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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.

Rodrigo de Azevedo
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rangus_steve
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1 Answers1

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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.

EditPiAf
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