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\title{\textsc{Differential Equations in Image Processing and Computer Vision \\ \large Assignment T2}}
\author{\shortstack[c]{Jan Hendrik Dithmar\\ \scriptsize 2031259} \\ Group 2: Thu, 12-14 (Markus Mainberger)}% \and \shortstack[c]{name\\ \scriptsize matr}}
\date{}

\pagestyle{fancy}

\lhead{Assignment T2}
\chead{}
\rhead{Dithmar (2031259)}
\lfoot{DIC}
\cfoot{Page \thepage}
\rfoot{SS 2008}

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\exercise{Isotropic Nonlinear Diffusion: Diffusivities}
\begin{enumerate}
	\item
		\begin{eqnarray*}
			g(s^2) & = & \frac{1}{1 + \frac{s^2}{\lambda^2}} \\
			\Phi(s) & = & s \cdot g(s^2) \\
			& = & s \cdot \frac{1}{1 + \frac{s^2}{\lambda^2}} = \frac{s}{1 + \frac{s^2}{\lambda^2}} \\
			\Phi'(s) & = & \frac{1 \cdot \left(1 + \frac{s^2}{\lambda^2}\right) - s \cdot \frac{1}{\lambda^2} \cdot 2s}{\left(1 + \frac{s^2}{\lambda^2}\right)^2} \\
			& = & \frac{1}{1 + \frac{s^2}{\lambda^2}} - \frac{2s^2}{\left(1 + \frac{s^2}{\lambda^2}\right)^2 \cdot \lambda^2} \\
			& = & \frac{\left(1 + \frac{s^2}{\lambda^2}\right) \cdot \lambda^2 - 2s^2}{\left(1 + \frac{s^2}{\lambda^2}\right)^2 \cdot \lambda^2} \\
			\Phi'(s) & = & \frac{\lambda^2 - s^2}{\left(1 + \frac{s^2}{\lambda^2}\right)^2 \cdot \lambda^2} \\[0.5cm]
			\Rightarrow \Phi'(s) > 0 & \Leftrightarrow & |s| < \lambda \quad \mbox{(forward diffusion)} \\
			\Phi'(s) < 0 & \Leftrightarrow & |s| > \lambda \quad \mbox{(backward diffusion)}
		\end{eqnarray*}
	\item
		\begin{eqnarray*}
			g(s^2) & = & \frac{1}{\sqrt{1 + \frac{s^2}{\lambda^2}}} \\
			\Phi(s) & = & \frac{s}{\sqrt{1 + \frac{s^2}{\lambda^2}}} \\
			\Phi'(s) & = & \frac{\left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{2} - s \cdot \frac{1}{2} \cdot \left(1 + \frac{s^2}{\lambda^2}\right) \cdot \frac{1}{\lambda^2} \cdot 2s}{1 + \frac{s^2}{\lambda^2}} \\
			& = & \frac{\left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{2} - \frac{s^2}{\lambda^2} \cdot \left(1 + \frac{s^2}{\lambda^2}\right)}{1 + \frac{s^2}{\lambda^2}} \\
			& = & \frac{1}{\left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{2} - \frac{s^2}{\lambda^2}} \\
			\Phi'(s) & = & \frac{\lambda^2 - s^2 \cdot \left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{2}}{\lambda^2 \cdot \left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{2}} \\[0.5cm]
			\Rightarrow \Phi'(s) > 0 & \Leftrightarrow & |s| < \frac{\lambda}{\left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{4}} \quad \mbox{(forward diffusion)} \\
			\Phi'(s) < 0 & \Leftrightarrow & |s| > \frac{\lambda}{\left(1 + \frac{s^2}{\lambda^2}\right)^\frac{1}{4}} \quad \mbox{(backward diffusion)}
