Home > Documents > Parabolic Partial Differential Equations - Transforming Numerical

Parabolic Partial Differential Equations - Transforming Numerical

Date post: 03-Feb-2022
Category:
View: 10 times
46
9/15/2011 http://numericalmethods.eng.usf.edu 1 Parabolic Partial Differential Equations http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates
Transcript

9/15/2011 http://numericalmethods.eng.usf.edu 1

Parabolic Partial Differential Equations

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM Undergraduates

Defining Parabolic 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 parabolic

For example, examine the heat-conduction equation given by

Then

thus allowing us to classify this equation as parabolic.

02

22

2

2

=+∂∂

+∂∂

∂+

∂∂ D

yuC

yxuB

xuA

042 =− ACB

, where

0)0)((4042

=−=− αACB

tT

xT

∂∂

=∂∂

2

2

α 1,0,0, −==== DCBA α

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.

tT

xT

∂∂

=∂∂

2

2

α

Discretizing the Parabolic PDE

Schematic diagram showing interior nodes

x

1−i i 1+i

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 1+nnLx =∆

t∆

jiT i

( )( )xix ∆= ( )( )tjt ∆=

The 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.

nLx =∆

i

x

1−i i 1+i

x∆ x∆

( )211

,2

2 2x

TTTxT j

ij

ij

i

ji ∆+−

≅∂∂ −+

j

The Explicit Method

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

x

1−i i 1+i

x∆ x∆

tTT

tT j

ij

i

ji ∆−

≅∂∂ +1

,

The 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,

.

( ) tTT

xTTT j

ij

ij

ij

ij

i

∆−

=∆

+− +−+

1

211 2

α

1+j

( )ji

ji

ji

ji

ji TTT

xtTT 112

1 2)( −+

+ +−∆∆

+= α

2)( xt

∆∆

=αλ

( )ji

ji

ji

ji

ji TTTTT 11

1 2 −++ +−+= λ

The Explicit Method

•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.

( )ji

ji

ji

ji

ji TTTTT 11

1 2 −++ +−+= λ

1+jj

0=j

1=j2=j

Example 1: Explicit MethodConsider 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 .

C°100C°25 m05.0

0=t 9=tmx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρ KkgJC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100

Example 1: Explicit MethodRecall,

therefore,

Then,

Ckρ

α =

490780054×

sm /104129.1 25−×=

( )2xt

∆∆

= αλ

( )25

01.03104129.1 −×=

4239.0= .

Number of time steps,

Boundary Conditions

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

ttt initialfinal

−=

309 −

=

3=

.

.

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j

C°20sec0=t .

1,2,3,4 allfor ,200 =°= iCTi

Example 1: Explicit MethodNodal temperatures when , :

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

sec0=t

CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

0=j

( )ji

ji

ji

ji

ji TTTTT 11

1 2 −++ +−+= λ

Example 1: Explicit MethodNodal temperatures when (Example Calculations)

setting

Nodal temperatures when , :

sec3=t

ConditionBoundary10010 −°= CT

nodesInterior

120.22

20

20

912.53

14

13

12

11

°=

°=

°=

°=

CTCTCT

CT

ConditionBoundary2515 −°= CT

ConditionBoundary10010 −°= CT

( )( )( )

C

TTTTT

°=+=+=

+−+=+−+=

912.53912.3320

804239.020100)20(2204239.020

2 00

01

02

01

11 λ ( )

( )( )

C

TTTTT

°=+=+=

+−+=+−+=

20020

04239.02020)20(2204239.020

2 01

02

03

02

12 λ

0=i

1=i 2=i

sec3=t 1=j

0=j

Example 1: Explicit MethodNodal temperatures when (Example Calculations)

setting ,

Nodal temperatures when , :

sec6=t

ConditionBoundary10020 −°= CT

nodesInterior

442.22

889.20

375.34

073.59

24

23

22

21

°=

°=

°=

°=

CTCTCTCT

ConditionBoundary2525 −°= CT

ConditionBoundary10020 −°= CT0=i

1=i 2=i

sec6=t 2=j

( )( )( )

C

TTTTT

°=+=+=

+−+=+−+=

073.591614.5912.53

176.124239.0912.53100)912.53(2204239.0912.53

2 10

11

12

11

21 λ ( )

( )( )

C

TTTTT

°=+=+=

+−+=+−+=

375.34375.1420

912.334239.020912.53)20(2204239.020

2 11

12

13

12

22 λ

1=j

Example 1: Explicit MethodNodal temperatures when (Example Calculations)

setting ,

Nodal temperatures when , :

sec9=t

ConditionBoundary10030 −°= CT

nodesInterior

872.22

266.27

132.39

953.65

34

33

32

31

°=

°=

°=

°=

CTCTCTCT

ConditionBoundary2535 −°= CT

2=jConditionBoundary1003

0 −°= CT0=i

1=i 2=i

sec9=t 3=j

( )( )( )

C

TTTTT

°=+=+=

+−+=+−+=

953.658795.6073.59

229.164239.0073.59100)073.59(2375.344239.0073.59

2 20

21

22

21

31 λ ( )

