\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 93, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/93\hfil Analytic solution] {Analytic solution of an initial-value problem from Stokes flow with free boundary} \author[X. Xie\hfil EJDE-2008/93\hfilneg] {Xuming Xie} \address{Xuming Xie \newline Department of Mathematics\\ Morgan State University\\ Baltimore, MD 21251, USA} \email{xuming.xie@morgan.edu} \thanks{Submitted March 5, 2008. Published July 2, 2008.} \thanks{Supported by grant DMS-0500642 from National Science Foundation} \subjclass[2000]{35Q72, 74F05, 80A22} \keywords{Stokes flow; free boundary problem; analytic solution; \hfill\break\indent abstract Cauchy-Kovalevsky problem; initial-value problem} \begin{abstract} We study an initial-value problem arising from Stokes flow with free boundary. If the initial data is analytic in disk $\mathcal{R}_r$ containing the unit disk, it is proved that unique solution, which is analytic in $\mathcal{R}_s$ for $s\in (1,r)$, exists locally in time. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newcommand{\abs}[1]{\lvert#1\rvert} \newcommand{\norm}[1]{\lVert#1\rVert} \section{Introduction} The study of the deformation and breakup of bubbles in a slow viscous flow is of importance in many practical applications such as the rheology and mixing in multiphase viscous system. There has been a lot of research on this subject, the review article by Stone \cite{s1} summarizes the state of affair in the early nineties. This problem has been recently studied by some investigators. Tanveer and Vasconcelos \cite{t1,t2} obtained polynomial exact solutions; Cummings et al \cite{c5} also obtained explicit solutions and some conserved quantities. Crowdy and Siegel \cite{c2} obtained new conserved quantities and exact solution based on Cauchy transform approach. Nie et al \cite{n1} numerically studied the singularity formation of the Stokes flow. Prokert \cite{p1} obtained existence result of solutions in Sobolev space for a similar problem. In this paper, we are going to establish a local existence result for an initial-value problem arising from free evolving bubble in Stokes flow. We first derive the initial-value problem using complex variables theory \cite{c1}, then obtain the local existence result based on a Nirenberg theorem \cite{n2,n3} on abstract Cauchy-Kowalewski problem in properly chosen Banach spaces. The same techniques have been used for other problems \cite{x1,x2}. \section{Stokes flow with free boundary} We consider the quasi-steady evolution of a bubble in an ambient Stokes flow \cite{t1,t2}. The fluid inside the bubble has a negligible viscosity and is at a constant pressure, which is set to be zero. The fluid outside the bubble has a viscosity $\mu$ and is incompressible. Under the assumption of no inertial effects, gravitational or other body forces, the fluid motion is governed by Stokes equation and the incompressibility condition \begin{gather}\label{e1} \mu \Delta u= \nabla p,\\ \label{e2} \nabla\cdot u=0 \end{gather} The above equations hold in the fluid region outside of the bubble. On the bubble boundary, we have stress condition $$\label{e3} -p n_j +2 \mu e_{jk} n_k = \tau \kappa n_j$$ where $n=(n_1,n_2)$ is a unit normal vector pointing outward from the bubble, $\tau$ is the surface tension coefficient, $\kappa$ is the curvature and $$\label{e4} e_{jk}=\frac{1}{2}\big( \frac{\partial u_j}{\partial x_k} +\frac{\partial u_k}{\partial x_j}\big)$$ is the rate of strain tensor. The kinematic condition on the free boundary is $$\label{e5} u\cdot n=V_n,$$ where $V_n$ is the normal component of the free surface motion. Introducing streaming function $\psi(x,y)$ such that $$\label{e6} u=\nabla^{\perp }\psi$$ then $\psi (x,y)$ satisfies the biharmonic equation $$\label{e7} \nabla^4 \psi =0\,.$$ Here $\psi(x,y)$ can be expressed as $$\label{e8} \psi=\mathop{\rm Im} [z^*f(z,t)+g(z,t)]$$ where $z=x+iy$ and $*$ denotes complex conjugate. Here Goursat functions $f(z,t)$ and $g(z,t)$ are analytic functions in the fluid region. In terms of Goursat functions, the physical quantities are established $$\label{e9} \begin{gathered} \frac{p}{\mu}-i\omega =4 f'(z,t), \\ u=u_1+iu_2=-f(z,t)+z[f'(z)]^*+[g( z)]^*,\\ e_{11}+ie_{12}=z[f''(z)]^*+[g''(z)]^*. \end{gathered}$$ where $*$ denotes the complex conjugate and $\omega$ is the vorticity. Defining $s$ to be the arc-length traversed in a counterclockwise direction around the bubble boundary, then the stress condition can be written as $$\label{e10} f(z,t)+z[f'(z,t)]^*+[g'(z,t)]^*=-i\frac{z_s}{2},$$ the kinematic equation can be written as $$\label{e11} \mathop{\rm Im} [(z_t+2f)z^*_s] =-\frac{1}{2}.$$ Equations \eqref{e10} and \eqref{e1} will be supplemented by far field conditions on $f$ and $g$ at infinity. \section{An initial-value problem} We consider the conformal mapping $z(\xi,t)$ that maps the interior of the unit circle $|\xi|<1$ in the $\xi$ plane to the fluid region in $z$-plane such that the $\xi=0$ is mapped to the point $z=\infty$. So $\xi z(\xi,t)$ is analytic in $|\xi|<1$. The kinematic condition can be written as $$\label{e12} \mathop{\rm Re} \big[\frac{z_t+2f(z,t)}{\xi z_\xi}\big]=\frac{\tau}{2|z_\xi|}.$$ as in Tanveer and Vasconcelos \cite{t1}, we use the far field condition $$\label{e13} f(z)\sim az+b+O(1/z) \quad \text{as } |z|\to \infty$$ where $a$ and $b$ are functions of $t$ only. In particular, we choose $f(z)= az +b$ where $a$ and $b$ are constants. From Poisson's formula, \eqref{e12} becomes $$\label{e14} z_t+2(az+b)=\xi z_\xi I_-(\xi,t) \text{ for } |\xi|<1;$$ where $I_-(\xi,t)$ is defined by $$\label{e15} I_-(\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'} [ \frac{\xi'+\xi}{\xi'-\xi}] \frac{1}{|z_\xi|}.$$ Let $h(\xi,t)=\xi z(\xi,t)$, then $h(\xi,t)$ is analytic in $|\xi|<1$. If $h(\xi,t)$ can be analytically extended to some region where $|\xi|>1$, then $$\label{e16} h_t+2ah+2b\xi =[\xi h_\xi-h]I_+[h](\xi,t) +\frac{\tau (\xi h_\xi-h)^{1/2}}{(\xi \bar h_\xi -\bar h)^{1/2}},$$ where $\bar h(\xi,t)$ is defined as $$\label{e17} \bar h(\xi,t)=[ h(\frac{1}{\xi^*})]^*,$$ and $$\label{e18} I_+[h](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'} [ \frac{\xi'+\xi}{\xi'-\xi}] \frac{1}{|\xi h_\xi- h|} \quad \text{for } |\xi|>1.