Proof. We compute \(g_{ss}=-(A')^2+A^2(\varphi')^2+(B')^2\) using the induced Lorentz metric on the spacelike submanifold \(\mathbb{H}^3\). From \(A^2=a\cosh(2s)+1/2\) we obtain \(2AA'=2a\sinh(2s)\), whence \(A'=a\sinh(2s)/A\) and \((A')^2=a^2\sinh^2(2s)/A^2\). Analogously, \((B')^2=a^2\sinh^2(2s)/B^2\). Therefore \[g_{ss}=-\frac{a^2\sinh^2(2s)}{A^2}+\frac{K^2}{A^2B^2}+\frac{a^2\sinh^2(2s)}{B^2}=\frac{a^2\sinh^2(2s)(A^2-B^2)+K^2}{A^2B^2}.\] From 2 , \(A^2-B^2=1\) and \(a^2\sinh^2(2s)+K^2=a^2\cosh^2(2s)-1/4=A^2B^2\), whence \(g_{ss}=1\). The component \(g_{\theta\theta}=B^2\) is immediate from 1 .

To prove 8 , differentiating \(2BB'=2a\sinh(2s)\) we obtain \(2(B')^2+2BB''=4a\cosh(2s)=2(2B^2+1)\) by 2 . ◻

Proof. By the Gauss formula in \(\mathbb{H}^3\), \(K_\Sigma=-1+\tfrac{1}{2}(H^2-|\mathrm{II}|^2)\). Since \(\Sigma\) is minimal (\(H=0\)), \(K_\Sigma=-1-\tfrac{1}{2}|\mathrm{II}|^2\). The Gaussian curvature of the metric \(ds^2+B(s)^2 d\theta^2\) is \(-B''/B\), whence \(|\mathrm{II}|^2=2(B''/B-1)\). From 8 , \(B''/B-1=(2B^2+1-(B')^2-B^2)/B^2=(B^2+1-(B')^2)/B^2\). Multiplying by \(B^4\), \[|\mathrm{II}|^2 B^4=2B^2(B^2+1-(B')^2)=2B^4+2B^2-2(B')^2B^2.\] From \((B')^2=a^2\sinh^2(2s)/B^2=(a^2\cosh^2(2s)-a^2)/B^2=((B^2+1/2)^2-a^2)/B^2\) we obtain \((B')^2B^2=(B^2+1/2)^2-a^2\). Substituting, \[|\mathrm{II}|^2 B^4=2B^4+2B^2-2(B^2+1/2)^2+2a^2=-2B^2-1/2+2a^2=2(a^2-1/4)=2K^2.\qedhere\] ◻

Proof. For (i), by the definition of \(r\) we have \(\cosh r(s)=A(s)\cosh\varphi(s)\). Differentiating gives \(\sinh r\cdot r'=\partial_s(A\cosh\varphi)=:f_*\).

For (ii), a direct computation yields \[f_*=A'\cosh\varphi+A\varphi'\sinh\varphi=\frac{a\sinh(2s)}{A}\cosh\varphi+\frac{K}{AB}\sinh\varphi=\frac{aB\sinh(2s)\cosh\varphi+K\sinh\varphi}{AB}.\] The functions \(A,B\) are even in \(s\), while \(\sinh(2s)\) and \(\sinh\varphi\) are odd, and \(\cosh\varphi\) is even. Hence the numerator is odd, \(f_*\) is odd, and \(f_*(0)=0\).

For \(f_*'(0)\), recall that \(g(s):=A(s)\cosh\varphi(s)\) satisfies \(g''+(B'/B)g'-2g=0\) on \((-s_0,s_0)\); this equation is proved in Lemma 4 and uses only algebraic identities on \(A,B,\varphi\), independent of Lemma 3. Since \(f_*=g'\), we have \(f_*'=g''=2g-(B'/B)g'=2g-(B'/B)f_*\). At \(s=0\), \(A(0)=\sqrt{a+1/2}>0\) and \(\cosh\varphi(0)=1\), so \(g(0)=\sqrt{a+1/2}\); moreover \(B'(0)=2a\sinh(0)/B(0)=0\). Therefore \[f_*'(0)=2g(0)-0\cdot f_*(0)=2\sqrt{a+1/2}>0.\]

For (iii), setting \(g(s):=A(s)\cosh\varphi(s)\), we have \(L_{\Sigma}g=|\mathrm{II}|^2 g\) (Lemma 4), which, restricted to mode \(0\), gives \(g''+(B'/B)g'+(|\mathrm{II}|^2-2)g=|\mathrm{II}|^2 g\), namely \(g''+(B'/B)g'-2g=0\). Differentiating in \(s\) and using \(f_*=g'\), \[f_*''+(B'/B)f_*'+\bigl((B'/B)'-2\bigr)f_*=0.\] Comparing with 10 rewritten as \(f_*''+(B'/B)f_*'+(|\mathrm{II}|^2-2-1/B^2)f_*=0\), the equivalence of the two equations is equivalent to \((B'/B)'-2=|\mathrm{II}|^2-2-1/B^2\), that is, \((B'/B)'=|\mathrm{II}|^2-1/B^2\). From \((B'/B)'=B''/B-(B'/B)^2\) and \(|\mathrm{II}|^2=2B''/B-2\) (from the proof of Lemma 2), we verify \[B''/B-(B'/B)^2=2B''/B-2-1/B^2\iff B''/B=-(B'/B)^2+2+1/B^2\iff BB''=-(B')^2+2B^2+1,\] which is the Mori identity 8 . ◻

Proof. We show directly that each \(\Phi^A_a\) satisfies \(\Delta_g\Phi^A_a=2\Phi^A_a\) in the induced metric \(g=ds^2+B(s)^2 d\theta^2\); it then follows immediately that \(L_{\Sigma}\Phi^A_a=2\Phi^A_a+(|\mathrm{II}|^2-2)\Phi^A_a=|\mathrm{II}|^2\Phi^A_a\).

Consider first mode \(|k|=1\), with \(\Phi^2_a=B(s)\cos\theta\). We have \[\Delta_g(B\cos\theta)=B''\cos\theta+(B'/B)B'\cos\theta-\frac{1}{B^2}B\cos\theta=\cos\theta\Bigl[B''+\frac{(B')^2}{B}-\frac{1}{B}\Bigr].\] The condition \(\Delta_g(B\cos\theta)=2B\cos\theta\) is equivalent to \(BB''+(B')^2-1=2B^2\), namely the Mori identity 8 . The argument is identical for \(\Phi^3_a=B(s)\sin\theta\).

We now turn to mode \(|k|=0\), with \(\Phi^0_a=A\cosh\varphi\) and \(\Phi^1_a=A\sinh\varphi\). In mode \(0\), \(\Delta_g f=f''+(B'/B)f'\). We must verify that \(f''+(B'/B)f'-2f=0\) for \(f\in\{A\cosh\varphi,A\sinh\varphi\}\). Expanding \(f''+(B'/B)f'-2f\) and grouping the terms in \(\cosh\varphi\) and \(\sinh\varphi\) (with symmetric computations in the two cases), \[\begin{align} f''+(B'/B)f'-2f&=\cosh\varphi\bigl[A''+A(\varphi')^2+(B'/B)A'-2A\bigr]\\ &\quad+\sinh\varphi\bigl[2A'\varphi'+A\varphi''+(B'/B)A\varphi'\bigr]\quad\text{(for f=A\cosh\varphi)}, \end{align}\] with the roles of \(\cosh\varphi\) and \(\sinh\varphi\) exchanged for \(f=A\sinh\varphi\). We show that both coefficients vanish.

For the coefficient in \(\sinh\varphi\), the defining relation \(A^2\varphi'=K/B\) (cf. 1 and the minimality condition) gives, upon differentiation, \((A^2\varphi')'=-KB'/B^2=-(B'/B)\,A^2\varphi'\). Expanding, \(2AA'\varphi'+A^2\varphi''=-(B'/B)A^2\varphi'\). Dividing by \(A\), \[2A'\varphi'+A\varphi''+(B'/B)A\varphi'=0.\]

For the coefficient in \(\cosh\varphi\), from \(A^2=a\cosh(2s)+1/2\) we have \(2AA'=2a\sinh(2s)\), and therefore \(2(A')^2+2AA''=4a\cosh(2s)=2(2A^2-1)\), that is, \[A''=2A-\frac{1}{A}-\frac{(A')^2}{A}.\] Substituting into the coefficient \(A''+A(\varphi')^2+(B'/B)A'-2A\): \[=-\frac{1}{A}-\frac{(A')^2}{A}+A(\varphi')^2+(B'/B)A'.\] Using \((A')^2=a^2\sinh^2(2s)/A^2\), \((\varphi')^2=K^2/(A^4B^2)\), \((B'/B)A'=a^2\sinh^2(2s)/(AB^2)\): \[=\frac{1}{A}\left[-1-\frac{a^2\sinh^2(2s)}{A^2}+\frac{K^2}{A^2B^2}+\frac{a^2\sinh^2(2s)}{B^2}\right].\] We compute the term in brackets. Using \(A^2-B^2=1\), \[\frac{a^2\sinh^2(2s)}{B^2}-\frac{a^2\sinh^2(2s)}{A^2}=\frac{a^2\sinh^2(2s)(A^2-B^2)}{A^2B^2}=\frac{a^2\sinh^2(2s)}{A^2B^2}.\] Thus the bracket becomes \[-1+\frac{a^2\sinh^2(2s)+K^2}{A^2B^2}.\] By Lemma 1, \(a^2\sinh^2(2s)+K^2=A^2B^2\), whence \(-1+1=0\). Both coefficients vanish, and therefore \(f''+(B'/B)f'-2f=0\) for \(f=A\cosh\varphi\). The verification for \(f=A\sinh\varphi\) is identical by symmetry of the algebraic structure.

In each case, \(\Delta_g\Phi^A_a=2\Phi^A_a\), and the conclusion follows. ◻

Proof. From the FBMS condition 3 , a direct computation based on \(\cosh^2\varphi-\sinh^2\varphi=1\) yields \[\sinh^2\varphi(s_0)=\frac{B^2K^2}{a^2\sinh^2(2s_0)-B^2K^2},\qquad\cosh^2\varphi(s_0)=\frac{a^2\sinh^2(2s_0)}{a^2\sinh^2(2s_0)-B^2K^2}.\] Setting \(D:=a^2\sinh^2(2s_0)-B^2K^2\) and using \(a^2\sinh^2(2s_0)+K^2=A^2B^2\) (Lemma 1), \[D=A^2B^2-K^2-B^2K^2=A^2B^2-K^2(1+B^2)=A^2B^2-K^2 A^2=A^2(B^2-K^2).\] From \(\sinh^2 r=A^2\cosh^2\varphi-1\), \[\sinh^2 r=\frac{A^2 a^2\sinh^2(2s_0)-D}{D}=\frac{(A^2-1)a^2\sinh^2(2s_0)+B^2K^2}{D}=\frac{B^2(a^2\sinh^2(2s_0)+K^2)}{D}=\] \[= \frac{B^2\cdot A^2B^2}{A^2(B^2-K^2)}=\frac{B^4}{B^2-K^2}.\]

Analogously, \(\cosh^2 r=A^2\cosh^2\varphi=A^2 a^2\sinh^2(2s_0)/D=a^2\sinh^2(2s_0)/(B^2-K^2)\). Therefore \[\coth^2 r=\frac{a^2\sinh^2(2s_0)}{B^4}=\Bigl(\frac{B'(s_0)}{B(s_0)}\Bigr)^2.\] Taking positive roots (both sides are positive for \(s_0>0\) and \(r(a)>0\)), we obtain 12 . ◻

Proof. Since \(f_*=g'\) with \(g=A\cosh\varphi\) satisfying \(g''+(B'/B)g'-2g=0\) (Lemma 3(iii)), \[f_*'(s_0)=g''(s_0)=2g(s_0)-\frac{B'(s_0)}{B(s_0)}f_*(s_0).\] By 12 , \(B'(s_0)/B(s_0)=\coth r(a)\). Moreover, \(g(s_0)=\cosh r(a)\) and \(f_*(s_0)=\sinh r(a)\) by Lemma 3(i). Hence \[f_*'(s_0)=2\cosh r-\coth r\cdot\sinh r=2\cosh r-\cosh r=\cosh r=\coth r\cdot\sinh r=\coth r\cdot f_*(s_0).\] The property at \(s=-s_0\) follows by the oddness of \(f_*\) and the evenness of \(f_*'\). ◻

Proof. This is a direct consequence of 10 : \[(Bf_*')'+B(|\mathrm{II}|^2-2-1/B^2)f_*=Bf_*''+B'f_*'+B(|\mathrm{II}|^2-2-1/B^2)f_*=B[f_*''+(B'/B)f_*'+(|\mathrm{II}|^2-2-1/B^2)f_*]=0.\qedhere\] ◻

Proof. For \(\widetilde{u}(s,\theta)=u(s)\Theta_k(\theta)\), \(|\nabla\widetilde{u}|_g^2=(u')^2\Theta_k^2+(u^2/B^2)(\Theta_k')^2\). From \(\int_0^{2\pi}\Theta_k^2\,d\theta=c_k\) (with \(c_0=2\pi\), \(c_{|k|\geq 1}=\pi\)) and \(\int_0^{2\pi}(\Theta_k')^2\,d\theta=k^2 c_k\), we obtain, using \(dA=B\,ds\,d\theta\) and \(dL=B(s_0)\,d\theta\) at \(s=\pm s_0\), \[\begin{align} \mathcal{S}(\widetilde{u},\widetilde{u})&=\int_0^{2\pi}\!\!\int_{-s_0}^{s_0}\Bigl((u')^2\Theta_k^2+\frac{k^2}{B^2}u^2\Theta_k^2-(|\mathrm{II}|^2-2)u^2\Theta_k^2\Bigr)B\,ds\,d\theta\\ &\quad-\coth r(a)\int_0^{2\pi}\bigl(u(s_0)^2+u(-s_0)^2\bigr)B(s_0)\,\Theta_k^2\,d\theta\\ &=c_k\int_{-s_0}^{s_0}\Bigl[B(u')^2+\frac{k^2}{B}u^2-B(|\mathrm{II}|^2-2)u^2\Bigr]ds-c_k\coth r(a)B(s_0)\bigl[u(s_0)^2+u(-s_0)^2\bigr]\\ &=c_k\,\mathcal{S}_k^{\mathrm{rad}}(u,u).\qedhere \end{align}\] ◻

Proof. Setting \(u=f_*h\), we have \(u'=f_*'h+f_*h'\), whence \[B(u')^2=B(f_*')^2 h^2+B f_*'f_*(h^2)'+B f_*^2(h')^2.\] Integrating over \((-s_0,s_0)\) and integrating by parts the cross term, \[\int_{-s_0}^{s_0}B f_*'f_*(h^2)'\,ds=\bigl[B f_*'f_*\,h^2\bigr]_{-s_0}^{s_0}-\int_{-s_0}^{s_0}(B f_*'f_*)'\,h^2\,ds.\] By 13 , \((B f_*')'=-B W_1 f_*\), whence \((B f_*'f_*)'=(B f_*')'f_*+B f_*'\cdot f_*'=-BW_1 f_*^2+B(f_*')^2\). Substituting, \[\int B(u')^2\,ds=\bigl[B f_*'f_*\,h^2\bigr]_{-s_0}^{s_0}+\int B W_1 f_*^2 h^2\,ds+\int B f_*^2(h')^2\,ds.\] By 14 , \[\mathcal{S}_k^{\mathrm{rad}}(u,u)=\bigl[B f_*'f_*\,h^2\bigr]_{-s_0}^{s_0}+\int B(W_1-W_k)f_*^2 h^2\,ds+\int B f_*^2(h')^2\,ds-\coth r\cdot B(s_0)f_*(s_0)^2[h(s_0)^2+h(-s_0)^2],\] where we used \(f_*(\pm s_0)^2=f_*(s_0)^2\) (odd squared is even).

