IRSTI 27.35.63 https://doi.org/10.26577/ijmph.2020.v11.i2.01 S.N. Kharin1,3*, T.A.Nauryz1,2,3,4, B. Miedzinski5
1Institute of Mathematics and Mathematical Modeling, Almaty, Kazakhstan
2Al-Farabi Kazakh National University, Almaty, Kazakhstan
3Kazakh-British Technical University, Almaty, Kazakhstam
4Satbayev University, Almaty, Kazakhstan
4Wroclaw University, Wroclaw, Poland
*e-mail: staskharin@yahoo.com
TWO PHASE SPHERICAL STEFAN INVERSE PROBLEM SOLUTION WITH LINEAR COMBINATION OF RADIAL HEAT POLYNOMIALS AND INTEGRAL ERROR FUNCTIONS IN ELECTRICAL CONTACT PROCESS
Abstract. In this research the inverse Stefan problem in spherical model where heat flux has to be determined is considered. This work is continuing of our research in electrical engineering that when heat flux passes through one material to the another material at the point where they contact heat distribution process takes the place. At free boundary
α
( )t contact spot starts to boiling and atβ
( )t stars to melting and there appear two phase: liquid phase and solid phase. Our aim to calculate temperature in liquid and solid phase, then find heat flux entering into contact spot. The exact solution of problem represented in linear combination of series for radial heat polynomials and integral error functions. The recurrent formulas for determine unknown coefficients are represented. The effectiveness of method is checked by test problem and approximate problem in which exact solution and approximate solution of heat flux is compared. The coefficients of heat at liquid and solid phases and heat flux are found. The heat flux equation is checked by testing problem by using Mathcad program.Key words: Stefan problem, radial heat polynomials, Faa-di Bruno, collocation method.
Introduction
Heat flux entering in electrical contact materials from electrical arc distributes radially and axially.
Spherical model is most convenient, introduced by Holm R. [1], in the problem of heat distribution in electrical materials. In this problem generalized heat equation can be used. The generalized heat equation of the form
2 1
1 v
a v r
t r x x
θ θ
∂∂ = ∂∂ ∂∂
have the fundamental solution with delta-function containing initial condition by using Laplace transform can be represented as
2 2
( , , ) ( ) 4 ,
2 2
x y
v t
C xy
G x y t xy e I
t t
β β
− +
=
where
We can consider the heat potentials related to this solution in form [2]
1 2
,
0
( , ) 2 ( 1) ( , , ) n v Q x tn v = −βΓ β+ −
∫
∞G x y t y dy+and by using integration by parts method we have the following explicit formula of heat polynomials
2( ) 2
, 0
! ( 1) ( , ) 2
!( )! ( 1 )
n k k
n k
n v k
n x t
Q x t
k n k n k
β β
−
=
Γ +
=
∑
− Γ + + −For applications it is convenient to multiply both sides of this equation by ( 1 )
( 1) β n
β Γ + +
Γ + and we get the following solution
2( ) 2
,
0
! ( 1 )
( , ) 2
!( )! ( 1 )
n k k
n k
n v k
n n x t
Q x t
k n k n k
β β
−
=
Γ + +
=
∑
− Γ + + −IRSTI 27.35.63 https://doi.org/10.26577/ijmph.2020.v11.i2.01 S.N. Kharin1,3*, T.A.Nauryz1,2,3,4 , B. Miedzinski5
1Institute of Mathematics and Mathematical Modeling, Almaty, Kazakhstan
2Al-Farabi Kazakh National University, Almaty, Kazakhstan
3Kazakh-British Technical University, Almaty, Kazakhstam
4Satbayev University, Almaty, Kazakhstan
5Wroclaw University, Wroclaw, Poland
*e-mail: staskharin@yahoo.com
TWO PHASE SPHERICAL STEFAN INVERSE PROBLEM SOLUTION WITH LINEAR COMBINATION OF RADIAL HEAT POLYNOMIALS AND INTEGRAL ERROR FUNCTIONS IN ELECTRICAL CONTACT PROCESS
In this research we consider v=2 which allow to transform to generalized heat equation to spherical heat equation [3]. The similar problems are considered in [4]-[7].
