45 lines
1.6 KiB
TeX
45 lines
1.6 KiB
TeX
\section*{Problem 4}
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% Show that each iteration of the discretized version naturally creates a matrix equation.
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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
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\begin{align*}
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- v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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\vdots & \\
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- v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
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\end{align*}
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Rearranging the first and last equation, moving terms of known boundary values to the RHS
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\begin{align*}
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2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
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- v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
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\vdots & \\
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- v_{m-2} + 2 v_{m-1} &= h^{2} \cdot f_{m-1} + v_{m} \\
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\end{align*}
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We now have a number of linear eqations, corresponding to the number of unknown values, which can be represented as an augmented matrix
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\begin{align*}
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\left[
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\begin{matrix}
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2v_{1} & -v_{2} & 0 & \dots & 0 \\
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-v_{1} & 2v_{2} & -v_{3} & 0 & \\
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0 & -v_{2} & 2v_{3} & -v_{4} & \\
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\vdots & & & \ddots & \vdots \\
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0 & & & -v_{m-2} & 2v_{m-1} \\
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\end{matrix}
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\left|
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\,
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\begin{matrix}
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g_{1} \\
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g_{2} \\
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g_{2} \\
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\vdots \\
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g_{m-1} \\
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\end{matrix}
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\right.
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\right]
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\end{align*}
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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 $n=m-2$.
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