sci.math.num-analysis

SUNA41, 1-D introduction to FEM
===============================

A good, readable book to start with is: Norrie D.H. & de Vries G.;
"An Introduction to Finite Element Analysis"; Acad. Press 1978.
However, only the first few chapters of it are truly usefull, IMHO.

The following example could easily be worked out in a FEM class.

Consider the equation:    dQ/dx = 0   where    Q = C(x).dT/dx

It describes for example heat conduction in a rod:
 x = coordinate , Q = heat flow , T = temperature , C = conductivity .

Some understanding of variational principles is required in the first place,
because the standard approach to Finite Elements is always via such principles.

For the problem at hand:

     /
    | C(x).(dT/dx)^2.dx = minimum  ; integral is taken over the whole 1-D grid
   /                                 *---*---*---*---*---*---*---*---*---*---*
                                     1   2   3   4   5   6   7   8   ...

Next split it in pieces over linear elements, with shape functions (h, 1-h),
where  x = (1-h).x1 + h.x2 .                     1*---*2
A little algebra gives: h = (x-x1)/(x2-x1)
The same for temperatures: T = (1-h).T1 + h.T2  and  C(h) = (1-h).C1 + h.C2
This simplifies calculation of the integrals:

    /1
   | C(h).(dT/dh.dh/dx)^2.(x2-x1).dh = (C1+C2)/2 . {(T2-T1)/(x2-x1)}^2.(x2-x1)
 0/       _______constant________

Here we neglected the boundary conditions. The _SUM over_ all these algebraic
expressions for all elements must reach a minimum, as a function of T1 and T2.
Differentiate (partial) to the latter, everywhere on the grid:

   d/dT1 ->  + (C1+C2)/(x2-x1) . T1  -  (C1+C2)/(x2-x1) . T2
   d/dT2 ->  + (C1+C2)/(x2-x1) . T2  -  (C1+C2)/(x2-x1) . T1

Rearranging:
                                  | +1   -1 || T1 |
         SUM    (C1+C2)/(x2-x1) . |         ||    | = 0     (: matrix!)
       elements                   | -1   +1 || T2 |
                                                       1   2   3 : global
Assemble TWO of these for neighbouring elements;       *-A-*-B-*
don't forget to change local into global numbers:      1  2+1  2 : local

- (C1+C2)/(x2-x1).T1 + {(C1+C2)/(x2-x1)+(C2+C3)/(x3-x2)}.T2 - (C2+C3)/(x3-x2).T3
         A                     A               B                     B
For a point off the boundary, this expression must be zero: an "F.D." scheme.
If specialized for  (x2-x1) = (x3-x2) = 1 , then:               ^^^^

 - (C1 + C2).T1 + (C1 + 2.C2 + C3).T2 - (C2 + C3).T3 = 0

Which EXACTLY matches the (equivalent) Finite Volume scheme for this problem.

Which is NO coincidence at all ...
-
* Han de Bruijn; Applications&Graphics | "A little bit of Physics * No
* TUD Computing Centre; P.O. Box 354   | would be NO idleness in  * Oil
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