Finish report

latex/sections/appendices.tex

@@ -47,7 +47,7 @@ the left hand side, and the terms containing $n$ time step on the right hand sid
     &= u_{\ivec, \jvec}^{n} + \frac{i \Delta t}{2 \Delta x^{2}} \big[ u_{\ivec+1, \jvec}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec-1, \jvec}^{n} \big] \\
     & \quad + \frac{i \Delta t}{2 \Delta y^{2}} \big[ u_{\ivec, \jvec+1}^{n} - 2u_{\ivec, \jvec}^{n} + u_{\ivec, \jvec-1}^{n} \big] - \frac{i \Delta t}{2} v_{\ivec, \jvec} u_{\ivec, \jvec}^{n} 
 \end{align*}
-Since we will use an equal step size $h$ in both $x$ and $y$ direction, we can use 
+Since we use an equal step size $h$ in both $x$ and $y$ direction, we can use 
 \begin{align*}
     \frac{i \Delta t}{2 \Delta h^{2}} = \frac{i \Delta t}{2 \Delta x^{2}} = \frac{i \Delta t}{2 \Delta y^{2}} \ ,
 \end{align*}
@@ -65,7 +65,7 @@ Now, the discretized Schrödinger equation can be rewritten as
 
 
 \section{Matrix structure}\label{ap:matrix_structure}
-For $u$ vector of length $(M-2) = 3$, the matrices $A$ and $B$ have size 
+For a $u$ vector of length $(M-2) = 3$, the matrices $A$ and $B$ have size 
 $(M-2)^{2} \times (M-2)^{2} = 9 \times 9$ given by 
 \begin{align*}
     A =