The estimate at the free edge

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The lesson on the differentiated height equation led to a linear equation $u_{tt}+A\,u_{ss}+B\,u_t+C\,u_s=f$ on a rectangle, in which the coefficient $A$ changes its sign: the equation is elliptic where $A>0$ and hyperbolic where $A<0$. This lesson proves an estimate for the first derivatives of $u$ by the mean squares of $u$ and $f$, on a rectangle with three edges in the elliptic region and a fourth edge, the free edge, at which nothing is assumed: $A$ may be negative there, and no condition on $u$ is imposed. What makes this possible is the structure of $C$ found in that lesson, $C=(l+1)A_s$ up to a multiple of $A$, together with a sign condition on the derivatives of $A$ where $A$ is small.

The identity obtained by multiplying the equation by $m\,u_s+n\,u_t$ and integrating by parts (Lemma 2.1).
With a weight that vanishes at the free edge, the resulting quadratic form is positive where $A$ is negative or small (Proposition 3.1).
Where $A$ is positive, an elliptic estimate with a weaker weight (Proposition 4.1).
The estimate (Theorem 5.1).

The lesson follows [OpenAI-I, Section 3, the lemma on the free cap edge], formulated for a general equation with these properties; all constants are tracked. It uses only integration by parts on a rectangle and the inequality $2xy\le\delta x^2+y^2/\delta$.

1. The assumptions

Let $T>0$, $s_0<b$, and $\Omega=[-T,T]\times[s_0,b]$ in the plane with the coordinates $t,s$. Let $d=b-s$, the distance to the edge $s=b$, and $d_{\max}=b-s_0$. Let $A,B,C,f,u$ be smooth functions on a neighbourhood of $\Omega$ with $u_{tt}+A\,u_{ss}+B\,u_t+C\,u_s=f .$ We assume, with numbers $l\ge0$, $M_0\ge1$, $0<T_1<T$, $s_0<s_1<b$, $\alpha_0>0$, $0<\varepsilon\le1$, $a_0>0$ and $\theta_0>0$:

(A1) $C=(l+1)A_s+A\omega$ with a continuous function $\omega$, and $|A|,|A_t|,|A_s|,|B|,|B_s|,|\omega|\le M_0$ on $\Omega$.

(A2) $A\ge\alpha_0$ at the points of $\Omega$ with $|t|\ge T_1$ or $s\le s_1$.

(A3) $16\,\varepsilon^2T^2M_0\le1$, and at the points of $\Omega$ with $|A|\le a_0$, $\Theta:=-(l+\tfrac12)A_s+\tfrac\varepsilon2\,t\,A_t\ \ge\ \theta_0 .$

Assumption (A2) says that the equation is elliptic near the edges $t=\pm T$ and $s=s_0$. Nothing is assumed at the edge $s=b$. Assumption (A3) concerns the curve where $A$ changes its sign: across it, $A$ decreases in the direction of increasing $s$, or it increases in the direction away from $t=0$.

Let $\Omega_0=[-T_1,T_1]\times[s_1,b_0]$ with $s_1<b_0<b$. We write $\|w\|^2=\iint_\Omega w^2\,dt\,ds$ and $|\nabla u|^2=u_t^2+u_s^2$.

2. The multiplier identity

Lemma 2.1. Let $m,n$ be smooth functions on a neighbourhood of $\Omega$ that vanish on the edge $s=b$ and in a neighbourhood of the other three edges. Then $\begin{gathered} \iint_\Omega f\,(m\,u_s+n\,u_t)=\iint_\Omega Q[u],\\ Q[u]=Q_{tt}u_t^2+Q_{ss}u_s^2+Q_{ts}u_tu_s, \end{gathered}$ where $\begin{aligned} Q_{tt}&=\tfrac12(m_s-n_t)+Bn,\\ Q_{ss}&=-\tfrac12(Am)_s+\tfrac12(An)_t+Cm,\\ Q_{ts}&=-m_t+Bm-(An)_s+Cn . \end{aligned}$

