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I am trying to solve convection diffusion type equations by a modification: $$-\epsilon \Delta u+ b \nabla u+cu=f_1,$$ with the following boundary conditions: $$u=g~ on ~\Gamma_D,~~~~~~~~~~~~~~ \epsilon \frac{\partial u}{\partial n}=g_N ~~on ~~\Gamma_N.$$

The idea is to solve another equation: $$-\nu\Delta u+ \nabla u+\frac{c}{b}u=f_2,$$ $$u=g~ on ~\Gamma_D,~~~~~~~~~~~~~~ \epsilon \frac{\partial u}{\partial n}=g_N ~~on ~~\Gamma_N.$$

where $\nu=\frac{\epsilon}{b}$ and then use $\nu+\delta$ instead of $\nu.$ so we should solve $$$$ $$-(\nu+ \delta) \Delta u+ \nabla u+\frac{c}{b}u=f_3,$$ $$u=g~ on ~\Gamma_D,~~~~~~~~~~~~~~ \epsilon \frac{\partial u}{\partial n}=g_N ~~on ~~\Gamma_N.$$ Here $(\nu+\delta)=\frac{h}{2}*coth(\frac{h}{2*\nu})$, where $h$ is the mesh size.

Now the question is that what the relations are between $f_1,f_2,f_3.$ I used $f_2=f_3$ and $f_2=\frac{f_1}{b}$. Is it true?

Also if $b$ be a vector I used $||b||_\infty$, but the results are acceptable just for some examples. Am I wrong in this way?

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    $\begingroup$ Err, you can't divide by a vector. If you divide by $\|b\|_\infty$, then the second term still looks like $\frac{b}{\|b\|_\infty }\cdot \nabla u$ which is still a directional derivative. $\endgroup$ Sep 5, 2016 at 21:23
  • $\begingroup$ Similarly, what do you mean by "where $\nu=...$ and then use $\nu+\delta$ instead of $\nu$"? Why would you do this? $\endgroup$ Sep 5, 2016 at 21:23
  • $\begingroup$ I wanted to use $\nu+\delta$ for stabilization of the convection diffusion, for increasing Peclet number. But this method does not work for one of my examples. I had to use $\nu+h/2$ instead of $\nu$. In fact using $h/2 \Delta u$ as an artificial diffusion. $\endgroup$
    – Rosa
    Sep 8, 2016 at 7:42
  • $\begingroup$ Well, but you still haven't answered my first question. $\endgroup$ Sep 11, 2016 at 18:44
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    $\begingroup$ I don't see what you edited, but at the very least, it didn't get any better: you just can't divide by a vector. $\endgroup$ Sep 13, 2016 at 19:28

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