		\end{eqnarray*}
	\item
		\begin{eqnarray*}
			g(s^2) & = & \exp\left(\frac{-s^2}{2\lambda^2}\right) \\
			\Phi(s) & = & s \cdot \exp \left(\frac{-s^2}{2\lambda^2}\right) \\
			\Phi'(s) & = & \exp\left(-\frac{s^2}{2\lambda^2}\right) + s \cdot \exp\left(-\frac{s^2}{2\lambda^2}\right) \cdot \left(-\frac{2s}{2\lambda^2}\right) \\
			& = & \exp\left(-\frac{s^2}{2\lambda^2}\right) - \frac{s^2}{\lambda^2} \cdot \exp\left(-\frac{s^2}{2\lambda^2}\right) \\
			\Phi'(s) & = & \left(1 - \frac{s^2}{\lambda^2}\right) \cdot \underbrace{\exp\left(-\frac{s^2}{2\lambda^2}\right)}_{>0} \\[0.5cm]
			\Rightarrow \Phi'(s) > 0 & \Leftrightarrow& |s| < \lambda \quad \mbox{(forward diffusion)} \\
			\Phi'(s) < 0 & \Leftrightarrow & |s| > \lambda \quad \mbox{(backward diffusion)}
		\end{eqnarray*}
\end{enumerate}

\newpage
\exercise{Isotropic Nonlinear Diffusion: Stencil Notation of Explicit Scheme}
\begin{eqnarray*}
	\partial_t u & = & \mbox{div} \left( g(|\nabla u|^2) \nabla u\right) \\
	& = & \partial_x \left(g(|\nabla u|^2)\partial_x u\right) + \partial_y \left(g(|\nabla u|^2) \partial_y u\right)
\end{eqnarray*}
We know from the lecture (Lecture 5, slide 3) that:
\begin{eqnarray*}
	\frac{du_{i,j}}{dt} & = & \frac{1}{h_1} \left(\frac{g_{i+1,j} + g_{i,j}}{2} \frac{u_{i+1,j} - u_{i,j}}{h_1} - \frac{g_{i,j} + g_{i-1,j}}{2} \frac{u_{i,j} - u_{i-1,j}}{h_1}\right) \\
	& & + \frac{1}{h_2} \left(\frac{g_{i,j+1} + g_{i,j}}{2} \frac{u_{i,j+1} - u_{i,j}}{h_2} - \frac{g_{i,j} + g_{i,j-1}}{2} \frac{u_{i,j} - u_{i,j-1}}{h_2}\right)
\end{eqnarray*}
Furthermore we know that
\begin{eqnarray*}
	\frac{du}{dt} & = & A(u)u \\
	\Leftrightarrow \frac{u_{i,j}^{k+1} - u_{i,j}^k}{\tau} & = & \frac{1}{2h_1^2} \left((g_{i+1,j} + g_{i,j}) (u_{i+1,j} - u_{i,j}) - (g_{i,j} + g_{i-1,j}) (u_{i,j} - u_{i-1,j})\right) \\
	& & + \frac{1}{2h_2^2} \left((g_{i,j+1} + g_{i,j}) (u_{i,j+1} - u_{i,j}) - (g_{i,j} + g_{i,j-1}) (u_{i,j} - u_{i,j-1})\right) \\
	\Leftrightarrow u_{i,j}^{k+1} & = & \frac{\tau}{2h_1^2} (g_{i+1,j}u_{i+1,j} - g_{i+1,j}u_{i,j} + g_{i,j}u_{i+1,j} - g_{i,j}u_{i,j}) \\
	& & - \frac{\tau}{2h_1^2} (g_{i,j}u_{i,j} - g_{i,j}u_{i-1,j} + g_{i-1,j}u_{i,j} - g_{i-1,j}u_{i-1,j}) \\
	& & + \frac{\tau}{2h_2^2} (g_{i,j+1}u_{i,j+1} - g_{i,j+1}u_{i,j} + g_{i,j}u_{i,j+1} - g_{i,j}u_{i,j}) \\
	& & - \frac{\tau}{2h_2^2} (g_{i,j}u_{i,j} - g_{i,j}u_{i,j-1} + g_{i,j-1}u_{i,j} - g_{i,j-1}u_{i,j-1}) \\
	& & + u_{i,j} \\
	& = & u_{i,j} \left(1 - \tau \frac{g_{i+1,j} + 2g_{i,j} + g_{i-1,j}}{2h_1^2} - \tau \frac{g_{i,j+1} + 2g_{i,j} + g_{i,j-1}}{2h_2^2}\right) \\
	& & + u_{i+1,j} \left(\tau \frac{g_{i+1,j} + g_{i,j}}{2h_1^2}\right) + u_{i-1,j} \left(\tau \frac{g_{i,j} + g_{i-1,j}}{2h_1^2}\right) \\
