# NS × X 積分 × 24/72 範式實戰
## Round 15 — Pure Continuous Dynamic p-Hodge Gauge / Gauge-Hessian Distortion Route

- 日期：2026-08-16
- 版本：v0.1
- 狀態：Proof-Route Experiment / Continuous-Only Dynamic Nonlinear-Gauge Branch
- 前一輪：`NS_X72_Round14_PureContinuous_CriticalOneForm_GaugeCurvature_v0.1_2026-08-16.md`
- canonical source：UTF-8 Markdown
- canonical math delimiters：inline `$...$`；display `$$...$$`

---

# 0. Round 14 handoff

令

$$
Q(t)=\mathfrak Q_3[u(t)]
=
\inf_q\|u+\nabla q\|_3,
$$

並令 unique optimal representative

$$
v=u+\nabla q,\qquad r=|v|,\qquad n=\frac v{|v|}.
$$

Euler–Lagrange gauge：

$$
\boxed{\operatorname{div}(|v|v)=0.}
$$

Round 14 得到：

$$
\boxed{
\frac13\frac d{dt}Q^3
+
\nu D
=
I_Q,
}
\tag{0.1}
$$

其中

$$
D
=
\mathfrak D_3(v)
=
\int r\left(|\nabla v|^2+|\nabla r|^2\right)dx,
$$

以及

$$
I_Q
=
\int r^3\kappa_Q\,dx,
\qquad
\kappa_Q
=
n^\top\nabla^2q\,n.
$$

本輪直接研究 nonlinear gauge 對 $\nabla^2q$ 的限制。

---

# 1. Gauge divergence identities

由

$$
\operatorname{div}(r^2n)=0
$$

得到

$$
\boxed{
\operatorname{div}n
=
-2\,n\cdot\nabla\log r.
}
\tag{1.1}
$$

又因

$$
v=rn,
$$

所以

$$
\boxed{
\operatorname{div}v
=
-
n\cdot\nabla r.
}
\tag{1.2}
$$

而

$$
\operatorname{div}v
=
\Delta q
$$

因 $\operatorname{div}u=0$，故

$$
\boxed{
\Delta q
=
-
n\cdot\nabla r.
}
\tag{1.3}
$$

---

# 2. Nonlinear elliptic trace relation

因

$$
n\cdot\nabla r
=
n^\top S_v n
=
n^\top S_u n
+
n^\top\nabla^2q\,n,
$$

所以

$$
\boxed{
\Delta q
+
\kappa_Q
+
n^\top S_u n
=
0.
}
\tag{2.1}
$$

等價於

$$
\boxed{
(I+n\otimes n):\nabla^2q
=
-
n^\top S_u n.
}
\tag{2.2}
$$

在 $r>0$ 區域，$I+n\otimes n$ 的 eigenvalues 為 $1,1,2$。

---

# 3. Curvature-payment dichotomy

令

$$
P_\perp=I-n\otimes n,
$$

$$
\tau_\perp
=
\operatorname{tr}(P_\perp\nabla^2q),
$$

以及

$$
\gamma_Q
=
-
n^\top S_un.
$$

因

$$
\Delta q=\kappa_Q+\tau_\perp,
$$

(2.1) 給

$$
\boxed{
2\kappa_Q+\tau_\perp
=
\gamma_Q.
}
\tag{3.1}
$$

因此

$$
\boxed{
\kappa_Q^+
\le
\frac12
\left[
\gamma_Q^+
+
(-\tau_\perp)^+
\right].
}
\tag{3.2}
$$

