4.4 divergence and curl of a Vector Field

Since \(\nabla \) has the character of a vector, we can consider the dot and cross product on which it appears as one factor. If \[F = F_1 \textbf {i} + F_2 \textbf {j} + F_3\textbf {k}\] is a vector whose components are differentiable functions of \(x, y, z\)

1.
The dot product \begin {align*} \nabla \cdot F & = \Bigg (\textbf {i}\frac {\partial }{\partial x} + \textbf {j}\frac {\partial }{\partial y} + \textbf {k}\frac {\partial }{\partial z}\Bigg )\cdot \Bigg (F_1\textbf {i} + F_2\textbf {j} + F_3\textbf {k}\Bigg )\\\\ \implies \hspace {0.5cm}\nabla \cdot F & = \frac {\partial F_1}{\partial x} + \frac {\partial F_2}{\partial y} + \frac {\partial F_3}{\partial z} \end {align*}

This is called the divergence of the vector \(F\) written as \(\dive \, F\).

2.
The cross product \begin {align*} \nabla \times F & = \Bigg (\textbf {i} \frac {\partial }{\partial x} + \textbf {j} \frac {\partial }{\partial y} + \textbf {k} \frac {\partial }{\partial z}\Bigg ) \times \Big ( F_1\textbf {i} + F_2\textbf {j} + F_3\textbf {k}\Big )\\\\ & = \Bigg ( \frac {\partial F_3}{\partial y}-\frac {\partial F_2}{\partial z}\Bigg )\textbf {i} - \Bigg (\frac {\partial F_3}{\partial x}-\frac {\partial F_1}{\partial z}\Bigg ) \textbf {j} + \Bigg (\frac {\partial F_2 }{\partial x}-\frac {\partial F_1}{\partial y}\Bigg ) \textbf {k} \end {align*}

is called the curl of \(F\) and written \(\curl \, F\).

Note that \[\curl F = \nabla \times F = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\\\ \dfrac {\partial }{\partial x} & \dfrac {\partial }{\partial y} & \dfrac {\partial }{\partial z}\\\\ F_1 & F_2 & F_3\\ \end {vmatrix} \]

3.
If \(f(x,y,z)\) is a twice differentiable function, then \[\grad \, f = \nabla f = \frac {\partial f}{\partial x}\textbf {i} + \frac {\partial f}{\partial y}\textbf {j} + \frac {\partial f}{\partial z}\textbf {k}\] and \[\dive \big (\grad \, f\big ) = \frac {\partial ^2 f}{\partial x^2} + \frac {\partial ^2 f}{\partial y^2}+ \frac {\partial ^2f}{\partial ^2z} = \nabla ^2f\]

Here \(\displaystyle {\nabla ^2 = \frac {\partial ^2}{\partial x^2}+ \frac {\partial ^2}{\partial y^2} + \frac {\partial ^2}{\partial z^2}}\) is called the Laplacian operator.

Any function \(f\) which satisfies the equation \(\nabla ^2f=0\) is called a Harmonic function.

Properties

If \(\textbf {U}\) and \(\textbf {V}\) are vector point functions and \(f\) is a scalar function, then

1.
\(\nabla \cdot \big (\textbf {U}\pm \textbf {V}\big ) = \nabla \cdot \textbf {U} \pm \nabla \cdot \textbf {V}\) or \(\dive \big (\textbf {U}\pm \textbf {V}\big ) = \dive \big (\textbf {U}\big ) \pm \dive \big (\textbf {V}\big )\)
2.
\(\nabla \cdot \Big (f(\textbf {U})\Big ) = \Big (\nabla \cdot \textbf {U}\Big )f + \textbf {U}\cdot \nabla f\)
3.
\(\nabla \times (\textbf {U}+\textbf {V}) = \nabla \times \textbf {U} + \nabla \times \textbf {V}\)
4.
\(\curl (f\textbf {V}) = f\, \curl \textbf {V} + (\grad \, f)\times \textbf {V} = f(\nabla \times \textbf {V}\big ) + \big (\nabla f\big )\times \textbf {V}\)

To prove, let \(\textbf {U} = U_1\textbf {i} + U_2\textbf {j} + U_3\textbf {k}\) and \(\textbf {V} = V_1\textbf {i} + V_2\textbf {j} + V_3\textbf {k}\).

Example 4.4.1.