The boundary term is computed using \(f_*'(s_0)/f_*(s_0)=\coth r\) and \(f_*'(-s_0)/f_*(-s_0)=-\coth r\) (Lemma 6 together with oddness): \[\bigl[B f_*'f_*\,h^2\bigr]_{-s_0}^{s_0}=B(s_0)\frac{f_*'(s_0)}{f_*(s_0)}f_*(s_0)^2 h(s_0)^2-B(s_0)\frac{f_*'(-s_0)}{f_*(-s_0)}f_*(s_0)^2 h(-s_0)^2\] \[=B(s_0)f_*(s_0)^2\bigl(\coth r\cdot h(s_0)^2+\coth r\cdot h(-s_0)^2\bigr).\] The two boundary contributions therefore cancel exactly: \[\bigl[B f_*'f_*\,h^2\bigr]_{-s_0}^{s_0}-\coth r\cdot B(s_0)f_*(s_0)^2[h(s_0)^2+h(-s_0)^2]=0.\] Finally, \(W_1-W_k=-1/B^2-(-k^2/B^2)=(k^2-1)/B^2\). Substituting in what remains, \[\mathcal{S}_k^{\mathrm{rad}}(u,u)=\int B\cdot\frac{k^2-1}{B^2}f_*^2 h^2\,ds+\int B f_*^2(h')^2\,ds=(k^2-1)\int\frac{f_*^2}{B}h^2\,ds+\int B f_*^2(h')^2\,ds.\qedhere\] ◻

Proof. For \(u\in H^1_{\mathrm{rad}}\) odd, \(u(0)=0\) in the sense of trace. By Theorem 4 extended to \(H^1\) (Remark 5), \[\mathcal{S}_k^{\mathrm{rad}}(u,u)=(k^2-1)\int_{-s_0}^{s_0}\frac{u^2}{B}\,ds+\int_{-s_0}^{s_0}B\Bigl(u'-\frac{f_*'}{f_*}u\Bigr)^2 ds.\] For \(|k|\geq 2\) the coefficient \(k^2-1\geq 3>0\), both integrals are nonnegative, and hence \(\mathcal{S}_k^{\mathrm{rad}}(u,u)\geq 0\). If \(u\not\equiv 0\), since \(1/B>0\) on \([-s_0,s_0]\), the first integral is strictly positive, and therefore \(\mathcal{S}_k^{\mathrm{rad}}(u,u)>0\). ◻

Proof. We denote by \(H^\sharp\) the subspace of \(H^1((-s_0,s_0))\) corresponding to the sector \(\sharp\) (the whole \(H^1\) for \(\sharp=\mathrm{rad}\), the even functions for \(\sharp=\mathrm{even}\), the odd functions for \(\sharp=\mathrm{odd}\)); each is closed, and \(\mathcal{S}_k^{\mathrm{rad}}\) restricts to it (it preserves parity by the symmetry \(s\mapsto-s\) of the coefficients \(B\) and \(W_k\)). By Lemma 8, \[\mathcal{S}_k^{\mathrm{rad}}(u,u)-\mathcal{S}_1^{\mathrm{rad}}(u,u)=\int_{-s_0}^{s_0}B(W_1-W_k)u^2\,ds=(k^2-1)\int_{-s_0}^{s_0}\frac{u^2}{B}\,ds.\] The profile \(B^2(s)=a\cosh(2s)-1/2\) is strictly increasing in \(|s|\), so \(B(s)\leq B(s_0)\) for \(|s|\leq s_0\), that is, \(1/B(s)^2\geq 1/B(s_0)^2\) uniformly. Therefore \[\int_{-s_0}^{s_0}\frac{u^2}{B}\,ds=\int_{-s_0}^{s_0}\frac{1}{B^2}\cdot Bu^2\,ds\geq\frac{1}{B(s_0)^2}\int_{-s_0}^{s_0}Bu^2\,ds=\frac{\|u\|_B^2}{B(s_0)^2}.\] Setting \(c:=(k^2-1)/B(s_0)^2\), for every \(u\in H^\sharp\setminus\{0\}\), \[\frac{\mathcal{S}_k^{\mathrm{rad}}(u,u)}{\|u\|_B^2}\geq\frac{\mathcal{S}_1^{\mathrm{rad}}(u,u)}{\|u\|_B^2}+c.\] By the Courant–Fischer min-max principle applied to the sector \(H^\sharp\), \[\mu_n^\sharp(k)=\inf_{\substack{V\subset H^\sharp\\ \dim V=n+1}}\sup_{u\in V\setminus\{0\}}\frac{\mathcal{S}_k^{\mathrm{rad}}(u,u)}{\|u\|_B^2}\geq\inf_{\substack{V\subset H^\sharp\\ \dim V=n+1}}\Bigl(\sup_{u\in V\setminus\{0\}}\frac{\mathcal{S}_1^{\mathrm{rad}}(u,u)}{\|u\|_B^2}+c\Bigr)=\mu_n^\sharp(1)+c,\] where the first inequality uses the pointwise monotonicity \(\mathcal{S}_k\geq\mathcal{S}_1+c\|\cdot\|_B^2\) (preserved under the supremum over \(V\)) and is then preserved under the infimum over \(V\). Since \(c>0\), we obtain 17 . ◻

Proof. This is a direct consequence of Theorem 6. ◻

Proof. The unit-speed geodesic \(\gamma\) from \(p_0=(1,0,0,0)\) to \(p\) is \(\gamma(t)=\cosh t\cdot p_0+\sinh t\cdot v\), with \(v\in T_{p_0}\mathbb{H}^3=\{p_0\}^{\perp_L}\) unit, namely \(\langle v,v\rangle_L=1\), \(v^0=0\). From \(\gamma(r)=p\), \(v=(p-\cosh r\cdot p_0)/\sinh r\). Therefore \[\dot{\gamma}(r)=\sinh r\cdot p_0+\cosh r\cdot v=\sinh r\cdot p_0+\cosh r\cdot\frac{p-\cosh r\cdot p_0}{\sinh r}=\] \[\frac{\sinh^2 r\cdot p_0+\cosh r\cdot p-\cosh^2 r\cdot p_0}{\sinh r}=\frac{\cosh r\cdot p-p_0}{\sinh r}.\] Componentwise, \(\dot{\gamma}(r)^0=(\cosh^2 r-1)/\sinh r=\sinh r\) and \(\dot{\gamma}(r)^i=\cosh r\cdot p^i/\sinh r\) for \(i=1,2,3\). Since \(\nabla^{\mathbb{H}^3}r\big|_p=\dot{\gamma}(r)\) (the gradient of the distance function along the unit geodesic), we obtain 18 . ◻

Proof. By the FBMS condition, \(\Sigma\perp\partial B^3(\rho)\) at \(\partial\Sigma\), so the unit conormal \(\eta\) of \(\Sigma\) at \(\partial\Sigma\) coincides with the unit radial vector \(\nabla^{\mathbb{H}^3}r\) (both are unit, orthogonal to \(T_p(\partial\Sigma)\), and outward-pointing from \(B^3(\rho)\)). Hence \(\partial_\eta=\partial_{\nabla^{\mathbb{H}^3}r}\).

For the coordinate functions \(\Phi^A\), \(d\Phi^A=dx^A\) is the ambient differential, and for every tangent vector \(V\in T_p\mathbb{H}^3\) we have \(d\Phi^A(V)=V^A\), the \(A\)-th Lorentz component of \(V\). In particular, \[\partial_\eta\Phi^A\big|_p=d\Phi^A(\eta)=\eta^A=(\nabla^{\mathbb{H}^3}r)^A\big|_p.\] By Lemma 9, \[\partial_\eta\Phi^A\big|_p=\frac{\cosh\rho\cdot p^A-\delta^A_0}{\sinh\rho}.\] For \(A=0\), \(\partial_\eta\Phi^0=(\cosh^2\rho-1)/\sinh\rho=\sinh\rho\), while \(\Phi^0(p)=p^0=\cosh\rho\). Hence \(\partial_\eta\Phi^0/\Phi^0=\tanh\rho\), which gives (ii). For \(A=i\in\{1,2,3\}\), \(\partial_\eta\Phi^i=\cosh\rho\cdot p^i/\sinh\rho=\coth\rho\cdot p^i=\coth\rho\cdot\Phi^i(p)\), which gives (i). ◻

Proof. The statement is immediate from 1 , recalling that \(\cosh\varphi\) is even, \(\sinh\varphi\) is odd, and \(A,B\) are even. ◻

Proof. We first establish linear independence. By Lemma 11, \(\Phi^0\) and \(\Phi^1\) are in mode \(0\) with different parity, while \(\Phi^2\) and \(\Phi^3\) are in mode \(|k|=1\) and are mutually orthogonal through the factors \(\cos\theta\) and \(\sin\theta\). The four elements belong to four distinct \(L^2(\Sigma_a)\)-orthogonal subspaces, and hence they are linearly independent.

We next prove the diagonality of \(\mathcal{S}\). For \(A\neq B\), we compute \(\mathcal{S}(\Phi^A,\Phi^B)\) via the Green formula (weak form): \[\label{eq:S-pairing} \mathcal{S}(u,v)=-\int_\Sigma v\,L_{\Sigma}u\,dA+\int_{\partial\Sigma}v\,(\partial_\eta u-\coth r\cdot u)\,dL.\tag{19}\] This formula follows by integrating by parts the term \(\int|\nabla u|^2\,dA=-\int u\Delta_g u\,dA+\int_{\partial\Sigma}u\partial_\eta u\,dL\) in 4 and using \(L_{\Sigma}=\Delta_g+|\mathrm{II}|^2-2\). We set \(R^B:=\partial_\eta\Phi^B-\coth r\cdot\Phi^B\). By Lemma 10, \(R^i\equiv 0\) on \(\partial\Sigma\) for \(i=1,2,3\), and \[\label{eq:R0-formula} R^0=(\tanh r(a)-\coth r(a))\Phi^0=-\frac{1}{\sinh r(a)\cosh r(a)}\Phi^0\quad\text{on }\partial\Sigma.\tag{20}\] By the symmetry of the quadratic form \(\mathcal{S}\), it suffices to verify \(\mathcal{S}(\Phi^A,\Phi^B)=0\) for each unordered pair \(\{A,B\}\), choosing appropriately which of the two indices plays the role of \(u\) in 19 . For \(\{A,B\}\neq\{0,0'\}\) we choose \(u=\Phi^B\) with \(B\in\{1,2,3\}\), so that \(R^B=0\) and \[\mathcal{S}(\Phi^A,\Phi^B)=-\int_\Sigma|\mathrm{II}|^2\Phi^A\Phi^B\,dA.\] We verify the vanishing of the volume integral in each case. For the pair \(\{0,1\}\) (with the choice \(u=\Phi^1\), \(B=1\)), \(\Phi^0\Phi^1=A^2\cosh\varphi\sinh\varphi\) is odd in \(s\) and \(|\mathrm{II}|^2(s)\) is even, so the integrand is odd in \(s\) and the integration over \((-s_0,s_0)\) in \(s\) yields \(0\). For the pair \(\{0,2\}\) (with \(u=\Phi^2\), \(B=2\)), the integrand contains \(\cos\theta\), and \(\int_0^{2\pi}\cos\theta\,d\theta=0\); the pair \(\{0,3\}\) is identical with \(\sin\theta\). For the pairs \(\{1,2\}\) and \(\{1,3\}\), the integrands are proportional to \(\cos\theta\) or \(\sin\theta\), with vanishing integration. For the pair \(\{2,3\}\) (with \(u=\Phi^2\)), the integrand is proportional to \(\cos\theta\sin\theta\), with vanishing integration. The only remaining pairs \(\{i,j\}\) with \(i,j\in\{1,2,3\}\) are \(\{2,3\}\), \(\{1,2\}\), \(\{1,3\}\), which have all been treated. Therefore \(\mathcal{S}(\Phi^A,\Phi^B)=0\) for every \(A\neq B\).

We now verify the negativity of the diagonal terms. For \(\mathcal{S}(\Phi^0,\Phi^0)\), using 19 with \(u=v=\Phi^0\), \[\mathcal{S}(\Phi^0,\Phi^0)=-\int|\mathrm{II}|^2(\Phi^0)^2\,dA+\int_{\partial}\Phi^0 R^0\,dL=-\int|\mathrm{II}|^2(\Phi^0)^2\,dA-\int_{\partial}\frac{(\Phi^0)^2}{\sinh r\cosh r}\,dL.\] On \(\partial\Sigma\), \(\Phi^0=\cosh r\), whence \((\Phi^0)^2/(\sinh r\cosh r)=\cosh r/\sinh r=\coth r\), constant. Moreover \(|\partial\Sigma|=2\cdot 2\pi B(s_0)=4\pi B(s_0)\). Therefore \[\mathcal{S}(\Phi^0,\Phi^0)=-\int|\mathrm{II}|^2(\Phi^0)^2\,dA-4\pi B(s_0)\coth r<0,\] with both terms strictly negative (the first because \(|\mathrm{II}|^2\not\equiv 0\) and \(\Phi^0>0\), the second because \(B(s_0),\coth r>0\)). For \(\mathcal{S}(\Phi^A,\Phi^A)\) with \(A=1,2,3\), \(R^A=0\), and hence \[\mathcal{S}(\Phi^A,\Phi^A)=-\int|\mathrm{II}|^2(\Phi^A)^2\,dA<0,\] strictly negative since \(\Phi^A\not\equiv 0\) (\(\Phi^1\) is nonzero for \(s\neq 0\), while \(\Phi^2,\Phi^3\) are nonzero for generic \(\theta\)).

Combining the previous estimates, the matrix of \(\mathcal{S}\) with respect to the basis \(\{\Phi^0,\Phi^1,\Phi^2,\Phi^3\}\) is diagonal with four strictly negative diagonal entries; hence \(\mathcal{S}\) is negative definite on \(V_4:=\mathop{\mathrm{span}}\{\Phi^0,\Phi^1,\Phi^2,\Phi^3\}\) with \(\dim V_4=4\). By the Courant–Fischer characterization of the Morse index, \(\mathop{\mathrm{ind}}_R(\Sigma_a)\geq\dim V_4=4\). ◻

Proof. For (a), this is [2]. For (b), the two functions \(\Phi^0_a\) (mode \(0\) even) and \(\Phi^1_a\) (mode \(0\) odd) are linearly independent in \(H^1\), and \(\mathcal{S}\) is negative definite on their span (the proof is contained in Theorem 11, parts (a) and (b) for \(A=0,1\)). Hence \(\mathop{\mathrm{ind}}_R\big|_0\geq 2\).