( )( )

C

TTTTT

°=+=+=

+−+=+−+=

132.397570.4375.34

222.114239.0375.34073.59)375.34(2899.204239.0375.34

2 21

22

23

22

32 λ

Example 1: Explicit MethodTo 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 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 Method

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

1+j

tT

xT

∂∂

=∂∂

2

2

α

( )2

11

111

1,2

2 2x

TTTxT j

ij

ij

i

ji ∆+−

≈∂∂ +

−++

+

+

i

The Implicit Method

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

tT

xT

∂∂

=∂∂

2

2

α

1+j i

tTT

tT j

ij

i

ji ∆−

≈∂∂ +

+

1

1,

The Implicit Method

Substituting these approximations into the heat conduction equation yields

tT

xT

∂∂

=∂∂

2

2

α

( ) tTT

xTTT j

ij

ij

ij

ij

i

∆−

=∆

+− ++−

+++

1

2

11

111 2

α

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.

The Implicit Method

( ) tTT

xTTT j

ij

ij

ij

ij

i

∆−

=∆

+− ++−

+++

1

2

11

111 2

α

ji

ji

ji

ji TTTT =−++− +

+++

−1

111

1 )21( λλλ

( )2xt

∆∆

= αλ

Example 2: Implicit MethodConsider 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: , ,

The initial temperature of the rod is .

C°100C°25 m05.0

0=t 9=tmx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρ KkgJC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100

Example 2: Implicit MethodRecall,

therefore,

Then,

Ckρ

α =

490780054×

sm /104129.1 25−×=

( )2xt

∆∆

= αλ

( )25

01.03104129.1 −×=

4239.0= .

Number of time steps,

Boundary Conditions

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

ttt initialfinal

−=

309 −

=

3=

.

.

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j

C°20sec0=t .

1,2,3,4 allfor ,200 =°= iCTi

Example 2: Implicit MethodNodal 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.

sec0=t

CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

0=j

ji

ji

ji

ji TTTT =−++− +

+++

−1

111

1 )21( λλλ

Example 2: Implicit MethodNodal 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

sec3=tConditionBoundary1001

0 −°= CT0=i

1=i

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

=−

=−+−

=−×++×−

=−++−

TTTT

TTTTTT λλλ

2=i204239.08478.14239.0

)21(1

312

11

02

13

12

11

=−+−

=−++−

TTTTTTT λλλ

=

−−−

−−−

598.302020390.62

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

14

13

12

11

TTTT

0=j 4,3,2,1=i

Hence, the nodal temps at are

Example 2: Implicit Method

=

−−−

−−−

598.302020390.62

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

14

13

12

11

TTTT

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

=

477.21438.21792.24451.39

14

13

12

11

TTTT

sec3=t

=

25477.21438.21792.24451.39

100

15

14

13

12

11

10

TTTTTT

Example 2: Implicit MethodNodal temperatures when , (Example Calculations)

For the interior nodes setting and gives the following,

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

sec6=tConditionBoundary1002

0 −°= CT0=i

1=i

2=i841.814239.08478.1

451.394239.08478.139.42

451.394239.0)4239.021()1004239.0(

)21(

22

21

22

21

22

21

11

22

21

20

=−

=−+−

=−×++×−

=−++−

TTTT

TTTTTT λλλ

792.244239.08478.14239.0

)21(2

32

22

1

12

23

22

21

=−+−

=−++−

TTTTTTT λλλ

=

−−−

−−−

075.32438.21792.24841.81

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

24

23

22

21

TTTT

1=j 4,3,2,1=i

Hence, the nodal temps at are

Example 2: Implicit Method

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

sec6=t

=

−−−

−−−

075.32438.21792.24841.81

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

24

23

22

21

TTTT

=

836.22876.23669.30326.51

24

23

22

21

TTTT

=

25836.22876.23669.30326.51

100

25

24

23

22

21

20

TTTTTT

Example 2: Implicit MethodNodal temperatures when , (Example Calculations)

For the interior nodes setting and gives the following,

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

sec9=tConditionBoundary1003

0 −°= CT0=i

1=i

2=i716.934239.08478.1

326.514239.08478.139.42

326.51)4239.0()4239.021()1004239.0(

)21(

32

31

32

31

32

31

21

32

31

30

=−

=−+−

=−×++×−

=−++−

TTTT

TTTTTT λλλ

669.304239.08478.14239.0

)21(3

33

23

1

22

33

32

31

=−+−

=−++−

TTTTTTT λλλ

=

−−−

−−−

434.33876.23669.30716.93

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

34

33

32

31

TTTT

2=j 4,3,2,1=i

Hence, the nodal temps at are

Example 2: Implicit Method

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

sec9=t

=

−−−

−−−

434.33876.23669.30716.93

8478.14239.0004239.08478.14239.0004239.08478.14239.0004239.08478.1

34

33

32

31

TTTT

=

243.24809.26292.36043.59

34

33

32

31

TTTT

=

25243.24809.26292.36043.59

100

35

34

33

32

31

30

TTTTTT

Example 2: Implicit MethodTo 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 MethodWHY:

Using the implicit method our approximation of was of

accuracy, while our approximation of was of accuracy.2

2

xT

∂∂ 2)( xO ∆

tT∂∂ )( tO ∆

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

( ) ( )

∆+−

+∆

+−≈

∂∂ +

−++

+−+2

11

111

211

,2

2 222 x

TTTx

TTTxT j

ij

ij

ij

ij

ij

i

ji

α

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 ,1+j

tTT

tT j

ij

i

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 descritizethe physical problem. We then solve a system of simultaneous linear equations to find the temperature at every node at any point in time.