Proof. The product $(u_{tt}+Au_{ss}+Bu_t+Cu_s)(mu_s+nu_t)$ is equal to $\begin{aligned} &Q_{tt}u_t^2+Q_{ss}u_s^2+Q_{ts}u_tu_s\\ &\quad+\partial_t\bigl(m\,u_tu_s+\tfrac12n\,u_t^2-\tfrac12An\,u_s^2\bigr)\\ &\quad+\partial_s\bigl(-\tfrac12m\,u_t^2+\tfrac12Am\,u_s^2+An\,u_su_t\bigr). \end{aligned}$ This is checked by carrying out the differentiations: for example $\begin{aligned} u_{tt}\,m\,u_s&=\partial_t(mu_tu_s)-m_tu_tu_s\\ &\quad-\partial_s(\tfrac12mu_t^2)+\tfrac12m_su_t^2,\\ Au_{ss}\,n\,u_t&=\partial_s(Anu_su_t)-(An)_su_su_t\\ &\quad-\partial_t(\tfrac12Anu_s^2)+\tfrac12(An)_tu_s^2, \end{aligned}$ and similarly for the other products. Integrate over $\Omega$. The integrals of the two derivatives are integrals over the edges of expressions that contain $m$ or $n$ as a factor, and these vanish on all four edges. $\blacksquare$

3. Positivity where $A$ is negative or small

Fix $p=8$. Let $\lambda>0$, to be chosen, and $\begin{gathered} I(t,s)=\int_0^tB(r,s)\,dr,\\ W=d^{\,p}\,e^{-\lambda s+I},\qquad H_0=\lambda+\frac pd . \end{gathered}$ Then $W_t=BW$ and $W_s=-(H_0-I_s)W$, and $|I_s|\le TM_0$. Let $Q^W_{tt},Q^W_{ss},Q^W_{ts}$ be the functions of Lemma 2.1 for $m=-W$ and $n=\varepsilon tW$. A computation gives $\begin{aligned} Q^W_{tt}/W&=\tfrac12(H_0-I_s)-\tfrac\varepsilon2+\tfrac\varepsilon2tB,\\ Q^W_{ss}/W&=\Theta-\tfrac12H_0A+A\beta,\\ Q^W_{ts}/W&=\varepsilon t\,\bigl(lA_s+A\omega-AI_s+H_0A\bigr), \end{aligned}$ with $\beta=\tfrac12I_s+\tfrac\varepsilon2(1+tB)-\omega$. Indeed $-m_t+Bm=W_t-BW=0$; $-\tfrac12(Am)_s=\tfrac12A_sW-\tfrac12A(H_0-I_s)W$; $\tfrac12(An)_t=\tfrac\varepsilon2(tA_t+A+tAB)W$; and $Cm=-(l+1)A_sW-A\omega W$. The term $-(l+1)A_s$ of $C$, together with $\tfrac12A_s$, gives the term $-(l+\tfrac12)A_s$ of $\Theta$. Put $\begin{gathered} \beta_0=M_0(T+1)+1,\\ c_J=M_0(l+M_0+TM_0),\\ \Theta_1=(l+1+T)M_0, \end{gathered}$ so that $|\beta|\le\beta_0$, $|lA_s+A\omega-AI_s|\le c_J$ and $|\Theta|\le\Theta_1$.

Proposition 3.1. Assume (A1) and (A3), and let $\alpha_0>0$ be the number of (A2). Let $\begin{gathered} \lambda=\max\{\lambda_1,\lambda_2,\lambda_3,\lambda_4\},\\ \lambda_1=2+4TM_0,\qquad\lambda_2=8\beta_0,\\ \lambda_3=\frac{16T^2c_J^2}{\theta_0},\qquad\lambda_4=\frac{8(\Theta_1+\theta_0)}{a_0}, \end{gathered}$ and $\begin{gathered} c_1=\min\{\gamma_1,\gamma_2,\gamma_3\},\\ \gamma_1=\frac{a_0}{d_{\max}},\qquad\gamma_2=\frac{\alpha_0}{d_{\max}},\\ \gamma_3=\frac{\theta_0}{4(\lambda d_{\max}+p+d_{\max}\beta_0)}. \end{gathered}$ Then at every point of $\Omega$ with $s<b$ and $A\le c_1d$, and for all real $\xi,\eta$, $\begin{aligned} &Q^W_{tt}\xi^2+Q^W_{ss}\eta^2+Q^W_{ts}\xi\eta\\ &\quad\ge c_2W(\xi^2+\eta^2),\\ &c_2=\min\bigl\{\tfrac\lambda8,\tfrac{\theta_0}2\bigr\}. \end{aligned}$

