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\section{Wave Equation - Hyperbolic PDE}

\subsection{Preliminaries}

This section will illustrate the derivation and the 
numerical solution  
of a special hyperbolic PDE, the {\em Wave Equation}.

\begin{equation}
        \frac{\partial^2 y(x,t)}{\partial t^2}
        =   c^2  \frac{ \partial^2 y(x,t)}{\partial x^2}
\end{equation}

This PDE describes waves 
on a string. It is an example often seen in 
Freshman’s Physics or in more advanced mechanics courses. 

In this course, we will naturally compare the numerical
approach to solve the  {\em Wave Equation} to that to
solve the {\em Heat
Transfer} equation.  There are many common aspects
between these algorithms.    

\subsection{Model}

The {\em Wave Equation} is a model to describe 
transverse displacements of a 
string under tension.  Assume that the tension $T$ is
constant and that the string is made out of material 
of constant linear density, which guaranty a uniform string. 
Further, we will assume that the  tension $T$ is
sufficiently large to avoid the string sagging under gravity.

The solution of the {\em Wave Equation} is a function,
$y(x,t)$,  that gives the perpendicular displacement profile
of the string with respect to its rest position. 

The figure below illustrates an infinitesimal element
of the string undergoing a perpendicular 
displacement.  

\begin{rawhtml}
<img src="../pictures/grid3.GIF" \>
\end{rawhtml}

The wave equation follows from applying
 Newton’s equation to such a small element of the string.

\begin{equation}
    \sum_{}  F_y  = \rho \Delta_x \frac{ \partial^2 y}{ \partial t^2}
\end{equation}

\begin{equation}
   \sum F_y = T \sin{ \theta(x+\Delta_x) } - T sin{ \theta(x) }
\end{equation}


\begin{equation} 
     = T  \left. \frac{\partial y}{\partial x } \right|_{(x + \Delta_x)}- 
            \left.  T \frac{\partial y}{\partial x} \right|_x
\end{equation}

The {\em Wave Equation} follows

\begin{equation}
     =  \frac{ \partial^2 y(x,t)}{\partial x^2} =
        \frac{1}{c^2} \frac{\partial^2 y(x,t)}{\partial t^2}
\end{equation}

This model is over-simplified.  As is, it admits an 
analytic solution via a separation of variables.
Of course it can be made more realistic by adding, for instance,
friction, gravity, ... Then the numerical solution becomes the 
only feasible solution. 


\subsection{Numerical Solution}

As in the cases of the elliptic and parabolic PDEs, we 
derive a finite difference method to solve 
the {\em Wave Equation} by seeking
a solution on equally spaced grids.  

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<img src="../pictures/grid1.GIF" \>
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Two grid spacing's are needed, 
an $x$ spacing, $\Delta_x$, and a time spacing,
$\Delta_t$. The grid variables are specified by
  $  x_i , i = 0..{{N_x}-1}$ and $  t_j , j = 0..N_t-1 $
Note that the time grid extends to arbitrary
large $N_t$.  The notation for the displacement on
the grid  $y(x,t)$
is then $y_i^j $.
   
The algorithm follows from writing a finite difference
version of the {\em Wave Equation}.  First off, we refer back to 
the finite difference expressions for the
derivative.

\begin{equation}
  \left( { y^{j+1}_i + y^{j-1}_i - 2 y^j } \right)
=   \frac{1}{c^2} \frac{\Delta_t^2 }{\Delta_x^2 }
  \left( { y^j_{i+1} + y^j_{i-1} - 2 y^j_i } \right)
\end{equation}

Isolating $y(x,t)$ leads to the finite difference form
of the {\em Wave Equation}.

\begin{equation}
  y_i^{j+1} = 2 y_i^j + \frac{c^2}{c'^2} \left[
                 y_{i+1}^j + y_{i-1}^j -2 y_i^g \right]
\end{equation}


\subsection{Problem}

A major difference between the {\em Wave Equation} and the 
{\em Heat Transfer} equation 
is that the time derivative in the LHS is of second order 
instead of first order. This creates a problem that 
we address next.

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<img src="../pictures/grid2.GIF" \>
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In the language of {\em finite difference}, the second time derivative 
mixes the solution at three time slices.  That is, to predict the solution 
at the time $t+dt$ requires knowing the solution at two previous time slices,
namely at $t$ and at $t-dt$. But, the initial value of the displacement
is given only on a single time slice, not on two time slices as required.

The solution to this shortcoming is found within a clever 
use of the initial value.  We simply have to assume that the 
displacement at $t=0$ is stationary.  The first derivative,
written in finite difference form at the initial time, 

\begin{equation}
      \frac{ \partial y }{\partial t}(x,0) = 
           \frac{ y(x,0) - y(x,0-\Delta_t) }{\partial \Delta_t} = 0
\end{equation} 

then shows that the displacement at $t-dt$  is the 
same as that at $t=0$,  $ y_i^{-1} = y_i^0 $.  Therefore 
the solution becomes self starting, 
bootstraping the solution at later time based on
the initial displacement at $t=0$.


\subsection{Courant Condition}

The constant in the finite difference wave equation,
\begin{equation}
 \frac{\rho}{T} \frac{1}{c^2} \frac{\Delta_t^2 }{\Delta_x^2 } =
           \frac{c^2}{c'^2}
\end{equation}

where    
\begin{equation}
c  =  \frac{ \rho}{ T } 
\end{equation}  
and
\begin{equation}
c'  =  \frac{ \Delta_x}{ \Delta_t } 
\end{equation}.

It can be shown that the {\em Courant Condition},
$ c < c' = \Delta_x / \Delta_t  $
limits constants values that allow stable solutions.  More accuurate
solutions will result from smaller $\Delta_x$ {\em and} $\Delta_t$
provided the {\em Courrant Condition} applies.

\end{document}
