Hello,

Roe's approximate Riemann solver is a famous scheme in FVM fof Euler equations. But it "seems" that many people do not understand the importance of Roe-averaged density which is \sqrt{rho_L*rho_R}. I saw an article in a Japanese journal which claims that density average cannot be determined, and is arbitrary , so arithmetic average can also be used. Also, I found in the Dr. Laney's book (Computational Gasdynamics) an ambiguous statement; "It is convenient and traditional ...".

In my opinion, the square root average is CONSISTENT with the Roe linearization. In Roe's method, you first linearize the Euler equations using Roe-averages of u, and H (density is not necessary here). Then the linear hyperbolic system is solved for a Riemann problem "exactly" by using the fact that any jump in U can be represented as a combination of the right eigenvectors, When we project the jump into the space of eigenvectors, we find the amplitudes in each basis, which thus will be determined UNIQUELY. If you actually compute them, you'll find that the square root average will appear naturally (I actually did this).

What do you say?

Nishikawa

(1). I am not familiar with this area, but I will give you a short comment. (2). rho_L and rho_R are density. But what is (rho_L*rho_R)? it has density square in unit. What is density square? I don't know. (3). If you use (rho_L+rho_R), then the unit is still the density. (4). Or you can use (f_L*rho_L + f_R*rho_R), and still get the density back. In this case, it is weighted density. (5). I don't think sqrt(rho_L*rho_R) is somekind of average, it is a non-linear operation.

(1)I don't know either what is density squared. But after taking square root, I know the result is density. (2)If averaging operations had to be linear, you would be right; sqrt(a*b) is not any kind of average. (3)How about harmonic mean? (1/a+1/b)^{-1}. This is also not a linear operation, but I think this is a kind of average. (4)You mean that sqrt(a*b) is not an average but a mean? What is the difference between average and mean? (5)Or, are you saying that it is meaningless to use nonphysical quantities in any stages of a computation? If so, I don't agree. What is (1/rho) then, when you compute the speed of sound sqrt(gamma*p/rho) for perfect gases?

(1). (rho_L*rho_R) is the product of the two densities. If we use graphic representation, put rho_L on the horizontal axis, and rho_R on the vertical axis, then (rho_L* rho_R) represents the area. (2). sqrt(rho_L*rho_R) *sqrt(rho_L*rho_R) will give you back the same area, that is (rho_L*rho_R). (3). Let equi_density=sqrt(rho_L*rho_R), then (equi_density*equi_density) represents the equivalent area with the same density as the side length, which is a square in the graphic representation. That is find the side of the equivalent square. (4). So, equi_density is something which will produce the same area as (rho_L*rho_R), when it is squared by itself. (5).And actually, (-equi_density)*(-equi_density)=(rho_L*rho_R), the negative equi_density is also one of the solution. (5). So, strictly speaking, it is abs( sqrt( abs(rho_L) * abs(rho_R) ) ). Without this, the sqrt function will fail. It is hard to say that this complicated function represent the average or mean density operation.

To Dr. Chien

(1) In the previous message, 1/rho is a specific volume. I meant sqrt(1/rho). (2) Absolute signs in the complicated square function, abs( sqrt( abs(rho_L) * abs(rho_R) ) ), are not necessary because first, sqrt() is always positive, and second, density is always positive. Negative density is nonphysical and meaningless. (3) sqrt(a*b) can be used only for positive values of a and b, which is, I think, obvious. And the purpose is to compute sqrt(a*b) for given a and b (both > 0), and the result is clearly positive. No need to consider -sqrt(a*b). You said "square function will fail". I don't see why it will. (4) It looks like you define the operation sqrt(a*b) as the one which gives a*b when squared. This is clearly not well defined. I say sqrt(rho_L*rho_R) is the one which gives the "side" of the "area" of a square equivalent to (rho_L*rho_R), and therefore everything must be positive (negative side, negative area, or nageative density do not make sense).

To other people

My original question is "Do you agree that the Roe-averaged density in the form sqrt(rho_L*rho_R) is the unique choice in Roe's approximate Riemann solver?". Or nobody uses the scheme any more?

Nishikawa

(1). Apparently you have everything figured out. So, there is not much I can do here. (2). But I will tell you that each time you run a transient compressible Navier-Stokes code, you will run into the negative density problem. And when that happens, the code will stop. (3). This negative density problem is still one of the major issue in the compressible flow calculations. The code does not know that the density should be positive. As a matter of fact, average or mean should be applicable to both the positive and the negative numbers. (4). I can't answer your original question, because I will have to dig out the Roe-averaged density related papers and find out exactly which one you are referring to in the first place. If you can post the name of the reference paper on Roe-averaged density, I will try to get our library to look into it.

you wrote: "But I will tell you that each time you run a transient compressible Navier-Stokes code, you will run into the negative density problem. And when that happens, the code will stop. (3). This negative density problem is still one of the major issue in the compressible flow calculations. The code does not know that the density should be positive."

