#09 (09/17/2025)

Physical interpretation of divergence

div_physical.jpg
The amount of flow that goes out from the control volume is

u ·n,1
(1)
where n is the normal (outward) perpendicular to the boundary.
divergence.jpg
Thus, the flow that goes out from the side at x = x is

u ·n  d y = ⎛
⎜
⎜
⎝
ux(x, y)
uy(x, y)
⎞
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎝
−1
0
⎞
⎟
⎟
⎠ 		 d y = − ux(x, y) dy,
(2)
and the flow that goes out from the side x = x + dx is

u ·n  d y = ⎛
⎜
⎜
⎝
ux(x + d x, y)
uy(x + d x, y)
⎞
⎟
⎟
⎠ 		·⎛
⎜
⎜
⎝
1
0
⎞
⎟
⎟
⎠ 		 d y = ux(x + d x, y) dy,
(3)
so the net flow that goes out from the element in the x-direction is

(ux(x + dx, y) − ux(x, y)) dy
=
⎛
⎝ 		ux(x, y) + ∂ux(x, y)
∂x
 dx + …−ux(x, y)⎞
⎠ 		dy
=
∂ux
∂x
 dx dy.
(4)
Similarly, the net flow that goes out from the element in the y-direction is

∂uy
∂y
 dx dy.
(5)
Addition of eqs.(4, 5) is the net amount of flow that goes out from the element, i.e.

( ∂ux
∂x
 + ∂uy
∂y
 ) dx dy = ∇·u dx dy.
(6)

Heat conduction

The net heat that goes out of an element is

∇·h dx dy,
(7)
where h is the heat flux vector (heat per second). The rate of heat loss in the element is (note the minus sign)

−ρCp ∂T
∂t
 dx dy,
(8)
where ρ is the mass density and Cp is the specific heat so

∇·h dx dy = −ρCp ∂T
∂t
 dx dy.
(9)
The Fourier law of heat conduction stipulates that the heat flux is proportional to the gradient of the temperature, i.e.

h = − k ∇T,
(10)
where k is the thermal conductivity.
By combining Eq.(9) with Eq.(10), one obtains

∇·( k ∇T) = ρCp ∂T
∂t
 .
(11)
Equation (11) is known as the transient heat conduction equation.

Equation of continuity

The amount of mass that flows out of an element is

∇·(ρv),
(12)
which must be equal to the mass loss per unit time of

− ∂ρ
∂t
 ,
(13)
which yields

∂ρ
∂t
 + ∇·(ρv) = 0.
(14)
which is known as the equation of continuity.

Directional derivative (gradient)

The directional derivative, ∂u/∂n, of a function, u, in the direction of n is defined as

∂u
∂n
 ≡ ∇u ·n,
(15)
where the vector, n, is a unit vector which represents the direction of the change of rate of v. Note that if n is in the direction of the x axis, ∂u/∂n = ∂u/∂x.
Example: Compute the directional derivative of u=x2 y −3 x z in the direction of (1, −2, 3) at (x, y, z) = (4, 3, −2).
Solution:
n = (1, −2, 3)/
√
 
14
 
 ,
and ∇u = (2 x y − 3 z, x2, −3 x) at (4, 3, − 2) is (30, 16, − 12). So
∂u
∂n
 = 30 − 32 − 36

√
14
 = − 38

√
14
 .

Identities among differential operators


∇·(αu + βv )
=
α∇·u + β∇·v,
(16)
∇( αu + βv)
=
α∇u + β∇v,
(17)
∇×( αu + βv)
=
α∇×u+ β∇×v,
(18)
∇·(u v)
=
∇u ·v + u ∇·v,
(19)
∇×(u v)
=
∇u ×v + u ∇×v,
(20)
∇·( u ×v)
=
v ·∇×u − u ·∇×v,
(21)
∇×( u ×v)
=
u ∇·v − v ∇·u + ( v·∇)u − (u·∇) v,
(22)
∇(u ·v)
=
(u ·∇) v + (v ·∇) u+ u ×( ∇×v) + v ×( ∇×u),
(23)
∇·∇×v
=
0,
(24)
∇×∇u
=
0.
(25)
When proving various identities that involve vectors and their derivatives (the divergence, gradient and curl), it is important to remember that the nabla (∇) works as (1) a vector and as (2) a differential operator ((uv)′ = u′v + u v′).

  1. ∇×∇u = 0.
    The vector product of two identical vectors (∇×∇) is identically zero.

  2. ∇·∇×v = 0.
    Using the identities of scalar triple product,
    ∇·∇×v
    =
    v ·∇×∇
    =
    ∇×∇·v
    =
    0.
    		(26)

  3. ∇×(u ×v).
    The vector triple product identity is
    ∇×(u ×v)
    =
    (∇·v) u −(∇·u) v.
    		(27)
    Note that ∇ in the right hand side operates on both u and v so (u v)′ = u′v + u v′ needs to be recursively applied. The first term in the right hand side is expanded as
    u ∇·v + (v ·∇) u,
    and the second term is expanded as
    v ∇·u + (u ·∇) v,
    so the right hand side is
    u ∇·v + (v ·∇) u −v ∇·u − (u ·∇) v.

Footnotes:

1 u ·n = || u || ||n|| cos θ = u cos θ.

File translated from TEX by TTH, version 4.03.
On 16 Sep 2025, 20:59.