[Ford Prefect]

Yes that sounds like two good approaches.



How about the method I presented above, any thoughts about the validity?


(I do not consider a method wrong just because there are other methods that may be more fundamental and/or easier to apply. If the math is incorrect then I agree it is wrong though.)

[FMDenaro]

Originally Posted by Ford Prefect

Yes that sounds like two good approaches.



How about the method I presented above, any thoughts about the validity?


(I do not consider a method wrong just because there are other methods that may be more fundamental and/or easier to apply. If the math is incorrect then I agree it is wrong though.)

Sorry, I haven't read all the details of what you proposed. I agree that you can get the final result in different ways. For example, starting from the integral form you can get the limit for the measure of the volume going to zero (of course after dividing by the measure of the volume)

[LuckyTran]

Originally Posted by Ford Prefect

Not sure what you mean. The definition of the derivative is there, just as the additional term is. It is all part of taking the limit as the sides of the volume goes to zero. Is your point here that it is impossible to ONLY use the definition of the derivative (say if I would have skipped the extra term)?

I agree your idea to exploit the limit definition of the derivative is cute. Your steps are also correct. I mean that it is not simply the derivative of the flux that pops out but the divergence of the flux. And you need to do this this taking the difference in fluxes exercise. I.e. you cannot jump straight to eq 11 (or for that matter, write down the final navier-stokes equations). You need to go through the figure and write down the difference in fluxes first.

Originally Posted by Ford Prefect

Ok, I didn't know that was considered cheating. Do you mean that phi(r+delta_r) does not approach phi(r) when delta_r approaches zero?

If my example was successful, due to luck, then could you please show a counter-example instead where the above method does not work?

You've given two nice examples already. The cartesian example is the one I considered where you got lucky. Here you exploited the definition of the derivative and got away with not formally needing taylor series. The cylindrical one, you got unlucky because you still have this phi(r+delta_r) to figure out.

Proving that phi(r+delta_r) converges to phi(r) requires more work and more assumptions (i.e. continuous & differentiable). You cannot say any longer than phi can be any flux which contradicts your initial assumption that phi can be any flux. I hope you realize the lack of mathematical/logical rigor in this attempt. Furthermore, simply being differentiable does not yet prove that phi(r+dr) converges to phi(r) in the limit of dr=>0 (which is why you need the taylor approximations, or something else).

This is more non-trivial than it seems. the limit of x+dx as dx=>0 converges to x because you know you are dealing with x. You can do an epsilon delta proof using f(x) = x to prove this. But, you have no idea what phi is! It looks trivial because you've seen the solution before you've seen the method to arrive at the solution.

[Ford Prefect]

Thank you for an interesting discussion.

Originally Posted by LuckyTran

I agree your idea to exploit the limit definition of the derivative is cute. Your steps are also correct. I mean that it is not simply the derivative of the flux that pops out but the divergence of the flux. And you need to do this this taking the difference in fluxes exercise. I.e. you cannot jump straight to eq 11 (or for that matter, write down the final navier-stokes equations). You need to go through the figure and write down the difference in fluxes first.

Still not sure what you mean. I have just shown you that there are two methods that can be used to arrive at the differential expression for the equations. One based on Taylor series approximation and the other based on the definition of the derivative (in which part is to take the limit as the sizes go towards zero). I don't think I have said anything about the end result other than that both methods produce the same expression. Then I have tried to prove that.

Originally Posted by LuckyTran

You've given two nice examples already.

Are you giving my examples as counter-examples to my examples?

Originally Posted by LuckyTran

Proving that phi(r+delta_r) converges to phi(r) requires more work and more assumptions (i.e. continuous & differentiable). You cannot say any longer than phi can be any flux which contradicts your initial assumption that phi can be any flux. I hope you realize the lack of mathematical/logical rigor in this attempt. Furthermore, simply being differentiable does not yet prove that phi(r+dr) converges to phi(r) in the limit of dr=>0 (which is why you need the taylor approximations, or something else).