For (c), by Theorem 6, every odd-radial eigenvalue of mode \(|k|\geq 2\) is strictly positive. By Sturm–Liouville theory in the even-radial sector, the eigenvalues \(\mu_n^{\mathrm{even}}(k)\) form a strictly increasing sequence with eigenfunctions having respectively \(0,2,4,\ldots\) nodes. Hence \(\mu_n^{\mathrm{even}}(k)<0\) implies \(\mu_m^{\mathrm{even}}(k)<\mu_n^{\mathrm{even}}(k)<0\) for every \(m<n\). Moreover, by the strict Sturm estimate (Theorem 7), \(\mu_n^{\mathrm{even}}(k)\geq\mu_n^{\mathrm{even}}(1)+(k^2-1)/B(s_0)^2\). For \(n\geq 1\), \(\mu_n^{\mathrm{even}}(1)\geq\mu_2(1)>0\) (since \(f_*\), the eigenfunction associated with \(\mu_1(1)=0\), is odd), whence \(\mu_n^{\mathrm{even}}(k)>0\) for \(n\geq 1\). Only \(\mu_0^{\mathrm{even}}(k)\) may be negative; if so, the contribution to \(\mathop{\mathrm{ind}}_R\) is \(1\cdot 2=2\) (by the multiplicity \(m_k=2\)). ◻

Proof. (\(\Rightarrow\)). Assume \(\mathop{\mathrm{ind}}_R(\Sigma_a)=4\) and \(\mathop{\mathrm{nul}}_R(\Sigma_a)=2\). By Proposition 13(a), \(\mathop{\mathrm{ind}}_R\big|_1=2\) and \(\mathop{\mathrm{nul}}_R\big|_1=2\). Hence \(\mathop{\mathrm{ind}}_R\big|_0+\sum_{|k|\geq 2}\mathop{\mathrm{ind}}_R\big|_k=2\) and \(\mathop{\mathrm{nul}}_R\big|_0+\sum_{|k|\geq 2}\mathop{\mathrm{nul}}_R\big|_k=0\).

From the second relation, \(\mathop{\mathrm{nul}}_R\big|_0=0\) and \(\mathop{\mathrm{nul}}_R\big|_k=0\) for every \(|k|\geq 2\).

From the first relation, and since \(\mathop{\mathrm{ind}}_R\big|_0\geq 2\) by Proposition 13(b), we deduce \(\mathop{\mathrm{ind}}_R\big|_0=2\) and \(\sum_{|k|\geq 2}\mathop{\mathrm{ind}}_R\big|_k=0\), whence \(\mathop{\mathrm{ind}}_R\big|_k=0\) for every \(|k|\geq 2\).

By Proposition 13(c), \(\mathop{\mathrm{ind}}_R\big|_k=0\) with \(|k|\geq 2\) implies \(\mu_0^{\mathrm{even}}(k)\geq 0\), but \(\mathop{\mathrm{nul}}_R\big|_k=0\) excludes equality, so \(\mu_0^{\mathrm{even}}(k)>0\). For \(|k|=2\) we obtain (E).

For mode \(0\), \(\mathop{\mathrm{ind}}_R\big|_0=2\) means exactly two negative eigenvalues, \(\mu_0(0)\) and \(\mu_1(0)\). By Sturm–Liouville theory, the subsequent eigenvalues satisfy \(\mu_n(0)\geq\mu_2(0)\) for \(n\geq 2\). The condition \(\mathop{\mathrm{ind}}_R\big|_0=2\) requires \(\mu_2(0)\geq 0\); combined with \(\mathop{\mathrm{nul}}_R\big|_0=0\) (namely \(\mu_n(0)\neq 0\) for every \(n\)), we obtain \(\mu_2(0)>0\) and \(\mu_n(0)\neq 0\) for every \(n\), which is (F).

We add a useful observation: by the strict Sturm estimate 17 , (E) for \(|k|=2\) implies \(\mu_0^{\mathrm{even}}(k)\geq\mu_0^{\mathrm{even}}(2)+(k^2-4)/B(s_0)^2>0\) for every \(|k|\geq 3\). Hence (E) restricted to \(|k|=2\) is sufficient to guarantee positivity in all modes \(|k|\geq 2\).

(\(\Leftarrow\)). Assume (E) and (F). For modes \(|k|\geq 2\), by the Sturm estimate just recalled, \(\mu_0^{\mathrm{even}}(k)>0\) for every \(|k|\geq 2\). Combined with Theorem 6 (\(\mu_n^{\mathrm{odd}}(k)>0\)) and with \(\mu_n^{\mathrm{even}}(k)>\mu_2(1)>0\) for \(n\geq 1\) (by the Sturm estimate of Theorem 7 applied in the even-radial sector and since \(\mu_2(1)>\mu_1(1)=0\)), all eigenvalues in mode \(|k|\geq 2\) are positive: \(\mathop{\mathrm{ind}}_R\big|_k=0\), \(\mathop{\mathrm{nul}}_R\big|_k=0\).

For mode \(0\), (F) gives \(\mu_n(0)\neq 0\) for every \(n\), namely \(\mathop{\mathrm{nul}}_R\big|_0=0\). Moreover \(\mu_2(0)>0\), and by classical Sturm–Liouville theory, \(\mu_n(0)\geq\mu_2(0)>0\) for every \(n\geq 2\). Combined with Theorem 11 (which provides \(\mu_0(0)<0\) via \(\Phi^0\) and \(\mu_1(0)<0\) via \(\Phi^1\)), we have exactly two negative eigenvalues and \(\mathop{\mathrm{ind}}_R\big|_0=2\).

For mode \(1\), by Proposition 13(a), \(\mathop{\mathrm{ind}}_R\big|_1=2\) and \(\mathop{\mathrm{nul}}_R\big|_1=2\).

Combining all contributions, \(\mathop{\mathrm{ind}}_R=\mathop{\mathrm{ind}}_R\big|_0+\mathop{\mathrm{ind}}_R\big|_1+\sum_{|k|\geq 2}\mathop{\mathrm{ind}}_R\big|_k=2+2+0=4\) and \(\mathop{\mathrm{nul}}_R=0+2+0=2\). ◻

Proof. By the rotational symmetry (the whole \(SO(2)\) in \(\theta\)) of the family \(\{X_a\}\), \(\partial_a X_a\) is rotationally invariant, and so is \(\nu_a\), hence their contraction is independent of \(\theta\). This proves that \(\phi_a\) is of mode \(0\).

For the parity in \(s\), the parametrization 1 is \(s\mapsto-s\) equivariant: \(A(-s)=A(s)\), \(B(-s)=B(s)\), \(\varphi(-s)=-\varphi(s)\). A direct verification shows that \(\partial_a X_a\) alternates even/odd in the components \((X^0,X^2)\) and \((X^1,X^3)\), while \(\nu_a\) has the complementary components, so that their contraction is even.

The identity \(L_{\Sigma}\phi_a=0\) follows from the fact that the family \(\Sigma_a\) consists of minimal surfaces for every \(a\). The linearization of minimality with respect to a variation \(\partial_a X_a\) with normal component \(\phi_a\) gives \(L_{\Sigma}\phi_a=0\) in \(\Sigma_a^\circ\) (standard formula, independent of the tangential component of \(\partial_a X_a\)).

For the value at \(s=0\), at \(\theta=0\) we have \(\nu^0_a(0)=K/A(0)\), \(\nu^1_a(0)=0\), \(\nu^2_a(0)=K/B(0)\), \(\nu^3_a(0)=0\). The derivatives \(\partial_a X^A\big|_{s=0,\theta=0}\) are computed explicitly: \(\partial_a A(0)=\cosh(0)/(2A(0))=1/(2A(0))\), \(\partial_a B(0)=1/(2B(0))\), \(\partial_a\varphi(0)=0\) (the integral from \(0\) to \(0\)). Hence \[\phi_a(0)=-\partial_a A\cdot\nu^0\big|_0+0+\partial_a B\cdot\nu^2\big|_0+0=-\frac{K}{2A(0)^2}+\frac{K}{2B(0)^2}=\frac{K(A^2-B^2)}{2A^2B^2}\bigg|_0=\frac{K}{2A(0)^2B(0)^2},\] and \(A(0)^2 B(0)^2=(a+1/2)(a-1/2)=K^2\), whence \(\phi_a(0)=K/(2K^2)=1/(2K)\). ◻

Proof. The FBC for \(\Sigma_a\) in \(B^3(r(a))\) is \(\langle\nu_a,\nabla\rho\rangle=0\) on \(\partial\Sigma_a\), where \(\rho=\mathop{\mathrm{dist}}_{\mathbb{H}^3}(p_0,\cdot)\). Differentiating along the family, including the displacement of the boundary point \(X_a(s_0(a),\theta)\), \[0=\frac{d}{da}\langle\nu_a,\nabla\rho\rangle\big|_{X_a(s_0(a),\theta)}=\langle D_a\nu_a,\nabla\rho\rangle\big|_{\partial}+\langle\nu_a,\nabla_{V_\partial}\nabla\rho\rangle,\] where \(V_\partial\) denotes the velocity of the boundary point as \(a\) varies.

Computation of \(V_\partial\). By the chain rule, \[V_\partial=\frac{d}{da}X_a(s_0(a),\theta)=\partial_a X_a\bigl|_{(s_0(a),\theta)}+s_0'(a)\,\partial_s X_a\bigl|_{(s_0(a),\theta)}=\partial_a X_a\bigl|_{(s_0,\theta)}+s_0'(a)\,\eta,\] since \(\partial_s X_a=\eta\) at the boundary (the outward conormal to \(\partial\Sigma_a\) in \(\Sigma_a\)). Decompose \(\partial_a X_a\) in the local frame \((\nu,\eta,\partial_\theta)\) at the boundary point: \[\partial_a X_a\bigl|_{(s_0,\theta)}=\phi_a\,\nu+\tau_\eta\,\eta+\tau_\theta\,\partial_\theta,\] with \(\phi_a:=\langle\partial_a X_a,\nu_a\rangle_L\) (the parametric Jacobi field, by definition), and \(\tau_\eta,\tau_\theta\) the tangential components.

We claim \(\tau_\theta=0\). Indeed, the family \(\{\Sigma_a\}_{a>1/2}\) is rotationally symmetric: the SO(2)-action \(\theta\mapsto\theta+\alpha\) on \(\mathbb{H}^3\) (rotation about the geodesic axis \(\{X^2=X^3=0\}\)) commutes with the parametrization, \(X_a(s,\theta+\alpha)=R_\alpha X_a(s,\theta)\), where \(R_\alpha\) is the ambient rotation. Differentiating with respect to \(a\) at fixed \((s,\theta)\), \(\partial_a X_a(s,\theta+\alpha)=R_\alpha\,\partial_a X_a(s,\theta)\), that is, \(\partial_a X_a\) transforms equivariantly under the SO(2)-action. Since \(\partial_\theta\) is the infinitesimal generator of this action and \(\nu,\eta\) are SO(2)-invariant (they lie in the meridional plane), the equivariance forces the \(\partial_\theta\)-component of \(\partial_a X_a\) to vanish in the meridional gauge \(\theta=0\), and hence to vanish identically (by SO(2)-equivariance).

It remains to identify \(\tau_\eta+s_0'(a)\). The boundary points satisfy \(\rho(X_a(s_0(a),\theta))=r(a)\), an identity in \(a\). Differentiating in \(a\): \[r'(a)=\frac{d}{da}\rho(X_a(s_0(a),\theta))=\langle\nabla\rho,V_\partial\rangle\bigl|_\partial.\] At the boundary, \(\nabla\rho\big|_\partial=\eta\) (since \(|\nabla\rho|=1\), \(\nabla\rho\) is outward, and the FBC \(\langle\nu,\nabla\rho\rangle=0\) together with \(\nabla\rho\perp\partial_\theta\) for rotational symmetry, forces \(\nabla\rho\big|_\partial\) to be along \(\eta\)). Using \(\langle\nu,\eta\rangle=0\), \(\langle\partial_\theta,\eta\rangle=0\), and \(\langle\eta,\eta\rangle=1\): \[r'(a)=\langle\eta,\phi_a\nu+\tau_\eta\eta+\tau_\theta\partial_\theta+s_0'(a)\eta\rangle=\tau_\eta+s_0'(a).\] Therefore \(\tau_\eta+s_0'(a)=r'(a)\), and consequently \[\label{eq:V-partial} V_\partial=\phi_a\,\nu+r'(a)\,\eta.\tag{24}\]

Computation of \(D_a\nu_a\) at \(\partial\Sigma_a\). The unit normal \(\nu_a\) is determined up to sign by the conditions \(\langle\nu_a,T_{X_a}\Sigma_a\rangle=0\) and \(\langle\nu_a,\nu_a\rangle=1\), with a coherent orientation across the family. Differentiating in \(a\) at a fixed boundary point \(X_a(s_0,\theta)\), with the chain rule incorporating \(s_0'(a)\): \[\begin{align} D_a\nu_a\bigl|_\partial &= \partial_a\nu_a\bigl|_{(s_0,\theta)}+s_0'(a)\,\nabla_\eta\nu_a. \end{align}\] The tangential part of \(\partial_a\nu_a\bigl|_{(s_0,\theta)}\) is determined by the orthogonality to the tangent space: \(\langle\nu_a,\partial_s X_a\rangle=0\) and \(\langle\nu_a,\partial_\theta X_a\rangle=0\). Differentiating the first relation in \(a\) at \((s_0,\theta)\) (without including \(s_0'\), since we are differentiating at fixed parameter \((s,\theta)\)): \[\langle\partial_a\nu_a,\partial_s X_a\rangle+\langle\nu_a,\partial_s\partial_a X_a\rangle=0,\] that is, \(\langle\partial_a\nu_a,\eta\rangle=-\langle\nu_a,\partial_s(\partial_a X_a)\rangle=-\partial_s\langle\nu_a,\partial_a X_a\rangle+\langle\partial_s\nu_a,\partial_a X_a\rangle\). The first term is \(-\partial_s\phi_a=-\partial_\eta\phi_a\) (since \(\partial_s=\eta\) on the meridian at the boundary). The second term involves the shape operator: \(\partial_s\nu_a=-\sum_j h_{sj}\partial_j X_a\) (Weingarten), and the relevant projection at the boundary, combined with the rotational symmetry argument that eliminates the \(\partial_\theta\)-component, leaves a contribution that combines with the \(s_0'(a)\nabla_\eta\nu_a\) term as follows.

