5/6/2015 1 Parabolic Partial Differential Equations Transforming Numerical Methods.

04/18/23http://

numericalmethods.eng.usf.edu 1

Parabolic Partial Parabolic Partial Differential EquationsDifferential Equations

http://numericalmethods.eng.usf.edu

Transforming Numerical Methods Education for STEM Undergraduates

Defining Parabolic Defining Parabolic PDE’sPDE’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 parabolic

For example, examine the heat-conduction equation given by

Then

thus allowing us to classify this equation as parabolic.

02

22

2

2

Dy

uC

yx

uB

x

uA

042 ACB

, where

0

)0)((4042

ACB

t

T

x

T

2

2

1,0,0, DCBA

Physical Example of an Physical Example of an Elliptic PDEElliptic 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

x

T

2

2

Discretizing the Parabolic Discretizing the Parabolic PDEPDE

Schematic diagram showing interior nodes

x

1i i 1i

x x

For a rod of length divided into nodes

The time is similarly broken into time steps of

Hence corresponds to the temperature at node ,that is,

and time

L 1nn

Lx

t

jiT i

xix tjt

The Explicit MethodThe Explicit Method

If we define we can then write the finite central divided difference

approximation of the left hand side at a general interior node ( ) as

where ( ) is the node number along the time.

n

Lx

i

x

1i i 1i

x x

211

,

2

2 2

x

TTT

x

T ji

ji

ji

ji

j


The time derivative on the right hand side is approximated by the forward divided difference method as,

x

1i i 1i

x x

t

TT

t

T ji

ji

ji

1

,

The Explicit MethodThe Explicit MethodSubstituting these approximations into the governing equation yields

Solving for the temp at the time node gives

choosing,

we can write the equation as,

.

t

TT

x

TTT ji

ji

ji

ji

ji

1

211 2

1j

jiji

ji

ji

ji TTT

x

tTT 112

1 2)(

2)( x

t

jiji

ji

ji

ji TTTTT 11

1 2


•This equation can be solved explicitly because it can be written for each internal location node of the rod for time node in terms of the temperature at time node .

•In other words, if we know the temperature at node , 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 , and using these to find the temperature at the next time node, . This process continues until we reach the time at which we are interested in finding the temperature.

jiji

ji

ji

ji TTTTT 11

1 2

1jj

0j

1j2j

Example 1: Explicit Example 1: Explicit MethodMethod

Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the explicit method to find the temperature distribution in the rod from and seconds. Use , .

Given: , ,

The initial temperature of the rod is .

C100C25 m05.0

0t 9tmx 01.0 st 3

Km

Wk

54

37800

m

kg

Kkg

JC

490

C20

0i 1 2 3 4 5

m01.0

CT 25CT 100

Example 1: Explicit Example 1: Explicit MethodMethodRecall,

therefore,

Then,

C

k

4907800

54

sm /104129.1 25

2x

t

25

01.0

3104129.1

4239.0 .

Number of time steps,

Boundary Conditions

All internal nodes are at for This can be represented as,

t

tt initialfinal

3

09

3

.

.

3,2,1,0allfor25

100

5

0

j

CT

CTj

j

C20sec0t .

1,2,3,4 allfor ,200 iCTi

Example 1: Explicit Example 1: Explicit MethodMethodNodal temperatures when , :

We can now calculate the temperature at each node explicitly using the equation formulated earlier,

sec0t

CT 10000

nodesInterior

20

20

20

20

04

03

02

01

CT

CT

CT

CT

CT 2505

0j

jiji

ji

ji

ji TTTTT 11

1 2


Nodal temperatures when (Example Calculations)

setting

Nodal temperatures when , :

sec3t

ConditionBoundary10010 CT

nodesInterior

120.22

20

20

912.53

14

13

12

11

CT

CT

CT

CT



C

TTTTT

912.53

912.3320

804239.020

100)20(2204239.020

2 00

01

02

01

11

C

TTTTT

20

020

04239.020

20)20(2204239.020

2 01

02

03

02

12

0i

1i 2i

sec3t 1j

0j



setting ,


sec6t


nodesInterior

442.22

889.20

375.34

073.59

24

23

22

21

CT

CT

CT

CT


ConditionBoundary10020 CT0i

1i 2i

sec6t 2j

C

TTTTT

073.59

1614.5912.53

176.124239.0912.53

100)912.53(2204239.0912.53

2 10

11

12

11

21

C

TTTTT

375.34

375.1420

912.334239.020

912.53)20(2204239.020

2 11

12

13

12

22

1j



setting ,


sec9t


nodesInterior

872.22

266.27

132.39

953.65

34

33

32

31

CT

CT

CT

CT


2jConditionBoundary1003

0 CT0i

1i 2i

sec9t 3j

C

TTTTT

953.65

8795.6073.59

229.164239.0073.59

100)073.59(2375.344239.0073.59

2 20

21

22

21

31

C

TTTTT

132.39

7570.4375.34

222.114239.0375.34

073.59)375.34(2899.204239.0375.34

2 21

22

23

22

32


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 MethodThe Implicit MethodWHY:

