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 CFDUser February 11, 2009 03:18

SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Hello,

I am looking for a proof (a paper or anything else) which describes in a few sentences the transformation of the integral formulation of the conservation laws into the differential formulation. In other words: - I know that unsteady phenomena need the integral formulation of mass-, impulse- and energy-conservation. But in many literatures, you can find such a semi-discretized formulation:

dQ/dt + dE/dx = 0

Q = conservative values E = flux

(Let us assume that we only have a rectangular domain --> no metric-terms.)

To my mind it comes from:

\int(dQ/dt)dV + \int(E)dS = 0

\int(dQ/dt)dV = (volume integral of time derivative), \int(E)dS = (surface integral of the flux E)

--> dQ/dt*dV + E*dS = 0 (semi-descritized dV=a small control-volume, dS=surface area)

<-> dQ/dt + 1/dV*E*dS = 0

<-> dQ/dt + dE/dx = 0 (dx = dS/dV)

But what does this assumption mean?

Would you agree that the formulation (dQ/dt + dE/dx = 0) is necessary to derive an approximative Riemann solver (Roe)? The question came up to me because the Roe uses the matrix dE/dx for his linearisation dE/dx=(dE/dQ)*(dQ/dx)....

 Tom February 11, 2009 05:23

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Most elementary text books on fluid mechanics cover this! Basically you write you're balance law in integral form which is in the form, for some quantity q

rate of change of q in arbitrary volume V = surface integral of flux of q through the boundary of V. (1)

You then apply the Guass' divergence theorem to obtain a single integral over the volume

integral over V [(dq/dt - div(F) ]dV = 0 (2)

where F is the flux on the left-hand-side of (1) - Note we've assumed differentiability of the flux function here so no discontinuities are allowed.

Now since (2) must hold for all possible choices of V the integrand must vanish; i.e.

dq/dt - div(F) = 0.

(If the integrand was none zero we could find a volume V for which it was of one sign in which case the integral would not vanish).

The reason why you need the differential form is that you need to approximate the characteristics of the pde which can't be obtained from the integral form.

 CFDUser February 11, 2009 06:00

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

I know the Gauss-theorem for the transformation.

1. But if I transformed the integral to a differential formulation, how can we use it to solve flows with none-continous phenomena!?!?!?

"Note we've assumed differentiability of the flux function here so no discontinuities are allowed." But the schemes which I want to analyse are used for discontinuities! This is what gives me a headache...

"The reason why you need the differential form is that you need to approximate the characteristics of the pde which can't be obtained from the integral form."

2. Why can you find the characteristics only in the differntial formulation?

 Tom February 11, 2009 09:32

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

"1. But if I transformed the integral to a differential formulation, how can we use it to solve flows with none-continous phenomena!?!?!?"

Well there are two ways to do this. The first way is to solve the equations in each domain (separated by the disconinuity) separately using the pdes (the solution is continuous in each of the subdomains) and then apply the correct jump conditions, as derived by the failure to be able apply Gauss' theorem across the interface, to fix the solutions. This is not very practical!

The other way is what people actually do which is to solve the viscous regularized equations; i.e. the pdes with a viscous correction term (this is essentially what TVD/upwinding schemes do). The numerics don't really have a discontinuity - it's a rapid change.

"2. Why can you find the characteristics only in the differntial formulation?"

Because the theory of characteristics is about pdes - simple as that. If you think about what characteristics are and then try to apply the ideas in a rigourous manner to the integral representation you'll see why.

 CFDUser February 11, 2009 10:44

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

1) To my understanding - but where is the difference/advantage of the finite volume scheme to a finite difference scheme if you treat it like this: dQ/dt + dE/dx = 0 ? To my mind, there is not any if you treat conservation laws this way....?!?!?!

- Do you know a book where I can read that description again?

2) Ok

 Tom February 11, 2009 11:14

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

The only real advantage that the finite volume method has over finite differences is in that it gives you more freedom over the "shape" of your control volumes. On a cartesian mesh they are essentially the same thing (in that every FVM discretization can also be considered a FD one).

