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
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Lecture 37
Partial Differential Equations




CISE301_Topic9
Partial Differential Equations (PDEs).
What is a PDE?
Examples of Important PDEs.
Classification of PDEs.
2
Partial Differential Equations
A partial differential equation (PDE) is an
equation that involves an unknown function
and its partial derivatives.
Example :
 2 u ( x, t )  u ( x, t )

2
t
x
PDE involves two or more independen t variable s
(in the example x and t are independen t variable s)
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Notation
 2 u ( x, t )
u xx 
x 2
 2 u ( x, t )
u xt 
x t
Order of the PDE  order of the highest order derivative .
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Linear PDE
Classification
A PDE is linear if it is linear in the unknown
function and its derivative s
Example of linear PDE :
2 u xx  1 u xt  3 utt  4 u x  cos( 2t )  0
2 u xx  3 ut  4 u x  0
Examples of Nonlinear PDE
2 u xx  u xt   3 utt  0
2
u xx  2 u xt  3 ut  0
2 u xx  2 u xt ut  3 ut  0
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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
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T ( x1 , t1 )
T=5.2
t1
T=3.5
x1
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
6
Heat Equation
ice
ice
x
Thin metal rod insulated
everywhere except at the
edges. At t =0 the rod is
placed in ice
 2 T ( x, t )  T ( x, t )

0
2
x
t
T (0, t )  T (1, t )  0
Different curve is
used for each value
of t
Temperature
Temperature at
different x at t=0
Position x
Temperature at
different x at t=h
T ( x,0)  sin(  x)
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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
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Laplace Equation
 2u ( x , y , z )  2u ( x , y , z )  2u ( x , y , z )


0
2
2
2
x
y
z
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.
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Heat Equation
  2u  2u  2u
 u ( x, y , z , t )
   2  2  2
t
y
z
 x




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.
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Simpler Heat Equation
 T ( x, t )  T ( x, t )

t
x 2
2
x
T(x,t) is used to represent the temperature
at time t at the point x of the thin rod.
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Wave Equation
2
2
2

 2u ( x , y , z , t )

u

u

u
2


c
 2  2
2
2

t
y
z
 x




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.
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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.
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Linear Second Order PDEs
Classification
A second order linear PD E (2 - independen t variable s)
A u xx  B u xy  C u yy  D  0,
A, B, and C are functions of x and y
D is a function of x, y , u, u x , and u y
is classified based on (B 2  4 AC) as follows :
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B 2  4 AC  0
Elliptic
B 2  4 AC  0
Parabolic
B 2  4 AC  0
Hyperbolic
14
Linear Second Order PDE
Examples (Classification)
Laplace Equation
 2u ( x , y )  2u ( x , y )

0
2
2
x
y
A  1, B  0, C  1  B 2  4 AC  0
 Laplace Equation is Elliptic
One possible solution : u ( x, y )  e x sin y
u x  e x sin y , u xx  e x sin y
u y  e x cos y , u yy   e x sin y
u xx  u yy  0
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Linear Second Order PDE
Examples (Classification)
 2u ( x , t )  u ( x , t )
Heat Equation


0
2
t
x
A   , B  0, C  0  B 2  4 AC  0
 Heat Equation is Parabolic
______________________________________
2
2

u
(
x
,
t
)

u( x, t )
2
Wave Equation c

0
2
2
x
t
A  c 2  0, B  0, C  1  B 2  4 AC  0
 Wave Equation is Hyperbolic
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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.
t

 2u ( x, t ) u ( x, t )
Heat Equation : 

0
2
t
x
u (0, t )  0
region of
interest
u (1, t )  0
u ( x,0)  sin(  x )
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x
1
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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.
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Lecture 38
Parabolic Equations





