# Neumann Boundary Condition at r=0 in Polar Coordinates (Numerical BCs)

I have asked a question in this regard earlier. I am trying to solve the following equation in Polar Co-ordinates:

$$u_t - (u_{rr} + \frac{1}{r} * u_r + \frac{1}{\theta} * u_{\theta\theta} + bu) = f(r,\theta, t)$$

I am building a finite difference method to tackle the above pde. I have assumed the solution to be of the following form:

$$u(r,\theta, t) = e^{-t} r^2 * Sin(2\theta)$$

Clearly, $\frac{\partial u}{\partial r}|_{r=0} = 0$.

My question is, how do I impose this? I have seen some documents that assume that the solution is axially symmetric at $r=0$ and hence only solve along the radial line while assuming a phantom node to the left of the origin, $u_{-1}$ and finally eliminates that using the FD scheme. For me, even though $\frac{\partial u}{\partial r}|_{r=0}$, my assumed solution is not axially symmetric (about z-axis).

Could someone use this image as an example to explain?

Thanks!

There are no boundary conditions at $r=0$. Boundary conditions are things you can impose at (a part of) the boundary. In other words, there is freedom: you could choose different values for the boundary values and get different solutions. But in the case of cylindrical or spherical coordinate systems, you cannot impose $u(r=0,\theta)=\text{something}$ or $\frac{\partial u(r=0,\theta)}{\partial r}=\text{something}$. In the first case, this is because for elliptic partial differential equations (and many other equations as well) the solution is only in $H^1$ and taking point values is not well defined. In the latter case it is because you cannot impose anything other than a zero value.

In other words, the condition $\frac{\partial u(r=0,\theta)}{\partial r}=0$ is not a boundary condition. It is a compatibility condition that the solution has to satisfy, but not something you can impose. To be correct, you better choose a numerical scheme that also satisfies this compatibility condition.

• Could you give a recommendation for the above PDE that satisfies the compatibility condition?
– mod0
Dec 14, 2014 at 13:45
• Could you let me know where I can read more about compatibility condition? Thanks!
– mod0
Dec 14, 2014 at 21:17
• You misunderstand. It's not a property of the PDE. Every function that is continuous satisfies this condition when you express it in polar coordinates. It is a compatibility condition in the sense that not every function given in terms of $r,\theta$ corresponds to a continuous function when given in terms of $x,y$, but every function given in terms of $r,\theta$ must satisfy $\frac{\partial u(r=0,\theta)}{\partial r}=0$ if it corresponds to a function that is continuous when represented in terms of $x,y$. Dec 15, 2014 at 4:47
• @WolfgangBangerth: your statement that every continuous function u(x,y) has the compatibility condition $\frac{\partial u(r=0,\theta)}{\partial r}$ is not correct: a simple counter example is the plane, e.g. $u(x,y)=x$, which in polar coordinates becomes $u(r,\theta)=r \cos(\theta)$ with $u_r = \cos(\theta)$. Graphically, this is obvious too. The cone $u(x,y)=\sqrt{x^2+y^2}$ is also a counter-example, one that is rotation symmetric. It's a continuous function, though it has a cusp at the origin (not differentiable there) Jan 25, 2021 at 13:37
• @armando.sano Yes, it should have been "Every function that is continuously differentiable and has the requisite symmetry properties satisfies the condition automatically. Jan 25, 2021 at 17:00

An idea is not to discretise at the centre (r=0). You put points in pairs near the centre such that each pair forms a line through the centre. Then the finite difference stencils at these points just go through the centre; exactly like in your figure except that $u_0$ should not be put.

In this way, you do not need a boundary condition at $r=0$; the latter is hardly physical but only due to the polar coordinates.

Interestingly, FEM in polar coordinates encounters a difficulty; see How to do FEM in sector elements?

• So how do I approximate u0 in my grid, would there be no u0 then?
– mod0
Dec 14, 2014 at 13:44
• Unless there is singular data at $r=0$, we could assume $r=0$ is a Lebegue point, which is true in your case, so that $u_0$ is just average of the $u$ along the small circle centred at $r=0$. Dec 14, 2014 at 17:00