99
Numerical Solution of Poisson Equation Using Fuzzy Data by
finite Difference
Mhassin A. A.
Faculty of Education For Pure Science, Al-Anbar University, Iraq
[email protected]
ABSTRACT
In this paper, we have discussed fuzzification of elliptic partial differential equation taking Poisson Equation in two dimensions are discussed. The interval of fuzzy interval can be determined. Finite difference method applied of two different grids using five points, first for initial values and the second to solve Poisson equation numerically.
Keywords: Fuzzy membership function (f.m.f.), interval of confidence, triangular fuzzy number (t.f.n.),
cuts, five points finite difference, Poisson Equation.
1 – Introduction
The concept of Fuzzy differential equation was first introduced by Chang Zadeh [10]. Dubois and Prade[5] has given extension principle. Raphel and Mhassin [8,9], used five points in regular domain. Here implementing five-points for finite difference method to solve Poisson equation in two variables numerically, then fuzzified.
2- Definitions
A triangular Fuzzy number
is defined by three real numbers with base as the interval[ c
a
,
]
and b as thevertex of triangle. The membership function are defined as follows [8, ,]:
otherwise
c
x
b
where
c
b
c
x
b
x
a
where
a
b
a
x
x
;
0
;
;
)
)(
(
The cuts are defined by L(
)()a
(ba) and R(
)()c
(bc).A triangular Fuzzy number
f is defined by three real numbers with base as the interval[
f
a,
f
c]
andf
b asthe vertex of triangle. The membership function are defined as follows
otherwise f x f where f f f x f x f where f f f x x b c c b c b a a b a f ; 0 ; ; ) )( (
100
2.1 Finite difference using to solve Poisson equation
Poisson equation in two variables is defined by
uxx(x,y)uyy(x,y) f(x,y) (1)
This equation is encountered in many application, fluid mechanics, study state , electrostatics, mass transfer, and
for other areas of mechanics and physics. Replacing
u
xx andu
yy by the central difference formula the value of)
,
(
x
ix
ju
at any mesh point is the arithmetic mean of the values at four neighboring mesh to the left, right,above and below which is called standard five points formula Fig.1, use for finding the initial data respectively
as follows, replacing
u
xx andu
yy by finite difference method2 1 , 1 , , 1 , , 1
2
h
u
u
u
u
u
u
xx
i j
ij
i j
ij
ij and 2 1 , 1 , , 1 ,2
k
u
u
u
u
u
yy
ij
ij
ij
ij . Then (1) becomes j i j i j i j i j i j i j if
k
u
u
u
h
u
u
u
2 1 , , 1 , 2 , 1 , , 12
2
(2)Here we will take
h
k
in the square mesh, the value ofu
i,jat any point is the arithmetic mean of its values atthe four neighboring mesh points to the left, right, above, and below, then
ij
i j i j ij ij ij
f h u u u u u 2 , 1 , 1 , , 1 , 1 , 4 1 (3)Which is called Standard Five Points Formula (SFPF) as in Fig.1, Or
i j i j i j i j ij
j i u u u u h f u 2 , 1 , 1 1 , 1 1 , 1 1 , 1 , 2 4 1 (4)Which is called Diagonally Five Points Formula (DFPF) as in Fig.2.[young], we will use (4) wherever necessary.
101
From
c
1 toc
16 represents the boundary conditions of the square mesh with fuzzy interval as in table 1.
