# Divergence and Curl of a Vector Field • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## The Divergence and Curl

Definition. Let $\mathbf{V}$ be a given vector field. The divergence of $\mathbf{V}$ is defined by div $\mathbf{V}=\nabla \cdot \mathbf{V}$ and the curl of $\mathbf{V}$ is defined by curl $\mathbf{V}=\nabla \times \mathbf{V}$ where \begin{equation} \nabla =\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}k\end{equation} is the del operator.

For a discussion on the physical meaning of the divergence and curl of a vector field read this.

Example. Find the divergence and curl of a constant vector field.

Solution. Let ${V}=a {i}+b {j}+c {k}$ for constants $a,$ $b,$ and $c.$ Then \begin{equation} \mathop{div} {V}=\frac{\partial }{\partial x}(a)+\frac{\partial }{\partial y}(b)+\frac{\partial }{\partial z}(c) = {0} \end{equation} and \begin{equation} \mathop{curl} {V}=\left| \begin{array}{cc} \begin{array}{c} {i} \\ \begin{array}{c} \frac{\partial }{\partial x} \\ a \end{array} \end{array} & \begin{array}{cc} {j} & {k} \\ \begin{array}{c} \frac{\partial }{\partial y} \\ b \end{array} & \begin{array}{c} \frac{\partial }{\partial z} \\ c \end{array} \end{array} \end{array} \right|=0 {i}-0 {j}+0 {k}=0. \end{equation}

Example. Find the divergence and curl of the vector field \begin{equation} {V}(x,y,z)=x z {i}+x y z {j}-y^2 {k}. \end{equation}

Solution. The divergence of $\mathbf{V}$ is \begin{equation} \mathop{div} \mathbf{V}=\nabla \cdot \mathbf{V}=\frac{\partial }{\partial x}(x z)+\frac{\partial }{\partial y}(x y z)+\frac{\partial }{\partial z}\left(-y^2\right)=z+x z. \end{equation} and the curl of $\mathbf{V}$ is \begin{align} & \mathop{curl} \mathbf{V} =\left| \begin{array}{ccc} {i} & {j} & {k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ x z & x y z & -y^2 \end{array} \right| =\left[\frac{\partial }{\partial y}\left(-y^2\right)-\frac{\partial }{\partial z}(x y z)\right]{i} \\ & \qquad \quad \qquad -\left[\frac{\partial }{\partial x}\left(-y^2\right)-\frac{\partial }{\partial z}(x z)\right]{j}+\left[\frac{\partial }{\partial x}(x y z)-\frac{\partial }{\partial y}(x z)\right]{k} \\ & \qquad =(-2y-x y){i}-(0-x){j}+(y z-0){k} =-y(2+x) {i}+x {j}+y z {k}. \end{align}

Theorem. (Properties of Divergence and Curl) Let ${U}$ and ${V}$ be vector fields with component functions that have continuous first and second partial derivatives. Then

$(1) \quad \mathrm{div} (c {U})= c \, \mathrm{div} {U}$

$(2) \quad \mathrm{div} ({U}+{V})= \mathrm{div} {U}+ \mathrm{div} {V}$

$(3) \quad \mathrm{div} (f {U})= f \mathrm{div} {U}+(\nabla f\cdot {U})$

$(4) \quad \mathrm{div} (f\nabla g)=f \mathrm{div} \nabla g+\nabla f \cdot \nabla g$

$(5) \quad \mathrm{curl} (c{U})= c \mathrm{curl} {U}$

$(6) \quad \mathrm{curl} ({U}+{V})= \mathrm{curl} {U}+ \mathrm{curl} {V}$

$(7) \quad \mathrm{curl} (f {U})=f \mathrm{curl} {U}+(\nabla f\times {U})$

$(8) \quad \mathrm{curl} (\nabla f+ \mathrm{curl} {U})= \mathrm{curl}(\nabla f)+ \mathrm{curl}(\mathrm{curl} {U})$

$(9) \quad \nabla \times (\nabla f)=0$

$(10) \quad \mathrm{div} (\mathrm{curl} {U})=0$

Example. Show that the divergence of the curl of a vector field is $0.$

Solution. Let ${V}=u(x,y,z){i}+v(x,y,z){j}+w(x,y,z){k}$ be a vector field, then \begin{equation} \mathop{curl} {V}=\left| \begin{array}{ccc} {i} & {j} & {k} \\ \frac{ \partial }{\partial x} & \frac{ \partial }{\partial y} & \frac{ \partial }{\partial z} \\ u & v & w \end{array} \right| = \left(w_y-v_z\right) {i}-\left(w_x-u_z\right){j}+\left(v_x-u_y\right){k}\end{equation} Therefore \begin{equation} \nabla \cdot {V}=\partial _x \left(w_y-v_z\right)-\partial _y \left(w_x-u_z\right)+\partial _z \left(v_x-u_y\right)=0 \end{equation} as desired.

Example. Show that the curl of the gradient of a function is always ${0}.$

Solution. Let $\nabla f=f_x{i}+f_y{j}+f_z{k}$ then \begin{equation} \mathop{curl} \nabla f=\left| \begin{array}{ccc} {i} & {j} &{k} \\
\frac{ \partial }{\partial x} & \frac{ \partial }{\partial y} & \frac{ \partial }{\partial z} \\ f_x & f_y & f_z \end{array} \right| =\left(f_{z y}-f_{y z}\right){i}-\left(f_{x z}-f_{z x}\right){j}+\left(f_{y x}-f_{x y}\right){k}={0}. \end{equation}

Example. Let ${F}=\langle x^2y,y z^2,z y^2\rangle.$ Either find a vector field ${G}$ such that ${F}=\mathop{curl} {G}$ or show that no such ${G}$ exists.

Solution. If a vector field ${G}$ does exist then, then $\mathop{div} {F}=\mathop{div} (\mathop{curl} {G})$ but $$\mathop{div} {F}=\partial _x \left(x^2y\right)+\partial _y \left(y z^2\right)+\partial _y \left(z y^2\right)=2 x y+z^2+2 y z$$ and since the div of the curl of a vector field is always zero we see there can be no vector ${G}$ with ${F}=\mathop{curl} {G}$ for this vector field $F.$

Definition. Let $f(x,y,z)$ define a function with continuous first and second partial derivatives. Then the Laplacian of $f$is \begin{equation} \nabla ^2f=\nabla \cdot \nabla f=\frac{\text{ }\partial ^2f}{\partial x^2}+\frac{\text{ }\partial ^2f}{\partial y^2}+\frac{\text{ }\partial ^2f}{\partial z^2}.\end{equation} The equation $\nabla ^2f=0$ is called Laplacian’s equation and a function that satisfies it in a region $D$ is said to be harmonic on $D.$