\mbox{in}\,\, \mathbb R^2,\quad i\in I=\{1,..,n\}\\
\int_{\mathbb R^2}e^{u_i}<\infty.
\end{array}
\right.$$ Suppose $\phi=(\phi_1,..,\phi_n)$ satisfies $$-\Delta \phi_i=\sum_{j=1}^na_{ij}e^{u_j}\phi_j,\quad i\in I.$$

1.  $$\label{1027e1}
    |\phi_i(x)|\le C(1+|x|)^{\tau},\quad x\in \mathbb R^2,$$

    for some $\tau\in (0,1)$ and $$\phi_i(0)=0,\quad i\in I.$$ Then there exist $c_1,c_2\in \mathbb R$ such that $$\phi_i(x)=c_1u_i'(r)\frac{x_1}r+c_2u_i'(r)\frac{x_2}r \quad \mbox{in}\quad \mathbb R^2, \quad i\in I.$$

2.  If $
    |\phi_i(x)|\le C$ for all $x\in \mathbb R^2$, then there exist $c_0,c_1,c_2\in \mathbb R$ such that $$\phi_i(x)=c_0(ru_i'(r)+2)+c_1u_i'(r)\frac{x_1}r+c_2u_i'(r)\frac{x_2}r,\quad \mathbb R^2 ,\quad i\in I.$$

3.  If $\phi_i(x)=O(|x|^2)$ near $0$ and $|\phi_i(x)|\le C(1+|x|)^{2-\epsilon_0}$ for some $\epsilon_0>0$, then $\phi_i\equiv 0$.

Before the proof of Theorem \[uniqlin\] we first establish

\[1129lem1\] Let $A$ and $u$ be as in Theorem \[uniqlin\], let $\Phi=(\Phi_1,..,\Phi_n)$ solve $$\label{12e1}
\left\{\begin{array}{ll}
\Phi_i''(r)+\frac 1r\Phi_i'(r)-\frac 1{r^2}\Phi_i(r)+\sum_{j=1}^na_{ij}e^{u_j}\Phi_j=0,\quad 0<r<\infty, \\
\\
|\Phi_i(r)|\le Cr/(1+r)^{\epsilon_0} \quad \mbox{ for some }\epsilon_0\in (0,1),  \quad \forall i\in I.
\end{array}
\right.$$ Then there exists a constant $C$ such that $\Phi_i(r)=Cu_i'(r)$ for $i\in I$.

[**Proof of Lemma \[1129lem1\]:**]{}

The proof is in two steps. First we show that under the assumption of $\Phi_i$ at infinity we have the following sharper decay estimate: $$\label{phidecay}
|\Phi_i(r)|\le Cr(1+r)^{-2},\quad 0<r<\infty,\quad i\in I.$$

Indeed, let $\bar \psi_i(r)=\Phi_i(r)/r$. By direct computation we see that $\bar \psi=(\bar \psi_1,...,\bar \psi_n)$ satisfies $$\label{12e2}
\bar \psi_i''(r)+\frac 3r\bar \psi_i'(r)+\sum_ja_{ij}e^{u_j}\bar \psi_j=0, \quad 0<r<\infty.$$ Clearly in 