•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 MethodThe Implicit Method

The second derivative on the left hand side of the equation is approximated by the CDD scheme at time level at node ( ) as

1j

t

T

x

T

2

2

2

11

111

1,

2

2 2

x

TTT

x

T ji

ji

ji

ji

i


The first derivative on the right hand side of the equation is approximated by the BDD scheme at time level at node ( ) as

t

T

x

T

2

2

1j i

t

TT

t

T ji

ji

ji

1

1,


Substituting these approximations into the heat conduction equation yields

t

T

x

T

2

2

t

TT

x

TTT ji

ji

ji

ji

ji

1

2

11

111 2

From the previous slide,

Rearranging yields

given that,

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.


t

TT

x

TTT ji

ji

ji

ji

ji

1

2

11

111 2

ji

ji

ji

ji TTTT

1

111

1 )21(

2x

t

Example 2: Implicit Example 2: Implicit MethodMethod

Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the implicit method to find the temperature distribution in the rod from and seconds. Use , .

Given: , ,


C100C25 m05.0

0t 9tmx 01.0 st 3

Km

Wk

54

37800

m

kg

Kkg

JC

490

C20

0i 1 2 3 4 5

m01.0

CT 25CT 100

Example 2: Implicit Example 2: Implicit MethodMethodRecall,

therefore,

Then,

C

k

4907800

54

sm /104129.1 25

2x

t

25

01.0

3104129.1

4239.0 .


Boundary Conditions


t

tt initialfinal

3

09

3

.

.

3,2,1,0allfor25

100

5

0

j

CT

CTj

j

C20sec0t .


Example 2: Implicit Example 2: Implicit MethodMethodNodal temperatures when , :

We can now form our system of equations for the first time step by writing the approximated heat conduction equation for each node.

sec0t

CT 10000

nodesInterior

20

20

20

20

04

03

02

01

CT

CT

CT

CT

CT 2505

0j

ji

ji

ji

ji TTTT

1

111

1 )21(


Nodal temperatures when , (Example Calculations)

For the interior nodes setting and gives the following,

For the first time step we can write four such equations with four unknowns, expressing them in matrix form yields

sec3t

ConditionBoundary1001

0 CT0i

1i

390.624239.08478.1

204239.08478.139.42

20)4239.0()4239.021()1004239.0(

)21(

12

11

12

11

12

11

01

12

11

10

TT

TT

TT

TTTT

2i204239.08478.14239.0

)21(1

312

11

02

13

12

11

TTT

TTTT

598.30

20

20

390.62

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

14

13

12

11

T

T

T

T

0j 4,3,2,1i

Hence, the nodal temps at are


598.30

20

20

390.62

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

14

13

12

11

T

T

T

T

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

477.21

438.21

792.24

451.39

14

13

12

11

T

T

T

T

sec3t

25

477.21

438.21

792.24

451.39

100

15

14

13

12

11

10

T

T

T

T

T

T




For the second time step we can write four such equations with four unknowns, expressing them in matrix form yields

sec6t


0 CT0i

1i

2i

841.814239.08478.1

451.394239.08478.139.42

451.394239.0)4239.021()1004239.0(

)21(

22

21

22

21

22

21

11

22

21

20

TT

TT

TT

TTTT

792.244239.08478.14239.0

)21(2

32

22

1

12

23

22

21

TTT

TTTT

075.32

438.21

792.24

841.81

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

24

23

22

21

T

T

T

T

1j 4,3,2,1i




sec6t

075.32

438.21

792.24

841.81

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

24

23

22

21

T

T

T

T

836.22

876.23

669.30

326.51

24

23

22

21

T

T

T

T

25

836.22

876.23

669.30

326.51

100

25

24

23

22

21

20

T

T

T

T

T

T




For the third time step we can write four such equations with four unknowns, expressing them in matrix form yields

sec9t


0 CT0i

1i

2i

716.934239.08478.1

326.514239.08478.139.42

326.51)4239.0()4239.021()1004239.0(

)21(

32

31

32

31

32

31

21

32

31

30

TT

TT

TT

TTTT

669.304239.08478.14239.0

)21(3

33

23

1

22

33

32

31

TTT

TTTT

434.33

876.23

669.30

716.93

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

34

33

32

31

T

T

T

T

2j 4,3,2,1i




sec9t

434.33

876.23

669.30

716.93

8478.14239.000

4239.08478.14239.00

04239.08478.14239.0

004239.08478.1

34

33

32

31

T

T

T

T

243.24

809.26

292.36

043.59

34

33

32

31

T

T

T

T

25

243.24

809.26

292.36

043.59

100

35

34

33

32

31

30

T

T

T

T

T

T

Example 2: Implicit Example 2: Implicit MethodMethodTo 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 The Crank-Nicolson MethodMethodWHY:

Using the implicit method our approximation of

was of accuracy, while our approximation of was of

accuracy.

