order to show (\[phidecay\]) we only need to show $|\bar \psi_i(r)|\le Cr^{-2}$ for $r>1$ under the assumption that $|\bar \psi_i(r)|\le Cr^{-\epsilon_0}$ for $i\in I$ and $r>1$. Let $$\tilde \psi_i(t)=\bar \psi_i(e^t)\quad \mbox{ and }\quad \tilde u_i(t)=u_i(e^t)+2t,$$ it is easy to see that $\tilde \psi(t)=(\tilde \psi_1(t),..,\tilde
\psi_n(t))$ satisfies $$\label{psidc1}
\tilde \psi_i''(t)+2\tilde \psi_i'(t)+\sum_{j\in I}a_{ij}e^{\tilde u_j(t)}\tilde \psi_j(t)=0,\quad -\infty<t<\infty$$ and our goal is to show $$\label{psidc2}
\tilde \psi_i(t)=O(e^{-2t})$$ knowing $\tilde \psi_i(t)=O(e^{-\epsilon_0t})$ for $t>0$, $i\in I$. Set $$l_i=\frac 1{2\pi}\sum_{j\in I}a_{ij}\int_{\mathbb R^2}e^{u_j},\quad l=\min\{l_1,..,l_n\}.$$ By Lemma \[globalcpt\] below $l>2$. Let $$h_i(t)=-\sum_{j\in I}a_{ij}e^{\tilde u_j(t)}\tilde \psi_j(t)=O(e^{(2-l-\epsilon_0)t}),\quad t>0.$$ Then $$\tilde \psi_i(t)=C_0+C_1 e^{-2t}+\frac 12\int_0^th_i(s)ds-\frac 12e^{-2t}\int_0^te^{2s}h_i(s)ds.$$ Using the asymptotic rate of $h_i(t)$ at infinity we further have $$\tilde \psi_i(t)=(C_0+\frac 12\int_0^{\infty}h_i(s)ds)+C_1e^{-2t}+O(e^{(2-l-\epsilon_0)t}).$$ Since $\tilde \psi_i(t)$ tends to $0$ as $t$ goes to infinity we know $\tilde \psi_i(t)=O(e^{-2t})$ if $l+\epsilon_0\ge 4$, in which case (\[psidc2\]) is established. Otherwise we obtain $\tilde \psi_i(t)=O(e^{(2-l-\epsilon_0)t})$. In the latter case, we apply the same procedure to obtain a better decaying rate of $\tilde \psi_i(t)$ at infinity. After finite steps, (\[psidc2\]) is established.

In the second step we complete the proof of the Lemma \[1129lem1\]. By way of contradiction we suppose there is a solution $\Phi=(\Phi_1,..,\Phi_n)$ that satisfies (\[12e1\]) and $\Phi$ is not a multiple of $f=(u_1'(r),..,u_n'(r))$. Let $\psi_i^0=-u_i'(r)/r$, then clearly 