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Parabolic Equations
Heat Conduction Equation
Explicit Method
Implicit Method
Cranks Nicolson Method
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Parabolic Equations
A second order linear PDE (2 - independen t variable s x , y )
A u xx  B u xy  C u yy  D  0,
A, B, and C are functions of x and y
D is a function of x, y, u, u x , and u y
is parabolic if
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B 2  4 AC  0
20
Parabolic Problems
Heat Equation :
 T ( x, t )  2 T ( x, t )

t
x 2
T (0, t )  T (1, t )  0
T ( x,0)  sin(  x )
ice
ice
x
* Parabo lic problem
( B 2  4 AC  0)
* Boundary conditions are needed to uniquely specify a solution.
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Finite Difference Methods
Divide the interval x into sub-intervals,
each of width h
 Divide the interval t into sub-intervals,
each of width k
t
 A grid of points is used for
the finite difference solution
 Ti,j represents T(xi, tj)
 Replace the derivates by
finite-difference formulas

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x
22
Finite Difference Methods
Replace the derivative s by finite difference formulas
 2T
Central Difference Formula for
:
2
x
 2T ( x, t ) Ti 1, j  2Ti , j  Ti 1, j Ti 1, j  2Ti , j  Ti 1, j


2
2
x
( x )
h2
T
Forward Difference Formula for
:
t
T ( x, t ) Ti , j 1  Ti , j Ti , j 1  Ti , j


t
t
k
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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.
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Explicit Method
T ( x, t )  2T ( x, t )

t
x 2
T ( x, t  k )  T ( x, t ) T ( x  h, t )  2T ( x, t )  T ( x  h, t )

k
h2
k
T ( x, t  k )  T ( x, t )  2 T ( x  h, t )  2T ( x, t )  T ( x  h, t ) 
h
k
Define   2
h
T ( x, t  k )   T ( x  h, t )  (1  2  ) T ( x, t )   T ( x  h, t )
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Explicit Method
How Do We Compute?
T ( x, t  k )   T ( x  h, t )  (1  2  ) T ( x, t )   T ( x  h, t )
means
T(x,t+k)
T(x-h,t)
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T(x,t)
T(x+h,t)
26
Convergence and Stability
T ( x, t  k ) can be computed directly using :
T ( x, t  k )   T ( x  h, t )  (1  2  ) T ( x, t )   T ( x  h, t )
Can be unstable errors are magnified

1
h2
To guarantee stability, (1  2  )  0     k 
2
2
This means that k is much smaller th an h
This makes it slow .
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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 x and t approach zero.

Stability:
An algorithm is stable if the errors at each
stage of the computation are not magnified
as the computation progresses.
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Example 1: Heat Equation
Solve the PDE :
 2u(x,t) u(x,t)

0
2
t
 x
u(0, t )  u(1, t )  0
u( x,0)  sin(  x )
ice
ice
x
Use h  0.25, k  0.25 to find u( x, t ) for x  [0,1], t  [0,1]
k
  2 4
h
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Example 1
 2 u ( x, t )  u ( x, t )

0
2
t
x
u ( x  h , t )  2u ( x , t )  u ( x  h , t ) u ( x , t  k )  u ( x , t )

0
2
k
h
16u( x  h, t )  2u( x, t )  u( x  h, t )   4u( x, t  k )  u( x, t )   0
u ( x , t  k )  4 u ( x  h, t )  7 u ( x , t )  4 u ( x  h, t )
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Example 1
u ( x , t  k )  4 u ( x  h, t )  7 u ( x , t )  4 u ( x  h, t )
t=1.0
0
0
t=0.75
0
0
t=0.5
0
0
t=0.25
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
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Example 1
u(0.25,0.25)  4 u(0,0)  7 u(0.25,0)  4 u(0.5,0)
 0  7 sin(  / 4)  4 sin(  / 2)  0.9497
t=1.0
0
0
t=0.75
0
0
t=0.5
0
0
t=0.25
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
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Example 1
u(0.5,0.25)  4 u(0.25,0)  7 u(0.5,0)  4 u(0.75,0)
 4 sin(  / 4)  7 sin(  / 2)  4 sin( 3 / 4)  0.1716
t=1.0
0
0
t=0.75
0
0
t=0.5
0
0
t=0.25
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
33
Remarks on Example 1
The obtained results are probably not accurate
because :
1  2  7
For accurate results : 1  2  0
h 2 (0.25) 2
One needs to select k 

