# Singular Nonlocal Problem Involving Measure Data

• Sekhar Ghosh
• Debajyoti Choudhuri
• Ratan Kr. Giri
Article

## Abstract

In this paper we prove existence of solutions for a partial differential equation involving a singularity with a general nonnegative, Radon measure as source term which is given as
\begin{aligned} (-\Delta )^s u&= f(x)h(u)+\mu ~\text {in}~\Omega , \\ u&=0~\text {in}~\mathbb {R}^N{\setminus }\Omega , \\ u&> 0~\text {in}~\Omega , \end{aligned}
where $$\Omega$$ is a bounded domain of $$\mathbb {R}^N$$, f is a nonnegative function over $$\Omega$$.

## Keywords

Elliptic PDE Fractional Sobolev space Schauder fixed point theorem

35J35 35J60

## 1 Introduction

We prove the existence of nonnegative weak solution to the following nonlocal problem.
\begin{aligned} \begin{aligned} (-\Delta )^s u&= f(x)h(u)+\mu ~\text {in}~\Omega ,\\ u&=0~\text {in}~\mathbb {R}^N{\setminus }\Omega ,\\ u&> 0~\text {on}~\Omega , \end{aligned} \end{aligned}
(1.1)
where $$\Omega$$ is a bounded domain in $$\mathbb {R}^N$$ for $$N \ge 2$$, $$0<s<1$$, $$f > 0$$ and $$\mu$$ is a nonnegative Radon measure. Here,
\begin{aligned} (-\Delta )^su(x)= & {} C_{N,s}\int \limits _{\mathbb {R}^N}\frac{(u(x)-u(y))}{|x-y|^{N+2s}}dy \end{aligned}
where $$C_{N,s}$$ is a normalization constant. A rich amount of research work has been done to prove the existence of a solution to the singular problem involving a Laplacian
\begin{aligned} \begin{aligned} -\Delta u&= f(x)h(u)~\text {in}~\Omega ,\\ u&=0~\text {on}~\partial \Omega . \end{aligned} \end{aligned}
(1.2)
Of them, a few noteworthy articles can be found in Lazer and McKenna (1991), Giachetti et al. (2017a, b), Taliaferro (1979), Gatica et al. (1989), Arcoya et al. (2009), Oliva and Petitta (2018), Panda et al. (2017) and the references therein. However, to our knowledge, there are very few articles which either involves a singularity or a measure data with a nonlocal operator, Ghanmi and Saoudi (2016). In our study we consider both a singularity and a measure involving a fractional Laplacian which is a new consideration. Due to the nonlocal behaviour of the operator $$(-\Delta )^s$$ and the vanishing Dirichlet boundary condition it is natural to consider the fractional Sobolev space $$W_0^{s,p}(\Omega )$$. We denote fractional Sobolev space as $$W^{s,p}(\Omega )$$ (Di Nezza et al. 2012), where $$\Omega$$ is an open set of $$\mathbb {R}^N$$, consists of all locally summable functions $$u: \Omega \rightarrow \mathbb {R}$$ such that $$\frac{u(x)-u(y)}{|x-y|^{\frac{N+ps}{p}}}\in L^p(\Omega \times \Omega )$$. If $$u\in W^{s,p}(\Omega )$$, then its norm is defined as
\begin{aligned} \Vert u\Vert _{W^{s,p}(\Omega )}= \left( \int _{\Omega }|u|^p+\int _{\Omega }\int _{\Omega }\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}\right) ^{\frac{1}{p}}\,\,\text{ for }\,\, 1\le p<\infty , 0<s<1, \end{aligned}
where $$\displaystyle {[u]_{W^{s,p}(\Omega )}= \left( \int _{\Omega }\int _{\Omega }\frac{|u(x)-u(y)|^p}{|x-y|^{N+sp}}\right) ^{\frac{1}{p}}}$$ is the so-called Gagliardo seminorm. Two other spaces which we will be using in this article are the Hölder Space and the Marcinkiewicz space. The Hölder space (Evans 2010) is denoted by $$C^{k, \beta }(\bar{\Omega })$$ with $$0<\beta \le 1$$ consists of all functions $$u\in C^k(\bar{\Omega })$$ such that the norm $$\sum \nolimits _{|\alpha |\le k}\sup |D^{\alpha }u|+\sup \nolimits _{x\ne y} \left\{ \frac{|D^ku(x)-D^ku(y)|}{|x-y|^{\beta }}\right\}$$ is finite. We will denote the Marcinkiewicz space by $${M}^q(\Omega )$$ (Benilan et al. 1995) [or weak $$L^q(\Omega )$$] defined for every $$0< q <\infty$$, as the space of all measurable functions $$f:\Omega \rightarrow \mathbb {R}$$ such that the corresponding distribution functions satisfy an estimate of the form
\begin{aligned} m(\{x\in \Omega :|f(x)|>t\})\le \frac{C}{t^q},~t>0, ~C<\infty . \end{aligned}
Here m refers to the Lebesgue measure. Certainly, for bounded $$\Omega$$ we have $${M}^q\subset {M}^{\bar{q}}$$ if $$q\ge \bar{q}$$, for some fixed positive $$\bar{q}$$. We bring to mind that the following continuous embeddings hold
\begin{aligned} L^q(\Omega )\hookrightarrow {M}^q(\Omega )\hookrightarrow L^{q-\epsilon }(\Omega ), \end{aligned}
(1.3)
for every $$1<q<\infty$$ and $$0<\epsilon <q-1$$.
We further denote the space of all finite Radon measures on $$\Omega$$ as $$\mathcal {M}(\Omega )$$. If $$\mu \in \mathcal {M}(\Omega )$$, then its norm is given by
\begin{aligned} \Vert \mu \Vert _{\mathcal {M}(\Omega )}=\int \limits _{\Omega }d|\mu |. \end{aligned}

