Lemma: Lemma C: smooth uniqueness closes only a matched problem

Let $F_1,F_2$ be smooth scalar, vector, or tensor forcings on $M_T$. Suppose $u_1,u_2$ are smooth solutions of

with equal ordinary coordinate-time Cauchy data. Then

The statement holds componentwise over $\mathbb{C}$, without spatial decay or finite total energy.

For different operators, uniqueness of each separate problem does not by itself identify their solutions. A proposed common solution must satisfy both equations.

Proof. Equality of solutions implies equality of forcings by application of ${\square }_{+}$.

Conversely, if the forcings agree, the difference $z=u_1-u_2$ satisfies ${\square }_{+}z=0$ and zero data. It suffices to treat a real scalar component. Fix $0<t_{\ast }<T$, ${\mathbf{x}}_{\ast }\in {\mathbb{R}}^3$, and $\rho >0$. Set

These are compact integration regions, so $E(s)$ is finite. The wave equation gives

The boundary moves inward with speed one. Therefore

Zero initial position implies zero initial spatial gradient, and hence $E(0)=0$. Nonnegativity gives $E(s)=0$. Thus $z_t(s,{\mathbf{x}}_{\ast })=0$ for $0\le s\le t_{\ast }$, and

The point was arbitrary. Smoothness includes the endpoints. Real and imaginary parts, and then all components, give the stated extensions.

For different operators, the difference of two solutions need not satisfy one homogeneous equation. The preceding uniqueness argument cannot be applied until the equations have been matched or the candidate has been tested in both.