2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

2
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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