The capturing of discontinuities is really more about local adjustment of the discretizaiton scheme to ensure (1) the solution doen't produce spurious oscillations and (2) satisfies the relevant physical constraints across the discontinuity. This can be done with either method on a cartesian grid.

Try looking at Leveque's book "Finite Volume Methods for Hyperbolic Problems" Cambridge University press.

Basically FVM does not give you the ability to handle shocks by itself (but it makes it easier) - you need to add flux limiters as well.

 CFDUser February 25, 2009 18:16

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Which pages/chapter would you advice in your reference? Looking on page 17 - there is an explanation that the integrand must be equal to zero. But the author even says that q(x,t) must be a smooth function. But this is not the case if you consider the Riemann-problem on the surface of two neighbouring cells...... That's what I do not understand

cite: "Basically FVM does not give you the ability to handle shocks by itself (but it makes it easier)" - What do you mean?

 Tom February 26, 2009 07:51

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Because of his, in my opinion, weird choice of notation (it's actually probably quite good if you're just learning) it's quite difficult to skip around the book. Chapters 3 and 4 discuss hyperbolic equations (3's on the Riemann problem) while 4 covers and introduction to the finite volume method and Roe's method. Chapter 11 covers nonliner equations - weak solutions and the Rankine-Hugonoit relations and chapter 12 has the Lax-Wendroff theorem.

"cite: "Basically FVM does not give you the ability to handle shocks by itself (but it makes it easier)" - What do you mean?"

The jump conditions at a discontinuity (Rankine-Hugonoit relations) are more natural in integral (weak) form and are thus easier to enforce (and hence get the shock speed right). As I said, on a cartesian mesh, every FVM method is a FD method - howver the converse is not true; i.e. there are FD methods that will have problems getting the shock speed right.

 CFDUser February 26, 2009 18:45

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Anyway, I have to repeat my question from the first thread:

## I know/understand the mathematical way to transform the integral form of conservation laws into the differntial formulation.

## I noticed that FV-method do not automatically cover unsteady problems.

## Moreover, I noticed that the transformation of the surface integrals requires a "differentiable flux function".

1. QUESTION/CONCLUSION: --> So, may I conclude that differntiable flux functions do _not_ <exclude> jumps of the _conservative values_?

-------------------------------------------------

2. QUESTION: What exactly is the mathematical element in the conservation laws which makes jumps possible?

-------------------------------------------------

## You mentioned on a cartesian grid (= a grid in "body fitted" coordinates (=hexahedrons)??) a FV scheme is identical to a FDs, but not the inverse.

-------------------------------------------------

3. QUESTION: So, where is the benefit to formulate the conservation laws in the integral form? To my mind the terms in the differential and integral conservation laws are absolute the same, if DV/DX = DS.

DV = Volume of a Cell DS = Surface of a Cell Face DX = DeltaX between to Surfaces DS

-------------------------------------------------

4. QUESTION:

## At last, I think I understood that a transformation of the integral form into the differential formulation is necessary if a approx. Riemann-solver shell be used. The differential formulation of the Euler equation can be decoupled - which means that the 3 equations are independent. In this context the Jacobi matrix A comes up which is dE/dQ. dE/dQ is even though the linearisation and I could not imagine how to formulate this in integral form: dE/dx = (dE/dQ)*(dQ/dx) Do you agree?

-----------------------------------------------

I appologize for my besetting manner of asking questions - but I haven't found all these answers in many CFD books. Morover, many people I was talking to stoped arguing after a while.... So, many thanks in advance.

 Tom February 27, 2009 06:06

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

1. Yes and no - certain combinations of variables can be continuous while the variables themselves are not. So it depends on what you are solving for and whether the equation contains non-conservative terms. In most cases however the flux function is continuous but not differentiable.