所以 positive longitudinal gauge curvature 必須由：

- physical compression；
- transverse gauge concavity；

至少其中之一支付。

---

# 4. Weighted trace cancellation

由 (1.3)：

$$
\int r^3\Delta q\,dx
=
-
\int r^3n\cdot\nabla r\,dx.
$$

但

$$
r^2n=rv
$$

divergence-free，且

$$
r^3n\cdot\nabla r
=
(r^2n)\cdot\nabla\left(\frac12r^2\right).
$$

因此

$$
\boxed{
\int r^3\Delta q\,dx=0.
}
\tag{4.1}
$$

---

# 5. Only deviatoric gauge curvature drives critical growth

定義

$$
H_q^0
=
\nabla^2q-\frac13(\Delta q)I.
$$

則由 (4.1)：

$$
\boxed{
I_Q
=
\int
r^3
n^\top H_q^0n\,dx.
}
\tag{5.1}
$$

所以 Round 14 identity sharpen 成

$$
\boxed{
\frac13\frac d{dt}Q^3
+
\nu D
=
\int
r^3
n^\top H_q^0n\,dx.
}
\tag{5.2}
$$

因此：

$$
\boxed{
\textbf{isotropic optimal-gauge curvature is globally invisible to }Q^3\textbf{ growth}.
}
$$

危險部分是 anisotropic / deviatoric curvature。

---

# 6. Nonlinear-Hodge metric

map

$$
J(v)=|v|v
$$

的 Jacobian為

$$
\boxed{
M_v
=
|v|(I+n\otimes n).
}
\tag{6.1}
$$

對任意 $\xi$：

$$
r|\xi|^2
\le
\xi^\top M_v\xi
\le
2r|\xi|^2.
$$

---

# 7. Differentiate the gauge

由

$$
\operatorname{div}J(v)=0
$$

對 $x_\ell$ 微分：

$$
\boxed{
\operatorname{div}
\left(
M_v\partial_\ell v
\right)=0.
}
\tag{7.1}
$$

又

$$
\partial_\ell v
=
\partial_\ell u
+
\nabla\partial_\ell q.
$$

令 $q_\ell=\partial_\ell q$，測試 (7.1) 得：

$$
\boxed{
\int
\nabla q_\ell
\cdot
M_v
\partial_\ell v
\,dx
=
0.
}
\tag{7.2}
$$

所以 gauge-Hessian derivative 與 full optimal-representative derivative 在 nonlinear-Hodge metric 中精確正交。

---

# 8. Nonlinear Hodge Gradient Pythagorean Identity

定義

$$
\boxed{
H
=
\mathcal H_Q
=
\sum_{\ell=1}^3
\int
\nabla q_\ell\cdot
M_v
\nabla q_\ell\,dx.
}
\tag{8.1}
$$

展開：

$$
\boxed{
H
=
\int
r
\left(
|\nabla^2q|^2
+
|\nabla^2q\,n|^2
\right)dx.
}
\tag{8.2}
$$

另一方面

$$
\boxed{
D
=
\sum_\ell
\int
\partial_\ell v\cdot
M_v
\partial_\ell v\,dx.
}
\tag{8.3}
$$

利用

$$
\partial_\ell u
=
\partial_\ell v
-
\nabla q_\ell
$$

與 (7.2)，得到

$$
\boxed{
\mathcal E_U^{(M)}
=
D+H,
}
\tag{8.4}
$$

其中

$$
\boxed{
\mathcal E_U^{(M)}
=
\sum_\ell
\int
\partial_\ell u
\cdot
M_v
\partial_\ell u\,dx.
}
$$