### Definition 1.1

Let $$(\mu _n)$$ be the sequence of measurable functions in $$\mathcal {M}(\Omega )$$. We say $$(\mu _n)$$ converges to $$\mu \in \mathcal {M}(\Omega )$$ in the sense of measure (Folland 2013) i.e. $$\mu _n\rightharpoonup \mu$$ in $$\mathcal {M}(\Omega )$$, if
\begin{aligned} \int \limits _\Omega \phi d\mu _n\rightarrow \int \limits _\Omega \phi d\mu , \,\,\,\forall \,\phi \in C_0(\Omega ). \end{aligned}
We will use the truncation function for fixed $$k > 0$$,
\begin{aligned} T_k(z)=\max \{-k,\min \{k,z\}\} \end{aligned}
and
\begin{aligned} G_k(z)=(|z|-k)^+sign(z). \end{aligned}
with $$z\in \mathbb {R}$$. Observe that $$T_k(z)+G_k(z)=z$$ for any $$z\in \mathbb {R}$$ and $$k>0$$.
For our ease, we will also make use of the notation
\begin{aligned} \int \limits _\Omega f(x) dx =\int \limits _\Omega f. \end{aligned}
The paper has been organized in the following way. In Sect. 2 we state and prove the main results of existence of a weak solution to the problem (1.1). Here we have discussed the cases $$\gamma \le 1$$ and $$\gamma >1$$ separately. In Sect. 3 we make a few remark for the case $$\gamma <1$$. In Subsect. 3.1 we discuss the problem with a few relaxation on the assumptions made on f.

