Discretization of the 2D Diffusion equation

panagiat · January 23, 2025, 11:12

Hello,

During the discretization of the 2d Diffusion equation for the x-component of velocity

and diffusion coefficient $\mu$

$\dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) = 0$

First of all, I know I can approach the discretization using the Gauss theorem. However I am interested in the alternative approach, where we evaluate each integral sequentially.
We start by integrating over an arbitrary cell

$\int_{y=s}^{n} \int_{x=w}^{e} \Bigg[ \dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) \Bigg] dx dy = 0 \\$

$\implies \int_{y=s}^{n} \int_{x=w}^{e} \dfrac{\partial}{\partial x} \bigg (\mu \dfrac{\partial u}{\partial x} \bigg ) dx dy + \int_{x=w}^{e} \int_{y=s}^{n} \dfrac{\partial}{\partial y} \bigg (\mu \dfrac{\partial u}{\partial y} \bigg ) dy dx = 0 \\$

$\implies \int_{y=s}^{n} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] dy + \int_{x=w}^{e} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] dx = 0$

Then it is time to evaluate the second integral. Everyone seems to be assuming that the fluxes are constant throughout the cell faces and thus reaching the following:

$\implies \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] \Delta y + \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] \Delta x = 0$

(reaching the same equation as we would using the Gauss theorem)
(also can be seen here)

Is this the correct justification for the last step?

FMDenaro · January 23, 2025, 14:18

Quote:

Originally Posted by panagiat

Hello,

During the discretization of the 2d Diffusion equation for the x-component of velocity

and diffusion coefficient $\mu$

$\dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) = 0$

First of all, I know I can approach the discretization using the Gauss theorem. However I am interested in the alternative approach, where we evaluate each integral sequentially.
We start by integrating over an arbitrary cell

$\int_{y=s}^{n} \int_{x=w}^{e} \Bigg[ \dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) \Bigg] dx dy = 0 \\$

$\implies \int_{y=s}^{n} \int_{x=w}^{e} \dfrac{\partial}{\partial x} \bigg (\mu \dfrac{\partial u}{\partial x} \bigg ) dx dy + \int_{x=w}^{e} \int_{y=s}^{n} \dfrac{\partial}{\partial y} \bigg (\mu \dfrac{\partial u}{\partial y} \bigg ) dy dx = 0 \\$

$\implies \int_{y=s}^{n} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] dy + \int_{x=w}^{e} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] dx = 0$

Then it is time to evaluate the second integral. Everyone seems to be assuming that the fluxes are constant throughout the cell faces and thus reaching the following:

$\implies \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] \Delta y + \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] \Delta x = 0$

(reaching the same equation as we would using the Gauss theorem)
(also can be seen here)

Is this the correct justification for the last step?

That depends on the accuracy order you want. At second order the mean formula can be used. At higher order the formula must be different.

LuckyTran · January 23, 2025, 15:14

Consider that you can partition the east and west faces into sub-partitions each with constant fluxes to approximate a non-constant flux (e.g. e = e1+e2+e3+e3...ej and w=w1+w2+w3+...wk). When you recognize this as a Riemann sum (midpoint rule, mean rule, trapezoidal rule, etc.), then it is more clear that higher order approximations exist for the Riemann integral but at the same time, there isn't a fundamental issue that breaks the whole theory.

panagiat · January 24, 2025, 05:49

Quote:

Originally Posted by FMDenaro

That depends on the accuracy order you want. At second order the mean formula can be used. At higher order the formula must be different.

What values are going to be used for the mean formula you refer to (since we are talking about one face at a time)? I think I am a bit confused. Do we split the face into subfaces as LuckyTran mentioned and work with these? If so, how do we even come up with different face fluxes since we only have 1 way to calculate the value (constant velocity within the cell)?

