is the Laplacian just the gradient dotted with itself?

Question

I guess no but in the solution of this problem (see attached picture) at some point, it is stated that $\Delta v = D_xv \cdot D_xv $

but $D_xv = [v_{x_1} \cdots v_{x_n}]^T$, right ?

so shouldn't $D_xv \cdot D_xv=v^2_{x_1}+ \cdots + v^2_{x_n}$ ?

also I did notice that sometimes they write $D_xv$ and sometimes $D_x \cdot v $.

I suspect that I am missing something here...

Answer 1

The answer is for a function $f=f(x, y, z)$ then \begin{align} \Delta f &= \nabla^2 f \\ &= (\nabla \cdot \nabla) f \\ &= \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \end{align}

However, \begin{align} \nabla f \cdot \nabla f &= \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \cdot \left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\\ &= \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + \left(\frac{\partial f}{\partial z}\right)^2 \end{align} Whence $$\nabla^2 f \neq \nabla f \cdot \nabla f$$

Answer 2

The equation $\Delta v = \nabla v \cdot \nabla v$ is false. The solution shown in the question is incorrect.

For example, if $f:\mathbb R \to \mathbb R$ is defined by $f(x) = x$, then the Laplacian of $f$ is $0$, but $\nabla f(x) \cdot \nabla f(x) = 1$.