## 2 Results on Existence of Solutions

Let us consider the following boundary value problem
\begin{aligned} (-\Delta )^s u&= h(u)f+\mu ~\text {in}~\Omega ,\nonumber \\ u&= 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega , \nonumber \\ u&> 0 ~\text {in}~ \Omega , \end{aligned}
(2.1)
where $$\Omega$$ is an open bounded subset of $$\mathbb {R}^N$$, $$0<s<1$$, $$N>2$$, $$\mu$$ is a nonnegative, bounded Radon measure on $$\Omega$$, $$f\in L^m (\Omega )$$ for $$m>1$$, which could be a measure, and h is a non-increasing continuous function that may blow up at zero.
On the nonlinear function $$h : \mathbb {R}^+ \rightarrow \mathbb {R}^+$$ we assume it to be continuous, such that
\begin{aligned} \underset{z\rightarrow 0^+}{\lim }\,\,\, h(z)\in (0,\infty ] \,\, \text {and}\,\,\underset{z\rightarrow \infty }{\lim }\,\,\, h(z)=h(\infty )<\infty . \end{aligned}
(2.2)
We assume the growth condition near zero as
\begin{aligned} \exists ~C_1, \underline{K}>0\,\,\text {such that}\,\, h(z)\le \frac{C_1}{z^\gamma }\,\,\text {if}\,\,z<\underline{K}, \end{aligned}
(2.3)
with $$\gamma > 0$$. We will later observe that the regularity of the solution u is affected by the behavior of h at infinity. So, we need to assume the following
\begin{aligned} \exists ~C_2, \overline{K}>0 \,\,\text {such that}\,\, h(z)\le \frac{C_2}{z^\theta }\,\,\text {if}\,\,z>\overline{K}, \end{aligned}
(2.4)
for $$\theta >0$$.

### Definition 2.1

We define the space $$W_0^{s,2}(\Omega )$$ as $$W_0^{s,2}(\Omega )=\{u\in W^{s,2}(\Omega ):u=0~\text {in}~\mathbb {R}^N{\setminus }\Omega \}$$.

### Definition 2.2

If $$\gamma \le 1$$, then a weak solution to the problem (2.1) is a function $$u\in W_0^{s,1}(\Omega )$$ such that
\begin{aligned} \forall K \subset \subset \Omega ,\,\, \exists \,\,C_K : u\ge C_K>0, \end{aligned}
(2.5)
and
\begin{aligned} \int \limits _{\Omega }(-\Delta )^{s/2} u\cdot (-\Delta )^{s/2}\varphi =\int \limits _{\Omega }h(u)f\varphi +\int \limits _{\Omega }\varphi d\mu , \,\,\forall \varphi \in C_c^1(\Omega ). \end{aligned}
(2.6)
with $$h(u)f\in L_{loc}^1(\Omega ).$$ If $$\gamma >1$$, then a weak solution to the problem (2.1) is a function $$u\in W_{loc}^{s,1}(\Omega )$$ satisfying Eqs. (2.5) and (2.6) such that $$T_k^{\frac{\gamma +1}{2}}(u)\in W_0^{s,2}(\Omega )$$ for every fixed $$k>0$$.

For both the cases, $$\gamma \le 1$$ and $$\gamma >1$$, we will show the existence of weak solutions for the problem (2.1) in the subsequent Subsects. 2.1 and 2.2.

In an attempt to obtain an existence result to the problem in Eq. (2.1), let us consider the following sequence of problems
\begin{aligned} \begin{aligned} (-\Delta )^s u_n&= h_n\left( u_n+\frac{1}{n}\right) f_n+\mu _n ~\text {in}~\Omega ,\\ u_n&= 0 ~\text {on}~\mathbb {R}^N{\setminus }\Omega , \end{aligned} \end{aligned}
(2.7)
where ($$\mu _n$$) is a sequence of smooth nonnegative functions bounded in $$L^1(\Omega )$$ and converging weakly to $$\mu$$ in the sense of Definition 1.1 and $$h_n=T_n(h)$$, $$f_n=T_n(f)$$ the truncation at level n.
The weak formulation of Eq. (2.7) is
\begin{aligned} \int \limits _{\Omega } (-\Delta )^{s/2} u_n \cdot (-\Delta )^{s/2}\varphi = \int \limits _{\Omega } h_n\left( u_n+\frac{1}{n}\right) f_n\varphi + \int \limits _\Omega \mu _n\varphi , ~\forall \,\varphi \in C_c^1(\Omega ).\quad \quad \end{aligned}
(2.8)

### Lemma 2.3

Problem (2.7) admits a nonnegative weak solution $$u_n\in W_0^{s,2} (\Omega )\cap L^{\infty }(\Omega )$$.