Table –1-
The interior points due to the square grid are
u
1 tou
9.Now to find the initial value of
u
5(0) using standard five-points formula (4) as
( ) ( ) ( ) ( ) ( )
4 1 ) ( 5 2 15 11 7 3 ) 0 ( 5 c c c c h f u (5)We can write equation (4) in some details as
4 , 4 , 4 ) ( 5,3 2 3 , 15 3 , 11 3 , 7 3 , 3 2 , 5 2 2 , 15 2 , 11 2 , 7 2 , 3 1 , 5 2 1 , 15 1 , 11 1 , 7 1 , 3 ) ( ) 0 ( 5 f h l l l l f h l l l l f h l l l l u (6)Fuzzy membership functions (f.m.f) are respective
-cuts ofc
1,
c
3,c
5,c
7,c
9,c
11,c
13and
c
15 are respectively as
otherwise
l
x
l
where
l
l
l
x
l
x
l
where
l
l
l
x
x
i i i i i i i i i i Ci;
0
;
;
)
)(
(
,2 ,3 3 , 2 , 3 , 2 , 1 , 1 , 2 , 1 ,
Hence the cuts ofc
i is given by
) ( ), ( ) ( ,1 ,2 ,1 ,3 ,2 ,3 ) ( i i i i i i i l l l l l lc Where i1,2,,16. Then from equation (6) we have
1,1 1,2 1,3
1( )l ;l ;l c c2()
l2,1;l2,2;l2,3
c3()
l3,1;l3,2;l3,3
c4()
l4,1;l4,2;l4,3
5,1 5,2 5,3
5( )l ;l ;l c c6()
l6,1;l6,2;l6,3
c7()
l7,1;l7,2;l7,3
c8()
l8,1;l8,2;l8,3
9,1 9,2 9,3
9(
)
l
;
l
;
l
c
c10()
l10,1;l10,2;l10,3
c
11(
)
l
11,1;
l
11,2;
l
11,3
3 , 12 2 , 12 1 , 12 12(
)
l
;
l
;
l
c
13,1 13,2 13,3
13(
)
l
;
l
;
l
c
c14()
l14,1;l14,2;l14,3
c15()
l15,1;l15,2;l15,3
3 , 16 2 , 16 1 , 16 16(
)
l
;
l
;
l
c
102
4 ) ( 4 ) ( ) ( ) ( ) ( ) ( , 4 4 ) ( ) ( ) ( ) ( ) ( ) )( ( 3 , 5 2 3 , 15 3 , 11 3 , 7 3 , 3 3 , 5 2 , 5 2 3 , 15 2 , 15 3 , 11 2 , 11 3 , 7 2 , 7 3 , 3 2 , 3 1 , 5 2 1 , 15 1 , 11 1 , 7 1 , 3 1 , 5 2 2 , 5 2 1 , 15 2 , 15 1 , 11 2 , 11 1 , 7 2 , 7 1 , 3 2 , 3 ) 0 ( 5 f h l l l l f f h l l l l l l l l f h l l l l f h f h l l l l l l l l x u or 4 4 4 4 ) )( ( 3 , 5 3 , 5 2 , 5 1 , 5 1 , 5 2 , 5 ) 0 ( 5 H H H H H H x u Where
3 , 5 2 3 , 15 3 , 11 3 , 7 3 , 3 3 , 5 2 , 5 2 2 , 15 2 , 11 2 , 7 2 , 3 2 , 5 1 , 5 2 1 , 15 1 , 11 1 , 7 1 , 3 1 , 5f
h
l
l
l
l
H
f
h
l
l
l
l
H
f
h
l
l
l
l
H
Fig.3
103
let 4 4 1 , 5 1 , 5 2 , 5 1 H H H x
and 4 4 3 , 5 3 , 5 2 , 5 2 H H Hx . Solving for
, we have1 , 5 2 , 5 1 , 5 1
4
H
H
H
x
and 3 , 5 2 , 5 3 , 5 24
H
H
H
x
. Hence f.m.f. foru
5(0) is
otherwise
H
x
H
H
H
H
x
H
x
H
H
H
H
x
x
u;
0
4
1
4
1
;
4
4
1
4
1
;
4
)
)(
(
5,2 5,3 3 , 5 2 , 5 3 , 5 2 , 5 1 , 5 1 , 5 2 , 5 1 , 5 ) 0 ( 5
(7) Where
[0,1].But to find the initial values of
u
1,u
3,u
9 andu