I beg to disagree on these points. (1) each time you run a NS code, you will run into the negative density problem: No. It really depends on the problem you are solving. Solving a 2D compression ramp will not create such problems.. A 2D expansion ramp though might, and clipping of the pressure and temperature might be necessary. However, should the code be run in time accurate form, and low enough CFLs be used, the latter should not occur. (2) I don't understand how this 'negative density problem' relates to the original question. By the time we reach the Roe average section of the code, the density has already been clipped and must necessarily be positive. One important aspect of any code based on the Roe scheme is the determination of the eigenvalues in which the sound speed of the system needs to be found, which includes determining the square root of some properties based on density or temperature (sqrt(gamma*R*T) for laminar flows..). In other words, whether Roe average or arithmetic average be used, the density must not be allowed to go negative.

Lastly I'd like to mention that the Roe average really is an average even though it includes a square root. Take for example H^bar or c^bar (average states of H or c), then they can be written as H^bar=x*H_R+(1-x)*H_L. For an arithmetic average, x=0.5. For the Roe average, x is equal to a function of rho_L and rho_R and always lies between 0 or 1. More specifically, x=sqrt(rho_R/rho_L)/(sqrt(rho_R/rho_L)+1).

"My original question is "Do you agree that the Roe-averaged density in the form sqrt(rho_L*rho_R) is the unique choice in Roe's approximate Riemann solver?". Or nobody uses the scheme any more? "

I for one use this scheme. For an ideal gas I found that the way the density at the interface is determined does not affect the equality between Delta F = A Delta Q. But the mass fractions c, global enthalpy H and velocities u_k must be Roe-averaged to ensure that the above equality holds true. This is particularly important to ensure that the discretization of dF/dx involves only F terms, and not A * Delta Q terms.

For certain types of finite-difference TVD schemes (or other schemes based on the Roe approximate Riemann solver), the capability to capture free stream exactly is only guaranteed when the Roe average is performed. But for many problems, whether Roe or arithmetic average will result in the same engineering answer, which is probably why many researchers choose not to use Roe and its computationally expensive square roots (!). But just to be safe, I would suggest always applying the Roe averaging when making use of the Roe Riemann solver.

Also, if you plan on simulating real gases (non-ideal), the Roe average needs to be modified (Vinokur) and gets far more complicated ;-)..

-bernard

(1). I think, I agree with you that the density must not be allowed to go negative. So, some people have to invent method to clip it. (2). As long as the final solution is accurate, it really does not matter whether it must be clipped first, then averaged. (3). My feeling is that the method is very mechanical. (4). The situation is even worse when the negative density is problem dependent. In real life, it is difficult to find problems with only compression waves, and problems with only expansion waves. It is likely that both types exist at the same time in the different part of the flow field.

To Dr. Chien

(1)I actually half expected a reply of this kind. I agree that the negative density problem is a problem, and that the code does not know that density should be positive. But I don't think average or mean should be applicable to negative density. You said that the code would stop when it happens. Then it is meaningless to have an average operation for negative density.

(2)You wrote "As a matter of fact, average or mean should be applicable to both the positive and the negative numbers. " Do you know AM-GM inequality? (At least here in Japan, we ALL learn this inequality in high school and first learn a kind of average "GM".) Of course, AM stands for "arithmetic mean" (a+b)/2, and GM stands for "Geometric MEAN" sqrt(a*b) (of course a and b must be positive). Do American high schools teach this inequality? You makes me very curious about it. (4) Anyway, it was interesting to discuss it with you.

Thank you for your comment! (I finally found a kind of comment that I wanted.)

It is interesting that it does not change "engineering results" very much. So, the cheaper one can be chosen.

It is also interesting to know that the capability to capture free stream exactly is only guaranteed when the Roe average is performed. I didn't know that.

Thank you,

Nishikawa

Hi, Guys,

When you use the simple form of Swanson and Turkel for Roe's scheme, you will see the Roe-averaged density disappears. So, this average is useless for Roe's scheme. Perhpas for flows with chemical reactions. However, Roe's schme does not work for flows with very low density. In this case, negative density or pressure can appear because of numerical errors. For such problems, AUSM is superior.

Stein

Roes averaged method is covered very comprehensively in the text book by C.Hirsch (Vol 2).

I dont know if your question is not framed right, but the Riemmen problem is not solved exactly, but approximately (that is why the cell average) by using the Roes method. For the exact solution we need to use other method's like the Gudonouv's method.

Finally, the codes which I have used in the past incorparate the Roes upwind scheme, and running into negative density does not depend on the type of scheme, but the type of problem (as someone rightly pointed out).