This is more non-trivial than it seems. the limit of x+dx as dx=>0 converges to x because you know you are dealing with x. You can do an epsilon delta proof using f(x) = x to prove this. But, you have no idea what phi is! It looks trivial because you've seen the solution before you've seen the method to arrive at the solution.

I don't think it is more non-trivial than it seems. phi(r) is just as "unknown" as phi(r+dr). Consider the Taylor series example. You can approximate the flux value at phi(r+dr) using the gradient of phi(r). You can just as easily approximate the flux value at phi(r) using the gradient of phi(r+dr) (a backwards difference in numerical terms). Then you end up with an expression in which phi(r+dr) is given, plus an additional term. However this is fixed by taking the limit as dr-->0.


Also, arguments of continuous and differentiable should also apply to the Taylor series expansion, right? The integral form is another matter.

[LuckyTran]

Originally Posted by Ford Prefect

I don't think it is more non-trivial than it seems. phi(r) is just as "unknown" as phi(r+dr). Consider the Taylor series example. You can approximate the flux value at phi(r+dr) using the gradient of phi(r). You can just as easily approximate the flux value at phi(r) using the gradient of phi(r+dr) (a backwards difference in numerical terms). Then you end up with an expression in which phi(r+dr) is given, plus an additional term. However this is fixed by taking the limit as dr-->0.

Also, arguments of continuous and differentiable should also apply to the Taylor series expansion, right? The integral form is another matter.

You need to show that the lim phi(r+dr) as dr=> 0 is indeed phi(r). I.e. you need to prove that the limit exists and converges to phi(r), by taking the limit from the left and taking the limit from the right both need to converge to the same limit. This is very hard to prove when the flux can be any flux. It is much more obvious when this does not work (for any flux) when you remember there can be discontinuities in phi. E.g. across shockwaves the limits will not converge to the same limit.

You need to assume it is differentiable in order to apply Taylor series expansions, yes. But what you are trying to show is how to derive N-S without taylor series. I'm trying to show you the roadblock you are having because you are not using the taylor series.

When you say the flux at phi(r+dr) can be approximated using the gradient of phi(r) you are already using taylor approximations in the back of your head. Can it be approximated that? Under what conditions? Prove it! I'm not saying you are not capable of doing such a proof. But in a derivation of any equation, theses are important.

Assuming that the flux of phi at r+dr can be approximated means that you will in the end have derived an approximate navier-stokes equations. What you want to derive is much more than that. A 90% derivation is not a complete derivation.

[Ford Prefect]

Originally Posted by LuckyTran

It is much more obvious when this does not work (for any flux) when you remember there can be discontinuities in phi. E.g. across shockwaves the limits will not converge to the same limit.

I think you can agree that a Taylor series expansion across a shockwave will not work either..

Originally Posted by LuckyTran

You need to assume it is differentiable in order to apply Taylor series expansions, yes. But what you are trying to show is how to derive N-S without taylor series. I'm trying to show you the roadblock you are having because you are not using the taylor series.

Thank you I appreciate that. The best thing would be for you to provide a counter-example where limits do not work, but a Taylor series does work.

Originally Posted by LuckyTran

When you say the flux at phi(r+dr) can be approximated using the gradient of phi(r) you are already using taylor approximations in the back of your head.

Yes, I wrote "Consider the Taylor series example. You can approximate the flux value at phi(r+dr) using the gradient of phi(r)". Obviously I think about the Taylor series expansion in my Taylor series example.


Did you have any other comments on the example?

Originally Posted by LuckyTran

Assuming that the flux of phi at r+dr can be approximated means that you will in the end have derived an approximate navier-stokes equations. What you want to derive is much more than that. A 90% derivation is not a complete derivation.

I don't understand your point here. Do you mean that I arrive at an approximate solution that just happens to be the same solution as the Taylor series expansion yields?

Furthermore, is it your view that the Taylor series expansion derivation is exact?

[FMDenaro]

I am not sure to follow the meaning of the discussion, the Taylor serie is based on the existance of the derivatives and to guarantee that you have the the limits, whatever directions are, must converge to the same value.


Therefore, using the Taylor serie is the same as invoking the regularity of the function and its derivatives.