Proof. Since $H_0\ge\lambda\ge2+4TM_0$, we have $Q^W_{tt}/W\ge\tfrac12H_0-\tfrac12(1+2TM_0)\ge\tfrac14H_0$. By $|Q^W_{ts}\xi\eta|\le\tfrac12Q^W_{tt}\xi^2+(Q^W_{ts})^2\eta^2/(2Q^W_{tt})$, the quadratic form is at least $\tfrac12Q^W_{tt}\,\xi^2+\Bigl(Q^W_{ss}-\frac{(Q^W_{ts})^2}{2Q^W_{tt}}\Bigr)\eta^2 .$ The first coefficient is at least $\tfrac18H_0W\ge\tfrac\lambda8W$. For the second, $|Q^W_{ts}|/W\le\varepsilon T(c_J+H_0|A|)$, so $\begin{aligned} \frac{(Q^W_{ts})^2}{2Q^W_{tt}W}&\le\frac{2\varepsilon^2T^2(c_J+H_0|A|)^2}{H_0}\\ &\le4\varepsilon^2T^2\Bigl(\frac{c_J^2}{H_0}+H_0A^2\Bigr). \end{aligned}$ The first of these two terms is at most $4T^2c_J^2/\lambda\le\tfrac14\theta_0$. By (A3) and $|A|\le M_0$, the second is at most $\tfrac14H_0|A|$. Hence $\begin{aligned} &\frac1W\Bigl(Q^W_{ss}-\frac{(Q^W_{ts})^2}{2Q^W_{tt}}\Bigr)\\ &\quad\ge\Theta-\tfrac12H_0A-\tfrac14H_0|A|\\ &\qquad-|A|\beta_0-\tfrac14\theta_0 . \end{aligned}$ We show that the right side is at least $\tfrac12\theta_0$ where $A\le c_1d$.

Where $A\le0$. The right side is $\Theta+\tfrac14H_0|A|-|A|\beta_0-\tfrac14\theta_0$, and this is at least $\Theta+\tfrac18H_0|A|-\tfrac14\theta_0$, because $H_0\ge8\beta_0$. If $|A|\le a_0$, then $\Theta\ge\theta_0$ by (A3). If $A<-a_0$, then $\Theta+\tfrac18H_0|A|\ge-\Theta_1+\tfrac18\lambda a_0\ge\theta_0$. In both cases the right side is at least $\tfrac34\theta_0$.

Where $0<A\le c_1d$. Then $A\le c_1d_{\max}\le a_0$, so $\Theta\ge\theta_0$. Further $H_0A\le c_1(\lambda d+p)\le c_1(\lambda d_{\max}+p)$ and $A\beta_0\le c_1d_{\max}\beta_0$. The right side is at least $\begin{aligned} &\theta_0-\tfrac34c_1(\lambda d_{\max}+p)\\ &\quad-c_1d_{\max}\beta_0-\tfrac14\theta_0\ \ge\ \tfrac12\theta_0 \end{aligned}$ by the choice of $c_1$. $\blacksquare$

The order of the choices is: $\varepsilon$ in (A3), then $\lambda$, then $c_1$. Where $A$ is definitely negative the term $-\tfrac12H_0A$ is favourable and large; near the curve $A=0$ the positivity comes from $\Theta$; and on the side $A>0$ the unfavourable term $-\tfrac12H_0A$ is small only as long as $A\le c_1d$, because $H_0$ contains $p/d$.

4. An elliptic estimate where $A$ is positive

Let $T_2=\tfrac12(T_1+T)$, $T_3=\tfrac12(T_2+T)$, $s_2=\tfrac12(s_0+s_1)$, $s_3=\tfrac12(s_0+s_2)$. Fix smooth functions $\zeta,\chi$ of $(t,s)$ with values in $[0,1]$, products of a function of $t$ and a function of $s$, such that

$\zeta=1$ where $|t|\le T_1$ and $s\ge s_1$, and $\zeta=0$ where $|t|\ge T_2$ or $s\le s_2$;
$\chi=1$ where $|t|\le T_2$ and $s\ge s_2$, and $\chi=0$ where $|t|\ge T_3$ or $s\le s_3$.

Let $c_\zeta$ be a bound for the first derivatives of $\zeta$ and $\chi$; it depends only on $T-T_1$ and $s_1-s_0$. Fix a smooth function $\varphi$ on $\mathbb R$ with values in $[0,1]$, equal to $0$ on $(-\infty,\tfrac12]$ and to $1$ on $[1,\infty)$, with $|\varphi'|\le3$.

