= Solution
Assume $0<\varepsilon<1$ and set
$$
a=1-\varepsilon,
\qquad
S_\varepsilon=\operatorname{artanh}(a)
=\frac12\log\frac{2-\varepsilon}{\varepsilon}.
$$
The <travel-time coordinate for a one-dimensional variable-speed wave equation>
$$
s=\operatorname{artanh}y
$$
sends the initial interval $I_\varepsilon=(-a,a)$ to $(-S_\varepsilon,S_\varepsilon)$. By the <characteristic curves for speed one minus y squared>, the two characteristic coordinates are $s-x$ and $s+x$. The <finite propagation speed> and uniqueness theorem for <hyperbolic partial differential equations> therefore give the maximal characteristic diamond
$$
\boxed{D_\varepsilon
=\left\{(x,y):|x|+|\operatorname{artanh}y|<S_\varepsilon\right\}.}
$$
Equivalently,
$$
D_\varepsilon
=\left\{(x,y):|x|<S_\varepsilon,
\ |y|<\tanh(S_\varepsilon-|x|)\right\}.
$$
In the $(x,s)$ plane this is a diamond with vertices $(0,\pm S_\varepsilon)$ and $(\pm S_\varepsilon,0)$; transforming back bends its four sides into the characteristic curves found in part 3(c). Beyond any one of those sides, a point's backward characteristics meet $x=0$ outside $I_\varepsilon$, where no <Cauchy data> were prescribed, so uniqueness cannot be extended farther.
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