7 using five points diagonally (DFPF) i.e. equation (4), and to find The initial values ofu
2,u
6,u
8andu
4 by (SFPF) i.e. equation (3).The initial value ofu
1(0) we use equation (4) the interval of confidence 4 2 , 4 2 , 4 2 ) ( 1,3 2 3 , 15 ) 0 ( 3 , 5 3 , 3 3 , 1 2 , 1 2 2 , 15 ) 0 ( 2 , 5 2 , 3 2 , 1 1 , 1 2 1 , 15 ) 0 ( 1 , 5 1 , 3 1 , 1 ) 0 ( 1 f h l u l l f h l u l l f h l u l l u (8) Hence f.m.f. for
c
1 is otherwise l x l where l l l x l x l where l l l x x C ; 0 ; ; ) )( ( 1,2 1,3 3 , 1 2 , 1 3 , 1 2 , 1 1 , 1 1 , 1 2 , 1 1 , 1 1 , Hence
cuts
ofc
1
) ( ), ( ) ( 1,1 1,2 1,1 1,3 1,2 1,3 ) ( 1 l l l l l l c
.As well,
cuts
ofc
3,c
15, andu
5(0)are
) ( ), ( ) ( 3,1 3,2 3,1 3,3 3,2 3,3 ) ( 3 l l l l l l c ,
) ( ), ( ) ( 15,1 15,2 15,1 15,3 15,2 15,3 ) ( 15 l l l l l l c and
() ( ), ( 5(0,3)) ) 0 ( 2 , 5 ) 0 ( 3 , 5 ) 0 ( 1 , 5 ) 0 ( 2 , 5 ) 0 ( 1 , 5 ) ( ) 0 ( 5 u u u u u u u .104
4
4
4
4
)
(
3 , 1 3 , 1 2 , 1 1 , 1 1 , 1 2 , 1 ) 0 ( 1X
X
X
X
X
X
u
where
3 , 1 2 ) 0 ( 3 , 5 3 , 15 3 , 3 3 , 1 3 , 1 2 , 1 2 ) 0 ( 2 , 5 2 , 15 2 , 3 2 , 1 2 , 1 1 , 1 2 ) 0 ( 1 , 5 1 , 15 1 , 3 1 , 1 1 , 12
2
2
f
h
u
l
l
l
X
f
h
u
l
l
l
X
f
h
u
l
l
l
X
. Le t 4 4 1 , 1 1 , 1 2 , 1 1 X X X x and4
4
3 , 1 3 , 1 2 , 1 2X
X
X
x
, solve for
1 , 1 2 , 1 1 , 1 1 4 X X X x and 3 , 1 2 , 1 3 , 1 2 4 X X X x . Hence f.m.f. foru
1(0) is
otherwise
X
x
X
where
X
X
X
x
X
x
X
where
X
X
X
x
x
u;
0
4
1
4
1
4
4
1
4
1
4
)
)(
(
1,2 1,3 2 , 1 3 , 1 3 , 1 2 , 1 1 , 1 1 , 1 2 , 1 1 , 1 ) 0 ( 1
(9)This process also for
u
3,u
9andu
7. In a similar way we evaluateu
2,u
6,u
8 andu
4, foru
2(0)we find thecuts
ofc
3,u
3(0),u
5(0)andu
1(0)we get 4 , 4 , 4 ) ( 2,3 2 ) 0 ( 3 , 1 ) 0 ( 3 , 5 ) 0 ( 3 , 3 3 , 3 2 , 2 2 ) 0 ( 2 , 1 ) 0 ( 2 , 5 ) 0 ( 2 , 3 2 , 3 1 , 2 2 ) 0 ( 1 , 1 ) 0 ( 1 , 5 ) 0 ( 1 , 3 1 , 3 ) 0 ( 2 f h u u u l f h u u u l f h u u u l u (10) 4 4 4 4 ) ( 3 , 2 3 . 2 2 . 2 1 , 2 1 . 2 2 . 2 ) 0 ( 2 H H H H H H u where 3 , 2 2 ) 0 ( 3 , 1 ) 0 ( 3 , 5 ) 0 ( 3 , 3 3 , 3 3 , 2 2 , 2 2 ) 0 ( 2 , 1 ) 0 ( 2 , 5 ) 0 ( 2 , 3 2 , 3 2 , 2 1 , 2 2 ) 0 ( 1 , 1 ) 0 ( 1 , 5 ) 0 ( 1 , 3 1 , 3 1 , 2 f h u u u l H f h u u u l H f h u u u l HNext successive approximations with their f.m.f. as required be obtain from previous approximations and specified boundary conditions.
