I was having some lectures and I didn't quite understand the following: let's say you have a grid like $$G=\{ x \in \mathbb{R} : x = x_j = hj, \ j = 0,1,...,n,\ h=1/n\}.$$ And you write in difference form the following 1D diffusion equation (comma before index indicates derivative along the index): $$-(au_{,1})_{,1}=s, \qquad x \in \Omega := (0,1).$$

Giving: $$ \{ -(a_{j-1} + a_j)u_{j-1} + (a_{j-1}+2a_j+a_{j+1})u_j-(a_j+a_{j+1} )u_{j+1}\}/2h^2 = s_j$$ Now let's say you have a discontinuity of $a$ along some interface $\Gamma$. In our 1D problem, let: $$a(x) = \epsilon, \ \ 0<x\le x^*, \ \ a(x) = 1, \ \ x^*<x<1.$$

For the boundary conditions $u(0) = 0$ and $u(1) = 1$. This jump can be written: $$\epsilon \,\underset{x \uparrow x^*}{\lim u_{,1}} = \underset{x \downarrow x^*}{\lim u_{,1}}. $$

Here is what I don't understand: the lecturer claims that by proposing a linear piecewise solution for the diffusion equation it is found that: $$u= cx, \ \ 0\le x< x^*, \ \ u=\epsilon c x + 1 -\epsilon c, \ \ x^*\le x\le1.$$ Where: $$c= 1/(x^*-\epsilon x^* + \epsilon).$$

This is done with the difference equation, assuming $x_j < x^* \le x_{j+1}$, giving: $$u_j= \alpha j, \ \ 0\le j< k, \ \ u=\beta j + 1 -\beta n, \ \ k+1\le j\le n.$$ Where, $$ \beta =\epsilon \alpha, \ \ \alpha=\left( \epsilon\frac{1-\epsilon}{1+\epsilon} + \epsilon(n-k)+k\right)^{-1}.$$ Hence: $$u_k = x_k/(\epsilon h(1-\epsilon)/(1+\epsilon)+(1-\epsilon)x_k+\epsilon).$$

How are those Piecewise Linear Solution built?

Any input will be much appreciated.

  • $\begingroup$ Is the $s$ of the original DE equal to 0? $\endgroup$ – Joce Oct 12 '15 at 16:40
  • $\begingroup$ @Joce Not necessarily $\endgroup$ – Kbzon Oct 14 '15 at 8:04
  • $\begingroup$ Well then it's funny it doesn't appear in the sequel? Anyway, you can build such a solution by focusing on each subdomain where the DE is trivial--this is where you find that if $s=0$ then the solition is piecewise linear--and then verify the conditions at the interface point (continuity and the jump in $u_{,1}$ $\endgroup$ – Joce Oct 14 '15 at 9:22
  • $\begingroup$ It's true, nevertheless there is no reason to assume is zero. That's one of the reasons I find this explanation a bit odd. I found this same explanation in the book "Introduction to Multigrid" of P. Wesseling, but isn't of any help either. $\endgroup$ – Kbzon Oct 14 '15 at 9:59
  • $\begingroup$ It's false if $s\not=0$. Have a look at cel.archives-ouvertes.fr/cel-00573970 page 39 for exact solution when $s=1$ and a FE code doing it. $\endgroup$ – Joce Oct 14 '15 at 10:30

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