$O(r^{-2})$) we immediately observe that $$\int_0^{\infty}e^{u_i(r)}\psi_i(r)r^3dr<\infty.$$ In regard to (\[12e2\]) we have $$\label{12e3}
r^3\psi_i'(r)=-\int_0^r\sum_{j}a_{ij}e^{u_j}\psi_j\sigma^3 d\sigma, \quad 0<r<\infty.$$ From (\[12e4\]) and (\[12e3\]) we see that $\psi_i$ is non-increasing. Since we have known that $|\psi_i(r)|\le Cr^{-2}$ near infinity we have $$\label{12e6}
\psi_i(r)\ge 0, \quad \forall r>0, \quad i\in I.$$

It is not possible to have all $\psi_i(0)=0$ because this implies $\psi_i\equiv 0$, a contradiction to the assumption that $\bar \psi$ is not a multiple of $\psi^0$. Therefore without loss of generality we assume $\psi_1(0)>0$. Then we further claim that $\psi_i$ is strictly decreasing for all $i\in I$. Indeed, let $I_1=\{ j\in I| \, a_{1j}>0 \, \}$, for each $j\in I_1$ we use (\[12e6\]) and $\psi_1(0)>0$ to obtain $$r^3\psi_j'(r)\le -\int_0^ra_{1j}e^{u_1}\psi_1\sigma^3d\sigma<0, \quad 0<r<\infty.$$ Therefore for each $j\in I_1$, $\psi_j$ is strictly decreasing, which immediately implies that $\psi_j(0)>0$. We can further define $I_2=\{ i\in I| \, a_{ij}>0 \mbox{ for some }j\in I_1. \}$. Then the same argument shows that $\psi_i$ is strictly decreasing for each $i\in I_2$ as well. Since the matrix $A=(a_{ij})_{n\times n}$ is irreducible, this process exhausts all $i\in I$.

(\[12e3\]) yields $\psi_i'(r)\le -Cr^{-3}$ for $r>1$ and $i\in I$. Then by using $\lim_{r\to \infty}\psi_i(r)=0$ we further have $$\label{12e5}
\psi_i(r)\ge Cr^{-2},\quad r\ge 1.$$ Then 