2

2

x

T

2)( xO

t

T

)( tO

The Crank-Nicolson The Crank-Nicolson MethodMethod

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

2

11

111

211

,

2

2 22

2 x

TTT

x

TTT

x

T ji

ji

ji

ji

ji

ji

ji


The first derivative, on the right hand side of the heat equation, is approximated using the forward divided difference method at time level ,1j

t

TT

t

T ji

ji

ji

1

,

•Substituting these approximations into the governing equation for heat conductance yields

giving

where

•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.


t

TT

x

TTT

x

TTT ji

ji

ji

ji

ji

ji

ji

ji

1

2

11

111

211 22

2

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2

2x

t

Example 3: Crank-Example 3: Crank-NicolsonNicolson

Consider a steel rod that is subjected to a temperature of on the left end and on the right end. If the rod is of length ,use the Crank-Nicolson method to find the temperature distribution in the rod from to seconds. Use , .

Given: , ,


C100C25 m05.0

0t9t mx 01.0 st 3

Km

Wk

54

37800

m

kg

Kkg

JC

490

C20

0i 1 2 3 4 5

m01.0

CT 25CT 100

Example 3: Crank-Example 3: Crank-NicolsonNicolsonRecall,

therefore,

Then,

C

k

4907800

54

sm /104129.1 25

2x

t

25

01.0

3104129.1

4239.0 .


Boundary Conditions


t

tt initialfinal

3

09

3

.

.

3,2,1,0allfor25

100

5

0

j

CT

CTj

j

C20sec0t .


Example 3: Crank-Example 3: Crank-NicolsonNicolsonNodal temperatures when , :

We can now form our system of equations for the first time step by writing the approximated heat conduction equation for each node.

sec0t

CT 10000

nodesInterior

20

20

20

20

04

03

02

01

CT

CT

CT

CT

CT 2505

0j

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2



For the interior nodes setting and gives the following

For the first time step we can write four such equations with four unknowns, expressing them in matrix form yields

sec3t


0 CT0i

1i

478.8044.2339.424239.08478.239.42

20)4239.0(20)4239.01(2100)4239.0(4239.0)4239.01(2)1004239.0(

)1(2)1(2

12

11

12

11

02

01

00

12

11

10

TT

TT

TTTTTT

30.1164239.08478.2 12

11 TT

718.52

000.40

000.40

30.116

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

14

13

12

11

T

T

T

T

0j 4,3,2,1i




sec3t

718.52

000.40

000.40

30.116

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

14

13

12

11

T

T

T

T

607.21

797.20

746.23

372.44

14

13

12

11

T

T

T

T

25

607.21

797.20

746.23

372.44

100

15

14

13

12

11

10

T

T

T

T

T

T



For the second time step we can write four such equations with four unknowns, expressing them in matrix form yields


sec6t


0 CT0i

1i

066.10125.5139.424239.08478.239.42

746.23)4239.0(372.44)4239.01(2100)4239.0(

4239.0)4239.01(2)1004239.0(

)1(2)1(2

22

21

22

21

12

11

10

22

21

20

TT

TT

TTTTTT

1j 4,3,2,1i

971.1454239.08478.2 22

21 TT

908.54

187.43

985.54

971.145

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

24

23

22

21

T

T

T

T




sec6t

908.54

187.43

985.54

971.145

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

24

23

22

21

T

T

T

T

730.22

174.23

075.31

883.55

24

23

22

21

T

T

T

T

25

730.22

174.23

075.31

883.55

100

25

24

23

22

21

20

T

T

T

T

T

T




For the third time step we can write four such equations with four unknowns, expressing them in matrix form yields

sec9t


0 CT0i

1i2j 4,3,2,1i

173.13388.6439.424239.08478.239.42

075.31)4239.0(883.55)4239.01(2100)4239.0(

4239.0)4239.01(2)1004239.0(

)1(2)1(2

32

31

32

32

22

21

20

32

31

30

TT

TT

TTTTTT

34.1624239.08478.2 32

31 TT

210.57

509.49

318.69

34.162

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

34

33

32

31

T

T

T

T




sec9t

210.57

509.49

318.69

34.162

8478.24239.000

4239.08478.24239.00

04239.08478.24239.0

004239.08478.2

34

33

32

31

T

T

T

T

042.24

562.26

613.37

604.62

34

33

32

31

T

T

T

T

25

042.24

562.26

613.37

604.62

100

35

34

33

32

31

30

T

T

T

T

T

T

Example 3: Crank-Example 3: Crank-NicolsonNicolsonTo 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.

Node Explicit ImplicitCrank-

NicolsonAnalytic

al

34

33

32

31

T

T

T

T

042.24

562.26

613.37

604.62

243.24

809.26

292.36

043.59

872.22

266.27

132.39

953.65

Internal Temperatures at Internal Temperatures at 9 sec.9 sec.The table below allows you to compare the results from all three methods discussed in juxtaposition with the analytical solution.

610.23

844.25

084.37

510.62

THE END

5/6/2015 1 Parabolic Partial Differential Equations Transforming Numerical Methods.

Documents

time slide

time node

explicit method recall

internal temperature

temperature distribution

example calculations

temperature variation

initial temperature