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Advanced Numerical Analysis Chapter 6

6.4.2 Uniform meshes in 2D

We can compute fpx, yq “ ′?uex explicitly. Moreover, uex “ 0 on B?. Then, we consider the homogeneous

Dirichlet problem:

′ ?u “ f, ?; (6.30a)

u “ 0, B?. (6.30b)

It is immediate to see that uex satisfies the boundary condition. Therefore, by the construction of f, we see

that the exact solution of the above problem is uex.

Exercise 6.8. Show that u R H2

p?q. (You do not have to show that u P H1` 2

3 ′?

p?q for any small 0 ? ? ?

2

3

.)

Exercise 6.9. Show that ?pr

α sinpαθqq “ 0.

Exercise 6.10. Compute fpx, yq explicitly.

The 1st step of the FEM is to formulate Problem (6.30) in a weak form. Let V “ H1

0

p?q. Then the weak

formulation is to find u P V such that

apu, vq “ pf, vq @ v P V. (6.31)

The 2nd step is to define the triangulation Th of ?. We begin with a uniform triangulation with n and mesh

size h “

1

n

, which is composed of cubes of size h ? h. The smartest way of defining the vertex points of the

cubes which are in ? may be as follows:

pxj , ykq “ hpj ′ n, k ′ nq,

$

&

%

j “ 0, ¨ ¨ ¨ , 2n ; k “ 0, ¨ ¨ ¨ , n;

j “ 0, ¨ ¨ ¨ , n ; k “ n ` 1, ¨ ¨ ¨ , 2n.

(6.32)

Let use denote Rjk “ pxj′1, xj q ? pyk′1, ykq.

Exercise 6.11. If g P C

0

pRjkq, for numerical approximation of the integral ?

Rjk

gpx, yq dx use the 2?2 Gauss

quadrature rule. Give the explicit formula for this 2 ? 2 Gauss quadrature rule in terms of the coordinates

of Rjk “ pxj′1, xj q ? pyk′1, ykq and g.

The 3rd step is to define the finite element space Vh, associated with Th, which approximates V. Notice

that we are solving the Dirichlet boundary value problem, and therefore the basis functions associated with

Th can be given for every interior vertex points pxj , ykq. Denote by Vi

h

and Vb

h

the set of all interior and

boundary vertices of Th.

Let us use the Park-Sheen P1-nonconforming quadrilateral finite element, which consists of piecewise

?Dongwoo Sheen, Ph.D. (http://www.nasc.snu.ac.kr) 143

Advanced Numerical Analysis Chapter 6

linear polynomial on each cube that is continuous at the midpoint of interior edges. The benefit of the

Park-Sheen element is that the gradients of the basis function are a constant vector on each cube, so

that the integration on each element is trivial. Explicitly, let Vh be the the Park-Sheen P1-nonconforming

quadrilateral finite element space:

Vh “ tv P L

2

p?q | v|Q P P1pQq @ Q P Th; rrvssepmq “ 0 @ midpoint m of interior edge e in Th;

vpmq “ 0 @ midpoint m of boundary edge in Thu,

where rrvssepmq denotes the jump of v across the interface e of Ql and Qr at m.

A set of basis functions for Vh consists of those functions φjk associated with the interior vertices pxj , ykq P

? defined as follows. First, notice that there are four rectangles which share an interior vertex pxj , ykq in

Th. Let us call them as

RI

jk “ pxj , xj`1q ? pyk, yk`1q, RII

jk “ pxj′1, xj q ? pyk, yk`1q, RIII

jk “ pxj′1, xj q ? pyk′1, ykq, RIV

jk “

pxj , xj`1q ? pyk′1, ykq.

