elliptic partial differential equations - math for...

9/14/2011 http://numericalmethods.eng.usf.edu 1

Elliptic Partial Differential Equations

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Defining Elliptic PDE’s The general form for a second order linear PDE with two independent

variables ( ) and one dependent variable ( ) is

Recall the criteria for an equation of this type to be considered elliptic

For example, examine the Laplace equation given by

then

thus allowing us to classify this equation as elliptic.

02

22

2

2

=+∂∂

+∂∂

∂+

∂∂ D

yuC

yxuB

xuA

042 <− ACB

02

2

2

2

=∂∂

+∂∂

yT

xT

, where , , and 1=A 0=B 1=C 0=D

04)1)(1(4042

<−=−=− ACB

yx, u

Physical Example of an Elliptic PDE

bT

lT

tT

rT

L

W

x

y

Schematic diagram of a plate with specified temperature boundary conditions

The Laplace equation governs the temperature: 02

2

2

2

=∂∂

+∂∂

yT

xT

Discretizing the Elliptic PDEtT

rT

x

y

),( ji ),1( ji +

)1,( −ji

),1( ji −

)1,( +ji

)0,0()0,(m

),0( n

bT

lT ),( ji

x∆

y∆x∆y∆

If we define and ,we can then write the finite difference

approximation of the partial derivatives at a general interior node ( ) as

mLx =∆

nWy =∆

ji,

( )2,1,,1

,2

2 2x

TTTxT jijiji

ji ∆

+−≅

∂∂ −+

and ( )21,,1,

,2

2 2y

TTTyT jijiji

ji ∆

+−≅

∂∂ −+

Substituting these approximations into the Laplace equation yields:

if,

the Laplace equation can be rewritten as

Discretizing the Elliptic PDE

( ) ( )0

222

1,,1,2

,1,,1 =∆

+−+

∆

+− −+−+

yTTT

xTTT jijijijijiji

yx ∆=∆

04 ,1,1,,1,1 =−+++ −+−+ jijijijiji TTTTT

02

2

2

2

=∂∂

+∂∂

yT

xT

Once the governing equation has been discretized there are several numerical methods that can be used to solve the problem.

We will examine the:•Direct Method•Gauss-Seidel Method•Lieberman Method

Discretizing the Elliptic PDE


Example 1: Direct MethodConsider a plate that is subjected to the boundary conditions shown below. Find the temperature at the interior nodes using a square grid with a length of by using the direct method.

C°50

C°75

C°300

C°100

m4.2

m0.3

x

y

mm 0.34.2 ×

m6.0

Example 1: Direct MethodWe can discretize the plate by taking,

myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

Example 1: Direct MethodThe nodal temperatures at the boundary nodes are given by:

3,2,1,3003,2,1,50

4,3,2,1,100

4,3,2,1,75

5,

0,

,4

,0

==

==

==

==

iTiT

jTjT

i

i

j

j

C°300

C°100

C°50

C°75

Example 1: Direct Method

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

3,2T

Here we develop the equation for the temperature at the node (2,3)

04 3,22,24,23,13,3 =−+++ TTTTT04 3,34,23,22,23,1 =++−+ TTTTT

i=2 and j=3 04 ,1,1,,1,1 =−+++ −+−+ jijijijiji TTTTT

Example 1: Direct MethodWe can develop similar equations for every interior node leaving us with an equal number of equations and unknowns.

Question: How many equations would this generate?

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

Example 1: Direct MethodWe can develop similar equations for every interior node leaving us with an equal number of equations and unknowns.

Question: How many equations would this generate? Answer: 12

Solving yields: C

TTTTTTTTTTTT

°

=

446.182271.131389.104

9833.82512.198248.138302.103

5443.77355.173907.119

0252.938924.73

4,3

3,3

2,3

1,3

4,2

3,2

2,2

1,2

4,1

3,1

2,1

1,1

x

y

75

75

75

75

50 50 50

100

100

100

100

300300300

173

120

93

74

199

138

103

78

182

131

104

83

The Gauss-Seidel Method Recall the discretized equation

This can be rewritten as

For the Gauss-Seidel Method, this equation is solved iteratively for all interior nodes until a pre-specified tolerance is met.


41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

Example 2: Gauss-Seidel MethodConsider a plate that is subjected to the boundary conditions shown below. Find the temperature at the interior nodes using a square grid with a length of using the Gauss-Siedel method. Assume the initial temperature at all interior nodes to be .

C°50

C°75

C°300

C°100

m4.2

m0.3

x

y

mm 0.34.2 ×

m6.0C°0

Example 2: Gauss-Seidel MethodWe can discretize the plate by taking

myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

Example 2: Gauss-Seidel MethodThe nodal temperatures at the boundary nodes are given by:

3,2,1,3003,2,1,50

4,3,2,1,100

4,3,2,1,75

5,

0,

,4

,0

==

==

==

==

iTiT

jTjT

i

i

j

j

C°300

C°100

C°50

C°75

Example 2: Gauss-Seidel Method•Now we can begin to solve for the temperature at each interior node using

•Assume all internal nodes to have an initial temperature of zero.

