Is this 2-tensor symmetric? It satisfies these conditions

Question

I have some scalar field $p$ and a 2nd order tensor $\textbf{T}$ such that

$div( \ \textbf{T} \ \textbf{grad}(p) \ )=0$

$\textbf{curl}(\ \textbf{T} \ \textbf{grad}(p) \ )=\textbf{0} $

$\textbf{T}$ is positive definite, i.e. $\textbf{a} \cdot \textbf{T}\textbf{a} >0 \ \ \ \forall \ \textbf{a} \in E$

I am trying to show that $\textbf{T}=\textbf{T}^T$. Is is true? If yes, how can I prove it? I've been working on it for the past few days and keep running in circles, I need help.

If it isn't true, what if I assume $\textbf{T}$ is constant, then is it symmetric? I'd appreciate any input.

Answer 1

This isn't true. For example let $$\mathbf{T}=\begin{pmatrix}2&2&2\\ 0&2&2\\ 0&0&2 \end{pmatrix}$$ everywhere.

Then $\mathbf{a}\cdot\mathbf{T}\mathbf{a}=2a_1^2+2a_2^2+2a_3^2+2a_1a_2+2a_2a_3+2a_3a_1$ $=(a_1+a_2)^2+(a_2+a_3)^2+(a_3+a_1)^2>0$

Let $p$ be any scalar field with a constant gradient, for example $p(x,y,z)=x$. Then $\mathbf{T}\;\mathbf{grad}(p)$ is constant and so has zero divergence and curl.