cise301_topic91 cise301: numerical methods topic 9 partial differential equations (pdes) lectures...

CISE301_Topic9 1

CISE301: Numerical Methods

Topic 9 Partial Differential Equations

(PDEs)Lectures 37-39

KFUPM(Term 101)Section 04

Read 29.1-29.2 & 30.1-30.4

CISE301_Topic9 2

Lecture 37Partial Differential

Equations

Partial Differential Equations (PDEs). What is a PDE? Examples of Important PDEs. Classification of PDEs.

CISE301_Topic9 3

Partial Differential Equations

s)t variableindependen are and example the(in

st variableindependen moreor twoinvolves PDE

),(),(

:Example

2

2

tx

t

txu

x

txu

A partial differential equation (PDE) is an equation that involves an unknown function and its partial derivatives.

CISE301_Topic9 4

Notation

.derivativeorder highest theoforder PDE theofOrder

),(

),(

2

2

2

tx

txuu

x

txuu

xt

xx

CISE301_Topic9 5

Linear PDEClassification

0322

032

032

PDENonlinear of Examples

0432

0)2cos(4312

:linear PDE of Example

sderivative its and function

unknown theinlinear isit iflinear is PDEA

2

ttxtxx

txtxx

ttxtxx

xtxx

xttxtxx

uuuu

uuu

uuu

uuu

tuuuu

CISE301_Topic9 6

Representing the Solution of a PDE(Two Independent Variables) Three main ways to represent the solution

Different curves are used for different values of one of the independent variable

x1

t1

),( 11 txT

Three dimensional plot of the function T(x,t)

The axis represent the independent variables. The value of the function is displayed at grid points

T=3.5

T=5.2

CISE301_Topic9 7

Heat Equation

)sin()0,(

0),1(),0(

0),(),(

2

2

xxT

tTtTt

txT

x

txT

x

ice iceTemperature at different x at t=0

Temperature at different x at t=h

Temperature

Position x

Thin metal rod insulated everywhere except at the edges. At t =0 the rod is placed in ice

Different curve is used for each value

of t

CISE301_Topic9 8

Examples of PDEs PDEs are used to model many systems in

many different fields of science and engineering.

Important Examples: Laplace Equation Heat Equation Wave Equation

CISE301_Topic9 9

Laplace Equation

Used to describe the steady state distribution of heat in a body.

Also used to describe the steady state distribution of electrical charge in a body.

0),,(),,(),,(

2

2

2

2

2

2

z

zyxu

y

zyxu

x

zyxu

CISE301_Topic9 10

Heat Equation

2

2

2

2

2

2

),,,(

z

u

y

u

x

u

t

tzyxu

The function u(x,y,z,t) is used to represent the temperature at time t in a physical body at a point with coordinates (x,y,z)

is the thermal diffusivity. It is sufficient to consider the case = 1.

CISE301_Topic9 11

Simpler Heat Equation

2

2 ),(),(

x

txT

t

txT

T(x,t) is used to represent the temperature at time t at the point x of the thin rod.

x

CISE301_Topic9 12

Wave Equation

2

2

2

2

2

22

2

2

),,,(

z

u

y

u

x

uc

t

tzyxu

The function u(x,y,z,t) is used to represent the displacement at time t of a particle whose position at rest is (x,y,z) .

The constant c represents the propagation speed of the wave.

CISE301_Topic9 13

Classification of PDEs Linear Second order PDEs are important

sets of equations that are used to model many systems in many different fields of science and engineering.

Classification is important because: Each category relates to specific engineering

problems. Different approaches are used to solve these

categories.

CISE301_Topic9 14

Linear Second Order PDEsClassification

Hyperbolic04

Parabolic04

Elliptic04

:follows as 4 on based classified is

and ,, ,, of functiona is D

and of functions are C and B,A,

,0

s)t variableindependen-(2 Elinear PDorder second A

2

2

2

2

ACB

ACB

ACB

AC)(B

uuuyx

yx

DuCuBuA

yx

yyxyxx

CISE301_Topic9 15

Linear Second Order PDEExamples (Classification)