$$ Making the change of variable, $$\label{e19} v(\xi,t)=\frac{1}{(\xi h_\xi -h)^{1/2}},$$ and using \eqref{e16}, we obtain $$\label{e20} v_t=\xi v_\xi I_+[v] -\frac{1}{2}\xi v [I_+[v]]_\xi +\frac{1}{2}\tau \xi v \bar v v_\xi -\frac{1}{2}\tau \xi v^2 \bar v_\xi +\frac{1}{2}v I_+[v] + \tau v^2\bar v +av,$$ where $$\label{e21} I_+[v](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'} \big[ \frac{\xi'+\xi}{\xi'-\xi}\big] v(\xi',t)\bar v(\xi',t) \quad \text{for } |\xi|>1.$$ The analytic continuation of \eqref{e20} to $|\xi|<1$ is $$\label{e22} v_t= \xi v_\xi I_-[v]-\frac{1}{2}\xi v [I_-[v]]_\xi +\frac{1}{2}v I_-[v] +a v,$$ where $$\label{e23} I_-[v](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'} [ \frac{\xi'+\xi}{\xi'-\xi}] v(\xi',t)\bar v(\xi',t)\quad \text{for } |\xi|<1.$$ We will consider equation \eqref{e20} \eqref{e22} with the initial condition $$\label{e24} v(\xi,0)=v_0(\xi).$$ Let us first introduce a scale of Banach spaces which are spaces of bounded analytic functions in disks. \begin{definition} \label{de3.1} \rm Let $\mathcal{R}_s$ be the disk in complex $\xi$ plane with radius $s$; i.e., $\mathcal{R}_s=\{\xi,|\xi|1$, then there exists one and only one solution $v\in C^1([0,T),\mathcal{B}_s)$, $10,\delta>0$ and every pair of numbers $s,s'$ such that $r_1< s'r_1>r_0>1$, then $\mathcal{R}_{r_0}\subset\mathcal{R}_{r_1}\subset\mathcal{R}_{r}$ and $\mathbf{B}_{r}\subset\mathbf{B}_{r_1}\subset\mathbf{B}_{r_0}$. In this and the following sections, $C>0$ represents a generic constant, it may vary from line to line. $C$ may depend on $r_0, r_1$ and $r$; but it is always independent of $s$ and $s'$. \begin{lemma}\label{lem:3.1} If $f\in \mathbf{B}_s, r_10$, if $f\in\mathbf{B}_s$, then $\bar{f}$ is analytic in $|\xi|>\frac{1}{s}$ and $|\bar{f}|\leq \norm{f}_s$ for $|\xi|>\frac{1}{s}$. \end{remark} \begin{lemma}\label{lem:3.4} If $f\in \mathbf{B}_s, r_10$ is independent of $s,s'$ and $f$. \end{lemma} \begin{proof} Since $\mathop{\rm dist}(\partial\mathbf{B}_{s'},\partial\mathbf{B}_{s} )=s-s'$, for $\xi\in\mathbf{B}_{s'}$, we are able to find a disk $D(\xi)$ centered at $\xi$ with radius $s-s'$ such that $D(\xi)$ is contained in $\mathcal{R}_s$. Using Cauchy integral formula, we have \begin{equation*} f_\xi(\xi)=\frac{1}{2\pi i}\int_{|t-\xi|=s-s'}\frac{f(t)}{(t-\xi)^2}dt \end{equation*} so \begin{equation*} \begin{split} |f_\xi(\xi)| &\le\frac{1}{2\pi }\int_{|t-\xi|=s-s'}\frac{|f(t)|}{|t-\xi|^2}|dt|\\ &\le \frac{1}{2\pi }\int_0^{2\pi}\frac{|f(\xi+|s-s'| e^{i\theta})|}{s-s'}d\theta\\ &\le \frac{K_3\norm{f}_s}{s-s'}, \end{split} \end{equation*} which proves the lemma. \end{proof} \begin{definition} \label{def4.6} \rm We define the function $$\label{3.4} G_1[v](\xi,t)=v(\xi,t)\bar{v}(\xi,t).$$ \end{definition} \begin{remark}\label{rem:3.7} \rm If $v\in\mathbf{B}_s, s>1$, then $G_1[v]$ are analytic in $\frac{1}{s}\le |\xi|\le s$. \end{remark} \begin{lemma}\label{lem:3.8} If $v\in\mathbf{B}_s$ and $10$ is a constant independent of $s, g$ and $h$. \end{lemma} \begin{proof} Due to Remark \ref{rem:3.7}, the integrands of $I_+$ are analytic in $\frac{1}{r_0}\le |\xi|\le 1$, changing the contour of integration in the definitions of $I^+$ from $|\xi|=1$ to $|\xi|=\frac{1}{r_0}$ gives $$\label{3.5} I_+[v](\xi,t)=\frac{1}{4\pi i}\int_{|\xi '| =\frac{1}{r_0}}\frac{d\xi'}{\xi'}[\frac{\xi+\xi'}{\xi'-\xi}]G_1[v](\xi',t)$$ For $|\xi|>1$ and $|\xi '|=\frac{1}{r_0}$, from simple geometry, we