Matakuliah
Tahun
: S0262-Analisis Numerik
: 2010
Solving Partial Differential Equation
Numerically
Pertemuan 13
Material Outline
• Partial differential equation
– 2nd order partial differential equations
– Linear 2nd order PDE
– Elliptic equations: Laplace equations
PARTIAL DIFFERENTIAL EQUATIONS
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BASIC CONCEPTS
A PDE is an equation involving one or more partial
derivatives of dependent variables u.
The order of PDE is the highest partial derivative at
the PDE
The degree of the highest partial derivative is the
degree of the PDE
If all dependent and independent variables as well
as their partial derivatives in 1st degree then PDE
called linear PDE
The most important PDE in application is 2nd order
PDE
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PARTIAL DIFFERENTIAL EQUATIONS
2ND ORDER LINEAR PDE
For dependent variable u with 2 independent variables
x and y:
 2u
 2u
 2u
A 2 B
C 2  D  0
x
xy
y
PDE can be divided in to 3 categories:
B2 – 4 AC
Category
<0
Elliptic
=0
Parabolic
>0
Hyperbolic
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IMPORTANT 2nd ORDER PDE
2
 2u

u
2

c
t 2
x 2
One Dimensiona l Wave Eq
 2u  2u
 2 0
2
x
y
 2u  2u
 2  f(x, y)
2
x
y
Two Dimensiona l Laplace Eq
Two Dimensiona l Poisson Eq
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TWO DIMENSIONAL LAPLACE EQUATION
(ELLIPTIC EQUATION)
Elliptic equation (Laplace equation) in engineering field
typically used to characterize steady state, boundary value
problems. Steady state flow of heat in heated plate can be
written as
 2T  2T
 2  0  no source/sin k
2
x
y
 2T  2T
 2  f ( x, y )  f ( x, y ) : sink/sourc e of heat
2
x
y
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Solution Technique
 2T  2T
 2 0
2
x
y
The solution will be based on finite difference technique that is
PDE is transformed into difference equation. For this, the plate
is treated as a grid of discrete points
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Solution Technique
For this, the plate is treated as a grid of discrete points
y
0,n+1
i,j+1
i,j
i-1,j
i+1,j
x
0,0
m+1,0
i,j-1
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Solution Technique
Central Finite Difference:
T ( x, y )  Ti , j  i, j  Discretiza tion
in x and y direction
T T ( x  x, y )  T ( x  x, y ) Ti 1, j  Ti 1, j


x
2x
2x
T T ( x, y   y )  T ( x, y  y ) Ti , j 1  Ti , j 1


y
2y
2y
Ti 1, j  2Ti , j Ti 1, j
 2T



x 2
x 2
Ti , j 1  2Ti , j Ti , j 1
 2T



y 2
y 2
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Solution Technique
Central Finite Difference:
 2T  2T
 2 0
2
x
y
Ti 1, j  2Ti , j  Ti 1, j
x 
2

Ti , j 1  2Ti , j  Ti , j 1
y 
2
0
If x  y 
Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
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Solution Technique
Central Finite Difference Equation:
 2T  2T
 2 0
2
x
y
Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
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TWO DIMENSIONAL LAPLACE EQUATION
Example: A 4 × 4 heated plate, where all sides are kept at constant
temperature as given in the following figure: How the temperature
distributed?
T= 100oC
T= 75oC
T= 50oC
T= 0oC
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TWO DIMENSIONAL LAPLACE EQUATION
Solution:
Boundary conditions:
T=
100oC
T4,4
T0,4
T1, 0  T2, 0  T3, 0  0
T1, 4  T2, 4  T3, 4  100
T0,1  T0, 2  T0,3  75
T= 75oC
T= 50oC
T0,1
T4,1  T4, 2  T4,3  4
Corner points 
T0, 0  (0  75) / 2  37.5
T0, 4  (0  50) / 2  25
T4, 4  (100  50) / 2  75
T0,0
T1,0
T= 0oC
T4,0
T4, 0  (75  100) / 2  87.5
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TWO DIMENSIONAL LAPLACE EQUATION
Solution:
Ti 1, j  Ti 1, j  Ti , j 1  Ti , j 1  4Ti , j  0
T= 100oC
T4,4
T0,4
T= 75oC
T= 50oC
The above
equation will be
used to solve the
temperature of
inner points
T0,1
T0,0
T1,0
T= 0oC
T4,0
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