Why notate the solution to steady-state heat equation as $2\Phi$?

Question

In literature I read the following:

Based on what is said, I think that to obtain Eqn(3) one substitutes Eqn(1.17(1)) into the heat equantion, Eqn(B), and then consider the steady-state condition, $\partial v/\partial t=0$, leading to Eqn(C): $$\tag{1.17(1)} f_i=-K_{ij} \frac{\partial v}{\partial x_j}$$ $$\tag{B} \rho c \frac{\partial v}{\partial t}+\left(\frac{\partial f_x}{\partial x}+\frac{\partial f_y}{\partial y}+\frac{\partial f_z}{\partial z}\right)=0$$ $$\tag{C} K_{ij}\frac{\partial ^2 v}{\partial x_i \partial x_j}=0$$ In deriving Eqn(C) the $K_{ij}$ are constants, and thus are pulled from the derivative. So my question is this: if I correctly derived Eqn(C), and I am correct that it is equivalent to Eqn(3), why then would the authors equate Eqn(3) to $2\Phi$ instead of $0$?

I note that I have not seen the variable $\Phi$ used/defined in this text leading up to this section (Sec. 1.20), nor have I seen it used after.

Some thoughts to consider:

1. The 2nd-rank tensor $K_{ij}$ is usually taken as symmetric, $K_{ij}=K_{ji}$. Under this assumption the right-hand side of Eqn(3) may be written as $$K_{11}\partial_{xx}^2v+K_{22}\partial_{yy}^2v+K_{33}\partial_{zz}^2v+2K_{23}\partial_{yz}^2v+2K_{13}\partial_{xz}^2v+2K_{12}\partial_{xy}^2v \tag{D}$$
2. The scalar magnitude $K$ of the symmetric positive-definite 2nd-rank tensor $K_{ij}$ is $$K=K_{ij}l_il_j=K_{11}l_1^2+K_{22}l_2^2+K_{33}l_3^2+2K_{23}l_2l_3+2K_{13}l_1l_3+2K_{12}l_1l_2 \tag{E}$$ where $l_i$ is a unit vector in the direction in which $K$ is evaluated. So perhaps in some way $2\Phi$ is related to the scalar magnitude of $K_{ij}$?

Answer 1

The notation is a bit unusual, using $v$ for temperature, but they probably use $T$ for something else. If you start with the steady-state heat equation: $$-\nabla \cdot (K \nabla v)=Q$$

with a heat coefficient tensor $K$ (or you can choose to see it as a matrix) and source term $Q$. So the heat equation is not necessarily equal to zero since you can have a source term.

For anisotropic materials, I think some authors write the left-hand side of the heat equation as: $$-\frac{1}{2}\nabla \cdot (K \nabla v)-\frac{1}{2}\nabla \cdot (K^T \nabla v)=$$ $$=-\nabla \cdot (\frac{1}{2}(K+K^T) \nabla v)$$ If the diffusion tensor/matrix $K$ is symmetric, which it usually is, then you get: $$\frac{1}{2}(K+K^T)=K$$

Maybe the author has written: $$-\frac{1}{2}\nabla \cdot (K \nabla v)-\frac{1}{2}\nabla \cdot (K^T \nabla v)=Q$$

If we now define: $$\Phi=\frac{1}{2}\nabla \cdot (K \nabla v)$$ we get the heat equation as: $$-2\Phi=Q$$

This is a highly unusual way of writing things but it is my best guess.