in the above we obtain $$g(t)=c'_1g_1(t)+c'_2g_2(t)+O(e^{(k-1+\tau-\delta_0)t}),\quad \mbox{for }t>1$$ where $c'_1,c'_2$ are two constants. Since $g(t)=O(e^{(k-1+\tau )t})$ for $t\to \infty$, we see that $c'_1=0$ and therefore $g(t)=O(e^{(k-1+\tau-\delta_0)t})$ as $t\to \infty$. Equivalently $$\label{11mar29e3}
\phi_i^k(r)=O(r^{k-1+\tau-\delta_0}) \quad i\in I.$$ With (\[11mar29e3\]) we further obtain $$\sum_ja_{ij}e^{u_j}\phi^k_j=O(r^{k-1+\tau-2-2\delta_0}).$$ Consequently $\phi_i^k(x)=O(r^{k-1+\tau-2\delta_0})$. Keep doing this for finite steps we obtain that $\phi_i^k$ decays faster than $r^{-1}$ at infinity. The asymptotic theory of ODE can be similarly used to show that $\phi_i^k(r)=o(r)$ as $r\to 0$.

To get a contradiction, without loss of generality, we may assume that some of $\phi_i^k$, say $\phi_1^k(r)>0$ for some $r>0$ and $$\max_{\mathbb R^+}\bigg (\frac{\phi_1^k(r)}{f_1(r)}\bigg )
=\max_{1\le j\le n}\bigg (\max_{\mathbb R^+}\frac{\phi_j^k(r)}{f_j(r)}\bigg ).$$ By noting $\phi_1^k(r)=o(r)$ as $r\to 0$ and $\phi_1^k(r)=o(\frac 1r)$ as $r\to \infty$, $\phi_1^k/
f_1(r)$ attains its maximum at some point $r_0\in \mathbb R^+$. Let $w_1(r)=\phi_1^k(r)/f_1(r)$. By a direct computation, $w_1(r)$ satisfies $$\label{1019e1}
\Delta w_1+2\nabla w_1\cdot \frac{\nabla f_1}{f_1}+\frac{1-k^2}{r^2}w_1=\sum_{j=2}^na_{1j}e^{u_j}
\bigg (\frac{w_1f_j-\phi_j}{f_1}\bigg ).$$ Now we apply the maximum principle at $r=r_0$, and obtain $$\Delta w_1(r_0)\le 0,\quad \mbox{and}\quad \nabla w_1(r_0)=0.$$ Since $k>1$, (\[1019e1\]) yields $$\label{1019e2}
\sum_{j=2}^na_{1j}e^{u_j}\bigg (\frac{w_1f_j-\phi_j^k}{f_1}\bigg )(r_0)<0$$ because $w_1(r_0)>0$. On the other hand, for $j\ge 2$, $$w_1(r_0)f_j(r_0)-\phi_j^k(r_0)=f_j(r_0)(\frac{\phi_1^k(r_0)}{f_1(r_0)}-\frac{\phi_j^k(r_0)}{f_j(r_0)})\ge
0,$$ which obviously contradicts (\[1019e2\]). Therefore (\[11mar29e1\]) is established. When $k=1$, $\phi_i^1\equiv 0$ because by Lemma \[1129lem1\], $\phi_i^1(r)=Cu_i'(r)$. By the 