weighted residual method
TRANSCRIPT
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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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 83
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 84
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 85
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 86
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 87
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 89
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 90
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 91
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 92
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 93
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 94
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 95
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 96
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 97
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 98
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 99
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 100
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 101
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 102
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 103
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 104
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
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 105
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
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 106
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.
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 107
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:
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 108
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
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 109
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.
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 110
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
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 111
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.
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 112
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 113
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 114
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 115
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
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CHAPTER 6. WEIGHTED RESIDUAL METHODS 116
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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CHAPTER 6. WEIGHTED RESIDUAL METHODS 117
arbitrary often differentiable
Galerkin method computationally expensive for nonlinear problems
Pseudospectral method employs spectral space only to determine spatial derivatives
(much more efficient)