Initial-value Problems

Here the problem is defined over an infinite range in $x$. The general statement of the problem is

$$c{\partial u \over \partial x} + {\partial u \over \partial t} = 0, \qquad -\infty < x < \infty, \hbox{ } t>0.$$ (1.11)

$$u(x,0) = f(x).$$ (1.12)

The initial value problem, IVP, defined in Equations (1.11) and (1.12), is also referred to as a Cauchy problem and the solution is determined uniquely by the single condition at $t=0$. From (1.10) we have

$$u(x,t) = f(x - ct), \qquad (= f(x_{0}))$$

If $f$ is continuously differentiable, then ${\partial u \over \partial x} = f^{\prime}(x - ct)$ and ${\partial u \over \partial t} = - c f^{\prime}(x - ct)$ are continuous. Thus, $u = f(x-ct)$ is a classical solution of (1.11) and (1.12).

If $f$ is only piecewise continuous (derivatives are not continuous), e.g.

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

then $u = f(x-ct)$ is called a weak solution.

Example 1. .7

$$2{\partial u \over \partial x} + {\partial u \over \partial t} = 0, \qquad -\infty < x < \infty, \hbox{ } t>0,$$

with the initial condition

$$u(x, 0) = {1\over 1 + x^{2}},$$

The characteristic curves are given by the solutions to

$${dx\over dr} = 2. \qquad \Rightarrow \qquad x = 2r + x_{0},$$

and

$${dt\over dr} = 1, \qquad \Rightarrow \qquad t = r.$$

Hence,

$$x = 2t + x_{0}, \qquad \Rightarrow \qquad x_{0} = x - 2t.$$

The PDE becomes

$${du\over dr} = 0, \qquad \Rightarrow \qquad u = u(x_{0}).$$

At $t=0$, $x=x_{0}$ and $u = {1\over 1 + x_{0}^{2}}$. Therefore,

$$u(x, t) = {1\over 1 + (x - 2t)^{2}}.$$

Example 1. .8

$$-{\partial u \over \partial x} + {\partial u \over \partial t} = 0, \qquad -\infty < x < \infty, \hbox{ } t>0,$$

with the initial condition

$$u(x, 0) = \left \{ \begin{array}{cc} 1 - \vert x\vert & \vert x\vert \le 1 \\ 0 & \vert x\vert > 1 \end{array} \right.$$

The characteristic curves are given by the solutions to

$${dx\over dr} = -1, \qquad \Rightarrow \qquad x = -r + x_{0},$$

and

$${dt\over dr} = 1, \qquad \Rightarrow \qquad t = r.$$

Thus, eliminating $r$ and re-writing $x_{0}$ in terms of $x$ and $t$, we obtain

$$x_{0} = x + t.$$

The PDE reduces to the ODE

$${du\over dr} = 0, \qquad \Rightarrow \qquad u = u(x_{0})$$

Using the initial condition, $t=0$, $x=x_{0}$, we have

$$u(x_{0}, 0) = \left \{ \begin{array}{cc} 1 - \vert x_{0}... ...\vert \le 1 \\ 0 & \vert x_{0}\vert > 1 \end{array} \right.$$

and so

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

The solution is shown in Figure 1.5.

Figure 1.5: In this case the wave speed is negative and the wave moves to the left.

From Figure 1.5 it is clear that the wave moves to the left. This allows us to summarise the results of the first order wave equation.

$$\begin{array}{cc} c > 0 & \Rightarrow \qquad \hbox{wave mo... ... \Rightarrow \qquad \hbox{wave moves to the left.} \end{array}$$