17.05.2015 Views

Numerical Integration Over Polygonal Domains using Convex ... - Ijecs

Numerical Integration Over Polygonal Domains using Convex ... - Ijecs

Numerical Integration Over Polygonal Domains using Convex ... - Ijecs

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

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)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!