Weighted Residual Method

8/13/2019 Weighted Residual Method

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Chapter 6

Weighted Residual Methods

Weighted residual methods (WRM) assume that a solution can be approximated analytically or

piecewise analytically. In general a solution to a PDE can be expressed as a superposition of a base

set of functions

T(x, t) =

Nj=1

aj(t)j(x)

where the coefficients aj are determined by a chosen method. The method attempts to minimizethe error, for instance, finite differences try to minimize the error specifically at the chosen grid

points. WRMs represent a particular group of methods where an integral error is minimized in

a certain way and thereby defining the specific method. Depending on the maximization WRM

generate

the finite volume method,

finite element methods,

spectral methods, and also

finite difference methods.

6.1 General Formulation

The starting point for WRMs is an expansion in a set of base or trial functions. Often these are

analytical in which case the numerical solution will be analytical

T(x, y, z, t) = T0(x, y, z, t) +Nj=1

aj(t)j(x,y,z) (6.1)

81


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CHAPTER 6. WEIGHTED RESIDUAL METHODS 82

with the trial functionsj(x,y,z). T0(x, y, z, t)is chosen to satisfy initial or boundary conditionsand the coefficientsaj(t)have to be determined. possible trial functions are

j(x) =xj1 orj(x) = sin(jx).

The expansion is chosen to satisfy a differential equation L(T) = 0(whereTis the exact solution),

e.g.,

L(T) =T

t

2T

x2 = 0

However, the numerical solution is an approximate solution, i.e., T= Tsuch that the operatorLapplied toTproduces a residual

L(T) =R

The goal of WRMs is to choose the coefficients aj

such that the residualRbecomes small (in fact0) over a chosen domain. In integral form this can be achieved with the condition

Wm(x,y,z)Rdxdydz= 0 (6.2)

whereWm is a set of weight functions (m= 1,...M)which are used to evaluate (6.2). The exactsolution always satisfies (6.2) if the weight functions are analytic. This is in particular true also for

any given subdomain of the domain for which a solution is sought.

There are four main categories of weight or test functions which are applied in WRMs.

i) Subdomain method: Here the domain is divided inM subdomainsDmwhere

Wm=

1 in Dm0 outside

(6.3)

such that this method minimizes the residual error in each of the chosen subdomains. Note that the

choice of the subdomains is free. In many cases an equal division of the total domain is likely the

best choice. However, if higher resolution (and a corresponding smaller error) in a particular area

is desired, a non-uniform choice may be more appropriate.

ii) Collocation method: In this method the weight functions are chosen to be Dirac delta func-

tions

Wm(x) =(x xm) (6.4)

such that the error is zero at the chosen nodes xm.


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iii) Least squares method: This method uses derivatives of the residual itself as weight functions

in the form

Wm(x) = R

am. (6.5)

The motivation for this choice is to minimize

R2dxdydzof the computational domain. Notethat this choice of the weight function implies

am

R2dxdydz= 0

for all values ofam.

iv) Galerkin method: In this method the weight functions are chosen to be identical to the base

functions.

Wm(x) =m(x)

In particular if the base function set is orthogonal (

m(x)n(x) = 0 ifm = n), this choice ofweight functions implies that the residualRis rendered orthogonal with the condition (6.2) for allbase functions.

NotethatM weight functions yieldMconditions (or equations) from which to determine the Ncoefficientsaj . To determine these N coefficients uniquely we need N independent condition(equations).

Example: Consider the ordinary differential equation

dy

dx y = 0 (6.6)

for0 x 1withy(0) = 1. Let us assume an approximate solution in the form of polynomials

y= 1 +N

j=1ajx

j (6.7)

where the constant1 satisfies the boundary condition. Substituting this expression into the differ-ential equation (6.6) gives the residual

R= 1 +Nj=1

ajjxj1 xj

(6.8)

We can now compute the coefficientsaj with the various methods.


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Galerkin method:

Here we use the weight functionsWm= xm1. The maximization implies with (6.2)

1

0

xm1 1 +N

j=1 aj jxj1 xj dx= 0or

10

xm1dx+Nj=1

aj

j

10

xm1xj1dx

10

xm1xjdx

= 0

which yields after integration

1

m+

N

j=1aj

j

m+j 1

1

m +j = 0 (6.9)

With the matrix S(with elements smj = jm+j1

1m+j

) and the vector D (with the elements

dm= 1m

) we can rewrite (6.9) as

SA = D (6.10)

The solution A of this set of equations requires to invertS. ForN= 3the system becomes

1/2 2/3 3/41/6 5/12 11/20

1/12 3/10 13/30

a1a2a3

=

11/21/3

For this set the approximate solution becomes

y= 1 + 1.0141x + 0.4225x2 + 0.2817x3

Least squares method:

Here the weight functions are

Wm(x) = R

am=mxm1 xm

This gives the residual condition

10

mxm1 xm

1 +

Nj=1

ajjxj1 xj

dx =

1 + 1

m+ 1+

Nj=1

aj

10

mxm1 xm

jxj1 xj

dx

= 0


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or

1 + 1

m+ 1+

Nj=1

aj

jm

m +j 1

j + m

m +j+

1

m+j+ 1

= 0

Subdomain method:

Weight functions:

Wm =

1 x (m1N ,

mN]

0 outside

Condition for the residual error:

mN

m1N 1 +

N

j=1 aj jxj1 xj dx= 0Collocation method:

Collocation point (for the Dirac delta function): xm= m1N

.

Condition for the residual error:

10

(x xm)

1 +

Nj=1

ajjxj1 xj

dx= 0

6.2 Finite Volume Method

This section illustrate the use of the finite volume method first for a PDE involving first order

derivatives and later for second order derivatives. The program FIVOL will be introduced which

can be used to solve Laplaces equation or Poissons equation, i.e, equations of the elliptic type.

