Numerical Integration Over Polygonal Domains using Convex ... - Ijecs
Numerical Integration Over Polygonal Domains using Convex ... - Ijecs
Numerical Integration Over Polygonal Domains using Convex ... - Ijecs
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x y<br />
f x y dxdy<br />
(<br />
, )<br />
<br />
dxdy<br />
<br />
<br />
( , )<br />
<br />
<br />
xy xy xy<br />
xy xy xy x<br />
<br />
Tj<br />
0i<br />
Tk<br />
0 j Ti<br />
0k<br />
<br />
<br />
Tj<br />
0i<br />
Tk<br />
0 j Ti<br />
0k<br />
<br />
<br />
<br />
= <br />
<br />
(<br />
x,<br />
y)<br />
dy<br />
<br />
xy xy xy<br />
<br />
T j 0i<br />
Tk<br />
0 j T i 0k<br />
<br />
<br />
= (<br />
x,<br />
y)<br />
dy (<br />
x,<br />
y)<br />
dy (<br />
x,<br />
y)<br />
dy<br />
<br />
<br />
l j 0 l0i lij <br />
lk 0 l0<br />
j l jk <br />
li<br />
0 l j 0k<br />
lki<br />
<br />
<br />
= (<br />
x,<br />
y)<br />
dy<br />
=<br />
( x , y)<br />
dy = f ( x,<br />
y)<br />
dxdy<br />
(34)<br />
xy<br />
xy<br />
<br />
lij l jk lki<br />
<br />
T ijk<br />
T ijk<br />
xy<br />
Thus, we have proved that the triangle Tijk<br />
expands into three new triangles with respect to the origin. We<br />
may note that in deriving the above result of eqn.(34), the following fact is used.<br />
(<br />
x,<br />
y)<br />
dy (<br />
x,<br />
y)<br />
dy 0<br />
<br />
l0i<br />
li<br />
0<br />
(<br />
x,<br />
y)<br />
dy (<br />
x,<br />
y)<br />
dy <br />
l<br />
j 0 0 j<br />
l<br />
(<br />
x , y)<br />
dy (<br />
x,<br />
y)<br />
dy <br />
l<br />
k 0 0k<br />
l<br />
0<br />
0<br />
The general result of eqn.(33) can be readily proved on similar lines.<br />
Theorem 2. Let us denote the triangle spanned by vertices ( xi , yi<br />
), (0,0), ( xk<br />
, yk<br />
), ( k i 1)<br />
as<br />
the integral of a smooth function f over the region<br />
II<br />
xy<br />
i 0k<br />
T<br />
( f ) ( x<br />
k<br />
y x y )<br />
i<br />
i<br />
k<br />
1 1<br />
<br />
0 0<br />
rf ( r(<br />
x<br />
Proof: Let us consider the integral:<br />
i<br />
x<br />
ki<br />
s),<br />
r(<br />
y<br />
i<br />
xy<br />
Ti<br />
0 k<br />
y<br />
<br />
ki<br />
is expressible as<br />
s))<br />
drds<br />
xy<br />
Ti<br />
0 k<br />
. Then<br />
H. T. Rathod a IJECS Volume 2 Issue 8 August, 2013 Page No.2576-2610 Page 2584<br />
(35)<br />
II xy ( f ) f ( x, y)<br />
dxdy<br />
(36)<br />
Tijk<br />
xy<br />
Tijk<br />
The parametric equations of the oriented triangle in the xy - plane with vertices spanned by x , y ), ( x , y )<br />
(<br />
i i j j<br />
and ( xk<br />
, yk<br />
), ( k i 1)<br />
which map this arbitrary triangle into a unit right isosceles triangle in the uv – plane<br />
are<br />
x x x u x v, y y y u y v<br />
(37)<br />
i<br />
where,<br />
ji<br />
ki<br />
i<br />
ji<br />
ji<br />
j<br />
ki<br />
0 u,<br />
v 1, u v 1, x x x , x x x , y y y , y y y<br />
(38)<br />
We have then<br />
(<br />
x,<br />
y)<br />
dxdy dudv ( x<br />
ji<br />
yki<br />
xkiy<br />
ji<br />
) dudv<br />
(<br />
u,<br />
v)<br />
xy<br />
= ( 2<br />
dudv<br />
ijk<br />
)<br />
i<br />
ki<br />
k<br />
= ( 2<br />
area of the triangle T xy<br />
ijk<br />
) dudv<br />
(39)<br />
and thus, we define<br />
xy<br />
( x y x y )<br />
(40)<br />
2<br />
ijk ji ki ki ji<br />
Using the above eqns.(37) –(39) in eqn.(36) gives us<br />
II<br />
xy<br />
i 0k<br />
T<br />
2<br />
1 1u<br />
xy<br />
ijk <br />
0 0<br />
f (( x<br />
i<br />
x<br />
ji<br />
u x<br />
ki<br />
v),(<br />
y<br />
i<br />
y<br />
ji<br />
u y<br />
ki<br />
i<br />
ji<br />
v))<br />
dudv<br />
Using the transformations<br />
u 1 r,<br />
v rs<br />
(42)<br />
we can map the above integral in eqn.(36) into an equivalent integral over the rectangle ( r , s) / 0 r,<br />
s 1<br />
.<br />
This gives us<br />
j<br />
i<br />
ki<br />
k<br />
i<br />
(41)