(1). My personal feeling is that the "arithmetic mean" (a+b)/2, deals with the quantities with the same unit. And this derived mean quantity is half way between a and b. (2). On the other hand, the product of (a*b) is no longer in the original space as (a+b). That is, one is dealing with the length, and the other is dealing with the area. (3). As a matter of fact, the sqrt(a*b)=sqrt(a)*sqrt(b). When compared with (a/2) + (b/2) , one can easily see that sqrt(a) and sqrt(b) are now in different space, that is the unit has been changed. (4). If a=0, and b=9, then the average is (a+b)/2=(0+9)/2=4.5 (5). On the other hand, sqrt(a*b)=sqrt(0*9)=sqrt(0)=0. So, this is a special formula where the derived quantity is calculated based on the separate criterion outside its own space (in the area space). In this case, we are dealing with the product space of (a*b). It is fine to use "Roe-averaged density", as long as we can easily identify the formula he uses. (6). If your first year income is zero, and the second year income is 30,000 US dollars, people would be very confused when you use the "Roe-averaged" two-year income. The "Roe-averaged" income for the first two years would be "zero" instead of the conventional average of 15,000 dollars. (7). So, all I can say is that the "Roe-averaged density" was invented for the special purpose in his numerical scheme. As long as it can serve that purpose, it is all right with me, I guess.

Hi,

The book by Hirsch (Vol.2) explains the scheme, but does not explain the density average. Laney's book, Leveque's book, and many other books do not explain how to derive the density average. As I said in the first message, Laney says in his book "It is convenient and traditional to use sqrt(rho_L*rho_R)" .

The Riemann problem for Euler equations is nonlinear. Roe's method does not solve this exactly as you said, but it does solve a linearized Euler exactly. After the Roe linearization, you have a linear system of equations (Jacobian matrix A(U) is now constant) for two constant states, i.e. Riemann Problem for Linearized Euler equations. We can solve this EXACTLY. This is the Roe-scheme. It fails to recognize expansion fan because a linear hypabolic system of equations do not have such a solution.

Hi,

It is interesting that the density average disappears. But, I don't know what "the simple form of Swanson and Turkel for Roe's scheme" is. Could you explain what it is? Or could you post the names of the paper or a book?

Thank you, Nishikawa

(1)Your personal feeling is interesting. (2)You think that Roe invented the operation sqrt(a*b)? No, he did not. (BTW, a and b are positive, meaning nonnegative and "nonzero"). As I said, it is a geometric mean, an elementary mathematical operation which have been known for a long time just like arithmetic or harmonic means. (3)So, I suspect that you don't know the geometric mean. (4)The point of my original question is "Of many possible averaging operations for density, is the geometric mean the only choice consistent with the Roe linearization?". Now I realize that I shouldn't have called it "Roe-averaged density" which is confusing. I apologize, and hope this would help you understand my and in turn your misunderstanding. (6)The discussion was really interesting, anyway.

(1). Why not track this Mr. Roe down, and send him e-mail about this "geometric mean density" question? (2). If he is still around, he should be able to give you good answers. (3). I am just trying to keep this forum interesting but not enough to create anger. Good luck to you.

You wrote: [[In my opinion, the square root average is CONSISTENT with the Roe linearization. In Roe's method, you first linearize the Euler equations using Roe-averages of u, and H (density is not necessary here). Then the linear hyperbolic system is solved for a Riemann problem "exactly" by using the fact that any jump in U can be represented as a combination of the right eigenvectors, When we project the jump into the space of eigenvectors, we find the amplitudes in each basis, which thus will be determined UNIQUELY. If you actually compute them, you'll find that the square root average will appear naturally (I actually did this). ]]

Unfortunately, many places just mention such a things are possible then apply roe-avaergeing for density afterawrds! I have not seen such a prove myself yet. Would it be possible you explain more or inroduce me a reference having that proof, or e-mail me somthing attached to the e-mail with your derivations.

THanks

MJ Fax: 613-520-5715 (FAO: M Kermani)

Consider 1D Euler, U_t + F_x =0 , or U_t + A(U) U_x =0.

I guess you know how to define Roe-linearization (i.e. to find U' s.t. DF=A(U')DU ) which is described in many CFD books such as Hirsch's book. (No density-average yet)

Now you have a "Linear" hyperbolic system and want to solve a Riemann Problem (at cell interface). As to how to solve a Riemann problem for linear equations, see

"Numerical Methods for Conservation Laws" by Randall J. Leveque / Birkhauser Boston / January 1992

You'll see that you can solve it exactly by projecting the change in U onto eigenvectors with correct amplitudes.

How to compute the wave strengths (amplitudes) for Euler equations is not actually simple at first sight. But see a neat trick in the book,

"Numerical Approximation of Hyperbolic Systems of Conservation Laws" by Edwige Godlewski,E. Godlewski,With Pierre-Arnaud Raviart / Springer-Verlag New York, / January 1996

I suppose they explicitly derive the Roe-average density. I don't remember. But I do remember that the trick was very useful in deriving it.