Information Technology Reference
In-Depth Information
@t
D
k
@
2
c
@c
C
f.x;t/:
(7.17)
@x
2
Deriving a Three-Dimensional PDE
In three space dimensions the derivation is very similar, but we need to make use
of some results from vector calculus and surface integrals. Just a brief sketch is pre-
sented here. We consider an arbitrarily chosen volume V and state that conservation
of mass inside V must imply that the net flow of mass into V equals the increase in
mass inside V .Let
q
be the velocity of the substance. The net flow of mass into V
in a time interval t can be expressed as a surface integral of the normal component
of
q
(the tangential component does not bring the substance into V !):
Z
%
q n
t
dS
:
@V
The symbol @V denotes the surface of V ,and
n
is an outward unit normal vector
to @V .
n
V
∂
V
The minus sign originates from the fact that we seek the net
inflow
, and the fact that
n
points
outward
, i.e.,
q n
represents the outflow; so to get the inflow, we must
change the sign.
The inflow is to be balanced by the increase in concentration times the density
and the total injected mass in V , exactly as in the one-dimensional case:
Z
%
q n
t
dS
C
Z
%f t
dV
D
Z
%
@c
@t
t
dV
:
@V
V
V
The surface integral can be transformed via the divergence theorem
7
to a volume
integral:
Z
%
q n
t
dS
D
Z
r
.%
q
/t
dV
:
@V
V
Collecting the two volume integrals, assuming constant %, dividing by %t ,and
requiring that the integrand must vanish, we arrive at the equation
7
Also called Gauss' theorem or Gauss' divergence theorem, cf. your favorite topic on surface
integrals.