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T i=1 and j=14

0,12,11,01,21,1

TTTTT

+++=

4500750 +++

=

C°= 2500.31

i=1 and j=2 41,13,12,02,2

2,1

TTTTT

+++=

42500.310750 +++

=

C°= 5625.26

Iteration #1

41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

Example 2: Gauss-Seidel MethodAfter the first iteration, the temperatures are as follows. These will now be used as the nodal temperatures for the second iteration.

x

y

31 20 43

27 12 39

25 3.9 37

100 102 135

C°300

C°100

C°50

C°75

Example 2: Gauss-Seidel Method

i=1 and j=1

Iteration #2

40,12,11,01,2

1,1

TTTTT

+++=

4505625.26753125.20 +++

=

C°= 9688.42 %27.27

1009688.42

2500.319688.42

1001,1

1,11,11,1

=

×−

=

×−

= present

previouspresent

a TTT

ε


x

y

300 300 300

75

75

75

75

133 156 161 100

56 56 87 100

39 29 56 100

43 37 56 100

50 50 50x

y

%25 %34 %16

%54 %83 %58

%31 %62 %32

%27 %45 %24

The figures below show the temperature distribution and absolute relative error distribution in the plate after two iterations:

Temperature Distribution

Absolute Relative Approximate

Error Distribution

NodeTemperature Distribution in the Plate (°C)

Number of Iterations

1 2 10 Exact

31.2500 42.9688 73.0239

26.5625 38.7695 91.9585

25.3906 55.7861 119.0976

100.0977 133.2825 172.9755

20.3125 36.8164 76.6127

11.7188 30.8594 102.1577

9.2773 56.4880 137.3802

102.3438 156.1493 198.1055

42.5781 56.3477 82.4837

38.5742 56.0425 103.7757

36.9629 86.8393 130.8056

134.8267 160.7471 182.2278


1,1T

2,1T

3,1T4,1T

1,2T

2,2T3,2T

4,2T

1,3T

2,3T

3,3T

4,3T

The Lieberman Method Recall the equation used in the Gauss-

Siedel Method,

Because the Guass-Siedel Method is guaranteed to converge, we can accelerate the process by using over- relaxation. In this case,

41,1,,1,1

,−+−+ +++

= jijijijiji

TTTTT

oldji

newji

relaxedji TTT ,,, )1( λλ −+=

Example 3: Lieberman MethodConsider a plate that is subjected to the boundary conditions shown below. Find the temperature at the interior nodes using a square grid with a length of . Use a weighting factor of 1.4 in the Lieberman method. Assume the initial temperature at all interior nodes to be .

C°50

C°75

C°300

C°100

m4.2

m0.3

x

y

mm 0.34.2 ×

m6.0C°0

Example 3: Lieberman MethodWe can discretize the plate by taking

myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

Example 3: Lieberman MethodWe can also develop equations for the boundary conditions to define the temperature of the exterior nodes.

3,2,1,3003,2,1,50

4,3,2,1,100

4,3,2,1,75

5,

0,

,4

,0

==

==

==

==

iTiT

jTjT

i

i

j

j

C°300

C°100

C°50

C°75

Example 3: Lieberman Method•Now we can begin to solve for the temperature at each interior node using the rewritten Laplace equation from the Gauss-Siedel method.

•Once we have the temperature value for each node we will apply the over relaxation equation of the Lieberman method

•Assume all internal nodes to have an initial temperature of zero.

i=1 and j=14

0,12,11,01,21,1

TTTTT

+++=

4500750 +++

=

C°= 2500.31

Iteration #1

1,1 1,1 1,1(1 )relaxed new oldT T Tλ λ= + −

C°=−+=

7500.430)4.11()2500.31(4.1

Iteration #2

i=1 and j=24

1,13,12,02,22,1

TTTTT

+++=

475.430750 +++

=

C°= 6875.29


C°=−+=

5625.410)4.11()6875.29(4.1

Example 3: Lieberman MethodAfter the first iteration the temperatures are as follows. These will be used as the initial nodal temperatures during the second iteration.

x

y

44 33 64

42 26 67

41 23 66

146 164 221

C°300

C°100

C°50

C°75

Example 3: Lieberman Method

i=1 and j=1

Iteration #2

40,12,11,01,2

1,1

TTTTT

+++=

C°=

+++=

8438.494

505625.41758125.32

%32.16

1002813.52

7500.432813.52

1001,1

1,11,11,1

=

×−

=

×−

= present

previouspresent

a TTT

ε


C°=−+=

2813.5275.43)4.11()8438.49(4.1


x

y

%6.9 %24 %22

%53 %81 %57

%19 %55 %13

%16 %39 %5.7

The figures below show the temperature distribution and absolute relative error distribution in the plate after two iterations:

Temperature Distribution

x

y

300 300 300

75

75

75

75

161 216 181 100

87 122 155 100

51 58 76 100

52 54 69 100

50 50 50

Absolute Relative Approximate

Error Distribution


NodeTemperature Distribution in the Plate (°C)

Number of Iterations

1 2 9 Exact

43.7500 52.2813 73.7832

41.5625 51.3133 92.9758

40.7969 87.0125 119.9378

145.5289 160.9353 173.3937

32.8125 54.1789 77.5449

26.0313 57.9731 103.3285

23.3898 122.0937 138.3236

164.1216 215.6582 198.5498

63.9844 69.1458 82.9805

66.5055 76.1516 104.3815

66.4634 155.0472 131.2525

220.7047 181.4650 182.4230

1,1T

2,1T

3,1T4,1T

1,2T

2,2T3,2T

4,2T

1,3T

2,3T

3,3T

4,3T

Alternative Boundary Conditions In Examples 1-3, the boundary conditions on the plate had a

specified temperature on each edge. What if the conditions are different ? For example, what if one of the edges of the plate is insulated.

In this case, the boundary condition would be the derivative of the temperature. Because if the right edge of the plate is insulated, then the temperatures on the right edge nodes also become unknowns.

C°50

C°75

C°300

m4.2

m0.3

x

y

Insulated

Alternative Boundary Conditions The finite difference equation in this case for the right edge for

the nodes for

However the node is not inside the plate. The derivative boundary condition needs to be used to account for these additional unknown nodal temperatures on the right edge. This is done by approximating the derivative at the edge node as

1,..3,2 −= nj

04 ,1,1,,1,1 =−+++ +−−+ jmjmjmjmjm TTTTT

),1( jm +

),( jm

)(2,1,1

, xTT

xT jmjm

jm ∆

−≅

∂∂ −+

C°50

C°75

C°300

m4.2

m0.3

x

y

Insulated

),( jm

Alternative Boundary Conditions Rearranging this approximation gives us,

We can then substitute this into the original equation gives us,

Recall that is the edge is insulated then,

Substituting this again yields,

jmjmjm x

TxTT,

,1,1 )(2∂∂

∆+= −+

04)(22 ,1,1,,

,1 =−++∂∂

∆+ +−− jmjmjmjm

jm TTTxTxT

0,

=∂∂

jmxT

042 ,1,1,,1 =−++ +−− jmjmjmjm TTTT

Example 3: Alternative Boundary Conditions

A plate is subjected to the temperatures and insulated boundary conditions as shown in Fig. 12. Use a square grid length of . Assume the initial temperatures at all of the interior nodes to be . Find the temperatures at the interior nodes using the direct method.

mm 0.34.2 ×m6.0

C°0

C°50

C°75

C°300

m4.2

m0.3

x

y

Insulated


We can discretize the plate taking,

myx 6.0=∆=∆

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,2T 3,3T 3,4T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T


We can also develop equations for the boundary conditions to define the temperature of the exterior nodes.

C°300

C°50

C°75

4,3,2,1;0

4,3,2,1;3004,3,2,1;50

4,3,2,1;75

,4

5,

0,

,0

==∂∂

==

==

==

jxT

iTiTjT

j

i

i

j

Insulated


x

y

0,0T0,1T 0,2T 0,3T 0,4T

1,0T

2,0T

3,0T

4,0T

5,0T

1,1T 1,2T 1,3T 1,4T

2,1T 2,2T 2,3T 2,4T

3,1T 3,3T

4,1T 4,2T4,3T 4,4T

5,1T 5,2T 5,3T 5,4T

3,2T3,4T

Here we develop the equation for the temperature at the node (4,3), to show the effects of the alternative boundary condition.i=4 and j=3 042 3,44,42,43,3 =−++ TTTT

042 4,43,42,43,3 =+−+ TTTT


The addition of the equations for the boundary conditions gives us a system of 16 equations with 16 unknowns.

Solving yields: C

TTTTTTTTTTTTTTTT

°

=

738.232830.178617.130

7882.88060.232483.174426.127

2678.87021.218614.159335.117

8571.82410.180617.128

4444.998254.76

4,4

3,4

2,4

1,4

4,3

3,3

2,3

1,3

4,2

3,2

2,2

1,2

4,1

3,1

2,1

1,1

x

y

300 300 300

75

75

75

75

180 218 232 233

129 160 174 179

99 117 127 131

77 83 87 89

50 50 50

THE END

elliptic partial differential equations - math for...

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