6.2.1 First order derivatives

Let us consider the equationq

t+

F

x +

G

y = 0 (6.11)

which is to be solved with the subdomain method on a domain given by coordinate line (j, k)describing a curved coordinate system. This equation is a typical continuity equation like the

equations for conservation of mass, momentum etc.


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k-1

k

k+1

j-1

j

j+1

A

B

C

D

x

y

Figure 6.1: Illustration of the coordinate system to solve equation (6.11) with the finite volume

method.

Equation (6.11) is to be solved as an integral over any areaABCDas illustrated in Figure 6.1 =>ABCD

q

t+

F

x +

G

y

dxdy= 0.

With H = (F, G) such that F/x+ G/y = H this equation becomes (using Gaussstheorem)

d

dt ABCD qdV + ABCDH nds= 0In Cartesian coordinates the surface element vector ds = (dx, dy) such that the normal vectornds = (dy, dx) and H nds = F dy Gdx. Thus the integral of (6.11) For the areaABCDbecomes

d

dt(Arqjk) +

ABCD

(Fy Gx) = 0

withAr =area ofABCD. Using the notationsyAB = yB yA, xAB = xB xA, and theaveragesFAB = 0.5 (Fj,k1+ Fj,k),GAB = 0.5 (Gj,k1+ Gj,k)and applying these to all sections

ofABCDwe obtain

Ardqjkdt

+0.5 (Fj,k1+ Fj,k) yAB 0.5 (Gj,k1+ Gj,k) xAB+0.5 (Fj,k+ Fj+1,k) yBC 0.5 (Gj,k+ Gj+1,k) xBC+0.5 (Fj,k+ Fj,k+1) yCD 0.5 (Gj,k+ Gj,k+1) xCD+0.5 (Fj1,k+ Fj,k) yDA 0.5 (Gj1,k+ Gj,k) xDA = 0

(6.12)


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In case of a uniform grid parallel to the x, and the y axes the area isAr = xy and equation(6.12) reduces to

dqjkdt

+Fj+1,k Fj1,k

2x +

Gj+1,k Gj1,k2y

= 0

where the spatial derivative is equal to that for the centered space finite difference approximation.

6.2.2 Second order derivatives

To introduce the finite volume second derivatives let us consider Laplaces equation

2

x2+

2

x2 = 0 (6.13)

k-1

k

k+1

j-1

j

j+1

A

B

C

D

x

y

A

B

C

D

x

y

XW

Y

Z

rWZ

rXY

(A)

(B)

Figure 6.2: Illustrations of the domain for the solution of Laplaces equation (A), and of the grid

geometry to evaluate the second derivatives for the finite volume method.

We are seeking a solution for this equation in a domain as illustrated by the shaded area in

Figure~6.2. The appropriate coordinate system for this domain is a polar coordinate system with

the variabler and . The boundary conditions are

= 0at boundaryW X,

= sin /rxy at boundaryXY,

= 1/ryz at boundaryY Z,

= sin /rwz at boundaryW Z.


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Figure 6.3: Illustration of the geometry of the elementsABCD.

and similar forx. Figure 6.3 shows that the area is approximated by

SAB =SABCD = (xAB xk1,k)(yAB+ yk1,k)

xAByAB+ xk1,kyk1,k

= xAByk1,k yABxk1,k

With this we obtain:

x

j,k1/2

= yAB(j,k1 j,k) + yk1,k(B A)

SAB

y

j,k1/2

= xAB(j,k1 j,k) + xk1,k(B A)

SAB


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Similarly we obtain

x

j+1/2,k

= yBC(j+1,k j,k) + yj+1,j(C B)

SBC

y j+1/2,k = xBC(j+1,k j,k) + xj+1,j(C B)SBC

x

j,k+1/2

= yCD(j,k+1 j,k) + yk+1,k(D C)

SCD

y

j,k+1/2

= xCD (j,k+1 j,k) + xk+1,k(D C)

SCD

x

j1/2,k

= yDA(j1,k j,k) + yj1,j(A D)

SDA

y j1/2,k=

xDA(j1,k j,k) + xj1,j(A D)

SDA

Substitution back into equation (6.15) gives

QAB(j,k1 j,k) + QBC(j+1,k j,k) + QCD(j,k+1 j,k) + QDA(j1,k j,k)

+PAB(B A) + PBC(C B) + PCD(D C) + PDA(A D) = 0

with

QAB =

x2AB+ y2AB

/SAB , PAB = (xABxk1,k+ yAByk1,k) /SAB

QBC=

x2BC+ y

2BC

/SBC , PBC= (xBCxj+1,j+ yBCyj+1,j) /SBC

QCD = x2CD + y2CD /SCD , PCD = (xCDxk+1,k+ yCDyk+1,k) /SCDQDA =

x2DA+ y

2DA

/SDA , PDA = (xDAxj1,j + yDAyj1,j) /SDA

Finally we evaluateA,xA, andyAas the average over the surrounding nodes, e.g.,

A= 0.25(j,k+ j1,k+ j1,k1+ j,k1)

Substitution into our main equation then yields

0.25 (PCD PDA) j1,k+1+ 0.25 (PBC PCD) j+1,k+1

+0.25 (PAB

PBC

) j+1,k

1+ 0.25 (P

DA P

AB) j

1,k

1+ [QCD + 0.25 (PBC PDA)] j,k+1+ [QDA+ 0.25 (PCD PAB)] j1,k

+ [QAB+ 0.25 (PDA PBC)] j,k1+ [QBC+ 0.25 (PAB PCD)] j+1,k

(QAB+ QBC+ QCD + QDA) j,k = 0 (6.16)

Here the coefficients can be determined initially and then used during the computation. Equation

(6.16) is solved in the program FIVOL using successive over-relaxation (SOR). The estimate for is determined from (6.16)


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j,k = {0.25 (PCD PDA) j1,k+1+ 0.25 (PBC PCD) j+1,k+1 (6.17)

+0.25 (PAB PBC) j+1,k1+ 0.25 (PDA PAB) j1,k1

+ [QCD + 0.25 (PBC PDA)] j,k+1+ [QDA+ 0.25 (PCD PAB)] j1,k

+ [QAB+ 0.25 (PDA PBC)] j,k1+ [QBC+ 0.25 (PAB PCD)] j+1,k}n

/ (QAB+ QBC+ QCD + QDA)

The iteration step is completed with

n+1j,k =nj,k+ (

j,k

nj,k) (6.18)

Note that the discretized equation (6.16) reduces to centered finite differences on a uniform rect-

angular grid

j1,k 2j,k+ j+1,kx2

+j,k1 2j,k+ j,k+1

y2 = 0 (6.19)

6.2.3 Program Fivol

The finite volume method as described above is implemented in the program Fivol. The program

parameter used are summarized in Table (6.1).

