= Solution
For a defining function $\phi$ with $d\phi\ne0$, the <characteristic hypersurface> test is that the <principal symbol> vanish at $d\phi$.
For the <wave equation> with speed $c>0$,
$$
u_{tt}-c^2\Delta_xu=0,\qquad p(\tau,\xi)=\tau^2-c^2|\xi|^2.
$$
Thus its <characteristic hypersurfaces> satisfy $\phi_t^2=c^2|\nabla_x\phi|^2$. In one space dimension the two families are $x\pm ct=\text{constant}$; cones are characteristic away from their vertices.
For the <free Schrodinger equation>, in normalized units,
$$
i u_t+\Delta_xu=0,\qquad p(\tau,\xi)=|\xi|^2.
$$
Its total-order <characteristic hypersurfaces> satisfy $\nabla_x\phi=0$. Their normal is purely temporal, so locally they are constant-time hypersurfaces. Multiplying the equation by a nonzero constant or choosing the opposite sign convention does not change this test.
For the <Laplace equation>,
$$
\Delta_xu=0,\qquad p(\xi)=|\xi|^2.
$$
\b[There are no real <characteristic hypersurfaces> for the <Laplace equation>], since their normal cannot be zero. These statements concern the ordinary total-order <principal symbol>, not a weighted space-time grading.
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