 0.03125
2
2
k
For example, choose k  0.025, then   2  0.4
h
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Example 1 – cont’d
u( x, t  k )  0.4 u( x  h, t )  0.2 u( x, t )  0.4 u( x  h, t )
t=0.10
0
0
t=0.075
0
0
t=0.05
0
0
t=0.025
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
35
Example 1 – cont’d
u(0.25,0.025)  0.4 u(0,0)  0.2 u(0.25,0)  0.4 u(0.5,0)
 0  0.2 sin(  / 4)  0.4 sin(  / 2)  0.5414
t=0.10
0
0
t=0.075
0
0
t=0.05
0
0
t=0.025
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
36
Example 1 – cont’d
u(0.5,0.025)  0.4 u(0.25,0)  0.2 u(0.5,0)  0.4 u(0.75,0)
 0.4 sin(  / 4)  0.2 sin(  / 2)  0.4 sin( 3 / 4)  0.7657
t=0.10
0
0
t=0.075
0
0
t=0.05
0
0
t=0.025
0
0
t=0
0
0
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x=0.0
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x=0.25
x=0.5
x=0.75
x=1.0
37
Crank-Nicolson Method
The method involves solving a Tridiagona l system of linear equations.
The method is stable (No magnificat ion of error).
 We can use larger h, k (compared to the Explicit Method).
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Crank-Nicolson Method
Based on the finite difference method
1. Divide the interval x into subinterva ls of width h
2. Divide the interval t into subinterva ls of width k
3. Replace the first and second partial derivative s with their
backward and central difference formulas respective ly :
u ( x, t ) u ( x, t )  u ( x, t  k )

t
k
 2 u ( x , t ) u ( x  h , t )  2u ( x , t )  u ( x  h , t )

2
x
h2
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Crank-Nicolson Method
 2 u ( x, t )  u( x, t )
Heat Equat ion :

becomes
2
t
x
u ( x  h , t )  2u ( x , t )  u ( x  h , t ) u ( x , t )  u ( x , t  k )

2
k
h
k
u( x  h, t )  2u( x, t )  u( x  h, t )   u( x, t )  u( x, t  k )
2
h
k
k
k
 2 u ( x  h, t )  (1  2 2 ) u ( x, t )  2 u ( x  h, t )  u ( x, t  k )
h
h
h
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Crank-Nicolson Method
k
Define   2 then Heat equation becomes :
h
  u( x  h, t )  (1  2 ) u( x, t )   u( x  h, t )  u( x, t  k )
u(x-h,t)
u(x,t)
u(x+h,t)
u(x,t - k)
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Crank-Nicolson Method
The equation :
  u( x  h, t )  (1  2 ) u( x, t )   u( x  h, t )  u( x, t  k )
can be rewritten as :
  ui 1, j  (1  2 ) ui , j   ui 1, j  ui , j 1
and can be expanded as a system of equations (fix j  1) :
  u0,1  (1  2 ) u1,1   u2,1  u1,0
  u1,1  (1  2 ) u2,1   u3,1  u2,0
  u2,1  (1  2 ) u3,1   u4,1  u3,0
  u3,1  (1  2 ) u4,1   u5,1  u4,0
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Crank-Nicolson Method
  u( x  h, t )  (1  2 ) u( x, t )   u( x  h, t )  u( x, t  k )
can be expressed as a Tridiagona l system of equations :

1  2
  u1,1   u1,0   u0,1 

   1  2
 u  
u

2,0


  2,1   
u3,0

  1  2
   u3,1  


 u  u   u 
  1  2   4,1   4,0
5,1 

where u1,0 , u2,0 , u3,0 , and u4,0 are the initial temperatu re values
at x  x0  h, x0  2h, x0  3h, and x0  4h
u0,1 and u5,1 are the boundary v alues at x  x0 and x0  5h
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Crank-Nicolson Method
The solution of the tridiagon al system produces :
The temperatu re values u1,1 , u2,1 , u3,1 , and u4,1 at t  t0  k
To compute the temperatu re values at t  t0  2k
Solve a second tridiagon al system of equations ( j  2)

1  2

   1  2




  1  2
 



  1  2 

To compute u1,2 , u2,2 , u3,2 , and u4,2
 u1,2   u1,1   u0,2 
u  

u
2,1
 2, 2   

u3,1
u3,2  

  

u
u


u
5, 2 
 4,2   4,1
Repeat the above step to compute temperatu re values at t0  3k , etc.
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Example 2
Solve the PDE :
 2u( x, t ) u( x, t )
0

