38 lines
1.3 KiB
TeX
38 lines
1.3 KiB
TeX
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\section*{Problem 3}
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To derive the discretized version of the Poisson equation, we first need
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the Taylor expansion for $u(x)$ around $x$ for $x + h$ and $x - h$.
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\begin{align*}
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u(x+h) &= u(x) + u'(x) h + \frac{1}{2} u''(x) h^2 + \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
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\end{align*}
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\begin{align*}
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u(x-h) &= u(x) - u'(x) h + \frac{1}{2} u''(x) h^2 - \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
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\end{align*}
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If we add the equations above, we get this new equation:
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\begin{align*}
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u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
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u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
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u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
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u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
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\end{align*}
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We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
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\begin{align*}
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- \frac{d^2u}{dx^2} &= f(x) \\
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\frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= f_i \\
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\end{align*}
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And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to
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differentiate between the exact solution and the approximate solution,
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and get the discretized version of the equation:
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\begin{align*}
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\frac{ - v_{i+1} + 2 v_i - v_{i-1}}{h^2} &= 100 e^{-10x_i} \\
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\end{align*}
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