1.
Find the curl and divergence of \[F = x^2y\textbf {i} - xyz\textbf {j} + yz^2\textbf {k}\]
2.
Prove that \(\nabla ^2\bigg (\dfrac {1}{r}\bigg ) = 0\), where \( r^2 = x^2 + y^2 + z^2\) and \(\displaystyle {\nabla ^2 = \frac {\partial ^2}{\partial x^2} + \frac {\partial ^2}{\partial y^2} + \frac {\partial ^2}{\partial z^2}}\)

Solution. divergence of \(F\) is \(\nabla \cdot F\) \begin {align*} \nabla \cdot F & = \Bigg (\textbf {i}\frac {\partial }{\partial x} + \textbf {j}\frac {\partial }{\partial y} +\textbf {k} \frac {\partial }{\partial z}\Bigg ) \cdot \big (x^2y\textbf {i} - xyz\textbf {j} + yz^2\textbf {k}\big )\\\\ & = \frac {\partial }{\partial x}\big (x^2y\big ) + \frac {\partial }{\partial y}\big (-xyz\big ) + \frac {\partial }{\partial z}\big (yz^2\big )\\\\ & = 2xy - xz + 2yz\\ \end {align*}

curl of \(F\) is \(\nabla \times F\) \begin {align*} \nabla \times F & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\\\ \dfrac {\partial }{\partial x} & \dfrac {\partial }{\partial y} & \dfrac {\partial }{\partial z}\\\\ x^2y & -xyz & yz^2\\ \end {vmatrix}\\\\ & = \textbf {i}\Bigg (\frac {\partial }{\partial x}\big (yz^2\big ) + \frac {\partial }{\partial z}\big (xyz\big )\Bigg )-\textbf {j}\Bigg (\frac {\partial }{\partial x}\big (yz^2\big ) - \frac {\partial }{\partial z}\big (x^2y\big ) \Bigg ) + \textbf {k}\Bigg (\frac {\partial }{\partial x}\big (-xyz\big ) - \frac {\partial }{\partial y}\big (x^2y\big )\Bigg )\\\\ & = \big (z^2+xy\big )\textbf {i} - (0)\textbf {j} + \big (-yz-x^2\big )\textbf {k}\\\\ \therefore \hspace {0.5cm} \curl \, F & = \big (xy+z^2\big )\textbf {i} -\big (x^2+yz\big )\textbf {k}\\ \end {align*}

Let \(F(x,y,z)\) be a vector field. We say that \(F\) is irrotational if \(\curl \, F = \textbf {0}\) and \(F\) is incompressible if \(\dive F =0\).

Recall in \(\mathbb {R}^3\), a vector field \(F=P\textbf {i} + Q\textbf {j} + R\textbf {k}\) is the gradient of a function \(f\) if and only if \[\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}\hspace {0.5cm} ,\hspace {1cm} \frac {\partial R}{\partial x} = \frac {\partial P}{\partial z}\hspace {0.5cm} , \hspace {1cm} \frac {\partial Q}{\partial z} = \frac {\partial R}{\partial y}\]

From the definition of \(\curl \, F\) , we get following result.

Theorem 4.4.2.

In a simply connected region, the differentiable vector field \(F\) is the gradient of a function \(f\) if and only if \(\curl F = \textbf {0}\).

Example 4.4.3.

Show that the vector field \[ F = 2xy\textbf {i} + (x^2 + z^2)\textbf {j} + 2zy\textbf {k}\] is a gradient of a function \(f\). Hence, find a potential function for \(F(x,y,z)\).

Solution.

We need to show that \(\curl F = \textbf {0}\) \begin {align*} \nabla \times F & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\\\ \dfrac {\partial }{\partial x} & \dfrac {\partial }{\partial y} & \dfrac {\partial }{\partial z}\\\\ 2xy & (x^2+ z^2) & 2zy\\ \end {vmatrix}\\\\ & = (0)\textbf {i} - (0)\textbf {j} + (0)\textbf {k}\\ \therefore \hspace {0.5cm} \nabla \times F &= \textbf {0} \end {align*}

Hence conservative.

Since \(F\) is conservative, there is a function \(f\) such that \begin {align*} F & = \nabla f\\ \implies \hspace {0.5cm} 2xy\textbf {i} + (x^2+z^2)\textbf {j} +2zy\textbf {k} & = \frac {\partial f}{\partial x}\textbf {i} + \frac {\partial f}{\partial y}\textbf {j} + \frac {\partial f}{\partial z} \textbf {k} \end {align*}

\[\frac {\partial f}{\partial x} = 2xy\hspace {0.5cm},\hspace {1cm} \frac {\partial f}{\partial y}= x^2+z^2\hspace {0.5cm}, \hspace {1cm} \frac {\partial f}{\partial z} = 2yz\]

\begin {align*} \frac {\partial f}{\partial x} = 2xy & \implies f(x,y,z) = \int 2xy\,dx\\ & \implies f(x,y,z) = x^2y + h(y,z)\\ \end {align*}

\[\implies \frac {\partial }{\partial y}f = x^2 + \frac {\partial h}{\partial y} = x^2 + z^2\]

\begin {align*} \implies & \frac {\partial h}{\partial y} = z^2\\ \implies & h(y,z) = \int z^2\,dy = yz^2 + g(z) \end {align*}

\[\text {Hence},\hspace {0.3cm} f(x,y,z) = x^2y + yz^2 + g(z)\]

\begin {align*} \implies & \frac {\partial f}{\partial z} = 2yz + g'(z) = 2yz\\ \implies & g'(z) = 0\\ \implies & g(z) = C\\ \end {align*}

\[\therefore \hspace {0.5cm} f(x,y,z) = x^2y + yz^2 + C\]

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.