Theorem: Hadamard data do not ensure a decaying rod composition

The class ${\mathfrak{W}}_{\mathcal{B}}$ contains pairs of representatives for which the one-sided decaying rod composition cannot be simultaneously defined. Consequently the XX-B algebraic and microlocal conditions do not imply the Q2 spatial-tail requirement.

Proof. Fix $t_0\in I$. Because all coefficients of $\mathcal{B}$ depend only on time in a spatially homogeneous frame, the space of spatial polynomials of degree at most one is preserved by $\mathcal{L}$. The equation reduces on that space to a finite system of smooth ordinary differential equations.

Choose a real solution $U_{\mathrm{aff}}$ with Cauchy data

and any fixed smooth polynomial time-derivative data, for example zero coordinate derivatives. This gives a smooth global solution on $I$; no spatial decay is asserted.

At $t_0$,

The first rod equation in component $a=1$ would therefore require

There is no smooth solution tending to zero at spatial infinity. Its spherical mean satisfies

as follows by averaging the Poisson equation over spheres.

Choose a compact test $f$ with $(U_{\mathrm{aff}},f{)}_0\ne 0$. Both

satisfy the same Hadamard, field-equation, adjoint, and canonical conditions.

If both had linear decaying rod compositions after testing with $f$, their difference would supply a decaying solution of (eq:constantPoisson) times the nonzero constant $(U_{\mathrm{aff}},f{)}_0$. This is impossible. Thus at least one of the pair is outside that decaying composition domain.