have $$\label{3.7} |\frac{\xi+\xi'}{\xi'-\xi}|\le C$$ where $C$ depends on only $r_0$. The lemma now follows from (\ref{3.5}) and Lemma \ref{lem:3.8}. \end{proof} Similarly we have \begin{lemma}\label{lem:3.10} If $v\in\mathbf{B}_s$ and $11$ and $|\xi '|=\frac{1}{r_0}$, from simple geometry, we have $$\label{3.10} |\frac{\xi+\xi'}{(\xi'-\xi)^2}|\le C$$ where $C$ depends on only $r_0$. The lemma now follows from (\ref{3.10}) and Lemma \ref{lem:3.8}. \end{proof} \begin{lemma}\label{lem:3.11} If $v\in\mathbf{B}_s$ and $10$ is independent of $s,s'$ and $u$ and $v$. \end{lemma} The above lemma follows from applying Lemma \ref{lem:3.5} with $f=u-v$. \begin{lemma}\label{lem:4.2} If $u\in \mathbf{B}_s,v\in \mathbf{B}_s, r_10$ is independent of $s$, $u$ and $v$. \end{lemma} The above lemma follows from applying Lemma \ref{lem:3.4} with $f=u-v$. \begin{lemma}\label{lem:4.2.5} If $u\in \mathbf{B}_s,v\in \mathbf{B}_s,\norm{u}_s\le M,\norm{v}_s\le M$, $r_1\frac{1}{r_0}$, $$|G_1[v](\xi,t)-G_1[u](\xi,t)|\le C\norm{v-u}_s .$$ \end{lemma} \begin{proof} From (\ref{3.4}), we have $$G_1[v]-G_1[u]=(v-u)\bar{v}+u(\bar{v}-\bar{u})$$ which proves the lemma by using Remark \ref{rem:3.3}. \end{proof} \begin{lemma}\label{lem:4.3} If $u\in \mathbf{B}_s,v\in \mathbf{B}_s,\norm{u}_s\le M,\norm{v}_s\le M$, $r_11$, $L[v]$ is defined by $$\label{4.1} L[v](\xi,t)= \xi v_\xi I_+[v] -\frac{1}{2}\xi v [I_+[v]]_\xi +\frac{1}{2}\tau \xi v \bar v v_\xi -\frac{1}{2}\tau \xi v^2 \bar v_\xi +\frac{1}{2}v I_+[v] + \tau v^2\bar v +av,$$ The analytic continuation of $L[v]$ to $|\xi|<1$ is $$\label{4.2} L[v](\xi,t)= \xi v_\xi I_-[v] -\frac{1}{2}\xi v [I_-[v]]_\xi +\frac{1}{2}v I_-[v] +av\,.$$ \begin{lemma}\label{lem:4.7} If $u\in \mathbf{B}_s$, $v\in \mathbf{B}_s$, $\norm{u}_s\le M$, $\norm{v}_s\le M$, then for $|\xi|\ge 1$ and $r_10$ is independent of $s$ and $s'$. \end{lemma} \begin{proof} By (\ref{4.2}), for $|\xi|<1$, \label{4.5} \begin{aligned} L[v](\xi,t)-L[u](\xi,t) &=\xi(v_\xi-u_\xi)I_-[v]+\xi u_\xi \{I_-[v]-I_-[u]\}+a(v-u)\\ &\quad -\frac{1}{2}\xi(v-u)(I_-[v])_\xi-\frac{1}{2} \xi u\{(I_-[v])_\xi -(I_-[u])_\xi\}, \end{aligned} and for $|\xi|>1$, \label{4.6} \begin{aligned} L[v](\xi,t)-L[u](\xi,t) &=\xi(v_\xi-u_\xi)I_+[v]+\xi u_\xi \{I_+[v]-I_+[u]\}\\ &\quad +a(v-u)-\frac{1}{2}\xi(v-u)(I_+[v])_\xi-\frac{1}{2} \xi u\{(I_+[v])_\xi -(I_+[u])_\xi\}\\ &\quad +\frac{1}{2}\tau \xi \{(v-u)\bar{v}v_\xi+u(\bar{v} -\bar{u} v_\xi + u\bar{u} (v_\xi-u_\xi)\}\\ &\quad -\frac{1}{2}\tau \xi \{ (v-u)(v+u)\bar{v}_\xi + u^2(\bar{v}_\xi+\bar{u}_\xi)(\bar{v}_\xi-\bar{u}_\xi)\}\\ &\quad +\tau (v-u)(v+u)\bar{v}+\tau u^2(\bar{v}-\bar{u}), \end{aligned} By Lemmas \ref{lem:3.4}--\ref{lem:3.12} and \ref{lem:4.1}-- \ref{lem:4.4}, each term in equations (\ref{4.5}) and (\ref{4.6}) can bounded by $\frac{C}{s-s'}\norm{v-u}_s$; hence the proof is complete. \end{proof} Let $p=v-v_0$, then $v$ is a solution of initial problem \eqref{e20}, \eqref{e22} and \eqref{e24} if and only if $p$ solves the following initial problem $$\label{4.9} p_t=\mathcal{L}[p], p|_{t=0}=0.$$ where the operator $\mathcal{L}$ is defined by $$\label{4.10} \mathcal{L}[p]=L[p+v_0]$$ \begin{lemma}\label{lem:4.9} If $p\in B_s, u\in B_s, \norm{p}_s\le M$ and \$\norm{u}_s\le M, r_1