Chapter 2 Method of Weighted Residuals

1 Chapter 2 Method of WeightedResidualsPrior to development of the Finite Element Method , there existed anapproximation technique for solving differential equations called theMethod of Weighted Residuals (MWR). This Method will be presentedas an introduction, before using a particular subclass of MWR, theGalerkin Method of Weighted Residuals , to derive the element equa-tions for thefinite element we have a linear differential operatorDacting on a functionuto produce a (u(x)) =p(x).We wish to approximateuby a functions u, which is a linear combi-nation of basis functions chosen from a linearly independent set.

2 Thatis,u = u=nXi=1ai i( )Now, when substituted into the differential operator,D,theresultof the operations is not, in general,p(x). Hence a error orresidualwillexist:E(x)=R(x)=D( u(x)) p(x)6= notion in the MWR is to force the residual to zero in some average12 Chapter 2. Method OF Weighted Residuals sense over the domain. That isZXR(x)Widx=0i=1,2, .., n( )where the nuber ofweight functionsWiis exactly equal the number ofunknow constantsaiin u. The result is a set ofnalgebraic equations forthe unknown are (at least)five MWR sub-methods,according to the choices for theWi s.

3 Thesefive methods are:1. collocation Sub-domain Least Squares Galerkin Method of of these will be explained below. Two examples are then givenillustrating their Collocation MethodIn this Method , the weighting functions are taken from the family ofDirac functions in the domain. That is,Wi(x)= (x xi). The Dirac function has the property that (x xi)=(1x= the integration of the Weighted residual statement results in theforcing of the residual to zero at specific points in the domain. That is,integration of withWi(x)= (x xi)resultsinR(xi)= SUB-DOMAIN Sub-domain MethodThis Method doesn t use weighting factors explicity, so it is not,strictlyspeaking, a member of the Weighted Residuals family.)

4 However, itcanbeconsideredamodification of the collocation Method . The ideais to force the Weighted residual to zero not just atfixedpointsinthedomain, but over various subsections of the domain. To accomplish this,the weight functions are set to unity, and the integral over the entiredomain is broken into a number of subdomains sufficient to evaluateall unknown parameters. That isZXR(x)Widx=Xi ZXiR(x)dx =0i=1,2, .., Least Squares MethodIf the continuous summation of all the squared Residuals is minimized,the rationale behind the name can be seen.

5 In other words, a minimumofS=ZXR(x)R(x)dx=ZXR2(x) order to achieve a minimum of this scalar function, the derivativesofSwith respect to all the unknown parameters must be zero. Thatis, S ai=0=2 ZXR(x) R aidxComparing with , the weight functions are seen to beWi=2 R aihowever, the 2 can be dropped, since it cancels out in the the weight functions for the Least Squares Method are justthe dierivatives of the residual with respect to the unknown constants:Wi= R ai4 Chapter 2. Method OF Weighted Galerkin MethodThis Method may be viewed as a modification of the Least SquaresMethod.

6 Rather than using the derivative of theresidualwith respectto the unknownai, the derivative of the approximating function is is, if the function is approximated as in , then the weightfunctions areWi= u aiNote that these are then identical to the original basis functions ap-pearing in u ai= i(x) Method of MomentsIn this Method , the weight functions are chosen from the family ofpolynomials. That isWi=xii=0,1,2, .., n 1In the event that the basis functions for the approximation (the i s)were chosen as polynomial, then the Method of moments may be iden-tical to the Galerkin ExampleAs an example, consider the solution of the following mathematicalproblem.

7 Findu(x)thatsatisfiesd2udx2+u=1u(0) = 1u(1) = EXAMPLE5 Note that for this problem the differential operatorD(u(x)) andp(x)areD(u(x)) = (d2dx2+1)u(x)p(x)=1 For reference, the exact solution can be found and is, in general form,u(x)=C1sinx+C2cosx+1and for the given boundary conditions the constants can be evaluatedu(0) = 1 = C2=0u(1) = 0 = C1= 1/sin(1)So the exact solution isu(x)=1 sinxsin(1)Let s solve by the Method of Weighted Residuals using a polynomialfunction as a basis. That is, let the approximating functioneu(x)beeu(x)=a0+a1x+ of the boundary conditions revealseu(0) = 1 =a0eu(1) = 0 = 1 +a1+a2ora1= (1 +a2)and the approximating polynomial which also satisfies the boundaryconditions is theneu(x)=1 (1 +a2)x+a2x2=1 x+a2(x2 x).

8 Tofind the residualR(x), we need the second derivative of this func-tion, which is simplyd2eu/dx2= the residual isR(x)=d2eudx2+eu 1=2a2+(1 x+a2(x2 x)) 1= x+a2(x2 x+2)6 Chapter 2. Method OF Weighted Collocation MethodFor the collocation Method , the residual is forced to zero at a num-ber of discrete points. Since there is only one unknown (a2), only onecollocation point is needed. We choose (arbitrarily, but from symme-try considerations) the collocation pointx= Thus, the equationneeded to evaluate the unknowna2isR( ) = +a2( .5+2)=0 Soa2=+ = 2/7= Subdomain MethodSince we have one unknown constant, we choose a single subdomain which covers the entire range ofx.

9 Therefore, the relation to evalutatethe constanta2isZ101 R(x)dx=0Z10h x+a2(x2 x+2)idx=0" x22+a2(x33 x22+2x) 10=0 Soa2(13 12+2)=12anda2=3/11 = Least-Squares MethodThe weight functionW1is just the derivative ofR(x) with respect tothe unknowna2:W1(x)=dRda2=x2 x+ EXAMPLE7So the Weighted residual statement becomesZ10W1(x) R(x)dx=0Z10 x2 x+2 h x+a2(x2 x+2)idx=0 The math is considerably more involved than before, but nothing morethan integration of polynomial terms. Direct evaluation leads to thealgebraic relation 1112+10130a2=0 Soa2=1112 30101=165/606 = Galerkin MethodIn the Galerkin Method , the weight functionW1is the derivative of theapproximating functioneu(x) with respect to the unknown coefficienta2:W1(x)=deuda2=x2 xSo the Weighted residual statement becomesZ10W1(x) R(x)dx=0Z10 x2 x h x+a2(x2 x+2)idx=0 Again, the math is straightforward buttedious.

10 Direct evaluation leadsto the algebraic equation:112 310a2=0 Soa2=112 103=5/18 = 2. Method OF Weighted Method of MomentsSince we have only one unknown coefficient, the weight functionW1(x)is simplyW1(x)=x0= a result, the Method of moments degenerates into the subdomainmethod for this case. Hence,a2=3/11 = ComparisonA table of the tabulated values resulting from the different approxima-tions is shown in Table below, and a graphical comparison is seenin Figure Figure shows therelativeerrors for each Method , asa percentage of the exact solution.