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\lhead{Umeå University\\
Department of Physics\\
Report Computer lab 2}
\rhead{11 November 2019}
\title{Mathematics of Physical Models\\Computer Lab 2}
\author{Anton Morian}
\date{anmo0148@student.umu.se}
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\begin{document}
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\tableofcontents
\lhead{Anton Morian}
\rhead{Mathematics of Physical Models, 10.5hp\\Cpmputer lab 2}
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\section{Dimensional analysis}
\subsection{Problem description of dimensional analysis}
In the first section a short dimensional analysis of the two parameters $c$ and $S$ used in this lab. The first parameter $c$ can be found below:
\begin{equation}
\frac{\partial^2 u}{\partial t^{2}}=c^2 \triangledown u.
\end{equation}
The second parameter $S$  comes from the equation:
\begin{equation}
    \rho_{l}(x) \frac{\partial^2}{\partial t^2} v(x,t)=S \frac{\partial^2}{\partial x^2} v(x,t),\quad 0<x<L \quad
\end{equation}
\subsection{Result and discussion of dimensional analysis }
The dimension of $c$ can be obtain from the equation above. With some elementary operation its easy to show that $c$ needs to have the dimension, $[c] = \frac{L}{T}$, as same as the velocity.
With the result from above, the dimension of $S$ can be expressed as, $[S] =\frac{M}{L}\frac{L^2}{T^2} $. Thus $S$ most have the dimension, $[S] = \frac{ML}{T^2}$

\newpage
\cfoot{2}
\section{Exercise 1}
\subsection{Problem description of Exercise 1}
What kind of  boundary condition is specified for the PDE below and plot your numeric solution, in the same graph plot the difference between the numeric solution and the analytical solution. The PDE is:
\begin{equation}
    \frac{\partial^2}{\partial{t}^2}u(x,t = 9\frac{\partial^2}{\partial{x}^2}u(x,t), \quad 0 < x < \pi, \quad t > 0,
\end{equation}
\begin{equation}
\label{1.u}
    u(0,t) = u(\pi,t) = 0, \quad t > 0,
\end{equation}
\begin{equation}
    u(x,0) = 3\sin{2x} + 12\sin{13x}, \quad 0 < x < \pi,
\end{equation}
\begin{equation}
    \frac{\partial}{\partial{t}}u(x,0) = 0, \quad 0 < x < \pi,
\end{equation}
and the analytical solution is:
\begin{equation}
    u_{analytic} = 3cos(6t)sin(2x) + 12cos(39t)sin(13x)
\end{equation}
\subsection{Result and discussion of Exercise 1}
The boundary condition in equation \ref{1.u} are homogeneous. Thus both equals 0.
\begin{figure}[H]%
    \centering
    \subfloat[Line graph]{\includegraphics[width=6cm]{u_uan_u_lab_1.png}}%
    \qquad
    \subfloat[Point graph]{\includegraphics[width=6cm]{u_uan_u_lab_1_point.png}}%
    \caption{Both (a) and (b) are graph of the numeric solution $u$ and the difference between $u$ and the analytical solution $u_{an}$.}%
    \label{uan}%
\end{figure}









\newpage
\section{Exercise 2}
\subsection{Problem description of Exercise 2}
A vibrating string is described by the PDE:
\begin{equation}
    \frac{\partial^2}{\partial{t}^2}u(x,t) = \alpha^2\frac{\partial^2}{\partial{x}^2}u(x,t), \quad 0 < x < L, \quad  t > 0
    \end{equation}
    \begin{equation}
        u(0,t) = u(L,t) = 0, \quad t > 0
    \end{equation}
    \begin{equation}
            u(x,0) = f(x), \quad 0 < x < L
    \end{equation}
    \begin{equation}
        \frac{\partial}{\partial{t}}u(x,0) = g(x), \quad 0 < x < L
    \end{equation}


If the string is pulled into a inverse V-shape, with height h0 at the point x = a, and then released at t = 0 the initial conditions are:
\begin{equation}
    \begin{cases}
        h_0\frac{w}{a}, \quad 0 < x < a\\
        h_0 \frac{1-x}{1-a}, \quad a < x < L
    \end{cases}
\end{equation}
\begin{equation}
    g(x) = 0
\end{equation}
If L = 1 is chosen, the analytical solution is given by:
\begin{equation}
    u_{analytic} = \frac{2h_0}{\pi^2a(1-a)}\sum\limits_{n=1}^{\infty} \frac{1}{n^2}\sin(n\pi a)\sin(n\pi x)\cos(n\pi \alpha t)
\end{equation}


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