Theorem: Pure second-normal coefficient

The pure $u^2$ coefficient is

The $K_8$ contribution is zero.

Proof. In Cartesian response coordinates $(c,s,r,y)$, all row weights are independent of $r$. The subsequent coordinate change does not mix $r$ with other coordinates. Residual order zero therefore uses only source terms with first source multiindex zero.

Let ${\mathcal{Q}}^{\mu \nu }({\partial }_s)={\mathsf{Q}}^{I,J}(0,{\partial }_s{)}^{\mu \nu }$. Its divergence in either output index is zero. The residual-order-zero $K_8$ part is

The derivative-root part reduces to

Since $\mathsf{b}(\mathsf{P}Q)=-{\mathrm{tr}}_{\eta }Q/2$, the mass-root part is

The density part is

All remaining propagation expressions have order at most two. Their pure second-normal symbol, times $\mathfrak{m}$, is (eq:c02).