# Holder Space Theory for the Rotation Problem of a Two-Phase Drop
## 2 Setting of a problem
\begin{split}&\int_{\mathcal{G}^{\pm}}r_{0}^{\prime\prime}\, \mathrm{d}\mathcal{G}=\int_{\mathcal{G}^{\pm}}\left(r_{0}-\varphi(z,r_{0}) \right)\mathrm{d}\mathcal{G}\equiv l^{\pm},\\ &\rho^{-}\int_{\mathcal{G}^{-}}r_{0}^{\prime\prime}\mathbf{z}\, \mathrm{d}\mathcal{G}+[\bar{\rho}]|_{\mathcal{G}^{+}}\int_{\mathcal{G}^{+}}r_ {0}^{\prime\prime}\mathbf{z}\,\mathrm{d}\mathcal{G}=\rho^{-}\int_{\mathcal{G}^{-} }\left(r_{0}\mathbf{z}-\psi^{-}(z,r_{0})\right)\mathrm{d}\mathcal{G}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad+[\bar{\rho}]\big{|}_ {\mathcal{G}^{+}}\int_{\mathcal{G}^{+}}\left(r_{0}\mathbf{z}-\psi^{+}(z,r_{0}) \right)\,\mathrm{d}\mathcal{G}\equiv l,\\ &\int_{\mathcal{F}}\bar{\rho}\mathbf{u}_{0}^{\prime\prime}\,\mathrm{d}z= \int_{\mathcal{F}}\bar{\rho}\mathbf{u}_{0}\big{(}1-L(z,r_{0})\big{)}\,\mathrm{d}z \equiv m,\\ &\int_{\mathcal{F}}\bar{\rho}\mathbf{u}_{0}^{\prime\prime}\cdot\mathbf{ \eta}_{j}(z)\,\mathrm{d}z+\omega\Big{(}\rho^{-}\int_{\mathcal{G}^{-}}r_{0}^{ \prime\prime}\mathbf{\eta}_{3}\cdot\mathbf{\eta}_{j}\,\mathrm{d}\mathcal{G}+[\bar{ \rho}]\big{|}_{\mathcal{G}^{+}}\int_{\mathcal{G}^{+}}r_{0}^{\prime\prime}\mathbf{ \eta}_{3}\cdot\mathbf{\eta}_{j}\,\mathrm{d}\mathcal{G}\Big{)}\\ &=\omega\Big{(}\rho^{-}\int_{\mathcal{G}^{-}}r_{0}\mathbf{\eta}_{3} \cdot\mathbf{\eta}_{j}\,\mathrm{d}\mathcal{G}+[\bar{\rho}]\big{|}_{\mathcal{G}^{+}} \int_{\mathcal{G}_{i}}r_{0}\mathbf{\eta}_{3}\cdot\mathbf{\eta}_{j}\,\mathrm{d} \mathcal{G}-\int_{\bar{\Omega}_{0}}\rho^{\pm}\mathbf{\eta}_{3}(y)\cdot\mathbf{\eta}_{j}( y)\,\mathrm{d}y\Big{)}\\ &\qquad\qquad\qquad+\int_{\mathcal{F}}\bar{\rho}\mathbf{u}_{0}\cdot\mathbf{ \eta}_{j}(z)\big{(}1-L(z,r_{0})\big{)}\,\mathrm{d}z+\int_{\mathcal{F}}\bar{ \rho}\mathbf{\eta}_{3}(z)\cdot\mathbf{\eta}_{j}(z)\,\mathrm{d}z,\quad j=1,2,3,\\ &\
abla\cdot\mathbf{u}_{0}^{\prime\prime}=l_{2}(\mathbf{u}_{0},r_{0})\quad \text{in}\cup\mathcal{F}^{\pm},\\ &\left[\mathbf{u}_{0}^{\prime\prime}\right]_{\mathcal{G}^{+}}=0,\quad \left[\bar{\mu}\Pi\Omega_{\mathcal{G}}\mathbb{S}(\mathbf{u}_{0}^{\prime\prime}) \mathbf{N}\right]\big{|}_{\mathcal{G}^{+}}=I_{3}^{+}(\mathbf{u}_{0},r_{0}),\quad\mu^{ -}\Pi\Omega_{\mathcal{G}}\mathbb{S}(\mathbf{u}_{0}^{\prime\prime})\mathbf{N}\big{|}_{ \mathcal{G}^{-}}=I_{3}^{-}(\mathbf{u}_{0},r_{0}).\end{split}  \tag{64}\]  
We seek a solution to (44) in the form of the sum  
\[\mathbf{u}=\mathbf{u}^{\prime}+\mathbf{u}^{\prime\prime},\quad q=q^{\prime}+q^{\prime\prime}, \quad r=r^{\prime}+r^{\prime\prime},\]  
\[\int_{\mathcal{F}}\bar{\rho}\mathbf{u}_{0}^{\prime}\cdot\mathbf{\eta}_{j}\,\mathrm{d}z+ \omega\Big{(}\rho^{-}\int_{\mathcal{G}^{-}}r_{0}^{\prime}\mathbf{\eta}_{3}\cdot\mathbf{ \eta}_{j}\,\mathrm{d}\mathcal{G}+[\bar{\rho}]|_{\mathcal{G}^{+}}\int_{\mathcal{ G}^{+}}r_{0}^{\prime}\mathbf{\eta}_{3}\cdot\mathbf{\eta}_{j}\,\mathrm{d}\mathcal{S}\Big{)}=0.\]  
Theorem 3 guarantees for the solution \((\mathbf{u}^{\prime},q^{\prime},r^{\prime})\) of problem (65), the inequality:  
\[N\big{(}\mathbf{u}^{\prime}(\cdot,T),r^{\prime}(\cdot,T)\big{)}\,\equiv\,|\mathbf{u}^{ \prime}(\cdot,T)|_{\cup\mathcal{F}^{\pm}}^{(2+\alpha)}+|r^{\prime}(\cdot,T)|_{ \mathcal{G}}^{(3+\alpha)}\,\leqslant\,c_{1}\mathrm{e}^{-\beta T}\big{\{}|\mathbf{ u}_{0}|_{\cup\mathcal{F}^{\pm}}^{(2+\alpha)}+|r_{0}|_{\mathcal{G}}^{(3+ \alpha)}\big{\}}\]  
for any positive \(T\). Let \(T=T_{0}\) be so large that  
\[c_{1}e^{-\beta T_{0}}\leqslant\theta/2\ll 1/2,\qquad\beta>0.\]  
Next, problem (66) can be solved by iterations, similarly to [9] (Ch. 12), on the basis of Theorem 2 and estimate (58) of nonlinear terms (45):  
\[Z_{0,T}(\mathbf{u}^{\prime}+\mathbf{u}^{\prime\prime},q^{\prime}+q^{\prime\prime},r^{ \prime}+r^{\prime\prime})\leqslant c\big{\{}|f|_{\mathcal{Q}_{T_{0}}^{(1, \frac{1+\alpha-\gamma}{2})}}^{(1,\frac{1+\alpha-\gamma}{2})}+Y_{0,T}^{2}(\mathbf{u} ^{\prime}+\mathbf{u}^{\prime\prime},q^{\prime}+q^{\prime\prime},r^{\prime}+r^{ \prime\prime})\big{\}}.\]  
We observe that smallness inequality (57) of the zero-approximation is guaranteed by (67) and (60).