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

Initial-value Problems

Here the problem is defined over an infinite range in $x$. The general statement of the problem is
$\displaystyle c{\partial u \over \partial x} + {\partial u \over \partial t}$ $\textstyle =$ $\displaystyle 0, \qquad -\infty < x < \infty, \hbox{ } t>0.$ (1.11)
$\displaystyle u(x,0)$ $\textstyle =$ $\displaystyle f(x).$ (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

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

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.

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

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

Example 1. .7

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

with the initial condition

\begin{displaymath}
u(x, 0) = {1\over 1 + x^{2}},
\end{displaymath}

The characteristic curves are given by the solutions to

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

and

\begin{displaymath}
{dt\over dr} = 1, \qquad \Rightarrow \qquad t = r.
\end{displaymath}

Hence,

\begin{displaymath}
x = 2t + x_{0}, \qquad \Rightarrow \qquad x_{0} = x - 2t.
\end{displaymath}

The PDE becomes

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

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

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

Example 1. .8

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

with the initial condition

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

The characteristic curves are given by the solutions to

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

and

\begin{displaymath}
{dt\over dr} = 1, \qquad \Rightarrow \qquad t = r.
\end{displaymath}

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

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

The PDE reduces to the ODE

\begin{displaymath}
{du\over dr} = 0, \qquad \Rightarrow \qquad u = u(x_{0})
\end{displaymath}

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

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

and so

\begin{displaymath}
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.
\end{displaymath}

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.
\includegraphics [scale=0.7]{figpde5.ps}

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


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