The Crank-Nicolson Method

( ) ( ) tTT

xTTT

xTTT j

ij

ij

ij

ij

ij

ij

ij

i

∆−

=

∆+−

+∆

+− ++−

+++−+

1

2

11

111

211 22

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2 +−

++

++− +−+=−++− λλλλλλ

( )2xt

∆∆

= αλ

Example 3: Crank-NicolsonConsider 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: , ,

The initial temperature of the rod is .

C°100C°25 m05.0

0=t9=t mx 01.0=∆ st 3=∆

KmWk−

= 54 37800mkg

=ρ KkgJC−

= 490

C°20

0=i 1 2 3 4 5

m01.0

CT °= 25CT °=100

Example 3: Crank-NicolsonRecall,

therefore,

Then,

Ckρ

α =

490780054×

sm /104129.1 25−×=

( )2xt

∆∆

= αλ

( )25

01.03104129.1 −×=

4239.0= .

Number of time steps,

Boundary Conditions

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

ttt initialfinal

−=

309 −

=

3=

.

.

3,2,1,0allfor25

100

5

0 =

°=

°=j

CTCT

j

j

C°20sec0=t .

1,2,3,4 allfor ,200 =°= iCTi

Example 3: Crank-NicolsonNodal 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.

sec0=t

CT °=10000

nodesInterior

20

20

20

20

04

03

02

01

°=

°=

°=

°=

CTCTCTCT

CT °= 2505

0=j

ji

ji

ji

ji

ji

ji TTTTTT 11

11

111 )1(2)1(2 +−

++

++− +−+=−++− λλλλλλ

Example 3: Crank-NicolsonNodal 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

sec3=tConditionBoundary1001

0 −°= CT0=i

1=i

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

++=−+−

+−+=−++×−

+−+=−++−

TTTT

TTTTTT λλλλλλ

30.1164239.08478.2 12

11 =− TT

=

−−−

−−−

718.52000.40000.4030.116

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

14

13

12

11

TTTT

0=j 4,3,2,1=i

Hence, the nodal temps at are

Example 3: Crank-Nicolson

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

sec3=t

=

−−−

−−−

718.52000.40000.4030.116

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

14

13

12

11

TTTT

=

607.21797.20746.23372.44

14

13

12

11

TTTT

=

25607.21797.20746.23372.44

100

15

14

13

12

11

10

TTTTTT

Nodal temperatures when , (Example Calculations)

For the interior nodes setting and gives the following,

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

Example 3: Crank-Nicolsonsec6=t

ConditionBoundary10020 −°= CT0=i

1=i

066.10125.5139.424239.08478.239.42746.23)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

TTTTTTTT λλλλλλ

1=j 4,3,2,1=i

971.1454239.08478.2 22

21 =− TT

=

−−−

−−−

908.54187.43985.54971.145

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

24

23

22

21

TTTT

Hence, the nodal temps at are

Example 3: Crank-Nicolson

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

sec6=t

=

−−−

−−−

908.54187.43985.54971.145

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

24

23

22

21

TTTT

=

730.22174.23075.31883.55

24

23

22

21

TTTT

=

25730.22174.23075.31883.55

100

25

24

23

22

21

20

TTTTTT

Example 3: Crank-NicolsonNodal temperatures when , (Example Calculations)

For the interior nodes setting and gives the following,

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

sec9=tConditionBoundary1003

0 −°= CT0=i

1=i2=j 4,3,2,1=i

173.13388.6439.424239.08478.239.42075.31)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

TTTTTTTT λλλλλλ

34.1624239.08478.2 32

31 =− TT

=

−−−

−−−

210.57509.49318.6934.162

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

34

33

32

31

TTTT

Hence, the nodal temps at are

Example 3: Crank-Nicolson

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

sec9=t

=

−−−

−−−

210.57509.49318.6934.162

8478.24239.0004239.08478.24239.0004239.08478.24239.0004239.08478.2

34

33

32

31

TTTT

=

042.24562.26613.37604.62

34

33

32

31

TTTT

=

25042.24562.26613.37604.62

100

35

34

33

32

31

30

TTTTTT

Example 3: Crank-NicolsonTo 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 Implicit Crank-Nicolson Analytical

34

33

32

31

TTTT

042.24562.26613.37604.62

243.24809.26292.36043.59

872.22266.27132.39953.65

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.

610.23844.25084.37510.62

THE END

Recommended