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CISE301_Topic9 KFUPM 1
SE301: Numerical Methods
Topic 9 Partial Differential Equations
(PDEs)Lectures 37-39
KFUPM
Read 29.1-29.2 & 30.1-30.4
CISE301_Topic9 KFUPM 2
Lecture 37Partial Differential
Equations
Partial Differential Equations (PDEs). What is a PDE? Examples of Important PDEs. Classification of PDEs.
CISE301_Topic9 KFUPM 3
Partial Differential Equations
s)t variableindependen are and example (in the
st variableindependen moreor twoinvolves PDE
),(),(
:Examples
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 KFUPM 4
Notation
.derivativeorder highest theoforder PDE theofOrder
),(
),(
2
2
2
tx
txuu
x
txuu
xt
xx
CISE301_Topic9 KFUPM 5
Linear PDEClassification
0322
032
032
PDENonlinear of Examples
0432
0)2cos(4312
:PDElinear of Example
sderivative its andfunction
unknown in thelinear isit iflinear is PDEA
2
ttxtxx
txtxx
ttxtxx
xtxx
xttxtxx
uuuu
uuu
uuu
uuu
tuuuu
CISE301_Topic9 KFUPM 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 KFUPM 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 KFUPM 8
Heat Equation
)sin()0,(
0),1(),0(
0),(),(
2
2
xxT
tTtTt
txT
x
txT
x
ice iceTime t
Temperature T(x,t)
Position xx1
t1
),( 11 txT
CISE301_Topic9 KFUPM 9
Linear Second Order PDEsClassification
Elliptic04
Hyperbolic04
Parabolic04
:follows as 4on based classfied is
and ,, y, x,offunction a is D
y and x of functions are C and B, A,
,0
s)t variableindependen-(2 PDElinear order secondA
2
2
2
2
ACB
ACB
ACB
AC)(B
uuu
DuCuBuA
yx
yyxyxx
CISE301_Topic9 KFUPM 10
Linear Second Order PDEExamples (Classification)
Hyperbolicis
ACBCBA
t
txu
x
txu
Parabolicis
ACBCBkA
t
txT
x
txTk
Equation Wave
041,0,1
0),(),(
Equation Wave
______________________________________
EquationHeat
040,0,
0),(),(
EquationHeat
2
2
2
2
2
2
2
2
CISE301_Topic9 KFUPM 11
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 KFUPM 12
Examples of PDEs PDEs are used to model many systems in
many different fields of science and engineering.
Important Examples: Wave Equation Heat Equation Laplace Equation Biharmonic Equation
CISE301_Topic9 KFUPM 13
Heat Equation
t
tzyxu
z
tzyxu
y
tzyxu
x
tzyxu
),,,(),,,(),,,(),,,(2
2
2
2
2
2
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) .
CISE301_Topic9 KFUPM 14
Simpler Heat Equation
t
txu
x
txu
),(),(2
2
u(x,t) is used to represent the temperature at time t at the point x of the thin rod.
x
CISE301_Topic9 KFUPM 15
Wave Equation
2
2
2
2
2
2
2
2 ),,,(),,,(),,,(),,,(
t
tzyxu
z
tzyxu
y
tzyxu
x
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) .
Used to model movement of 3D elastic body.
CISE301_Topic9 KFUPM 16
Laplace Equation
0),,,(),,,(),,,(
2
2
2
2
2
2
z
tzyxu
y
tzyxu
x
tzyxu
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.
CISE301_Topic9 KFUPM 17
Biharmonic Equation
0),,(),,(
2),,(
4
4
22
4
4
4
y
tyxu
yx
tyxu
x
tyxu
Used in the study of elastic stress.
