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42f85c1eb7 Wrote a rough draft of problem 5 2023-09-02 15:50:13 +02:00
2 changed files with 12 additions and 38 deletions

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@@ -86,6 +86,7 @@
\section*{Problem 1} \section*{Problem 1}
% Do the double integral
\begin{align*} \begin{align*}
u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\ u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
&= \int \int -100 e^{-10x} dx^2 \\ &= \int \int -100 e^{-10x} dx^2 \\
@@ -125,40 +126,7 @@ Using the values that we found for $c_1$ and $c_2$, we get
\section*{Problem 3} \section*{Problem 3}
To derive the discretized version of the Poisson equation, we first need % Show how it's derived and where we found the derivation.
the taylor expansion for $u(x)$ around $x + h$ and $x - h$.
\begin{align*}
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)
\end{align*}
\begin{align*}
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)
\end{align*}
If we add the equations above, we get this new equation:
\begin{align*}
u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
\end{align*}
We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
\begin{align*}
- \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
\frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
\end{align*}
And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to
differentiate between the exact solution and the approximate solution,
and get the discretized version of the equation:
\begin{align*}
align* \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
\end{align*}
\section*{Problem 4} \section*{Problem 4}
@@ -168,8 +136,16 @@ align* \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
\subsection*{a)} \subsection*{a)}
% Phrase it better
$n = m - 2$, since $\textbf{A}$ is used to solve for all of the points in
between the end-points $0$ and $1$. For the complete solution, we need to add
$u(0)$ and $u(1)$.
\subsection*{b)} \subsection*{b)}
When solving for $\vec{v}$, we find the approximate solutions for $u(x)$
that are in between the end-points, but not the end-points themselves.
\section*{Problem 6} \section*{Problem 6}
\subsection*{a)} \subsection*{a)}
@@ -180,9 +156,7 @@ align* \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
% Figure it out % Figure it out
\section*{Problem 7} % Linelevant files on gh and possibly add some comments
% Link to relevant files on gh and possibly add some comments
\section*{Problem 8} \section*{Problem 8}
@@ -190,7 +164,7 @@ align* \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
\section*{Problem 9} \section*{Problem 9}
% Show the algorithm, then calculate FLOPs, then link to relevant files % Shon*{Proalgorithm, then calculate FLOPs, then link to relevant files
\section*{Problem 10} \section*{Problem 10}