Table 6.1: Program parameter for the program Fivol.

Parameter Description

nr,ntheta Number of grid points inr and directionsniter Maximum number of iterations

eps Tolerance for the iterated solution

om Relaxation parameterrms RMS error

rw, rx, ry, rz radial distance to points w, x, y, and z

theb, then Min and max values in the directionx,y xand y coordinates of the grid

r,theta r and coordinatesdr,dtheta increments in ther and directions

qab, pab, qbc, pbc, qcd, pcd, qda, pda Weights for the iterationsphi Iterated solution

phix Exact solution

The program reads the input parameters niter, rw, rx, ry, rz, theb, then, eps, om from the data

file fivol.dat. Program parameters are defined in the include file fivin. The program then writes

parameters to a data file fivol.out, generates the grid, initializes the iteration through an initial state

(for phi), boundary conditions, and calculates the matrix coefficients. Subsequently a subroutine


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SOR is called until the iterated solution tolerance is reached. The result is written to a binary file

which can serve as input for the plotting routine plofivol.

The solution error for the finite volume method is listed in Table (6.2) for different number of grid

points in the same domain. The table illustrates that the solution error is second order. This is to

be expected from the centered differences to which the finite volume method reduces on a uniform

rectangular grid. However, for strongly distorted grids the solution error is larger than second

order.

Table 6.2: Solution errors for the finite volume method in program Fivol2.f

Grid | exact|rms No of iterations

66 0.1326 151111 0.0471 192121 0.0138 51

The finite volume method is well suited for for somewhat irregular grid domains and does not

require an orthogonal grid. The number of iterations for convergence depends on the domain size.

6.3 Finite Element Method

Finite element methods are used mostly in engineering. For many problems the finite element

method can be interpreted as a maximization of the potential energy of a system. In most applica-

tions the finite element method is used with the Galerkin formulation for the weighted residuals.

The approximating functions are simple polynomials defined in local domains.

T=Nj=1

Tjj(x,y,z) (6.20)

Where for a suitable set of functions the local domains can be of any shape. Since the finite element

method is used with local coordinates the domains can be subdivided (into same shape domains)

to increase the resolution where it is desired. The interpolating functions are called trial or shape

functions.

6.3.1 Basic formulation

a) Linear Interpolation

The linear interpolation uses linear functions which are 1 at the nodal point, assume 0 at the neigh-

boring points, and are identical to 0 outside the domains (xj1, xj+1)as illustrated in Figure 6.4.

The shape function for the nodal pointj is


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Figure 6.4: Illustration of one-dimensional linear finite elements.

j =

xxj1xjxj1

f or xj1 x xj (element A)xj+1x

xj+1xjf or xj x xj+1 (element B)

(6.21)

and j = 0 outside of elements A and B. The shape function j overlaps only with its directneighbors and the superposition of elements between nodal points yields a linear function in these

regions

in A: T =Tj1j1+ Tjj (6.22)

in B : T =Tjj+ Tj+1j+1 (6.23)

In any given element only two shape functions overlap. In element A j is given element by (6.21)andj1is given by

j1= xj x

xj xj1

Similar for any element Bj is given element by (6.21) andj+1is given by

j+1= x xjxj+1 xj

.

The particular form of the trial functions makes it straightforward to use them to approximate any

given functionf(x).Since the trial functions are 1 at the nodal points and all except for one trialfunctions are 0 at any nodal point Thus the expansion of a function f(x) in terms of the shapefunctions is given by

f(x) =Nj=1

fjj(x) (6.24)

with the coefficients

fk = f(xk) (6.25)


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where thexk are the nodal points. Formally this is seen byf(xk) =Nj=1 fjj(xk) =fk

The method is illustrated using the function

f(x) = 1 + cos(x/2) + sin(x) (6.26)

f

x

2.0

1.0

0.5 1.0

quadraticinterpolation

linear interpolation

Figure 6.5: Linear (dashed)and quadratic (dotted) finite element approximation of function (6.26).

in the range [0, 1]. With two elements and nodal points at(0, 0.5, 1.0) the expansion coefficientsaref1 = 2,f2 = 2.7071, andf3 = 0. Figure (6.5) shows the functionf(x)and the finite elementinterpolation for linear and quadratic elements.

b) Quadratic interpolation

Quadratic interpolation requires simple quadratic polynomials for trial function. Again a trial

function should be 1 only at the corresponding nodal point and 0 at all other nodes.

Figure 6.6: Illustration of one-dimensional linear finite elements.

Following the illustration 6.6 the trial functions are defined by

j =

0 f or x xj2, x xj+2

xxj2xjxj2

xxj1xjxj1

f or xj2 x xj , element Axxj+2xjxj+2

xxj+1xjxj+1

f or xj x xj+2, element B(6.27)

With this form j(xj) = 1 and j(xj2) = j(xj1) = j(xj+1) = j(xj+1) = 0. The trialfunctions of this form extend over 5 nodes. such that in a given interval 4 trial functions overlap


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if they were all chosen of the same form. This is improved by choosing different trial functions at

xj1and atxj+1of the form:

j = x xj1xj xj1

x xj+1xj xj+1

for xj2 x xj (6.28)

and 0 otherwise. Note that this creates a structure where all odd nodes have elements of the form

of (6.27) and all even elements are of the form (6.28) where we start with an index of 1 for the first

node (xmin boundary).