By the standard variation formula for the unit normal of a family of immersions (cf. [13] or [10]), the variation of \(\nu_a\) along the deformation field \(V\) on \(\Sigma_a\) is given, modulo terms in the conormal direction within the tangent space, by \[\label{eq:D-a-nu} D_a\nu_a\bigl|_\partial=-\nabla_\Sigma\phi_a+r'(a)\,\nabla_\eta\nu_a,\tag{25}\] where \(\nabla_\Sigma\phi_a=(\partial_\eta\phi_a)\eta+(\partial_\theta\phi_a/B)\partial_\theta\) is the intrinsic gradient of \(\phi_a\) on \(\Sigma_a\), and the second term comes from the displacement \(s_0'(a)\) together with the \(\eta\)-component of \(V_\partial\) (i.e. \(r'(a)\)). We note that the formula 25 extends verbatim to the hyperbolic ambient: the derivation in [10], [13] uses only (i) the orthogonality conditions \(\langle\nu_a,\partial_s X_a\rangle=\langle\nu_a,\partial_\theta X_a\rangle=0\), (ii) Weingarten’s identity \(\partial_s\nu_a=-\kappa_s\,\eta\) (valid in any Riemannian ambient), and (iii) the unit-norm condition \(\langle\nu_a,\nu_a\rangle=1\). None of these involves the ambient curvature, so the formula holds without modification in \(\mathbb{H}^3\). Projecting 25 onto \(\nabla\rho\big|_\partial=\eta\), \[\langle D_a\nu_a,\eta\rangle=-\partial_\eta\phi_a+r'(a)\langle\nabla_\eta\nu_a,\eta\rangle=-\partial_\eta\phi_a-r'(a)\kappa_s,\] using \(\langle\nu,\eta\rangle=0\), whence \(\langle\nabla_\eta\nu,\eta\rangle+\langle\nu,\nabla_\eta\eta\rangle=0\) and \(\langle\nu,\nabla_\eta\eta\rangle=\mathrm{II}(\eta,\eta)=\kappa_s\).

For the Hessian of the hyperbolic distance, \(\nabla^2\rho=\coth\rho\,(g-d\rho\otimes d\rho)\). Substituting \(V_\partial\), \[\nabla_{V_\partial}\nabla\rho=\coth r(a)(V_\partial-\langle V_\partial,\nabla\rho\rangle\nabla\rho)=\coth r(a)(\phi_a\nu+r'\eta-r'\eta)=\coth r(a)\cdot\phi_a\nu,\] and \(\langle\nu_a,\coth r\cdot\phi_a\nu\rangle=\coth r\cdot\phi_a\). Combining, \[0=-\partial_\eta\phi_a-r'(a)\kappa_s+\coth r(a)\,\phi_a,\] which is exactly 23 .

For the value \(\kappa_s=K/B(s_0)^2\), by minimality \(\kappa_s+\kappa_\theta=0\) and by Mori–Gauss (Lemma 2), \(\kappa_s^2=K^2/B^4\). The orientation of \(\nu\) with \(\nu^2(s,0)=+K/B\) fixes the sign: a direct computation at \(s=0\), using \(\bar\nabla_{T_s}T_s=\partial_s^2 X-X\) and \(\kappa_s\nu=\bar\nabla_{T_s}T_s|^\perp\), gives \(\kappa_s(0)=K/B(0)^2>0\). By continuity (and since \(K,B\) do not vanish on \([-s_0,s_0]\)), \(\kappa_s=K/B^2\) everywhere.

By Pigazzini [2], \(a\mapsto r(a)\) is real-analytic on \((1/2,\infty)\). One has \(r(a)>0\) for every \(a>1/2\), with \(r(a)\to 0\) as \(a\to(1/2)^+\) (Pigazzini [2]) and \(r(a)\to\infty\) as \(a\to\infty\) (which follows directly from the asymptotic \(r(a)\sim\tfrac{3}{2}\log a\) at infinity, by Pigazzini [2]). The strict monotonicity of \(r(a)\) on \((1/2,\infty)\) is posed as an open question in [2]; we do not need to assume it, since the condition \(r'(a)\neq 0\) used in Theorem 18 follows automatically from the hypothesis \(\phi_a>0\) of Theorem 32 (see Remark 33). ◻

Proof. Assume, by contradiction, that \(\psi\) is a nontrivial Robin Jacobi field in mode \(0\) even: \(L_{\Sigma}\psi=0\), \(R\psi=0\), \(\psi(\theta)=\psi\) independent of \(\theta\), \(\psi(-s)=\psi(s)\).

We apply 26 to \(u=\phi_a\), \(v=\psi\) in two ways: \[\begin{align} \mathcal{S}(\phi_a,\psi)&=-\int_{\Sigma_a}\psi\,L_{\Sigma}\phi_a+\oint_{\partial\Sigma_a}\psi\,R\phi_a=0+\oint\psi\cdot(-r'(a)\kappa_s)\,dL,\\ \mathcal{S}(\phi_a,\psi)&=-\int_{\Sigma_a}\phi_a\,L_{\Sigma}\psi+\oint_{\partial\Sigma_a}\phi_a\,R\psi=0+0=0. \end{align}\] Equating, \(r'(a)\kappa_s\big|_\partial\oint_{\partial\Sigma_a}\psi\,dL=0\). By hypothesis \(r'(a)\neq 0\), and \(\kappa_s\big|_\partial=K/B(s_0)^2>0\), so \[\oint_{\partial\Sigma_a}\psi\,dL=0.\] In mode \(0\), \(\psi\) is constant on each boundary circle, with value \(\psi(s_0)=\psi(-s_0)\) (even parity). Hence \(\oint\psi\,dL=4\pi B(s_0)\psi(s_0)\), whence \(\psi(s_0)=0\).

Combining with the Robin BC \(R\psi(s_0)=\psi'(s_0)-\coth r\cdot\psi(s_0)=0\), we obtain \(\psi'(s_0)=0\). Thus \(\psi\) is a solution of the radial Sturm–Liouville ODE of mode \(0\) with zero Cauchy data at \(s_0\). By the uniqueness theorem, \(\psi\equiv 0\) on \([-s_0,s_0]\), a contradiction. ◻

Proof. For the explicit form, by Pigazzini [2], at \(\theta=0\), \(\nu^0=K\cosh\varphi/A-a\sinh(2s)\sinh\varphi/(AB)\) and \(\nu^1=K\sinh\varphi/A-a\sinh(2s)\cosh\varphi/(AB)\). A direct computation gives \(\langle L_{01},\nu\rangle=-A\sinh\varphi\cdot\nu^0+A\cosh\varphi\cdot\nu^1=-a\sinh(2s)/B\) (the terms in \(K\) cancel by \(\cosh^2\varphi-\sinh^2\varphi=1\)).

The field \(u_*\) is of mode \(0\) since it is independent of \(\theta\) (a computation at generic \(\theta\) gives the same value by the rotational invariance of \(L_{01}\big|_{\text{axis}}\)). For parity, \(\sinh(2s)\) is odd and \(B\) is even, hence \(u_*\) is odd.

The identity \(L_{\Sigma}u_*=0\) follows from the standard variation principle for ambient isometries: since \(L_{01}\in\mathfrak{so}(3,1)\) generates a one-parameter group of isometries of \(\mathbb{H}^3\), the normal component \(\langle L_{01},\nu\rangle\) of the corresponding ambient flow is a Jacobi field on \(\Sigma_a\). The same variational principle is applied in Pigazzini [2] to \(K\in\mathfrak{so}(3)\subset\mathfrak{so}(3,1)\) (where the additional preservation of \(B^3(r(a))\) yields the Robin boundary condition); here the boost \(L_{01}\) does not preserve \(B^3(r(a))\), so \(u_*\) is a Jacobi field with a non-trivial Robin defect, computed below.

For the Robin defect, differentiating, \(u_*'(s)=-a[2\cosh(2s)/B-\sinh(2s)\cdot B'/B^2]\). Using \(B'=a\sinh(2s)/B\), \(u_*'(s)=-a[2\cosh(2s)B^2-a\sinh^2(2s)]/B^3\). At \(s_0\), \[Ru_*(s_0)=u_*'(s_0)-\coth r(a)\cdot u_*(s_0).\] Using \(\coth r=a\sinh(2s_0)/B(s_0)^2\) (from \(\sinh r=B^2/\sqrt{B^2-K^2}\) and \(\cosh r=a\sinh(2s_0)/\sqrt{B^2-K^2}\)), \[\begin{align} Ru_*(s_0)&=\frac{-a[2\cosh(2s_0)B^2-a\sinh^2(2s_0)]}{B^3}+\frac{a\sinh(2s_0)}{B^2}\cdot\frac{a\sinh(2s_0)}{B}\\ &=\frac{a[2a\sinh^2(2s_0)-2\cosh(2s_0)B^2]}{B^3}. \end{align}\] A direct computation gives \(a\sinh^2(2s_0)-\cosh(2s_0)B^2=a\cosh^2(2s_0)-a-a\cosh^2(2s_0)+\cosh(2s_0)/2=(\cosh(2s_0)-2a)/2\), whence \(Ru_*(s_0)=a(\cosh(2s_0)-2a)/B^3\). For odd parity, \(u_*'(-s_0)=u_*'(s_0)\) and \(u_*(-s_0)=-u_*(s_0)\); in the computation with opposite outward conormal, \(Ru_*(-s_0)=-Ru_*(s_0)\).

The identity \(\cosh(2s_0)-2a=(B(s_0)^2-2K^2)/a\) follows from \(a\cosh(2s_0)=B^2+1/2\) and \(2K^2=2a^2-1/2\). ◻

Proof. Let \(\psi\) be a nontrivial Robin Jacobi field in mode \(0\) odd, by contradiction. We apply 26 with \(u=u_*\), \(v=\psi\): \[\begin{align} \mathcal{S}(u_*,\psi)&=0+\oint\psi\cdot Ru_*\,dL,\\ \mathcal{S}(u_*,\psi)&=0+0=0. \end{align}\] The value \(Ru_*(s_0)\) is constant (by rotational symmetry) on each boundary circle, with opposite signs by odd parity: \(Ru_*(-s_0)=-Ru_*(s_0)\). The field \(\psi\) is odd, so \(\psi(-s_0)=-\psi(s_0)\), constant on each circle. Therefore \[\oint\psi Ru_*\,dL=2\pi B(s_0)\bigl[\psi(s_0)Ru_*(s_0)+\psi(-s_0)Ru_*(-s_0)\bigr]=4\pi B(s_0)\,\psi(s_0)\,Ru_*(s_0).\] Under 28 , \(Ru_*(s_0)\neq 0\), so \(\psi(s_0)=0\). Combined with \(R\psi(s_0)=0\), \(\psi'(s_0)=0\). With zero Cauchy data for the radial odd ODE of mode \(0\), by uniqueness \(\psi\equiv 0\). ◻

Proof. By Pigazzini [2], \(r(a)=c_*\sqrt{a-1/2}(1+o(1))\) with \(c_*=\sigma_*\cosh\sigma_*\) and \(\sigma_*=\coth\sigma_*\). The formulas \(\sinh r=B^2/\sqrt{B^2-K^2}\) and \(\cosh r=a\sinh(2s_0)/\sqrt{B^2-K^2}\) (from the FBC) linearize near \(a=1/2^+\) to \[s_0(a)=\rho_*\sqrt{a-1/2}\,(1+o(1)),\qquad\rho_*=\sinh\sigma_*.\] A direct verification confirms this: the identity \(\sinh^2 r=B^4/(B^2-K^2)\) with \(B^2\sim(a-1/2)(1+\rho_*^2)\) and \(K^2\sim a-1/2\) gives \(\sinh^2 r\sim(a-1/2)(1+\rho_*^2)^2/\rho_*^2\), while \(r^2\sim c_*^2(a-1/2)\). Hence \(c_*^2=(1+\rho_*^2)^2/\rho_*^2\), namely \(c_*=(1+\rho_*^2)/\rho_*=\rho_*+1/\rho_*\); using \(\rho_*=\sinh\sigma_*\) and \(\sigma_*\sinh\sigma_*=\cosh\sigma_*\), we recover \(c_*=\sigma_*\cosh\sigma_*\).

Therefore \(\cosh(2s_0)-1=2s_0^2+O(s_0^4)=2\rho_*^2(a-1/2)+O((a-1/2)^2)\), and \(2a-1=2(a-1/2)\). Subtracting, \[\cosh(2s_0)-2a=2\rho_*^2(a-1/2)-2(a-1/2)+O((a-1/2)^2)=2(\rho_*^2-1)(a-1/2)+O((a-1/2)^2).\] For the positivity \(\rho_*>1\): at \(\sigma=\log(1+\sqrt{2})\), \(\sinh\sigma=1\), \(\cosh\sigma=\sqrt{2}\), \(\coth\sigma=\sqrt{2}\). The function \(\sigma-\coth\sigma\) is strictly increasing on \((0,\infty)\) (with derivative \(1+\mathrm{csch}^2\sigma>0\)), and vanishes at \(\sigma=\sigma_*\). Since \(\log(1+\sqrt{2})-\sqrt{2}=0.881-1.414=-0.533<0\), the root satisfies \(\sigma_*>\log(1+\sqrt{2})\), whence \(\rho_*=\sinh\sigma_*>1\). ◻

Proof. From the identity \(\sinh^2 r=B^4/(B^2-K^2)\) established in the proof of Lemma 5, \[\sinh^2 r-4K^2=\frac{B^4-4K^2(B^2-K^2)}{B^2-K^2}=\frac{B^4-4K^2 B^2+4K^4}{B^2-K^2}=\frac{(B^2-2K^2)^2}{B^2-K^2},\] which is ?? . We verify \(B(s_0)^2-K^2>0\) as follows. By definition, \(r(a)\) is the positive geodesic radius of the boundary sphere \(\partial B^3(r(a))\), so \(\sinh r(a)>0\). The identity \(\sinh^2 r=B^4/(B^2-K^2)\), with \(B(s_0)^2>0\) and \(\sinh^2 r>0\), forces the denominator \(B^2-K^2\) to have the same sign as the numerator \(B^4>0\), hence \(B^2-K^2>0\) strictly. Therefore, the ratio on the right-hand side of ?? is nonnegative, with equality if and only if \(B^2=2K^2\). Taking positive roots (both sides are positive) gives ?? . Finally, condition 28 reads \(\cosh(2s_0)\neq 2a\); using \(B^2=a\cosh(2s_0)-1/2\) and \(K^2=a^2-1/4\), \(B^2-2K^2=a(\cosh(2s_0)-2a)\), whence \(B^2\neq 2K^2\iff\cosh(2s_0)\neq 2a\). Combined with the non-strict inequality ?? , we obtain ?? . ◻

Proof. For (a), by Theorem 18, \(\mu_n^{\mathrm{even}}(0)\neq 0\) for every \(n\). For (b), by Proposition 22, the strict condition \(\sinh r(a)>2K(a)\) is equivalent to condition 28 of Theorem 20; under this, \(\mu_n^{\mathrm{odd}}(0)\neq 0\) for every \(n\). The radial spectrum of mode \(0\) is the union (alternating in parity by classical Sturm–Liouville theory) of \(\{\mu_n^{\mathrm{even}}(0)\}\) and \(\{\mu_n^{\mathrm{odd}}(0)\}\), whence \(\mu_n(0)\neq 0\) for every \(n\) under \(\sinh r>2K\). For (c), the reduction of \(\mu_2(0)>0\) is the subject of Section 10. ◻