Mathematical model
Let us consider the liquid phase described in domain α( )t < <r β( ),t t>0and solid phase in
( )t r , t 0
β < < ∞ > with spherical heat equations
2 2
2
1 , 1,2
i i
ai r i
t r r r
θ θ
∂∂ = ∂∂ ∂∂ = (1)
and each phase has initial condition as follows
1( ( ),0) 0,t (0) ( ) 0,t
θ α = α =β = (2)
2( ,0)r f r( ), f(0) m.
θ = =θ (3) Heat flux entering P t( ) into spherical domain from electrical arc with radius r0 in process pf heat transfer within electrical contact materials can be determined from condition
0 1
1 ( ).
r r
r P t λ θ
=
− ∂ =
∂ (4) Temperatures in liquid and solid phase at free boundaryα( )t is equal to melting temperature
( ( ), ) , 1,2.
i t t m i
θ β =θ = (5) Motion of the free boundary can be calculated at Stefan’s condition
1 2
1 2
( ) ( )
r t r t
L d
r β r β dt
θ θ β
λ λ γ
= =
∂ ∂
− = − +
∂ ∂ (6)
and temperature of solid zone at infinity turns to zero
2r 0.
θ =∞ = (7) Problem solution
The solution of problem (1)-(7) we represent as linear combination of series for radial heat equation and integral error functions
2 2( )
1
0 0
2 1
2 1 2 1
1
0 1 1
2 ! 3 ( , ) 2
!( )! 3 2
(2 ) ,
2 2
k n k k
n
n nk
n
n n
n n
n n r t
r t A
k n k n k
a t r r
B i erfc i erfc
r a t a t
θ
−
∞
= =
∞ +
+ +
=
Γ +
= +
− Γ + −
−
+ −
∑ ∑
∑
(8)
2 2( )
1 0 0
2 1
2 1 2 1
2
0 2 2
2 ! 3 ( , ) 2
!( )! 3 2
(2 ) .
2 2
k n k k
n
n nk
n n n
n n
n n r t
r t C
k n k n k
a t r r
D i erfc i erfc
r a t a t
θ
−
∞
= =
∞ +
+ +
=
Γ +
= +
− Γ + −
−
+ −
∑ ∑
∑
(9)
The equations (8) and (9) satisfy heat equation (1) and undetermined coefficients A B Cn, ,n n and Dn have to be founded to determine temperatures in phases.
The function at initial condition for θ2( , )r t is represented in expansion by Maclaurin series
( )
0
( ) (0)
!
n n
n
f r f r
n
∞
=
=
∑
and free boundaries can be considered in power series /2 10
( ) n n
n
t t
α ∞ α +
=
=
∑
and/2 1 0
( ) n n
n
t t
β ∞ β +
=
=
∑
. Heat flux which have to be determined from condition (4) can be written in1/2 3/2 /2
0 1 2 3
0
( ) ... n n .
n
P t p p t p t p t ∞ p t
=
= + + + =
∑
At first, we must find temperatures in liquid and solid zones, then by using property of integral error function to condition (3) we get
( )
2 2
0 0 0
2 (0)
(2 1)! !
n
n n n
n n
n n n
C r D r f r
n n
∞ ∞ ∞
= = =
+ =
∑ ∑
+∑
(10)By comparing the power of r in both sides (10) we obtain the following form
2 (2 )(0) (2 1)! (2 )!
n
n n
C D f
n n
+ =
+ (11) and from conditions (5) we have
( )
2 2( ) 2
0 0
2 1
2 1 2 1
1 0
2 ! 3 ( )
2
!( )! 3 2
(2 ) ( ( )) ( ( )) ,
( )
k n k k
n n
n k
n
n n
n m
n
n n
A
k n k n k
B a i erfc i erfc
β τ τ
τ υ τ υ τ θ
β τ
−
∞
= =
∞ +
+ +
=
Γ + +
− Γ + −
+ − − =
∑ ∑
∑
(12)
( )
2 2( ) 2
0 0
2 1 2 1 2 1
2 0
2 ! 3 ( )
2
!( )! 3 2
(2 ) ( ( )) ( ( )) ,
( )
k n k k
n
n nk
n n n
n m
n
n n
C
k n k n k
D a i erfc i erfc
β τ τ
τ υ τ υ τ θ
β τ
−
∞
= =
∞ +
+ +
=
Γ + +
− Γ + −
+ − − =
∑ ∑
∑
(13)
and from Stefan’s condition we obtain
( )
( )
2 2( ) 1 2
2 1
2 1 2 1
1 12
0 0 0
2 2
2 2
1
2
2 ! 32 2( ) ( ) (2 ) ( ( )) ( ( ))
3 ( )
!( )!
