Transcript mws_gen_pde_ppt_para..

Parabolic Partial
Differential Equations
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM Undergraduates
4/9/2015
http://numericalmethods.eng.usf.edu
1
Defining Parabolic PDE’s

The general form for a second order linear PDE with two independent
variables and one dependent variable is
 2u
 2u
 2u
A 2 B
C 2  D 0
x
xy
y


Recall the criteria for an equation of this type to be considered parabolic
B 2  4 AC  0
For example, examine the heat-conduction equation given by
 2T T
 2 
t
x
, where
A   , B  0, C  0, D  1
Then
B 2  4 AC  0  4( )(0)
0
thus allowing us to classify this equation as parabolic.
Physical Example of an Elliptic
PDE
The internal temperature of a metal rod exposed to two different
temperatures at each end can be found using the heat conduction
equation.
 T T
 2 
t
x
2
Discretizing the Parabolic PDE
x
x
x
i 1
i
i 1
Schematic diagram showing interior nodes
For a rod of length L divided into
n  1 nodes x 
L
n
The time is similarly broken into time steps of t
Hence
Ti j corresponds to the temperature at node i
x  i x 
and time
t   j t 
,that is,
The Explicit Method
x
i 1
If we define x 
x
x
L
n
i 1
i
we can then write the finite central divided difference
approximation of the left hand side at a general interior node ( i ) as
 2T
2
x

i, j
Ti j1  2Ti j  Ti j1
x 
2
where ( j ) is the node number along the time.
The Explicit Method
x
x
x
i 1
i
i 1
The time derivative on the right hand side is approximated by the
forward divided difference method as,
T
t
i, j
Ti j 1  Ti j

t
The Explicit Method
Substituting these approximations into the governing equation yields

Ti j1  2Ti j  Ti j1
x2
Ti j 1  Ti j

t
Solving for the temp at the time node j  1 gives
Ti j 1  Ti j  
choosing,
 
t
j
j
j
T

2
T

T
i 1
i
i 1
(x) 2

t
(x) 2
we can write the equation as,
Ti

j 1

 Ti   T  2Ti  T
j
j
i 1
j
j
i 1

.
The Explicit Method
j 1
j
j
j
j
Ti  Ti   Ti1  2Ti  Ti1 
•This equation can be solved explicitly because it can be written for each
internal location node of the rod for time node j  1 in terms of the
temperature at time node j .
•In other words, if we know the temperature at node j  0 , and the
boundary temperatures, we can find the temperature at the next time
step.
•We continue the process by first finding the temperature at all nodes j  1 ,
and using these to find the temperature at the next time node, j  2 .This
process continues until we reach the time at which we are interested in
finding the temperature.
Example 1: Explicit Method
Consider a steel rod that is subjected to a temperature of 100 C on the left
end and 25 C on the right end. If the rod is of length 0.05 m ,use the explicit
method to find the temperature distribution in the rod from t  0 and t  9
seconds. Use x  0.01m , t  3s .
Given: k  54
J
W
kg
,   7800 3 , C  490
kg  K
mK
m
The initial temperature of the rod is 20 C .
i0
1
2
T 100  C
3
4
5
T  25 C
0.01m
Example 1: Explicit Method
Recall,

Number of time steps,
t final  t initial

t
k
C

therefore,

54
7800  490
 3.
 1.4129105 m2 / s.
Then,
 
x 2
 0.4239 .
Boundary Conditions
T0 j  100C 
 for all j  0,1,2,3
j
T5  25C 
t
 1.4129 105
90
3
3
0.01
2
All internal nodes are at 20 C
for t  0 sec . This can be
represented as,
Ti 0  20C, for all i  1,2,3,4
Example 1: Explicit Method
Nodal temperatures when t  0 sec , j  0:
T00 100C
T10  20C 

T20  20C 
 Interiornodes
T30  20C 
T40  20C 
T50  25C
We can now calculate the temperature at each node explicitly using
the equation formulated earlier,
Ti
j 1

 Ti   T  2Ti  T
j
j
i 1
j
j
i 1

Example 1: Explicit Method
Nodal temperatures when t  3 sec (Example Calculations)
i0
T01  100C  Boundary Condition
setting
i 1
j0

T11  T10   T20  2T10  T00

 20  0.423920  2(20)  100
i2
 20  0.423980

T21  T20   T30  2T20  T10
 20  0.423920  2(20)  20
 20  0.42390 
 20  33.912
 20  0
 53.912C
 20C
Nodal temperatures when t  3 sec , j  1 :
T01  100C  Boundary Condition
T11  53.912C 

