I want to simplify the following formula (heat transfer context) :
$$f(\vec{v}, T) = v_x \frac{\partial T}{\partial x} + v_y \frac{\partial T}{\partial y} + v_z \frac{\partial T}{\partial z}$$
I'm very tempted to write:
$$f(\vec{v}, T) = \nabla (vT)$$ Yet, I believe this isn't thorough enough. Indeed:
$$\vec{div}(f) : \mathbb{R} \rightarrow \mathbb{R}^3 : f(x,y,z) \mapsto \frac{\partial f}{\partial x}\vec{1}_x + \frac{\partial f}{\partial y}\vec{1}_y + \frac{\partial f}{\partial z}\vec{1}_z$$
$$grad(\vec f) : \mathbb{R}^3 \rightarrow \mathbb{R} : (f_x, f_y, f_z) \mapsto \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_y}{\partial z}$$
The temperature $T(x, y, z)$ is scalar, but I want $f(\vec{v}, T)$ to be a scalar too. So my "$\nabla$" would be neither a gradient nor a divergence. Do you have an idea of what I can write ?
EDIT:
This comes from the following equation:
Is it what you want ? I am not sure. $$f(\vec{v}, T) = v_x \frac{\partial T}{\partial x} + v_y \frac{\partial T}{\partial y} + v_z \frac{\partial T}{\partial z}$$ I suppose that your notation $v_x$ doesn't means $\frac{\partial v(x,y,z)}{\partial x}$.
In order to avoid confusion I change of notation such as $v_x\equiv v_1$ : $$\vec{v}=\begin{bmatrix} v_1\\ v_2\\ v_3\end{bmatrix}$$ and vector gradiant : $$\vec{grad(T)}=\begin{bmatrix} \frac{\partial T}{\partial x}\\ \frac{\partial T}{\partial y}\\ \frac{\partial T}{\partial z}\end{bmatrix}$$ Dot product of vectors : $$\boxed{v_1 \frac{\partial T}{\partial x} + v_2 \frac{\partial T}{\partial y} + v_3 \frac{\partial T}{\partial z}=\vec{v}\:.\:\vec{grad(T)}}$$