2
2 ! 3
(2 ) ( ( )) ( ( ) 2
( )
k n k k
n n
n n
n n
n k n
k n
n n
n
n n n k a
A B i erfc i erfc
k n k n k
n n
a i erfc i erfc C
β τ τ
λ τ υ τ υ τ
β τ
τ υ τ υ τ λ
α τ
− −
∞ ∞ +
+ +
= = =
Γ + −
− − Γ + − + − − −
Γ +
− − + = −
∑ ∑ ∑
( ) ( )
2( ) 1 2
0 0
2 1 2
2 1 2 1 2 2
2 2
0 2
2( ) ( )
!( )! 3 2
(2 ) ( ( )) ( ( )) (2 ) ( ( )) ( ( )) ,
( ) ( )
n k k
n
n k
n n
n n n n
n n
n k
k n k n k
a a d
D i erfc i erfc i erfc i erfc L
d
β τ τ
τ υ τ υ τ τ υ τ υ τ γ β
β τ β τ τ
− −
∞
= =
∞ +
+ +
=
−
+
− Γ + −
+ − − − − − + +
∑ ∑
∑
(14)
where t =τ and
2
0 1 2
1 1 0
... 1
( ) 2 2
n n n
a a
β β τ β τ
υ τ ∞ υ τ
=
+ + +
= =
∑
.Firstly, we take l-th derivative both sides of (13) when τ =0using Leibniz rule for first and second term of (13)
2 2( )
(2 4 )
0
( ) ! [ ( )] ,
( 2 )!
l k n k
n k l l
l l k
τ
τ β τ τ β τ
−
− −
=
∂ =
∂ − (15)
2 1 2 1 2 1
0
2 1 2 1 2 1
( ( )) ( ( ))
! ( ( )) ( ( )) .
( 2 1)!
l k n n
l
n n l k
i erfc i erfc
l i erfc i erfc
l k
τ
τ υ τ υ τ
τ
υ τ υ τ
+ + +
=
+ + − −
∂ − − =
∂
= − − − −
(16)
Using Faa-di Bruno for (15) and (16) we get
2 1
1 2
( 2 ) 0
2 ( ) 1 2 2 1
1 0 1 2 2 1
! [ ( )]
( 2 )!
( 2 1)! ...
! ,
( 2 )! ! !... !
l n
i
l n
b
b b
l n m l n
m b l n
l l k
l k l
l k b b b
τ
β τ
β β β
β − +
−
=
−
− +
= − +
− =
− −
= −
∑ ∑
(17)
2
1 2
2 1 2 1 2 1
0
2 1 2 1 2 1 ( )
0 0
1
1 2 2
1 2 2
! ( ( )) ( ( ))
( 2 1)!
! ( ) ( )
( 2 1)!
( 2 1)! ... .
! !... !
l n
i
n n l k
l k n n m
m
b b b
l n
b l k
l i erfc i erfc
l k
l i erfc i erfc
l k
l k b b b
τ
υ τ υ τ
υ υ
υ υ υ −
+ + − −
=
− − + +
=
−
−
− − =
− −
= − − − − ×
− −
×
∑
∑
(18)
From system of equations (11) and (13) we determine the coefficients C Dn, n. Multiplying both sides of (13) by β τ( ) we have
( )
2 2( ) 1 2
0 0
2 1 2 1 2 1
0 2
2 ! 3 ( )
2
!( )! 3 2
(2 ) ( ( )) ( ( ))
( ),
k n k k
n
n nk
n n n
n n
m
n n
C
k n k n k
D a i erfc i erfc
β τ τ
τ υ τ υ τ
θ β τ
− +
∞
= =
∞ + + +
=
Γ + +
− Γ + −
+ − − =
=
∑ ∑
∑
Taking l-th derivative both sides of this expression and using (10) we have
(2 ) , ,
(2 1)! !(2 )! (0)
2(2 )! ,
n
m l n l
n
n l
n l n f
D n
θ β δ
ξ
+ −
= (19)
(2 )
(2 ) , ,
(0) (2 )!