Proposition 4.1. Assume (A1), and let $c_1>0$. Let $\Omega_e$ be the set of the points of $\Omega$ with $A\ge c_1d$ and $\chi=1$. There is a constant $C_e$, depending only on $l,M_0,d_{\max},c_1,c_\zeta$, such that $\iint_{\Omega_e}d^{\,p-1}\,|\nabla u|^2\le C_e\bigl(\|u\|^2+\|f\|^2\bigr).$

Proof. Let $v=d^{\,p-2}\chi^2\,\varphi(A/(c_1d))^2$ for $s<b$, and $v=0$ on the edge $s=b$. Where $v>0$ we have $A>\tfrac12c_1d$. The product $-vu\,(u_{tt}+Au_{ss}+Bu_t+Cu_s)$ is equal to $\begin{aligned} &v\,(u_t^2+Au_s^2)+u\,(v_t-vB)\,u_t\\ &\quad+u\,((Av)_s-vC)\,u_s\\ &\quad-\partial_t(vuu_t)-\partial_s(vAuu_s). \end{aligned}$ Integrate over $[-T,T]\times[s_0,b-\eta]$ and let $\eta\to0$. The integrals of the two derivatives vanish in the limit: $v=0$ near the three edges, and on the line $s=b-\eta$ the function $v$ is at most $\eta^{p-2}$. So $\begin{aligned} &\iint v\,(u_t^2+Au_s^2)\\ &\quad=-\iint vuf-\iint u\,(v_t-vB)\,u_t\\ &\qquad-\iint u\,((Av)_s-vC)\,u_s . \end{aligned}$ Derivatives of $v$. Where $\varphi'(A/(c_1d))\ne0$ we have $A\le c_1d$. The derivatives of $A/(c_1d)$ are $A_t/(c_1d)$ and $A_s/(c_1d)+A/(c_1d^2)$, so there they are at most $M_0/(c_1d)$ and $(M_0/c_1+1)/d$ in absolute value. Since $v$ is $d^{\,p-2}$ times the square of $\chi\varphi$, $|v_t|+|v_s|\le c_3\sqrt v\;d^{(p-2)/2}\,(1+d^{-1}),$ with $c_3$ depending on $p,c_\zeta,M_0,c_1$; the derivative of $d^{\,p-2}$ contributes $(p-2)v/d\le(p-2)\sqrt v\,d^{(p-2)/2}d^{-1}$.

The error terms. Where $v>0$, using $A>\tfrac12c_1d$, $|C|\le(l+2)M_0^2$ and $v\le d^{\,p-2}$, $\begin{aligned} &\frac{(v_t-vB)^2}v\\ &\quad\le2c_3^2d^{\,p-2}(1+d^{-1})^2+2M_0^2d^{\,p-2},\\ &\frac{((Av)_s-vC)^2}{Av}\\ &\quad\le\frac{3(A_s^2+C^2)\,v}A+\frac{3A\,v_s^2}v\\ &\quad\le\frac{6(l+3)^2M_0^4}{c_1}\,d^{\,p-3}\\ &\qquad+3M_0c_3^2\,d^{\,p-2}(1+d^{-1})^2 . \end{aligned}$ Since $p=8$, all these are bounded on $\Omega$ by a constant $c_4$ depending on $l,M_0,d_{\max},c_1,c_3$. By the inequality $2xy\le x^2+y^2$, $\begin{aligned} &\bigl|u\bigl((v_t-vB)u_t+((Av)_s-vC)u_s\bigr)\bigr|\\ &\quad\le\tfrac12v\,(u_t^2+Au_s^2)+c_4u^2 . \end{aligned}$ Also $|vuf|\le\tfrac12d_{\max}^{\,p-2}(u^2+f^2)$. Therefore $\begin{aligned} &\tfrac12\iint v\,(u_t^2+Au_s^2)\\ &\quad\le\bigl(c_4+\tfrac12d_{\max}^{\,p-2}\bigr)\bigl(\|u\|^2+\|f\|^2\bigr). \end{aligned}$ Where $\chi=1$ and $A\ge c_1d$, we have $v=d^{\,p-2}\ge d^{\,p-1}/d_{\max}$ and $vA\ge c_1d^{\,p-1}$. $\blacksquare$

The weight $d^{\,p-2}$ lets the ellipticity degenerate at the free edge at the rate $A\ge c_1d$.

