradient operator is expressed in the new coordinates.

EDIT Since in a symmetry plane no flow crosses this boundary we have that the BC is:

$$\vec{n}^T\vec{u}=0$$ Therefore, since $\vec{u}=\vec{grad}\,\phi$ you have the required zero normal gradient BC $$\vec{n}^T\vec{grad}\,\phi=0$$

The symmetry condition enforces a Neumann Boundary condition on the velocity field and reads $$n\cdot\nabla u=0$$. These are actually two equations. In any coordinate system this is done by the contraction of the covariant derivative of the velocity field with the normal vector of the boundary: $$n^i \nabla_i u^k=n^i \left(\frac{\partial u^k}{\partial \xi^i}+\Gamma^k_{il} u^l\right)=0$$. In cartesian coordinates where the coordinate lines align with the boundary, they are exactly the stated equations.

Since the velocity field $u^\alpha=g^{\alpha\beta}\frac{\partial\Phi}{\partial \xi^\beta}$ you have $$n^\gamma \nabla_\gamma g^{\alpha\beta}\frac{\partial\Phi}{\partial \xi^\beta}=n^\gamma g^{\alpha\beta}\left(\frac{\partial^2 \Phi}{\partial \xi^\beta\partial\xi^\gamma}-\Gamma^l_{\gamma \beta} \frac{\partial\Phi}{\partial \xi^l}\right)=0$$. These are two equations one for each part of the curved coordinate system with respect to the normal condition.

Further, since $\Phi=U_\infty x_1(\xi_1,\xi_2)+\phi$ you get for the derivatives of $\Phi$:$$\frac{\partial\Phi}{\partial\xi^\alpha}=U_\infty \frac{\partial x_1}{\partial \xi^\alpha}+\frac{\partial \phi}{\partial \xi^\alpha}$$.

So finally for $\phi$ this reads: $$n^\gamma g^{\alpha\beta}\left(\frac{\partial }{\partial \xi^\gamma}\left(U_\infty \frac{\partial x_1}{\partial \xi^\beta}+\frac{\partial \phi}{\partial \xi^\beta}\right)-\Gamma^l_{\gamma \beta} \left(U_\infty \frac{\partial x_1}{\partial \xi^l}+\frac{\partial \phi}{\partial \xi^l}\right)\right)=0$$.

Check this with cartesian coordinates, $x_1$ is independent of $x_2$ so these derivatives vanish. Christoffel symbols are all 0. Metric is diagonal 1, $n^1=0$, $n^2=1$: $$n^\gamma g^{\alpha\beta}\left(\frac{\partial }{\partial \xi^\gamma}\left(\frac{\partial \phi}{\partial \xi^\beta}\right)\right)=0$$ so for each component:$$n^2 g^{1\beta}\left(\frac{\partial }{\partial x^2}\left(\frac{\partial \phi}{\partial x^\beta}\right)\right)=\left(\frac{\partial }{\partial x^2}\left(\frac{\partial \phi}{\partial x^1}\right)\right)=0\\ n^2 g^{2\beta}\left(\frac{\partial }{\partial x^2}\left(\frac{\partial \phi}{\partial x^\beta}\right)\right)=\left(\frac{\partial }{\partial x^2}\left(\frac{\partial \phi}{\partial x^2}\right)\right)=0$$.

• The BC you gave is not correct for potential flow. The correct one should be: $\vec{n}^T\vec{v}=0$. You cannot impose more than one condition for a velocity that comes from a potential. In this case,the symmetry plane does not impose the zero normal gradient of the velocity but the normal zero flow. And also you cannot put these type of boundary conditions you proposed for the Laplace equation...
– HBR
Feb 21, 2018 at 21:44
• The wanted condition is symmetry on the flow field. If the flow field comes from a potential as in this case both conditions collapse to one, exactly to $\frac{\partial \phi}{\partial x^2}=const$. This is your gauge freedom and usually set to 0. So there is no contradiction.
– Bort
Feb 23, 2018 at 9:00