From c42426847d9f3b95125c387fd802ee238710b768 Mon Sep 17 00:00:00 2001 From: Janita Willumsen Date: Fri, 8 Sep 2023 09:39:53 +0200 Subject: [PATCH] Finished exercise 4 --- latex/problems/problem4.tex | 41 +++++++++++++++++++++++++++++++++++++ 1 file changed, 41 insertions(+) diff --git a/latex/problems/problem4.tex b/latex/problems/problem4.tex index 2f690f3..fea7255 100644 --- a/latex/problems/problem4.tex +++ b/latex/problems/problem4.tex @@ -1,3 +1,44 @@ \section*{Problem 4} % Show that each iteration of the discretized version naturally creates a matrix equation. + +The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we can approximate the value of $f(x_{i}) = f_{i}$. This will result in a set of equations +\begin{align*} + - v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\ + - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\ + \vdots & \\ + - v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\ +\end{align*} + +Rearranging the first and last equation, moving terms of known boundary values to the RHS +\begin{align*} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\ + - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\ + \vdots & \\ + - v_{m-2} + 2 v_{m-1} &= h^{2} \cdot f_{m-1} + v_{m} \\ +\end{align*} + +We now have a number of linear eqations, corresponding to the number of unknown values, which can be represented as an augmented matrix +\begin{align*} + \left[ + \begin{matrix} + 2v_{1} & -v_{2} & 0 & \dots & 0 \\ + -v_{1} & 2v_{2} & -v_{3} & 0 & \\ + 0 & -v_{2} & 2v_{3} & -v_{4} & \\ + \vdots & & & \ddots & \vdots \\ + 0 & & & -v_{m-2} & 2v_{m-1} \\ + \end{matrix} + \left| + \, + \begin{matrix} + g_{1} \\ + g_{2} \\ + g_{2} \\ + \vdots \\ + g_{m-1} \\ + \end{matrix} + \right. + \right] + \end{align*} + where $g_{i} = h^{2} f_{i}$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$. +