Again, the derivation of any physical conservation equation must start from the global balance in a volume of small but finite lenght (just think about the fundamental trasnport Reynolds theorem). Only if suitable regularity hypotesis are assumed the equations can be written in differential poitwise form.

[LuckyTran]

Originally Posted by Ford Prefect

You do not have to apply the Taylor series expansion do describe the value of (r+dr). You can just as well write sigma*Area(r+dr) - sigma*Area(r), etc. Then dividing by the volume and letting the sizes go to zero yields the pde.

From this I got the impression you were trying to prove you derive the n-s equations without using taylor series. You gave an example in cartesian where you were successful. Then you gave an example in cylindrical coordinates where you were not succesful (because you did not use Taylor series).

Originally Posted by Ford Prefect

I don't understand your point here. Do you mean that I arrive at an approximate solution that just happens to be the same solution as the Taylor series expansion yields?

Furthermore, is it your view that the Taylor series expansion derivation is exact?

If you assume that the flux on the right can be approximated by the gradient, then you derive a pde which is based on this assumption, and it's only an approximation (i.e. it is only 99% accurate). You need to show that the flux can always be written exactly... You missed that Taylor series can be exact under the same conditions that you called approximations. If you stop at the first derivative then you end up with an approximation... which is why you must write down the infinite taylor series with infinite higher order derivatives.

[FMDenaro]

Maybe, consider just a my idea, the confusion is on the exact relation


f(r+dr) - f(r) = Int [r; r+dr] (df/dr) dr

[Ford Prefect]

Originally Posted by LuckyTran

From this I got the impression you were trying to prove you derive the n-s equations without using taylor series. You gave an example in cartesian where you were successful. Then you gave an example in cylindrical coordinates where you were not succesful (because you did not use Taylor series).

Correct. I gave two examples using limits where I arrived at the same result as a Taylor series approach would give. I have asked you to give a counter-example where this method does not work, but you have not provided one. You just state that the approximation used in my method is wrong. If a counter-example does not exist and I always arrive at the same equation regardless of method then I conclude that both methods are valid.

Originally Posted by LuckyTran

If you assume that the flux on the right can be approximated by the gradient, then you derive a pde which is based on this assumption, and it's only an approximation (i.e. it is only 99% accurate). You need to show that the flux can always be written exactly... You missed that Taylor series can be exact under the same conditions that you called approximations. If you stop at the first derivative then you end up with an approximation... which is why you must write down the infinite taylor series with infinite higher order derivatives.

1. I think you confuse "i.e." with "e.g."


2. Accuracy only comes into play when you have a finite step-size.


3. "The Taylor expansion method" of deriving the flux equations always assume that the higher order terms can be truncated. This is OK due to point 2 above.


4. Later, the numerical method of solving the differential equation can use higher order difference approximations. This is for accuracy.

[LuckyTran]

Originally Posted by Ford Prefect

Correct. I gave two examples using limits where I arrived at the same result as a Taylor series approach would give. I have asked you to give a counter-example where this method does not work, but you have not provided one.

Your cylindrical one did not work. I'm saying your cylindrical example is a counter-example and I don't need to provide another one. You proved you could not do it. One counter-example is sufficient for a disproof. There is a glaring hole in your example which I tried to explain to you repeatedly. You can consider an integral approach like Filippo suggested or something else but you need to use more tools to complete the derivation of the pde. Is that not what this discussion is about? The navier-stokes equations is a pde (the equation is continuous and exact). You write it with an equal symbol (=, always equal) and not an approximately equal (not ~ or the double ~~) symbol.


If you say I can approximate the flux at r+dr using blah blah blah then you end up with a pde that is stuff ~ stuff and not stuff = stuff like you should have. Put in the work, prove that you can actually write down exactly the flux at r+dr and you can turn the ~ into a =.