The closure of these four rectangles is the support of φjk. That is, φjk is defined as zero on ?zYι“I,II,III,IV Rι

jk,

and it is the defined such that it is piecewise linear and it has value 1 at the four midpoints of interior edges

of these four rectangles and value 0 at the eight midpoints of exterior edges of these four rectangles. Notice

that the four midpoints of interior edges of these four rectangles are pxj` 1

2

, ykq pxj , yk` 1

2

q, pxj′ 1

2

, ykq, and

pxj , yk′ 1

2

q. The eight midpoints of exterior edges of these four rectangles are

pxj`1, yk` 1

2

q, pxj` 1

2

, yk`1q on BRI

jk;

pxj′ 1

2

, yk`1q, pxj′1, yk` 1

2

q on BRII

jk;

pxj′1, yk′ 1

2

q, pxj′ 1

2

, yk′1q on BRIII

jk ;

pxj` 1

2

, yk′1q, pxj`1, yk′ 1

2

q on BRIV

jk .

Exercise 6.12. Define the basis function φjkpx, yq on each Rι

jk, ι “ I, II, III, IV. Also, write the formula

for the gradient of the basis function, ?φjkpx, yq on each Rι

jk, ι “ I, II, III, IV.

By finishing defining the FEM basis functions, we have

Vh “ Spantφjk | pxj , ykq P Vi

h

u,

recalling that Vi

h

denotes the set of all interior vertices of Th.

The 4th step is to set up the linear system using the basis functions φjk in Vh. We will abuse the notation

as follows:

jk P Vi

h e? pxj , ykq P Vi

h

.

We will integrate on ? by integrating over each Q P Th and summing up them. That is,

ahpu, vq :“

?

QPTh

p?u, ?vqQ “

?

QPTh

?

Q

?u ¨ ?v dx .

The the finite element solution is to find uh “

?

jkPVi

h

Ujkφjk such that

ahpuh, vq “ pf, vq @ v P Vh.

?Dongwoo Sheen, Ph.D. (http://www.nasc.snu.ac.kr) 144

Advanced Numerical Analysis Chapter 6

This is equivalent to finding the coefficient vector pUjkqjkPVi

h

such that

?

jkPVi

h

ahpφjk, φlmqUjk “ pf, φlmq @ lm P Vi

h

. (6.33)

Exercise 6.13. For n “ 10, 20, 40, 80, 160 with h “

1

n

, use the CGM (COnjugate Gradient Method) to solve

(6.33) for pUjkqjkPVi

h

. You do not have to report the vectors pUjkqjkPVi

h

. Then, compute the error

Eh “

gffe

?

jkPVi

h

?

Rjk

|uhpx, yq ′ uexpx, yq|2 dx (6.34)

Then, compute the error reduction ratios

log2

E2h

Eh

, h “

1

10

,

1

20

,

1

40

,

1

80

,

1

160

.

Here, to appoximate ?

Rjk

gpx, yq dx use the 2 ? 2 Gauss quadrature rule.

Exercise 6.14. In the above exercise, we used the Park-Sheen P1-nonconforming FEM. But this time, repeat

the same exercise by using the conforming Q1-element (or bilinear element). For this, the space Vh in (6.33)

should be replaced by

Vh “ tv P C

0

p?q | v|Q P Q1pQq @ Q P Th; v “ 0 on B?u.

Here, and in what follows, for domain K ? R

2

, Q1pKq “ Spant1, x, y, xyu “ ta0 ` a1x ` a2y ` a3xy |

a0, a1, a2, a3 P Ru. Then, in this Q1-conforming element case, define the basis function φjkpx, yq P Vh on

each Rι

jk, ι “ I, II, III, IV so that

φjk|Rι

jk

P Q1pR

ι

jkq @ ι “ I, II, III, IV

and

φjkpxl

, ymq “ δjlδkm @ lm P Vi

h

.

Of course, you need to have the gradient ?φjk, too.

6.4.3 Graded meshes in 2D

Continuing from the previous subsection, now we proceed to work on graded meshes.

As in the 1D example, we attempt to introduce graded meshes on the corner singular elements.

This procedure is a recursive procedure.

They are three elements of size h ? h from Th that share their vertices with p0, 0q :

Rnn “ pxn′1, xnq ? pyn′1, ynq “: G

p0q

c

,

Rn`1,n “ pxn, xn`1q ? pyn′1, ynq “: G

p0q

r

,

Rn,n`1 “ pxn′1, xnq ? pyn, yn`1q “: G

p0q

t

.

