Finite Volume Method in 2D

FMDenaro · November 1, 2023, 13:17

Quote:

Originally Posted by hunt_mat

I think I am pretty comfortable with what I've done. Constructing the fluxes for the main equations will be messy, but not anything that I can't overcome.

Compute only one flux for each face, then it is sufficien to change the sign to have the fluxes for two adjacent cells.

hunt_mat · November 2, 2023, 05:27

Quote:

Originally Posted by FMDenaro

Compute only one flux for each face, then it is sufficien to change the sign to have the fluxes for two adjacent cells.

I don't understand what you mean here.

FMDenaro · November 2, 2023, 05:36

Quote:

Originally Posted by hunt_mat

I don't understand what you mean here.

you have adjacent cells sharing a face, the flux is unique, only the normal unit vector changes.

Thus, you compute for each face the flux and then us that to update the values in time just considering the correct sign.

For example, the flux at the est face of the cell i is equal to the west face of the cell i+1, just with opposite sign.

hunt_mat · November 3, 2023, 09:27

Quote:

Originally Posted by FMDenaro

you have adjacent cells sharing a face, the flux is unique, only the normal unit vector changes.

Thus, you compute for each face the flux and then us that to update the values in time just considering the correct sign.

For example, the flux at the est face of the cell i is equal to the west face of the cell i+1, just with opposite sign.

I see what you're saying. This would be one of the ways to reduce computational time right? As I don't need to use so many computations?

FMDenaro · November 3, 2023, 09:50

Quote:

Originally Posted by hunt_mat

I see what you're saying. This would be one of the ways to reduce computational time right? As I don't need to use so many computations?

Not only, this is mandatory to ensure conservation in discrete sense.

hunt_mat · November 6, 2023, 05:39

If I were to now discretise the equation in detail, I use the approximation:
$f(x)+f(x+\delta x)\approx 2f(x+\delta x/2)$
The equation then becomes:
$\int_{X}^{X+\delta X}C(s,Y+\delta Y/2)ds+\int_{Y}^{Y+\delta Y}D(X+\delta X/2,s)ds\approx 0$
Using the midpoint rule for integrals, the equation reduces to:
$C(X+\delta X/2,Y+\delta Y/2)\delta X+D(X+\delta X/2,Y+\delta Y/2)\delta Y\approx 0$
This looks reasonable to me, but that doesn't necessarily mean that it's correct.

FMDenaro · November 6, 2023, 06:58

Quote:

Originally Posted by hunt_mat

If I were to now discretise the equation in detail, I use the approximation:
$f(x)+f(x+\delta x)\approx 2f(x+\delta x/2)$
The equation then becomes:
$\int_{X}^{X+\delta X}C(s,Y+\delta Y/2)ds+\int_{Y}^{Y+\delta Y}D(X+\delta X/2,s)ds\approx 0$
Using the midpoint rule for integrals, the equation reduces to:
$C(X+\delta X/2,Y+\delta Y/2)\delta X+D(X+\delta X/2,Y+\delta Y/2)\delta Y\approx 0$
This looks reasonable to me, but that doesn't necessarily mean that it's correct.

I suggest thinking always to the discretization of the outer operator, in this case the integral.

In a more general second order discretization, you can use trapezoidal rule. You see a different and larger stencil.

hunt_mat · November 6, 2023, 07:20

Quote:

Originally Posted by FMDenaro

I suggest thinking always to the discretization of the outer operator, in this case the integral.

In a more general second order discretization, you can use trapezoidal rule. You see a different and larger stencil.

Which I did. If X and X+dX are very close to each other then the approximation I gave should be valid.

FMDenaro · November 6, 2023, 07:55

Quote:

Originally Posted by hunt_mat

Which I did. If X and X+dX are very close to each other then the approximation I gave should be valid.

No, the trapezoidal rule is not what you wrote. You wrote the mean value formula, is second order only on a regular grid.
The trapezoidal rule is [f(a)+f(b)]*h/2

hunt_mat · November 6, 2023, 10:00

Quote:

Originally Posted by FMDenaro

No, the trapezoidal rule is not what you wrote. You wrote the mean value formula, is second order only on a regular grid.
The trapezoidal rule is [f(a)+f(b)]*h/2

For small enough dX, and dY, then the approximations are as good as each other, and implementing makes things harder,

LuckyTran · November 6, 2023, 10:05

You still need to discretize the face flux (the integral) either way. You don't know C(x+dx/2,y+dy/2)*dx

I don't understand this obsession with midpoint rule or small dx and small dy. We are in the realm of discrete math. Delta-epsilon arguments don't work here. There is (almost) always a finite error that is the neglected higher order terms (almost unless you have a TVD scheme). This is just another rabbit hole. Just write out the entire discretized integral (i.e. write down the face flux already!).

Let me point out what is wrong with the logic. For small enough dx and dy f(x,y)=f(x+dx,y+dy)=f(x+2dx,y+2dy) are all equally good approximations of each other. Clearly there are practical issues with this implementation.

FMDenaro · November 6, 2023, 11:03

Quote:

Originally Posted by hunt_mat

For small enough dX, and dY, then the approximations are as good as each other, and implementing makes things harder,

No, for your problem you have to write properly the equation by dividing for the measure of the volume this way

1/(dx*dy) * Integral[S] v.f ds=0

and when dx,dy->0 the equation tend to Div.f=0.