0

sin ,cos

sin ,sin

sin),( :solution possible One

EquationLaplace

041,0,1

0),(),(

EquationLaplace

2

2

2

2

2

yyxx

xyy

xy

xxx

xx

x

uu

yeuyeu

yeuyeu

yeyxu

Ellipticis

ACBCBA

y

yxu

x

yxu

CISE301_Topic9 16

Linear Second Order PDEExamples (Classification)

Hyperbolicis

ACBCBcA

t

txu

x

txuc

Parabolicis

ACBCBA

t

txu

x

txu

Equation Wave

041 ,0 ,0

0),(),(

Equation Wave

______________________________________

EquationHeat

040 ,0 ,

0),(),(

EquationHeat

22

2

2

2

22

2

2

2

CISE301_Topic9 17

Boundary Conditions for PDEs To uniquely specify a solution to the PDE,

a set of boundary conditions are needed. Both regular and irregular boundaries are

possible.

)sin()0,(

0),1(

0),0(

0),(),(

:EquationHeat 2

2

xxu

tu

tut

txu

x

txu

region of interest

x1

t

CISE301_Topic9 18

The Solution Methods for PDEs Analytic solutions are possible for simple

and special (idealized) cases only.

To make use of the nature of the equations, different methods are used to solve different classes of PDEs.

The methods discussed here are based on the finite difference technique.

CISE301_Topic9 19

Lecture 38Parabolic Equations

Parabolic Equations Heat Conduction Equation Explicit Method Implicit Method Cranks Nicolson Method

CISE301_Topic9 20

Parabolic Equations

04 if parabolic is

and ,, y,x, of functiona is D

yandx of functions are C and B,A,

,0

) , st variableindependen-(2 linear PDEorder second A

2

ACB

uuu

DuCuBuA

yx

yx

yyxyxx

CISE301_Topic9 21

Parabolic Problems

solution.a specify uniquely toneeded are conditionsBoundary *

)04( problem lic Parabo*

)sin()0,(

0),1(),0(

),(),(:EquationHeat

2

2

2

ACB

xxT

tTtTx

txT

t

txT

x

ice ice

CISE301_Topic9 22

Finite Difference Methods

t

x

Divide the interval x into sub-intervals, each of width h

Divide the interval t into sub-intervals, each of width k

A grid of points is used forthe finite difference solution

Ti,j represents T(xi, tj) Replace the derivates by

finite-difference formulas

CISE301_Topic9 23

Finite Difference Methods

k

TT

t

TT

t

txT

t

T

h

TTT

x

TTT

x

txT

x

T

jijijiji

jijijijijiji

,1,,1,

2,1,,1

2,1,,1

2

2

2

2

),(

:for Formula DifferenceForward

2

)(

2),(

:for Formula DifferenceCentral

formulas difference finiteby sderivative theReplace

CISE301_Topic9 24

Solution of the Heat Equation

• Two solutions to the Parabolic Equation (Heat Equation) will be presented:

1. Explicit Method:

Simple, Stability Problems.

2. Crank-Nicolson Method:

Involves the solution of a Tridiagonal system of equations, Stable.

CISE301_Topic9 25

Explicit Method

),(),()21(),(),(

),(),(2),(),(),(

),(),(2),(),(),(

),(),(

2

2

2

2

2

thxTtxTthxTktxTh

kDefine

thxTtxTthxTh

ktxTktxT

h

thxTtxTthxT

k

txTktxTx

txT

t

txT

CISE301_Topic9 26

Explicit MethodHow Do We Compute?

means

thxTtxTthxTktxT ),(),()21(),(),(

T(x-h,t) T(x,t) T(x+h,t)

T(x,t+k)

CISE301_Topic9 27

Convergence and Stability

.slowit makes This

ansmaller th much is that means This2

2

10)21( stability, guarantee To

magnified are errors unstable be Can

),(),()21(),(),(

:usingdirectly computed be can),(

2

hk

hk

thxTtxTthxTktxT

ktxT

CISE301_Topic9 28

Convergence and Stability of the Solution Convergence

The solutions converge means that the solution obtained using the finite difference method approaches the true solution as the steps approach zero.