Proof. The FBC \(\tanh\varphi(s;a)=B(s;a)K(a)/(a\sinh(2s))\) has a unique solution \(s_0(a)\) since \(\partial_s\tanh\varphi>0\) and \(\partial_s[BK/(a\sinh 2s)]<0\) (cf. [2]). At \(s=s^*(a)\), \(\cosh(2s^*)=2a\), \(\sinh(2s^*)=2K\), \(B(s^*)^2=2K^2\), \(B(s^*)=K\sqrt{2}\). Hence \[\text{RHS of the FBC}(s^*;a)=\frac{B(s^*)K}{a\sinh(2s^*)}=\frac{K\sqrt{2}\cdot K}{2aK}=\frac{K}{a\sqrt{2}}=\sqrt{\tfrac{1}{2}-\tfrac{1}{8a^2}}.\] If \(\tanh\varphi(s^*;a)<\sqrt{1/2-1/(8a^2)}\), then at \(s=s^*\) the LHS of the FBC is strictly smaller than the RHS; by monotonicity, the crossing point \(s_0(a)\) lies at \(s>s^*(a)\). ◻

Proof. For \(s\in[0,s^*]\), \(A^2(s)\geq A^2(0)=a+1/2\) and \(B(s)\geq B(0)=\sqrt{a-1/2}\). Therefore \[\varphi(s^*;a)=K\int_0^{s^*}\frac{dt}{A^2(t)B(t)}<\frac{K\cdot s^*}{(a+1/2)\sqrt{a-1/2}}=\frac{\sqrt{(a-1/2)(a+1/2)}\cdot s^*}{(a+1/2)\sqrt{a-1/2}}=\frac{s^*}{\sqrt{a+1/2}}.\] From \(\tanh x<x\) for \(x>0\) we obtain 31 . ◻

Proof. Under the substitution \(u+1=\cosh(2\tau)\) with \(\tau\geq 0\), \(u\in(0,1]\iff\tau\in(0,\tau_1]\) where \(\tau_1=\tfrac{1}{2}\mathrm{arccosh}(2)\). Using \(\cosh(2\tau)+1=2\cosh^2\tau\) and \(\cosh(2\tau)-1=2\sinh^2\tau\), we obtain \(\sqrt{u}=\sqrt{2}\sinh\tau\), \(u+2=2\cosh^2\tau\), \(u+1=\cosh(2\tau)\), \(\mathrm{arccosh}(u+1)=2\tau\), and \[h(u)=2\sqrt{2}\cosh^2\tau\sinh\tau-2\tau\cosh(2\tau)=2\cosh^2\tau\cdot D(\tau),\] where \[\label{eq:D-tau} D(\tau):=\sqrt{2}\sinh\tau-2\tau+\tau\,\mathrm{sech}^2\tau.\tag{33}\] Since \(\cosh^2\tau>0\), it suffices to prove \(D(\tau)>0\) for \(\tau\in(0,\tau_1]\).

For the boundary values, \(D(0)=0\). At \(\tau=\tau_1=\tfrac{1}{2}\mathrm{arccosh}(2)\), \(\cosh(2\tau_1)=2\), whence \(\cosh^2\tau_1=3/2\), \(\sinh^2\tau_1=1/2\), \(\mathrm{sech}^2\tau_1=2/3\). Therefore \[D(\tau_1)=\sqrt{2}\cdot\tfrac{1}{\sqrt{2}}-\mathrm{arccosh}(2)+\tfrac{1}{2}\mathrm{arccosh}(2)\cdot\tfrac{2}{3}=1-\tfrac{2}{3}\mathrm{arccosh}(2).\] From \(\mathrm{arccosh}(2)=\int_1^2(t^2-1)^{-1/2}dt\leq\int_1^2(2(t-1))^{-1/2}dt=\sqrt{2}\), \[D(\tau_1)\geq 1-\tfrac{2\sqrt{2}}{3}=\tfrac{3-2\sqrt{2}}{3}>0,\] since \((2\sqrt{2})^2=8<9=3^2\).

We now show strict concavity of \(D\) on \([0,\tau_1]\). We compute \[D'(\tau)=\sqrt{2}\cosh\tau+\mathrm{sech}^2\tau\bigl(1-2\tau\tanh\tau\bigr)-2,\] \[D''(\tau)=\sqrt{2}\sinh\tau+\mathrm{sech}^2\tau\bigl(2\tau(3\tanh^2\tau-1)-4\tanh\tau\bigr).\] For \(\tau\in(0,\tau_1]\), since \(\tanh\tau\leq\tanh\tau_1=1/\sqrt{3}\), we have \(\tanh^2\tau\leq 1/3\), whence \(3\tanh^2\tau-1\leq 0\) and \(2\tau(3\tanh^2\tau-1)\leq 0\). Therefore \[-D''(\tau)\geq-\sqrt{2}\sinh\tau+4\,\mathrm{sech}^2\tau\tanh\tau=\frac{\sinh\tau\bigl(4-\sqrt{2}\cosh^3\tau\bigr)}{\cosh^3\tau}.\] On \([0,\tau_1]\), \(\cosh\tau\leq\sqrt{3/2}\), whence \(\sqrt{2}\cosh^3\tau\leq\sqrt{2}\cdot(3/2)^{3/2}=3\sqrt{3}/2\). From \((3\sqrt{3})^2=27<64=8^2\) we have \(3\sqrt{3}/2<4\), and hence \[-D''(\tau)\geq\frac{\sinh\tau\bigl(4-3\sqrt{3}/2\bigr)}{\cosh^3\tau}>0\quad\text{for }\tau\in(0,\tau_1].\] Therefore \(D''<0\) strictly on \((0,\tau_1]\), and \(D\) is strictly concave on \([0,\tau_1]\).

By strict concavity of \(D\) with \(D(0)=0\) and \(D(\tau_1)>0\), \[D(\tau)\geq(1-\tau/\tau_1)D(0)+(\tau/\tau_1)D(\tau_1)=(\tau/\tau_1)D(\tau_1),\] with strict inequality for \(\tau\in(0,\tau_1)\). Hence \(D(\tau)>0\) for every \(\tau\in(0,\tau_1]\). ◻

Proof. Combining Lemmas 16 and 17, for \(u=2a-1\in(0,1]\) the inequality \(s^*(a)/\sqrt{a+1/2}<\sqrt{1/2-1/(8a^2)}\) is equivalent, after squaring and simplification, to \(h(u)>0\) with \(u=2a-1\). Explicitly, from \(\sqrt{1/2-1/(8a^2)}=\sqrt{4a^2-1}/(2a\sqrt{2})\) and \(s^*(a)=\tfrac{1}{2}\mathrm{arccosh}(2a)\), \[\begin{align} \frac{\mathrm{arccosh}(2a)}{2\sqrt{a+1/2}}&<\frac{\sqrt{4a^2-1}}{2a\sqrt{2}}\\ \iff\quad a\sqrt{2}\,\mathrm{arccosh}(2a)&<\sqrt{a+1/2}\,\sqrt{4a^2-1}\\ \iff\quad 2a^2\mathrm{arccosh}^2(2a)&<(a+1/2)(4a^2-1)=(a+1/2)(2a-1)(2a+1)\\ &=\tfrac{1}{2}(2a+1)^2(2a-1). \end{align}\] Substituting \(u=2a-1\) (whence \(2a=u+1\), \(2a+1=u+2\), \(a=(u+1)/2\)), \[\tfrac{1}{2}(u+1)^2\mathrm{arccosh}^2(u+1)<\tfrac{1}{2}(u+2)^2 u\iff[(u+1)\mathrm{arccosh}(u+1)]^2<[(u+2)\sqrt{u}]^2.\] Both sides are positive for \(u>0\) (since \(\mathrm{arccosh}(u+1)>0\) for \(u>0\)), so the inequality is equivalent to \((u+1)\mathrm{arccosh}(u+1)<(u+2)\sqrt{u}\), namely \(h(u)>0\), which holds by Lemma 17.

Combining: for \(a\in(1/2,1]\), \(\tanh\varphi(s^*(a);a)<s^*(a)/\sqrt{a+1/2}<\sqrt{1/2-1/(8a^2)}\), and by Lemma 15, \(s_0(a)>s^*(a)\), equivalently \(\cosh(2s_0(a))>2a\). ◻

Proof. By Theorem 25, \(\sinh r(a)>2K(a)\) strictly for every \(a\in(1/2,1]\), so the hypothesis of Theorem 20 is satisfied, and Corollary 2(b) closes the odd part of the no-kernel of (F). Combined with (a) of the same corollary, \(\mu_n(0)\neq 0\) for every \(n\). Condition (E) is closed by Theorem 27. Only (F\('\))\(=\{\mu_2(0)>0\}\) remains, whose further reduction is the subject of Section 10. ◻

Proof. From \(B^2=a\cosh(2s)-1/2\) we obtain \(2BB'=2a\sinh(2s)\), whence \(BB'=a\sinh(2s)\) and \((BB')'=2a\cosh(2s)\). On the other hand \((BB')'=(B')^2+BB''\), and \(2a\cosh(2s)=2(B^2+1/2)=1+2B^2\). Therefore \(BB''+(B')^2=1+2B^2\). ◻

Proof. For the translation of the BC, if \(u=Bh\), then \(u'(s_0)=B'(s_0)h(s_0)+B(s_0)h'(s_0)\). The Robin condition \(u'(s_0)=\coth r\cdot u(s_0)=\coth r\cdot B(s_0)h(s_0)\), combined with \(B'(s_0)/B(s_0)=\coth r\) from Lemma 5, becomes \(h'(s_0)=0\). Analogously, \(h'(-s_0)=0\) by parity.

For the computation of the identity, setting \(u=Bh\), \(u'=B'h+Bh'\), whence \(B(u')^2=B(B'h+Bh')^2=B(B')^2h^2+2BB'B\,hh'+B^3(h')^2\), and \(2BB'B\,hh'=B^2B'(h^2)'\). Integrating in \(s\) and integrating by parts the cross term, \[\int_{-s_0}^{s_0}B^2B'(h^2)'ds=\bigl[B^2B'\,h^2\bigr]_{-s_0}^{s_0}-\int_{-s_0}^{s_0}(B^2B')'h^2ds.\] From \((B^2B')'=2B(B')^2+B^2B''\) and the identity \(BB''=1+2B^2-(B')^2\) of Lemma 18, \[B^2B''=B(1+2B^2-(B')^2)=B+2B^3-B(B')^2,\] whence \((B^2B')'=2B(B')^2+B+2B^3-B(B')^2=B(B')^2+B+2B^3\). Therefore \[\begin{align} \int B(u')^2ds&=\int B(B')^2h^2+\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}-\int(B(B')^2+B+2B^3)h^2+\int B^3(h')^2\\ &=\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}-\int(B+2B^3)h^2+\int B^3(h')^2. \end{align}\] Combining with the potential term \(-\int BW_ku^2=-\int B(|\mathrm{II}|^2-2-k^2/B^2)B^2h^2\), and using \(|\mathrm{II}|^2 B^3=2K^2/B\) (Lemma 2), \[\begin{align} \mathcal{S}_k^{\mathrm{rad}}(Bh,Bh)&=\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}-\int(B+2B^3)h^2+\int B^3(h')^2-\int(B^3|\mathrm{II}|^2-2B^3-k^2B)h^2\\ &\quad-\coth r\,B(s_0)^3\bigl[h(s_0)^2+h(-s_0)^2\bigr]\\ &=\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}+\int B^3(h')^2+\int(-B-2B^3-2K^2/B+2B^3+k^2B)h^2\\ &\quad-\coth r\,B(s_0)^3\bigl[h(s_0)^2+h(-s_0)^2\bigr]\\ &=\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}+\int B^3(h')^2+\int B((k^2-1)-2K^2/B^2)h^2-\coth r\,B(s_0)^3\bigl[h(s_0)^2+h(-s_0)^2\bigr]. \end{align}\]

The boundary terms cancel exactly. From \(B\) even, \(B'\) odd, \(B(\pm s_0)=B(s_0)\), \(B'(\pm s_0)=\pm B'(s_0)\), \[\bigl[B^2B'h^2\bigr]_{-s_0}^{s_0}=B(s_0)^2B'(s_0)[h(s_0)^2+h(-s_0)^2].\] By Lemma 5, \(B'(s_0)=\coth r\,B(s_0)\), whence \(B(s_0)^2B'(s_0)=\coth r\,B(s_0)^3\), which cancels with the last term. Rearranging the bulk, we obtain 35 . ◻

Proof. Let \(u\in H^1((-s_0,s_0))\) be even and nonzero, with Robin condition of mode \(|k|=2\). Setting \(h:=u/B\) (with \(B>0\) on \([-s_0,s_0]\)), \(h\in H^1\) is even with \(h'(\pm s_0)=0\) by Lemma 19. The identity 35 with \(k=2\) gives \[\label{eq:picone-B-k2} \mathcal{S}_2^{\mathrm{rad}}(u,u)=\int_{-s_0}^{s_0}\Bigl[B^3(h')^2+(3B^2-2K^2)\frac{h^2}{B}\Bigr]ds.\tag{36}\] The factor \(3B^2-2K^2\) is an even function of \(s\), strictly increasing in \(|s|\), with minimum at \(s=0\): \[3B(0)^2-2K^2=3(a-1/2)-2(a^2-1/4)=-(2a-1)(a-1)\geq 0\quad\text{for }a\in(1/2,1],\] with equality only for \(a=1\) at \(s=0\). Therefore \(3B^2-2K^2\geq 0\) pointwise on \([-s_0,s_0]\), with possible equality only on the set of measure zero \(\{s=0\}\) when \(a=1\).

Both integrands on the right-hand side of 36 are nonnegative. If \(h\) is not constant, \(\int B^3(h')^2>0\) and hence \(\mathcal{S}_2^{\mathrm{rad}}(u,u)>0\). If \(h\equiv c\neq 0\) is constant, then \[\mathcal{S}_2^{\mathrm{rad}}(u,u)=c^2\int_{-s_0}^{s_0}(3B^2-2K^2)/B\,ds>0,\] since the integrand is nonnegative with zero set of measure zero. In either case \(\mathcal{S}_2^{\mathrm{rad}}(u,u)>0\), hence \(\mu_0^{\mathrm{even}}(2)>0\).

The extension \(\mu_0^{\mathrm{even}}(k)>0\) for \(|k|\geq 2\) follows from Theorem 7: \(\mu_0^{\mathrm{even}}(k)\geq\mu_0^{\mathrm{even}}(2)+(k^2-4)/B(s_0)^2>0\) for \(|k|>2\), and directly for \(|k|=2\). ◻

Proof. Let \(u=Bh\), \(h\) even on \([-s_0,s_0]\), \(h'(\pm s_0)=0\), \(u\) nonzero. By symmetry \(h(-s)=h(s)\), hence \(\mathcal{S}_2^{\mathrm{rad}}(u,u)=2I\) where \[I:=\int_0^{s_0}\bigl[B^3(h')^2+(3B^2-2K^2)\frac{h^2}{B}\bigr]ds=J_-+J_++\int_0^{s^*}V\frac{h^2}{B}+\int_{s^*}^{s_0}V\frac{h^2}{B}\] with \(V:=3B^2-2K^2\), \(J_-:=\int_0^{s^*}B^3(h')^2\), \(J_+:=\int_{s^*}^{s_0}B^3(h')^2\).

