 8

For the particular equation, write $D_x=x{\partial }_x$ and $D_y=y{\partial }_y$. Its left side is $(D_x^2-D_y^2)u$. Choose $\varphi =xy$ and $\psi =x/y$; these are smooth invertible coordinates in the positive quadrant, with inverse $x=\sqrt{\varphi \psi }$, $y=\sqrt{\varphi /\psi }$. In them $D_x=\varphi {\partial }_{\varphi }+\psi {\partial }_{\psi }$ and $D_y=\varphi {\partial }_{\varphi }-\psi {\partial }_{\psi }$. Subtraction of their squares leaves $4\varphi \psi u_{\varphi \psi }$, so the canonical equation is $u_{\varphi \psi }=0$. Integrating it gives $u=F(\varphi )+G(\psi )$.

Parameterize the data curve by $(x,y)=(s,s^2)$, $s\ge 1$. Its coordinate values are $(\varphi ,\psi )=(s^3,s^{-1})$, soDifferentiate the first identity along the curve to get $3s^2{F}^{\prime }(s^3)-s^{-2}{G}^{\prime }(s^{-1})=6s^5-s^{-2}$. Adding gives $F^{\prime }(s^3)=2s^3$, and then $G^{\prime }(s^{-1})=1$. Thus $F(\varphi )={\varphi }^2+C$ for $\varphi \ge 1$ and $G(\psi )=\psi -C$ for $0<\psi \le 1$. ThereforeThe coordinate rectangle gives exactly the regionIts boundary consists of $xy=1$ up to $(1,1)$ and the ray $y=x$ from that point onward. It is unbounded upward. The original data curve lies inside this region. The sketch displays both characteristic boundaries and that data arc.

To establish the exact extent of uniqueness, outside this region either $\varphi <1$ or $\psi >1$. A smooth function supported in that unprescribed coordinate range can be added to $F$ or $G$ without changing either datum on the arc. Hence the displayed global formula is one extension, but data alone do not uniquely determine values outside the shaded region. Boundary values are fixed by continuity. The following classification criteria are pointwise and assume a nonzero principal coefficient matrix.