14
1 Preliminaries
∂ϕ
∂ N
= G · n = |G| cos(G, n) = |G| = grad ϕ · n = ∇ϕ · n
= |grad ϕ|n · n =
grad ϕ · grad ϕ
|grad ϕ|
=
∇ϕ · ∇ϕ
|∇ϕ|
= |grad ϕ| = |∇ϕ| (1.3.24)
Equation (1.3.24) shows that the directional derivative of the function ϕ along
the gradient direction derivative is constant greater than or equal to zero, namely the
gradient always points to the direction of the function ϕ increase. Obviously there is
cos(G, n) = cos 0 = 1, that is to say that the gradient direction of ϕ is the same as
the normal direction of the contour surface of ϕ.
In the rectangular coordinate system, the directional derivative ϕ along the
direction N can also be written in the following form
∂ϕ
∂ N
= ∇ϕ · n =
∂ϕ
∂ x
n x +
∂ϕ
∂ y
n y +
∂ϕ
∂z
n z = |∇ϕ| =
∂ϕ
∂ x
2
+
∂ϕ
∂ y
2
+
∂ϕ
∂z
2
(1.3.25)
Substituting Eq. (1.3.24) into Eq. (1.3.21), we obtain
grad ϕ = ∇ϕ =
∂ϕ
∂ N
n
(1.3.26)
From Eqs. (1.3.22) to (1.3.24), we obtain
∂
∂ N
= n · ∇ = l
∂
∂ x
+ m
∂
∂ y
+ n
∂
∂z
= n x
∂
∂ x
+ n y
∂
∂ y
+ n z
∂
∂z
(1.3.27)
where,
∂
∂ N
is called the normal derivative operator or differential operator.
Let the gradient of each point in the scalar field correspond to each point in the
scalar field, a vector field could be obtained, this vector field is called the gradient
field produced by the scalar field.
Example 1.3.1 Let dϕ = a · dr, prove the vector identity
a = ∇ϕ
(1.3.28)
Proof The total differential of the function ϕ is
dϕ =
∂ϕ
∂ x
dx +
∂ϕ
∂ y
dy +
∂ϕ
∂z
dz =
∂ϕ
∂ x
i +
∂ϕ
∂ y
j +
∂ϕ
∂z
k
· (dx i + dy j + dzk) = ∇ϕ · dr
Substituting the above expression into the given condition, we have
dϕ = a · dr = ∇ϕ · dr
or