4. Numerical example
Let us consider the Poisson equation 4 2 2
)
(
2
)
(
)
(
u
x
y
e
u
xx yy (11)105
In the domain
0
x
4
,
0
y
4
with boundary conditions corresponding to the points shown in table -2-.Table -2-
Leibmann's process will be applied to solve equation (12).
Solution:
The boundary conditions are given, the initial values of
u
i
1
,
2
,
3
,...,
9
may be calculated with the help of standard five points and diagonal five points formulas, to get the approximate solution. From the equations (5) and (7) we have
) ( ) ( ) ( ) ( ) ( 4 1 ) ( 5 2 15 11 7 3 ) 0 ( 5 c c c c h f u and
otherwise
H
x
H
H
H
H
x
H
x
H
H
H
H
x
x
u;
0
4
1
4
1
;
4
4
1
4
1
;
4
)
)(
(
5,2 5,3 3 , 5 2 , 5 3 , 5 2 , 5 1 , 5 1 , 5 2 , 5 1 , 5 ) 0 ( 5
Where
3 , 5 2 3 , 15 3 , 11 3 , 7 3 , 3 3 , 5 2 , 5 2 2 , 15 2 , 11 2 , 7 2 , 3 2 , 5 1 , 5 2 1 , 15 1 , 11 1 , 7 1 , 3 1 , 5f
h
l
l
l
l
H
f
h
l
l
l
l
H
f
h
l
l
l
l
H
0
.
292
,
0
.
293
,
0
.
294
)
(
) 0 ( 5
u
(12)To find f.m.f. and respective interval of confidence these eight
c
i'
s
as follows: otherwise x where x x where x x i C ; 0 001 . 0 0 ; 001 . 0 0 001 . 0 0 001 . 0 ; 001 . 0 0 001 . 0 ) )( (
,
(
)
0
.
001
0
.
001
,
0
.
001
0
.
001
) (
ic
for i1,9,10,11,12,13,14,15and
c
16are the same f.m.f. .0
1
c
c
2
0
.
293
c
3
1
.
172
c
4
2
.
637
c
5
4
.
688
0
16
c
u
1u
2u
3c
6
2
.
637
0
15
c
u
4u
5u
6c
7
1
.
172
0
14
c
u
7u
8u
9c
8
0
.
293
0
13
c
c
12
0
c
11
0
c
10
0
c
9
0
106
otherwise x where x x where x x C ; 0 294 . 00 293 . 0 ; 294 . 0 293 . 0 294 . 0 293 . 0 292 . 0 ; 292 . 0 293 . 0 292 . 0 ) )( ( 2
,
( )
0.001 0.292, 0.001 0.294
) ( 2
c ,
otherwise
x
where
x
x
where
x
x
C;
0
173
.
0
172
.
0
;
173
.
0
172
.
0
173
.
0
172
.
0
171
.
0
;
171
.
0
172
.
0
171
.
0
)
)(
(
3
,
( )
0.001 0.171, 0.001 0.173
) ( 3
c ,
otherwise
x
where
x
x
where
x
x
C;
0
638
.
2
637
.
2
;
638
.
2
637
.
2
638
.
2
637
.
2
636
.
2
;
636
.
2
637
.
2
636
.
2
)
)(
(
4
,
( )
0.001 2.636, 0.001 2.638
) ( 4
c ,
otherwise
x
where
x
x
where
x
x
C;
0
690
.
4
689
.
4
;
960
.
4
690
.
4
690
.
4
689
.
4
688
.
4
;
688
.
4
689
.
4
688
.