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Advanced Numerical Analysis Chapter 6

Let us denote the above three rectangles by G

p0q

c , Gp0q

r , Gp0q

t and their centers by g

p1q

c , g

p1q

r , g

p1q

t

, respectively.

(“c,r,t” mean center, right, top.) Also the superscript pkq denotes the k-th level of mesh grading. Using the

centers g

p1q

ι , ι “ c, r, t, we subdive each of the rectangles G

p0q

ι , ι “ c, r, t, into two quadrilaterals Q

p1q

ι,1

, Qp1q

ι,2

,

and a cube G

p1q

ι of size h{2. The cubes G

p1q

ι are supposed to have common vertex p0, 0q and edges are parallel

to the x- and y-axes.

In this procedure, there are two new vertices, denoted by g

p1q

h “ hp′1

2

, 0q and g

p1q

v “ hp0, ′

1

2

q (“h,v”

represent horizontal and vertical).

By this, we have 5 more interior vertices g

p1q

ι , ι “ c, r, t, h, v and three original rectangles of size h are

subdivided to 6 quadrilaterals and 3 cubes G

p1q

ι of size h

2

.

We repeat this procedure for G

p1q

ι to get 3 cubes G

p2q

ι of size h

2

2 with 5 more vertices g

p2q

ι , ι “ c, r, t, h, v.

Repeating this procedure for s steps, we have 5s new graded vertices g

pkq

ι , ι “ c, r, t, h, v for k “ 1, 2, ¨ ¨ ¨ , s

in addition to the uniform vertices pxj , ykq

1

s.

For each of these graded vertices, we can creat new basis functions ψ

pkq

ι , ι “ c, r, t, h, v for k “ 1, 2, ¨ ¨ ¨ , s.

Hence we will have modified finite element space associated with the graded meshes:

V

psq

h “ Spantφjk,pj, kq P Vi

h

; ψ

pkq

ι

, ι “ c, r, t, h, v for k “ 1, 2, ¨ ¨ ¨ , su.

Here, observe that the five basis functions φn′1,n`1, φn′1,n, φn′1,n′1, φn,n′1, φn`1,n′1 are modified from

those associated with the uniform mesh due to the modification of the origianl uniform mesh, and they have

new supporting quadrilaterals.

The finite element solution is then to find uh P V

psq

h

of the form

uh “

?

jkPVi

h

Ujkφjk `

?s

k“1

?

ι“c,r,t,h,v

Gkιψ

pkq

ι

.

which satisfies

?

jkPVi

h

ahpφjk, φlmqUjk `

?s

k“1

?

ι“c,r,t,h,v

ahpψ

pkq

ι

, φlmqGkι “ pf, φlmq @ lm P Vi

h

, (6.35a)

?

jkPVi

h

ahpφjk, ψppq

q

qUjk `

?s

k“1

?

ι“c,r,t,h,v

ahpψ

pkq

ι

, ψppq

q

qGkι “ pf, ψppq

q

q (6.35b)

@ p “ 1, ¨ ¨ ¨ , s, q “ c, r, t, h, v.

In (6.35), observe those terms without common support are zeros. So the above system is sparse.

Exercise 6.15. You can choose to use either the Park-Sheen P1-nonconforming element or the conforming

Q1 element. Let the grading step number to be s “ 5.

1. Give the explicit formula of the basis functions and their gradients φn′1,n`1, φn′1,n, φn′1,n′1, φn,n′1, φn`1,n′1.

Also give the explicit formula of the basis functions and their gradients ψ

pkq

ι , ι “ c, r, t, h, v for k “

1, 2, ¨ ¨ ¨ , su.

?Dongwoo Sheen, Ph.D. (http://www.nasc.snu.ac.kr) 146

Advanced Numerical Analysis Chapter 6

2. Use the Conjugate Gradient Method to solve (6.35) in order to find the coefficients the coefficients

pUjkqjkPVi

h

and pGkιqι“c,r,t,h,v;k“1,2,¨¨¨ ,s.

Finally compute the error reduction ratios