CISE301_Topic9 KFUPM 18
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),(),(
:EquationHeat2
2
xxu
tu
tut
txu
x
txu
region of interest
x1
t
CISE301_Topic9 KFUPM 19
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 KFUPM 20
Lecture 38Parabolic Equations
Parabolic Equations Heat Conduction Equation Explicit Method Implicit Method Cranks Nicolson Method
CISE301_Topic9 KFUPM 21
Parabolic Equations
04 if parabolic is
and ,, y, x,offunction a is D
y and x of functions are C and B, A,
,0
y) , x st variableindependen-(2 PDElinear order secondA
2
ACB
uuu
DuCuBuA
yx
yyxyxx
CISE301_Topic9 KFUPM 22
Parabolic Problems
solution. aspecify uniquely toneeded are conditionsBoundary *
)04( problem Parabolic *
)sin()0,(
0),1(),0(
0),(),(
:EquationHeat
2
2
2
ACB
xxT
tTtTt
txT
x
txT
x
ice ice
CISE301_Topic9 KFUPM 23
First Order Partial Derivative Finite Difference
x
TT
x
T
x
TT
x
T
x
TT
x
T jijijijijiji
,1,,,1,1,1 ,
2
ForwardDifferenceMethod
Central Difference Method
Backward Difference
Method
CISE301_Topic9 KFUPM 24
Finite Difference Methods
t
TT
t
txT
x
TT
x
txT
t
TTT
t
txu
x
TTT
x
txT
x
TT
x
txT
jijijiji
jijiji
jijiji
jiji
,1,,,1
2
1,,1,
2
2
2
,1,,1
2
2
,1,1
),(,
),(
:Formula DifferenceForward
)(
2),(
)(
2),(
2
),(
: FormulasDifferenceCentral
t]and x offunction is T [ formula difference finiteby sderivative theReplace
CISE301_Topic9 KFUPM 25
Finite Difference MethodsNew Notation
t
TT
t
txT
x
TT
x
txT
t
TTT
t
txT
x
TTT
x
txT
x
TT
x
txT
li
li
li
li
li
li
li
li
li
li
li
li
11
2
11
2
2
211
2
2
11
),(,
),(
:Formula DifferenceForward
)(
2),(
)(
2),(
2
),(
: FormulasDifferenceCentralSuperscript for t-axis
andSubscript for x-axisTil-1=Ti,j-1=T(x,t-∆t)
CISE301_Topic9 KFUPM 26
Solution of the PDEs
points. grid at the t)T(x, of value theDetermine
:meansSolution
)sin()0,(
0),1(),0(
0),(),(
EquationHeat2
2
xxT
tTtTt
txT
x
txT
t
x
CISE301_Topic9 KFUPM 27
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 KFUPM 28
Explicit Method
),(),()21(),(),(
0),(),(1
),(),(2),(1
),(),(),(),(2),(
0),(),(
2
2
2
2
2
thxutxuthxuktxuh
kDefine
txuktxuk
thxutxuthxuh
k
txuktxu
h
thxutxuthxut
txu
x
txu
CISE301_Topic9 KFUPM 29
Explicit MethodHow Do We Compute?
means
thxutxuthxuktxu ),(),()21(),(),(
u(x-h,t) u(x,t) u(x+h,t)
u(x,t+k)
CISE301_Topic9 KFUPM 30
Explicit MethodHow Do We Compute?
means
thxutxuthxuktxu ),(),()21(),(),(
CISE301_Topic9 KFUPM 31
Explicit Method
.slowit makes This
.2
0)21( stability, guarantee To
.magnified are errors unstable beCan
),(),()21(),(),(
:usingdirectly computed becan ),(
2h kor
thxutxuthxuktxu
ktxu
CISE301_Topic9 KFUPM 32
Crank-Nicolson Method
),(),(),(),(
2,
0),(),(1
),(),(2),(1
),(),(),(),(2),(
0),(),(
2
2
2
2
2
thxutxurthxuktxus
srk
hsDefine
ktxutxuk
thxutxuthxuh
k
ktxutxu
h
thxutxuthxut
txu
x
txu
CISE301_Topic9 KFUPM 33
Explicit MethodHow Do We Compute?
means
thxutxurthxuktxus ),(),(),(),(
u(x-h,t) u(x,t) u(x+h,t)
u(x,t - k)
CISE301_Topic9 KFUPM 34
Crank-Nicolson Method
4
3
2
1
4
3
2
1
1
11
11
1
:equations of system lTridiagona a as expressed becan
),(),(),(),(
:equation The
b
b
b
b
u
u
u
u
r
r
r
r
thxutxurthxuktxus
CISE301_Topic9 KFUPM 35
Crank-Nicolson Method
Method).Explicit the to(comparedk h,larger usecan We
error). ofion magnificat (No stable is method The
equations.linear of system lTridiagona a solving involves method The
CISE301_Topic9 KFUPM 36
Examples
Explicit method to solve Parabolic PDEs.
Cranks-Nicholson Method.
CISE301_Topic9 KFUPM 37
Heat Equation
solution. aspecify uniquely toneeded are conditionsAuxiliary *
)04( problem Parabolic *
)sin()0,(
0),1(),0(
0),(),(
2
2
2
ACB
xxu
tutut
txu
x
txu
x
ice ice
CISE301_Topic9 KFUPM 38
Example 1
4
]1,0[],1,0[for x)u(t, find to25.0,25.0
)sin()0,(
0),1(),0(
0
:PDE theSolve
2
2
2
h
k
txkhUse
xxu
tutut
u(x,t)
x
u(x,t)
CISE301_Topic9 KFUPM 39
Example 1 (Cont.)