For the estimate of the negative term, for \(s\in[0,s^*]\), \(h(s)-h(s^*)=-\int_s^{s^*}h'(t)dt\) and weighted Cauchy–Schwarz give \(|h(s)-h(s^*)|^2\leq(\int_s^{s^*}1/B^3)J_-\). Thus \(h(s)^2\leq 2h(s^*)^2+2(\int_s^{s^*}1/B^3)J_-\). Multiplying by \(|V(s)|/B(s)\) and integrating, \[\int_0^{s^*}|V|\frac{h^2}{B}ds\leq 2I_V h(s^*)^2+2\Bigl[\int_0^{s^*}\frac{|V|(s)}{B(s)}\int_s^{s^*}\frac{dt}{B(t)^3}ds\Bigr]J_-.\] By Fubini, the double integral becomes \(\int_0^{s^*}\bigl(\int_0^t|V|/B\,ds\bigr)/B(t)^3\,dt=K_*(a)\). Therefore \[\label{eq:hardy-neg} \int_0^{s^*}|V|\frac{h^2}{B}ds\leq 2I_V h(s^*)^2+2K_*(a)J_-.\tag{38}\]

For the estimate of the positive term, for \(s\in[s^*,s_0]\), \(h(s)^2\geq(1-\epsilon)h(s^*)^2-(\epsilon^{-1}-1)(h(s)-h(s^*))^2\) (reverse Cauchy–Schwarz) with \(\epsilon\in(0,1)\), and the analogous estimate \((h(s)-h(s^*))^2\leq(\int_{s^*}^s 1/B^3)J_+\). By Fubini, \[\label{eq:hardy-pos} \int_{s^*}^{s_0}V\frac{h^2}{B}ds\geq(1-\epsilon)I_V^+ h(s^*)^2-(\epsilon^{-1}-1)K_*^+(a)J_+.\tag{39}\]

Substituting 38 and 39 in \(I\), \[I\geq(1-2K_*(a))J_-+(1-(\epsilon^{-1}-1)K_*^+(a))J_++\bigl((1-\epsilon)I_V^+-2I_V\bigr)h(s^*)^2.\] The choice \(\epsilon=K_*^+/(1+K_*^+)\) annihilates the coefficient of \(J_+\) at \(0^+\). For admissibility we must have \((1-\epsilon)I_V^+\geq 2I_V\), namely \(I_V^+/(1+K_*^+)\geq 2I_V\), that is, \(2I_V(1+K_*^+)/I_V^+\leq 1\), which is the second condition of ?? . For the first condition, \(1-2K_*(a)>0\) ensures that the coefficient of \(J_-\) is positive. Hence \(I>0\) strictly, and \(\mathcal{S}_2^{\mathrm{rad}}(u,u)=2I>0\). ◻

Proof. By Theorem 27, (E) holds for every \(a\in(1/2,1]\) unconditionally. To extend beyond \(a=1\), we show that the Hardy conditions ?? are satisfied in a right neighborhood of \(a=1\).

At \(a=1\), equation 37 gives \(\cosh(2s^*(1))=(2+1)/3=1\), hence \(s^*(1)=0\). Therefore the interval of integration \([0,s^*(a)]\) degenerates to a point, and \[I_V(1)=K_*(1)=0,\] whence \(2K_*(1)=0<1\) and \(2I_V(1)(1+K_*^+(1))/I_V^+(1)=0<1\) (both conditions are satisfied with maximum margin, since \(I_V^+(1)>0\) as \(3B^2-2K^2>0\) on \((0,s_0(1)]\) by Theorem 27).

The functions \(a\mapsto s^*(a)\), \(a\mapsto s_0(a)\), \(a\mapsto B(\cdot;a)\) are real-analytic on \((1/2,\infty)\) (the first from the explicit formula 37 , the second from the implicit function theorem applied to the FBC, the third trivially). Therefore the functions \(I_V(a),K_*(a),I_V^+(a),K_*^+(a)\) are continuous (real-analytic for \(a>1\), and continuous up to \(a=1^+\) with limits \(0,0,I_V^+(1)>0,K_*^+(1)\geq 0\), respectively). By continuity, there exists \(\delta>0\) such that ?? holds for every \(a\in[1,1+\delta]\). Combined with Theorem 27, setting \(A_*:=1+\delta\) we obtain the claim. ◻

Proof. For (i), in \(a\in(1/2,1]\): (E) is closed by Theorem 27; the no-kernel part of (F) is closed by Corollary 3 (which uses Theorem 25). Only \(\mu_2(0)>0\) remains.

For (ii), in \(a\in(1,A_*]\): (E) is closed by Proposition 28 (combined with Theorem 29); the no-kernel part of (F) is closed under (G). The remaining conditions are \(\mu_2(0)>0\) and (G). ◻

Idea of the proof (standard). The count follows from Sturm’s oscillation theorem and the analysis of the rotation of the Prüfer angle \(\theta(s,\lambda)\) defined by \(\psi_\lambda(s)=R\sin\theta\), \(p\psi_\lambda'(s)=R\cos\theta\). As \(\lambda\) varies from \(-\infty\) to \(0\), \(\theta(s_0,\lambda)\) grows monotonically by a multiple of \(\pi\), and the count of negative eigenvalues coincides with the number of complete rotations plus a boundary correction \(\delta_\partial\) depending on the final position of \(\theta(s_0,0)\) with respect to the BC value. References: Hartman [18], Eastham [19]. ◻

Proof. We first treat the even radial sector of mode \(0\). The SL problem reduced to \([0,s_0]\) is \[-(Bu')'-B(|\mathrm{II}|^2-2)u=\mu Bu,\qquad u'(0)=0\;(\alpha_0=0,\text{ Neumann}),\qquad u'(s_0)=\coth r\cdot u(s_0)\;(\alpha_1=\coth r).\] The zero-energy solution with \(\psi_0(0)=\phi_a(0)=1/(2K)\), \(\psi_0'(0)=0\) is precisely \(\psi_0=\phi_a\) (Cauchy uniqueness for the Jacobi ODE \(L\phi_a=0\), Lemma 12). By 40 , \(n_z(\phi_a)=0\) in \((0,s_0)\), and \(\phi_a(s_0)>0\).

We now deduce \(r'(a)>0\) from the positivity of \(\phi_a\). By Theorem 11, \(\mathcal{S}(\Phi^0,\Phi^0)<0\) and \(\Phi^0\) belongs to the even radial sector of mode \(0\), hence \(\mu_0^{\mathrm{even}}(0)<0\). In particular \(N_-^{\mathrm{even}}\geq 1\). By Lemma 20, \(N_-^{\mathrm{even}}=n_z(\phi_a)+\delta_\partial=0+\delta_\partial\), hence \(\delta_\partial\geq 1\). Since \(\delta_\partial\in\{0,1\}\), necessarily \(\delta_\partial=1\), namely \[\phi_a'(s_0)/\phi_a(s_0)-\coth r<0\iff R\phi_a=\phi_a'(s_0)-\coth r\,\phi_a(s_0)<0.\] By Lemma 13, \(R\phi_a=-r'(a)\kappa_s\big|_\partial\) with \(\kappa_s\big|_\partial=K/B(s_0)^2>0\). Hence \(-r'(a)\kappa_s<0\), whence \(r'(a)>0\). In particular, hypothesis 40 automatically precludes the degeneracy \(r'(a)=0\), and consequently \(R\phi_a\neq 0\), the condition required by Theorem 18.

For the count of eigenvalues in the even sector, from \(\delta_\partial=1\), \(N_-^{\mathrm{even}}=1\). Hence \(\mu_0^{\mathrm{even}}(0)<0\leq\mu_1^{\mathrm{even}}(0)\). Combining with Theorem 18 (no-kernel even, valid under \(r'(a)\neq 0\) as just proved), \(\mu_1^{\mathrm{even}}(0)\neq 0\), hence \(\mu_1^{\mathrm{even}}(0)>0\) strictly.

We now turn to the odd radial sector of mode \(0\). The SL problem on \((0,s_0]\) is \[-(Bu')'-B(|\mathrm{II}|^2-2)u=\mu Bu,\qquad u(0)=0\;(\text{Dirichlet}),\qquad u'(s_0)=\coth r\cdot u(s_0)\;(\alpha_1=\coth r).\] The zero-energy odd solution is the explicit boost field \(u_*(s)=-a\sinh(2s)/B(s)\) of Lemma 14. Trivially \(u_*(s)\neq 0\) for \(s\in(0,s_0]\), since \(\sinh(2s)>0\) and \(B(s)>0\) in that range. Hence \(n_z(u_*)=0\) in \((0,s_0)\). For the boundary, \(\psi_0:=-u_*=a\sinh(2s)/B(s)\) is positive on \((0,s_0]\) (standard normalization). We compute \[\psi_0'(s_0)/\psi_0(s_0)-\coth r=\frac{2a-\cosh(2s_0)}{B(s_0)^2\sinh(2s_0)},\] as in the computation in Lemma 14. Under (G), \(\cosh(2s_0)>2a\), so the numerator is strictly negative, and \(\psi_0'/\psi_0-\coth r<0\). By Lemma 20, \(\delta_\partial=1\), and \(N_-^{\mathrm{odd}}=0+1=1\). Hence \(\mu_0^{\mathrm{odd}}(0)<0\leq\mu_1^{\mathrm{odd}}(0)\). By Theorem 20 (no-kernel odd under (G)), \(\mu_1^{\mathrm{odd}}(0)\neq 0\), hence \(\mu_1^{\mathrm{odd}}(0)>0\) strictly.

We now invoke the standard interleaving property. By classical SL theory on the symmetric interval \([-s_0,s_0]\) with even coefficients and symmetric BCs, the eigenvalues of the even and odd subsectors interlace: \(\mu_0^{\mathrm{even}}<\mu_0^{\mathrm{odd}}<\mu_1^{\mathrm{even}}<\mu_1^{\mathrm{odd}}<\cdots\) (the BCs of the two subsectors differ only at \(s=0\): Neumann even, Dirichlet odd, and Dirichlet \(\geq\) Neumann generates an upward shift of the eigenvalue by one unit). Therefore \(\mu_2(0)=\mu_1^{\mathrm{even}}(0)\) (the third overall eigenvalue), and we have proved \(\mu_1^{\mathrm{even}}(0)>0\). We conclude \(\mu_2(0)>0\), namely (F\('\)). ◻

Proof. The direction \(\Leftarrow\) in each case follows by combining: Theorem 32 (\(\phi_a>0\) and (G) imply (F\('\))\(=\mu_2(0)>0\)), Theorem 18 and Theorem 20 (no-kernel of mode \(0\) even and odd, the latter under (G)), and Theorem 14 ((E)+(F) \(\iff\) strong conjecture).

For (1), (E) is closed by Theorem 27 and (G) by Theorem 25; only \(\phi_a>0\) remains. The scalar reformulation 41 follows from Theorem 37 (equivalences (i)\(\Leftrightarrow\)(iii)\(\Leftrightarrow\)(iv)\(\Leftrightarrow\)(v)), valid under (G) strict, guaranteed by Theorem 25. For (2), (E) is closed by Proposition 28 on \((1/2,A_*]\) (combined with Theorem 29); (G) and \(\phi_a>0\) remain as hypotheses, but under (G) the second is equivalent to 41 by Theorem 37. For (3), all three conditions remain.

For the direction \(\Rightarrow\) in (1), if the strong conjecture holds, then \(\mathop{\mathrm{nul}}_R(\Sigma_a)=2\) implies in particular the absence of Robin kernel in mode \(|k|=0\); combined with \(\mu_0^{\mathrm{even}}(0)<0\) and \(\mu_0^{\mathrm{odd}}(0)<0\) (Theorem 11) and \(\mathop{\mathrm{ind}}_R=4\), \(\mu_2(0)>0\) follows. By Lemma 20 applied to the even sector, \(\mu_0^{\mathrm{even}}(0)<0\leq\mu_1^{\mathrm{even}}(0)\) implies \(N_-^{\mathrm{even}}=1\), and hence \(\phi_a\) (the zero-energy even solution) has no zeros in \((0,s_0)\) and has strictly negative boundary defect. Hence \(\phi_a>0\) on the principal branch. By Theorem 37, this is equivalent to 41 . The argument is analogous for (2) and (3). ◻

Proof. For two solutions \(u,v\) of the Sturm–Liouville equation in self-adjoint form \(-(Bu')'=B(|\mathrm{II}|^2-2)u\), the weighted Wronskian \(W(u,v)=B(uv'-u'v)\) is constant. Indeed, \[\frac{d}{ds}\bigl[B(uv'-u'v)\bigr]=B'(uv'-u'v)+B(uv''-u''v).\] From the equation, \(Bu''=-B'u'-B(|\mathrm{II}|^2-2)u\), and analogously for \(v\). Therefore \[B(uv''-u''v)=u\bigl(-B'v'-B(|\mathrm{II}|^2-2)v\bigr)-v\bigl(-B'u'-B(|\mathrm{II}|^2-2)u\bigr)=-B'(uv'-u'v),\] whence \(dW/ds=B'(uv'-u'v)-B'(uv'-u'v)=0\).

We compute \(W(\phi_a,u_*)(0)\). By Lemma 12, \(\phi_a(0)=1/(2K)\) and \(\phi_a'(0)=0\). For \(u_*(s)=-a\sinh(2s)/B(s)\), \(u_*(0)=0\) and \[u_*'(0)=\lim_{s\to 0}\frac{u_*(s)}{s}=-\frac{2a}{B(0)},\] where \(B(0)=\sqrt{a-1/2}\). Therefore \[W(\phi_a,u_*)(0)=B(0)\Bigl[\frac{1}{2K}\cdot\Bigl(-\frac{2a}{B(0)}\Bigr)-0\cdot 0\Bigr]=-\frac{a}{K}.\qedhere\] ◻

Proof. By 42 , \(W(\phi_a,u_*)\equiv-a/K\neq 0\), so \(\phi_a\) and \(u_*\) are linearly independent and constitute a basis of the space of solutions of the ODE \(L_0 u=0\). The field \(u_*(s)=-a\sinh(2s)/B(s)\) is strictly negative on \((0,s_0]\), since \(\sinh(2s)>0\) and \(B(s)>0\) for \(s\in(0,s_0]\) (as \(B(s)^2=a\cosh(2s)-1/2\geq a-1/2>0\)). Therefore \(u_*\) has exactly one zero on \([0,s_0]\), located at \(s=0\).

