Fourier heat-flux constitutive model

For the field record of Rectangular local-equilibrium field data, adopt $j_Q(x,t)=-K(x,T(x,t))\mathrm{\nabla }T(x,t)\mathrm{.}$ Here $K$ has entries in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $j_Q$ in $\mathrm{W}{\mathrm{m}}^{-2}$, so the equation is dimensionally consistent. Write $K_s=(K+K^{\mathsf{T}})\mathrm{/}2$ for the symmetric part and assume $z^{\mathsf{T}}{K}_s(x,T)z\ge 0$ for every $z\in {\mathbb{R}}^3$. Define the thermal quadratic density by ${\sigma }_Q(x,t):=-\frac{j_Q(x,t)\cdot \mathrm{\nabla }T(x,t)}{T(x,t{)}^2}=\frac{(\mathrm{\nabla }T(x,t){)}^{\mathsf{T}}{K}_s(x,T(x,t))\mathrm{\nabla }T(x,t)}{T(x,t{)}^2},$ with units $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. For the antisymmetric part $K_a=(K-K^{\mathsf{T}})\mathrm{/}2$, the scalar $z^{\mathsf{T}}{K}_{a}z$ equals its own negative by transposition, so it is zero. Thus $K_a$ may contribute to $j_Q$ but not to ${\sigma }_Q$, and ${\sigma }_Q\ge 0$. Interpreting this density as entropy production additionally requires a local entropy map $s\in C^1(\mathcal{T},\mathbb{R})$ in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ with $s^{\mathrm{\prime }}(T)=u^{\mathrm{\prime }}(T)\mathrm{/}T$, the energy balance ${\mathrm{\partial }}_{t}u+\mathrm{div}{j}_Q=0$, and regularity permitting the product and chain rules; under those additional assumptions the entropy balance is ${\mathrm{\partial }}_{t}s(T)+\mathrm{div}(j_Q\mathrm{/}T)={\sigma }_Q$. This postulate alone adopts neither that caloric entropy relation nor the energy balance.

This is an adopted constitutive model in the stated continuum regime. It is neither a consequence of equilibrium entropy alone nor a universal law for all media. In particular, it does not assert a finite propagation speed or an empirical value for $K$.