Stability: An algorithm is stable if the errors at each

stage of the computation are not magnified as the computation progresses.

tx and

CISE301_Topic9 29

Example 1: Heat Equation

4

]1,0[],1,0[for ),( find to25.0,25.0

)sin()0,(

0),1(),0(

0

: PDE theSolve

2

2

2

h

k

txtxukhUse

xxu

tutut

u(x,t)

x

u(x,t)

x

ice ice

CISE301_Topic9 30

Example 1

),(4),(7),(4),(

0),(),(4),(),(2),(16

0),(),(),(),(2),(

0),(),(

2

2

2

thxutxuthxuktxu

txuktxuthxutxuthxuk

txuktxu

h

thxutxuthxut

txu

x

txu

CISE301_Topic9 31

Example 1),(4),(7),(4),( thxutxuthxuktxu

t=0

t=0.25

t=0.5

t=0.75

t=1.0

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

CISE301_Topic9 32

Example 1

9497.0)2/sin(4)4/sin(70

)0,5.0(4)0,25.0(7)0,0(4)25.0,25.0(

uuuu

t=0

t=0.25

t=0.5

t=0.75

t=1.0

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

CISE301_Topic9 33

Example 1

1716.0)4/3sin(4)2/sin(7)4/sin(4

)0,75.0(4)0,5.0(7)0,25.0(4)25.0,5.0(

uuuu

t=0

t=0.25

t=0.5

t=0.75

t=1.0

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

CISE301_Topic9 34

Remarks on Example 1

4.0 then0.025,choose example,For

0.031252

(0.25)

2select toneeds One

021 :results accurateFor

721 :because

accuratenot probably are results obtained The

2

22

h

kk

hk

CISE301_Topic9 35

Example 1 – cont’d),(4.0),(2.0),(4.0),( thxutxuthxuktxu

t=0

t=0.025

t=0.05

t=0.075

t=0.10

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

CISE301_Topic9 36

Example 1 – cont’d

t=0

t=0.025

t=0.05

t=0.075

t=0.10

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

0.5414)2/sin(4.0)4/sin(2.00

)0,5.0(4.0)0,25.0(2.0)0,0(4.0)025.0,25.0(

uuuu

CISE301_Topic9 37

Example 1 – cont’d

t=0

t=0.025

t=0.05

t=0.075

t=0.10

x=0.25 x=0.5x=0.0 x=0.75 x=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

0.7657)4/3sin(4.0)2/sin(2.0)4/sin(4.0

)0,75.0(4.0)0,5.0(2.0)0,25.0(4.0)025.0,5.0(

uuuu

CISE301_Topic9 38

Crank-Nicolson Method

Method). Explicit the to(compared ,larger use can We

error). of ionmagnificat (No stable is method The

equations.linear of system lTridiagonaa solving involves method The

kh

CISE301_Topic9 39


22

2 ),(),(2),(),(

),(),(),(

:lyrespective formulas and

their withsderivative partial second andfirst the Replace3.

widthof lssubinterva into interval the Divide2.

widthof lssubinterva into interval the Divide1.

method difference finite theon Based

h

thxutxuthxu

x

txu

k

ktxutxu

t

txu

fferencecentral dibackward

kt

hx

CISE301_Topic9 40


),(),(),( )21(),(

),(),(),(),(2),(

),(),(),(),(2),(

becomes ),(),(

:ionHeat Equat

222

2

2

2

2

ktxuthxuh

ktxu

h

kthxu

h

k

ktxutxuthxutxuthxuh

kk

ktxutxu

h

thxutxuthxu

t

txu

x

txu

CISE301_Topic9 41


),(),(),()21(),(

:becomes equation Heat then Define 2

ktxuthxutxuthxuh

k

u(x-h,t) u(x,t) u(x+h,t)

u(x,t - k)

