thermodynamics - Squared d'Alembert Operator

I was investigating an equation suggested by Arun and Sivaram (but with some investigation) does not originate with them,

$$\frac{\partial R}{\partial t} = \alpha \nabla^2R$$

From this, I wondered if a Minkowski space analogue would exist such that the wave equation is non-linear

$$\frac{\partial R}{\partial t} = \alpha \Box R = \alpha \partial^{\mu}\partial_{\mu}R = \alpha g^{\mu \nu}\partial_{\nu}\partial^{\mu}R$$

For those interested, the Ricci flow [is] the heat equation for a Riemannian manifold. I was interested enough that I searched for similar approaches and did in fact find one:

https://en.wikipedia.org/wiki/Relativistic_heat_conduction

Except they have used a notation to ''highlight the scalar property'' of the d'Alembert operator in this context by squaring the operator $$\Box^2$$ . Doing so allows you to write it in the hyperbolic form

$$\frac{1}{c^2} \frac{\partial^2 R}{\partial t^2} + \frac{1}{\alpha}\frac{\partial R}{\partial t} = \nabla^2 R$$

This allows us to write

$$\frac{\partial R}{\partial t} = \alpha \Box^2 R = -\frac{\alpha}{c^2}\frac{\partial^2 R}{\partial t^2} + \alpha \nabla^2 R$$

I am not very fond though of this notation where we square the operator. The question really revolves around the use of this notation. Is it really pertinent? By understanding the generalization of a covariant derivative in terms of the Laplacian is the d'Alembert operator allows us to look at the RHS operator ''as though it was like'' the presence of that covariant derivative. The Ricci scalar curvature is just

$$R \equiv R_{ijkl}g^{ik}g^{jl}$$

It's not impossible to consider it as I had first expected as a form satisfying the covariant derivative, but in a simpler more precise form is

$$^{g}\nabla R = \nabla R$$

Which is a notation for the covariant derivative acting on $$R$$ with respect to some Riemannian metric $$g$$ . https://arxiv.org/pdf/1504.02910.pdf