命名：

$$
\boxed{
\textbf{Nonlinear Hodge Gradient Pythagorean Identity}.
}
$$

意義：optimal gauge curvature 具有一個 exact nonnegative distortion energy $H$。

---

# 9. Gauge curvature has an exact weighted cost

由

$$
|\kappa_Q|
\le
|\nabla^2q\,n|
$$

得到

$$
\boxed{
\int
r|\kappa_Q|^2dx
\le
H.
}
\tag{9.1}
$$

因此 positive gauge curvature不能在 weighted $L^2$ 層級免費形成。

---

# 10. Growth bound by gauge distortion

Cauchy–Schwarz：

$$
|I_Q|
\le
H^{1/2}
\left(
\int r^5dx
\right)^{1/2}.
$$

Interpolation：

$$
\|v\|_5
\le
\|v\|_3^{2/5}
\|v\|_9^{3/5}
$$

給

$$
\|v\|_5^{5/2}
\le
Q\,
\|v\|_9^{3/2}.
$$

又令

$$
W=r^{3/2}.
$$

由

$$
D
\ge
\frac49\|\nabla W\|_2^2
$$

與 Sobolev：

$$
\|W\|_6^2
\le
C\|\nabla W\|_2^2
$$

得到

$$
\boxed{
\|v\|_9^3
\le
C D.
}
\tag{10.1}
$$

所以

$$
\boxed{
|I_Q|
\le
C
Q
H^{1/2}
D^{1/2}.
}
\tag{10.2}
$$

---

# 11. Dimensionless gauge-distortion ratio

定義

$$
\boxed{
\Xi_Q
=
\frac{
Q^2H
}{
\nu^2D
}
}
\tag{11.1}
$$

當 $D>0$。

由 (10.2)：

$$
\boxed{
|I_Q|
\le
C\nu D\sqrt{\Xi_Q}.
}
\tag{11.2}
$$

所以若

$$
\Xi_Q<C^{-2},
$$

則

$$
\frac d{dt}Q^3<0.
$$

反過來：

$$
\boxed{
\frac d{dt}Q^3>0
\Longrightarrow
\Xi_Q>C^{-2}.
}
\tag{11.3}
$$

命名：

$$
\boxed{
\textbf{Gauge-Distortion Necessity}.
}
$$

critical quotient norm要上升，gauge-Hessian distortion相對 quotient dissipation必須跨過一個 dimensionless threshold。

---

# 12. Young-form inequality

由 (10.2)：

$$
C QH^{1/2}D^{1/2}
\le
\frac\nu2D
+
\frac{C_\ast}{\nu}Q^2H.
$$

因此

$$
\boxed{
\frac13\frac d{dt}Q^3
+
\frac\nu2D
\le
\frac{C_\ast}{\nu}Q^2H.
}
\tag{12.1}
$$

所以真正缺的只剩：

$$
\boxed{
H
\stackrel{?}{\lesssim}
\frac{\nu^2}{Q^2}D
}
\tag{12.2}
$$

或一個足夠可積分的 weaker replacement。

---

# 13. Dynamic gauge-maintenance equation

若 $v(t)$ 始終保持 nonlinear optimal gauge，則代表 equation：

$$
v_t
=
\nu\Delta v
-
\mathcal L_u^{(1)}v
+
\nabla\chi.
$$

時間微分

$$
\operatorname{div}J(v)=0
$$

得到

$$
\boxed{
\operatorname{div}(M_v\nabla\chi)
=
\operatorname{div}
\left[
M_v
\left(
\mathcal L_u^{(1)}v-\nu\Delta v
\right)
\right].
}
\tag{13.1}
$$

以 $\chi$ 測試：

$$
\boxed{
\int
\nabla\chi\cdot M_v\nabla\chi
\le
\int
F\cdot M_vF,
}
\tag{13.2}
$$

其中

$$
F
=
\mathcal L_u^{(1)}v-\nu\Delta v.
$$

所以維持 optimal gauge本身也需要一個 continuous weighted elliptic feedback。

---

# 14. Why the standard weighted shortcut is unavailable for free

本輪 natural scalar weight是

$$
\boxed{
w=|v|.
}
$$

而

$$
M_v
\simeq
wI
$$

只在 weighted sense elliptic。

標準 degenerate-elliptic Calderón–Zygmund / Kato 類理論通常需要對 weight class（例如 Muckenhoupt $A_2$）有控制。

