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2
FINITE DIFFERENCE METHODS
3
Introduction
In general real life EM problems cannot be solved by using the analytical methods, because:
The PDE is not linear, The solution region is complex, The boundary conditions are of mixed types, The boundary conditions are time dependent, The medium is inhomogeneous or anisotropic.
4
Introduction
For such complicated problems numerical methods must be employed.
Finite Difference Method (FDM) is one of the available numerical methods which can easily be applied to solve PDE’s with such complexity.
FDM, is easy to understand and apply.
5
Introduction
FD techniques are based upon approximations of differential equations by finite difference equations.
Finite difference approximations: have algebraic forms, Relate the value of the dependent variable
at a point in the solution region to the values at some neighboring points.
6
Introduction
Steps of finite difference solution: Divide the solution region into a grid of
nodes, Approximate the given differential
equation by finite difference equivalent, Solve the differential equations subject
to the boundary conditions and/or initial conditions.
7
Introduction
Rectangular Grid Pattern
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Finite Difference Schemes
The derivative of a given function f(x) can be approximated as:
0 0 0
0 0 0
0 0 0
( ) ( ) ( )(1)
( ) ( ) ( )(2)
( ) ( ) ( )(3)
2
df x f x x f xForward Difference Formula
dx xdf x f x f x x
Backword Difference Formuladx x
df x f x x f x xCentral Difference Formula
dx x
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Finite Difference Schemes
0x x
f(x)
x
A
P B
0x x 0x
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Finite Difference Schemes
Estimate of second derivative of f(x) at P:
0 0
2
2
0 0 0 0
0 0 02
( ) ( )2 2
1 ( ) ( ) ( ) ( )
( ) 2 ( ) ( )(4)
( )
x xf x f x
f x xx x
f x x f x f x f x x
x x x
f x x f x f x x
x
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Finite Difference Schemes
Or, by using Taylor Series:
2 3320 0 0
0 0 2 3
2 3320 0 0
0 0 2 3
( ) ( ) ( )1 ( )( ) ( ) ( ) ... (5)
2! 3!
( ) ( ) ( )1 ( )( ) ( ) ( ) ... (6)
2! 3!
f x f x f xxf x x f x x x
x x x
f x f x f xxf x x f x x x
x x x
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Finite Difference Schemes
Adding these equations:
Assuming that higher order terms are neglected:
22 40
0 0 0 2
( )( ) ( ) 2 ( ) ( ) ( )
f xf x x f x x f x x O x
x
20 0 0
2 2
( ) 2 ( ) ( )
( )
f f x x f x f x x
x x
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Finite Difference Schemes
is the error introduced by truncating the series.
It represents the terms that are not greater than .
4( )O x
4( )x
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Finite Difference Schemes
Subtracting Eq.6 from Eq. 5 and neglecting terms of the order
Error is second order. Eq. 3- Eq.4 are second order, Eq.1 and
Eq. 2 are first order.
3( )x
0 0 0( ) ( ) ( )
2
f x f x x f x x
x x
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Finite Difference Schemes
As long as the derivatives of f are well behaved and the step size is not too large, the central difference should be the most accurate of the other two. i.e. Backward and forward difference approximations.
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Finite Difference Schemes
The error in central difference decreases quadratically as the step size decreases, whereas the decrease is only linear for the other two formula.
In general, CDF is to be preferred. Situations where the data is not available on both sides of the point where the numerical derivative is to be calculated are exceptions.
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Finite Difference Schemes
To apply the difference method to find the solution of a function the solution region is divided into rectangles:
( , )x t
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Let the coordinates (x, t) of a typical grid point or a node be:
The value of at a point:
Finite Difference Schemes
, 0,1,2,...
, 0,1,2,...
x i x i
t j t j
( , ) ( , )i j i x j t
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Finite Difference Schemes
Using this notation, the central difference approximations of at (i, j) are: First derivative:
( , )
( , )
( 1, ) ( 1, )
2
( , 1) ( , 1)
2
i j
i j
i j i j
x x
i j i j
t t
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Finite Difference Schemes
Second derivative:
2
2 2( , )
2
2 2( , )
( 1, ) 2 ( , ) ( 1, )
( )
( , 1) 2 ( , ) ( , 1)
( )
i j
i j
i j i j i j
x x
i j i j i j
t t
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Finite Difference Schemes
One dimensional example: Solve
Subject to the boundary conditions:
Assume
2
24 0 , 0 1.0
ff x
x
(0) 0
(1) 1
f
f
0.5x
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Finite Difference Schemes
Solution: Discretization of the equation:
2
2
( 1) 2 ( ) ( 1)4 ( ) 0
( )
(2) 2 (1) (0)4 (1) 0
(0.5)
f i f i f if i
x
f f ff
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Finite Difference Schemes
Since there is one unknown, one equation is enough to find this unknown.
(1) (0) (2)
(1) 1
f f f
f
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Finite Difference Schemes
If 0.25x
2
2
2
(2) 2 (1) (0)4 (1) 0
(0.25)
(3) 2 (2) (1)4 (2) 0
(0.25)
(4) 2 (3) (2)4 (3) 0
(0.25)
f f ff
f f ff
f f ff
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Finite Difference Schemes
f(1), f(2) and f(3) are unknowns. Therefore, three equations are written to find three unknowns.
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Finite Difference Schemes
For we have (N-1) unknowns and we need (N-1) equations to solve.
The equation of the i th node is:
1 N x
2
( 1) 2 ( ) ( 1)4 ( ) 0
( )
f i f i f if i
x
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Finite Differencing of Parabolic PDE’s
Consider the diffusion equation:
Where k is a constant. Discretized equation:
2
2kt x
2
( , 1) ( , ) ( 1, ) 2 ( , ) ( 1, )
( )
i j i j i j i j i jk
t x
28
Finite Difference Schemes
Where
Forward difference for t and central difference for x is used.
Let:
, 0,1,2,...,
, 0,1,2,...,
x i x i n
t j t j n
2( )
trk x
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Finite Difference Schemes
We can write:
This equation can be calculated in terms of boundary and initial conditions.
This explicit formula can be used to find
explicitly in terms of .
( , 1) ( 1, ) (1 2 ) ( , ) ( 1, )i j r i j r i j r i j
( , )x t t ( , )x t
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Stability
Stability is important for the numerical solutions.
A stable solution is only possible when (1-2r) is nonnegative. Or:
Choose r=1/2. Then our equation becomes:
10
2r
1( , 1) ( 1, ) ( 1, )
2i j i j i j
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Stability
Stability increases by applying an implicit formula. The formula is simple to apply but it is computationally slow. Crank and Nicholson’s formula:
2
2
( , 1) ( , ) 1 ( 1, ) 2 ( , ) ( 1, ){2 ( )
( 1, 1) 2 ( , 1) ( 1, 1)}
( )
i j i j i j i j i jk
t x
i j i j i j
x
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Example (Cont.)
Rewrite:
Three known values on the left hand side and three unknown values on the right hand side.
( 1, 1) 2(1 ) ( , 1) ( 1, 1)
( 1, ) 2(1 ) ( , ) ( 1, )
r i j r i j r i j
r i j r i j r i j
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The formula is valid for all finite values of r, chose r=1:
Example (cont.)
( 1, 1) 4 ( , 1) ( 1, 1) ( 1, ) ( 1, )i j i j i j i j i j
