The divergence of a vector field in the space.

Definition 6.37   Let $\overrightarrow{F} (x,y,z)= M(x,y,z) \overrightarrow{i} + N((x,y,z) \overrightarrow{j} + P(x,y,z) \overrightarrow{k}$be a vector filed in the 3-dimensional space. The divergence of the vector field $\overrightarrow{F}$is the scalar function

$$\text{div} \overrightarrow{F} = \frac {\partial M}{\partial x... ...rac {\partial N}{\partial y} +\frac {\partial P}{\partial z} .$$

It has the same physical interpretation as the divergence of a vector field in the plane (v.s. Remark 6.34).

Remark 6.38   The divergence of the vector field $\overrightarrow{F}$can be expressed as the formal scalar product

$$\text{div} \overrightarrow{F} = \nabla \cdot \overrightarrow{F}$$

where

$$\nabla = \frac {\partial }{\partial x} \overrightarrow{i} +\f... ...ghtarrow{j} +\frac {\partial }{\partial z}\overrightarrow{k} .$$

Remark 6.39   Let f(x,y,z) be a function of three real variables defined on an open domain $\mathcal{D}$in $\mathbb{R} ^3$. Then, as we saw in 6.18, the gradient of f is:

$$\overrightarrow{\nabla f} = \frac {\partial f}{\partial x} \o... ...htarrow{j} +\frac {\partial f}{\partial z}\overrightarrow{k} .$$

Apply the divergence operator to this vector field (=the gradient field); we have:

$$\nabla \cdot ( \overrightarrow{\nabla f} )= \frac {\partial^... ...rtial^2 f }{\partial y^2} +\frac {\partial^2 f }{\partial z^2}$$

i.e. $\nabla^2$is Laplace's operator.

More on this topic will be studied in Section 5, about the Divergence Theorem.