但 nonlinear gauge

$$
\operatorname{div}(|v|v)=0
$$

本身不推出 uniform $A_2$ control。

---

# 15. Smooth gauge witness with non-$A_2$ natural weight

令

$$
\rho=\sqrt{x^2+y^2}
$$

並取 smooth axisymmetric swirl

$$
\boxed{
v
=
\eta(\rho,z)
\rho^{2k}
(-y,x,0),
\qquad
k\ge1,
}
\tag{15.1}
$$

其中 $\eta$ smooth，且在軸附近 $\eta=1$。

此 field純 azimuthal且無 $\theta$ dependence，因此

$$
\operatorname{div}v=0
$$

及

$$
\boxed{
\operatorname{div}(|v|v)=0.
}
\tag{15.2}
$$

但軸附近：

$$
|v|
\sim
\rho^{2k+1}.
$$

因此：

$$
|v|^{-1}
\sim
\rho^{-(2k+1)}.
$$

transverse measure為 $\rho\,d\rho\,d\theta$，故

$$
\int_0^\varepsilon
|v|^{-1}\rho\,d\rho
\sim
\int_0^\varepsilon
\rho^{-2k}d\rho
=
\infty.
$$

所以 $|v|^{-1}$ 甚至不 local integrable near the axis，從而

$$
\boxed{
|v|\notin A_2.
}
\tag{15.3}
$$

因此：

$$
\boxed{
\textbf{critical nonlinear gauge does not imply }A_2\textbf{ regularity of the natural weight.}
}
$$

這不是 NS singularity example；它只是 nonlinear-gauge structural witness。

---

# 16. What has been learned

Round 14 的 Boss 是：

$$
\text{positive optimal gauge curvature}.
$$

Round 15 把它縮成：

1. isotropic Hessian trace globally cancels；
2. dangerous part必須是 anisotropic；
3. anisotropic curvature必須支付 gauge-Hessian energy $H$；
4. positive critical growth requires
   $$
   \Xi_Q\gtrsim1;
   $$
5. 尚未證明
   $$
   H
   \lesssim
   Q^{-2}\nu^2D.
   $$

所以真正 frontier：

$$
\boxed{
\textbf{weighted nonlinear-Hodge distortion versus quotient dissipation}.
}
$$

---

# 17. STOP-C19

$$
\boxed{
\textbf{STOP-C19:
Weighted Gauge-Hessian / Quotient-Dissipation Gap}.
}
$$

其 diagnostic：

$$
\boxed{
\begin{aligned}
\text{critical carrier}&=Q,
\\
\text{quotient dissipation}&=D,
\\
\text{gauge distortion}&=H,
\\
\text{exact decomposition}&=\mathcal E_U^{(M)}=D+H,
\\
\text{growth necessity}&=\Xi_Q\gtrsim1,
\\
\text{standard weighted shortcut}&=\text{not automatic},
\\
T_{\mathsf C\to\mathsf D}&=\text{NOT REACHED}.
\end{aligned}
}
$$

---

# 18. 24/72 Ledger — Round 15

| Step | object | $B$ | $U$ | $O$ | $L$ | status |
|---|---|---|---|---|---|---|
| C166 | gauge divergence identities | $\mathsf C$ | differential | relational | $\mathsf F$ | EXACT |
| C167 | nonlinear elliptic trace relation | $\mathsf C$ | elliptic | relational | $\mathsf F$ | EXACT |
| C168 | curvature-payment dichotomy | $\mathsf C$ | geometry | targeted | $\mathsf F$ | PROVED |
| C169 | weighted trace cancellation | $\mathsf C$ | global pairing | scalar | $\mathsf F$ | EXACT |
| C170 | deviatoric-curvature reduction | $\mathsf C$ | geometry | targeted | $\mathsf F$ | EXACT |
| C171 | nonlinear-Hodge metric $M_v$ | $\mathsf C$ | variational | $\mathsf X$ | $\mathsf F$ | FORM |
| C172 | differentiated gauge | $\mathsf C$ | elliptic | relational | $\mathsf F$ | EXACT |
| C173 | weighted orthogonality | $\mathsf C$ | variational | relational | $\mathsf F$ | EXACT |
| C174 | nonlinear Hodge Pythagorean identity | $\mathsf C$ | geometric | $\mathsf X$ | $\mathsf F$ | EXACT |
| C175 | distortion growth bound | $\mathsf C$ | interpolation | scalar | $\mathsf F$ | PROVED |
| C176 | $\Xi_Q$ | $\mathsf C$ | recognition | scalar | $\mathsf F$ | FORM |
| C177 | positive growth $\Rightarrow\Xi_Q\gtrsim1$ | $\mathsf C$ | necessity | scalar | $\mathsf F$ | PROVED |
| C178 | dynamic gauge-maintenance PDE | $\mathsf C$ | weighted elliptic | $\mathsf X$ | $\mathsf F$ | EXACT |
| C179 | gauge $\Rightarrow A_2$ | $\mathsf C$ | weighted geometry | scalar | $\mathsf F$ | REFUTED |
| C180 | $H\lesssim Q^{-2}\nu^2D$ | $\mathsf C$ | weighted nonlinear-Hodge | targeted | $\mathsf F$ | OPEN |