(2 1)! !(2 )! (0)
2 ,
(2 1)! 2(2 )!
n n
n
m l n l
n l
C f n
n l n f
n n
θ β δ
ξ
= −
+ −
− +
(20)
where
2 1
1 2
2
,
2 ( ) 1 2 2 1
1 0 1 2 2 1
2 ! 32 !
3 ( 2 )!
!( )!
2
( 2 1)! ... ,
! !... !
l n
i k
n l
b
b b
l n m l n
m b l n
n n
l l k
k n k n k
l k b b b δ
β β β
β − +
−
− +
= − +
Γ +
= ×
−
− Γ + −
− −
×
∑ ∑
2 1 2
2 1 2 1 2 1 2 1 ( )
1
, 0 0
1
1 2 2
1 2 2
(2 ) ! ( ) ( )
( 2 1)!
( 2 1)! ... .
! !... !
l n
i
n l k n n m
n l m
b b b
l n
b l k
a l i erfc i erfc
l k
l k b b b
ξ υ υ
υ υ υ −
+ − −
+ +
=
−
−
= − − − − ×
− −
×
∑
∑
Multiplying β τ( ) both sides of (12) and (14) we have
( )
2 2( ) 1 2
0 0
2 1 2 1 2 1
0 1
2 ! 3 ( )
2
!( )! 3 2
(2 ) ( ( )) ( ( ))
( ),
k n k k
n
n nk
n n n
n n
m
n n
A
k n k n k
B a i erfc i erfc
β τ τ
τ υ τ υ τ
θ β τ
− +
∞
= =
∞
+ + +
=
Γ +
+
− Γ + −
+ − − =
=
∑ ∑
∑
(21)( )
2 2( ) 2
2 2 2
1 1
0 0 0
2 2( ) 2
2 0 0
2 ! 3 2( ) ( )
2 (2 ) ( ( )) ( )
!( )! 3 2
2 ! 3 2( ) ( )
2
!( )! 3 2
k n k k
n n n n
n n
n k n
k n k k
n
n nk
n n n k
A B a i erfc i erfc
k n k n k
n n n k
C D
k n k n k
β τ τ
λ τ υ τ υ τ
β τ τ
λ
−
∞ ∞
= = =
−
∞
= =
Γ + −
− − − + =
− Γ + −
Γ + −
= − −
− Γ + −
∑ ∑ ∑
∑ ∑
2 2(
2 2)
0
2 1
(2 ) ( ( )) ( ( ))
( ) ( ) ,
2
n n n
n n
m m
a i erfc i erfc
L
τ υ τ υ τ
λ λ θ γν β
∞
=
− +
+ − +
∑
(22)where '( ) ( ) 1 2( ) 2
d
β τ β τ = drβ τ and
2 2
0 0
0
( ) ( )m n, ( ) ,
n
β τ ∞ ν β τ ν β β
=
=
∑
=0 1
( )m 1 m (3 ) k ( ) ,m k 1
k
k m m
ν β m β ν β
β = −
=
∑
− ≥Taking l-th derivative both sides of equations (21) and (22) at τ =0 we get
2 1
1 2
2
2 ( ) 1 2 2 1
0 0 1 0 1 2 2 1
2 1 2 1 2 1 2 1 ( )
1
0 0
0 1
2 ! 3
( 2 1)! ...
2 !
3 ( 2 )! ! !... !
!( )!
2
(
(2 ) ! ( ) ( )
( 2 1)!
l n
i k
b
b b
l n l n
m l n
n
n k m b l n
l n l k n n m
n n m
n n
l k A l
l k b b b
k n k n k
l a l
B i erfc i erfc
l k
β β β β
υ υ
− − +
− +
= = = − +
+ − −
+ +
= =
Γ + − −
−
− Γ + −
+ − − − −
∑ ∑ ∑ ∑
∑ ∑
11 22 221 2 2
2 1)! ... !
! !... !
l n
i
b b b
l n m l
b l k
k l
b b b
υ υ υ−− θ β
−
− −
∑
=(23)
and
2 1
1 2
2
2 ( ) 1 2 2 1
1 0
0 0 1 1 2 2 1
2 1 2 2
1
0 0
0
2 ! 3 2( )
( 2 )! ...
2 !
2 !( )! 3 ( 2 )! ! !... !