So, I suppose that what I wrote in my post is considered first-order? How would I search for this type of approximation? Do you have any resources?

panagiat · January 24, 2025, 05:55

Quote:

Originally Posted by LuckyTran

Consider that you can partition the east and west faces into sub-partitions each with constant fluxes to approximate a non-constant flux (e.g. e = e1+e2+e3+e3...ej and w=w1+w2+w3+...wk). When you recognize this as a Riemann sum (midpoint rule, mean rule, trapezoidal rule, etc.), then it is more clear that higher order approximations exist for the Riemann integral but at the same time, there isn't a fundamental issue that breaks the whole theory.

I find the approach of thinking of it as a Riemann sum is helpful. If I am not mistaken, this is also the approach when considering the whole domain split in individual cells as when they get smaller and smaller they tend to converge to the analytical solution. So, would this mean that within each cell, we consider every quantity to be constant (whether this is the value of the variable u or the flux of each face)? Isn't this the whole idea behind FVM?

FMDenaro · January 24, 2025, 07:04

Quote:

Originally Posted by panagiat

What values are going to be used for the mean formula you refer to (since we are talking about one face at a time)? I think I am a bit confused. Do we split the face into subfaces as LuckyTran mentioned and work with these? If so, how do we even come up with different face fluxes since we only have 1 way to calculate the value (constant velocity within the cell)?

So, I suppose that what I wrote in my post is considered first-order? How would I search for this type of approximation? Do you have any resources?

Not a first order ... As you can see, your approach to discretize in a FV manner leads to the same algebric form of the second order accurate FD scheme.
That happens only for linear equations at second order when using the mean formula on stuctured grid. If you really want to see the difference between FV and FD methods, just think about what happens by using the trapezoidal rule for the integrals.

LuckyTran · January 24, 2025, 18:47

Quote:

Originally Posted by panagiat

So, would this mean that within each cell, we consider every quantity to be constant (whether this is the value of the variable u or the flux of each face)? Isn't this the whole idea behind FVM?

No, not at all. To again use the Riemann sum example, you can discretize a function and use a particular value in an interval, but you never assume that the function is constant over the interval. Although the area of each partition is equivalent to a constant function over the interval, it is not an explicit presumption that the function is constant. If the function were constant, the integral would be trivial.

In your case, you have to now define how to calculate the fluxes on the east and west faces!

$\Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] \Delta y + \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] \Delta x = 0$

You will very quickly discover that obviously u cannot (in general) be constant over the cell because then the east and west fluxes would be equal and the integral becomes the trivial 0=0, which is not useful!

panagiat · January 28, 2025, 09:03

Quote:

Originally Posted by LuckyTran

Although the area of each partition is equivalent to a constant function over the interval, it is not an explicit presumption that the function is constant.

I am a bit confused regarding this. Don't we consider velocity u to be constant inside the cell area and equal to the representative value of the cell (the centroid)?
How is the last step on the discretization process I have in my post (solving the second integral) justified?

Quote:

Originally Posted by LuckyTran

You will very quickly discover that obviously u cannot (in general) be constant over the cell because then the east and west fluxes would be equal and the integral becomes the trivial 0=0, which is not useful!

Sorry I think I have not explained my thoughts correctly. What I meant by saying that we consider every quantity to be constant is that for example throughout an individual cell face the flux is constant, not that all the faces on the cell have the same constant flux and thus the integral does not become 0=0. is this true or am I still mistaken?

Isn't this the next step in the discretization process which includes choosing a scheme? To my understanding (at least in a colocated cartesian grid), we have the unknown velocities at the center of the cell (representative values for the entirety of the cell) and then in order to calculate the fluxes on the cell faces, one must either interpolate (like central differencing) or think of another scheme.

LuckyTran · January 28, 2025, 11:31

If it makes you feel better, you can add $\lim_{\Delta y \to 0}$ and $\lim_{\Delta x \to 0}$ and replace the equality sign with $\approx$

Quote:

Originally Posted by panagiat

How is the last step on the discretization process I have in my post (solving the second integral) justified?

