#+TITLE: Numerical solution of diffusion equation in 2D with ADI Scheme
#+LaTeX_CLASS_OPTIONS: [a4paper,10pt]
#+LATEX_HEADER: \usepackage{fullpage}
#+LATEX_HEADER: \usepackage{amsmath, systeme}
#+OPTIONS: toc:nil


* Diffusion in 1D

** Finite differences with nodes as cells' centres

The 1D diffusion equation is:

#+NAME: eqn:1
\begin{align}
\frac{\partial C }{\partial t} & = \frac{\partial}{\partial x} \left(\alpha \frac{\partial C }{\partial x} \right) \nonumber \\
   & = \alpha \frac{\partial^2 C}{\partial x^2}
\end{align}

We aim at numerically solving [[eqn:1]] on a spatial grid such as:

[[./images/grid_pqc.pdf]]

The left boundary is defined on $x=0$ while the center of the first
cell - which are the points constituting the finite difference nodes -
is in $x=dx/2$, with $dx=L/n$.


** The explicit FTCS scheme (as in PHREEQC)

We start by discretizing [[eqn:1]] following an explicit Euler scheme and
specifically a Forward Time, Centered Space finite difference. 

For each cell index $i \in 1, \dots, n-1$ and assuming constant
$\alpha$, we can write:

#+NAME: eqn:2
\begin{equation}\displaystyle
   \frac{C_i^{t+1} -C_i^{t}}{\Delta t} = \alpha\frac{\frac{C^t_{i+1}-C^t_{i}}{\Delta x}-\frac{C^t_{i}-C^t_{i-1}}{\Delta x}}{\Delta x}
\end{equation}

In practice, we evaluate the first derivatives of $C$ w.r.t. $x$ on
the boundaries of each cell (i.e., $(C_{i+1}-C_i)/\Delta x$ on the
right boundary of the i-th cell and $(C_{i}-C_{i-1})/\Delta x$ on its
left cell boundary) and then repeat the differentiation to get the
second derivative of $C$ on the the cell centre $i$.

This discretization works for all internal cells, but not for the
domain boundaries ($i=0$ and $i=n$). To properly treat them, we need
to account for the discrepancy in the discretization.

For the first (left) cell, whose center is at $x=dx/2$, we can
evaluate the left gradient with the left boundary using such distance,
calling $l$ the numerical value of a constant boundary condition:

#+NAME: eqn:3
\begin{equation}\displaystyle
\frac{C_0^{t+1} -C_0^{t}}{\Delta t} = \alpha\frac{\frac{C^t_{1}-C^t_{0}}{\Delta x}-
\frac{C^t_{0}-l}{\frac{\Delta x}{2}}}{\Delta x}
\end{equation}

This expression, once developed, yields:

#+NAME: eqn:4
\begin{align}\displaystyle
C_0^{t+1} & =  C_0^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^t_{1}-C^t_{0}- 2 C^t_{0}+2l \right) \nonumber \\
          & =  C_0^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^t_{1}- 3 C^t_{0} +2l \right)
\end{align}


In case of constant right boundary, the finite difference of point
$C_n$ - calling $r$ the right boundary value - is:

#+NAME: eqn:5
\begin{equation}\displaystyle
\frac{C_n^{t+1} -C_n^t}{\Delta t} = \alpha\frac{\frac{r - C^t_{n}}{\frac{\Delta x}{2}}-
\frac{C^t_{n}-C^t_{n-1}}{\Delta x}}{\Delta x}
\end{equation}

Which, developed, gives
#+NAME: eqn:6
\begin{align}\displaystyle
C_n^{t+1} & =  C_n^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2 r - 2 C^t_{n} -C^t_{n} + C^t_{n-1} \right) \nonumber \\
          & =  C_n^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2 r - 3 C^t_{n} + C^t_{n-1} \right)
\end{align}

If on the right boundary we have closed or Neumann condition, the left derivative in eq. [[eqn:5]]
becomes zero and we are left with:


#+NAME: eqn:7
\begin{equation}\displaystyle
C_n^{t+1} = C_n^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot (C^t_{n-1} - C^t_n)
\end{equation}


A similar treatment can be applied to the BTCS implicit scheme.

