Let $\alpha >0$, $\alpha \notin \mathbb{N}$, and let the data of problem (4), (5) be such that condition (26) holds. Then for an arbitrary β satisfying the estimate with small enough ε, there exists a unique solution $(R^{\pm },\omega ,c^{\pm })\in {\tilde{C}}^{2+\alpha }\left(S_1\right)\times {\tilde{C}}^{2+\alpha }\left(S_1\right)\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$ to system (11)–(13). It obeys the inequality

In order to linearize system (11), we apply the formula for the first variation of a functional ${\delta }_{0}R\left[r\right]=\frac{\mathrm{d}}{\mathrm{d}s}R\left[sr\right]{|}_{s=0}$. According to [14,15], the first variation of twice the mean curvature of ${\mathcal{G}}^{\pm }$ with respect to the double curvature of the sphere $S_{R_0^{\pm }}$ is where ${\Delta }_{S_1}$ is the Laplace – Beltrami operator on $S_1$ and $r^{\pm }\equiv R^{\pm }-R_0^{\pm }$; ${\mathcal{G}}^{\pm }=\{x=R_0^{\pm }\xi +r^{\pm }\left(\xi \right)\mathit{N},\xi \in S_1\}$, $\mathit{N}$ is the outward normal to $S_1$. This yields where $k^{-}=\frac{2{\sigma }^{-}}{R_0^{-}}-{\mathcal{P}}^{-}\left(c^{-}\right)\equiv {\mathcal{P}}^{-}\left({\rho }^{-}\right)-{\mathcal{P}}^{-}\left(c^{-}\right)$, $k^{+}=\frac{2{\sigma }^{+}}{R_0^{+}}-{\mathcal{P}}^{+}\left(c^{+}\right)+{\mathcal{P}}^{-}\left(c^{-}\right)\equiv {\mathcal{P}}^{+}\left({\rho }^{+}\right)-{\mathcal{P}}^{+}\left(c^{+}\right)+{\mathcal{P}}^{-}\left(c^{-}\right)-{\mathcal{P}}^{-}\left({\rho }^{-}\right)$ in view of (2);

We integrate equations (16) over $S_1$ by parts, then we get The integrals ${\int }_{S_1}{r}^{\pm }\mathrm{d}\xi$ can be expressed in terms of the differences of the volumes $|\mathcal{F}|-|B_{R_0^{-}}|$ and $|{\mathcal{F}}^{+}|-|B_{R_0^{+}}|$, $|B_{R_0^{\pm }}|=\frac{4\pi }{3}{R_0^{\pm }}^3$. For example, and hence where $Q\left[r^{\pm }\right]\equiv -\frac{1}{R_0^{\pm }}{\int }_{S_1}{r^{\pm }}^2\mathrm{d}\xi -\frac{1}{3{R_0^{\pm }}^2}{\int }_{S_1}{r^{\pm }}^3\mathrm{d}\xi$.

We rewrite (12) as follows where $|{\xi }^{\prime }{|}^2={\xi }_1^2+{\xi }_2^2$. On the other hand, $m^{+}={\rho }^{+}|B_{R_0^{+}}|$, $m^{-}={\rho }^{-}(|B_{R_0^{-}}|-|B_{R_0^{+}}|)$. That’s why We express $|{\mathcal{F}}^{\pm }|-|B_{R_0^{\pm }}|$ from (20) and substitute in (19). Then equalities (18) imply that and Finally, we have

In order to prove the solvability of system (16), (13), (21), (22), we use implicit function theorem. We represent this system as a nonlinear vector equation: where $\mathsf{\Phi }\left(\mathit{\phi }\right)=({\mathsf{\Phi }}_1^{\pm },{\mathsf{\Phi }}_2,{\mathsf{\Phi }}_3^{\pm })$, $\mathit{\phi }\equiv (r^{\pm },\omega ,{\lambda }^{\pm })$, ${\lambda }^{\pm }\equiv c^{\pm }-{\rho }^{\pm }$,