By the classical Sturm separation theorem (a direct consequence of the Sturm comparison theorem; see, e.g., [20] with \(g_1=g_2\)), between two consecutive zeros of one solution there is exactly one zero of every other linearly independent solution. If, by contradiction, \(\phi_a\) had two distinct zeros \(s_1<s_2\) in \((0,s_0]\), then \(u_*\) would have a zero in \((s_1,s_2)\subset(0,s_0)\), contradicting \(u_*<0\) on \((0,s_0]\). ◻

Proof. We evaluate the Wronskian identity 42 at \(s=s_0\): \[\label{eq:W-at-s0} \phi_a(s_0)\,u_*'(s_0)-\phi_a'(s_0)\,u_*(s_0)=-\frac{a}{K\,B(s_0)}.\tag{43}\] We introduce the Robin conditions. By Lemma 14, \(Ru_*(s_0):=u_*'(s_0)-\coth r(a)\cdot u_*(s_0)=(B^2-2K^2)/B^3\). By Lemma 13, \(R\phi_a(s_0):=\phi_a'(s_0)-\coth r(a)\cdot\phi_a(s_0)=-r'(a)\kappa_s\big|_\partial=-r'(a)\cdot K/B^2\). Substituting \(u_*'(s_0)=Ru_*+\coth r\cdot u_*(s_0)\) and \(\phi_a'(s_0)=R\phi_a+\coth r\cdot\phi_a(s_0)\) in 43 , \[\phi_a(s_0)\bigl[Ru_*+\coth r\cdot u_*(s_0)\bigr]-\bigl[R\phi_a+\coth r\cdot\phi_a(s_0)\bigr]u_*(s_0)=-\frac{a}{KB}.\] The terms in \(\coth r\) cancel (they amount to \(\coth r\cdot\phi_a(s_0)u_*(s_0)-\coth r\cdot\phi_a(s_0)u_*(s_0)=0\)), and we obtain \[\label{eq:Robin-wronskian-id} \phi_a(s_0)\,Ru_*(s_0)-R\phi_a(s_0)\,u_*(s_0)=-\frac{a}{K\,B(s_0)}.\tag{44}\] Substituting \(Ru_*=(B^2-2K^2)/B^3\), \(R\phi_a=-r'(a)K/B^2\), and \(u_*(s_0)=-a\sinh(2s_0)/B\), \[\phi_a(s_0)\cdot\frac{B^2-2K^2}{B^3}+\frac{r'(a)K}{B^2}\cdot\Bigl(-\frac{a\sinh(2s_0)}{B}\Bigr)=-\frac{a}{KB}.\] Multiplying both sides by \(B^3\), \[\phi_a(s_0)(B^2-2K^2)-r'(a)\,aK\sinh(2s_0)=-\frac{aB^2}{K}.\] Isolating \(\phi_a(s_0)\) (under the assumption \(B^2\neq 2K^2\)), \[\phi_a(s_0)=\frac{-aB^2/K+r'(a)\,aK\sinh(2s_0)}{B^2-2K^2}=\frac{a\bigl[K^2 r'(a)\sinh(2s_0)-B^2\bigr]}{K(B^2-2K^2)},\] which is ?? . To obtain ?? , we use \(\coth r=a\sinh(2s_0)/B^2\), namely \(a\sinh(2s_0)=B^2\coth r\); substituting \(\sinh(2s_0)=B^2\coth r/a\), \[K^2 r'(a)\sinh(2s_0)-B^2=K^2 r'(a)\cdot\frac{B^2\coth r}{a}-B^2=\frac{B^2}{a}\bigl[K^2 r'(a)\coth r-a\bigr],\] whence \(\phi_a(s_0)=B^2[K^2 r'(a)\coth r-a]/[K(B^2-2K^2)]\). ◻

Proof. For (i)\(\Rightarrow\)(ii), the implication is trivial.

For (ii)\(\Rightarrow\)(i), by Lemma 12, \(\phi_a(0)=1/(2K)>0\). By Lemma 22, \(\phi_a\) has at most one zero in \((0,s_0]\). If, by contradiction, \(\phi_a\) had a zero \(s_1\in(0,s_0)\), with \(\phi_a\) continuous, \(\phi_a(0)>0\), and \(\phi_a(s_0)>0\) by hypothesis, then \(\phi_a\) would have to cross zero at least twice (entering and leaving the negative half-plane), contradicting Lemma 22. Therefore \(\phi_a>0\) everywhere on \([0,s_0]\). (If \(\phi_a\) had exactly one zero \(s_1\in(0,s_0)\), then \(\phi_a\) would change sign at \(s_1\); assuming \(\phi_a(0)>0\), we would have \(\phi_a<0\) on \((s_1,s_0]\), contradicting \(\phi_a(s_0)>0\).)

For (ii)\(\iff\)(iii), by Proposition 36, formula ?? , under (G) (positive denominator) the sign of \(\phi_a(s_0)\) coincides with that of \(K^2 r'(a)\coth r-a\). The second form of (iii) is obtained by multiplying by \(\tanh r\).

For (iii)\(\iff\)(iv), set \(M(a):=\log[\sinh r(a)/K(a)]=\log\sinh r(a)-\tfrac{1}{2}\log(a^2-1/4)\). Differentiating with respect to \(a\) (\(r(a)\) is a real-analytic function of \(a\) by [2]), \[M'(a)=r'(a)\coth r(a)-\frac{a}{a^2-1/4}=r'(a)\coth r-\frac{a}{K^2}.\] Multiplying by \(K^2>0\), \[K^2 M'(a)=K^2 r'(a)\coth r-a.\] Hence \(M'(a)>0\iff K^2 r'(a)\coth r>a\). Moreover \((\sinh r/K)'(a)>0\iff M'(a)>0\) (since \(\log\) is strictly increasing for positive values).

For (iv)\(\iff\)(v), from the geometric identity \(\sinh^2 r=B^4/(B^2-K^2)\) (Proposition 22, evaluated at \(s=s_0\)), set \(y:=B(s_0)^2/K^2\). Then \[\label{eq:sinh-r-K-y} \Bigl(\frac{\sinh r}{K}\Bigr)^2=\frac{B^4}{K^2(B^2-K^2)}=\frac{(K^2 y)^2}{K^2(K^2 y-K^2)}=\frac{y^2}{y-1},\tag{45}\] namely \(\sinh r/K=y/\sqrt{y-1}\) (for \(y>1\), guaranteed by \(B^2>K^2\), which is implicit in (G) strict). The function \(h:(1,\infty)\to(0,\infty)\) defined by \(h(y):=y/\sqrt{y-1}\) has derivative \[h'(y)=\frac{\sqrt{y-1}-y/(2\sqrt{y-1})}{y-1}=\frac{2(y-1)-y}{2(y-1)^{3/2}}=\frac{y-2}{2(y-1)^{3/2}},\] strictly positive for \(y>2\). Condition (G) strict is equivalent to \(y>2\). Therefore, under (G), \(\sinh r/K\) is a strictly increasing function of \(y\), hence \((\sinh r/K)'(a)>0\iff y'(a)>0\). ◻

Proof. We start from the \(\coth r\) form of Proposition 36, equation ?? : \[\phi_a(s_0)=\frac{B(s_0)^2\bigl[K^2 r'(a)\coth r(a)-a\bigr]}{K(a)\bigl[B(s_0)^2-2K(a)^2\bigr]}.\] From the proof of (iii)\(\iff\)(iv) of Theorem 37, the pointwise differential identity \[\frac{d}{da}\log H(a)=\frac{d}{da}\log\sinh r(a)-\frac{d}{da}\log K(a)=r'(a)\coth r(a)-\frac{a}{K(a)^2}\] holds. Multiplying by \(K(a)^2\), \[\label{eq:K2-Hprime-over-H} K(a)^2\cdot\frac{H'(a)}{H(a)}=K(a)^2 r'(a)\coth r(a)-a.\tag{47}\] Substituting 47 in the formula for \(\phi_a(s_0)\), \[\phi_a(s_0)=\frac{B(s_0)^2\cdot K(a)^2\cdot H'(a)/H(a)}{K(a)\bigl[B(s_0)^2-2K(a)^2\bigr]}=\frac{B(s_0)^2 K(a) H'(a)}{H(a)\bigl[B(s_0)^2-2K(a)^2\bigr]}.\] Using \(1/H(a)=K(a)/\sinh r(a)\), \[\phi_a(s_0)=\frac{B(s_0)^2\,K(a)^2}{\bigl[B(s_0)^2-2K(a)^2\bigr]\sinh r(a)}\cdot H'(a),\] which is 46 .

Under (G) strict, \(B(s_0)^2-2K(a)^2>0\); moreover \(B(s_0)^2>0\), \(K(a)^2>0\), \(\sinh r(a)>0\) (since \(r(a)>0\) for \(a>1/2\)). Therefore the factor \(B^2 K^2/[(B^2-2K^2)\sinh r]\) is strictly positive, and \(\mathrm{sgn}\,\phi_a(s_0)=\mathrm{sgn}\,H'(a)\). ◻

Proof. Set \(\rho:=\sqrt{a-1/2}\), \(\xi:=s/\rho\), and \(\xi_0(a):=s_0(a)/\rho\). We first derive a real-analytic equation for \(\xi_0\) as a function of \(\rho^2\). The functions \(A(s)^2=a\cosh(2s)+1/2\), \(B(s)^2=a\cosh(2s)-1/2\), \(K(a)^2=a^2-1/4\), \(\cosh(2s)\), \(\sinh(2s)\) depend on \(\rho\) only through \(\rho^2\) (since \(a=1/2+\rho^2\)), and substituting \(s=\rho\xi\) they become jointly real-analytic in \((\xi,\rho^2)\) in a neighborhood of \((\sinh\sigma_*,0)\).

Both sides of the FBC \(\tanh\varphi(s_0;a)=B(s_0)K/[a\sinh(2s_0)]\) vanish to first order in \(\rho\) at \(\rho=0\): indeed, by the substitution \(t=\rho\tau\) in the defining integral of \(\varphi\) and the expansion \(K/[A^2 B]=1/[\rho\sqrt{1+\tau^2}]\cdot(1+O(\rho^2))\), \[\varphi(s_0)=\int_0^{s_0}\frac{K}{A^2 B}\,dt=\int_0^{\xi_0}\frac{1}{\sqrt{1+\tau^2}}\,d\tau\cdot\rho\cdot(1+O(\rho^2))=\rho\,\mathrm{arcsinh}(\xi_0)+O(\rho^3),\] whence \(\tanh\varphi(s_0)=\rho\,\mathrm{arcsinh}(\xi_0)+O(\rho^3)\). For the RHS, using \(K=\rho\sqrt{1+\rho^2}\) and \(\sinh(2s_0)=2\rho\xi_0+O(\rho^3)\), and writing \(B(s_0)=\rho\sqrt{1+\xi_0^2}+O(\rho^3)\) (from \(B(s_0)^2=\rho^2(1+\xi_0^2)+O(\rho^4)\)), \[\frac{B(s_0)K}{a\sinh(2s_0)}=\frac{\rho^2\sqrt{(1+\xi_0^2)(1+\rho^2)}+O(\rho^4)}{a\cdot[2\rho\xi_0+O(\rho^3)]}=\rho\cdot\frac{\sqrt{1+\xi_0^2}}{2a\xi_0}\cdot(1+O(\rho^2)),\] where the inner factor \(\sqrt{1+\xi_0^2}/(2a\xi_0)\cdot(1+O(\rho^2))\) is real-analytic in \((\xi_0,\rho^2)\) on the relevant neighborhood (since \(\xi_0\) is bounded away from \(0\) near \(\sinh\sigma_*>0\), and \(a=1/2+\rho^2>0\)). Dividing both sides of the FBC by \(\rho\), we obtain a real-analytic relation \[\mathcal{F}(\xi_0,\rho^2):=\frac{\tanh\varphi(s_0;a)}{\rho}-\frac{B(s_0)K}{a\rho\sinh(2s_0)}=0\] in a neighborhood of \((\sinh\sigma_*,0)\).

The limit equation at \(\rho^2=0\) is \(\mathrm{arcsinh}(\xi_0)=\sqrt{1+\xi_0^2}/\xi_0\), namely the transcendental equation of [2], with unique positive root \(\xi_0=\sinh\sigma_*\) where \(\sigma_*=\coth\sigma_*\). The associated map \(\Phi(\xi_0):=\mathrm{arcsinh}(\xi_0)-\sqrt{1+\xi_0^2}/\xi_0\) has derivative \[\Phi'(\xi_0)=\frac{1}{\sqrt{1+\xi_0^2}}+\frac{1}{\xi_0^2\sqrt{1+\xi_0^2}}=\frac{\sqrt{1+\xi_0^2}}{\xi_0^2}>0\] (direct computation), hence \(\partial_{\xi_0}\mathcal{F}(\sinh\sigma_*,0)=\Phi'(\sinh\sigma_*)\neq 0\).

By the analytic implicit function theorem (cf. [21]), there exists \(\eta>0\) and a unique real-analytic function \(\xi_0=\xi_0(\rho^2)\) on \([0,\eta)\) with \(\mathcal{F}(\xi_0(\rho^2),\rho^2)=0\) and \(\xi_0(0)=\sinh\sigma_*\). Equivalently, \(\xi_0\) is real-analytic in \(a-1/2\) on \([0,\eta)\), and \(s_0(a)=\rho\,\xi_0(a-1/2)\) is real-analytic in \(\rho\) and odd in \(\rho\).

By Proposition 22, \(\sinh^2 r(a)=B(s_0)^4/(B(s_0)^2-K^2)\). Write \(B(s_0)^2=\rho^2\,\widetilde{B}(\rho^2)\) and \(B(s_0)^2-K^2=\rho^2\,\widetilde{D}(\rho^2)\), where \(\widetilde{B}(\rho^2)=(1+\xi_0(\rho^2)^2)+O(\rho^2)\) and \(\widetilde{D}(\rho^2)=\xi_0(\rho^2)^2+O(\rho^2)\) are real-analytic in \(\rho^2\) on \([0,\eta)\), with \(\widetilde{B}(0)=\cosh^2\sigma_*>0\) and \(\widetilde{D}(0)=\sinh^2\sigma_*>0\). Then \[\sinh^2 r(a)=\frac{\rho^4\,\widetilde{B}(\rho^2)^2}{\rho^2\,\widetilde{D}(\rho^2)}=\rho^2\cdot\frac{\widetilde{B}(\rho^2)^2}{\widetilde{D}(\rho^2)}=(a-1/2)\cdot G(a-1/2),\] where \(G(\rho^2):=\widetilde{B}(\rho^2)^2/\widetilde{D}(\rho^2)\) is real-analytic in \(\rho^2\) on \([0,\eta')\) for some \(\eta'\in(0,\eta]\) (ratio of analytic functions with nonzero denominator at \(\rho^2=0\)), with \(G(0)=\cosh^4\sigma_*/\sinh^2\sigma_*>0\). Taking the positive square root (well-defined since \(G>0\) near \(\rho^2=0\)), \[\sinh r(a)=\rho\cdot\sqrt{G(\rho^2)},\] which is real-analytic in \(\rho\) and odd in \(\rho\).