---

# 19. Continuous-versus-discrete status

本輪進入 degenerate weighted elliptic geometry後，仍全部是 continuous field / variational / elliptic structure。

因此：

$$
\boxed{
T_{\mathsf C\to\mathsf D}
=
\text{NOT YET REACHED}.
}
$$

---

# 20. Next round — Distortion Feedback / Continuous Level-Set Route

下一輪直接追：

$$
\boxed{
\Xi_Q
=
\frac{Q^2H}{\nu^2D}.
}
$$

主問題：

1. $H$ 增大是否必迫使 physical weighted gradient energy
   $$
   \mathcal E_U^{(M)}
   $$
   增大；
2. NS energy/enstrophy能否控制這個 weighted quantity；
3. 若沒有 uniform $A_2$，能否用 PDE-specific cancellation；
4. 若需分解 weight，先使用 continuous layer-cake：
   $$
   |v|
   =
   \int_0^\infty
   \mathbf 1_{\{|v|>\lambda\}}d\lambda;
   $$
5. 即使進入 level sets，$\lambda$ 仍是 continuous parameter，因此仍不提前宣告 $\mathsf C\to\mathsf D$。

---

# 21. External primary-source anchors

1. Thomas H. Otway, *Nonlinear Hodge maps*, arXiv:math-ph/9908030.
   - nonlinear Hodge variational systems與 nonuniform ellipticity背景。

2. Tadele Mengesha, Tuoc Phan, *Weighted $W^{1,p}$- estimates for weak solutions of degenerate elliptic equations with coefficients degenerate in one variable*, arXiv:1612.07371.
   - $A_2$-weighted degenerate elliptic Calderón–Zygmund-type estimates背景。

3. Pascal Auscher, Li Chen, José María Martell, Cruz Prisuelos-Arribas, *The regularity problem for degenerate elliptic operators in weighted spaces*, arXiv:2106.14422.
   - degenerate elliptic operators與 Muckenhoupt weighted framework背景。

本輪 weighted trace cancellation、Pythagorean identity、distortion ratio與 non-$A_2$ smooth gauge witness均為本文直接推導。

---

# 22. Commit state

$$
\boxed{
\begin{aligned}
\text{Route}
&=
\mathrm{Pure\ Continuous\ Dynamic\ }p\mathrm{\text{-}Hodge\ Gauge},
\\
\text{Essential }\mathsf C\to\mathsf D
&=
\mathrm{Not\ reached},
\\
\text{Dangerous curvature}
&=
\mathrm{deviatoric},
\\
\text{Optimal metric}
&=
M_v=|v|(I+n\otimes n),
\\
\text{Gauge-Hessian energy}
&=
H,
\\
\text{Exact decomposition}
&=
\mathcal E_U^{(M)}=D+H,
\\
\text{Growth necessity}
&=
\Xi_Q\gtrsim1,
\\
\text{Automatic }A_2
&=
\mathrm{false},
\\
\text{STOP-C19}
&=
\mathrm{Weighted\ Gauge\text{-}Hessian/Quotient\text{-}Dissipation\ Gap},
\\
\text{Next}
&=
\mathrm{Distortion\ Feedback/Continuous\ Level\text{-}Set}.
\end{aligned}
}
$$