147
APPENDIX C
We want to show that:
, m m s m .
S S
dS dS
φ ξ φ ξ φ ξ
∇ = ∇ ⋅ ∫∫ = − ∫∫ ∇ ⋅ (C.1) In order to do that, it is convenient to choose a Cartesian reference system with the x and y axis belonging to the plane defined by the generic face S of the triangle domain and with the origin in one of the triangle vertex (Fig. C.1).
Figure C.1. Cartesian reference system belonging to the triangle face S.
Noting that with the chosen reference system the RWG basis does not present any z-component, we obtain:
( ) ( )
( ) , ( ) ( ) , ( )
, , , ,
.
, , , , , ,
m m
S
m x m y
S
x y z x y dxdy
x y z x y x y z x y dxdy
x y
φ ξ φ ξ
φ ξ φ ξ
∇ = ∇ ⋅ =
∂ + ∂
∂ ∂
∫∫
∫∫ (C.2)
By using the definition (2.20), the (C.2) can be written as:
VECTOR IDENTITY
148
( ) ( )
, , , , , .
2 2
m m
m
m m
S S
l x l x
x y z dxdy x y z dxdy
x A y A
φ ξ φ φ
±
±
±±
∂ ∂
∇ = ± ±
∂ ∂
∫∫ ∫∫ (C.3)
The first integral in (C.3) can be evaluated analytically as follow (in a similar way we can solve the second one):
( ) ( )
( ) ( )
( )
0
, , , ,
2 2
, , , , ,
2 2
, , 2
C
B
C
bx x
m m
m ax m
S cx d x
m m
m m
ax x S
m S m
l x l x
x y z dxdy x y z dy
x A A
l x l
x y z dy x y z dxdy
A A
x y z l dxdy A
φ φ
φ φ
φ
±
±
±
± ±
− +
± ±
±
∂ =
∂
+ − =
−
∫∫ ∫
∫ ∫∫
∫∫
(C.4)
since:
( )
( ) ( )
, , 2
.
, , , , 0
2 2
B
B
C C
C C
cx d
m B B
ax m
B B
bx cx d
m C m C
C C
m m
ax ax
C C
x y z l x dy A
ax cx d
l x l x
x y z dy x y z dy
A A
bx cx d
φ
φ φ
− +
±
− +
± ±
= − +
− =
= − +
∫
∫ ∫
(C.5)
For the relation (C.5) and (2.21), the (C.1) becomes:
( )
( ) ( )
( )
, , ,
, , , , .
, , ,
m m
S m
m m
m m
S S
s m s m
S
x y z l dS A
l l
x y z dxdy x y z dxdy
A A
x y z dS
φ ξ φ
φ φ
φ ξ φ ξ
±
+ −