### Proof

This proof is derived from Schauder’s fixed point argument in Oliva and Petitta (2016). For a fixed $$n\in \mathbb {N}$$, let us define a map
\begin{aligned} G:L^2(\Omega )\rightarrow L^2(\Omega ), \end{aligned}
such that, for any $$v\in L^2(\Omega )$$ gives the weak solution w to the following problem
\begin{aligned} \begin{aligned} (-\Delta )^s w&= h_n\left( |v|+\frac{1}{n}\right) f_n+\mu _n ~\text {in}~\Omega ,\\ w&= 0 ~\text {in}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned} \end{aligned}
(2.9)
The existence of a unique $$w\in W_0^{s,2}(\Omega )$$ corresponding to a $$v\in L^2(\Omega )$$ is guaranteed by Petitta (2016). Thus we can choose w as a test function in the weak formulation (2.8) with the test function space $$W_0^{s,2}(\Omega )$$. Thus
\begin{aligned} \lambda _1\int \limits _{\Omega }|w|^2&\le \int \limits _\Omega |(-\Delta )^{s/2} w|^2 = \int \limits _\Omega (-\Delta )^{s/2} w \cdot (-\Delta )^{s/2} w \nonumber \\&= {\int \limits _\Omega h_n\left( |v|+\frac{1}{n}\right) f_n w} +\int \limits _\Omega w{\mu }_n \nonumber \\&\le { C_1 \int \limits _{\left( |v|+\frac{1}{n}\le \underline{K}\right) }\frac{f_n w}{\left( |v|+\frac{1}{n}\right) ^\gamma }} + \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le (|v|+\frac{1}{n})\le \overline{K}\right) } f_n w \nonumber \\&\quad + C_2 \int \limits _{\left( |v|+\frac{1}{n} \ge \overline{K}\right) }\frac{f_n w}{\left( |v|+\frac{1}{n}\right) ^\theta }+ C (n) \int _\Omega |w| \nonumber \\&\le { C_1 n^{1+\gamma }\int \limits _{\left( |v|+\frac{1}{n}\le \underline{K}\right) }|w| + n \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le \left( |v|+\frac{1}{n}\right) \le \overline{K}\right) } |w|} \nonumber \\&\quad {+ C_2n^{1+\theta } \int _{\left( |v|+\frac{1}{n} \ge \overline{K}\right) } |w| } + C(n) \int \limits _\Omega |w|\nonumber \\&\le C(n,\gamma ) \int \limits _\Omega |w| \end{aligned}
(2.10)
\begin{aligned}&\le C(n,\gamma ) \left( \int \limits _\Omega |w|^2\right) ^{1/2} \end{aligned}
(2.11)
By using the Poincaré inequality (Di Nezza et al. 2012) on the left hand side and Hölder’s inequality on the right side of the inequality in Eq. (2.10) we get
\begin{aligned} \int \limits _\Omega |w|^2 \le C'\cdot C(n,\gamma )\bigg (\int \limits _\Omega |w|^2\bigg )^\frac{1}{2}. \end{aligned}
Therefore
\begin{aligned} \left\| w \right\| _{L^2(\Omega )}\le C'\cdot C(n,\gamma ), \end{aligned}
(2.12)
where, $$C'$$ and $$C(n,\gamma )$$ are independent of v. We now prove that the map G is continuous on $$L^2(\Omega )$$. Consider a sequence ($$v_k$$) that converges to v in $$L^2(\Omega )$$. Then by the dominated convergence theorem,
\begin{aligned} \left\| {\left( h_n\left( v_k +\frac{1}{n}\right) f_n+{\mu }_n\right) -\left( h_n\left( v +\frac{1}{n}\right) f_n+{\mu }_n\right) } \right\| _{L^2(\Omega )}\longrightarrow 0, \end{aligned}
as $$k\rightarrow \infty$$. Hence, by the uniqueness of the weak solution, we can say that $$w_k=G(v_k)$$ converges to $$w=G(v)$$ in $$L^2(\Omega )$$. Thus G is continuous over $$L^2(\Omega )$$.
What finally needs to be checked is that $$G(L^2(\Omega ))$$ is relatively compact in $$L^2(\Omega )$$. We proved in Eq. (2.10) that
\begin{aligned} \int \limits _\Omega |(-\Delta )^{s/2} w|^2=\int \limits _\Omega |(-\Delta )^{s/2} G(v)|^2 \le C(n,\gamma ), \end{aligned}
for any $$v\in L^2(\Omega )$$, so that $$G(L^2(\Omega ))$$ is relatively compact in $$L^2(\Omega )$$ by Rellich–Kondrachov theorem. Therefore, we proved that $$G(L^2(\Omega ))$$ is relatively compact in $$L^2(\Omega )$$.