2
t
 x
u(0, t )  u(1, t )  0
u( x,0)  sin(  x )
Solve using Crank - Nicolson method
Use h  0.25, k  0.25 to find u( x, t ) for x  [0,1], t  [0,1]
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Example 2
Crank-Nicolson Method
 2 u ( x, t )  u ( x, t )

0
2
t
x
u ( x  h , t )  2u ( x , t )  u ( x  h , t ) u ( x , t )  u ( x , t  k )

2
k
h
16u( x  h, t )  2u ( x, t )  u ( x  h, t )   4u( x, t )  u( x, t  k )   0
k
Define   2  4
h
 4 u ( x  h, t )  9 u ( x , t )  4 u ( x  h, t )  u ( x , t  k )
 4 ui 1, j  9 ui , j  4 ui 1, j  ui , j 1
CISE301_Topic9
46
Example 2
 sin(  / 4)
 4u0,1  9u1,1  4u2,1  u1,0  9u1,1  4u2,1
 4u1,1  9u2,1  4u3,1  u2,0  4u1,1  9u2,1  4u3,1  sin(  / 2)
 4u2,1  9u3,1  sin( 3 / 4)
 4u2,1  9u3,1  4u4,1  u3,0 
t4=1.0
0
t3=0.75
0
t2=0.5
0
u1,4
u2,4
u3,4
u1,3
u2,3
u3,3
u1,2
u2,2
u3,2
0
0
0
u1,1
u2,1
u3,1
t1=0.25
0
0
t0=0
0
0
CISE301_Topic9
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x0=0.0
x1=0.25
x2=0.5
x3=0.75
x4=1.0
47
Example 2
Solution of Row 1 at t1=0.25 sec
The Solution of the PDE at t1  0.25 sec is the solution
of the following tridiagon al system of equations :
 9 4
  u1,1  sin( 0.25 )
  4 9  4 u    sin( 0.5 ) 

  2,1  

 4 9  u3,1  sin( 0.75 )

 u1,1   0.21151
  
 u2,1   0.29912


u3,1   0.21151
CISE301_Topic9
48
Example 2:
Second Row at t2=0.5 sec
 4u0,2  9u1,2  4u2,2  u1,1  9u1,2  4u2,2
 0.21151
 4u1,2  9u2,2  4u3,2  u2,1  4u1,2  9u2,2  4u3,2  0.29912
 4u2,2  9u3,2  4u4,2  u3,1 
t4=1.0
0
t3=0.75
0
t2=0.5
0
 4u2,2  9u3,2  0.21151
u1,4
u2,4
u3,4
u1,3
u2,3
u3,3
u1,2
u2,2
u3,2
0
0
0
u1,1
u2,1
u3,1
t1=0.25
0
0
t0=0
0
0
CISE301_Topic9
Sin(0.25π) Sin(0. 5π) Sin(0.75π)
x0=0.0
x1=0.25
x2=0.5
x3=0.75
x4=1.0
49
Example 2
Solution of Row 2 at t2=0.5 sec
The Solution of the PDE at t2  0.5 sec is the solution
of the following tridiagon al system of equations :
 9 4
  u1, 2   u1,1   0.21151
  4 9  4 u   u   0.29912

  2,2   2,1  

 4 9  u3,2  u3,1   0.21151

 u1,2  0.063267
  
 u2,2   0.089473


u3,2  0.063267
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50
Example 2
Solution of Row 3 at t3=0.75 sec
The Solution of the PDE at t3  0.75 sec is the solution
of the following tridiagon al system of equations :
 9 4
  u1,3   u1,2  0.063267
  4 9  4 u   u    0.089473

  2,3   2, 2  

 4 9  u3,3  u3,2  0.063267

 u1,3  0.018924
 
 u2,3    0.026763


u3,3  0.018924
CISE301_Topic9
51
Example 2
Solution of Row 4 at t4=1 sec
The Solution of the PDE at t4  1 sec is the solution
of the following tridiagon al system of equations :
 9 4
  u1,4   u1,3  0.018924
  4 9  4 u   u    0.026763

