both $\psi^0=(\psi_1^0,..,\psi_n^0)$ and $\bar \psi=(\bar \psi_1,..,\bar \psi_n)$ satisfy (\[12e2\]). We verify by direct computation that $$\int_0^re^{u_j}\psi_j^0(\sigma)\sigma^3ds=-\int_0^r(e^{u_j})'\sigma^2d\sigma
=-e^{u_j(r)}r^2+2\int_0^re^{u_j}\sigma d\sigma>0.$$ Note that the last inequality is justified by $u'_i(r)<0$ for $r>0$ and $i\in I$. Also, since $e^{u_i}\le Cr^{-2-\delta}$ for some $\delta>0$ and $r>1$, $\displaystyle{\int_0^{\infty}e^{u_i}\psi_i^0(\sigma)\sigma^3d\sigma<\infty}$. Based on the computation above we set $$S=\bigg \{ t|\, \psi_i^t=\psi_i^0+t(\psi_i^0-\bar \psi_i) \, \mbox{and }\,
\int_0^re^{u_i}\psi_i^t(\sigma )\sigma^3d\sigma>0,\,\, \forall r>0,\,\, i\in I\,  \bigg \}.$$ Let $\bar \psi=(\bar \psi_1,..,\bar \psi_n)$ and $\psi^0=(\psi_1^0,..,\psi_n^0)$. We first observe that $\psi_i^0(0)>0$. Suppose $\bar \psi \not\equiv \psi^0$, we can assume that $\bar \psi_1(0)\neq 0$ and $|\bar \psi_i(0)|<\psi_i^0(0)$ for all $i\in I$.

From the definition of $S$ we immediately see that $0\in S$. Moreover, since $|\bar \psi_i(r)|\le Cr^{-2}$ near infinity, we can choose $|t|<\delta$ with $\delta$ small so that all $|t|<\delta$ belong to $S$. Another immediate observation is that $S$ has a lower bound. Indeed, for $T$ sufficiently negative, $\psi_1^T(0)<0$, which is impossible to have $\int_0^re^{u_1}\psi_1^T(\sigma)\sigma^3d\sigma>0$ for $r$ small.

Let $T=\inf S$ and let $t_m\to T^+$. The sequence $\psi^{t_m}_i$ obviously converges to a function $\psi_i$, which is just $\psi_i^0+T(\psi_i^0-\bar \psi_i)$. $\psi=(\psi_1,..,\psi_n)$ satisfies the property $$\label{12e4}
\int_0^re^{u_i(\sigma)}\psi_i(\sigma)\sigma^3d\sigma\ge 0,\quad \forall r>0,\quad i=1,..,n.$$ On the other hand, from the behavior of $\bar \psi$ and $\psi^0$ at infinity (both are 