Now, applying the Schauder fixed point theorem to obtain that G has a fixed point $$u_n\in L^2(\Omega )$$ that is a solution to Eq. (2.7) in $$W_0^{s,2}(\Omega )$$. Furthermore, $$u_n$$ belongs to $$L^{\infty }(\Omega )$$ by Canino et al. (2017).

Since, $$\left( h_n\left( u_n+\frac{1}{n}\right) f_n+{\mu }_n\right) \ge 0$$ then by the maximum principle $$u_n\ge 0$$ and this concludes the proof. $$\square$$

The next step is to prove that ($$u_n$$) is uniformly bounded from below on compact subsets of $$\Omega$$.

### Lemma 2.4

The sequence ($$u_n$$) is such that for every $$K\subset \subset \Omega$$ there exists $$C_K$$ (independent of n) such that $$u_n(x)\ge C_K >0$$, a.e. in K, and for every $$n\in \mathbb {N}$$.

### Proof

Let us consider the problem
\begin{aligned} \begin{aligned} (-\Delta )^s w_n&= h_n\left( w_n+\frac{1}{n}\right) f_n ~\text {in}~\Omega , \\ w_n&= 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned} \end{aligned}
(2.13)
We will show the existence of a weak solution $$w_n$$ of Eq. (2.13) such that
\begin{aligned} \forall K \subset \subset \Omega , \exists \,\,C_K \,\,\,\text {such that}\,\,\, w_n \ge C_K >0, \end{aligned}
for almost every x in K and $$C_K$$ is independent of n. Before establishing this we will show that the sequence ($$w_n$$) is monotonically increasing. Towards this we consider
\begin{aligned} (-\Delta )^s w_n&=f_n h_n\left( w_n+\frac{1}{n}\right) \nonumber \\&\le f_{n+1}h_n\left( w_n+\frac{1}{n+1}\right) . \end{aligned}
(2.14)
We also know that $$w_{n+1}$$ is a weak solution to
\begin{aligned} (-\Delta )^s w_{n+1}&= f_{n+1}h_{n+1}\left( w_{n+1}+\frac{1}{n+1}\right) ~\text {in}~\Omega ,\nonumber \\ w_{n+1}&= 0 ~\text {in}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned}
(2.15)
The difference between the weak formulations of the problems in (2.14), (2.15) with the choice of a test function as $$(w_n-w_{n+1})^{+}$$ we obtain
\begin{aligned}&\int \limits _{\Omega }(-\Delta )^s(w_n-w_{n+1})\cdot (-\Delta )^s(w_n-w_{n+1})^{+}=\int \limits _{\Omega }|(-\Delta )^s(w_n-w_{n+1})^{+}|^2 \nonumber \\&\quad \le \int \limits _{\Omega }f_{n+1}\left[ h_n\left( w_n+\frac{1}{n+1}\right) -h_{n+1}\left( w_{n+1}+\frac{1}{n+1}\right) \right] (w_n-w_{n+1})^{+}\nonumber \\&\quad = \int \limits _{\Omega }f_{n+1}\bigg [\left\{ h_n\left( w_n+\frac{1}{n+1}\right) \right. \nonumber \\&\qquad \left. -h_{n}\left( w_{n+1}+\frac{1}{n+1}\right) \right\} \chi _{[w_n\le w_{n+1}]}(w_n-w_{n+1})^{+}\nonumber \\&\qquad +\left\{ h_n\left( w_n+\frac{1}{n+1}\right) -h_{n}\left( w_{n+1}+\frac{1}{n+1}\right) \right\} \chi _{[w_n> w_{n+1}]}(w_n-w_{n+1})^{+}\bigg ]\nonumber \\&\quad = \int \limits _{\Omega }f_{n+1}\left[ \left\{ h_n\left( w_n+\frac{1}{n+1}\right) -h_{n}\left( w_{n+1}+\frac{1}{n+1}\right) \right\} \right] \nonumber \\&\qquad \chi _{[w_n>w_{n+1}]}(w_n-w_{n+1})^{+}\nonumber \\&\qquad \le 0. \end{aligned}
