Question: Can the Forchheimer $\beta$-factor be a 2nd rank tensor (like permeability, $k$)?
In most cases in literature, the quadratic Forchheimer equation is applied only to one-dimensional flow. However I have seen in some instances claims that the Forchheimer eqn is applicable in multiple dimensions as a vector equation. The proposed representation of the vector Forchheimer equation in Cartesian coordinates is (using Einstein's summation convention (or double index convention)): $$\tag{1} -\frac{\partial p}{\partial x_j} = \frac{\mu}{k_{ij}}q_i+\beta_{ij} \rho q_i|q_i|$$
In this case, the quadratic term is simply the magnitude of the specific discharge vector $|q_i|$ (the 'superficial velocity') times the directional specific discharge vector $q_i$.
If I underscore the variables in the equation with the letters 's', 'v', and 't' to represent the type of physical quantity they are, that is, scalar, vector, and 2nd rank tensor, respectively, I show:
$$\tag{2} \underbrace{-\frac{\partial p}{\partial x_j}}_{\text{v}} = \underbrace{\frac{\mu}{k_{ij}}}_{\text{t}}\underbrace{q_i}_{\text{v}}+\underbrace{\beta_{ij}}_{\text{t}} \underbrace{\rho}_{\text{s}} \underbrace{q_i}_{\text{v}}\underbrace{|q_i|}_{\text{s}}$$
I see that all terms on the righ-hand side of (2), when multiplied together, would result in a vector (consistent with the left-hand side).
But what it I wanted to rearrange (1) to solve for $q_i$?
One way I see how this may be down is as so:
Multiply through by the permeability tensor, $k_{ij}$ $$\tag{3} -k_{ij}\frac{\partial p}{\partial x_j} = \mu q_i+\beta_{ij}k_{ij} \rho q_i|q_i|$$
Extract the product $\mu q_i$ from the right-hand side of (3) $$\tag{4} -k_{ij}\frac{\partial p}{\partial x_j} = \mu q_i \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)$$
Divide through by $\mu \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)$ $$\tag{5} -\frac{k_{ij}}{\mu \left(1+\frac{\beta_{ij}k_{ij} \rho |q_i|}{\mu}\right)}\frac{\partial p}{\partial x_j} = q_i $$
And here is where I am unsure. In the first term on the LHS, I have one tensor in the numerator and two tensors in the denominator along with scalar quantities.
Questions: For those who are familiar with the theory of tensor analysis, is what I have done correct? If not, what have I done wrong? If I have done nothing wrong, does this mean that the Forchheimer $\beta$-factor cannot be a 2nd rank tensor?