Therefore, $A_0>0$. Since $$\lim_{|x|\to\infty}u_k(x)=0\quad{\rm and} \quad\lim_{|x|\to\infty}\tilde u_k(x)=0,$$ for $R>0$ large enough, $$\epsilon_1:=\sup_{|x|\ge R}u_k(x)\le \frac{A_0}2\quad {\rm and}\quad\epsilon_2:=\sup_{|x|\ge R}\tilde u_k(x)\le \frac{A_0}2.$$ Since $u_k$ and $\tilde u_k$ are weak solutions of (\[eq1.3\]), by Lemma \[lm 3.1\], there exist weak solutions $u_{k,R}$ and $\tilde u_{k,R}$ to (\[3.11\]) such that $$\label{3.1.2}
u_k-\epsilon_1\le  u_{k,R}\le u_k\quad {\rm in}\quad B_R(0)\setminus\{0\}$$ and $$\label{3.1.1}
\tilde u_k-\epsilon_2\le \tilde u_{k,R}\le \tilde u_k\quad {\rm in}\quad B_R(0)\setminus\{0\}.$$ Moreover, by Proposition \[pr 2.1\] we obtain $$u_{k,R}\equiv\tilde  u_{k,R},$$ which, together with (\[3.1.1\]) and (\[3.1.2\]), implies that $$|u_k-\tilde u_k|\le \max\{\epsilon_1,\epsilon_2\} \quad {\rm in }\quad B_R(0)\setminus\{0\}.$$ Thus, $$\limsup_{x\to0}|u_k-\tilde u_k|(x)\le \max\{\epsilon_1,\epsilon_2\}<A_0.$$ This contradicts to the definition of $A_0$. As a consequence, problem (\[eq1.3\]) has a unique weak solution. $\Box$

Now we estimate the singularity rate of weak solution to (\[eq1.3\]) at the origin.

\[pr 3.2\] Let $k>0$, $0<p<\frac{N}{N-2\alpha}$ and $u_k$ be the weak solution of (\[eq1.3\]). Then $$\label{3.1.3}
  \lim_{x\to0}u_k(x)|x|^{N-2\alpha}=c_{N,\alpha}k.$$

[**Proof.**]{} On the one hand, we have that $$\label{eq 3.1--1}
  u_k(x)\le \mathbb{G}_{{\mathbb{R}}^N}[k\delta_0](x)=c_{N,\alpha}k|x|^{-N+2\alpha},\quad x\in{\mathbb{R}}^N\setminus\{0\}.$$ On the other hand, from the proof of Theorem \[teo 1\] and uniqueness of weak solution to (\[eq1.3\]), we know $u_k=\lim_{R\to\infty} u_{k,R},$ where $u_{k,R}$ is the weak solution of (\[3.11\]). By [@CV3 Lemma 2.1] and [@CV3 Proposition1.1], we have that $$\lim_{x\to0}u_{k,R}(x)|x|^{N-2\alpha}=c_{N,\alpha}k.$$ Then together with (\[eq 3.1–1\]) and 