2 Discretization of the Simple Extrinsic Diffusion Model

We take the functions $w(\mathbf{x})$, $v(\mathbf{x})$ and $u(\mathbf{x})$ defined on bounded open domen $\Omega$. Starting from Green's theorem, a simple relationship can be proved,

$$\int\limits_\Omega \nabla v \nabla w u d\Omega + \int\limits_\O... ...w v d\Omega=\int\limits_\Gamma v u \frac{\partial w}{\partial n} d\Gamma,$$ (141)

where $\Gamma$ is the boundary of the domain $\Omega$. We write (3.33) as

$$\nabla (h(C_A)D) \nabla C_A + D h(C_A) \triangle C_A = \frac{\partial C_A}{\partial t}.$$ (142)

By multiplying (3.79) with the basis function $\varphi_i = \varphi_i(\mathbf{x})$ and integrating over the domen $\Omega$ we have,

$$\int\limits_\Omega \nabla (h(C_A)D) \nabla C_A \varphi_i d\O... ...mega = \int\limits_\Omega \frac{\partial C_A}{\partial t} \varphi_i d\Omega.$$ (143)

Applying (3.78) on (3.80) gives,

$$D\int\limits_\Gamma h(C_A) \varphi_i \frac{\partial C_A}{\part... ...Omega = \int\limits_\Omega \frac{\partial C_A}{\partial t} \varphi_i d\Omega.$$ (144)

Assuming zero-Neumann boundary conditions on $\Gamma$ we obtain the weak formulation of the equation (3.33)

$$\int\limits_\Omega \frac{\partial C_A}{\partial t} \varphi_i d\Omega+D\int\limits_\Omega h(C_A) \nabla C_A \nabla \varphi_i d\Omega = 0.$$ (145)

By introducing time discretization with time step $\Delta t$ and writing the last equation for the single element $T\in T_h(\Omega)$ we obtain,

$$\int\limits_T (C^n_A- C^{n-1}_A) \varphi_i d\Omega+ D\Delta t \int\limits_T h(C_A^n) \nabla C_A^n \nabla \varphi_i d\Omega = 0.$$ (146)

The scalar functions $C_A(\mathbf{x},t_n)$, $\nabla C_A(\mathbf{x},t_n)$, and $h(C_A(\mathbf{x},t_n))$, for $t_n=n\cdot \Delta t$, are linearly approximated on the element $T\in T_h(\Omega)$,

$$C_A^n=C_A(\mathbf{x},t_n) = \sum_{i=1}^{4} \varphi_i C^{n}_{A,i},$$ (147)

$$\nabla C_A^n = \nabla C_A(\mathbf{x},t_n) =\sum_{i=1}^{4} \nabla \varphi_i C^{n}_{A,i},$$ (148)

$$h(C_A^n)= h(C_A(\mathbf{x},t_n))= h(C^{n}_{A,m}).$$ (149)

The $C_1^n, C_2^n, C_3^n, C_4^n$, are the values of the concentration at the nodes of the element $T$, $C^{n}_{m}$ is the concentration value at some point inside the $T$. Normally, for $C^{n}_{A,m}$ we use the following simple approximation,

$$C^{n}_{A,m} = \frac{1}{4} \sum_{i=1}^{4} C^{n}_{A,i}.$$ (150)

We can define discrete operator $\ell^e$ for each $T\in T_h(\Omega)$. Since we have to deal only with a single partial differential equation and not with a system of equations, we can omit the second index in (2.27) by defining the operator,

$$\ell_{i}^e(C_{A,1}^n,C_{A,2}^n,C_{A,3}^n,C_{A,4}^n)=\frac{1}{\Del... ...{ik} C_{A,i}^n+D h(C_{A,m}^n)\sum_{i=1}^4 K_{ik} C_{A,i}^n,\quad k=1,2,3,4.$$ (151)

The nucleus matrix is than defined as

$$\mathbf{\Pi}(T)=\frac{\partial (\ell_{1}^e,\ell_{2}^e,\ell_{3}^e,\ell_{4}^e)}{\partial (C_{A,1}^n,C_{A,2}^n,C_{A,3}^n,C_{A,4}^n)}.$$ (152)

and the residuum vector as,

$$R_k^e=\ell_{k}^e(C_{A,1}^n,C_{A,2}^n,C_{A,3}^n,C_{A,4}^n)-\frac{1}{\Delta t}\sum_{i=1}^4 M_{ik} C_{A,i}^{n-1},\quad k=1,2,3,4.$$ (153)

The nucleus matrix for this simple case can be also calculated analytically. In this case is also $rank(\mathbf{G})=N$.