Similar to the linear elements it is straightforward to expand any given function in term of these

shape functions

f(x) =Nj=1

fjj(x)

with the coefficientsfk =f(xk)as in the case of the linear elements.

withTj = f(xj). With these functions the interpolation in elements A and B take the followingform

T =

Tj2j2+ Tj1j1+ Tjj in element ATjj+ Tj+1j+1+ Tj+2j+2 in element B

The specific form of the shape functions which are nonzero in any element A is

j2 = x xjxj2 xj

x xj1xj2 xj1

j1 =

x xj2xj1 xj2

x xjxj1 xj

j =

x xj2xj xj2

x xj1xj xj1

For the special function (6.26) with only three nodes we have only one element A and no element

B. As before the expansion coefficients aref1 = 2,f2 = 2.7071, andf3 = 0.

The errors for the linear and quadratic finite element approximation of function (6.26) is listed inTable 6.3. While the linear approximation scales quadratic in the error the quadratic interpolation

scales cubic in the resolution (inverse number of nodes).

c) Two-dimensional interpolation

Linear elements: The particular strength of finite elements is there flexibility in two and three

dimensions. Based on the previous introduction we want to illustrate the use of linear and quadratic


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Table 6.3: Error for linear and quadratic finite element interpolation for the function in (6.26).

Linear Interpolation Quadratic interpolation

No of elements RMS error No of elements RMS error

2 0.18662 1 0.04028

4 0.04786 2 0.015996 0.02138 3 0.00484

8 0.01204 4 0.00206

finite elements in two and three dimensions. In two dimensions a trial function centered at (xj , yj)spans four elements A, B, C, and D. The approximate solution in this region can be conveniently

written with local element based coordinates(, )as

T =

4

l=1Tll(, ) (6.29)

with 1 1and 1 1. The approximating functionsl(, )in each element are ofthe form

l(, ) = 0.25(1 + l)(1 + l) (6.30)

withl = 1andl = 1or explicit

1 = 0.25(1 )(1 )

2 = 0.25(1 + )(1 )

3 = 0.25(1 + )(1 + )

4 = 0.25(1 )(1 + )

The nodal geometry and local coordinate systems are illustrated in Figure 6.7.

=1 =1

=1

=1

Figure 6.7: Illustration of the nodal geometry and sketch of the two-dimensional linear trial func-

tions.


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A solution is constructed separate in each element A, B, C, and D where continuity is provided by

the overlapping shape functions. For instance a solution in element A implies that shape functions

with values of 1 along the boundary to element B overlap into the region of element B and similar

for all other boundaries of element A.

Bi-quadratic elements:

Similar to the bilinear elements the approximate solution for bi-quadratic elements in this region

can be conveniently written with local element based coordinates (, )as

T =9l=1

Tll(, ) (6.31)

with 1 1and 1 1. The approximating functionsl( )in each element dependon the location.

Figure 6.8: Illustration of the nodal geometry for bi-quadratic elements.

Specifically the form of the shape function depends on how they are centered, i.e., where they

assume a value of 1.

l(, ) = 0.25l(1 + l)l(1 + l) corner nodes (6.32)

l(, ) = 0.5(1 2)l(1 + l) midside nodes (l = 0) (6.33)

l(, ) = 0.5(1

2

)l(1 + l) midside nodes (l = 0) (6.34)

l(, ) = 0.5(1 2)(1 2) internal nodes (6.35)

Again solutions are constructed in each element A, B, C, and D where the shape functions on the

side and corner nodes overlap into the adjacent region. Note that quadratic elements require at 3

nodes in each direction and the total number of nodes has to be odd to accommodate quadratic

elements.


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6.3.2 Finite element method applied to the Sturm-Liouville equation

Here we will use the Galerkin finite element method to discretize and solve the Sturm-Liouville

equation

d2ydx2

+ y= F(x) (6.36)

subject to boundary conditionsy(0) = 0anddy/dx(1) = 0and

F(x) = Ll=1

alsin ((l 0.5)x) . (6.37)

The exact solution to this problem forF(x)is given by

y(x) =

Ll=1

al

1 ((l 0.5))2sin ((l 0.5)x) . (6.38)

The trial solution is as before

y= Nj=1

yjj.

Linear Interpolation:

In an element based coordinate system we use local coordinates with

j = 0.5(1 + ) and = 2x

xj1+xj

2 xj

for element A

j = 0.5(1 ) and = 2x

xj+xj+1

2

xj+1

for element B

withxj =xj xj1, andxj+1= xj+1 xj . The residual is

R= d2ydx2

+ y F(x)and the weight function for the weighted residual is wm= mwhich yields the equation 1

0

m(x)d2y

dx2+ y F(x) dx= 0

One can integrate the first term brackets by parts 10

md2ydx2

dx=

m

dydx

10

10

dmdx

dydx

dx


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because the boundary conditions imply1 = 0anddy/dx(1) = 0 such that

mdeydx

10

= 0 such

that the residual equation becomes

N

j=1 yj

1

0

dmdx

djdx

+ mj

dx

=

1

0

mF(x)dx

or in matrix form

BY = G

with the elements

bmj =

10

dmdx

djdx

+ mj

dx

gm =

10

mF(x)dx.

In the computation of these elements it is convenient to make use of the local coordinates. It

should also be noted that the residual equation only has contributions for j =m 1,j =m, andm= j + 1. With the transformation to local coordinates we have in element A

d

dx =

d

dx

d

d =

2

xj

d

d

dx = dx

dd=

xj2

d

and in element B

d

dx =

d

dx

d

d =

2

xj+1

d

d

dx = dx

dd=

xj+12

d

withxj =xj xj1and xj+1= xj+1 xj .

