From d54fe25cace10879e7846f67cedc20b3d8955db1 Mon Sep 17 00:00:00 2001
From: Max Luebke <mluebke@gfz-potsdam.de>
Date: Fri, 22 Apr 2022 09:49:53 +0200
Subject: [PATCH] Update documentation of implicit BTCS

---
 doc/ADI_scheme.org | 41 +++++++++++++++++++++++++++--------------
 1 file changed, 27 insertions(+), 14 deletions(-)

diff --git a/doc/ADI_scheme.org b/doc/ADI_scheme.org
index d47f3cb..7a0b0c7 100644
--- a/doc/ADI_scheme.org
+++ b/doc/ADI_scheme.org
@@ -195,23 +195,36 @@ A similar treatment can be applied to the BTCS implicit scheme.
 
 *** implicit BTCS
 
-\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}
+First, we define the Backward time difference:
+
+\begin{equation}
+    \frac{\partial C }{\partial t} = \frac{C^j_i - C^{j-1}_i}{\Delta t}
 \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 $i$.
+Second the spatial derivative approximation:
 
-This discretization works for all internal cells, but not for the
-boundaries. To properly treat them, we need to account for the
-discrepancy in the discretization.
+\begin{equation}
+    \frac{\partial^2 C }{\partial t} = \frac{\frac{C^{j}_{i+1}-C^{j}_{i}}{\Delta x}-\frac{C^{j}_{i}-C^{j}_{i-1}}{\Delta x}}{\Delta x}
+\end{equation}
 
-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:
+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} -C_i^{j-1}}{\Delta t} = \alpha\frac{\frac{C^{j}_{i+1}-C^{j}_{i}}{\Delta x}-\frac{C^{j}_{i}-C^{j}_{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:
+
+\begin{align}\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} \nonumber \\
+                                        & = \alpha\frac{C^{j+1}_{i-1} - 2C^{j+1}_{i} + C^{j+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^{j+1} -C_0^{j}}{\Delta t} = \alpha\frac{\frac{C^{j+1}_{1}-C^{j+1}_{0}}{\Delta x}-
@@ -225,7 +238,7 @@ C_0^{j+1} & =  C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C
           & =  C_0^{j} + \frac{\alpha \cdot \Delta t}{\Delta x^2} \cdot \left( C^{j+1}_{1}- 3 C^{j+1}_{0} +2l \right)
 \end{align}
 
-Now we define variable $s_x$ as following:
+Now we define variable $s_x$ as followed:
 
 \begin{equation}
     s_x = \frac{\alpha \cdot \Delta t}{\Delta x^2}