5. The estimate

Theorem 5.1. Assume (A1), (A2) and (A3), and let $s_1<b_0<b$. There is a constant $C_*$, depending only on the numbers $l,M_0,T,T_1,s_0,s_1,b,b_0,\alpha_0,\varepsilon,a_0,\theta_0$, such that $\iint_{\Omega_0}\bigl(u_t^2+u_s^2\bigr)\ \le\ C_*\iint_\Omega\bigl(u^2+f^2\bigr).$

Proof. Take $\lambda,c_1,c_2$ from Proposition 3.1, and apply Lemma 2.1 with $m=-\zeta W$ and $n=\varepsilon t\zeta W$; they vanish at $s=b$ because $W$ does, and near the other edges because $\zeta$ does. Differentiating the products shows that the functions of Lemma 2.1 are $\begin{aligned} Q_{tt}&=\zeta Q^W_{tt}-\tfrac12(\zeta_s+\varepsilon t\zeta_t)W,\\ Q_{ss}&=\zeta Q^W_{ss}+\tfrac12A(\zeta_s+\varepsilon t\zeta_t)W,\\ Q_{ts}&=\zeta Q^W_{ts}+(\zeta_t-\varepsilon tA\zeta_s)W . \end{aligned}$ Let $\Omega_-$ be the set of the points of $\Omega$ with $A<c_1d$ and $\Omega_+$ the set of those with $A\ge c_1d$. The derivatives of $\zeta$ vanish outside the set where $|t|\ge T_1$ or $s\le s_1$; there $A\ge\alpha_0\ge c_1d_{\max}$ by (A2) and the choice of $c_1$. So the terms with derivatives of $\zeta$ live in $\Omega_+$. On $\Omega$, $W\le E_1d^{\,p}$ with $E_1=e^{\lambda(|s_0|+|b|)+TM_0}$, and $H_0W\le(\lambda d_{\max}+p)E_1d^{\,p-1}$. With the formulas of Section 3, all of $|Q^W_{tt}|,|Q^W_{ss}|,|Q^W_{ts}|$ and the terms with derivatives of $\zeta$ are at most $c_5d^{\,p-1}$, with a constant $c_5$ depending on the numbers of the statement. Lemma 2.1 and Proposition 3.1 give $\begin{aligned} &c_2\iint_{\Omega_-}\zeta W|\nabla u|^2\\ &\quad\le\iint_\Omega|f|\,\zeta W(1+T)|\nabla u|\\ &\qquad+3c_5\iint_{\Omega_+,\ \zeta>0}d^{\,p-1}|\nabla u|^2 . \end{aligned}$ In the first integral on the right, the function $|f|W(1+T)|\nabla u|$ is at most $\begin{gathered} \tfrac12c_2W|\nabla u|^2+\frac{(1+T)^2Wf^2}{2c_2}\quad\text{on }\Omega_-,\\ d^{\,p-1}|\nabla u|^2+\frac{(1+T)^2W^2f^2}{d^{\,p-1}}\quad\text{on }\Omega_+, \end{gathered}$ where $W^2/d^{\,p-1}\le E_1^2d_{\max}^{\,p+1}$. Since $\chi=1$ where $\zeta>0$, Proposition 4.1 bounds the integrals over $\Omega_+$. So $\tfrac12c_2\iint_{\Omega_-}\zeta W|\nabla u|^2\le c_6\bigl(\|u\|^2+\|f\|^2\bigr).$ On $\Omega_0$ we have $\zeta=\chi=1$, $d\ge b-b_0$ and $W\ge w_0=(b-b_0)^pe^{-\lambda(|s_0|+|b|)-TM_0}$. Hence $\begin{aligned} &\iint_{\Omega_0}|\nabla u|^2\\ &\quad\le\frac1{w_0}\iint_{\Omega_-}\zeta W|\nabla u|^2\\ &\qquad+(b-b_0)^{1-p}\iint_{\Omega_e}d^{\,p-1}|\nabla u|^2, \end{aligned}$ and both terms are bounded by a constant times $\|u\|^2+\|f\|^2$. $\blacksquare$

6. Exercises

6.1. Verify the formulas for $Q^W_{tt}$, $Q^W_{ss}$ and $Q^W_{ts}$ of Section 3.

6.2. Let $A=t^2-\tfrac14-s$ on $\Omega=[-2,2]\times[-1,\tfrac12]$, so that the equation is elliptic below the parabola $s=t^2-\tfrac14$ and hyperbolic above it, and let $B=0$, $C=(l+1)A_s=-(l+1)$. Show that (A1), (A2) and (A3) hold, with $T_1=\tfrac32$, $s_1=-\tfrac12$ and suitable numbers $M_0,\alpha_0,\varepsilon,a_0,\theta_0$.