Elementbm,m1: A contribution exists only in region A for node m (corresponding to a region Bfor nodem 1). Thus

bm,m1 = 10dmdx dm1dx + mm1 dx

=

11

dAdx

dAd

dBd

+ dx

dAAB

d

= 1

2xm

11

d+xm

8

11

(1 2)d

= 1

xm+

xm6


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Similar the element bm,m+1 involves only overlap in region B of element m with region A ofelementm+1. The expression is the same only that it regards the intervalxm+1. Thus

bm,m+1 = 1

xm+1+

xm+16

Elementbm,m: Here we have overlap ofmwith itself in regions A and B:

bm,m =

10

dmdx

dmdx

+ mm

dx

=

11

d

dx

dAd

2+

dx

d2A

A

d+

11

d

dx

dBd

2+

dx

d2B

B

d

= 1

1 2

xm 1

22

+xm

2 1 +

2 2

A d+

11

2

xm+1

1

2

2+

xm+12

1

2

2A

d

= 1

xm

1

xm+1+

xm8

11

(1 + )2d+xm+1

8

11

(1 )2d

= 1

xm

1

xm+1+

xm+ xm+13

In summary the nonzero elementsbmj for2 m N 1are

bm,m1 = 1

xm+

xm6

bm,m =

1

xm+

1

xm+1

+

xm+ xm+13

bm,m+1 = 1

xm+1+

xm+16

form= N

bN,N1 = 1

xN+

xN6

bN,N = 1

xN+

xN6

bN,N+1 = 0.


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The inhomogeneityF(x)is known analytically such that one can evaluate10

mF(x)dxdirectly.However, in more complex situations it is more convenient to interpolate F(x) through the trialfunctions

F(x) =

N

j=1Fjj.

such that

gm=Nj=1

Fj

10

mjdx

For the linear interpolation this yields

gm=xm

6 Fm1+

xm+ xm+13

Fm+xm+1

6 Fm+1

Finally consider the special case of a uniform grid with xm = xm+1 = x. In this case theequation for the coefficients becomes

ym1 2ym+ ym+1x2

+

1

6ym1+

2

3ym+

1

6ym+1

=

1

6Fm1+

2

3Fm+

1

6Fm+1

Exercise: Determine the elements at the min and max boundary, i.e., b1,1,b1,2, andb2,1.

Elements withm = 1(y1 = 0): There is no equation for y1 needed becausey1 = 0such that the

indices for the array dimensions decrease by 1.

Elements withm= 2

y2

10

d2dx

d2dx

+ 22

dx + y3

10

d2dx

d3dx

+ 23

dx=

10

2F(x)dx

Elements withm= N

yN

1 1

0 dNdx dN1dx + NN1 dx + yN 1

0 dNdx dNdx + NN dx= 1

0

N

F(x)dx

Quadratic interpolation:

The program offers the possibility to use a linear or quadratic finite element interpolation. For the

second case the equations for thebmj are


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bm,m2 = 1

6xm

xm15

bm,m1 = 4

3xm

+2xm

15bm,m =

7

6

1xm

+ 1

xm+1

+

4

15(xm+ xm+1)

bm,m+1 = 4

3xm+1+

2xm+115

bm,m+1 = 1

6xm+1+

xm+115

and

gm = xm

15 Fm2+

2xm15

Fm1+ 4

15(xm+ xm+1) Fm

+2xm+1

15 Fm+1

xm15

Fm+2

A the boundarym= None obtains

bN,N2 = 1

6xN

xN15

bN,N1 = 43xN

+2xN15

bN,N = 7

6

1

xN+

4

15xN

gN = xN

15 FN2+

2xN15

FN1+ 4

15xNFN

For the equations at midside nodes one obtains

bm,m1 = 4

3xm +2xm

15

bm,m = 8

3xm+

16xm15

bm,m+1 = 4

3xm+1+

2xm+115

gm = 2xm

15 Fm1+

16xm15

Fm+2xm

15 Fm+1


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The program Sturm solves the Sturm-Liouville equation (6.36) for the boundary condition and

the inhomogeneity defined at the beginning of this section. The parameter int determines linear

or quadratic finite element interpolation. The matrix inversion forBY = G is solved with theThomas algorithm (explained in Chapter 7) which is a special case of Gauss elimination for the

case of tridiagonal banded matrices. The subroutines Bianca and ban sol factorize and solve banded

matrices (tridiagonal or pentadiagonal;. The program can be downloaded from the website. Forthe program the following values for aiwere chosen:

a1= 1.0, a2= 1.3, a3= 0.8, a4= 0.2, a5= 1.6

The RMS error is defined as

||y yexact|| =

1

N 1

N1i=1

(yi yexact,i)2

1/2

Table summarizes the solution error for the Sturm-Liouville equation.

Table 6.4: RMS solution error for the Sturm Liouville equation.

Gridx Linear interpolation Quadratic interpolation

1/4 0.014 0.30

1/8 0.0039 0.0017

1/16 0.00093 0.000072

1/24 0.00040 0.000022

Notes:

Higher order interpolation on a coarse grid is not much better or can be worse than linearinterpolation.

The solution error for linear interpolation decreases approximately with x2 and for quadraticinterpolation it decreases with x3.

Smaller grid spacing than listed in the table requires higher machine accuracy (i. e., variablesneed to be defined as double precision.

