\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 135, pp. 1--6.\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-2008/135\hfil Quantization effects] {Quantization effects for a variant of the Ginzburg-Landau type system} \author[L. Ma\hfil EJDE-2008/135\hfilneg] {Li Ma} \address{Li Ma \newline Institute of Science, PLA University of Science and Technology\\ Nanjing, 211101, China} \email{mary96@126.com} \thanks{Submitted September 9, 2008. Published October 9, 2008.} \subjclass[2000]{35J55, 35Q40} \keywords{Quantization; Ginzburg-Landau type functional} \begin{abstract} The author uses Pohoaev's identity to research the quantization for a Ginzburg-Landau type functional. Under the logarithmic growth condition which is different assumption from that of in \cite{BMR}, the author obtain the analogous quantization results. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \section{Introduction} In \cite{BMR} and \cite{Sh;94}, the authors have studied the quantization effects for the system $$ -\Delta u=u(1-|u|^2) \quad \text{in } \mathbb{R}^2, $$ which is associated with the Ginzburg-Landau functional $$ F(u)=\int_{\Omega}[\frac{1}{2}|\nabla u|^2 +\frac{1}{4\varepsilon^2}(1-|u|^2)^2]dx, $$ where $\Omega\subset \mathbb{R}^2$ is a bounded domain with smooth boundary, and $\varepsilon>0$ is a small parameter \cite{BBH;94}. Lassoued and Lefter have investigated the asymptotic behavior of minimizers $u_\varepsilon\in H_g^1(B_1,\mathbb{R}^2)$ to the Ginzburg-Landau type energy $$ E_\varepsilon(u,\Omega)=\frac{1}{2}\int_{\Omega}|\nabla u|^2dx +\frac{1}{4\varepsilon^2}\int_{\Omega}|u|^2(1-|u|^2)^2dx, $$ when $\varepsilon \to 0$, where $g:\partial \Omega\to S^1$ is a smooth map \cite{LL;98}. In view of \cite[(1.4)]{LL;98}, the Euler-Lagrange system of the minimizer $u_\varepsilon$ is $$ -\Delta u=\frac{1}{\varepsilon^2}u|u|^2(1-|u|^2) -\frac{1}{2\varepsilon^2}u(1-|u|^2)^2 \quad\text{in } \Omega. $$ Let $\Omega_\varepsilon=\frac{1}{\varepsilon}\Omega$. Then we have \begin{equation} -\Delta u=u|u|^2(1-|u|^2)-\frac{1}{2}u(1-|u|^2)^2 \label{e1.1} \end{equation} in $\Omega_\varepsilon$. In a natural way, we shall study the system \eqref{e1.1} in $\mathbb{R}^2$. In view of \cite[Propositions 2.1 and 2.2]{LL;98}, we have \begin{gather} |u|\leq 1, \quad\text{in } \mathbb{R}^2; \label{e1.2}\\ \|\nabla u\|_{L^{\infty}(\mathbb{R}^2)}< +\infty. \label{e1.3} \end{gather} Regarding the boundary condition $u_\varepsilon|_{\partial B_1}=g$, we assume that \begin{equation} |u(x)|\to 1,\quad as \quad |x|\to \infty. \label{e1.4} \end{equation} Then, $\deg(u,\partial B_r)$ is well defined for $r$ large \cite{BMR}. We denote $d=|\deg(u,\partial B_r)|$. In virtue of \eqref{e1.4}, we see that there exists $R_0>0$, such that \begin{equation} |u(x)|\geq \sqrt{\frac{2}{3}}, \quad \text{for } |x|=R\geq R_0. \label{e1.5} \end{equation} Thus, there is a smooth single-valued function $\psi(x)$, defined for $|x|\geq R_0$, such that \begin{equation} u(x)=\varrho(x)e^{i(d\theta+\psi(x))}, \label{e1.6} \end{equation} where $\varrho=|u|$. If denote $\phi(x)=d\theta+\psi$, then $\phi$ is well defined and smooth locally on the set $|x|\geq R_0$. In this paper, we investigate the quantization of the energy functional $E_\varepsilon(u,\Omega)$, by an argument as in \cite{BMR} for the systems \eqref{e1.1}. \begin{theorem} \label{thm1.1} Assume that $u$ solves \eqref{e1.1}. If $u$ satisfies \eqref{e1.4}, and there exists an absolute constant $C>0$, such that for any $r>1$, \begin{equation} \int_{B_r}|\nabla u|^2dx+\int_{B_r}|u|^2(1-|u|^2)^2dx\leq C(\ln r+1). \label{e1.7} \end{equation} Then \begin{equation} \int_{\mathbb{R}^2}|u|^2(1-|u|^2)^2dx=2\pi d^2. \label{e1.8} \end{equation} \end{theorem} If $u$ is a solution of \eqref{e1.1}, and under the assumption \begin{equation} \int_{\mathbb{R}^2} |\nabla u|^2dx < +\infty, \label{e1.9} \end{equation} instead of \eqref{e1.2}-\eqref{e1.4} and \eqref{e1.7}, then there holds the following stronger conclusion. \begin{theorem} \label{thm1.2} Assume $u$ solves \eqref{e1.1} and satisfies \eqref{e1.9}, then either $u(x)\equiv 0$ or $u\equiv C$ with $|C|=1$ on $\mathbb{R}^2$. \end{theorem} \section{Preliminaries} \begin{proposition}[Pohozaev identity] \label{prop2.1} If $u$ solves \eqref{e1.1}. Then for any $r>0$, there holds \begin{equation} \int_{B_r}|u|^2(1-|u|^2)^2dx =\frac{1}{2}\int_{\partial B_r} |u|^2(1-|u|^2)^2|x|ds +\int_{\partial B_r}|x|(|\partial_{\tau} u|^2 -|\partial_{\nu}u|^2)ds. \label{e2.1} \end{equation} \end{proposition} \begin{proof} Multiply \eqref{e1.1} with $(x\cdot \nabla u)$, and integrate over a bounded domain $\Omega$ with smooth boundary. Noting \begin{equation} \begin{aligned} \int_{\Omega}(x\cdot\nabla u)\Delta udx &=\int_{\partial \Omega}\partial_{\nu}u(x\cdot\nabla u)ds-\int_{\Omega}\nabla(x\cdot\nabla u)\nabla u\,dx\\ &=\int_{\partial \Omega}(x\cdot \nu) ||\partial_{\nu}u|^2ds-\frac{1}{2}\int_{\Omega}x\cdot\nabla (|\nabla u|^2)dx-\int_{\Omega}|\nabla u|^2dx\\ &=\int_{\partial \Omega}(x\cdot \nu)|\partial_{\nu}u|^2ds-\frac{1}{2}\int_{\partial \Omega}(x\cdot \nu) |\nabla u|^2ds, \end{aligned} \label{e2.2} \end{equation} and \begin{align*} &\int_{\Omega}(x\cdot\nabla u)u|u|^2(1-|u|^2)dx -\frac{1}{2}\int_{\Omega}(x\cdot\nabla u)u(1-|u|^2)^2dx\\ &=\frac{1}{2}\int_{\Omega}|u|^2(1-|u|^2)^2dx -\frac{1}{4}\int_{\Omega}\mathop{\rm div}[x|u|^2(1-|u|^2)^2]dx\\ &=\frac{1}{2}\int_{\Omega}|u|^2(1-|u|^2)^2dxdy -\frac{1}{4}\int_{\partial \Omega}|u|^2(1-|u|^2)^2(x\cdot \nu)ds, \end{align*} we obtain \begin{equation} \begin{aligned} &\int_{\Omega}|u|^2(1-|u|^2)^2dx\\ &=\frac{1}{2}\int_{\partial \Omega}|u|^2(1-|u|^2)^2(x\cdot \nu)ds +\int_{\partial \Omega}(x\cdot \nu)|\nabla u|^2ds -2\int_{\partial \Omega}(x\cdot \nu)|\partial_{\nu}u|^2ds. \end{aligned} \label{e2.3} \end{equation} Thus, \eqref{e2.1} can be seen by taking $\Omega=B_r$ in the identity above. The proof is complete. \end{proof} \section{Proof of Theorem \ref{thm1.1}} \begin{proposition} \label{prop3.1} Assume $u$ solves \eqref{e1.1}. If $u$ satisfies \eqref{e1.4} and \eqref{e1.7}, then \begin{equation} \int_{\mathbb{R}^2}(1-|u|^2)^2dx < +\infty. \label{e3.1} \end{equation} \end{proposition} \begin{proof} Denote $f(t)=\int_{\partial B_t}[|\nabla u|^2+|u|^2(1-|u|^2)^2]ds$. Applying \cite[Proposition 2.2]{L;07}, from \eqref{e1.7} we are led to $$ \frac{1}{2}\inf\{tf(t);t\in[\sqrt{r},r]\ln r\leq \int_{\sqrt{r}}^r\frac{tf(t)}{t}dt\leq E(u,B_r)\leq C\ln r, $$ which implies $\inf\{tf(t);t\in[\sqrt{r},r]\leq C$. Thus, there exists $t_m\to \infty$ such that \begin{equation} t_mf(t_m)\leq O(1). \label{e3.2} \end{equation} Taking $r=t_j\to \infty$ in \eqref{e2.1}, and substituting \eqref{e3.2} into it, we obtain \begin{equation} \int_{\mathbb{R}^2}|u|^2(1-|u|^2)^2dx < +\infty. \label{e3.3} \end{equation} Noting \eqref{e1.5} we can see the conclusion of the proposition. \end{proof} Substituting \eqref{e1.6} into \eqref{e1.1} yields \begin{gather} -\Delta \varrho+\varrho|\nabla \phi|^2=\varrho^3(1-\varrho^2) -\frac{1}{2}\varrho(1-\varrho^2)^2, \quad\text{in } \mathbb{R}^2\setminus B_{R_0}, \label{e3.4}\\ -\mathop{\rm div}(\varrho^2\nabla\phi)=0\quad\text{in } \mathbb{R}^2\setminus B_{R_0}. \label{e3.5} \end{gather} By an analogous argument of Steps 1 and 2 in the proof of \cite[Proposition 1]{BMR}, we also derive from \eqref{e3.5} that \begin{equation} \int_{\mathbb{R}^2\setminus B_{R_0}}|\nabla \psi|^2dx<+\infty. \label{e3.6} \end{equation} In addition, we also deduce the following proposition. \begin{proposition} \label{prop3.2} Under the assumption of Proposition \ref{prop3.1}, we have \begin{equation} \int_{\mathbb{R}^2\setminus B_{R_0}}|\nabla\varrho|^2dx<+\infty. \label{e3.7} \end{equation} \end{proposition} \begin{proof} Let $\eta\in C^{\infty}(\mathbb{R}^2,[0,1])$ satisfy $\eta(x)=1$ for $|x|\leq 1/2$, and $\eta(x)=0$ for $|x|\geq 1$. Set $\eta_t(x)=\eta(\frac{x}{t})$ for $tR_0. \end{aligned} \label{e3.10} \end{equation} Using H\"older's inequality, from \eqref{e3.1} and \eqref{e3.6}, we deduce that \begin{equation} \begin{aligned} \int_{B_r\setminus B_{R_0}}|\nabla \phi|^2\varrho(1-\varrho)\eta_t^2 dx & \leq \Big(\int_{B_r\setminus B_{R_0}}\frac{d^4}{|x|^4}dx\Big)^{1/2} \Big(\int_{\mathbb{R}^2}(1-\varrho)^2 dx\Big)^{1/2}\\ &\quad +\int_{\mathbb{R}^2\setminus B_{R_0}}|\nabla \psi|^2dx<+\infty. \end{aligned} \label{e3.11} \end{equation} At last, \eqref{e1.5} implies \begin{equation} \int_{B_r\setminus B_{R_0}}[\varrho^3(1-\varrho^2) -\frac{1}{2}\varrho(1-\varrho^2)^2] (1-\varrho)\eta_t^2 dx\geq 0. \label{e3.12} \end{equation} Substituting \eqref{e3.9}-\eqref{e3.12} into \eqref{e3.8}, and letting $t \to \infty$, we can deduce \eqref{e3.7}. The proof is complete. