assumption $\phi_i^1(x)=O(|x|^2)$ near $0$, $C=0$. The third statement of Theorem \[uniqlin\] is established.

Again by Lemma \[1129lem1\] the first statement of Theorem \[uniqlin\] is established.

Finally, the second statement of Theorem \[uniqlin\] is an immediate consequence of Lemma 3.1 of [@linzhang1]. Theorem \[uniqlin\] is established. $\Box$

A Pohozaev identity for global solutions
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\[globalcpt\] Let $u=(u_1,...,u_n)$ be an entire, radial solution of $$\left\{\begin{array}{ll}-\Delta u_i=\sum_{j=1}^na_{ij}e^{u_j}, \quad \mbox{in}\quad \mathbb R^2, \\
\int_{\mathbb R^2}e^{u_i}<\infty
\end{array}
\right.$$ where $A$ is a constant matrix that satisfies $(H1)$. Let $$c_i=u_i(0)+\frac 1{2\pi}\int_{\mathbb R^2}\log |\eta |(\sum_{j=1}^n a_{ij}e^{u_j(\eta)})d\eta,$$ $$\sigma_{i}=\frac 1{2\pi}\int_{\mathbb R^2}e^{u_i},\quad l_i=\sum_{j=1}^na_{ij}\sigma_{j},\,\, l=\min\{l_{1},...,l_{n}\}$$ and $$\sigma_{iR}=\frac 1{2\pi}\int_{B_R}e^{u_i}.$$ Then for some $\delta_0>0$ $$\label{12apr29e1}
e^{u_i(r)}=e^{c_i}r^{-l_i}(1+o(r^{-\delta_0})), \quad r>1,$$ $$\label{1105e1}
4\sum_{i\in I}\sigma_{iR}=\sum_{i,j\in I}a_{ij}\sigma_{iR}\sigma_{jR}+2\sum_{i\in I}e^{c_{i}}R^{2-l_{i}}+O(R^{2-l-\delta_0}).$$

[**Proof of Lemma \[globalcpt\]:**]{}

It is well known that $$\label{11may27e2}
u_i(x)=-\frac 1{2\pi}\int_{\mathbb R^2}\log |x-\eta |(\sum_{j=1}^n a_{ij}e^{u_j(\eta)})d\eta+c_i.$$ Indeed, let $w_i$ be the function defined by the right hand side of (\[11may27e2\]). Then $w_i-u_i$ is a harmonic function. Since they both have logarithmic growth at infinity, $w_i-u_i=c$. Evaluating both functions at $0$ we have $c=c_i$.

Clearly $$\label{12apr24e1}
u_i(x)+l_{i}\log |x|=-\frac 1{2\pi}\int_{\mathbb R^2}\log \frac{|x-\eta |}{|x|}\sum_ja_{ij}e^{u_j(\eta)}d\eta+c_{i}.$$ Using $\sum_j a_{ij}e^{u_j(r)}=O(r^{-2-\delta_0})$ for some $\delta_0>0$ and $r$ large, we obtain, by elementary estimates, $$u_i(r)=-l_i \log r+c_i+o(r^{-\delta_0}),$$ which leads to $$\label{11may30e1}
\sigma_{i}=\sigma_{iR}+\frac{e^{c_{i}}}{l_{i}-2}R^{2-l_{i}}+O(R^{2-l_{i}-\delta}).$$ We arrive at (\[1105e1\]) by using (\[11may30e1\]) 