Page 96 - Process Modelling and Simulation With Finite Element Methods

Partial Differential Equations and the Finite Element Method 83

Using integration by parts, one can simplify above equation to obtain

zl4
(2.32)
dx dx

To advance further we need to make some crucial assumptions. Since we are free
to assign any function to U(x) and $(x) as far as they agree with the boundary
conditions, we assume U(x) = Nx). This is known as Galerkin’s method. If
U(x) # fix), then it gives the Rayleigh-Ritz formulation. We have to select an
algebraic function of x to satisfy the boundary conditions u(0) = u(7d4) = 0.
N
q(X) = u(x) = @(X) = Clql + c2q2 +. . . . +cN(pN = cjqi (2.33)
i=O
We assume N functions as follows.

N
n
qN=X (--x)
4
Therefore
N ?T
q(x) = c CiXi (- - x) (2.34)
i=l 4
This selection satisfies the boundary conditions regardless the number of terms
included in the series. Since U(x) = Nx), the weighted residual becomes
z14 z14
R(x) = L[ 4q2 -[ zIldx-8 j@x’dx (2.35)
0 2 0
By substituting (2.34) in to (2.35) and evaluating the integral we obtain an
expression for R independent of x. However, that expression contains N
.
unknowns, ci We have to evaluate values of the ci so that the weighted residual
is minimum.
From (2.34)

(2.36)