Proposition: Same-operator infrared non-determination

The conditions (eq:W) alone do not ensure a decaying rod composition after compact testing.

Proof. In a spatially homogeneous frame all coefficients of $\mathcal{L}$ depend only on time. Spatial polynomials of degree at most one are preserved by the operator, reducing the Cauchy problem on that subspace to smooth finite-dimensional ordinary differential equations.

Choose a real solution $U$ with data at $t_0\in I$

and any fixed polynomial first time-derivative data. Then $\mathcal{B}U=0$, and

A decaying first rod in component $a=1$ would solve

Its spherical mean is

which cannot tend to zero.

Both $W$ and $W+U\otimes U$ satisfy (eq:W): their difference is a smooth real symmetric bisolution. Choose compact $g$ with $(U,g{)}_0\ne 0$. If both compact-tested covariances had linear decaying rod compositions, subtracting them would yield a decaying rod for the displayed multiple of $U$, a contradiction. At least one representative lies outside that composition domain.