(2.16)
Therefore, $$(w_n-w_{n+1})^{+}=0$$ almost everywhere in $$\Omega$$, thus implying that $$w_n \le w_{n+1}$$. The right hand side of the problem
\begin{aligned} (-\Delta )^s w_1=f_1h_1(w_1+1)\in L^{\infty }(\Omega ) \end{aligned}
(2.17)
due to the definition of $$f_1$$ and $$h_1$$. We use the Summability theorem given in Lemma 2.2, page 6 of Canino et al. (2017) to obtain $$w_1\in L^{\infty }(\Omega )$$. Let $$\Vert w_1\Vert _{L^{\infty }}\le L$$. Thus we have
\begin{aligned} (-\Delta )^s w_1&=f_1h_1(w_1+1)\nonumber \\&\ge f_1h_1(\Vert w_1\Vert _{\infty }+1)\nonumber \\&\ge f_1h_1(L+1)\nonumber \\&>0. \end{aligned}
(2.18)
Since $$f_1h_1(L+1)$$ is identically not equal to zero, hence by the strong maximum principle over $$(-\Delta )^s$$ we have $$w_1>0$$. Since we have considered a relatively compact subset K of $$\Omega$$, there exists a constant $$L_K$$ such that $$w_1\ge L_K>0$$.
Now taking the difference between the weak formulations of Eqs. (2.7) and (2.13) respectively we have
\begin{aligned}&\int \limits _{\Omega }((-\Delta )^{s/2} u_n-(-\Delta )^{s/2} w_n)\cdot (-\Delta )^{s/2}\varphi \nonumber \\&\quad =\int \limits _{\Omega } \bigg (h_n\left( u_n +\frac{1}{n}\right) -h_n\left( w_n +\frac{1}{n}\right) \bigg )f_n(x)\varphi \nonumber \\&\qquad + \int \limits _\Omega \mu _n\varphi ,\,\,\forall \varphi \in C_c^1(\Omega ). \end{aligned}
(2.19)
It is easy to show that $$u_n\ge w_n$$ a.e. in K, for if not, i.e. there exists a subset of K which is of non zero Lebesgue measure on which $$u_n<w_n$$. On taking $$(u_n-w_n)^-$$ as a test function in the Eq. (2.19), we get
\begin{aligned} -\int \limits _\Omega |(-\Delta )^{s/2}(u_n-w_n)^-|^2&=\int \limits _\Omega (-\Delta )^{s/2} (u_n-w_n)\cdot (-\Delta )^{s/2} (u_n-w_n)^{-} \\&=\int \limits _\Omega \bigg (h_n\left( u_n +\frac{1}{n}\right) -h_n\left( w_n +\frac{1}{n}\right) \bigg )f_n(x) (u_n-w_n)^-\\&\quad + \int \limits _\Omega \mu _n(u_n-w_n)^{-} \\&\ge 0. \end{aligned}
This implies $$u_n\ge w_n$$ a.e. in $$\Omega$$ and hence in K, and so $$\forall K \subset \subset \Omega$$ there exists $$C_K$$ such that $$u_n \ge C_K >0$$ a.e. in K. $$\square$$

