6 Examples involving grad, div, curl and the Laplacian

The vector differential operators can be combined in several ways as the following examples show.

Example 16

If $\underset{̲}{A}=2yz\underset{̲}{i}-x^{2}y\underset{̲}{j}+x{z}^2\underset{̲}{k}$ , $\underset{̲}{B}=x^2\underset{̲}{i}+yz\underset{̲}{j}-xy\underset{̲}{k}$ and $\varphi =2x^{2}y{z}^3$ , find

1. $\left(\underset{̲}{A}\cdot \underset{̲}{\nabla }\right)\varphi$
2. $\underset{̲}{A}\cdot \underset{̲}{\nabla }\varphi$
3. $\underset{̲}{B}\times \underset{̲}{\nabla }\varphi$
4. ${\nabla }^2\varphi$

Solution

1. $$\begin{array}{rcll}\left(\underset{̲}{A}\cdot \underset{̲}{\nabla }\right)\varphi & =& \left[\left(2yz\underset{̲}{i}-x^{2}y\underset{̲}{j}+x{z}^2\underset{̲}{k}\right)\cdot \left(\frac{\partial }{\partial x}\underset{̲}{i}+\frac{\partial }{\partial y}\underset{̲}{j}+\frac{\partial }{\partial z}\underset{̲}{k}\right)\right]\varphi & \\ & =& \left[2yz\frac{\partial }{\partial x}-x^{2}y\frac{\partial }{\partial y}+x{z}^2\frac{\partial }{\partial z}\right]2x^{2}y{z}^3& \\ & =& 2yz\frac{\partial }{\partial x}\left(2x^{2}y{z}^3\right)-x^{2}y\frac{\partial }{\partial y}\left(2x^{2}y{z}^3\right)+x{z}^2\frac{\partial }{\partial z}\left(2x^{2}y{z}^3\right)& \\ & =& 2yz\left(4xy{z}^3\right)-x^{2}y\left(2x^2{y}^3\right)+x{z}^2\left(6x^{2}y{z}^2\right)& \\ & =& 8x{y}^2{z}^4-2x^4{y}^4+6x^{3}y{z}^4& \end{array}$$
2. $$\begin{array}{rcll}\underset{̲}{\nabla }\varphi & =& \frac{\partial }{\partial x}\left(2x^{2}y{z}^3\right)\underset{̲}{i}+\frac{\partial }{\partial y}\left(2x^{2}y{z}^3\right)\underset{̲}{j}+\frac{\partial }{\partial z}\left(2x^{2}y{z}^3\right)\underset{̲}{k}& \\ & =& 4xy{z}^3\underset{̲}{i}+2x^2{z}^3\underset{̲}{j}+6x^{2}y{z}^2\underset{̲}{k}& \end{array}$$
   $$\begin{array}{rcll}\text{So}\underset{̲}{A}\cdot \underset{̲}{\nabla }\varphi & =& \left(2yz\underset{̲}{i}-x^{2}y\underset{̲}{j}+x{z}^2\underset{̲}{k}\right)\cdot \left(4xy{z}^3\underset{̲}{i}+2x^2{z}^3\underset{̲}{j}+6x^{2}y{z}^2\underset{̲}{k}\right)& \\ & =& 8x{y}^2{z}^4-2x^{4}y{z}^3+6x^{3}y{z}^4& \end{array}$$
3. $\underset{̲}{\nabla }\varphi =4xy{z}^3\underset{̲}{i}+2x^2{z}^3\underset{̲}{j}+6x^{2}y{z}^2\underset{̲}{k}$ so $$\begin{array}{rcll}\underset{̲}{B}\times \underset{̲}{\nabla }\varphi & =& \left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill \\ \hfill x^2\hfill & \hfill yz\hfill & \hfill -xy\hfill \\ \hfill \hfill \\ \hfill 4xy{z}^3\hfill & \hfill 2x^2{z}^3\hfill & \hfill 6x^{2}y{z}^2\hfill \end{array}\right|& \\ & & & \\ & =& \underset{̲}{i}\left(6x^2{y}^2{z}^3+2x^{3}y{z}^3\right)+\underset{̲}{j}\left(-4x^2{y}^2{z}^3-6x^{4}y{z}^2\right)+\underset{̲}{k}\left(2x^4{z}^3-4x{y}^2{z}^4\right)& \end{array}$$
4. ${\nabla }^2\varphi =\frac{{\partial }^2}{\partial x^2}\left(2x^{2}y{z}^3\right)+\frac{{\partial }^2}{\partial y^2}\left(2x^{2}y{z}^3\right)+\frac{{\partial }^2}{\partial z^2}\left(2x^{2}y{z}^3\right)=4y{z}^3+0+12x^{2}yz$

Example 17

For each of the expressions below determine whether the quantity can be formed and, if so, whether it is a scalar or a vector.

1. grad(div $\underset{̲}{A}$ )
2. grad(grad $\varphi$ )
3. curl(div $\underset{̲}{F}$ )
4. div [ curl ( $\underset{̲}{A}\times$ grad $\varphi$ ) ]

Solution

1. $\underset{̲}{A}$ is a vector and div $\underset{̲}{A}$ can be calculated and is a scalar. Hence, grad(div $\underset{̲}{A}$ ) can be formed and is a vector.
2. $\varphi$ is a scalar so grad $\varphi$ can be formed and is a vector. As grad $\varphi$ is a vector, it is not possible to take grad(grad $\varphi$ ).
3. $\underset{̲}{F}$ is a vector and hence div $\underset{̲}{F}$ is a scalar. It is not possible to take the curl of a scalar so curl(div $\underset{̲}{F}$ ) does not exist.
4. $\varphi$ is a scalar so grad $\varphi$ exists and is a vector. $\underset{̲}{A}\times$ grad $\varphi$ exists and is also a vector as is curl $\underset{̲}{A}\times$ grad $\varphi$ . The divergence can be taken of this last vector to give
   
   div [ curl ( $\underset{̲}{A}\times$ grad $\varphi$ ) ] which is a scalar.