	& & + u_{i,j+1} \left(\tau \frac{g_{i,j+1} + g_{i,j}}{2h_2^2}\right) + u_{i,j-1} \left(\tau \frac{g_{i,j} + g_{i,j-1}}{2h_2^2}\right)
\end{eqnarray*}
With this knowledge, we can give the stencil notation:
{\scriptsize
\begin{center}
	\begin{tabular}{|c|c|c|}
		\hline
		0 & $\displaystyle \tau \frac{g_{i,j+1} + g_{i,j}}{2h_2^2}$ & 0 \\ \hline
		$\displaystyle \tau \frac{g_{i,j} + g_{i-1,j}}{2h_1^2}$ & $\displaystyle 1 - \tau \left(\frac{g_{i+1,j} + 2g_{i,j} + g_{i-1,j}}{2h_1^2} + \frac{g_{i,j+1} + 2g_{i,j} + g_{i,j-1}}{2h_2^2}\right)$ & $\displaystyle \tau \frac{g_{i+1,j} + g_{i,j}}{2h_1^2}$ \\ \hline
		0 & $\displaystyle \tau \frac{g_{i,j} + g_{i,j-1}}{2h_2^2}$ & 0 \\ \hline
	\end{tabular}
\end{center}
}

\newpage
\exercise{Isotropic Nonlinear Diffusion: Matrix Notation of Semi-Implicit Scheme}
For a 1-D signal of length 4 we get a $4 \times 4$ matrix $A$ which has the following entries:
\begin{eqnarray*}
	a_{11} & = & -\left(\frac{g_1 + g_0}{2h^2} + \frac{g_1 + g_2}{2h^2}\right) = -\frac{g_0 + 2g_1 + g_2}{2h^2} \\
	a_{22} & = & -\left(\frac{g_2 + g_1}{2h^2} + \frac{g_2 + g_3}{2h^2}\right) = -\frac{g_1 + 2g_2 + g_3}{2h^2} \\
	a_{33} & = & -\left(\frac{g_3 + g_2}{2h^2} + \frac{g_3 + g_4}{2h^2}\right) = -\frac{g_2 + 2g_3 + g_4}{2h^2} \\
	a_{44} & = & -\left(\frac{g_4 + g_3}{2h^2} + \frac{g_4 + g_5}{2h^2}\right) = -\frac{g_3 + 2g_4 + g_5}{2h^2} \\[0.5cm]
	a_{12} & = & \frac{g_1 + g_2}{2h^2} \\
	a_{13} & = & 0 \\
	a_{14} & = & 0 \\[0.5cm]
	a_{21} & = & \frac{g_1 + g_2}{2h^2} \\
	a_{23} & = & \frac{g_2 + g_3}{2h^2} \\
	a_{24} & = & 0 \\[0.5cm]
	a_{31} & = & 0 \\
	a_{32} & = & \frac{g_2 + g_3}{2h^2} \\
	a_{34} & = & \frac{g_3 + g_4}{2h^2} \\[0.5cm]
	a_{41} & = & 0 \\
	a_{42} & = & 0 \\
	a_{43} & = & \frac{g_3 + g_4}{2h^2} \\
\end{eqnarray*}
In the case of a 2-D image of size $3 \times 4$, we get a $12 \times 12$ matrix $A$ with the following entries (where all the entries not stated here are 0):
\begin{eqnarray*}
	a_{1,1} & = & - \left(\frac{g_1 + g_1}{2h_1^2} + \frac{g_1 + g_2}{2h_1^2} + \frac{g_1 + g_1}{2h_2^2} + \frac{g_1 + g_4}{2h_2^2}\right) \\
	& = & -\frac{3g_1 + g_2}{2h_1^2} - \frac{3g_1 + g_4}{2h_2^2} \\
	a_{2,2} & = & - \left(\frac{g_2 + g_1}{2h_1^2} + \frac{g_2 + g_3}{2h_1^2} + \frac{g_2 + g_2}{2h_2^2} + \frac{g_2 + g_5}{2h_2^2}\right) \\
	& = & -\frac{g_1 + 2g_2 + g_3}{2h_1^2} - \frac{3g_2 + g_5}{2h_2^2} \\
	a_{3,3} & = & -\frac{g_2 + 3g_3}{2h_1^2} - \frac{3g_3 + g_6}{2h_2^2} \\
	a_{4,4} & = & -\frac{3g_4 + g_5}{2h_1^2} - \frac{g_1 + 2g_4 + g_7}{2h_2^2} \\
	a_{5,5} & = & -\frac{g_4 + 2g_5 + g_6}{2h_1^2} - \frac{g_2 + 2g_5 + g_8}{2h_2^2} \\
	a_{6,6} & = & - \frac{g_5 + 3g_6}{2h_1^2} - \frac{g_3 + 2g_6 + g_9}{2h_2^2} \\
