# Holder Space Theory for the Rotation Problem of a Two-Phase Drop
## 2 Setting of a problem
Thus, if \(\varepsilon\) is small enough, by inequalities (55) and (56), we obtain  
\[Y_{0,T_{0}}(\mathbf{u}^{\prime\prime},q^{\prime\prime},r^{\prime\prime}) \leqslant c_{2}(\varepsilon)\big{(}|f|_{\mathcal{Q}_{T_{0}}}^{(1, \frac{1+\alpha-\gamma}{2})}+|\mathbf{u}_{0}|_{\cup\mathcal{F}^{\pm}}^{(2+\alpha)}+ |r_{0}|_{\mathcal{G}}^{(3+\alpha)}\big{)},\]  
\[N\big{(}\mathbf{u}(\cdot,T_{0}),r(\cdot,T_{0})\big{)} \leqslant N\big{(}\mathbf{u}^{\prime}(\cdot,T_{0}),r^{\prime}(\cdot,T _{0})\big{)}+N\big{(}\mathbf{u}^{\prime\prime}(\cdot,T_{0}),r^{\prime\prime}( \cdot,T_{0})\big{)}\] \[\leqslant(\theta/2+\theta)\big{(}|\mathbf{u}_{0}|_{\cup\mathcal{F}^{ \pm}}^{(2+\alpha)}+|r_{0}|_{\mathcal{G}}^{(3+\alpha)}\big{)}+c|f|_{\mathcal{Q} _{T_{0}}}^{(1,\frac{1+\alpha-\gamma}{2})}.\]  
We set \(\lambda\equiv\theta/2+\theta<1\), due to (60), (61), which implies  
\[ \begin{split} Y_{0,T_{0}}(\mathbf{u},q,r)&\leqslant c \Big{(}|f|_{\mathcal{Q}_{T_{0}}^{(1,\frac{1+\alpha-\gamma}{2})}}^{(1,\frac{1+ \alpha-\gamma}{2})}+|\mathbf{u}_{0}|_{\mathcal{G}}^{(3+\alpha)}\Big{)}\leqslant c \varepsilon,\\ N\big{(}\mathbf{u}(\cdot,T_{0}),r(\cdot,T_{0})\big{)}& \leqslant\lambda\big{(}|\mathbf{u}_{0}|_{\cup\mathcal{F}^{\pm}}^{(2+ \alpha)}+|r_{0}|_{\mathcal{G}}^{(3+\alpha)}\big{)}+c|f|_{\mathcal{Q}_{T_{0}}} ^{(1,\frac{1+\alpha-\gamma}{2})}\leqslant C\varepsilon.\end{split}  \tag{68}\]  
In view of inequalities (68), we can extend the solution \((\mathbf{u},q,r)\) into the intervals \((T_{0},2T_{0}),\,...,\,(kT_{0},(k+1)T_{0}),...\) up to the infinite interval \(t>0\) by means of the repeated applications of the obtained local result and to complete the proof of Theorem 5 by analogy with [9] (Ch. 12). Thus, let us suppose that the solution has already been found for \(t\leqslant kT_{0}\). Then we can define it for \(t\in(kT_{0},(k+1)T_{0}]\) as a solution of the problem with the initial conditions \(\mathbf{u}(z,kT_{0})\equiv\mathbf{u}_{k}(z)\) and \(r(z,kT_{0})\equiv r_{k}(z)\). We consider the case \(k=1\). From (54) and (55), it follows that  
\[N_{1}\equiv N(\mathbf{u}_{1},r_{1})\leqslant C\varepsilon;\]  
hence, by replacing \(\varepsilon\) with \(C^{-1}\varepsilon\), we see that this problem is solvable in the time interval \((T_{0},2T_{0}]\), and by (68), the estimates  
\[ \begin{split} Y_{1}(\mathbf{u},q,r)\leqslant c\Big{\{}N_{1}+|f|_{ \mathcal{Q}_{T_{0}},2T_{0}}^{(1,\frac{1+\alpha-\gamma}{2})}\Big{\}},\\ N_{2}\leqslant\lambda N_{1}+c|f|_{\mathcal{Q}_{T_{0}},2T_{0}}^{(1, \frac{1+\alpha-\gamma}{2})}\leqslant C\varepsilon\end{split} \]  
are satisfied, where  
\[N_{k}\equiv N(\mathbf{u}_{k},r_{k}),\qquad Y_{k}(\mathbf{u},q,r)\equiv Y_{kT_{0},(k+1)T _{0}}(\mathbf{u},q,r).\]If the solution is found for \(t\leqslant kT_{0}\) and the inequalities  
\[N_{j} \leqslant\lambda N_{j-1}+c|f|_{Q_{(j-1)}T_{0}/T_{0}}^{\left(\frac{ 1,\frac{1+\alpha-\gamma}{2}}{2}\right)},\quad\lambda<1, \tag{69}\] \[Y_{j} \leqslant c\Big{\{}N_{j}+|f|_{Q_{(j+1)}T_{0}}^{\left(\frac{1+ \alpha-\gamma}{2}\right)}\Big{\}},\quad j=1,...,k,\]  
are proved, then for \(\lambda_{0}=\mathrm{e}^{-bT_{0}}<\lambda\)  
\[N_{j} \leqslant...\leqslant\lambda^{j}N_{0}+c\sum_{l=0}^{j-1}\lambda^{ j-1-i}|f|_{Q_{T_{0}(i+1)}T_{0}}^{\left(\frac{1,\frac{1+\alpha-\gamma}{2}}{2} \right)} \tag{70}\] \[\leqslant\lambda^{j}N_{0}+c\lambda^{j-1}|\mathrm{e}^{bT_{0}}f|_{ Q_{0}/T_{0}}^{\left(\frac{1+\alpha-\gamma}{2}\right)}\sum_{l=0}^{j-1}\frac{ \lambda_{0}^{i}}{\lambda^{l}}\leqslant c\lambda^{j}\Big{(}N_{0}+\frac{| \mathrm{e}^{bT_{0}}f|_{Q_{\infty}}^{\left(\frac{1,\frac{1+\alpha-\gamma}{2} }{2}\right)}}{\lambda-\lambda_{0}}\Big{)}\leqslant c\lambda^{j}\varepsilon\]  
with the constants \(c\) independent of \(j\). We have used inequalities (61) for \(\mathbf{f}\). Since \(\lambda^{j}\to 0\) as \(j\to\infty\), the right-hand side of (70) is less than \(\varepsilon\) for \(j\geqslant j_{0}\), and the replacement of \(\varepsilon\) with \(\mathrm{C}^{-1}\varepsilon\) can be done only a finite number of times. Let \(\lambda_{1}>\lambda\) (\(\lambda_{1}=\mathrm{e}^{-aT_{0}}\), \(a<b\)).