T21  20C

 Interiornodes
1
T3  20C

T41  22.120C 
T51  25C  Boundary Condition

Example 1: Explicit Method
Nodal temperatures when t  6 sec (Example Calculations)
i0
T02  100C  Boundary Condition
setting
i 1

j 1 ,
i  2 T22  T21   T31  2T21  T11 
 53.912  0.423920  2(53.912)  100
 20  0.423920  2(20)  53.912
 53.912  0.423912.176
 20  0.423933.912
T12  T11   T21  2T11  T01

 53.912  5.1614
 20  14.375
 59.073C
 34.375C
Nodal temperatures when t  6 sec , j  2:
T02  100C  Boundary Condition
T12  59.073C 

T22  34.375C 
 Interiornodes
2
T3  20.889C 
T42  22.442C 
T52  25C  Boundary Condition
Example 1: Explicit Method
Nodal temperatures when t  9 sec (Example Calculations)
i0
i 1
T03  100C  Boundary Condition
setting j  2 ,

T13  T12   T22  2T12  T02

 59.073 0.423934.375 2(59.073)  100
 59.073 0.423916.229
i2

T23  T22   T32  2T22  T12
 34.375 0.423920.899  2(34.375)  59.073
 34.375 0.423911.222
 59.073 6.8795
 34.375 4.7570
 65.953C
 39.132C
Nodal temperatures when t  9 sec , j  3 :
T03  100C  BoundaryCondition
T13  65.953C 

T23  39.132C 
 Interiornodes
3
T3  27.266C 
T43  22.872C 
T53  25C  Boundary Condition

Example 1: Explicit Method
To better visualize the temperature variation at different
locations at different times, the temperature distribution along
the length of the rod at different times is plotted below.
The Implicit Method
WHY:
•Using the explicit method, we were able to find the temperature at
each node, one equation at a time.
•However, the temperature at a specific node was only dependent
on the temperature of the neighboring nodes from the previous
time step. This is contrary to what we expect from the physical
problem.
•The implicit method allows us to solve this and other problems by
developing a system of simultaneous linear equations for the
temperature at all interior nodes at a particular time.
The Implicit Method
 T
T


2
t
x
2
The second derivative on the left hand side of the equation is
approximated by the CDD scheme at time level j  1 at node (i )
as
 2T
x 2

i , j 1
Ti j11  2Ti j 1  Ti j11
x 
2
The Implicit Method
 T
T


2
t
x
2
The first derivative on the right hand side of the equation is
approximated by the BDD scheme at time level j  1 at node ( i )
as
j 1
j
i
i
T
t

i , j 1
T
T
t
The Implicit Method
 T
T


2
t
x
2
Substituting these approximations into the heat conduction
equation yields

Ti j11  2Ti j 1  Ti j11
x
2
Ti j 1  Ti j

t
The Implicit Method
From the previous slide,

Ti j11  2Ti j 1  Ti j11
x2
Ti j 1  Ti j

t
Rearranging yields
 Ti j11  (1  2)Ti j 1  Ti j11  Ti j
given that,
 
t
x 
2
The rearranged equation can be written for every node during each time
step. These equations can then be solved as a simultaneous system of
linear equations to find the nodal temperatures at a particular time.
Example 2: Implicit Method
Consider a steel rod that is subjected to a temperature of 100 C on the left
end and 25 C on the right end. If the rod is of length 0.05 m ,use the implicit
method to find the temperature distribution in the rod from t  0 and t  9
seconds. Use x  0.01m , t  3s .
Given: k  54
J
W
kg
,   7800 3 , C  490
kg  K
mK
m
The initial temperature of the rod is 20 C .
i0
1
2
T 100  C
3
4
5
T  25 C
0.01m
Example 2: Implicit Method
Recall,

Number of time steps,
t final  t initial

t
k
C

therefore,

54
7800  490
 3.
 1.4129105 m2 / s.
Then,
 
x 2
 0.4239 .
Boundary Conditions
T0 j  100C 
 for all j  0,1,2,3
j
T5  25C 
t
 1.4129 105
90
3
3
0.01
2
All internal nodes are at 20 C
for t  0 sec . This can be
represented as,
Ti 0  20C, for all i  1,2,3,4
Example 2: Implicit Method
Nodal temperatures when t  0 sec , j  0:
T00 100C
T10  20C 