	a_{7,7} & = & - \frac{3g_7 + g_8}{2h_1^2} - \frac{g_4 + 2g_7 + g_{10}}{2h_2^2} \\
	a_{8,8} & = & - \frac{g_7 + 2g_8 + g_9}{2h_1^2} - \frac{g_5 + 2g_8 + g_{11}}{2h_2^2} \\
	a_{9,9} & = & - \frac{g_8 + 3g_9}{2h_1^2} - \frac{g_6 + 2g_9 + g_{12}}{2h_2^2} \\
	a_{10,10} & = & - \frac{3g_{10} + g_{11}}{2h_1^2} - \frac{g_7 + 3g_{10}}{2h_2^2} \\
	a_{11,11} & = & - \frac{g_{10} + 2g_{11} + g_{12}}{2h_1^2} - \frac{g_8 + 3g_{11}}{2h_2^2} \\
	a_{12,12} & = & - \frac{g_{11} + 3g_{12}}{2h_1^2} - \frac{g_9 + 3g_{12}}{2h_2^2}
\end{eqnarray*}
\begin{eqnarray*}
	a_{1,2} = \frac{g_1 + g_2}{2h_1^2} \quad \quad
	a_{1,4} & = & \frac{g_1 + g_4}{2h_2^2} \quad \quad
	a_{2,1} = \frac{g_1 + g_2}{2h_1^2} \\
	a_{2,3} = \frac{g_2 + g_3}{2h_1^2} \quad \quad
	a_{2,5} & = & \frac{g_2 + g_5}{2h_2^2} \quad \quad
	a_{3,2} = \frac{g_2 + g_3}{2h_1^2} \\
	a_{3,6} = \frac{g_3 + g_6}{2h_2^2} \quad \quad
	a_{4,1} & = & \frac{g_1 + g_4}{2h_2^2} \quad \quad
	a_{4,5} = \frac{g_4 + g_5}{2h_1^2} \\
	a_{4,7} = \frac{g_4 + g_7}{2h_2^2} \quad \quad
	a_{5,2} & = & \frac{g_2 + g_5}{2h_2^2} \quad \quad
	a_{5,4} = \frac{g_4 + g_5}{2h_1^2} \\
	a_{5,6} = \frac{g_5 + g_6}{2h_1^2} \quad \quad
	a_{5,8} & = & \frac{g_5 + g_8}{2h_2^2} \quad \quad
	a_{6,3} = \frac{g_3 + g_6}{2h_2^2} \\
	a_{6,5} = \frac{g_5 + g_6}{2h_1^2} \quad \quad
	a_{6,9} & = & \frac{g_6 + g_9}{2h_2^2} \quad \quad
	a_{7,4} = \frac{g_4 + g_7}{2h_2^2} \\
	a_{7,8} = \frac{g_7 + g_8}{2h_1^2} \quad \quad
	a_{7,10} & = & \frac{g_7 + g_{10}}{2h_2^2} \quad \quad
	a_{8,5} = \frac{g_5 + g_8}{2h_2^2} \\
	a_{8,7} = \frac{g_7 + g_8}{2h_1^2} \quad \quad
	a_{8,9} & = & \frac{g_8 + g_9}{2h_1^2} \quad \quad
	a_{8,11} = \frac{g_8 + g_{11}}{2h_2^2} \\
	a_{9,6} = \frac{g_6 + g_9}{2h_2^2} \quad \quad
	a_{9,8} & = & \frac{g_8 + g_9}{2h_1^2} \quad \quad
	a_{9,12} = \frac{g_9 + g_{12}}{2h_2^2} \\
	a_{10,7} = \frac{g_7 + g_{10}}{2h_2^2} \quad \quad
	g_{10,11} & = & \frac{g_{10} + g_{11}}{2h_1^2} \quad \quad
	g_{11,8} = \frac{g_8 + g_{11}}{2h_2^2} \\
	g_{11,10} = \frac{g_{10} + g_{11}}{2h_1^2} \quad \quad
	g_{11,12} & = & \frac{g_{11} + g_{12}}{2h_1^2} \quad \quad
	g_{12,9} = \frac{g_9 + g_{12}}{2h_2^2} \\
	g_{12,11} & = & \frac{g_{11} + g_{12}}{2h_1^2}
\end{eqnarray*}

\newpage
\exercise{Linear Systems of Equations and Thomas' Algorithm}
\begin{enumerate}
	\item	We want to decompose the matrix
		$$A = \left(\begin{array}{ccccc}\gamma_1 & \delta_1& \varepsilon_1 & 0 & 0 \\ \beta_1 & \gamma_2 & \delta_2 & \varepsilon_2 & 0 \\ \alpha_1 & \beta_2 & \gamma_3 & \delta_3 & \varepsilon_3 \\ 0 & \alpha_2 & \beta_3 & \gamma_4 & \delta_4 \\ 0 & 0 & \alpha_3 & \beta_4 & \gamma_5\end{array}\right)$$
		into
