Lemma: Smooth zero-data uniqueness without spatial decay

Let $\mu \ge 0$ and let $z$ be a real smooth function on $M_T$. If

then $z=0$.

The same conclusion holds componentwise for complex functions, vectors, and tensors. Consequently, a smooth solution of $LZ=F$ with prescribed initial data is unique whenever it exists. No spatial decay or finite total energy is required.

Proof. Fix $0<t_0<T$, ${\mathbf{x}}_0\in {\mathbb{R}}^3$, and $\rho >0$. For $0\le s\le t_0$, let

and define the local energy

Every integration region lies in a fixed compact set. Thus $\mathcal{E}(s)$ is finite even if the total energy on ${\mathbb{R}}^3$ is infinite.

Writing $e=({z_t}^2+|\nabla z{|}^2+{\mu }^2{z}^2)/2$, the equation gives

Since the boundary of $B_s$ moves inward at speed one, the divergence theorem and differentiation of the moving region give

At each boundary point,

Hence ${\mathcal{E}}^{\prime }(s)\le 0$.

The zero initial data imply $\mathcal{E}(0)=0$: the initial spatial gradient also vanishes because $z(0,\cdot )$ is identically zero. Nonnegativity gives $\mathcal{E}(s)=0$ for every $s\le t_0$. Smoothness then implies $z_t=0$ throughout each $B_s$. In particular $z_t(s,{\mathbf{x}}_0)=0$ for $0\le s\le t_0$, so

The point was arbitrary. Continuity includes $t=T$.

Real and imaginary parts satisfy the same real equation. The argument therefore applies componentwise.

If $LZ=0$, then $\square (\mathcal{P}Z)=0$. The tensor $\mathcal{P}Z$ has zero initial data whenever $Z$ does. The componentwise result gives $\mathcal{P}Z=0$, and ${\mathcal{P}}^2=\mathrm{id}$ gives $Z=0$. Apply this to the difference of two solutions.