I was looking at a book of FEM on problems of Diffusion-Transport.

$$-div(\mu \nabla u) + b \cdot \nabla u + \gamma u = f \qquad in~\Omega$$ $$u = 0 \qquad in~\partial\Omega\text{ (in the boundaries)}$$

It says that if $\displaystyle \frac{|b|}{\mu} \gg 1$ then the problem is a problem dominated by transport.

Taking some stuff from the previous chapter, more precisely the Galerking analysis of stability and convergence, and the approximation of the error, we have

$$\displaystyle M \cong \mu + |b| \qquad \textrm{continuity constant}$$ $$\alpha = \mu \qquad \textrm{coercivity constant}$$

Then it has $$\displaystyle \frac{M}{\alpha} \cong1+\frac{|b|}{\mu} \gg 1$$

As a consequence of that, it concludes that, the estimation of the error is $$\displaystyle ||u -u_h|| ≤C\frac{M}{\alpha}h^r|u|_{H^{r+1}(\Omega)}$$

which tell us that the Galerkin's method could give unsatisfactory results if the space step $h$ isn't small enough.

I Can't understand how can it conclude such thing. Any idea? Please, do not give complex explanations, I'm just trying to understand this FEM stuff, I'm a computer science student more than a math one.

  • $\begingroup$ I'm not exactly clear what your question is? Which of the steps you state is not clear to you? $\endgroup$ Jan 22, 2013 at 2:24

1 Answer 1


If you follow the analysis until the last equation, then you can see the problem. You write:

$|| u-u_h|| \le C \frac{M}{\alpha} h^r |u|_{H^{r+1}(\Omega)}$

which from the expression just above it could also be written as

$|| u-u_h|| \le C \left(1+\frac{|b|}{\mu}\right) h^r |u|_{H^{r+1}(\Omega)}$

Now say you want to solve a problem for a fixed mesh size. In the first problem $|b|=\mu$, then you have

$|| u-u_h|| \le 2C h^r |u|_{H^{r+1}(\Omega)}$

Now you move to the next problem and the transport is strong compared to diffusion, $|b|=10^4 \mu$.

$|| u-u_h|| \le 10^4 C h^r |u|_{H^{r+1}(\Omega)}$

So on a mesh with fixed size, the second solution is orders of magnitude worse. The only way to improve it is to decrease $h$ to an amount that resolves the transport scale (counteracting the huge constant).

  • $\begingroup$ I can see that the second solution is worse, so I suppose we decide to reduce the step size $h$ because is the only parameter that we can manipulate to make that result better, is that true? $\endgroup$
    – BRabbit27
    Jan 22, 2013 at 7:37
  • $\begingroup$ Yes, that is why $h$ must be 'small enough'. $\endgroup$ Jan 22, 2013 at 8:19

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