Divergence and Curl

Divergence and Curl of a Vector Field

Master of Science in Mathematics
Lecture Notes. Accessed on: 2019-10-21 18:39:35

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 $D.$

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