I am trying to model a diffusion phenomenon with anisotropic and heterogeneous diffusion.
$$\nabla \cdot (D\nabla c)$$
where
$$D = \begin{bmatrix} D_{xx}(x,y)&D_{xy}(x,y)\\ D_{yx}(x,y)&D_{yy}(x,y) \end{bmatrix}$$
and $D_{xy} = D_{yx}$. From Wikipedia, I get
$$\nabla \cdot (D\nabla c) = \nabla \cdot[D(x,y)] \nabla c(x,y) + \mbox{tr} \left( D(x,y) \nabla \nabla ^T c(x,y) \right)$$
If I am correct, then
$$\mbox{tr} \left( D(x,y)\nabla \nabla ^Tc(x,y) \right) = D_{xx}\frac{\partial^2 c}{\partial x^2}+D_{yy}\frac{\partial^2 c}{\partial y^2}$$
but I am unsure how to expand the first term. Perhaps,
$$\left( \frac{\partial c}{\partial x} \hat i + \frac{\partial c}{\partial y} \hat j \right) (\nabla D_{xx}+2\nabla D_{xy}+\nabla D_{yy})$$
where $\nabla (\cdot) = \left(\frac{\partial (\cdot)}{\partial x} \hat i+\frac{\partial (\cdot)}{\partial y} \hat j \right)$ completely FOIL'd out?
Any help would be appreciated, as I am trying to use this for implementing a model. I also apologize for my notation.
The expansion of the divergence is explained in this post, where intrinsic expressions are given. We might want to expand the divergence explicitly, noting that \begin{aligned} \nabla\cdot(D\nabla c) &= (D_{xx}c_{,x} + D_{xy}c_{,y})_{,x} + (D_{xy}c_{,x} + D_{yy}c_{,y})_{,y} \\ &= D_{xx,x}c_{,x} + D_{xy,x}c_{,y} + D_{xy,y}c_{,x} + D_{yy,y}c_{,y} \\ &\quad + D_{xx}c_{,xx} + 2\, D_{xy}c_{,xy} + D_{yy}c_{,yy} . \end{aligned} where we have used differential rules, the symmetry property $D_{yx}=D_{xy}$, and the definition $\nabla c = [c_{,x}, c_{,y}]^T$ of the gradient (indices after the comma denote partial differentiation). In other words, we get $$ \nabla\cdot(D\nabla c) = (\nabla\cdot D) \cdot \nabla c + D : \nabla \nabla c \tag{*} $$ as was found in the related post, where $$ \nabla\cdot D = \begin{bmatrix} D_{xx,x}+D_{xy,y}\\ D_{xy,x}+D_{yy,y} \end{bmatrix}, \qquad \nabla \nabla c = \begin{bmatrix} c_{,xx} & c_{,xy}\\ c_{,xy} & c_{,yy} \end{bmatrix}, $$ and the colon denotes the Frobenius inner product (which is analogous to the vector dot product). In Eq. (*), the first term corresponds to the first line of the expanded expression above, and the second term corresponds to the second line.
Note: in this answer, $\nabla$ is viewed as the gradient operator, not as a vector of scalar differential operators.