),(4),(7),(4),(
0),(),(4),(),(2),(16
0),(),(),(),(2),(
0),(),(
2
2
2
thxutxuthxuktxu
txuktxuthxutxuthxu
k
txuktxu
h
thxutxuthxu
t
txu
x
txu
CISE301_Topic9 KFUPM 40
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 KFUPM 41
Example 1
9497.00)4/sin(7)2/sin(4
)0,0(4)0,25(.7)0,5(.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 KFUPM 42
Example 1
1716.0)4/sin(4)2/sin(7)4/3sin(4
)0,25.0(4)0,5.0(7)0,75.0(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 KFUPM 43
Remarks on Example 1
0.025kLet
0.03125kselect toneeds One
021 :results accurateFor
721 :because
accuratenot probably are results obtained The
CISE301_Topic9 KFUPM 44
Example 1),(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 KFUPM 45
Example 1
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)
0.54140)4/sin(2.)2/sin(4.0
)0,0(4.0)0,25.0(2.0)0,5.0(4.0)025.0,25.0(
uuuu
CISE301_Topic9 KFUPM 46
Example 1
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/sin(4.0)2/sin(2.)4/3sin(4.0
)0,25.0(4.0)0,5.0(2.0)0,75.0(4.0)025.0,5.0(
uuuu
CISE301_Topic9 KFUPM 47
Example 2
]1,0[],1,0[for x)u(t, find to25.0,25.0
methodNicolson -Crank using Solve
)sin()0,(
0),1(),0(
0
:PDE theSolve
2
2
txkhUse
xxu
tutut
u(x,t)
x
u(x,t)
CISE301_Topic9 KFUPM 48
Example 2Crank-Nicolson Method
),(),(25.2),(),(25.0
25.22,25.0
0),(),(4),(),(2),(16
),(),(),(),(2),(
0),(),(
2
2
2
2
thxtxuthxuktxu
srk
hsDefine
ktxutxuthxutxuthxu
k
ktxutxu
h
thxutxuthxu
t
txu
x
txu
CISE301_Topic9 KFUPM 49
Example 2
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)
2125.20)25.0sin(25.0
)25.0,5.0()25.0,25.0(25.2)25.0,0()25.0,0.0(25.0
uu
uuuu
u1 u2 u3
CISE301_Topic9 KFUPM 50
Example 2
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)
321 25.2)5.0sin(25.0
)25.0,75.0()25.0,5(.25.2)25.0,25.0()5.0,0(25.0
uuu
uuuu
u1 u2 u3
CISE301_Topic9 KFUPM 51
Example 2
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)
025.2)75.0sin(25.0
)25.0,1()25.0,75(.25.2)25.0,5.0()0,75.0(25.0
32
uu
uuuu
u1 u2 u3
CISE301_Topic9 KFUPM 52
Example 2Crank-Nicolson Method
21151.0
2912.0
21152.0
)75.0sin(25.0
)5.0sin(25.0
)25.0sin(25.0
25.21
125.21
1252
:system al tridiagonfollowing
theofsolution a toconverted is PDE theofsolution The
3
2
1
3
2
1
u
u
u
u
u
u.
CISE301_Topic9 KFUPM 53
Example 2Second Row
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)
2125.202115.0
)5.0,5.0()5.0,25.0(25.2)5.0,0()25.0,25.0(25.0
uu
uuuu
0.2115 0.2991 0.2115
u1 u2 u3
CISE301_Topic9 KFUPM 54
Example 2The process is continued until the values of u(x,t) on the desired grid are computed.
CISE301_Topic9 KFUPM 55
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 KFUPM 56
Lecture 39Elliptic
Equations Elliptic Equations Laplace Equation Solution
CISE301_Topic9 KFUPM 57
Elliptic Equations
04 if Elliptic is
and ,, y, x,offunction a is D
y and x of functions are C and B, A,
,0
y) , x st variableindependen-(2 PDElinear order secondA
2
ACB
uuu
DuCuBuA
yx
yyxyxx
CISE301_Topic9 KFUPM 58
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)f(x,
y)(x,point at re temperatustatesteady :
),(),(),(
2
2
2
2
T
yxfy
yxT
x
yxT
CISE301_Topic9 KFUPM 59
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 KFUPM 60
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 KFUPM 61
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 KFUPM 62
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 KFUPM 63
Solution Technique
jiT ,1jiT ,jiT ,1
1, jiT
1, jiT
04 ,1,1,,1,1 jijijijiji TTTTT
CISE301_Topic9 KFUPM 64
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 KFUPM 65
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 KFUPM 66
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 KFUPM 67
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 KFUPM 68
Solution The Rest of the Equations
150
100
175
50
0
75
50
0
75
4101
14101
014001
1004101
1014101
1014001
100410
10141
1014
33
23
13
32
22
12
31
21
11
T
T
T
T
T
T
T
T
T
CISE301_Topic9 KFUPM 69
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