4
)
)(
(
5
,
( )
0.001 4.688, 0.001 4.690
) ( 5 c ,
c6()()
0.001
2.636,0.001
2.638
,
c7()()
0.001
0.171,0.001
0.173
and
8( )()
0.001
0.292,0.001
0.294
c ,with
[
0
,
1
]
. To findu
5(0) using equation (7)Let 0.001
0.292x1 and 0.001
0.294x2, then solving for
we get001 . 0 292 . 0 1 x and 001 . 0 294 . 0 2 x
, hence f.m.f. foru
5(0)is otherwise x where x x where x x u ; 0 294 . 0 293 . 0 ; 294 . 0 293 . 0 294 . 0 293 . 0 292 . 0 ; 292 . 0 293 . 0 292 . 0 ) )( ( ) 0 ( 5
107
otherwise
x
where
x
x
where
x
x
u;
0
184
.
0
183
.
0
;
184
.
0
183
.
0
184
.
0
183
.
0
182
.
0
;
182
.
0
183
.
0
182
.
0
)
)(
(
) 0 ( 1
, otherwise x where x x where x x u ; 0 502 . 1 501 . 1 ; 502 . 1 501 . 1 502 . 1 501 . 1 500 . 1 ; 50 . 1 501 . 1 50 . 1 ) )( ( ) 0 ( 3
otherwise
x
where
x
x
where
x
x
u;
0
037
.
0
036
.
0
;
037
.
0
036
.
0
037
.
0
036
.
0
035
.
0
;
035
.
0
036
.
0
035
.
0
)
)(
(
) 0 ( 7
, otherwise x where x x where x x u ; 0 185 . 0 184 . 0 ; 185 . 0 184 . 0 185 . 0 184 . 0 183 . 0 ; 183 . 0 184 . 0 183 . 0 ) )( ( ) 0 ( 9 otherwise x where x x where x x u ; 0 668 . 0 667 . 0 ; 668 . 0 667 . 0 668 . 0 668 . 0 667 . 0 ; 667 . 0 668 . 0 667 . 0 ) )( ( ) 0 ( 2 , otherwise x where x x where x x u ; 0 083 . 0 082 . 0 ; 083 . 0 082 . 0 083 . 0 082 . 0 081 . 0 ; 081 . 0 082 . 0 081 . 0 ) )( ( ) 0 ( 4 otherwise x where x x where x x u ; 0 668 . 0 667 . 0 ; 668 . 0 667 . 0 668 . 0 668 . 0 667 . 0 ; 667 . 0 668 . 0 667 . 0 ) )( ( ) 0 ( 6 and otherwise x where x x where x x u ; 0 083 . 0 082 . 0 ; 083 . 0 082 . 0 083 . 0 082 . 0 081 . 0 ; 081 . 0 082 . 0 081 . 0 ) )( ( ) 0 ( 8 108
In the following there are f.m.f of the fifth approximations using nine-points by the method of Lebmann's iteration process applied to equation (4) have been found as
otherwise x where x x where x x u ; 0 54 6562366215 0.1 554 1655236621 . 0 ; 557129 2.06693458 554 1655236621 . 0 54 6562366215 0.1 554 1655236621 . 0 21554 0.16542366 ; 21554 0.16542366 554 1655236621 . 0 21554 0.16542366 ) )( ( ) 5 ( 1 otherwise x where x x where x x u ; 0 02214 0.66015798 214 6600579802 . 0 ; 02214 0.66015798 214 6600579802 . 0 02214 0.66015798 214 6600579802 . 0 214 6599579802 . 0 ; 21554 6599579802 . 0 214 6599579802 . 0 214 6600579802 . 0 ) )( ( ) 5 ( 2 otherwise x where x x where x x u ; 0 350 4840162514 . 1 350 4839162514 . 1 ; 350 4840162514 . 1 214 6600579802 . 0 350 4840162514 . 1 350 4839162514 . 1 350 4838162514 . 1 ; 350 4839162514 . 1 350 4838162514 . 1 350 4839162514 . 1 ) )( ( ) 5 ( 3 otherwise x where x x where x x u ; 0 56564 0.07405753 56564 0.07395753 ; 56564 0.07405753 56564 0.07385753 56564 0.07405753 56564 0.07395753 56564 0.07385753 ; 564 0738575356 . 0 56564 0.07395753 564 0738575356 . 0 ) )( ( ) 5 ( 4 otherwise x where x x where x x u ; 0 50375 0.29384922 75 2937492503 . 0 ; 50375 0.29384922 56564 0.07385753 50375 0.29384922 75 2937492503 . 0 2250375 0.0.293649 ; 75 2937492503 . 0 50375 0.29364922 75 2937492503 . 0 ) )( ( ) 5 ( 5 otherwise x where x x where x x u ; 0 601 6598130042 . 0 42601 0.65971300 ; 601 6598130042 . 0 42601 0.65971300 601 6598130042 . 0 42601 0.65971300 42601 0.65961300 ; 42601 0.65961300 42601 0.65971300 42601 0.65971300 ) )( ( ) 5 ( 6
otherwise
x
where
x
x
where
x
x
u;
0
525
0187651402
.