** Implicit BTCS scheme

First, we define the Backward Time difference:

\begin{equation}
    \frac{\partial C^{t+1} }{\partial t} = \frac{C^{t+1}_i - C^{t}_i}{\Delta t}
\end{equation}

Second the spatial derivative approximation, evaluated at time level $t+1$:

#+NAME: eqn:secondderivative
\begin{equation}
    \frac{\partial^2 C^{t+1} }{\partial x^2} = \frac{\frac{C^{t+1}_{i+1}-C^{t+1}_{i}}{\Delta x}-\frac{C^{t+1}_{i}-C^{t+1}_{i-1}}{\Delta x}}{\Delta x}
\end{equation}

Taking the 1D diffusion equation from [[eqn:1]] and substituting each term by the
equations given above leads to the following equation:


# \begin{equation}\displaystyle
#    \frac{C_i^{j+1} -C_i^{j}}{\Delta t} = \alpha\frac{\frac{C^{j+1}_{i+1}-C^{j+1}_{i}}{\Delta x}-\frac{C^{j+1}_{i}-C^{j+1}_{i-1}}{\Delta x}}{\Delta x}
# \end{equation}

# Since we are not able to solve this system w.r.t unknown values in $C^{j-1}$ we
# are shifting each j by 1 to $j \to (j+1)$ and $(j-1) \to j$ which leads to:

#+NAME: eqn:1DBTCS
\begin{align}\displaystyle
\frac{C_i^{t+1} - C_i^{t}}{\Delta t}    & = \alpha\frac{\frac{C^{t+1}_{i+1}-C^{t+1}_{i}}{\Delta x}-\frac{C^{t+1}_{i}-C^{t+1}_{i-1}}{\Delta x}}{\Delta x} \nonumber \\
                                        & = \alpha\frac{C^{t+1}_{i-1} - 2C^{t+1}_{i} + C^{t+1}_{i+1}}{\Delta x^2}
\end{align}

This only applies to inlet cells with no ghost node as neighbor. For the left
cell with its center at $\frac{dx}{2}$ and the constant concentration on the
left ghost node called $l$ the equation goes as followed:

\begin{equation}\displaystyle
\frac{C_0^{t+1} -C_0^{t}}{\Delta t} = \alpha\frac{\frac{C^{t+1}_{1}-C^{t+1}_{0}}{\Delta x}-
\frac{C^{t+1}_{0}-l}{\frac{\Delta x}{2}}}{\Delta x}
\end{equation}

This expression, once developed, yields:

\begin{align}\displaystyle
C_0^{t+1} & =  C_0^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^{t+1}_{1}-C^{t+1}_{0}- 2 C^{t+1}_{0}+2l \right) \nonumber \\
          & =  C_0^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^{t+1}_{1}- 3 C^{t+1}_{0} +2l \right)
\end{align}

Now we define variable $s_x$ as followed:

\begin{equation}
    s_x = \frac{\alpha \cdot \Delta t}{\Delta x^2}
\end{equation}

Substituting with the new variable $s_x$ and reordering of terms leads to the equation applicable to our model:

\begin{equation}\displaystyle
    -C^t_0 = (2s_x) \cdot l + (-1 - 3s_x) \cdot C^{t+1}_0 + s_x \cdot C^{t+1}_1
\end{equation}

The right boundary follows the same scheme. We now want to show the equation for the rightmost inlet cell $C_n$ with right boundary value $r$:

\begin{equation}\displaystyle
\frac{C_n^{t+1} -C_n^{t}}{\Delta t} = \alpha\frac{\frac{r-C^{t+1}_{n}}{\frac{\Delta x}{2}}-
\frac{C^{t+1}_{n}-C^{t+1}_{n-1}}{\Delta x}}{\Delta x}
\end{equation}