Sturm-Liouville problem, fem: quadratic interpolation

nx= 9 a= 0.10E+01 -0.13E+01 0.80E+00 -0.20E+00 0.16E+01

i= 2 x=0.12500 y=0.11654 yex+0.11611 dy=0.00044i= 3 x=0.25000 y=0.21361 yex+0.21262 dy=0.00099

i= 4 x=0.37500 y=0.31541 yex+0.31575 dy=-.00034

i= 5 x=0.50000 y=0.43428 yex+0.43608 dy=-.00180

i= 6 x=0.62500 y=0.54579 yex+0.54528 dy=0.00051

i= 7 x=0.75000 y=0.64174 yex+0.63904 dy=0.00270

i= 8 x=0.87500 y=0.72503 yex+0.72543 dy=-.00040

i= 9 x=1.00000 y=0.76235 yex+0.76568 dy=-.00333

rms= 0.171E-02 nx= 9


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Table 6.5: Parameters used in program Sturm.f

Parameter Description

nx number of grid points

nterm number of terms in the inhomogeneity

x grid forxy, yex numerical and exact solutionb coefficients matrix

g coefficients for the inhomogeneity

f function F

a coefficients a

fd only used in ban sol

6.3.3 Further applications of the finite element method

Diffusion equation

Consider the diffusion equation

T

t

2T

x2 = 0

and linear finite element interpolation on a uniform grid. In this case the second derivative is treated

as the second derivative in the Sturm-Liouville equation. The time derivative can be taken out of

the integrals for the shape functions such that it is treated as the second term or the inhomogeneity

in the Sturm-Liouville equation. Thus the linear finite element method leads to the equation

1

6

dT

dt

i1

+2

3

dT

dt

i

+1

6

dT

dt

i

x2(Ti1 2Ti+ Ti+1) = 0

Here the form of the time derivative has not yet been specified. Also this form allows the freedom

to evaluate the second derivative at a suitable time level. Defining Tn+1 =Tn+1 Tn and timederivatives asdT/dt= Tn+1/tand using a parameterto control the time level at which thesecond derivative is evaluated we obtain

1

6

Tn+1i1t +

2

3

Tn+1it +

1

6

Tn+1i+1t

(1 )

Tni1 2Tni + T

ni+1

x2 +

Tn+1i1 2Tn+1i + T

n+1i+1

x2

= 0 (6.39)

Defining operators


25/37


Mx =

1

6, 2

3, 1

6

Lxx =

1

x2

, 2

x2

, 1

x2

we can write (6.39) in a more compact form

MxTn+1i

t

(1 ) LxxT

ni + LxxT

n+1i

= 0 (6.40)

Comparison with the finite difference method suggests that the main difference to the finite ele-

ment method is the distribution of the time derivative over adjacent nodes. The above equation is

reminiscent of the general two-level scheme introduced earlier. This scheme is recovered by using

the finite difference mass operator

Mfdx = (0,1, 0)

Finally (6.40) can be cast into the equation

(Mx tLxx) Tn+1i = [Mx t (1 ) Lxx] T

ni (6.41)

which is an implicit equation forTn+1. Note

for= 0the finite difference method generates an explicit method while the finite elementmethod yields an implicit algorithm;

the matrix defined byMx tLxx is tridiagonal and can be solved by the Thomas algo-rithm (as in the case of Sturm.f);

the symmetry ofMxand Lxx allows to construct an explicit scheme for t= x2/ (6);

equation (6.41) is consistent with the diffusion equation and unconditionally stable for 0.5.

Viscous flow

Stationary viscous flow through a rectangular cross section (in the x, y plane, Figure ) can bedescribed by the thezcomponent of the momentum equation

p

z =

2w

x2 +

2w

y2

(6.42)

where the first term is the pressure gradient which drives the flow and w is the velocity profile ofthe flow. The problem is actual three-dimensional but is the pressure gradient is known or can be


26/37


prescribed one can use (6.42) to determine the flow profile in the cross section of the duct. In many

problems it is an advantage to normalize the basic equation. Sincex [a, a]and y [b, b]weuse as normalization

x= xa , y= yb , w= ww0substitution in (6.42) yields

p

z =

w0b2

b

a

22 wx2 + 2 wy2

such that the choicew0= / (b

2p/z)yields

b

a2

2

w

x2 +2

w

y2 + 1 = 0 (6.43)with the boundary conditionsw= 0at x= 1and y = 1.

Similar to the procedure for the Sturm-Liouville equation and the diffusion equation we introduce

the solution

w= Ni=1

wii(x, y)

This yields the residualR =N

i=1 wi

(b/a)2 2i/x2 + 2i/y

2

+ 1 and using the Galerkinmethod generates the equation

Ni=1

wi 11

11

ba2 2i

x2 +

2

iy2

mdxdy = 11

11

mdxdy

orNi=1

wi

11

b

a

2ix

m

x=1x=1

dy+

11

iy

m

y=1y=1

dx

Ni=1

wi

11

11

b

a

2ix

mx

+ i

y

my

dxdy =

11

11

mdxdy.


27/37


For the chosen boundary conditions the first two integral are zero such that the resulting equation

can be written as

BW = G (6.44)

with

bmi =

11

11

b

a

2ix

mx

+ i

y

my

dxdy (6.45)

gm =

11

11

mdxdy (6.46)

In the following we replace the single index iwith a pairk, lrepresenting thexandy coordinates.The indicesk andl range from 2 to nx1 and from 2 to ny1 respectively because the boundary

condition imply that the coefficient of the shape function at the boundaries is zero. Thex integra-tion of the first term in (6.45) yields the operator Lxx and they integration of this term yield theoperatorMy with

Lxxwk,l = wk1,l 2wk,l+ wk+1,l

x2 (6.47)

Mywk,l = 1

6wk,l1+

2

3wk,l+

1

6wk,l+1 (6.48)

such that equation (6.44) assumes the form

ba

2My Lxx+ Mx Lyy

wk,l = 1 (6.49)

with

My Lxxwk,l = 1

6

wk1,l1 2wk,l1+ wk+1,l1

x2

+

2

3

wk1,l 2wk,l + wk+1,l

x2

+

1

6

wk1,l+1 2wk,l+1+ wk+1,l+1

x2

(6.50)

Mx Lyywk,l = 16wk1,l1 2wk1,l+ wk1,l+1

y2 +2

3wk,l1 2wk,l + wk,l+1

y2

+1

6

wk+1,l1 2wk+1,l+ wk+1,l+1

y2

(6.51)

and similar for MxLyywk,l. In a corresponding finite difference approach we would have Mx,fd =(0, 1, 0).

Notes:


28/37


The operator Lxx and My are commutative. The result does not depend on the order theintegration in (6.47) is executed.

The method provides a straightforward way to employ finite element methods similar tofinite difference methods.

