Mathematics of Physics and Engineering

154 The Three Theorems

the flux across dG, we have

dM

dt HIpudV - ifpv-da; (3.2.7)

G 8G

recall that the positive direction of the flux is out of G, which reduces
the mass. On the other hand, M = fffp dV. Assuming that we can
G
differentiate under the integral, dM/dt = JJf(dp/dt) dV. If the functions
G
p and v are continuously differentiable, then, by Gauss's Theorem (3.2.4),
Jfpv-dcT = fff div(pv)dV. As a result, (3.2.7) implies
dG G

///(

-£ + div(pv) - pi> ) dV = 0.

Given a point in space, this equality holds for every domain G containing
that point, and we therefore conclude that the equation of continuity

-£+div(pv)=pu (3.2.8)

holds at every point of the space. Conversely, after integration, (3.2.8)
implies (3.2.7).

EXERCISE 3.2.5.C (a) Convince yourself that (3.2.7) implies (3.2.8). Hint:

assume that the left-hand side of (3.2.8) is bigger than up at one point. By conti-

nuity, the inequality will hold in some neighborhood of that point. Integrate over

that neighborhood and get a contradiction, (b) Verify that (3.2.8) is equivalent

to

-?-+pdivv = up. (3.2.9)

dt

Hint: Consider a trajectory r = r(t) of a moving particle so that v = r ' and

p = p(r(t),t). Then dp/dt = Vp • v + dp/dt. On the other hand, by (3.1.30),

div(pu) = Vp • v + p divv.

A continuum is called incompressible if the density p does not change

in time and space: p(t, x) = const. For an incompressible continuum, equa-

tion (3.2.8) becomes divw = u; the velocity v of an incompressible contin-

uum without sources or sinks satisfies divt; = 0.

We can also state Gauss's Theorem in a two dimensional domain G.

Let the functions P, Q and the domain G satisfy the conditions of Green's