6.3. In the situation of Exercise 6.2, replace $C$ by $+(l+1)$. Which of the assumptions fails, and where does the proof of Proposition 3.1 break down?

6.4. Show that in Lemma 2.1 the assumption that $m$ and $n$ vanish on the edge $s=b$ cannot simply be dropped: for $A=-1$, $B=C=0$, find the terms on the edge that appear, and say why they have no sign.

7. Solutions

6.1. With $m=-W$: $m_s=(H_0-I_s)W$ and $m_t=-BW$. With $n=\varepsilon tW$: $n_t=\varepsilon W+\varepsilon tBW$. So $\begin{aligned} Q_{tt}&=\tfrac12(H_0-I_s)W-\tfrac\varepsilon2(1+tB)W\\ &\quad+\varepsilon tBW . \end{aligned}$ Next, $\begin{aligned} -\tfrac12(Am)_s&=\tfrac12A_sW-\tfrac12A(H_0-I_s)W,\\ \tfrac12(An)_t&=\tfrac\varepsilon2(tA_t+A+tAB)W, \end{aligned}$ and $Cm=-CW$ with $C=(l+1)A_s+A\omega$; the sum of the three is $W$ times $\begin{aligned} &-(l+\tfrac12)A_s+\tfrac\varepsilon2tA_t-\tfrac12H_0A\\ &\quad+A\bigl(\tfrac12I_s+\tfrac\varepsilon2(1+tB)-\omega\bigr). \end{aligned}$ Finally $-m_t+Bm=0$, $-(An)_s=-\varepsilon t(A_s-A(H_0-I_s))W$ and $Cn=\varepsilon tCW$; the sum is $\varepsilon tW$ times $lA_s+A\omega+A(H_0-I_s)$.

6.2. (A1): $\omega=0$, and $|A|\le4-\tfrac14+1<5$, $|A_t|=|2t|\le4$, $A_s=-1$, so $M_0=5$ works. (A2): for $|t|\ge\tfrac32$, $A\ge\tfrac94-\tfrac14-\tfrac12=\tfrac32$; for $s\le-\tfrac12$, $A\ge t^2+\tfrac14\ge\tfrac14$; so $\alpha_0=\tfrac14$. (A3): $\Theta=(l+\tfrac12)+\varepsilon t^2\ge l+\tfrac12$ at every point, so $\theta_0=l+\tfrac12$ with any $a_0$, for every $\varepsilon$ with $16\,\varepsilon^2\cdot4\cdot5\le1$, for example $\varepsilon=\tfrac1{18}$.

6.3. For $A=t^2-\tfrac14-s$ and $C=+(l+1)=-(l+1)A_s$, the decomposition (A1) would need $A\omega=2(l+1)$, and $\omega$ is unbounded near the curve $A=0$. In the proof, the term $Cm$ then contributes $-(l+1)W$ instead of $+(l+1)W$ to $Q^W_{ss}$, and $Q^W_{ss}/W$ is $-(l+\tfrac32)+\varepsilon t^2$ plus terms that vanish with $A$: it is negative on the part of the curve $A=0$ with $t^2<(l+\tfrac32)/\varepsilon$. The sign of the first-order term across the curve decides.

6.4. For $A=-1$, $B=C=0$ the equation is the wave equation $u_{tt}-u_{ss}=f$. The integral of $\partial_s(-\tfrac12mu_t^2+\tfrac12Amu_s^2+Anu_su_t)$ contributes on $s=b$ the integral over $t$ of $-\tfrac12m(u_t^2+u_s^2)-n\,u_su_t$. Since $|u_su_t|\le\tfrac12(u_t^2+u_s^2)$, this has the sign of $-m$ if $|n|\le|m|$, and no sign otherwise; and $u$ and its derivatives on that edge are not controlled by anything. In the hyperbolic region an edge can be free only if the multiplier is adapted to it. The weight $d^{\,p}$ avoids the question by vanishing there.

References

[OpenAI-I] OpenAI, A smooth metric with no local isometric immersion into three-space, OpenAI Math Release preprint, 24 September 2026, Section 3. https://github.com/openai/math/tree/main/preprints/A-Smooth-Metric-with-No-Local-Isometric-Immersion-into-Three-Space-September-24-2026