T20  20C 
 Interiornodes
T30  20C 
T40  20C 
T50  25C
We can now form our system of equations for the first time step by
writing the approximated heat conduction equation for each node.
 T
j 1
i 1
 (1  2)Ti
j 1
 T
j 1
i 1
 Ti
j
Example 2: Implicit Method
Nodal temperatures when t  3 sec ,
i0
T  100C  Boundary Condition
i 1
 T01  (1  2 )T11  T21  T10
(Example Calculations)
1
0
For the interior nodes setting j  0 and i 1, 2, 3, 4 gives the following,
(0.4239100)  (1  2  0.4239)T11  (0.4239T21 )  20
 42.39  1.8478T11  0.4239T21  20
1.8478T11  0.4239T21  62.390
i2
 T11  (1  2 )T21  T31  T20
 0.4239T11  1.8478T21  0.4239T31  20
For the first time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 1.8478  0.4239
 T11  62.390
 0.4239 1.8478  0.4239
  1   20 
0

 T2   

1

0
 0.4239 1.8478  0.4239 T3   20 

 1  

0
0

0
.
4239
1
.
8478
30
.
598
T



 4  

Example 2: Implicit Method
0
0
 1.8478  0.4239
 T11  62.390
 0.4239 1.8478  0.4239
  1   20 
0

 T2   

1

0
 0.4239 1.8478  0.4239 T3   20 

 1  

0
0

0
.
4239
1
.
8478
30
.
598
T

  4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T11   39.451
 1 

24
.
792
T
2
 

1
T3   21.438
 1 

T4  21.477
Hence, the nodal
temps at t  3 sec are
T01   100 
 1 

39
.
451
T
 1 

1
T2  24.792
 1  

21
.
438
T
 3 

T 1  21.477
 41  

25
T
 5  

Example 2: Implicit Method
Nodal temperatures when t  6 sec ,
i0
(Example Calculations)
T  100C  Boundary Condition
2
0
For the interior nodes setting j  1 and i 1, 2, 3, 4 gives the following,
i 1
 T02  (1  2 )T12  T22  T11
(0.4239 100)  (1  2  0.4239)T12  0.4239T22  39.451
 42.39  1.8478T12  0.4239T22  39.451
1.8478T12  0.4239T22  81.841
i2
 T12  (1  2 )T22  T32  T21
 0.4239T12  1.8478T22  0.4239T32  24.792
For the second time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 1.8478  0.4239
 T12   81.841
 0.4239 1.8478  0.4239
  2  24.792
0

 T2   


0
 0.4239 1.8478  0.4239 T32  21.438

 2  

0
0

0
.
4239
1
.
8478
32
.
075

 T4  

Example 2: Implicit Method
0
0
 1.8478  0.4239
 T12   81.841
 0.4239 1.8478  0.4239
  2  24.792
0

 T2   

2

0
 0.4239 1.8478  0.4239 T3  21.438

 2  

0
0

0
.
4239
1
.
8478
32
.
075
T



 4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T12  51.326
 2 

30
.
669
T
 2 

2
T3  23.876
 2 

T4  22.836
Hence, the nodal
temps at t  6 sec are
T02   100 
 2 

51
.
326
T
 1  

2
T2  30.669
 2  

23
.
876
T3  

T 2  22.836
 42  

T5   25 
Example 2: Implicit Method
Nodal temperatures when t  9 sec ,
i0
(Example Calculations)
T  100C  Boundary Condition
3
0
For the interior nodes setting j  2 and i 1, 2, 3, 4 gives the following,
i 1
 T03  (1  2 )T13  T23  T12
(0.4239 100)  (1  2  0.4239)T13  (0.4239T23 )  51.326
 42.39  1.8478T13  0.4239T23  51.326
1.8478T13  0.4239T23  93.716
i2
 T13  (1  2 )T23  T33  T22
 0.4239T13  1.8478T23  0.4239T33  30.669
For the third time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 1.8478  0.4239
 T13  93.716
 0.4239 1.8478  0.4239
  3  30.669
0

 T2   

3

0
 0.4239 1.8478  0.4239 T3  23.876

 3  

0
0

0
.
4239
1
.
8478
33
.
434
T

  4  

Example 2: Implicit Method
0
0
 1.8478  0.4239
 T13  93.716
 0.4239 1.8478  0.4239
  3  30.669
0

