Given a uniform polar grid, as in the figure below: enter image description here

and a FV discretization of a gradient for example:

$\frac{\partial p}{\partial \varphi} = 0$

$\Delta r \frac{p_e - p_w}{\Delta \varphi} = 0$

My Question revolves around the value of $\Delta \varphi$

As the Grid is uniform, $\Delta r$ does not change across the grid. But what about $\Delta \varphi$? From an angle point of view, it does not change either.

$\Delta \varphi = 2\pi / NCells = $ constant

From an arc point of view, the length between two cells will change for different radii.

$arc = \Delta \varphi \cdot r_i$

Which one is correct? Textbooks rarely concern themselves with anything other than Cartesian grids or jump directly into more complex ones.

Edit 1:

The actual PDE is more complicated. I compared the results of the simulation against the exact solution for a constant $\Delta\varphi$, but which only exists for a simplified problem. They match pretty well.

As I was moving onto more complex situations, I was starting to doubt that assumption and before jumping into the code, I would like to clarify the situation.


1 Answer 1


Your cells are not infinitesimally small so it will be a bit more complicated than either formulation you have there. (The ``width'' of your finite volume cell varies nontrivially over the cell, so we cannot just approximate it as $\Delta s \Delta r$.)

Let's say the piece you're integrating is defined by $$ r_1 < r < r_2 $$ $$ \varphi_1 < \varphi < \varphi_2 $$

We begin with

$$ \frac{\partial p}{\partial \varphi} = 0$$

When we integrate in polar coordinates, we need a factor of $r$ there. $$ \int_{\varphi_1}^{\varphi_2}\int_{r_1}^{r_2} \frac{\partial p}{\partial \varphi} r dr d\varphi = 0 \\ \int_{r_1}^{r_2} \left[\int_{\varphi_1}^{\varphi_2} \frac{\partial p}{\partial \varphi} d\varphi \right] r dr = 0 \\ \int_{r_1}^{r_2} \left[p(r,\varphi_2) - p(r,\varphi_1) \right] r dr = 0 \\ (rp)_e - (rp)_w = 0$$

In the last line, I have tried to express the integral result in the notation you used, but I'm not exactly sure if that's what you intended with your notation. Note that the function $p$ must be weighted by $r$ before integrating over the east and west borders of your finite volume cell.

If $p$ is constant with respect to $r$, the previous result simplifies to

$$ \left[p(\varphi_2) - p(\varphi_1) \right] \int_{r_1}^{r_2} r dr = 0 \\ \left[p(\varphi_2) - p(\varphi_1) \right] \frac{r_2^2-r_1^2}{2} = 0 $$

Multiplying it by $\Delta r$ and $\Delta \varphi$ is unnecessary/arbitrary unless you have other terms in your differential equation. When you do have other terms, I recommend you do the integration as I have done to see what the coefficients are.

  • $\begingroup$ I do have other terms in the equation. I would like to point towards Patankars book (p. 72/73), where a similar, but a more detailed situation is depicted. The integration over the control volume is straight forward, I'm just having trouble wrapping my head around the geometrical implications. From your answer, I'm assuming, that I do need to consider the different size of the control volume in the circumferential direction for different radii. Thus geometrically, if p is cell centred, I need to calculate the arc length arc(radius) to the cell faces, which lie midway between two cells. $\endgroup$
    – Rover
    Commented Mar 14, 2017 at 12:01
  • $\begingroup$ Yes, it is definitely trickier with noncartesian coordinates. Actually, now that I think about it, I suppose there is nothing really stopping you from just integrating without the factor of $r$ in the integrand. The result won't be "incorrect." However, your unknowns ($p_e, p_w$, etc.) will likely not have as much physical relevance as you would like and, if you're trying to preserve something physical in your solution, it probably wouldn't be preserved in that case. Perhaps a good way to think about it is to think about what you want the unknowns in your solution to represent. $\endgroup$
    – nukeguy
    Commented Mar 14, 2017 at 13:38
  • $\begingroup$ p is supposed to be the pressure, the actual PDE is a bit more involved. In fact, I compared my current result with a constant $\Delta\varphi$ to the exact solution and they match, for a simplified problem. I recently started to doubt that assumption due to other reasons and would like to try to clarify the situation before jumping into the code. Even though your answer suggests taking the different arc lengths into account, the control volume integral uses radians. In that case, $\Delta\varphi$ is a constant for a uniform grid. $\endgroup$
    – Rover
    Commented Mar 14, 2017 at 14:08
  • $\begingroup$ It really depends on how you're connecting your control volumes together. A finite volume scheme needs two things: (1) balance equations obtained by integrating the PDE over your defined control volume and (2) something to estimate the "fluxes" between the control volumes. Whether or not you include the Jacobian ($r$ in this case) in obtaining (1) will affect the difficulty of obtaining (2). As long as (2) is consistent with (1), however, it should be okay. The problem is that (2) may be really complex if you leave out the Jacobian. $\endgroup$
    – nukeguy
    Commented Mar 14, 2017 at 18:37
  • $\begingroup$ Just to clarify something: $\Delta \varphi$ and $\Delta r$ are constant on your grid -- you can see that from just looking at the grid. It's $\Delta s$ (where s is the arclength) that will vary. However, I would not expect your finite volume discretization to yield a balance equation of the form $\Delta r \frac{p_e - p_w}{\Delta \varphi} = 0$ as you have suggested, unless the FV method is doing something special to obtain the inter-volume fluxes... $\endgroup$
    – nukeguy
    Commented Mar 14, 2017 at 19:11

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