		$$L := \left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\ b_1 & 1 & 0 & 0 & 0 \\ a_1 & b_2 & 1 & 0 & 0 \\ 0 & a_2 & b_3 & 1 & 0 \\ 0 & 0 & a_3 & b_4 & 1\end{array}\right)$$
		and
		$$R := \left(\begin{array}{ccccc}c_1 & d_1 & e_1 & 0 & 0 \\ 0 & c_2 & d_2 & e_2 & 0 \\ 0 & 0 & c_3 & d_3 & e_3 \\ 0 & 0 & 0 & c_4 & d_4 \\ 0 & 0 & 0 & 0 & c_5\end{array}\right).$$
		The matrix $A$ can be written as follows: \\[0.5cm]
		$A = LR$
		$$= \left(\begin{array}{ccccc}c_1 & d_1 & e_1 & 0 & 0 \\ b_1c_1 & b_1d_1+c_2 & b_1e_1+d_2 & e_2 & 0 \\ a_1c_1 & a_1d_1+b_2c_2 & a_1e_1+b_2d_2+c_3 & b_2e_2+d_3 & e_3 \\ 0 & a_2c_2 & a_2d_2+b_3c_3 & a_2e_2+b_3d_3+c_4 & b_3e_3+d_4 \\ 0 & 0 & a_3c_3 & a_3d_3+b_4c_4 & a_3e_3+b_4d_4+c_5\end{array}\right)$$ \\
		\textit{LR-Decomposition:} \\
		We know that $e_i = \varepsilon_i, i \in \{1, 2, 3\}$. The coefficients $c_i, b_i$ and $a_i$ can be computed by: \\
		\texttt{
		c$_1$ := $\gamma_1$ \\
		b$_1$ := $\beta_1$ / c$_1$ \\
		a$_1$ := $\alpha_1$ / c$_1$ \\
		d$_1$ := $\delta_1$ \\[0.3cm]
\newpage
		for i = 2, ..., 5: \\
		\phantom{xxx} if i $\leq$ 4: \\
		\phantom{xxxxx} d$_i$ := $\delta_i$ - b$_{i-1}$e$_{i-1}$ \\ \\
		\phantom{xxx} c$_i$ := $\gamma_i$ - b$_{i-1}$d$_{i-1}$ - a$_{i-2}$e$_{i-2}$ \\ \\
		\phantom{xxx} if i $\leq$ 4: \\
		\phantom{xxxxx} b$_i$ := $\beta_i$ / c$_i$ - a$_{i-1}$d$_{i-1}$ \\ \\
		\phantom{xxx} if i $\leq$ 3: \\
		\phantom{xxxxx} a$_i$ := $\alpha_i$ / c$_i$
		}
	\item \textit{Forward Elimination:} \\
		We solve $L\textbf{y} = \textbf{u}$ for $\textbf{y}$:
		$$\left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\ b_1 & 1 & 0 & 0 & 0 \\ a_1 & b_2 & 1 & 0 & 0 \\ 0 & a_2 & b_3 & 1 & 0 \\ 0 & 0 & a_3 & b_4 & 1\end{array}\right) \left(\begin{array}{c}y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5\end{array}\right) = \left(\begin{array}{c}u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5\end{array}\right)$$ \\
		\texttt{
		y$_1$ := u$_1$ \\
		y$_2$ := u$_2$ - b$_1$y$_1$ \\
		for i = 3, .., 5: \\
		\phantom{xxx} y$_i$ := u$_i$ - a$_{i-2}$y$_{i-2}$ - b$_{i-1}$y$_{i-1}$
		} \\[0.5cm]
		\textit{Backward Substitution:} \\
		We solve $R\textbf{x} = \textbf{y}$ for $\textbf{x}$:
		$$\left(\begin{array}{ccccc}c_1 & d_1 & e_1 & 0 & 0 \\ 0 & c_2 & d_2 & e_2 & 0 \\ 0 & 0 & c_3 & d_3 & e_3 \\ 0 & 0 & 0 & c_4 & d_4 \\ 0 & 0 & 0 & 0 & c_5\end{array}\right) \left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{array}\right) = \left(\begin{array}{c}y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5\end{array}\right)$$ \\
		\texttt{
		x$_5$ := y$_5$ / c$_5$ \\
		x$_4$ := (y$_4$ - d$_4$x$_5$) / c$_4$ \\
		for i = 3, ..., 1: \\
		\phantom{xxx} x$_i$ := (y$_i$ - d$_i$x$_{i+1}$ - e$_i$x$_{i+2}$) / c$_i$
		}
\end{enumerate}

\end{document}