This expression, once developed, yields:

\begin{align}\displaystyle
C_n^{t+1} & =  C_n^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2r - 2C^{t+1}_{n} - C^{t+1}_{n} + C^{t+1}_{n-1} \right) \nonumber \\
          & =  C_0^{t} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( 2r - 3C^{t+1}_{n} + C^{t+1}_{n-1} \right)
\end{align}

Now rearrange terms and substituting with $s_x$ leads to:

\begin{equation}\displaystyle
    -C^t_n = s_x \cdot C^{t+1}_{n-1} + (-1 - 3s_x) \cdot C^{t+1}_n + (2s_x) \cdot r
\end{equation}

*TODO*
- Tridiagonal matrix filling

#+LATEX: \clearpage

* Diffusion in 2D: the Alternating Direction Implicit scheme


In 2D, the diffusion equation (in absence of source terms) and
assuming homogeneous but anisotropic diffusion coefficient
$\alpha=(\alpha_x,\alpha_y)$ becomes:

#+NAME: eqn:2d
\begin{equation}
\displaystyle  \frac{\partial C}{\partial t} = \alpha_x \frac{\partial^2 C}{\partial x^2} + \alpha_y\frac{\partial^2 C}{\partial y^2}
\end{equation}

** 2D ADI using BTCS scheme

The Alternating Direction Implicit method consists in splitting the
integration of eq. [[eqn:2d]] in two half-steps, each of which represents
implicitly the derivatives in one direction, and explicitly in the
other. Therefore we make use of second derivative operator defined in
equation [[eqn:secondderivative]] in both $x$ and $y$ direction for half
the time step $\Delta t$.

Denoting $i$ the grid cell index along $x$ direction, $j$ the index in
$y$ direction and $t$ as the time level, the spatially centered second
derivatives can be written as:

\begin{align}\displaystyle
\frac{\partial^2 C^t_{i,j}}{\partial x^2} &= \frac{C^{t}_{i-1,j} - 2C^{t}_{i,j} + C^{t}_{i+1,j}}{\Delta x^2} \\
\frac{\partial^2 C^t_{i,j}}{\partial y^2} &= \frac{C^{t}_{i,j-1} - 2C^{t}_{i,j} + C^{t}_{i,j+1}}{\Delta y^2}
\end{align}

The ADI scheme is formally defined by the equations:

#+NAME: eqn:genADI
\begin{equation}
\systeme{ 
  \displaystyle  \frac{C^{t+1/2}_{i,j}-C^t_{i,j}}{\Delta t/2}     = \displaystyle \alpha_x \frac{\partial^2 C^{t+1/2}_{i,j}}{\partial x^2} + \alpha_y \frac{\partial^2 C^{t}_{i,j}}{\partial y^2},
  \displaystyle  \frac{C^{t+1}_{i,j}-C^{t+1/2}_{i,j}}{\Delta t/2} = \displaystyle \alpha_x \frac{\partial^2 C^{t+1/2}_{i,j}}{\partial x^2} + \alpha_y \frac{\partial^2 C^{t+1}_{i,j}}{\partial y^2}
}
\end{equation}

\noindent The first of equations [[eqn:genADI]], which writes implicitly
the spatial derivatives in $x$ direction, after bringing the $\Delta t
/ 2$ terms on the right hand side and substituting $s_x=\frac{\alpha_x
\cdot \Delta t}{2 \Delta x^2}$ and $s_y=\frac{\alpha_y\cdot\Delta
t}{2\Delta y^2}$ reads:

\begin{equation}\displaystyle
C^{t+1/2}_{i,j}-C^t_{i,j} = s_x (C^{t+1/2}_{i-1,j} - 2C^{t+1/2}_{i,j} + C^{t+1/2}_{i+1,j}) + s_y (C^{t}_{i,j-1} - 2C^{t}_{i,j} + C^{t}_{i,j+1})
\end{equation}

\noindent Separating the known terms (at time level $t$) on the left
hand side and the implicit terms (at time level $t+1/2$) on the right
hand side, we get:

#+NAME: eqn:sweepX
\begin{equation}\displaystyle
-C^t_{i,j} - s_y (C^{t}_{i,j-1} - 2C^{t}_{i,j} + C^{t}_{i,j+1}) = - C^{t+1/2}_{i,j} + s_x (C^{t+1/2}_{i-1,j} - 2C^{t+1/2}_{i,j} + C^{t+1/2}_{i+1,j}) 
\end{equation}

\noindent Equation [[eqn:sweepX]] can be solved with a BTCS scheme since
it corresponds to a tridiagonal system of equations, and resolves the
$C^{t+1/2}$ at each inner grid cell.