 T2   

3

0
 0.4239 1.8478  0.4239 T3  23.876

 3  

0
0

0
.
4239
1
.
8478
33
.
434
T



 4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T13  59.043
 3 

36
.
292
T
 2

3
T3  26.809
 3 

24
.
243
T
 4  

Hence, the nodal
temps at t  9 sec are
T03   100 
 3 

59
.
043
T
1
  

3
T2  36.292
 3  

26
.
809
T
 3 

3
T   24.243
 43  

25

T5  
Example 2: Implicit Method
To better visualize the temperature variation at different
locations at different times, the temperature distribution along
the length of the rod at different times is plotted below.
The Crank-Nicolson Method
WHY:
 2T
Using the implicit method our approximation of
was of
x 2
accuracy, while our approximation of
T
t
O(x) 2
was of O(t ) accuracy.
The Crank-Nicolson Method
One can achieve similar orders of accuracy by approximating the
second derivative, on the left hand side of the heat equation, at the
midpoint of the time step. Doing so yields
 2T
x 2

i, j
  Ti j1  2Ti j  Ti j1

2
x 
2
Ti j11  2Ti j 1  Ti j11 


2
x 

The Crank-Nicolson Method
The first derivative, on the right hand side of the heat equation, is
approximated using the forward divided difference method at time
level j  1 ,
T
t

i, j
Ti
j 1
 Ti
t
j
The Crank-Nicolson Method
•Substituting these approximations into the governing equation for
heat conductance yields
  Ti j1  2Ti j  Ti j1 Ti j11  2Ti j 1  Ti j11  Ti j 1  Ti j

2
x 
2

x 
2


t
giving
 Ti j11  2(1  )Ti j 1  Ti j11  Ti j1  2(1  )Ti j  Ti j1
where
 
t
x 2
•Having rewritten the equation in this form allows us to descritize
the physical problem. We then solve a system of simultaneous linear
equations to find the temperature at every node at any point in
time.
Example 3: Crank-Nicolson
Consider a steel rod that is subjected to a temperature of 100 C on the left
end and 25 C on the right end. If the rod is of length 0.05 m ,use the CrankNicolson method to find the temperature distribution in the rod from t  0
to t  9 seconds. Use x  0.01m , t  3s .
Given: k  54
J
W
kg
,   7800 3 , C  490
kg  K
mK
m
The initial temperature of the rod is 20 C .
i0
1
2
T 100  C
3
4
5
T  25 C
0.01m
Example 3: Crank-Nicolson
Recall,

Number of time steps,
t final  t initial

t
k
C

therefore,

54
7800  490
 3.
 1.4129105 m2 / s.
Then,
 
x 2
 0.4239 .
Boundary Conditions
T0 j  100C 
 for all j  0,1,2,3
j
T5  25C 
t
 1.4129 105
90
3
3
0.01
2
All internal nodes are at 20 C
for t  0 sec . This can be
represented as,
Ti 0  20C, for all i  1,2,3,4
Example 3: Crank-Nicolson
Nodal temperatures when t  0 sec , j  0:
T00 100C
T10  20C 

T20  20C 
 Interiornodes
T30  20C 
T40  20C 
T50  25C
We can now form our system of equations for the first time step by
writing the approximated heat conduction equation for each node.
 Ti j11  2(1  )Ti j 1  Ti j11  Ti j1  2(1  )Ti j  Ti j1
Example 3: Crank-Nicolson
Nodal temperatures when t  3 sec ,
i0
(Example Calculations)
T01  100C  Boundary Condition
For the interior nodes setting j  0 and i 1, 2, 3, 4 gives the following
i 1
 T01  2(1   )T11  T21  T00  2(1   )T10  T20
(0.4239 100)  2(1  0.4239)T11  0.4239T21  (0.4239)100 2(1  0.4239)20  (0.4239)20
 42.39  2.8478T11  0.4239T21  42.39  23.044 8.478
2.8478T11  0.4239T21  116.30
For the first time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 2.8478  0.4239
 T11  116.30
 0.4239 2.8478  0.4239
  1  40.000
0

 T2   

1
 0
 0.4239 2.8478  0.4239 T3  40.000

 1  

0
0

0
.
4239
2
.
8478
52
.
718

 T4  

Example 3: Crank-Nicolson
0
0
 2.8478  0.4239
 T11  116.30
 0.4239 2.8478  0.4239
  1  40.000
0