CISE301_Topic9 42


0,41,51,41,3

0,31,41,31,2

0,21,31,21,1

0,11,21,11,0

1,,1,,1

)21(

)21(

)21(

)21(

:)1(fix equations of systema as expanded be can and

)21(

:as rewritten be can

),(),(),()21(),(

:equation The

uuuu

uuuu

uuuu

uuuu

j

uuuu

ktxuthxutxuthxu

jijijiji

CISE301_Topic9 43


hxxxuu

hxhxhxhxx

uuuu

uu

u

u

uu

u

u

u

u

ktxuthxutxuthxu

5 and at aluesboundary v theare and

4 and ,3 ,2 ,at

valuesre temperatuinitial theare and , , , where

21

21

21

21

:equations of system lTridiagonaa as expressed be can

),(),(),()21(),(

001,51,0

0000

0,40,30,20,1

1,50,4

0,3

0,2

1,00,1

1,4

1,3

1,2

1,1

CISE301_Topic9 44


etc. ,3at valuesre temperatucompute tostep above Repeat the

and , , , compute To

21

21

21

21

)2( equations of system al tridiagonseconda Solve

2at valuesre temperatu thecompute To

at and , , , valuesre temperatuThe

:produces system al tridiagon theof solution The

0

2,42,32,22,1

2,51,4

1,3

1,2

2,01,1

2,4

2,3

2,2

2,1

0

01,41,31,21,1

kt

uuuu

uu

u

u

uu

u

u

u

u

j

ktt

kttuuuu

CISE301_Topic9 45

Example 2

]1,0[],1,0[for ),( find to25.0,25.0

method Nicolson-Crank using Solve

)sin()0,(

0),1(),0(

0),(),(

: PDE theSolve

2

2

txtxukhUse

xxu

tutut

txu

x

txu

CISE301_Topic9 46

Example 2Crank-Nicolson Method

1,,1,,1

2

2

2

2

4 9 4

),(),( 4),( 9),( 4

4

0),(),(4),(),(2),(16

),(),(),(),(2),(

0),(),(

jijijiji uuuu

ktxuthxutxuthxuh

kDefine

ktxutxuthxutxuthxuk

ktxutxu

h

thxutxuthxut

txu

x

txu

CISE301_Topic9 47

Example 2

)4/3sin(94 494

)2/sin(494494

)4/sin( 49 494

1,31,20,31,41,31,2

1,31,21,10,21,31,21,1

1,21,10,11,21,11,0

uuuuuu

uuuuuuu

uuuuuu

u1,1 u2,1 u3,1

t0=0

t1=0.25

t2=0.5

t3=0.75

t4=1.0

x1=0.25 x2=0.5x0=0.0 x3=0.75 x4=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

u1,4 u2,4 u3,4

u1,3 u2,3 u3,3

u1,2 u2,2 u3,2

CISE301_Topic9 48

Example 2Solution of Row 1 at t1=0.25 sec

21151.0

29912.0

21151.0

)75.0sin(

)5.0sin(

)25.0sin(

94

494

49

:equations of system al tridiagonfollowing theof

solution theis sec 25.0at PDE theof Solution The

1,3

1,2

1,1

1,3

1,2

1,1

1

u

u

u

u

u

u

t

CISE301_Topic9 49

Example 2: Second Row at t2=0.5 sec

21151.094 494

29912.0494494

21151.0 49 494

2,32,21,32,42,32,2

2,32,22,11,22,32,22,1

2,22,11,12,22,12,0

uuuuuu

uuuuuuu

uuuuuu

u1,1 u2,1 u3,1

t0=0

t1=0.25

t2=0.5

t3=0.75

t4=1.0

x1=0.25 x2=0.5x0=0.0 x3=0.75 x4=1.0

0

0

0

0

0

0

0

0

0

0

Sin(0.25π) Sin(0. 5π) Sin(0.75π)

u1,4 u2,4 u3,4

u1,3 u2,3 u3,3

u1,2 u2,2 u3,2

CISE301_Topic9 50


063267.0

089473.0

063267.0

21151.0

29912.0

21151.0

94

494

49



2,3

2,2

2,1

1,3

1,2

1,1

2,3

2,2

2,1

2

u

u

u

u

u

u

u

u

u

t

CISE301_Topic9 51


018924.0

026763.0

018924.0

063267.0

089473.0

063267.0

94

494

49



3,3

3,2

3,1

2,3

2,2

2,1

3,3

3,2

3,1

3

u

u

u

u

u

u

u

u

u

t

CISE301_Topic9 52

Example 2Solution of Row 4 at t4=1 sec

0056606.0

0080053.0

0056606.0

018924.0

026763.0

018924.0

94

494

49


solution theis sec 1at PDE theof Solution The

4,3

4,2

4,1

3,3

3,2

3,1

4,3

4,2

4,1

4

u

u

u

u

u

u

u

u

u

t

CISE301_Topic9 53

RemarksThe Explicit Method:

•One needs to select small k to ensure stability.

