4
1 2 1 2
(c) (d)
3 3
5 6 5
1 4 2 1 4 2
z„
¥ ¢
D T
¡
Original triangle (a) and three possible re¬ne
ment scenarios.
! ©¤( &% &¡ ( £ ¥
£
©
%
c

The ¬nite volume method is geared toward the solution of conservation laws of the form:
0 ¤ ‚ — ¢ Y
t
V ¨ ¢ & ¥
(
¡
B¤ ¢
¡
In the above equation, is a certain vector function of and time, possibly nonlinear.
V V
0
C
This is called the “¬‚ux vector.” The source term is a function of space and time. We now
apply the principle used in the weak formulation, described before. Multiply both sides by
a test function , and take the integral
£¢ £¢ £¢
h 9 ¤ ‚
x9 0
V
—
9¢
¦ ¦ ¦
¡
Then integrate by part using formula (2.28) for the second term on the lefthand side to
obtain
£¢ £¢ ¥¢ £¢
¤ b ¤˜ ¤
x9 0
V
—
9¢ 9 9
3
¦ ¦ ¦
¡ ¡
¤
Consider now a control volume consisting, for example, of an elementary triangle in
the twodimensional case, such as those used in the ¬nite element method. Take for a
–
function whose value is one on the triangle and zero elsewhere. The second term in the
” ” ”n
uu u ™
n
¢
¤§ ¥ §¢
¡
¡© £ £ ¥ £ ¤ § ¨¡ ¢
§¥ © ©
above equation vanishes and the following relation results:
¢ ¦¢
¥
g 9 0 R ¥ ¢ 9 ¤˜ ¤ Y
t
9 ¢V ¨ ¢¢ & ¥
— 2
¦ ¦
¡
¤
R¥ R