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Near a non-characteristic plane $x=0$, use the small cap ${\omega }_{\epsilon }=\{x>0,x+y^2<\epsilon \}$. Its curved face is non-characteristic by continuity of the principal symbol. A polynomial coordinate change flattens that face. Uniform real analytic coefficient bounds and the uniform Cauchy radius for polynomial forcing solve $P^{\ast }v=g$ throughout a common cap, with zero boundary jet on the curved face. The complete boundary-jet condition for formal adjoints then gives ${\int }_{{\omega }_{\epsilon }}ug=0$ for every polynomial $g$. Uniform approximation of $\bar{u}$ gives $\int |u{|}^2=0$. Repeating on the other side of the plane proves two-sided local uniqueness.

Fix the coefficient normal form and a positive analyticity exponent ${\zeta }_0$ once and for all. If $g$ is a polynomial, its pullback ${\tilde{g}}_{\epsilon }=g(\tilde{x}-{\tilde{y}}^2+\epsilon ,\tilde{y})$ is again a polynomial, with degree bounded independently of $\epsilon$. On a fixed closed coordinate ball, its finitely many nonzero derivatives are uniformly bounded as the parameter ranges over a fixed compact small interval. Hence there is a finite $A_g\ge 1$ such thatAll higher derivatives vanish, so arbitrarily high polynomial degree causes no failure of this bound; it only changes $A_g$.

The normalized forcing belongs to one common majorant class. The Cauchy-Kovalevskaya theorem therefore gives a radius $r_{\ast }>0$ independent of $g$. The problem is linear with zero data, so multiplying the normalized solution by $A_g$ solves the original forcing without changing its radius. Choose one $\epsilon$ with the cap inside this radius. ThusThis is the uniform Cauchy radius for polynomial forcing. It asserts a common domain, not a uniform solution-amplitude bound over all polynomials.