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Next: Method of Characteristics : Up: Method of Characteristics Previous: Initial-value Problems

Initial-boundary-value Problems

This time the infinite $x$ plane is replace by the semi-infinite plane, with a boundary condition imposed on $x=0$.
\begin{displaymath}
\left.
\begin{array}{cc}
c{\partial u \over \partial x} ...
..., 0 ) = f(x), & \\
u(0, t) = g(t). &
\end{array} \right \}
\end{displaymath} (1.13)

Consider the case $c > 0$, so that the waves are travelling from small $x$ to large $x$. The characteristics are again given by

\begin{displaymath}
{dx\over dr} = c, \qquad {dt\over dr} = 1, \qquad \Rightarrow \qquad
x = ct + x_{0}.
\end{displaymath}

The characteristic through the origin, splits the $x-t$ plane into two regions, as shown in Figure 1.6.

Figure 1.6: The characteristic, $x = ct$, splits the $x-t$ plane into two regions
\includegraphics [scale=0.7]{figpde6.ps}

In region $R_{1}$, $ x > ct$,the solution is determined by the initial condition at $t=0$. In this region the characteristics originate from the $x$-axis.

In region $R_{2}$, $x < ct$, the solution is given by the boundary condition at $x=0$. Here the characteristics originate from the $t$-axis. Remembering that the solution is constant along the characteristics, this implies that the solution is determined by where the characteristics come from. Thus, the solution is

\begin{displaymath}
u(x,t) = \left \{
\begin{array}{cc}
f(x -ct) & x > ct, \\
g(t - x/c) & x < ct.
\end{array}\right.
\end{displaymath} (1.14)

Note that the characteristics are

\begin{displaymath}
x - ct = x_{0}, \qquad \Rightarrow \qquad t - {x\over c} =
-{x_{0}\over c}.
\end{displaymath}

It is clear that (1.14) satisfies the initial condition, since $t=0$ implies that $x >0$ and $u(x,0) = f(x)$, and the boundary condition since $x=0$ implies that $t > 0$ and so $u(0,t) = g(t)$. In addition, it is easily shown that $u(x,t)$ also satisfies the PDE.

Example 1. .9

\begin{displaymath}
2{\partial u \over \partial x} + {\partial u \over \partial t} = 0,
\end{displaymath}

with

\begin{displaymath}
u(x,0) = \left \{
\begin{array}{cc}
1 - x & 0 < x \le 1, ...
... 0 & x >1,
\end{array}\right.
u(0, t) = e^{-t}, \qquad t >0.
\end{displaymath}

The characteristics are given by

\begin{displaymath}
x_{0} = x - 2t.
\end{displaymath}

Therefore the plane is split into three regions and the solution is

\begin{displaymath}
u(x,t) = \left \{
\begin{array}{cc}
0 & x - 2t >1, \\
1...
... \le 1, \\
e^{-(t - x/2)} & x - 2t < 0.
\end{array} \right.
\end{displaymath}

A little bit of thought about this last example will soon illustrate a major problem. If the wave speed is negative, i.e. $c < 0$, then there is in general no solution to the initial-boundary-value problem. This is because the characteristics intersect both boundaries and $u$ cannot be defined by two different values (resulting from values of $f(x)$ and $g(t)$) on the same characteristics. This is illustrated in Figure 1.7.

Figure 1.7: The characteristic, with $c < 0$, intersects both the $x$-axis and the $t$-axis.
\includegraphics [scale=0.7]{figpde7.ps}


next up previous
Next: Method of Characteristics : Up: Method of Characteristics Previous: Initial-value Problems
Prof. Alan Hood
2000-10-30