Trace of product of gradient

Question

Consider vector field $v$. We know, that $Tr(\nabla v)=div ~ v$. Is is true that $$Tr(\nabla v\nabla v)=(div ~ v)^2$$ Thanks for any hint.

Answer 1

By considering $\nabla v$ to be the Jacobian of the vector field $v$. One can show that:

$$\nabla v \nabla v = \begin{bmatrix} \nabla v_1^T\\ \dots\\ \nabla v_n^T \end{bmatrix} \begin{bmatrix} \partial_{x_1}v & \dots & \partial_{x_n}v \end{bmatrix} = \begin{bmatrix} \left< \nabla v_1, \partial_{x_1}v\right>& \dots & \dots\\ \dots & \dots & \dots \\ \dots & \dots & \left< \nabla v_n, \partial_{x_n}v\right> \end{bmatrix}$$

Hence,

$$Tr(\nabla v \nabla v) = \sum_{i=1}^{n} \left< \nabla v_i, \partial_{x_i} v\right > \;\;(1)$$

And $$\left< \nabla v_i, \partial_{x_i} v\right > = \sum_{k=1}^n \frac{\partial v_i}{\partial x_k} \frac{\partial v_k}{\partial x_i}\;\;(2)$$

Therefore, from (1) and (2):

$$Tr(\nabla v \nabla v) = \sum_{i=1}^n\sum_{k=1}^n \frac{\partial v_i}{\partial x_k} \frac{\partial v_k}{\partial x_i} \;\;(3)$$

Now $$\text{div}(v) = \sum_{i=1}^n \frac{\partial v_i}{\partial x_i}$$

$$\left ( \text{div}(v) \right )^2 = \left (\sum_{i=1}^n \frac{\partial v_i}{\partial x_i} \right )\left (\sum_{k=1}^n \frac{\partial v_k}{\partial x_k} \right ) = \sum_{i=1}^n\sum_{k=1}^n \frac{\partial v_i}{\partial x_i} \frac{\partial v_k}{\partial x_k}\;\;(4)$$

From (3) and (4) we deduce that $Tr(\nabla v \nabla v) \neq \left ( \text{div}(v) \right )^2$.

Sorry for the computational "proof". I hope I find a more geometric one in the future.

Answer 2

A simple counterexample in $\mathbb R^n$.

Take the identity vector field $J: x \mapsto x$. Then the Jacobian $\nabla J$ is the identity matrix $I_n$. So

$$Tr(\nabla J \cdot \nabla J)= Tr(I_n)= n \neq n^2= (div \ I)^2$$