2. The jumps are caused by the intersection of a characteristics from the same family. This means that at the point of intersection the solution becomes multivalued and some method is required to make the solution single valued. A discontinuity is then inserted into the solution to maintain a unique value - the viscous equations don't have this problem so the conditions across the jump can be determined by taking the inviscid limit of the viscous solution.

Also if you start with the physical problem in integral form you can determine the jump conditions across the discontinuity by assuming the variables are differentiable everywhere except a surface of discontinuity - apply the divergence theorem in the two separate halves leaving a surface integral around the jump (texts on continuum mechanics cover this).

The reason the converse (re. not all FD schemes are FVM ones) isn't true is that the FD formulation is not necessarily conservative.

4. Yes you need to be in differential form to use the method of characteristics to solve the Riemann problem.

 CFDUser February 27, 2009 06:24

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

1. Non-conservative terms would be source-terms, right? I am actually talking about the standard (compressible) NS-equations. The vector of interest is Q with the conservative variables. So, would you agree that in this particular case Q can be non-continuous but the total flux is?

 Tom February 27, 2009 07:08

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

If the flux wasn't continuous you wouldn't conserve - the flux leaving one cell is equal to the flux entering the neighbour(s).

 CFDUser February 28, 2009 07:20

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Ok, and since you start talking about conservation there is one last thing which comes into my mind:

I was looking for an example which describes a conservative and a none-conservative discretisation scheme. I found an example in Charles Hirsch's book. Unfortunately, I havent it next to me but I think he was talking about a 1D-convection problem:

dQ/dt + dE/dx = 0 (1)

To explain what a conservative discretisation scheme is C. Hirsch wrote (1) for 3 small control volumes and a big one which is equivalent to the three small ones. For the convection terme he used to approaches:

1) dE/dx = E_(i+1) - E_(i-1)/(2Dx) 2) dE/dx = A*(dQ/dx), A=f(Q)

Hirsch prooves that the second approach is not conservative but he mentioned that it is possible to modify the scheme in a way that it is conservative again. I really did not understand that. Even though I am confused to see that the approach which looks quite similar to the Riemann problem is not conservative itself.

# Could you explain what I will have to do to make it conservative/where I can read a more detailed description?

# Where is the difference concerning conservation between FD-schemes and FV-schemes? Is there really any? This topic is most confusing since we already found out that a FV-scheme is mostly a finite difference scheme on a cartesian grid....

 CFDUser March 3, 2009 03:26

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Let me put my question in another way:

Would you agree if I said that it is possible to write consistent FD formulas. But there is no natural way to check conservation for finite differences because there is no flux evaluation on a surface necessary?

 Phi09 March 5, 2009 02:49

jump of solution

From one earlier statement of Tom I would conclude that a shock or jump of the solution can only appear on a cell boundary/surfaces.

 pzhang March 5, 2009 22:33

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

If you can write your scheme (FV or FD, whatever) in the form dQ_i/dt = [ E_(i+1) - E_(i-1) ] / (2Dx), then your scheme is conservative because the sum of dQ_i/dt over a grid depends only on the boundary data.

Good luck.

 pzhang March 5, 2009 22:44

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

To me, this is the most natural transformation:

Integral form, d/dt*\int(Q)dV + \int(E)dS = 0,

Divergence theorem gives

d/dt*\int(Q)dV + \int(dE/dx)dV = 0,

Factoring gives

\int(dQ/dt + dE/dx)dV = 0,

If this holds for arbitrary control volumes, we must have

dQ/dt + dE/dx = 0.

So, the assumption on the differential form has been introduced in the divergence theorem, i.e., dE/dx exists within the volume. Theoretically, you can always satisfy this by taking a control volume in which there are no discontinuities (possibly on the boundary). Shock waves has measure zero, anyway. Don't worry.

 Phi09 March 7, 2009 04:03

Re: SEARCH A PROOF FOR DERIVING DIFF. FORM OF CL

Hello,

I thought the main advantage between finite differences and finite volume schemes is the property of conservation, which you do not have for finite differences (?!?)

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