0
525
0186651402
.
0
;
525
0187651402
.
0
02525
0.01856514
525
0187651402
.
0
525
0186651402
.
0
02525
0.01856514
;
525
0186651402
.
0
02525
0.01856514
525
0186651402
.
0
)
)(
(
) 5 ( 7
109
otherwise
x
where
x
x
where
x
x
u;
0
645
0737125597
.
0
645
0736125597
.
0
;
645
0737125597
.
0
97645
0.07351255
645
0737125597
.
0
645
0736125597
.
0
97645
0.07351255
;
645
0736125597
.
0
97645
0.07351255
645
0736125597
.
0
)
)(
(
) 4 ( 8
otherwise x where x x where x x u ; 0 562 1651157520 . 0 20562 0.16501575 ; 562 1651157520 . 0 20562 0.16491575 562 1651157520 . 0 20562 0.16501575 20562 0.16491575 ; 20562 0.16501575 20562 0.16491575 20562 0.16501575 ) )( ( ) 5 ( 9
5. CONCLUSIONFor the given initial values the fourth approximations to solve the above example numerically is very significant results in comparison with example solved in [8] using five points only. However may increased the accuracy as desired if we take more iterations. As well, using nine-points is more accurate than five points.
6. REFERENCES
[1] Baruah, Hemanta K. Set Superimposition and its applications to the theory of Fuzzy sets. Journal of Assam Science Society , Vol. 40. Nos. 1 & 2, 25-31 (1999).
[2] Baruah, Hemanta K. Construction of the membership function of Fuzzy number, ICIC express letters (2010a).
[3] Baruah, Hemanta K. The mathematics of Fuzziness Myths and realities, Lambert Academic Publishing, Saarbruken, Germany (2010b).
[4] Chang, S. L., Zadeh, L. A. On Fuzzy mapping and control, IEEE Trans. Systems man cyber net, 2 (1972) pp.30-34.
[5] Dubois, D., Prade, H. Towards Fuzzy differential calculus, Fuzzy Sets and system, Part 3, 8 (1982) pp. 225-233.
[6] Grewal, B. S. Numerical methods in engineering and science with program in C & C++, Khanna Publishers, New Delhi-110002,(2010) pp. 343-348.
[7] Kaufmann, A. and Gupta, M.M. Introduction to Fuzzy arithmetic. Theory and applications, Van Nostrand Reinhold Co. Inc., Wokingham, Berkshire (1984).
[8] Mhassin A. A. Numerical Solution of Laplace Equation Using Fuzzy Data by Nine Points Finite Difference, IJMA
Vol.-5, Issue-6,June-2014.
[9] Raphel Kumar Saikia, Numerical Solution of Laplace Equation using Fuzzy Data, International Journal of computer applications (0975-8887) Vol. 38, No. 8, January, (2012).
[10] Sastry, S.S. Introductory methods of numerical analysis, PHI Learning private limited, New Delhi 110001, (2009).
[11] Zadeh, L. A. Probability measure of Fuzzy events, Journal Mathematical Analysis and Applications, Vol. 23 No.2, August, (1968) pp. 421-427.
Mhassin A. A. Math. Department, Faculty of Education for Pure Science, Al-Anbar-University, Ramadi, Iraq, E-Mail [email protected], Tel: 00964-7818483667.
