IDOS Energy, Coherence, and Supporting Dynamics — Part 77

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$Y_R=(f_R,-L\nabla f_R/8)$. The first linear support residual cancels exactly. For $z_R=\sqrt{\mu_s}Y_R$, the remaining squared residuals give $\|\mathscr A_s z_R\|_2\le CR^{-3}$ on that annulus, whereas $R^3\int w'(W_0)f_R\,dx$ tends to a strictly positive constant. A physical $L^2$ solution would instead make this pairing at most $CR^{-3}\|Y_{\rm phys}\|_{L^2(R<|x|<2R)}=o(R^{-3})$. This contradiction excludes the common-phase partner. No other kernel source enters that support equation and can cancel it.

The Remaining Coupled Spatial Range Problem

Physical translation already has the exact boost partner

$$k_a:\quad Y=-D_sa,\quad\theta_k=-\xi_k\cdot a; \qquad z_{1,a}:\quad v=a,\quad\chi=-\Omega a\cdot x, \tag{N6}$$

with other entries zero. Its combined kinetic and static energy is $Q\Omega|a|^2/2$, so its complete inertia is $Q\Omega I$. Unused-frame rotations have zero corrector and inertia $\kappa_EQ$ per orthonormal rotation generator. Subtract these boosts before solving the remaining range equation; its translation parameter is then zero.

Let $\mathsf H$ be the actual radial Hessian, defined by $Q_{\rm rad}(u)=\mathbb E[u\mathsf Hu]/2$ on $\mathbb E[u]=0$. For $a_0=N/(d+1)$ and $\Delta_i u(k)=u_0+u_k-u_{e_i}-u_{k-e_i}$, it is fixed explicitly by

$$\begin{aligned} Q_{\rm rad}/\Omega={}&-4\mathbb E[u^2] +4\lambda\mathbb E[(u_0-u_k)^2] +4\lambda\mathbb E\!\sum_{i:k_i>0}(u_{k-e_i}-u_k)^2\\ &+4\lambda a_0^2\sum_{i\ne j\le d}\sum_{k:k_i>0} (u_k-u_{k-e_i+e_j})^2 +\lambda\eta\mathbb E\!\sum_{i:k_i>0}(\Delta_i u(k))^2. \end{aligned} \tag{N7}$$

The exchange term is an ordinary sum. The anchor identity implies $\mathsf H\ge8\Omega(\lambda-1)I$; no covariance or scalar substitute is made for its inverse. Define

$$T=\mu_sD_s^TD_s/w,\quad R(x)=\nu I-\varepsilon C_\xi-T(x),\quad R_\infty=\nu I-\varepsilon C_\xi.$$

The actual support gives $T\ge0$, $\|T(x)\|\le C/(1+|x|^2)$ and $R(x)\ge(223/424)\Omega I$. For example, with $y=x/L$, $t=|y|^2$, $S=1+t$,

$$T=\frac{C_s}{L^2}\left\{S^{-2}I+ [64S^{-2}-20S^{-3}+100tS^{-4}]yy^T\right\}.$$

With any allowed centered $\alpha_h$ and antisymmetric physical rotation $F$, put $\gamma_k(x)=\alpha_{h,k}+(F\xi_k)\cdot x$. The full reduced problem, retaining all ten flow components, is

$$\begin{aligned} \mathsf Hu-\frac\varepsilon4w^{-1}\nabla\cdot(w\nabla u) +2\kappa B^*L_0&=2\gamma,\\ -\kappa\nabla\cdot(wL_0)&=0,\\ -g w^{-1}\nabla\cdot(w\nabla v)+\kappa\nu L_0+R(x)v &=\varepsilon B\gamma,\\ L_0&=\nabla\chi+2Bu+\nu v. \end{aligned} \tag{N8}$$

Its unresolved sources were the $d$ relative registrations and $45-r_d$ active rotations. The following proof treats all of them and also the allowed constant nonaffine phases when $\eta=0$.

A Range Estimate Without A Gap

Set $X=(u,v)$, with its finite inner product given by $\mathbb E$ on $u$ and the ordinary Euclidean product on $v$, and define

$$PX=2Bu+\nu v,\quad D_g=\operatorname{diag}(\varepsilon I/4,gI),\quad M(x)=\operatorname{diag}(\mathsf H,R(x)),\quad M_\infty=\operatorname{diag}(\mathsf H,R_\infty),\quad V=M_\infty-M=\operatorname{diag}(0,T).$$

The symmetric form for (N8) is

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