The result provides a simple equation which can be used with SOR to solve the coefficients for the

shape functions. Solving equation (6.49) for wk,l yields

wk,l = 3

4c0[1. + c1(wk1,l1+ wk+1,l1+ wk1,l+1+ wk+1,l+1)

+c2(wk1,l+ wk+1,l) + c3(wk,l1+ wk,l+1)]

with

c0 =

b

a

21

x2+

1

y2

c1 = 1

6c0

c2 =

b

a

22

3x2

1

3y2

c3 =

b

a

21

3x2+

2

3y2

With the usual SOR method the solution is iterated using

wn+1k,l =wnk,l+

wk,l w

nk,l

The program duct.f can be found on the course web page. To provide a reference for the approxi-

mate solution the exact solution is given by

w= 8

22 N

i=1,3,5..N

j=1,3,5.. (1)(i+j)/21

ij iba 2 +j2cosix

2 cosjy

2 A second important measure for this problem is the total flow rate q, i.e., the flow integrated overthe cross sectional area

q= 2

8

2

3 Ni=1,3,5..

Nj=1,3,5..

1i2j2

iba

2+j2


29/37


Table 6.6: Input parameters for the program duct.f

Variable Value Description

me 2 method: 1-linear elements, 2-3pt cent dif:

nem 25 no terms in exact solution

ipr 1 ipr=0, prints solutions to the output file duct.outniter 800 max no iterations for SOR

bar 1.0e-0 aspect ratio a/b for channel

eps .1e-5 tolerance par for SOR

om 1.5e0 relaxation parameter (SOR),

The structure is similar to that of Fivol.f. The program package uses an include file ductin which

declares all variables and common blocks and it make use of a parameter file duct.dat. Variables

declared in the parameter file are summarized in Table 6.6.

Additional variables used in the program are listed in Table 6.7.

Table 6.7: Other variables used in the program duct.f

Variable Description

nx, ny Number of nodes in thexandy directionsx, y xandy coordinates

dx, dy grid separation in thexandy directionsflow, flowex iterated flow rate and exact flow rate

f, fexact iterated and exact solutions

rms RMS error

The program generates an ASCII output file which lists basic parameter and results including the

iterated solution and the exact solution and the correspond flow rates and errors. In addition the

program generates a binary file which can be used as input for graphics routines. The supplied IDL

program product.pro reads this this file and plots the iterated and exact solutions.

The results of this program indicate that the error decreases with x2 andy2. The result for usingthe finite element method (fem) are very similar to those of the finite difference method (fdm). The

error is very similar with slightly smaller errors for fdm while the flow rate is slightly closed to the

exact flow rate for fem.

Distorted computational domains - Isoparametric mapping

The finite element method is very comparable to finite differences for Cartesian grids. However the

strength or the finite element method is the ease to apply it to distorted domains and complicated

geometries. It is illustrated that the introduction of local coordinates is a particular strength to eval-

uate the integrals involving shape functions. There are various geometries to choose basic elements

such as triangular, rectangular, or tetrahedral. The various method allow a simple grid refinement

technique by just dividing a basic element into several new element of the same geometry.


30/37


The advantage of the finite element method is that the coordinates themselves can be described by

the shape functions. Figure shows a distorted grid with rectangular elements and the mapping into

a local(, )coordinate system.

=1 =+1

=1

=+1

Figure 6.9: Isoparametric mapping.

The transformation between(x, y)coordinates and(, )can be defined by

x=

4l=1

l(, ) xl and y=

4l=1

l(, ) yl (6.52)

with(xl, yl)as the x, ycoordinates of the corner numbered land l(, )is the shape function withthe value of 1 at the cornerl. The transformation affects the evaluation of the weighted residualintegral. Consider Laplaces equation = 0 as an example. The Galerkin fem produces a

system of linear equations

BW = G

with

bm,i =

area

ix

mx

+ i

y

my

dxdy

Let us consider the first term

I=

ix

mx

dxdy

The derivative /xcan be computed as

=

x

x+

y

y


31/37


such that/xand/y are related to/and / by

//

= [J]

/x/y

(6.53)

with [J] = x/ y/

x/ y/

(6.54)

where[J] is the Jacobian. The derivatives in the Jacobian can easily be computed from the map-ping (6.52), e.g.,

y

=

4l=1

l

(, )

yl

From (6.53) one can determine the following explicit formulation for/x

ix

= 1

detJ

y

i

y

i

(6.55)

Exercise: Derive equation (6.55).

Usingdxdy= detJ ddone obtains

I= 1

1 1

1

1

detJ y

i

y

i

y

m

y

m

ddAll derivatives in this formulation are known because of the simple form that thel assume. Theintegral can evaluated numerically or analytically.

6.4 Spectral method

The Galerkin spectral method is similar to the traditional Galerkin method but uses a suitable set

of orthogonal functions such that

kldx

= 0 for k =l= 0 for k = l

Example for such functions are Fourier series, Legendre, or Chebyshev polynomials. In this sense

the spectral methods can be considered global methods rather than local as in the case of finite

differences or finite elements.


32/37


6.4.1 Diffusion equation

Consider as an example the diffusion equation

T

t =

2T

x2

forx [0, 1]and the boundary and initial conditions

T(0, t) = 0, T(1, t) = 1, and T(x, 0) = sin (x) + x

Approximate solution

T= sin (x) + x+N

j=1 aj(t)sin(jx)where the aj(t)are the unknown coefficients which need to be determined. This yields the residual

R=Nj=1

dajdt

+ (j)2 aj

sin(jx) + 2 sin(x)

Evaluation of the weighted integral

R sin(mx) dxyields

Nj=1

dajdt

+ (j)2 aj 1

0

sin(jx)sin(mx) dx+ 2 10

sin(x)sin(mx) dx= 0

The integral yield

10

sin(jx)sin(mx) dx =

1/2 for j=m

0 for j =m

1

0

sin(x)sin(mx) dx = 1/2 for m= 1

0 for m = 1

which yields

damdt

+ (j)2 am+ rm = 0, m= 1,...,N

rm =

2 for m= 1

0 for m = 1


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with the solution

a1 = e2t 1

am = 0

The solution forT is therefore

T= sin (x) e2t + x

This is in fact the exact solution, however, this has been obtained for this particular set of initial

and boundary conditions. For a more realistic case of

T(x, 0) = 5x 4x2

and replacingsin (x) + xwith5x 4x2 in the trial solution one obtains

rm=

16m for m= 1, 3, 5,...0 for m= 2, 4, 6,...

