approach the weak solution.

Here we develop the latter method to attain weak solution of (\[eq 1.1\]) with $\epsilon=-1$ and the main results state as follows.

\[teo 3\] Let $\epsilon=-1$, $k>0$ and $\nu\in\mathfrak{M}^b_{\partial\Omega}(\bar\Omega)$ nonnegative with ${\|\nu\|}_{\mathfrak{M}^b(\bar\Omega)}=1$.

$(i)$ Suppose that $$\label{06-08-2}
g(s)\le c_3s^{p_0}+\epsilon,\quad \forall s\ge0,$$ for some $p_0\in(0,1]$, $c_3>0$ and $\epsilon>0$. Assume more that $c_3$ is small enough when $p_0=1$.

Then problem (\[eq 1.1\]) admits a nonnegative weak solution $u_\nu$ satisfying $$\label{1.5}
 u_\nu(x)\ge \mathbb{G}_\alpha[\frac{\partial^\alpha\nu}{\partial\vec{n}^\alpha}](x),\qquad \forall x\in\Omega.$$

$(ii)$ Suppose that $$\label{1.1+++}
 g(s)\le c_4s^{p_*}+\epsilon,\quad \forall s\in[0,1]$$ and $$\label{1.4}
g_\infty:=\int_1^{\infty} g(s)s^{-1-\frac{N+\alpha}{N-\alpha}}ds<+\infty,$$ where $c_4,\epsilon>0$ and $p_*>1$.

Then there exist $k_0,\epsilon_0>0$ depending on $c_4, p_* $ and $ g_\infty$ such that for $k\in[0,k_0)$ and $\epsilon\in(0,\epsilon_0)$, problem (\[eq 1.1\]) admits a nonnegative weak solution $u_\nu$ satisfying (\[1.5\]).

We remark that $(i)$ it does not require any restrictions on parameters $c_3, \epsilon, k$ when $p_0\in(0,1)$ or on parameters $ \epsilon, \sigma$ when $p_0=1$; $(ii)$ the integral subcritical condition (\[1.4\]) has the same critical value with (\[g1\]).

The rest of the paper is organized as follows. In Section 2 we study the properties of $\frac{\partial^\alpha \nu}{\partial\vec{n}^\alpha}$. Section 3 is devoted to prove Theorem \[teo 1\]. In Section 4 we analyse the properties of the weak solution for problem (\[eq 