Chapter2
Method of Weighted Residuals
Prior to development of the Finite Element Method,there existed an approximation technique for solving differential equations called the Method of Weighted Residuals(MWR).This method will be prented as an introduction,before using a particular subclass of MWR,the Galerkin Method of Weighted Residuals,to derive the element equa-tions for thefinite element method.
Suppo we have a linear differential operator D acting on a function u to produce a function p.
D(u(x))=p(x).
We wish to approximate u by a functions˜u,which is a linear combi-nation of basis functions chon from a linearly independent t.That is,
u∼=˜u=n X i=1a iϕi(2.1)
Now,when substituted into the differential operator,D,the result of the operations is not,in general,p(x).Hence a error or residual will exist:
E(x)=R(x)=D(˜u(x))−p(x)=0.
The notion in the MWR is to force the residual to zero in some average
1
2CHAPTER2.METHOD OF WEIGHTED RESIDUALS n over the domain.That is
Z X R(x)W i dx=0i=1,2,...,n(2.2) where the nuber of weight functions W i is exactly equal the number of unknow constants a i in˜u.The result is a t of n algebraic equations for the unknown constants a i.There are(at least)five MWR sub-methods, according to the choices for the W i’s.Thefive methods are:
2.Sub-domain method.
3.Least Squares method.
4.Galerkin method.
5.Method of moments.
Each of the will be explained below.Two examples are then given illustrating their u.
2.1Collocation Method
In this method,the weighting functions are taken from the family of Diracδfunctions in the domain.That is,W i(x)=δ(x−x i).The Dirac δfunction has the property that
δ(x−x i)=(1x=x i
.
0otherwi
Hence the integration of the weighted residual statement results in the forcing of the residual to zero at specific points in the domain.That is, integration of2.2with W i(x)=δ(x−x i)results in
R(x i)=0
2.2.SUB-DOMAIN METHOD3 2.2Sub-domain Method
This method doesn’t u weighting factors explicity,so it is not,strictly speaking,a member of the Weighted Residuals family.However,it can be considered a modification of the collocation method.The idea is to force the weighted residual to zero not just atfixed points in the domain,but over various subctions of the domain.To accomplish this, the weight functions are t to unity,and the integral over the entire domain is broken into a number of subdomains sufficient to evaluate all unknown parameters.That is
Z X R(x)W i dx=X iµZ X i R(x)dx¶=0i=1,2,...,n
2.3Least Squares Method
If the continuous summation of all the squared residuals is minimized, the rationale behind the name can be en.In other words,a minimum of
S=Z X R(x)R(x)dx=Z X R2(x)dx.
In order to achieve a minimum of this scalar function,the derivatives of S with respect to all the unknown parameters must be zero.That is,
∂S
i
=0
=2Z X R(x)∂R∂a i dx
Comparing with2.2,the weight functions are en to be
W i=2
∂R
i
however,the“2”can be dropped,since it cancels out in the equation. Therefore the weight functions for the Least Squares Method are just the dierivatives of the residual with respect to the unknown constants:
W i=∂R ∂a i
4CHAPTER2.METHOD OF WEIGHTED RESIDUALS 2.4Galerkin Method
This method may be viewed as a modification of the Least Squares Method.Rather than using the derivative of the residual with respect to the unknown a i,the derivative of the approximating function is ud. That is,if the function is approximated as in2.1,then the weight
functions are
W i=∂˜u ∂a i
Note that the are then identical to the original basis functions ap-pearing in2.1
W i=∂˜u
∂a i
=ϕi(x)
2.5Method of Moments
In this method,the weight functions are chon from the family of polynomials.That is
W i=x i i=0,1,2,...,n−1
In the event that the basis functions for the approximation(theϕi’s) were chon as polynomial,then the method of moments may be iden-tical to the Galerkin method.
2.6Example
As an example,consider the solution of the following mathematical problem.Find u(x)that satisfies
d2u dx2+u=1 u(0)=1 u(1)=0.
2.6.EXAMPLE5 Note that for this problem the differential operator D(u(x))and p(x) are
D(u(x))=(d2
+1)u(x)
p(x)=1
For reference,the exact solution can be found and is,in general form,
u(x)=C1sin x+C2cos x+1
and for the given boundary conditions the constants can be evaluated
u(0)=1=⇒C2=0
u(1)=0=⇒C1=−1/sin(1)
So the exact solution is
u(x)=1−sin x sin(1)
Let’s solve by the Method of Weighted Residuals using a polynomial function as a basis.That is,let the approximating function e u(x)be
e u(x)=a0+a1x+a2x2.
Application of the boundary conditions reveals
e u(0)=1=a0
e u(1)=0=1+a1+a2
or
a1=−(1+a2)
and the approximating polynomial which also satisfies the boundary conditions is then
e u(x)=1−(1+a2)x+a2x2
=1−x+a2(x2−x).
Tofind the residual R(x),we need the cond derivative of this func-tion,which is simply d2e u/dx2=2a2.So the residual is
R(x)=
d2e u dx2+e u−1
=2a2+(1−x+a2(x2−x))−1
=−x+a2(x2−x+2)
6CHAPTER 2.METHOD OF WEIGHTED RESIDUALS
2.6.1Collocation Method
For the collocation method,the residual is forced to zero at a num-ber of discrete points.Since there is only one unknown (a 2),only one collocation point is needed.We choo (arbitrarily,but from symme-try considerations)the collocation point x =0.5.Thus,the equation needed to evaluate the unknown a 2is
R (0.5)=−0.5+a 2(0.25−.5+2)=0
So
a 2=+0.5/1.75=2/7=0.285714
2.6.2Subdomain Method
Since we have one unknown constant,we choo a single “subdomain”which covers the entire range of x .Therefore,the relation to evalutate the constant a 2is
Z 1
1·R (x )dx =0Z 10h −x +a 2(x 2−x +2)i dx =0
"−x 22+a 2(x 33−x 22+2x )¯¯¯¯¯1
0=0So
a 2(13−12+2)=12
and a 2=3/11=0.272727.
2.6.3Least-Squares Method
The weight function W 1is just the derivative of R (x )with respect to the unknown a 2:W 1(x )=dR da 2=x 2−x +2
