Initial-boundary-value Problems

This time the infinite $x$ plane is replace by the semi-infinite plane, with a boundary condition imposed on $x=0$.

$$\left. \begin{array}{cc} c{\partial u \over \partial x} ... ..., 0 ) = f(x), & \\ u(0, t) = g(t). & \end{array} \right \}$$ (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

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

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

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

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

Note that the characteristics are

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

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

$$2{\partial u \over \partial x} + {\partial u \over \partial t} = 0,$$

with

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

The characteristics are given by

$$x_{0} = x - 2t.$$

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

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

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.