= Solution
Expanding the equation gives
$$
u_{xx}-(1-y^2)^2u_{yy}+4y(1-y^2)u_y=0,
$$
so its <principal symbol> is
$$
p(y,\xi)=\xi_x^2-(1-y^2)^2\xi_y^2.
$$
If a <characteristic curve> is locally a graph $y=y(x)$, its conormal is proportional to $(-y',1)$. The characteristic equation is therefore
$$
(y')^2-(1-y^2)^2=0,
\qquad
\frac{dy}{dx}=\pm(1-y^2).
$$
On each region separated by $y=\pm1$, separation of variables gives
$$
\frac12\log\left|\frac{1+y}{1-y}\right|=\pm x+C.
$$
Thus all the characteristic curves are
$$
\boxed{
\begin{cases}
y=\tanh(\pm x+C),&|y|<1,\\
y=\coth(\pm x+C),&|y|>1,\\
y=1\text{ or }y=-1.&
\end{cases}}
$$
The inner curves approach the horizontal characteristics $y=\pm1$ as $x\to\pm\infty$; the outer hyperbolic-cotangent branches have vertical asymptotes and also approach $y=\pm1$. This describes the requested sketch.
The initial line $x=0$ has conormal $(1,0)$, and $p(y,1,0)=1$, so it is a <non-characteristic hypersurface> at every point. Since the coefficients and prescribed data are real analytic, the <Cauchy-Kovalevskaya theorem> gives a unique real analytic solution in a neighborhood of each point of $\{x=0\}$, hence in a neighborhood of that line.
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