We linearize (23) at zero and show that the derivative of ${\mathsf{\Phi }}^{\prime }$ at this point is not equal to zero. Thus, where ${\mathsf{\Phi }}^{\prime }\left(0\right)=({{\mathsf{\Phi }}_1^{\pm }}^{\prime },{\mathsf{\Phi }}_2^{\prime },{{\mathsf{\Phi }}_3^{\pm }}^{\prime })\left(0\right)$ is the Fréchet derivative of $\mathsf{\Phi }$, that is, and $\mathit{\psi }\left(\mathit{\phi }\right)=\mathsf{\Phi }\left(\mathit{\phi }\right)-{\mathsf{\Phi }}^{\prime }\left(0\right)\mathit{\phi }\equiv ({\mathit{\psi }}_1^{\pm },{\mathit{\psi }}_2,{\mathit{\psi }}_3^{\pm })$ with

Let $\mathcal{X}={\tilde{C}}^{2+\alpha }\times {\tilde{C}}^{2+\alpha }\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$ and $\mathcal{Y}={\tilde{C}}^{\alpha }\times {\tilde{C}}^{\alpha }\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$. Using fixed point theorem, we will show that equation (23) is solvable in $\mathcal{X}.$

In [16] (see the corollary to Theorem 2.1), it was proved that for any $f\in {\tilde{C}}^{\alpha }\left(S_1\right)$ the equation has a unique solution $r\in {\tilde{C}}^{2+\alpha }\left(S_1\right)$ and Moreover, due to the assumptions of the theorem, the determinant of the matrix in the left hand sides of two equations ${{\mathsf{\Phi }}_3^{\pm }}^{\prime }\left(0\right)\mathit{\phi }=-{\mathit{\psi }}_3^{\pm }\left(\mathit{\phi }\right)$ is not equal to zero: (Here we have taken into account relations (8).) Hence, this matrix is invertible. Thus, the whole operator ${\mathsf{\Phi }}^{\prime }\left(0\right):\mathcal{X}\to \mathcal{Y}$ is also invertible and where ${\parallel \mathit{\phi }\parallel }_{\mathcal{X}}=|r^{+}{|}_{C^{2+\alpha }\left(S_1\right)}+|r^{-}{|}_{C^{2+\alpha }\left(S_1\right)}+|\omega |+|{\lambda }^{+}|+|{\lambda }^{-}|$ and

Therefore, equation (24) can be written in the form:

Let us now estimate the non-linear operator $\mathit{\psi }$. The term $f^{-}$ given by (17) can be written as where ${\mathcal{H}}^{-}\left[sr\right]$ is twice the mean curvature of the surface ${\mathcal{G}}_s^{-}$; ${\mathcal{G}}_s^{\pm }=\{x=R_0^{\pm }\xi +s{r}^{\pm }\left(\xi \right)\mathit{N},\xi \in S_1\}$, $\mathit{N}\equiv \frac{\xi }{|\xi |}$; ${\overline{c}}^{-}$ is some point of the interval $[c^{-},\frac{{\omega }^2}{2}{R^{+}}^2\left(\xi \right)|{\xi }^{\prime }{|}^2+c^{-}]$. One can write a similar formula for $f^{+}$. The terms $f^{\pm }$ have the second order of smallness with respect to $\mathit{\phi }$. After simple calculations for two values of the arguments $\mathit{\phi }=(r^{\pm },\omega ,{\lambda }^{\pm })$ and ${\mathit{\phi }}^{\prime }=({r^{\pm }}^{\prime },{\omega }^{\prime },{{\lambda }^{\pm }}^{\prime })\in \mathcal{X}$ such that it can be shown that the estimates hold. This implies that $\mathcal{A}\mathit{\psi }\left(\mathit{\phi }\right)$ maps the ball $\{\mathit{\phi }\in \mathcal{X}:\parallel \mathit{\phi }{\parallel }_{\mathcal{X}}\le \eta \}$ into itself and it is a contraction operator if These inequalities are satisfied for $\eta$ such that So if then, by fixed point theorem, equation (27) has a unique solution in the ball $\{\parallel \mathit{\phi }{\parallel }_{\mathcal{X}}\le \eta \}$ with $\eta$ satisfying (29). Thus, equation (23) is also uniquely solvable for $\beta$ obeying inequality (30), which gives the estimate of $\epsilon$ in condition (14).

We now prove estimate (15). Since then for $c_2{c}_3{\eta }^{\alpha }\le \frac{1}{2}$ which coincides with inequality (15). □