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Method of Characteristics

The first order wave equation
\begin{displaymath}
c{\partial u\over \partial x} + {\partial u \over \partial t} = 0,
\end{displaymath} (1.3)

describes the movement of a wave in one direction with no change of shape. A solution is shown in Figure 1.3 as a surface plot and a contour plot.

Figure 1.3: The left figure shows a surface plot and the right figure shows a contour plot. In the contourplot the lines are for a given value of $u$.
\includegraphics [scale=0.7]{figpde3.ps}

In the contour plot, in Figure 1.3, the straight lines are given by the lines that have the same value of $u$. $u$ is constant along these lines.

The aim of the method of characteristics is to solve the PDE by finding curves in the $x-t$ plane that reduce the equation to an ODE. In general, any curve in the $x-t$ plane can be expressed in parametric form by

\begin{displaymath}
x = x(r), \qquad t = t(r),
\end{displaymath}

where the parameter, $r$, gives a measure of the distance along the curve. The curve starts at the initial point, $x=x_{0}$, $t=0$, when $r=0$. Assuming that we can solve the resulting ODE means that $u$ is known everywhere along this curve, i.e. along the curve picked out by the value of $x_{0}$. Another choice for $x_{0}$ gives another curve and the value of $u$ is determined along this curve. In this manner, $u$ can be determined at any point in the $x-t$ plane by choosing the curve, defined by $x_{0}$, that passes through this point and taking the correct value of $r$, the distance along the curve. Hence, we can ervaluate $u(x,t)$. These curves are illustrated in Figure 1.4

Figure 1.4: The parametric representation of a curve in the $x-t$ plane. The solid curve starts at $r=0$, $t=0$ and $x=x_{0}$. Choosing another value of $x_{0}$ gives the dashed curve.
\includegraphics [scale=0.7]{figpde4.ps}

Therefore, we have $u(x,t) = u(x(r), t(r))$ and so $u$ is a function of $r$. Hence, the derivative of $u$ with respect to to $r$ is

\begin{displaymath}
{du\over dr} = {dx\over dr}{\partial u\over \partial x} + {dt\over
dr}{\partial u \over \partial t}.
\end{displaymath} (1.4)

Compare (1.4) with the left hand side of (1.3) and we can convert (1.3) into an ordinary derivative of $u$ with respect to $r$, i.e.
\begin{displaymath}
{du\over dr} = 0,
\end{displaymath} (1.5)

provided the parametric representation of the curve satisfies
\begin{displaymath}
{dx\over dr} = c,
\end{displaymath} (1.6)

and
\begin{displaymath}
{dt\over dr} = 1.
\end{displaymath} (1.7)

(1.6) and (1.7) give the characteristic curves. (1.5) shows that
\begin{displaymath}
u = \hbox{constant},
\end{displaymath} (1.8)

along a characteristic curve but the constant may be different on different characteristic curves. As $x_{0}$ gives us a different characteristic curve, this implies that $u = F(x_{0})$.

Solving (1.7), we have

\begin{displaymath}
t = r, \qquad t = 0, \hbox{ at }r=0,
\end{displaymath}

and (1.6) gives

\begin{displaymath}
x = cr + x_{0} = ct + x_{0}, \qquad \hbox{since }x = x_{0} \hbox{
at } r=0.
\end{displaymath}


\begin{displaymath}
\Rightarrow \qquad x_{0} = x - ct.
\end{displaymath} (1.9)

$x_{0}$ defines which characteristic curve you are on and from (1.8), $u$ is constant on a characteristic curve that depends on $x_{0}$. Hence,
\begin{displaymath}
u = F(x_{0}) = F(x - ct),
\end{displaymath} (1.10)

as shown before, where the arbitrary function, $F$, is determined by the initial condition.

Subsections
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Prof. Alan Hood
2000-10-30