The intermediate value theorem guarantees that I can find a flux that will make the integral exact. So is it justifed or not? It can be. There is no mathematical meaning to asking whether or not it is justified until you complete the discretization.

Don't miss the forest for the trees.

$\dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) = 0$

First of all, I know I can approach the discretization using the Gauss theorem. However I am interested in the alternative approach, where we evaluate each integral sequentially.
We start by integrating over an arbitrary cell

$\int_{a}^{b} f(x)dx \approx f(a \leq x' \leq b)*(b-a)$
f(x') is single valued but that is not an assumption that f is constant over the interval a to b. That was my point and I'm sorry that wasn't articulated well

panagiat · January 31, 2025, 14:08

Thank you, this makes things a lot clearer!

marek_ac · February 17, 2025, 10:31

Hi,
there is a detailed description how to create linear equations to resolve diffusion equation in 3D dynamic with convection:
https://marek-ac.meri.pl/
Marek

January 23, 2025, 11:12	Discretization of the 2D Diffusion equation	#1
panagiat New Member Panagiotis Iatrou Join Date: Oct 2024 Posts: 11 Rep Power: 2	Hello, During the discretization of the 2d Diffusion equation for the x-component of velocity and diffusion coefficient $\mu$ $\dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) = 0$ First of all, I know I can approach the discretization using the Gauss theorem. However I am interested in the alternative approach, where we evaluate each integral sequentially. We start by integrating over an arbitrary cell $\int_{y=s}^{n} \int_{x=w}^{e} \Bigg[ \dfrac{\partial}{\partial x}\bigg(\mu \dfrac{\partial u}{\partial x}\bigg) + \dfrac{\partial}{\partial y}\bigg(\mu \dfrac{\partial u}{\partial y}\bigg) \Bigg] dx dy = 0 \\$ $\implies \int_{y=s}^{n} \int_{x=w}^{e} \dfrac{\partial}{\partial x} \bigg (\mu \dfrac{\partial u}{\partial x} \bigg ) dx dy + \int_{x=w}^{e} \int_{y=s}^{n} \dfrac{\partial}{\partial y} \bigg (\mu \dfrac{\partial u}{\partial y} \bigg ) dy dx = 0 \\$ $\implies \int_{y=s}^{n} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] dy + \int_{x=w}^{e} \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] dx = 0$ Then it is time to evaluate the second integral. Everyone seems to be assuming that the fluxes are constant throughout the cell faces and thus reaching the following: $\implies \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial x}\bigg)_{w} \Bigg] \Delta y + \Bigg[ \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{e} - \bigg(\mu \dfrac{\partial u}{\partial y}\bigg)_{w} \Bigg] \Delta x = 0$ (reaching the same equation as we would using the Gauss theorem) (also can be seen here) Is this the correct justification for the last step?

February 17, 2025, 10:31	the x,y,z - component of velocity and diffusion coefficient	#11
marek_ac New Member Marek Chodorski Join Date: Aug 2013 Location: Poland Posts: 16 Rep Power: 13	Hi, there is a detailed description how to create linear equations to resolve diffusion equation in 3D dynamic with convection: https://marek-ac.meri.pl/ Marek Last edited by marek_ac; February 17, 2025 at 10:42. Reason: typo

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January 23, 2025, 15:14		#3
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,783 Rep Power: 66	Consider that you can partition the east and west faces into sub-partitions each with constant fluxes to approximate a non-constant flux (e.g. e = e1+e2+e3+e3...ej and w=w1+w2+w3+...wk). When you recognize this as a Riemann sum (midpoint rule, mean rule, trapezoidal rule, etc.), then it is more clear that higher order approximations exist for the Riemann integral but at the same time, there isn't a fundamental issue that breaks the whole theory.

January 31, 2025, 14:08		#10
panagiat New Member Panagiotis Iatrou Join Date: Oct 2024 Posts: 11 Rep Power: 2	Thank you, this makes things a lot clearer!