 T2   

1
 0
 0.4239 2.8478  0.4239 T3  40.000

 1  

0
0

0
.
4239
2
.
8478
52
.
718
T



 4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T11   44.372 Hence, the nodal
 1 

23
.
746
T
 2
 temps at t  3 sec are
T31  20.797
 1 

21
.
607
T
 4  

T01   100 
 1 

44
.
372
T
1
  

1
T2  23.746
 1  

20
.
797
T
 3 

T 1  21.607
 41  

T5   25 
Example 3: Crank-Nicolson
Nodal temperatures when t  6 sec ,
i0
(Example Calculations)
T02  100C  Boundary Condition
For the interior nodes setting j  1 and i 1, 2, 3, 4 gives the following,
i 1
 T02  2(1   )T12  T22  T01  2(1   )T11  T21
(0.4239 100)  2(1  0.4239)T12  0.4239T22 
(0.4239)100 2(1  0.4239)44.372 (0.4239)23.746
 42.39  2.8478T12  0.4239T22  42.39  51.125 10.066
2.8478T12  0.4239T22  145.971
For the second time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 2.8478  0.4239
 T12  145.971
 0.4239 2.8478  0.4239
  2   54.985 
0

 T2   

2
 0
 0.4239 2.8478  0.4239 T3   43.187 

 2  

0
0

0
.
4239
2
.
8478
54
.
908

 T4  

Example 3: Crank-Nicolson
0
0
 2.8478  0.4239
 T12  145.971
 0.4239 2.8478  0.4239
  2   54.985 
0

 T2   

2
 0
 0.4239 2.8478  0.4239 T3   43.187 

 2  

0
0

0
.
4239
2
.
8478
54
.
908
T



 4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T12  55.883
 2 

31
.
075
T
 2 

2
T3  23.174
 2 

22
.
730
T
 4  

Hence, the nodal
temps at t  6 sec are
T02   100 
 2 

55
.
883
T
 1  

2
T2  31.075
 2  

T3  23.174
T 2  22.730
 42  

25
T


 5  
Example 3: Crank-Nicolson
Nodal temperatures when t  9 sec ,
i0
(Example Calculations)
T  100C  Boundary Condition
3
0
For the interior nodes setting j  2 and i 1, 2, 3, 4 gives the following,
i 1
 T03  2(1   )T13  T23  T02  2(1   )T12  T22
(0.4239 100)  2(1  0.4239)T23  0.4239T23 
(0.4239)100 2(1  0.4239)55.883 (0.4239)31.075
 42.39  2.8478T13  0.4239T23  42.39  64.388 13.173
2.8478T13  0.4239T23  162.34
For the third time step we can write four such equations with four
unknowns, expressing them in matrix form yields
0
0
 2.8478  0.4239
 T13  162.34
 0.4239 2.8478  0.4239
  3  69.318
0

 T2   

3

0
 0.4239 2.8478  0.4239 T3  49.509

 3  

0
0

0
.
4239
2
.
8478
57
.
210
T



 4  

Example 3: Crank-Nicolson
0
0
 2.8478  0.4239
 T13  162.34
 0.4239 2.8478  0.4239
  3  69.318
0

 T2   

3

0
 0.4239 2.8478  0.4239 T3  49.509

 3  

0
0

0
.
4239
2
.
8478
57
.
210
T

  4  

The above coefficient matrix is tri-diagonal. Special algorithms
such as Thomas’ algorithm can be used to solve simultaneous
linear equation with tri-diagonal coefficient matrices.The
solution is given by
T13  62.604 Hence, the nodal
 3 

37
.
613
T
 2
 temps at t  9 sec are
T33  26.562
 3 

24
.
042
T
 4  

T03   100 
 3 

T1  62.604
T23  37.613
 3  

26
.
562
T
 3 

3
T  24.042
 43  

T5   25 
Example 3: Crank-Nicolson
To better visualize the temperature variation at different
locations at different times, the temperature distribution along
the length of the rod at different times is plotted below.
Internal Temperatures at 9 sec.
The table below allows you to compare the results from all three
methods discussed in juxtaposition with the analytical solution.
Node
Explicit
Implicit
CrankNicolson
Analytical
T13
T23
T33
T43
65.953
59.043
62.604
62.510
39.132
27.266
36.292
26.809
37.613
26.562
37.084
25.844
22.872
24.243
24.042
23.610
THE END