[Ford Prefect]

Originally Posted by LuckyTran

Your cylindrical one did not work. I'm saying your cylindrical example is a counter-example and I don't need to provide another one. You proved you could not do it. One counter-example is sufficient for a disproof. There is a glaring hole in your example which I tried to explain to you repeatedly. You can consider an integral approach like Filippo suggested or something else but you need to use more tools to complete the derivation of the pde. Is that not what this discussion is about? The navier-stokes equations is a pde (the equation is continuous and exact). You write it with an equal symbol (=, always equal) and not an approximately equal (not ~ or the double ~~) symbol.


If you say I can approximate the flux at r+dr using blah blah blah then you end up with a pde that is stuff ~ stuff and not stuff = stuff like you should have. Put in the work, prove that you can actually write down exactly the flux at r+dr and you can turn the ~ into a =.

As "proof" I would like to use the derivation presented by Bird, Stewart, Lightfoot "Transport Phenomena, 2nd edition"


You can find the derivation using limits for cylindrical coordinates under §2.3


The book is one of the most referenced books when it comes to transport phenomena.

[LuckyTran]

We all know what navier-stokes looks like in cylindrical coordinates. It is freely available on wikipedia and can be seen by anyone at any time. I'm not saying that the equation there or in any particular text is wrong. I'm saying you personally did not prove this limit converges to the limit that you wanted because you missed the conditions needed / you skipped some steps.

The only reason I bring up this point was because you questioned from a pedagogical perspective why some steps were necessary.

[Ford Prefect]

Originally Posted by LuckyTran

We all know what navier-stokes looks like in cylindrical coordinates. It is freely available on wikipedia and can be seen by anyone at any time. I'm not saying that the equation there or in any particular text is wrong.

Yes we all know the equations are available, but that was not the discussion here.

Originally Posted by LuckyTran

I'm saying you personally did not prove this limit converges to the limit that you wanted because you missed the conditions needed / you skipped some steps. The only reason I bring up this point was because you questioned from a pedagogical perspective why some steps were necessary.

So this is what you must mean:

Originally Posted by Ford Prefect

Thank you! This makes it a bit clearer.



I guess the "extra" term that I missed is the (sigma_theta_theta / r) in equation 11.


What is the origin of this term? Is it because the surfaces are not orthogonal, which means that there is a contribution to the force in the r-direction from the stress in the theta direction? It would seem that way by looking at equation 10.



Also from a pedagogical perspective I find it a bit confusing when mixing the Taylor expansion approach (figure 1) with the methodology described in equations 10 and 11, i.e. the figure does not convey what is done in the derivation of the equations.

My argument here is that there are two methods of deriving the flux equations in differential form. One is based on Taylor expansion (common in text books on CFD) and the other is based on limits (as presented by Bird, Stewart, Lightfoot).


In essence they are the same thing, but for students new to the subject it can be confusing to mix them.



In the text provided by Prof. Denaro it seems that both methods are used without explanation.


My explanation between the steps would be that when dividing by volume and taking the limits as the control volume shrinks to zero yields an approximation that truncates the Taylor series at the second derivative (i.e. write out at least the second derivative in the Taylor expansion and then keep it until the limit is taken).

Originally Posted by LuckyTran

Okay I understand better now what you mean.

If you correctly label the flux at x+dx as phi(x+dx) then yes you end up with the same result. So in this case you would remake the original figure and only label the fluxes without applying taylor approximations.

But it only looks nice when you do it in cartesian because your area is not changing. In cylindrical and spherical you also get an area change and here it is not so obvious how to reduce it.


p.s. I got ahead of myself when I said the high order terms in taylor series cancel out. This is very dependent on the problem you are working with and depends where you center your coordinate system. In this example they disappear when you take the limit. If you don't take the limit and keep it discrete, the terms remain. If you don't take the limit you of course don't get any closer to the continuous navier-stokes, so it's a different exercise.

After this statement I have tried to show you that the "limits" approach to deriving the differential flux equations work for both Cartesian and cylindrical coordinates.


The assumptions are the same as in the Taylor series approach.


We end up with the same differential expression.



To end the discussion about the phi(r+dr) term being equal to phi(r), would you accept that this approximation is of the same order as the approximation used for the first derivative? You can easily see this from a Taylor series expansion.