•Computation per point is very simple but many points are needed.

Cranks Nicolson:

• Requires the solution of a Tridiagonal system.

• Stable (Larger k can be used).

CISE301_Topic9 54

Lecture 39Elliptic

Equations Elliptic Equations Laplace Equation Solution

CISE301_Topic9 55

Elliptic Equations

04 if Ellipticis

and ,, ,, of functiona is D

and of functions are C and B,A,

,0

) , st variableindependen-(2 linear PDEorder second A

2

ACB

uuuyx

yx

DuCuBuA

yx

yx

yyxyxx

CISE301_Topic9 56

Laplace Equation Laplace equation appears in several

engineering problems such as: Studying the steady state distribution of heat in a

body. Studying the steady state distribution of electrical

charge in a body.

sink)heat (or sourceheat :),(

y)(x,point at re temperatustatesteady :

),(),(),(

2

2

2

2

yxf

T

yxfy

yxT

x

yxT

CISE301_Topic9 57

Laplace Equation

EllipticACB

CBA

yxfy

yxT

x

yxT

044

1,0,1

),(),(),(

2

2

2

2

2

Temperature is a function of the position (x and y)

When no heat source is available f(x,y)=0

CISE301_Topic9 58

Solution Technique A grid is used to divide the region of

interest. Since the PDE is satisfied at each point in

the area, it must be satisfied at each point of the grid.

A finite difference approximation is obtained at each grid point.

2

1,,1,

2

2

2

,1,,1

2

2 2),(,

2),(

y

TTT

y

yxT

x

TTT

x

yxT jijijijijiji

CISE301_Topic9 59

Solution Technique

0

22

:by edapproximat is

0),(),(

2),(

,2),(

2

1,,1,

2

,1,,1

2

2

2

2

2

1,,1,

2

2

2

,1,,1

2

2

y

TTT

x

TTT

y

yxT

x

yxT

y

TTT

y

yxT

x

TTT

x

yxT

jijijijijiji

jijiji

jijiji

CISE301_Topic9 60

Solution Technique

04

:

) (

022

,1,1,,1,1

2

1,,1,

2

,1,,1

jijijijiji

jijijijijiji

TTTTT

hyxAssume

EquationDifferenceLaplacian

y

TTT

x

TTT

CISE301_Topic9 61

Solution Technique

jiT ,1jiT ,jiT ,1

1, jiT

1, jiT

04 ,1,1,,1,1 jijijijiji TTTTT

CISE301_Topic9 62

Example It is required to determine the steady state

temperature at all points of a heated sheet of metal. The edges of the sheet are kept at a constant temperature: 100, 50, 0, and 75 degrees.

50

0

100

75

The sheet is divided to 5X5 grids.

CISE301_Topic9 63

Example1004,1 T 1004,2 T 1004,3 T

503,4 T

502,4 T

501,4 T

753,0 T

752,0 T

751,0 T

00,1 T 00,2 T 00,3 T

Known

To be determined

3,1T

2,1T

1,1T

3,2T

2,2T

1,2T

3,3T

2,3T

1,3T

CISE301_Topic9 64

First Equation1004,1 T 1004,2 T

753,0 T

752,0 T

Known

To be determined

3,1T

2,1T

3,2T

2,2T

0410075

04

3,13,22,1

3,13,22,14,13,0

TTT

TTTTT

CISE301_Topic9 65

Another Equation1004,1 T 1004,2 T 1004,3 T

Known

To be determined

3,1T

2,1T

3,2T

2,2T

3,3T

2,3T

04100

04

3,22,23,33,1

3,22,23,34,23,1

TTTT

TTTTT

CISE301_Topic9 66

Solution The Rest of the Equations

150

100

175

50

0

75

50

0

75

4101

14101

014001

1004101

1014101

1014001

100410

10141

1014

3,3

3,2

3,1

2,3

2,2

2,1

1,3

1,2

1,1

T

T

T

T

T

T

T

T

T

cise301_topic91 cise301: numerical methods topic 9 partial differential equations (pdes) lectures...

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