Unformatted text preview: f the di usivity p(x y),
then the standard methods that we have been studying work ne. This, however, is not
the case in many applications and, as indicated by the following example, standard nite
element methods can produce spurious results.
Example 9.5.1 1]. Consider the steady, one-dimensional, convection-di usion equation ; u00 + u0 = 0 0<x<1 (9.5.2a) with Dirichlet boundary conditions u(0) = 1 u(1) = 2: (9.5.2b) 28 Parabolic Problems The exact solution of this problem is ;
u(x) = 1 + e 1 ; e;1=e
;(1;x)= ;1= : (9.5.2c)
If 0 <
1 then, as shown by the solid line in Figure 9.5.1, the solution features
a boundary layer near x = 1. At points removed from an O( ) neighborhood of x = 1,
the solution is smooth with u 1. Within the boundary layer, the solution rises sharply
from its unit value to u = 2 at x = 1.
2 N odd 1 0 −1 −2 −3 N even
0 0.2 0.4 0.6 0.8 1 Figure 9.5.1: Solutions of (9.5.2) with = 10;3. The exact solution is shown as a solid
line. Piecewise-linear Galerkin solutions with 10- and 11-element meshes are shown as
dashed and dashed-dotted lines, respectively 1].
The term u00 is di usive while the term u0 is convective. With a small di usivity
, convection dominates di usion outside of the narrow O( ) boundary layer. Within
this layer, di usion cannot be neglected and is on an equal footing with convection.
This simple problem will illustrate many of the di culties that arise when nite element
methods are applied to convection-di usion problems while avoiding the algebraic and
geometric complexities of more realistic problems.
Let us divide 0 1] into N elements of width h = 1=N . Since the solution is slowly
varying over most of the domain, we would like to choose h to be signi cantly larger than 9.5. Convection-Di usion Systems 29 the boundary layer thickness. This could introduce large errors within the boundary layer
which we assume can be reduced by local mesh re nement. This strategy is preferable to
the alternative of using a ne mesh everywhere when the solution is only varying rapidly
within the boundary layer.
Using a piecewise-linear basis, we write the nite element solution as U (x) =
j =0 cj j (x) 8 x;x ;
> x ;x ;
< x ;x
k (x) =
> 0 ;x
k c0 = 1 cN = 2 (9.5.3a) if xk;1 < x xk
if xk < x xk+1 :
otherwise (9.5.3b) The coe cients c0 and cN are constrained so that U (x) satis es the essential boundary
The Galerkin problem for (9.5.2) consists of determining U (x) 2 S0 such that
( 0i U 0 ) + ( i U 0) = 0 i = 1 2 : : : N ; 1: (9.5.4a) Since this problem is similar to Example 9.2.1, we'll omit the development and just write
the inner products
( 0i U 0 ) = (ci;1 ; 2ci + ci+1)
h (9.5.4b) ( i U 0) = ci+1 ; ci;1 :
2 (9.5.4c) Thus, the discrete nite element system is h
(1 ; 2 )ci+1 ; 2ci + (1 + 2 )ci;1 = 0 i = 1 2 : : : N ; 1: (9.5.4d) The solution of this second-order, constant-coe cient di erence equation is
ci = 1 + 1 ; N
i = 0 1 ::: N
= 1 ; h=2 :
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This document was uploaded on 03/16/2014 for the course CSCI 6860 at Rensselaer Polytechnic Institute.
- Spring '14
- The Land