Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I understand mathematically how one can obtain the conservation equations in both the conservative $${\partial\rho\over\partial t}+\nabla\cdot(\rho \textbf{u})=0$$

$${\partial\rho{\textbf{u}}\over\partial t}+\nabla\cdot(\rho \textbf{u})\textbf{u}+\nabla p=0$$

$${\partial E\over\partial t}+\nabla\cdot(\textbf{u}(E+p))=0$$

and non-conservative forms. However, I am still confused, why do we call them conservative and non-conservative forms? can any one explain from a physical and mathematical point of view?

Many off-site threads deal with this question (here and here), but none of them provides a good enough answer for me!

If any one can provide some hints, I will be very grateful.

share|improve this question

migrated from stackoverflow.com Jul 9 '13 at 23:07

This question came from our site for professional and enthusiast programmers.

2 Answers 2

up vote 9 down vote accepted

What does it mean?

The reason they are conservative or non-conservative has to do with the splitting of the derivatives. Consider the conservative derivative:

$$ \frac{\partial \rho u}{\partial x} $$

When we discretize this, using a simple numerical derivative just to highlight the point, we get:

$$ \frac{\partial \rho u}{\partial x} \approx \frac{(\rho u)_i - (\rho u)_{i-1}}{\Delta x} $$

Now, in non-conservative form, the derivative is split apart as:

$$ \rho \frac{\partial u}{\partial x} + u \frac{\partial \rho}{\partial x} $$

Using the same numerical approximation, we get:

$$ \rho \frac{\partial u}{\partial x} + u \frac{\partial \rho}{\partial x} = \rho_i \frac{u_i - u_{i-1}}{\Delta x} + u_i \frac{\rho_i - \rho_{i-1}}{\Delta x} $$

So now you can see (hopefully!) there are some issues. While the original derivative is mathematically the same, the discrete form is not the same. Of particular difficulty is the choice of the terms multiplying the derivative. Here I took it at point $i$, but is $i-1$ better? Maybe at $i-1/2$? But then how do we get it at $i-1/2$? Simple average? Higher order reconstructions?

Those arguments just show that the non-conservative form is different, and in some ways harder, but why is it called non-conservative? For a derivative to be conservative, it must form a telescoping series. In other words, when you add up the terms over a grid, only the boundary terms should remain and the artificial interior points should cancel out.

So let's look at both forms to see how those do. Let's assume a 4 point grid, ranging from $i=0$ to $i=3$. The conservative form expands as:

$$ \frac{(\rho u)_1 - (\rho u)_0}{\Delta x} + \frac{(\rho u)_2 - (\rho u)_1}{\Delta x} + \frac{(\rho u)_3 - (\rho u)_2}{\Delta x} $$

You can see that when you add it all up, you end up with only the boundary terms ($i = 0$ and $i = 3$). The interior points, $i = 1$ and $i = 2$ have canceled out.

Now let's look at the non-conservative form:

$$ \rho_1 \frac{u_1 - u_0}{\Delta x} + u_1 \frac{\rho_1 - \rho_0}{\Delta x} + \rho_2 \frac{u_2 - u_1}{\Delta x} + u_2 \frac{\rho_2 - \rho_1}{\Delta x} + \rho_3 \frac{u_3 - u_2}{\Delta x} + u_3 \frac{\rho_3 - \rho_2}{\Delta x} $$

So now, you end up with no terms canceling! Every time you add a new grid point, you are adding in a new term and the number of terms in the sum grows. In other words, what comes in does not balance what goes out, so it's non-conservative.

You can repeat the analysis by playing with altering the coordinate of those terms outside the derivative, for example by trying $i-1/2$ where that is just the average of the value at $i$ and $i-1$.

How to choose which to use?

Now, more to the point, when do you want to use each scheme? If your solution is expected to be smooth, then non-conservative may work. For fluids, this is shock-free flows.

If you have shocks, or chemical reactions, or any other sharp interfaces, then you want to use the conservative form.

There are other considerations. Many real world, engineering situations actually like non-conservative schemes when solving problems with shocks. The classic example is the Murman-Cole scheme for the transonic potential equations. It contains a switch between a central and upwind scheme, but it turns out to be non-conservative.

At the time it was introduced, it got incredibly accurate results. Results that were comparable to the full Navier-Stokes results, despite using the potential equations which contain no viscosity. They discovered their error and published a new paper, but the results were much "worse" relative to the original scheme. It turns out the non-conservation introduced an artificial viscosity, making the equations behave more like the Navier-Stokes equations at a tiny fraction of the cost.

Needless to say, engineers loved this. "Better" results for significantly less cost!

share|improve this answer
    
Simple and to the point @tpg2114 !...Thank you very very much for taking the time to write such an explanation. You exaclty gave me what I was looking for. Cheers –  user2536125 Jul 10 '13 at 6:40
    
@user2536125 Glad I could help. This was a question on my PhD qualifying exams a few years back :) Don't forget to accept this answer if you're satisfied it answers your question. You just click the check mark on the left side under the voting arrows! –  tpg2114 Jul 10 '13 at 6:47
    
Agreed, nice and simple answer to something which quickly gets very complex. –  ccook Jul 10 '13 at 12:39

You show the Euler Equations, reduced forms of Navier-Stokes, which are conservation laws for mass, momentum, and energy along with Stoke's hypothesis. Now, conservative/non-conservative has nothing to do with conservation laws.

In conservative form, you can directly integrate the derivatives once on a control volume and conserve the flux quantities through the control surfaces (i.e., finite volume method). That's conservative vs. non-conservative form (the flux).

In CFD algorithms the form has important implications with shock placements and propagation speeds. For compressible flows, you want to utilize the conservative forms (where you have shocks).

share|improve this answer
    
thanks for the reply. It would be very helpfull if you can be more generous about your last comment concerning the CFD algorithms. Could you please specify why one would use conservative forms for compressible flows? –  user2536125 Jul 10 '13 at 0:09
    
Yes, but I will need to do so when I have more time on my hands to do it proper. –  ccook Jul 10 '13 at 12:39
    
@user2536125 The primitive flow variables vary discontinuously across a shock wave, and so the gradient operator employed in the non-conservation form loses all physical (and computational) significance. Simply put, we cannot solve a matrix equation that has infinite values in it. The conservation form handles this problem nicely, because the conservation variables are continuous through the shockwave, even if the pressure, temperature, and density are not. –  Bryson S. Jun 26 at 15:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.