It is ok to use the midpoint rule but be aware that is only first order accurate on an irregular grid.

hunt_mat · November 6, 2023, 11:11

Quote:

Originally Posted by LuckyTran

You still need to discretize the face flux (the integral) either way. You don't know C(x+dx/2,y+dy/2)*dx

I don't understand this obsession with midpoint rule or small dx and small dy. We are in the realm of discrete math. Delta-epsilon arguments don't work here. There is (almost) always a finite error that is the neglected higher order terms (almost unless you have a TVD scheme). This is just another rabbit hole. Just write out the entire discretized integral (i.e. write down the face flux already!).

Let me point out what is wrong with the logic. For small enough dx and dy f(x,y)=f(x+dx,y+dy)=f(x+2dx,y+2dy) are all equally good approximations of each other. Clearly there are practical issues with this implementation.

The logic is fine. I like the mid-point rule because this yields the approximant as the centre of the cell, which is what I want. The fluxes have been computed as the flux F=(D,-C), has been computed, I have computed the integral using the mid-point rule. If I used the trapezium rule, I would then have to approximate the function on the cell edges. The mid-point rule gets around that.

hunt_mat · November 6, 2023, 11:13

Quote:

Originally Posted by FMDenaro

No, for your problem you have to write properly the equation by dividing for the measure of the volume this way

1/(dx*dy) * Integral[S] v.f ds=0

and when dx,dy->0 the equation tend to Div.f=0.

It is ok to use the midpoint rule but be aware that is only first order accurate on an irregular grid.

I'm using the reference configuration, so it will be a regular grid. Can you explain why what you wrote is relevant to my question? I don't understand what you're saying.

FMDenaro · November 6, 2023, 11:28

Quote:

Originally Posted by hunt_mat

I'm using the reference configuration, so it will be a regular grid. Can you explain why what you wrote is relevant to my question? I don't understand what you're saying.

if you work only on the equation Int[S] n.f dS=0, this equation would be erroneously verified for |S|->0, even for a wrong flux.

To be consistent to the limit for vanishing dx and dy you have to divide the integral.

As far as the discretization of the integral is concerned, the mid-point rule has a poor spectral representation compared to the trapezoidal rule.

hunt_mat · November 6, 2023, 12:16

Quote:

Originally Posted by FMDenaro

if you work only on the equation Int[S] n.f dS=0, this equation would be erroneously verified for |S|->0, even for a wrong flux.

To be consistent to the limit for vanishing dx and dy you have to divide the integral.

As far as the discretization of the integral is concerned, the mid-point rule has a poor spectral representation compared to the trapezoidal rule.

What are you trying to get at?

FMDenaro · November 6, 2023, 12:23

Quote:

Originally Posted by hunt_mat

What are you trying to get at?

just answered to your doubts ...

hunt_mat · November 6, 2023, 12:34

Quote:

Originally Posted by FMDenaro

just answered to your doubts ...

You've not actually answered in a way that makes sense to me. You don't seem to like the midpoint approximation for some reason, the trapezium introduces complications for obvious reasons.

FMDenaro · November 6, 2023, 13:00

Quote:

Originally Posted by hunt_mat

You've not actually answered in a way that makes sense to me. You don't seem to like the midpoint approximation for some reason, the trapezium introduces complications for obvious reasons.

You need to study the spectral analysis of the formulas and look at the resulting error in the wavenumber space to understand why.

But I think you are not interested in that, so use the midpoint rule that is very easy to implement.

hunt_mat · November 7, 2023, 05:03

Quote:

Originally Posted by FMDenaro

You need to study the spectral analysis of the formulas and look at the resulting error in the wavenumber space to understand why.

But I think you are not interested in that, so use the midpoint rule that is very easy to implement.

I'm not doing Fourier analysis here. It's a very nonlinear problem.

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November 6, 2023, 05:39		#26
hunt_mat Senior Member Matthew Join Date: Mar 2022 Location: United Kingdom Posts: 178 Rep Power: 4	If I were to now discretise the equation in detail, I use the approximation: $f(x)+f(x+\delta x)\approx 2f(x+\delta x/2)$ The equation then becomes: $\int_{X}^{X+\delta X}C(s,Y+\delta Y/2)ds+\int_{Y}^{Y+\delta Y}D(X+\delta X/2,s)ds\approx 0$ Using the midpoint rule for integrals, the equation reduces to: $C(X+\delta X/2,Y+\delta Y/2)\delta X+D(X+\delta X/2,Y+\delta Y/2)\delta Y\approx 0$ This looks reasonable to me, but that doesn't necessarily mean that it's correct.

November 6, 2023, 10:05		#31
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,728 Rep Power: 66	You still need to discretize the face flux (the integral) either way. You don't know C(x+dx/2,y+dy/2)*dx I don't understand this obsession with midpoint rule or small dx and small dy. We are in the realm of discrete math. Delta-epsilon arguments don't work here. There is (almost) always a finite error that is the neglected higher order terms (almost unless you have a TVD scheme). This is just another rabbit hole. Just write out the entire discretized integral (i.e. write down the face flux already!). Let me point out what is wrong with the logic. For small enough dx and dy f(x,y)=f(x+dx,y+dy)=f(x+2dx,y+2dy) are all equally good approximations of each other. Clearly there are practical issues with this implementation.