What would be the rigorous mathematical expression of the fact that a conservation law discretized with a Finite Element method with Galerkin discretization does not result in a conservative scheme ?

  • 1
    $\begingroup$ in other words, are you asking what does it mean for a Galerkin FEM scheme to be conservative? if so, you might find this answer to be relevant. $\endgroup$
    – GoHokies
    Commented Sep 4, 2016 at 18:57
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    $\begingroup$ You should perhaps define what exactly you mean by "conservative" and clarify which problems you are interested in. Some methods are conservative in a global sense if properly discretized and for local conservation it is usually possible to obtain conservative fluxes by solving local problems on patches. So even when a method is not locally conservative in a straightforward fashion, this property is often not lost in the sense that it can be rather inexpensively recovered using only local information. $\endgroup$ Commented Sep 4, 2016 at 19:52
  • $\begingroup$ @GoHokies Thanks, the second answer given in the linked post seems to clarify what definition of "conservation" should be used. But the author states that for the example given (diffusion), "The CG and non-conforming are non-conservative". Would it be possible to have the mathematical expression of this fact (in the case of CG) ? And why the example of diffusion (in the sense, would it be different with advection) ? $\endgroup$
    – Jack
    Commented Sep 4, 2016 at 21:12

2 Answers 2


Most pdes coming from physics have a divergence structure

$$ u_t + \nabla\cdot F =0 \qquad \textrm{in} \quad \Omega $$

Then for any arbitrary control volume $D \subset \Omega$, we have

$$ \frac{d}{dt}\int_D u dx + \oint_{\partial D} F \cdot n ds = 0 $$

i.e., the total quantity inside $D$ changes due to fluxes on the boundary of $D$.

A numerical method that has the same property would be called conservative.

Here one can distinguish between locally conservative and globally conservative methods.

A locally conservative method would satisfy this integral equation on every cell in the mesh. Examples are finite volume methods and DG methods.

A locally conservative method is also globally conservative.

Continuous Galerkin methods may not be locally conservative but are globally conservative. You can see this if you apply a CG method to heat equation with Neumann boundary conditions. Taking the unit test function, you will see that the integral equation is satisfied if $D$ = the whole domain.

Finite volume method: If $F=(f,g)$ then

$$ \Delta x \Delta y \frac{d u_{i,j}}{dt} + [f_{i+1/2,j}-f_{i-1/2,j}] \Delta y + [g_{i,j+1/2}-g_{i,j-1/2}]\Delta x = 0 $$

This is a discrete analogue of the conservation law.

DGM: Let $V_h$ be space of discontinuous, piecewise polynomials. The DG scheme can be written as

$$ \frac{d}{dt}(u_h, \phi_h) - (F(u_h), \nabla\phi_h) + \sum_e \int_e \hat{F}(u_h^-,u_h^+,n) \phi_h ds + \sum_{e\in\partial\Omega}\int_e \hat{F}_b \phi_h ds = 0 \qquad \forall \phi_h \in V_h $$

where $\hat{F}$, $\hat{F}_b$ are numerical flux functions for interior and boundary faces. If $K$ is a cell in the mesh (for simplicity, assume it is away from the boundary), let us take the test function

$$ \phi_h = \begin{cases} 1 & (x,y) \in K \\ 0 & \textrm{otherwise} \end{cases} $$

which is in $V_h$. The DGM scheme gives

$$ \frac{d}{dt}\int_K u_h dx + \sum_{e \in \partial K}\int_e \hat{F}(u_h^-,u_h^+,n) ds =0 $$ which is a discrete approximation to the conservation law.

Continuous Galerkin (CG) method: Let $W_h$ be space of continuous piecewise polynomials. The CG scheme is (we are not concerned with how good this scheme is) $$ \frac{d}{dt}(u_h, \phi_h) - (F(u_h), \nabla\phi_h) + \sum_{e\in\partial\Omega}\int_e \hat{F}_b \phi_h ds = 0 \qquad \forall \phi_h \in W_h $$ We cannot take a test function with support inside one single cell as we did in case of DG scheme. We can take $$ \phi(x,y)=1 \qquad (x,y) \in \Omega $$ and the CG scheme becomes $$ \frac{d}{dt} \int_\Omega u_h dx + \sum_{e\in\partial\Omega}\int_e \hat{F}_b ds = 0 $$ which shows global conservation property.

  • $\begingroup$ Thanks. But why not just write it completely so that the answer is clear. $\endgroup$
    – Jack
    Commented Sep 6, 2016 at 21:25
  • $\begingroup$ Since there are different types of pde and schemes, I cannot give a general answer. I have added examples for finite volume and DG to my answer. $\endgroup$
    – cfdlab
    Commented Sep 8, 2016 at 3:06
  • $\begingroup$ Could you please write the example for the CG method (showing the non local conservation but global conservation), so that I can accept your answer ? $\endgroup$
    – Jack
    Commented Sep 8, 2016 at 19:28
  • $\begingroup$ I have added a CG scheme in the answer. $\endgroup$
    – cfdlab
    Commented Sep 9, 2016 at 7:52

A mathematical expression which shows that a scheme is not conservative is usually an inequality. For example, many methods for the Advection Equation (and generalizations) do not conserve energy (this is known as numerical dissipation). This is usually shown in a theorem like:

For $u_n$ being the approximation of $u(n\Delta t)$, for method M we have that

$$ \Vert u_{n+1} \Vert_{2} < \Vert u_{n} \Vert_{2} $$

In this kind of equation, energy is measured as the $L^2$ norm. Real solutions are travelling waves which don't dissipate energy. However, the numerical approximation adds a little bit of diffusion to the wave, and so the diffusing wave loses energy the same way the diffusion equation does (indeed, you can show that the numerical method actually solves a diffusion+advection equation where the diffusion constant is related to the local truncation error). The previous statement captures this fact by basically saying that the peak of the numerical wave is guaranteed to dissipate at least a little bit at each time step.

  • $\begingroup$ That's not a useful answer because the scheme would also be non-conservative if $\|u_{n+1}\|_2>\|u_n\|_2$. But the question is also about conservation laws, which are first-order differential equations for which the $l_2$ norm is not a conserved quantity. $\endgroup$ Commented Sep 5, 2016 at 21:29

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