1/30/2016 1 elliptic partial differential equations – lieberman method – part 1 of 2 elliptic...
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Elliptic Partial Elliptic Partial Differential Equations Differential Equations – Lieberman Method – – Lieberman Method – Part 1 of 2Part 1 of 2
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Physical Example of an Elliptic Physical Example of an Elliptic PDEPDE
bT
lT
tT
rT
L
W
x
y
02
2
2
2
yT
xT
Discretizing the Elliptic Discretizing the Elliptic PDEPDE
tT
rT
x
y
),( ji ),1( ji
)1,( ji
),1( ji
)1,( ji
)0,0()0,(m
),0( n
bT
lT ),( yx
x
yxy
mLx
nWy
04 ,1,1,,1,1 jijijijiji TTTTT
The Gauss-Seidel The Gauss-Seidel MethodMethodRecall 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.
04 ,1,1,,1,1 jijijijiji TTTTT
41,1,,1,1
,
jijijijiji
TTTTT
The Lieberman MethodThe Lieberman MethodRecall the equation used in the
Gauss-Siedel Method
If 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(
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This instructional power point brought to you byNumerical Methods for STEM undergraduatehttp://numericalmethods.eng.usf.eduCommitted to bringing numerical methods to the undergraduate
AcknowledgementAcknowledgement
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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Elliptic Partial Elliptic Partial Differential Equations Differential Equations – Lieberman Method – – Lieberman Method – Part 2 of 2Part 2 of 2
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
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Click on KeywordClick on Elliptic Partial Differential
Equations
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Example: Lieberman Example: Lieberman MethodMethodConsider 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 guess at all interior nodes to be 0oC.
C50
C75
C300
C100
m4.2
m0.3
x
y
mm 0.34.2
m6.0
Example: Lieberman Example: Lieberman MethodMethodWe can discretize the plate by taking
myx 6.0
C50
C75
C300
C100
m4.2
m0.3
x
y
x
y
0,0T 0,1T 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 4,2T 4,3T 4,4T
5,1T 5,2T 5,3T 5,4T
Example: Lieberman Example: Lieberman MethodMethodWe can also develop equations for the boundary conditions to
define the temperature of the exterior nodes.
3,2,1,300
3,2,1,50
4,3,2,1,100
4,3,2,1,75
5,
0,
,4
,0
iT
iT
jT
jT
i
i
j
j
C300
C100
C50
C75
Example: Lieberman Example: Lieberman MethodMethod
•Solve for the temperature at each interior node using the rewritten discretized Laplace equation from the Gauss-Siedel method.
•Apply the over relaxation equation using temperatures from previous iteration.
i=1 and j=1
40,12,11,01,2
1,1
TTTTT
4500750
C 2500.31
Iteration #1
1,1 1,1 1,1(1 )relaxed new oldT T T
C
7500.430)4.11()2500.31(4.1
%00.100
1007500.43
07500.43
1001,1
1,11,11,1
present
previouspresent
a TTT
Example: Lieberman Example: Lieberman MethodMethod
Iteration #1
i=1 and j=2
41,13,12,02,2
2,1
TTTTT
475.430750
C 6875.29
C
5625.410)4.11()6875.29(4.1
oldnewrelaxed TTT 2,12,12,1 )1( %00.100
1005625.43
05625.41
1002,1
2,12,12,1
present
previouspresent
a TTT
Example: Lieberman Example: Lieberman MethodMethodAfter the first iteration the temperatures are as follows.
These will be used as the nodal temperatures during the second iteration.
x
y
44 33 64
42 26 67
41 23 66
146 164 221
C300
C100
C50
C75
Example: Lieberman Example: Lieberman MethodMethod
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
1,1 1,1 1,1(1 )relaxed new oldT T T
C
2813.5275.43)4.11()8438.49(4.1
Example: Lieberman Example: Lieberman MethodMethod
Iteration #2
i=1 and j=2
41,13,12,02,2
2,1
TTTTT
C
5274.484
2813.527969.40750313.26
1,2 1,2 1,2(1 )relaxed new oldT T T
C
3133.515625.41)4.11()5274.48(4.1
%00.19
1003133.51
5625.413133.51
1002,1
2,12,12,1
present
previouspresent
a TTT
Example: Lieberman Example: Lieberman MethodMethod
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
Example: Lieberman Example: Lieberman MethodMethod
NodeTemperature Distribution in the Plate
(°C)
Number of Iterations1 2 9
43.7500 52.2813 73.783241.5625 51.3133 92.975840.7969 87.0125 119.9378
145.5289 160.9353 173.393732.8125 54.1789 77.544926.0313 57.9731 103.328523.3898 122.0937 138.3236
164.1216 215.6582 198.549863.9844 69.1458 82.980566.5055 76.1516 104.381566.4634 155.0472 131.2525
220.7047 181.4650 182.4230
1,1T
2,1T
3,1T
4,1T
1,2T
2,2T3,2T
4,2T
1,3T
2,3T
3,3T
4,3T
THE ENDTHE ENDhttp://numericalmethods.eng.usf.edu
This instructional power point brought to you byNumerical Methods for STEM undergraduatehttp://numericalmethods.eng.usf.eduCommitted to bringing numerical methods to the undergraduate
AcknowledgementAcknowledgement
For instructional videos on other topics, go to
http://numericalmethods.eng.usf.edu/videos/
This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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