The second of equations [[eqn:genADI]] can be treated the same way and
yields:

#+NAME: eqn:sweepY
\begin{equation}\displaystyle
-C^{t + 1/2}_{i,j} - s_x (C^{t + 1/2}_{i-1,j} - 2C^{t + 1/2}_{i,j} + C^{t + 1/2}_{i+1,j}) = - C^{t+1}_{i,j} + s_y (C^{t+1}_{i,j-1} - 2C^{t+1}_{i,j} + C^{t+1}_{i,j+1}) 
\end{equation}

This scheme only applies to inner cells, or else $\forall i,j \in [1,
n-1] \times [1, n-1]$. Following an analogous treatment as for the 1D
case, and noting $l_x$ and $l_y$ the constant left boundary values and
$r_x$ and $r_y$ the right ones for each direction $x$ and $y$, we can
modify equations [[eqn:sweepX]] for $i=0, j \in [1, n-1]$

#+NAME: eqn:boundXleft
\begin{equation}\displaystyle
-C^t_{0,j} - s_y (C^{t}_{0,j-1} - 2C^{t}_{0,j} + C^{t}_{0,j+1}) = - C^{t+1/2}_{0,j} + s_x (C^{t+1/2}_{1,j} - 3C^{t+1/2}_{0,j} + 2 l_x) 
\end{equation}

\noindent Similarly for  $i=n, j \in [1, n-1]$:
#+NAME: eqn:boundXright
\begin{equation}\displaystyle
-C^t_{n,j} - s_y (C^{t}_{n,j-1} - 2C^{t}_{n,j} + C^{t}_{n,j+1}) = - C^{t+1/2}_{n,j} + s_x (C^{t+1/2}_{n-1,j} - 3C^{t+1/2}_{n,j} + 2 r_x) 
\end{equation}

\noindent For $i=j=0$:
#+NAME: eqn:bound00
\begin{equation}\displaystyle
-C^t_{0,0} - s_y (C^{t}_{0,1} - 3C^{t}_{0,0} + 2l_y) = - C^{t+1/2}_{0,0} + s_x (C^{t+1/2}_{1,0} - 3C^{t+1/2}_{0,0} + 2 l_x) 
\end{equation}

Analogous expressions are readily derived for all possible
combinations of $i,j \in 0\times n$. In practice, wherever an index
$i$ or $j$ is $0$ or $n$, the centered spatial derivatives in $x$ or
$y$ directions must be substituted in relevant parts of the sweeping
equations \textbf{in both the implicit or the explicit sides} of
equations [[eqn:sweepX]] and [[eqn:sweepY]] by a term

#+NAME: eqn:bound00
\begin{equation}\displaystyle
 s(C_{forw} - 3C + 2 bc) 
\end{equation}
\noindent where $bc$ is the boundary condition in the given direction,
$s$ is either $s_x$ or $s_y$, and $C_{forw}$ indicates the contiguous
cell opposite to the boundary. Alternatively, noting the second
derivative operator as $\partial_{dir}^2$, we can write in compact
form:

\begin{equation}
\systeme{
  \displaystyle  \partial_x^2 C_{0,j} = 2l_x - 3C_{0,j} + C_{1,j} ,
  \displaystyle  \partial_x^2 C_{n,j} = 2r_x - 3C_{n,j} + C_{n-1,j} ,
  \displaystyle  \partial_y^2 C_{i,0} = 2l_y - 3C_{i,0} + C_{i,1} ,
  \displaystyle  \partial_y^2 C_{i,n} = 2r_y - 3C_{i,n} + C_{i,n-1}
}
\end{equation}



#+LATEX: \clearpage