The Divergence of a Vector Field

# The Divergence of a Vector Field

Recall that if $w = f(x, y, z)$ is a three variable real-valued function, then the gradient of $f$ denoted $\nabla f$ is given by:

(1)
\begin{align} \quad \nabla f(x, y, z) = \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right ) = \frac{\partial f}{\partial x} \vec{i} + \frac{\partial f}{\partial y} \vec{j} + \frac{\partial f}{\partial z} \vec{k} \end{align}

Thus in a sense, the gradient of a function (or rather, the gradient for a scalar field) provided us information about the function itself. We will now look at an important underlying function known as the divergence of a vector field which also makes use of the valuable information stored in partial derivatives.

 Definition: If $\mathbf{F} (x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a vector field on $\mathbb{R}^3$ and $\frac{\partial P}{\partial x}$, $\frac{\partial Q}{\partial y}$ and $\frac{\partial R}{\partial z}$ all exist then the Divergence of $\mathbf{F}$ is a scalar field (function) given by $\mathrm{div} (\mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$.

The notation $\nabla \cdot \mathbf{F}$ is often used to represent the divergence of a vector field. We note that in $\mathbb{R}^3$, $\nabla = \frac{\partial}{\partial x} \vec{i} + \frac{\partial}{\partial y} \vec{j} + \frac{\partial}{\partial z} \vec{k}$ and $\mathbf{F} = P \vec{i} + Q \vec{j} + R \vec{k}$ so the dot product is $\mathrm{div}(\mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$.

Geometrically, the divergence of a vector field $\mathbf{F}$ at a point $P$ can be considered as follows. Suppose that $\mathbf{F}$ represents a velocity field for fluid flow. Then the divergence of $\mathbf{F}$ at $P$ represents the net rate of change an object/particle moves away or diverges from $P$.

If $\mathrm{div} (\mathbf{F}) = 0$ at a point $P$, then we give $\mathbf{F}$ a special term at $P$ which we define below.

 Definition: Let $\mathbf{F}$ be a vector field on $\mathbb{R}^3$. If $\mathrm{div} (\mathbf{F}) = 0$ at a point $P$ at $(x, y, z)$ then $\mathbf{F}$ is said to be Incompressible at $P$.

The term Solenoidal at $P$ or Divergence Free at $P$ mean the same thing as Incompressible at $P$.

 Theorem 1: Let $\mathbf{F}(x, y, z) = P(y, z) \vec{i} + Q(x, z) \vec{j} + R(x, y) \vec{k}$. Then $\mathbf{F}$ is incompressible for every $(x, y, z) \in D(\mathbf{F})$.
• Proof: Let $\mathbf{F}(x, y, z) = P(y, z) \vec{i} + Q(x, z) \vec{j} + R(x, y) \vec{k}$. Then the divergence of $\mathbf{F}$ is:
(2)
\begin{align} \quad \mathrm{div} (\mathbf{F}) = \frac{\partial}{\partial x} P(y, z) + \frac{\partial}{\partial y} Q(x,z) + \frac{\partial}{\partial z} R(x, y) = 0 \end{align}
• Since $\mathrm{div} (\mathbf{F}) = 0$ then we have that $\mathbf{F}$ is incompressible/solenoidal for all $(x, y, z) \in D(\mathbf{F})$. $\blacksquare$

Let's look at some examples of computing the divergence of a vector field.

## Example 1

Find the divergence of the vector field $\mathbf{F} = (2x - 2y)\vec{i} + y\sin x \vec{j} + xy^2z^2 \vec{k}$.

Applying the formula above and we get that:

(3)
\begin{align} \quad \mathrm{div} (\mathbf{F}) = \frac{\partial}{\partial x} (2x - 2y) + \frac{\partial}{\partial y} (y \sin x) + \frac{\partial}{\partial z} (xy^2z^2) \\ \quad \mathrm{div} (\mathbf{F}) = 2 + \sin x + 2xy^2z \end{align}

## Example 2

Find the divergence of the vector field $\mathbf{F} = xe^xyz \vec{i} + 2y^3(x + z) \vec{j} + \cos(xz) \vec{k}$.

Applying the formula above and we get that:

(4)
\begin{align} \quad \mathrm{div} (\mathbf{F}) = \frac{\partial}{\partial x} (xe^xyz) + \frac{\partial}{\partial y} (2y^3(x + z)) + \frac{\partial}{\partial z} (\cos (xz)) \\ \quad \mathrm{div} (\mathbf{F}) = yz(xe^x + e^x) + 6y^2(x + z) + -x \sin (xz) \end{align}