2
(2 ) ! ( 1) ( ) ( )
( 2 )!
l n
i k
b
b b
l n l n
m l n
n nk m b l n
l n
m n m n
n n
n n n k l l k
A k n k n k l k b b b
a l
B i erfc i erfc
l n
β β β
λ β
υ υ
− − +
− +
= = = − +
+
−
=
Γ + −
−
− −
−
− Γ + −
− − − − −
∑ ∑ ∑ ∑
∑
1 2 2 12 1
1 2
2 ( ) 1 2 2 1
1 1 2 2 1
2
2 ( ) 1 2 2 1
2 0
0 0 1 1 2 2 1
( 2 )! ...
! !... !
2 ! 3 2( )
( 2 )! ...
2 !
2 !( )! 3 ( 2 )! ! !... !
2
l n
i
l n
i
b b b
l n m l n
m b l n
k
b
b b
n l n
m l n
n
n k m b l n
l n b b b
n n n k
l k C l
l k b b b
k n k n k
υ υ υ
β β β
λ β
− +
− +
−
− +
= − +
−
− +
= = = − +
−
=
Γ + − −
= − − Γ + − −
∑ ∑
∑ ∑ ∑
2 1 1 2
2 1 2 2 2 ( ) 1 2 2 1
2
0 0 1
0 1 1 2 2 1
( 2 )! ...
(2 ) ! ( 1) ( ) ( ) ! ( )
( 2 )! ! !... ! 2
l n
i l
b b b
l n l n m n m n m l n
n l
n m b l n
l n
a l L
D i erfc i erfc l
l n b b b
υ υ υ γ
υ υ − + ν β
+ −
− − +
+
= = − +
−
−
− − − − − +
∑
∑ ∑ ∑
(24)
From recurrent equations (23) and (24) we can determined the coefficients An and Bnas free boundary is known.
,
, 1
, ,
, ,
1 , ,
,
2 !
! ,
2
m l n l
n l
m l n n l n l
n n
n l n l
n l n l n l
l
A l B B
θ β ϑ
χ λ
θ β η ω
ω λ η ϑ ζ
ω
− +
= =
+
(25)
where
2 1 2
2 1 2 1 2 1 2 1 ( ) 1 2 2
1
, 0 0
1 1 2 2
( 2 1)! ...
(2 ) ! ( ) ( ) ,
( 2 1)! ! !... !
l n
i
b b b
n l k n n m l n
n l
m b l k
l k
a l i erfc i erfc
l k b b b
υ υ υ
η + − − + υ + υ −−
= −
− −
= − −
∑
− − ×∑
2 1
1 2
2
2 ( ) 1 2 2 1
, 0
0 1 1 2 2 1
2 ! 3
( 2 1)! ...
2 ! ,
3 ( 2 )! ! !... !
!( )!
2
l n
i k
b
b b
n l n
m l n
n l k m b l n
n n l l k
l k b b b
k n k n k
β β β
ω − β − +− +
= = − +
Γ + − −
= ×
−
− Γ + −
∑ ∑ ∑
2 1
1 2
2
2 ( ) 1 2 2 1
, 2 0
0 0 1 1 2 2 1
2 1
2 2
2 0 0
2 ! 3 2( )
( 2 )! ...
2 !
2 !( )! 3 ( 2 )! ! !... !
2
(2 ) ! ( 1) ( ) (
( 2 )!
l n
i k
b
b b
l n l n
m l n
n l n
n k m b l n
l n
m n m n
n n
n n n k l l k
C k n k n k l k b b b
a l
D i erfc i erfc
l n
β β β
χ λ β
υ υ
− − +
− +
= = = − +
+
−
=
Γ + −
−
= − −
−
− Γ + −
− − − −
−
∑ ∑ ∑ ∑
∑
2 0 ( ) 11 22 2 12 1 11 1 2 2 1
( 2 )! ...
) ! ( ) ,
! !... ! 2
l n
i
b b b
l n m l n
m b l n l
l n L l
b b b
υ υ υ − + γ ν β
−
− +
+
= − +
− +
∑ ∑
2 1 1 2
2 1 2 2 2 ( ) 1 2 2 1
1
, 0 0
1 1 2 2 1
( 2 )! ...
(2 ) ! ( 1) ( ) ( ) .