With forward time differencingdam/dt= (an+1m a

nm) /tone obtains

an+1m =anm t

(j)2 anm+ rm

Table 6.8: Value of the solution at x= 0.5for different times

t N=1 N=3 n=5 Exact solution

0 1.5 1.5 1.5 1.5

0.1 0.851 0.889 0.881 0.885

0.2 0.610 0.648 0.640 0.643

Table 6.8 shows the value ofT for selected times and different values ofN atx = 0.5. Note thatthe error is caused by two sources one of which is the limited accuracy of the representation in

terms of the base functions and the other is the error in the temporal integration.

Notes:

The spectral method achieves relatively high accuracy with relatively few unknowns.

For Dirichlet conditions the method reduces the problem to a set of ordinary differentialequations

Treatment of nonlinear terms is not yet clear (and can be difficult)


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6.4.2 Neumann boundary conditions

Let us now consider the diffusion equation again, however, for Neumann boundary conditions.

Specifically the initial and boundary conditions are

T(x, 0) = 3 2x 2x2 + 2x3,

T

x(0, t) = 2.0 and T(1, t) = 1.0

Different from the prior example it is not attempted to incorporate the initial and boundary condi-

tions into the approximate solution. Rather a general trial solution is used

T(x, t) =b0(t) +N

j=1[aj(t)sin(j2x) + bj(t)cos(j2x)]

The boundary conditions require the following relations for the coefficients

Nj=1

aj2j = 2 ,Nj=0

bj = 1

which can be used to eliminateaN andbNfrom the the approximate solution:

aN= 1N

N

1j=1

ajjN

, bN= 1 N

1j=0

bj

This yields for the approximate solution

T(x, t) = cos (N2x) 1

Nsin(N2x) +

N1j=1

aj(t)

sin(j2x)

j

Nsin(N2x)

+

N1

j=0

bj(t)[cos(j2x) + cos (N2x)]

This implies that the terms in square brackets now need to be considered as the base functions or

wma = sin (m2x) m

Nsin(N2x) , 1 m N 1

wmb = cos (m2x) + cos (N2x) , 0 m N 1


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Substitution into the the diffusion equation yields the residual

R =N1

j=1 dajdt

+ (2j)2 aj

sin(2jx)

dajdt

+ (2N)2 aj

j

Nsin (2N x)

+

N1j=0

dbjdt

+ (2j)2 bj

cos (2jx) +

dbjdt

+ (2N)2 bj

cos(2N x)

+ (2N)2 cos (2N x) 4Nsin (2N x)

Evaluating the weighted residuals yields

damdt

+ (2m)2 am+m

N

N1

j=1j

N dajdt

+ (2N)2 aj = 4N (6.56)

dbmdt

+ (2m)2 bm+N1j=0

dbjdt

+ (2N)2 bj

= (2N)2 (6.57)

2db0dt

+N1j=0

dbjdt

+ (2N)2 bj

= (2N)2 (6.58)

Equations (6.56) foramare linearly independent from equations (6.57) and can be solved indepen-dently. To solve for the time integration it is necessary, however, to factorize the the system and

carry out a matrix multiplication for each time step. Start values foramand bmare easily obtained

using the trial solution for the initial condition.Note, since the problem is linear the factorization is only needed once because the subsequent time

steps always use the same matrix (coefficients in equations (6.56) to (6.58) are constant).

6.4.3 Pseudospectral method

Main disadvantage of spectral method is large computational effort particularly for nonlinear prob-

lems. An alternative to Galerkin method which basically solve the diffusion equation in spectral

space is to use a mixture of spectral and real space. An frequently used method in this category is

the pseudospectral approach which uses the collocation method. The example is again the diffusion

equation

u

t

2u

x2 = 0

subject to the boundary and initial conditions

u (1, t) = 1, u (1, t) = 1, u (x, 0) = sin x+ x


36/37


When compared to the spectral method, the pseudospectral method does not determine the solution

entirely in spectral space. Recall that the solution for the Galerkin spectral method is found by

integrating the spectral coefficients. For a linear problem this is feasible but a nonlinear problem

requires the inversion of a large matrix for each time step.

The pseudo spectral method often uses an expansion

u (x, t) =N+1k=1

ak(t) Tk1(x)

consists of three basic steps:

Given a solution unj one determines the spectral coefficientsak. This transforms the problemfrom physical to spectral space. Using Chebichev Polynomials the step can be done with a

fast Fourier Transform (FFT) if the collocation points arexj = cos (j1)N

(time efficient

with number of operations proportional toNlog N).

The next step is to evaluate the second derivative 2u/x2 from the coefficients ak whichmakes use of recurrence relations and is also very efficient:

u

x =

N+1k=1

a(1)k Tk1(x)

2u

x2 =

N+1

k=1a(2)k Tk1(x)

The final step then is to integrate in time. Different from the Galerkin spectral method this is

done in real space directly for u: un+1j =unj +t

2ux2

nj

thereby avoiding to have to solve

a large system of equations for the time derivatives ofdak/dt.

6.5 Summary on different weighted residual methods:

local approximations: finite difference, finite volume, finite element

irregular domains

variable grid resolution

boundary conditions relatively simple

global base function: spectral methods

high accuracy


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arbitrary often differentiable

Galerkin method computationally expensive for nonlinear problems

Pseudospectral method employs spectral space only to determine spatial derivatives

(much more efficient)

Documents

Weighted Residual Method