Data on one characteristic line do not determine a one-dimensional wave

Reading the general solution on the line $\xi =0$: if $u=F(\xi )+G(\eta )$, then $u(0,\eta )=F(0)+G(\eta )$, ${\mathrm{\partial }}_{x}u(0,\eta )=F^{\mathrm{\prime }}(0)+G^{\mathrm{\prime }}(\eta )$ and ${\mathrm{\partial }}_{t}u(0,\eta )=-c{F}^{\mathrm{\prime }}(0)+c{G}^{\mathrm{\prime }}(\eta )$. Thus the line data determine the function $G$ up to an additive constant and the number $F^{\mathrm{\prime }}(0)$; $F(0)$ retains the common additive-shift freedom, but leave the function $F$ away from $\xi =0$ completely free; by [F1] every $C^2$ solution near the line has this form.

The witness pair. Both $u_1=0+\sin (\eta )$ and $u_2={\xi }^3+\sin (\eta )$ are $C^2$ sums of a function of $\xi$ and a function of $\eta$, hence $C^2$ solutions by [F1]. On $\xi =0$ their traces agree: $u_1(0,\eta )=\sin \eta =u_2(0,\eta )$, ${\mathrm{\partial }}_x{u}_1=\cos \eta$ and ${\mathrm{\partial }}_x{u}_2=3{\xi }^2+\cos \eta$ coincide at $\xi =0$, and ${\mathrm{\partial }}_t{u}_1=c\cos \eta$ and ${\mathrm{\partial }}_t{u}_2=-3c{\xi }^2+c\cos \eta$ coincide there as well.

However $u_2-u_1={\xi }^3$ is nonzero for every $\xi \mathrm{\ne }0$, so the two solutions differ at points of every neighbourhood of the line $\xi =0$; hence data on the characteristic line do not determine the solution. This exhibits failure of uniqueness for these characteristic line data.

Therefore the displayed statement is refuted: the same values of $u,{\mathrm{\partial }}_{x}u,{\mathrm{\partial }}_{t}u$ on the single characteristic line are shared by two $C^2$ solutions that disagree on every neighbourhood of it.