50 lines
1.8 KiB
TeX
50 lines
1.8 KiB
TeX
\section*{Problem 6}
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\subsection*{a)}
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% Use Gaussian elimination, and then use backwards substitution to solve the equation
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Renaming the sub-, main-, and supdiagonal of matrix $\boldsymbol{A}$
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\begin{align*}
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\vec{a} &= [a_{2}, a_{3}, ..., a_{n-1}, a_{n}] \\
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\vec{b} &= [b_{1}, b_{2}, b_{3}, ..., b_{n-1}, b_{n}] \\
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\vec{c} &= [c_{1}, c_{2}, c_{3}, ..., c_{n-1}] \\
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\end{align*}
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Following Thomas algorithm for gaussian elimination, we first perform a forward sweep
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\begin{algorithm}[H]
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\caption{Foreward sweep}\label{algo:foreward}
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\begin{algorithmic}
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\State Create new vector $\vec{\hat{b}}$ of length n.
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\State $\hat{b}[0] = b[0]$ \Comment{Handle first element in main diagonal outside loop}
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\For{$i = 1, ..., n-1$}
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\State $d = \frac{a[i-1]}{b[i-1]}$
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\State $b -= d*(*sup_diag)(i-1);$
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$(*g_vec)(i) -= d*(*g_vec)(i-1);$
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\EndFor
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\While{Some condition}
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\State Do something more here
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\EndWhile
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\State Maybe even some more math here, e.g $\int_0^1 f(x) \dd x$
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\end{algorithmic}
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\end{algorithm}
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Here the index i does not refer to the element in the vector, ie. $b_{i}$, but to the index in the vector where the element is found ie. $b[i] = b_{i+1}$.
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\begin{align*}
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\hat{b}_{1} &= b_{1} \\
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\hat{b}_{2} &= b_{2} - \frac{a_{2}}{\hat{b}_{1}} \cdot c_{1} \\
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\vdots & \\
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\hat{b}_{i} &= b_{i} - \frac{a_{i}}{\hat{b}_{i-1}} \cdot c_{i-1} \\
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\vdots & \\
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\end{align*}
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% as g_hat also make use of the ratio (ai/bi-1_hat) it only needs to be calculated once
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% for n steps we have n+1 points, that is values of x to evaluate u(x) at, when boundary values are known
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% we have to perform (n+1)-2 = n-1 calculations (equal number of col in matrix). Dealing
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% with a quadratic matrix number of col = number of rows
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\subsection*{b)}
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% Figure it out
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