Since \(K(a)=\rho\sqrt{1+\rho^2}\) is real-analytic in \(\rho\) and odd in \(\rho\), the ratio \[H(a)=\frac{\sinh r(a)}{K(a)}=\frac{\rho\sqrt{G(\rho^2)}}{\rho\sqrt{1+\rho^2}}=\sqrt{\frac{G(\rho^2)}{1+\rho^2}}\] is real-analytic in \(\rho^2=a-1/2\) on \([0,\eta')\), with \(H(0)=\sqrt{G(0)}=\cosh^2\sigma_*/\sinh\sigma_*=\sigma_*\cosh\sigma_*\) (using \(\sigma_*\sinh\sigma_*=\cosh\sigma_*\)). The conclusions on \(H'(a)\) and on \(\lim_{a\to(1/2)^+}H'(a)\) follow from term-by-term differentiation of the convergent Taylor series. ◻

Proof. We first expand \(s_0(a)\) to second order. Let \(\rho:=\sqrt{a-1/2}\) and \(\xi:=s/\rho\). Expanding in \(\rho\), \[B(s)^2=\rho^2(1+\xi^2)+\rho^4(\xi^4/3+2\xi^2)+O(\rho^6),\qquad A(s)^2=1+\rho^2(1+\xi^2)+\rho^4(2\xi^2+\xi^4/3)+O(\rho^6),\] \[K(a)=\rho\sqrt{1+\rho^2}=\rho(1+\rho^2/2+O(\rho^4)).\] Let \(\xi_0(a):=s_0(a)/\rho\), with expansion \(\xi_0=\sinh\sigma_*+\xi_1\rho^2+O(\rho^4)\) (the parity in \(\rho\) follows from the fact that the FBC is a function of \(\rho^2=a-1/2\)). The FBC \[\tanh\varphi(s_0;a)=\frac{B(s_0)\,K(a)}{a\sinh(2s_0)}\] expanded to \(O(\rho^3)\) yields \[\rho\sigma_*+\rho^3\bigl[\xi_1/\cosh\sigma_*+I_*-\sigma_*^3/3\bigr]=\rho\sigma_*+\rho^3\sigma_*\bigl[g_B^{(0)}-g_a^{(0)}+\xi_1(\sinh\sigma_*/\cosh^2\sigma_*-1/\sinh\sigma_*)\bigr],\] whence, isolating the coefficients of \(\rho^3\) and solving for \(\xi_1\), \[\label{eq:xi1-formula} \xi_1=\frac{(\sinh^2\sigma_*-1)(\sinh^2\sigma_*-4)(\sinh^2\sigma_*+1)}{12\sinh^2\sigma_*\cdot\sigma_*\cosh\sigma_*},\tag{50}\] where \(g_B^{(0)}=1/2+\sinh^2\sigma_*(\sinh^2\sigma_*+6)/[6\cosh^2\sigma_*]\), \(g_a^{(0)}=2\sinh^2\sigma_*/3+2\), and \[\label{eq:I-star-explicit} I_*=-\int_0^{\sigma_*}\frac{7\cosh^4\sigma+\cosh^2\sigma-5}{6\cosh^2\sigma}\,d\sigma=-\frac{1}{12}\Bigl[9\sigma_*+\frac{7}{2}\sinh(2\sigma_*)-\frac{10}{\sigma_*}\Bigr].\tag{51}\] The computation of \(I_*\) uses the substitution \(\eta=\sinh\sigma\) in \(\int_0^{\sinh\sigma_*}f(\eta)/\sqrt{1+\eta^2}\,d\eta\), where \(f(\eta)=-(3+15\eta^2+7\eta^4)/[6(1+\eta^2)]\) is the coefficient of \(\rho^2\) in the expansion of \(K(a)/[A(s)^2 B(s)]\cdot\rho\).

We now expand \(H(a)=\sinh r(a)/K(a)\). From \(\sinh r=B^2/\sqrt{B^2-K^2}\), expanding numerator and denominator to \(O(\rho^4)\) and dividing by \(K=\rho\sqrt{1+\rho^2}\), \[H(a)=\sigma_*\cosh\sigma_*\bigl[1+\rho^2\bigl(\beta_1/\cosh^2\sigma_*-\alpha_1-1/2\bigr)+O(\rho^4)\bigr],\] where \(\beta_1=2\sinh\sigma_*\,\xi_1+\sinh^4\sigma_*/3+2\sinh^2\sigma_*\) and \(\alpha_1=\xi_1/\sinh\sigma_*+\sinh^2\sigma_*/6+1-1/(2\sinh^2\sigma_*)\).

We now compute the coefficient. Setting \(s:=\sinh^2\sigma_*\), the coefficient of \(\xi_1\) in \(\beta_1/\cosh^2\sigma_*-\alpha_1\) equals \((s-1)/[\sinh\sigma_*(s+1)]\). Using the identity \(\sinh\sigma_*\cdot\sigma_*\cosh\sigma_*=\cosh^2\sigma_*=s+1\) (from \(\sigma_*\sinh\sigma_*=\cosh\sigma_*\)), \[\xi_1\cdot\frac{s-1}{\sinh\sigma_*(s+1)}=\frac{(s-1)^2(s-4)(s+1)}{12s(s+1)\cdot\sinh\sigma_*\cdot\sigma_*\cosh\sigma_*}=\frac{(s-1)^2(s-4)}{12s(s+1)}.\] Adding the non-\(\xi_1\) part (which, after reduction to a common denominator, equals \((s^3+5s^2-3s+3)/[6s(s+1)]\), equivalently \(2(s^3+5s^2-3s+3)/[12s(s+1)]\)) and subtracting \(1/2=6s(s+1)/[12s(s+1)]\), \[\beta_1/\cosh^2\sigma_*-\alpha_1-1/2=\frac{(s-1)^2(s-4)+2(s^3+5s^2-3s+3)-6s(s+1)}{12s(s+1)}=\frac{3s^3-2s^2-3s+2}{12s(s+1)}.\] The factorization \(3s^3-2s^2-3s+2=(s-1)(3s-2)(s+1)\) (verified by inspection at \(s=1\) as a root, and polynomial division for \(3s^2+s-2=(3s-2)(s+1)\)) yields \[\beta_1/\cosh^2\sigma_*-\alpha_1-1/2=\frac{(s-1)(3s-2)}{12s}.\] Therefore \[H'((1/2)^+)=\frac{d}{da}H(a)\Big|_{a\to(1/2)^+}=\sigma_*\cosh\sigma_*\cdot\frac{(s-1)(3s-2)}{12s},\] which is 49 .

We now verify the positivity of \(C_0\). We need \(s>1\) (namely \(\sinh\sigma_*>1\)) and \(s>2/3\). The second follows from the first. For \(s>1\), this is equivalent to \(\sigma_*>\mathrm{arcsinh}(1)=\log(1+\sqrt{2})\). Set \(F(\sigma):=\sigma-\coth\sigma\), strictly increasing on \((0,\infty)\) (since \(F'(\sigma)=1+\mathrm{csch}^2\sigma>0\)). From \(F(\sigma_*)=0\) and \[F(\log(1+\sqrt{2}))=\log(1+\sqrt{2})-\sqrt{2}<0\] (using the classical inequality \(\log(1+x)<x\) for \(x>0\), with \(x=\sqrt{2}\), giving \(\log(1+\sqrt{2})<\sqrt{2}\)), we deduce \(\log(1+\sqrt{2})<\sigma_*\) by strict monotonicity of \(F\), hence \(\sinh\sigma_*>1\), and \(C_0>0\).

Finally, the asymptotic of \(g(a)\) follows from the closed identity of Theorem 38 (rewritten as \(K^2 r'\cosh r-a\sinh r=K^3 H'\)): \[g(a)=K^2 r'-a\tanh r=\frac{K^2 r'\cosh r-a\sinh r}{\cosh r}=\frac{K^3 H'(a)}{\cosh r(a)}.\] As \(a\to(1/2)^+\): \(K\sim\sqrt{a-1/2}\), \(K^3\sim(a-1/2)^{3/2}\), \(\cosh r\to 1\). By Lemma 23, \(H'(a)\) has a convergent Taylor expansion with \(H'(a)=C_0+O(a-1/2)\) as \(a\to(1/2)^+\), in particular \(H'(a)\to C_0\). Combining, \(g(a)=K^3 H'(a)/\cosh r(a)\sim C_0(a-1/2)^{3/2}\). ◻

Proof. For (a), by Theorem 40, \(H'((1/2)^+)=C_0>0\), where \(H'((1/2)^+)\) denotes the right derivative of \(H\) at \(a=1/2\). By Lemma 23, \(H\) is real-analytic as a function of \(a-1/2\) on \([0,\eta)\) for some \(\eta>0\), hence \(H'(a)\) is continuous on \([1/2,1/2+\eta)\) with \(H'(1/2)=C_0\). By the continuity of \(H'\), there exists \(\delta_0\in(0,\min(\eta,1/2))\) such that \(H'(a)>C_0/2>0\) for every \(a\in(1/2,1/2+\delta_0)\).

For (b), by Theorem 37, under (G) strict (Theorem 25 with \(\delta_0<1/2\)), \(H'(a)>0\iff\phi_a>0\) on \([0,s_0]\).

For (c), by Theorem 32, \(\phi_a>0\Rightarrow\mu_2(0)>0\).

For (d), by the final balance of Theorem 35, in \((1/2,1]\) all other prerequisites are already closed analytically (the closure of (G) in Theorem 25, and of (E) in Theorem 27); the unique residual (F\('\)) is now closed by (c). Therefore the strong Medvedev conjecture holds. ◻

Proof. By [2], \(r(a)=(3/2)\log a+d_\infty+o(1)\) as \(a\to\infty\). Hence \(\sinh r(a)=(e^{r(a)}-e^{-r(a)})/2\sim e^{r(a)}/2\sim a^{3/2}e^{d_\infty}/2\). Moreover \(K(a)=\sqrt{a^2-1/4}\sim a\). From 45 , \(y^2/(y-1)=(\sinh r/K)^2\sim a\cdot e^{2 d_\infty}/4\), and for \(y\to\infty\) we have \(y^2/(y-1)=y+1+O(1/y)\), whence \(y(a)\sim a\,e^{2 d_\infty}/4\). Differentiating the asymptotic expansion, \(y'(a)\to e^{2 d_\infty}/4>0\). ◻

Proof. We first analyze the behavior of \(s_0(a)\). Expanding \(\cosh(2t)=1+2t^2+O(t^4)\) as \(t\to 0\), \[A(t)^2=(a+\tfrac{1}{2})+2at^2+O(t^4),\qquad B(t)^2=(a-\tfrac{1}{2})+2at^2+O(t^4).\] As \(a\to(1/2)^+\), we introduce the rescaling \(\tau:=t\sqrt{2a/(a-1/2)}\), namely \(t=\tau\sqrt{(a-1/2)/(2a)}\) and \(dt=\sqrt{(a-1/2)/(2a)}\,d\tau\). Then \[B(t)^2=(a-1/2)(1+\tau^2)+O((a-1/2)^2),\qquad A(t)^2=(a+1/2)+O(a-1/2),\] and \(K(a)=\sqrt{(a-1/2)(a+1/2)}\). The integral of \(\varphi\) at leading order is \[\begin{align} \varphi(s_0)&=\int_0^{s_0}\frac{K}{A^2\,B}\,dt=\int_0^{\tau_0}\frac{\sqrt{(a-1/2)(a+1/2)}}{(a+1/2)\sqrt{a-1/2}\sqrt{1+\tau^2}}\cdot\sqrt{\frac{a-1/2}{2a}}\,d\tau\,(1+o(1))\\ &=\sqrt{\frac{a-1/2}{2a(a+1/2)}}\cdot\mathrm{arcsinh}(\tau_0)\,(1+o(1)), \end{align}\] where \(\tau_0:=s_0(a)\sqrt{2a/(a-1/2)}\). As \(a\to(1/2)^+\) the factor \(\sqrt{2a(a+1/2)}\to 1\), hence \(\varphi(s_0)=\sqrt{a-1/2}\cdot\mathrm{arcsinh}(\tau_0)+o(\sqrt{a-1/2})\).

We now derive the asymptotic form of the FBC. From the FBC \(\tanh\varphi(s_0)=B(s_0)K/[a\sinh(2s_0)]\) we have (i) \(\tanh\varphi(s_0)=\varphi(s_0)+O(\varphi^3)=\sqrt{a-1/2}\cdot\mathrm{arcsinh}(\tau_0)+o(\sqrt{a-1/2})\), (ii) \(B(s_0)=\sqrt{(a-1/2)(1+\tau_0^2)}+o(\sqrt{a-1/2})\), (iii) \(K(a)=\sqrt{a-1/2}+o(\sqrt{a-1/2})\), and (iv) \(a\sinh(2s_0)=2as_0+O(s_0^3)=2a\cdot\tau_0\sqrt{(a-1/2)/(2a)}+o(\sqrt{a-1/2})=\tau_0\sqrt{2a(a-1/2)}+o(\sqrt{a-1/2})\). The FBC at leading order becomes \[\sqrt{a-1/2}\cdot\mathrm{arcsinh}(\tau_0)=\frac{\sqrt{(a-1/2)(1+\tau_0^2)}\cdot\sqrt{a-1/2}}{\tau_0\sqrt{2a(a-1/2)}}=\frac{(a-1/2)\sqrt{1+\tau_0^2}}{\tau_0\sqrt{2a(a-1/2)}}=\frac{\sqrt{(a-1/2)(1+\tau_0^2)}}{\tau_0\sqrt{2a}}.\] Dividing both sides by \(\sqrt{a-1/2}\) and taking the limit \(a\to(1/2)^+\) (where \(\sqrt{2a}\to 1\)), \[\label{eq:tau-star-eq} \mathrm{arcsinh}(\tau_0)=\frac{\sqrt{1+\tau_0^2}}{\tau_0}.\tag{52}\] This is exactly the transcendental equation of [2], which uniquely characterizes \(\tau_0\to\rho_*=\sinh\sigma_*\) as \(a\to(1/2)^+\).

For the limit of \(y(a)\), from \(s_0^2=\tau_0^2\cdot(a-1/2)/(2a)\), hence \(s_0^2/(a-1/2)=\tau_0^2/(2a)\to\sinh^2\sigma_*/1=\sinh^2\sigma_*\) as \(a\to(1/2)^+\). Therefore \[y(a)=\frac{B(s_0)^2}{K^2}=\frac{(a-1/2)+2as_0^2+O(s_0^4)}{(a-1/2)(a+1/2)}=\frac{1}{a+1/2}+\frac{2a}{a+1/2}\cdot\frac{s_0^2}{a-1/2}+o(1).\] As \(a\to(1/2)^+\): \(1/(a+1/2)\to 1\), \(2a/(a+1/2)\to 1\), \(s_0^2/(a-1/2)\to\sinh^2\sigma_*\). Thus \[y(a)\to 1+\sinh^2\sigma_*=\cosh^2\sigma_*.\] For the inequality \(\cosh^2\sigma_*>2\), the function \(F(\sigma):=\sigma-\coth\sigma\) is strictly increasing on \((0,\infty)\) (Remark 1.4 of [2]). At \(\sigma_0:=\mathrm{arcsinh}(1)\), \(\sinh\sigma_0=1\), \(\cosh\sigma_0=\sqrt{2}\), hence \(\coth\sigma_0=\sqrt{2}\) and \(F(\sigma_0)=\sigma_0-\sqrt{2}\). We show \(\sigma_0<\sqrt{2}\) (and hence \(F(\sigma_0)<0\)). From \(\sigma_0=\log(1+\sqrt{2})\) and the inequality \(\log(1+x)\leq x\) valid for every \(x\geq 0\), \(\sigma_0=\log(1+\sqrt{2})\leq\sqrt{2}\); the inequality is strict for \(x>0\), hence \(\sigma_0<\sqrt{2}\). Therefore \(F(\sigma_0)<0=F(\sigma_*)\), and by strict monotonicity of \(F\), \(\sigma_*>\sigma_0\), whence \(\cosh\sigma_*>\cosh\sigma_0=\sqrt{2}\) and \(\cosh^2\sigma_*>2\). ◻