( 2 )! ! !... !
l n
i
b b b
n l n m n m n m l n
n l
m b l n
l n
a l i erfc i erfc
l n b b b
υ υ υ
ζ + − − υ υ − +− +
= − +
= −
∑
− − − ×∑
−From condition at heat flux entering we have the following equation
( ) ( )
2 1 2
2 1 2 1 2 2
1 1
1 2
0
2
2( ) 1 2 1
0 0 0
(2 ) ( ( )) ( ( )) (2 ) ( ( ) ( ( ))
( ) ( )
2 ! 3 2( )
2 ( )
!( )! 3 2
n n
n n n n
n n
k
l n
n k n n
n n
n k n
a a
B i erfc i erfc i erfc i erfc
n n n k
A p
k n k n k
τ τ
λ ϕ τ ϕ τ ϕ τ ϕ τ
α τ α τ
α τ τ τ
∞ +
+ +
=
− − − ∞
= = =
− − − − − − + +
Γ + −
+ =
− Γ + −
∑
∑ ∑ ∑
(26)
Multiplying both sides by α τ2( )we obtain the next equation
( ) ( )
(
2 1 2 1 2 1 2 2 2)
1 1 1
0
2
2( ) 1 2 1 2
0 0 0
(2 ) ( ( )) ( ( )) (2 ) ( ) ( ( ) ( ( ))
2 ! 3 2( )
2 ( ) ( )
!( )! 3 2
n n n n n n
n n
k
l n
n k n n
n n
n k n
B a i erfc i erfc a i erfc i erfc
n n n k
A p u
k n k n k
λ τ ϕ τ ϕ τ τ α τ ϕ τ ϕ τ
α τ τ τ τ
∞ + + +
=
− + − ∞
= = =
− − − − − − + +
Γ + −
+ =
− Γ + −
∑
∑ ∑ ∑
(27)where 0 1 2 2
1 1 0
... 1
( ) 2a 2a n n n
α α τ α τ
ϕ τ ∞ ϕ τ
=
+ + +
= =
∑
and2 2
0 0
0 0 1
( ) ( )m n, ( ) , ( )m 1 m (3 ) k ( ) ,m k 1
n u u u k k m u m
α τ α τ α α α m α α
α
∞
−
= =
=
∑
= =∑
− ≥ .Analogously, taking l-th derivative of both sides of equation (27) we have
( )
1 2 2 11 2
2 1 2 1
2 1 2 1 1 2 2 1
1
1 0 0
0 1 1 2 2 1
2 2
( ) 1 2
1 0
1
( 2 )! ...
(2 ) ! ( 1) ( ) ( )
( 2 1)! ! !... !
2 ( 2 )! ...
(2 ) ! [ ]
( 2 )!
l n
i
b
b b
l n l n
m n m n m l n
n
n m b l n
b b
n l n
m l
m
l n a l
B i erfc i erfc
l n b b b
l n l n
a l l n m
ϕ ϕ ϕ
λ ϕ ϕ
α α α α
+ − − − +
+ − + − − +
= = − +
−
=
−
− − − − − − − −
− −
− −
∑ ∑ ∑
∑
2 11 2 2 1
2 2
2 1
0
1 2 2 1 1
2
2 1 2 2 1
0 1 2 2 1 0 0
( 1) ( ( )
! !... !
2 ! 3 2( )
( 2 )! ... 2 !
( ))
! !... ! !( )! 3 ( 2 1)!
2
l n
i
l n
i
b l n m
p n p
n
b l n p
k b
b b l n
n p l n
b l n n nk
i erfc b b b
n n n k
l n l
i erfc A
b b b k n k n k l n
ϕ
ϕ ϕ ϕ ϕ
− +
− +
− − −
− +
− + =
− − +
= =
− +
− − +
Γ + −
−
+ − −
− +
− Γ + −
∑ ∑
∑ ∑
2 2
1 2
2 1 4 2 1 2 2 2
0 0 1
1 1 2 2 2 0
( 2 1)! ... !
(2 2 1)! ! ( )
! !... ! 2( )!
l n
i
b b b
l n l
n k l l n
n l
m b l n n
l n l
n k p l u
b b b l n
ϕ ϕ ϕ
α − + α
− + − − − +
+
= − + =
− +
⋅ − + = −
∑
∑ ∑ ∑
(28)
