[Math] How to show that $\nabla \cdot (\nabla f) = \nabla ^{2} f$ for this function

Question

I've been given a homework question that asks me to show that $\nabla \cdot (\nabla f) = \nabla ^{2} f$, where $f(x,y,z)=xyz-z^{2}sin(y)$. However, $\nabla \cdot (\nabla f)$ is a scalar quantity, whilst $\nabla ^{2} f$ is a vector quantity, so they cannot be equal.
What am I missing here?
Thanks.

Answer 1

The Laplacian operator, $\nabla ^2$ can be defined to operate on both scalar fields and vector fields. If $f$ is a scalar field, then

$$\bbox[5px,border:2px solid #C0A000]{\nabla ^2 f(\vec r)\equiv \nabla \cdot \nabla f(\vec r)}$$

which in Cartesian coordinates can be written

$$\begin{align} \nabla \cdot \nabla f(\vec r)&=\sum_{i=1}^{3}\sum_{j=1}^{3}\left(\hat x_i\frac{\partial}{\partial x_i}\right)\cdot \hat x_j\frac{\partial f(\vec r)}{\partial x_j}\\\\ &=\sum_{i=1}^{3}\frac{\partial^2f(\vec r)}{\partial x_i^2} \end{align}$$

If $\vec f(\vec r)$ is a vector field, then the

$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 \vec f(\vec r)\equiv \nabla \left(\nabla \cdot \vec f(\vec r)\right)-\nabla \times \nabla \times \vec f(\vec r)}$$

which in Cartesian coordinates can be written

$$\begin{align} \nabla (\nabla \cdot \vec f(\vec r)-\nabla \times \nabla \times \vec f(\vec r) &=\sum_{i=1}^{3}\sum_{j=1}^{3}\hat x_i\frac{\partial}{\partial x_i}\frac{\partial f_j(\vec r)}{\partial x_j}-\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{k=1}^{3}\left(\hat x_i\times(\hat x_j\times \hat x_k)\frac{\partial^2 f_k(\vec r)}{\partial x_i \partial x_j}\right) \\\\ &=\sum_{i=1}^{3}\frac{\partial^2 \vec f(\vec r)}{\partial x_i^2} \end{align}$$

We note that while in Cartesian coordinates, the forms for the Laplacian on a scalar and a vector are identical, this is not the case in other curvilinear coordinates.