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] First, we have \begin{equation} \begin{aligned} |\partial_{\tau}u|^2 &=|\partial_{\tau}\varrho|^2+\varrho^2(\frac{d}{|x|} +\partial_{\tau}\psi)^2\\ &=\frac{d^2}{|x|^2}+|\partial_{\tau}\varrho|^2 +(\varrho^2-1)\frac{d^2}{|x|^2} +2\varrho^2\frac{d}{|x|}\partial_{\tau}\psi +\varrho^2|\partial_{\tau}\psi|^2, \end{aligned} \label{e3.13} \end{equation} Obviously, \eqref{e3.1}, \eqref{e3.3}, \eqref{e3.6} and \eqref{e3.7} imply \begin{align*} &\int_{B_r\setminus B_{R_0}}[|u|^2(1-|u|^2)^2+|\partial_{\tau}\varrho|^2 +(1-\varrho^2)\frac{d^2}{|x|^2}\\ &+2\varrho^2\frac{d}{|x|}|\partial_{\tau}\psi| +\varrho^2|\partial_{\tau}\psi|^2+|\partial_{\nu}u|^2]dx\leq C, \end{align*} where $C$ is independent of $r$. Similar to the derivation of \eqref{e3.2}, by using \cite[Proposition 2.2]{L;07}, it also follows that $$ \inf\{F(r_j);r_j\in[\sqrt{r},r]\}\leq C(\ln r)^{-1}, $$ where \begin{align*} F(r_j)&:=r_j\int_{\partial (B_{r_j}\setminus B_{R_0})}[|u|^2(1-|u|^2)^2+|\partial_{\tau}\varrho|^2 +(1-\varrho^2)\frac{d^2}{|x|^2}\\ &\quad +2\varrho^2\frac{d}{|x|}|\partial_{\tau}\psi| +\varrho^2|\partial_{\tau}\psi|^2+|\partial_{\nu}u|^2]ds. \end{align*} Thus, we see that there exists $r_j\to \infty$, such that $F(r_j)\leq o(1)$. Combining this with \eqref{e3.13}, we can see \eqref{e1.8} since $$ \int_{\partial B_r}|x|\frac{d^2}{|x|^2}ds=2\pi d^2. $$ The proof is complete. \end{proof} \section{Proof of Theorem \ref{thm1.2}} First, we shall prove \eqref{e1.2}. Similar to the derivation of \eqref{e3.8} in \cite{BMR}, we also have $$ \Delta h \geq |u|(1+|u|)h(3|u|^2-1)/2, \quad h=(|u|-1)^+. $$ Write $G=\{x\in \mathbb{R}^2;|u(x)|>\sqrt{1/3}\}$. In the argument of Step 1 in the proof of \cite[Theorem 2]{BMR}, we replace $\mathbb{R}^2$ by $G$ to be the integral domain. Applying \eqref{e1.9} we also deduce that $$ |u|h(3|u|^2-1)\equiv 0, \quad \text{on } G. $$ This implies \eqref{e1.2}. Next, \eqref{e1.1} leads to \begin{equation} \Delta |u|^2=2|\nabla u|^2+|u|^2(|u|^2-1)(3|u|^2-1),\quad \text{on } B_r. \label{e4.1} \end{equation} Multiplying this equality by $\eta_t$ and integrating over $B_r$, we have \begin{equation} \begin{aligned} &\int_{B_r}|u|^2(1-|u|^2)(3|u|^2-1)\eta_tdx\\ &=2\int_{B_r}|\nabla u|^2\eta_tdx -\int_{\partial B_r} \eta_t\partial_{\nu}|u|^2ds+2\int_{B_r}u\nabla u\nabla\eta_tdx. \end{aligned} \label{e4.2} \end{equation} From \eqref{e4.2} with $t