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Appendix: All Solutions to the Problems
˝
V ( f g − g f )dV =
S
f
∂g
∂n
− g
∂ f
∂n
dS = 0 namely
S f
∂g
∂n
dS =
S g
∂ f
∂n
dS
Quod erat demonstrandum.
1.12 Let a surface equation be ϕ(x, y, z, t) = 0, where, x, y and z are all the functions
of time, prove that the normal velocity of the motion surface is v n = v·
∇ϕ
|∇ϕ| = −
ϕ t
|∇ϕ| .
Proof Taking the differential to the equation ϕ(x, y, z, t) = 0, to get ϕ x dx +ϕ y dy +
ϕ z dz + ϕ t dt = 0, or
ϕ x
dx
dt
+ ϕ y
dy
dt
+ ϕ z
dz
dt
dz + ϕ t = 0
Obviously,
dx
dt
,
dy
dt
and
dz
dt
are the three components of velocity vector v at the
moment t at a point M(x, y, z) on the surface, and ϕ x , ϕ y and ϕ z are the three
components of the gradient ∇ϕ of ϕ. Therefore the above equation can be written as
v · ∇ϕ = −ϕ t
It can be known from Eq. (1.3.21) that n =
∇ϕ
|∇ϕ| , so the normal velocity is
v n = v · n = v ·
∇ϕ
|∇ϕ|
= −
ϕ t
|∇ϕ|
Quod erat demonstrandum.
1.13 Prove
˝
V (∇ × a · ∇ × b − b · ∇ × ∇ × a)dV =
S (b × ∇ × a) · ndS.
The formula is called the Green(’s) first vector formula.
Proof Let c = b × ∇ × a, taking the divergence, to get ∇ · c = ∇ · (b × ∇ ×
a) = ∇ × a · ∇ × b − b · ∇ × ∇ × a, substituting it into the Gauss formula, to
get
˝
V (∇ × a · ∇ × b − b · ∇ × ∇ × a)dV =
S (b × ∇ × a) · ndS. Quod erat
demonstrandum.
1.14 Prove the Green(’s) second vector formula
˚
V
(b · ∇ × ∇ × a − a · ∇ × ∇ × b)dV =
S
(a × ∇ × b − b × ∇ × a) · ndS
=
S
[(n × a) · ∇ × b − (n × b) · ∇ × a]dS
Proof 1 Let c = b × ∇ × a, taking the divergence, to get ∇ · c = ∇ · (b × ∇ × a) =
∇ × a · ∇ × b − b · ∇ × ∇ × a, substituting it into the Gauss formula, to get