  2, 4   2,3  

 4 9  u3,4  u3,3  0.018924

 u1, 4  0.0056606
 
 u2,4   0.0080053


u3, 4  0.0056606
CISE301_Topic9
52
Remarks
The 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
53
Lecture 39
Elliptic Equations



CISE301_Topic9
Elliptic Equations
Laplace Equation
Solution
54
Elliptic Equations
A second order linear PDE (2 - independen t variable s x , y )
A u xx  B u xy  C u yy  D  0,
A, B, and C are functions of x and y
D is a function of x, y , u, u x , and u y
is Elliptic if
CISE301_Topic9
B 2  4 AC  0
55
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.
 T ( x, y )  T ( x, y )

 f ( x, y )
2
2
x
y
T : steady state temperatu re at point (x, y)
2
2
f ( x, y ) : heat source (or heat sink)
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56
Laplace Equation
 2 T ( x, y )  2 T ( x, y )

 f ( x, y )
2
2
x
y
A  1, B  0, C  1
B 2  4 AC  4  0 Elliptic

Temperature is a function of the position (x and y)

When no heat source is available f(x,y)=0
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57
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 T ( x, y ) Ti 1, j  2Ti , j  Ti 1, j

,
2
2
x
x 
CISE301_Topic9
 2 T ( x, y ) Ti , j 1  2Ti , j  Ti , j 1

2
y
y 2
58
Solution Technique
 2 T ( x, y ) Ti 1, j  2Ti , j  Ti 1, j

,
2
2
x
x 
 2 T ( x, y ) Ti , j 1  2Ti , j  Ti , j 1

2
2
y
y 
 2 T ( x, y )  2 T ( x, y )


0
2
2
x
y
is approximat ed by :
Ti 1, j  2Ti , j  Ti 1, j
x 
2
CISE301_Topic9

Ti , j 1  2Ti , j  Ti , j 1
y 
2
0
59
Solution Technique
Ti 1, j  2Ti , j  Ti 1, j
x 
2

Ti , j 1  2Ti , j  Ti , j 1
y 
2
0
( Laplacian Difference Equation)
Assume : x  y  h
 Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
CISE301_Topic9
60
Solution Technique
Ti , j 1
Ti 1, j
Ti , j
Ti 1, j
Ti , j 1
Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
CISE301_Topic9
61
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.
100
75
50
The sheet is divided
to 5X5 grids.
CISE301_Topic9
62
Known
Example
To be determined
T1, 4  100
T2, 4  100
T3, 4  100
T0,3  75
T1, 3
T2 ,3
T3,3
T4,3  50
T0, 2  75
T1, 2
T2, 2
T3, 2
T4, 2  50
T0,1  75
T1,1
T2 ,1
T3,1
T4,1  50
T1, 0  0
CISE301_Topic9
T2, 0  0
T3, 0  0
63
Known
First Equation
T1, 4  100
T1, 3
T0,3  75
T0, 2  75
T1, 2
To be determined
T2, 4  100
T2 ,3
T2, 2
T0,3  T1, 4  T1, 2  T2,3  4T1,3  0
75  100  T1, 2  T2,3  4T1,3  0
CISE301_Topic9
64
Known
Another Equation
T1, 4  100
To be determined
T2, 4  100
T3, 4  100
T1, 3
T2 ,3
T3,3
T1, 2
T2, 2
T3, 2
T1,3  T2,4  T3,3  T2,2  4T2,3  0
T1,3  100  T3,3  T2,2  4T2,3  0
CISE301_Topic9
65
Solution
The Rest of the Equations
 4 1 0 1
  T1,1   75 
 



1 4 1 0 1
  T2,1   0 
 0 1 4
  T   50 
0
0 1

  3,1  

0
4 1 0 1
1 0
  T1,2   75 

 T    0 
1 0 1 4 1 0 1

  2, 2  

1 0 1 4
0
0  1  T3,2   50 


  T   175 

1
0
0
4

1
0

  1,3  


 1 0  1 4  1  T2,3  100 
 



T

1
0

1
4
150

  3,3  

CISE301_Topic9
66