We are now in a position to prove the existence of a solutions to the problem (2.1). In order to do this we differentiate between the following two cases.

### 2.1 When $$\gamma \le 1$$

In this subsection, we consider the problem in Eq. (2.7) for the case of $$\gamma \le 1$$.

### Lemma 2.5

Let $$u_n$$ be a solution of Eq. (2.7), where h satisfy Eqs. (2.3) and (2.4), with $$\gamma \le 1$$. Then $$(u_n)$$ is bounded in $$W_0^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$.

### Proof

We will prove that ($$(-\Delta )^{s/2} u_n$$) is bounded in $${M}^{\frac{N}{N-s}}(\Omega )$$. For this, we take $$\varphi =T_k(u_n)$$ as a test function in the weak formulation (2.7) and get
\begin{aligned} \int \limits _\Omega |(-\Delta )^{s/2} T_k(u_n)|^2 = \int \limits _\Omega h_n\left( u_n +\frac{1}{n}\right) T_k(u_n)f_n + \int \limits _\Omega T_k(u_n) \mu _n. \end{aligned}
(2.20)
Now, $$\frac{T_k(u_n)}{(u_n+\frac{1}{n})^{\gamma }}\le \frac{u_n}{(u_n+\frac{1}{n})^{\gamma }}=\frac{u_n^{\gamma }}{(u_n+\frac{1}{n})^{\gamma }u_n^{\gamma -1}}\le u_n^{1-\gamma }$$.
Using Eqs. (2.3) and (2.4) in the right hand side of Eq. (2.20) we have,
\begin{aligned}&{\int \limits _\Omega h_n\left( u_n +\frac{1}{n}\right) f_n T_k(u_n)}\\&\quad \le { C_1 \int \limits _{\left( u_n +\frac{1}{n}\le \underline{K}\right) }\frac{f_n T_k(u_n)}{\left( u_n +\frac{1}{n}\right) ^\gamma }} + \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le \left( u_n+\frac{1}{n}\right) \le \overline{K}\right) } f_n T_k(u_n) \\&\qquad + C_2 \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) }\frac{f_n T_k(u_n)}{\left( u_n+\frac{1}{n}\right) ^\theta } \\&\quad \le C_1\underline{K}^{1-\gamma } \int \limits _{\left( u_n+\frac{1}{n}\le \underline{K}\right) }f + k \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le \left( u_n+\frac{1}{n}\right) \le \overline{K}\right) }f + \frac{C_2 k}{\overline{K}^{\theta }} \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) } f \\&\quad \le C \end{aligned}
and
\begin{aligned} \int \limits _\Omega T_k(u_n) \mu _n \le k\parallel \mu _n\parallel _{L^1(\Omega )} \le Ck. \end{aligned}
Combining the above two inequalities in this lemma, we obtain
\begin{aligned} \int \limits _\Omega |(-\Delta )^{s/2} T_k(u_n)|^2 \le Ck. \end{aligned}
(2.21)
Since
\begin{aligned} \left\{ |(-\Delta )^{s/2} u_n|\ge t\right\}&= \left\{ |(-\Delta )^{s/2} u_n|\ge t,u_n< k\right\} \cup \left\{ |(-\Delta )^{s/2} u_n| \ge t,u_n \ge k\right\} \\&\subset \left\{ |(-\Delta )^{s/2} u_n|\ge t,u_n <k\right\} \cup \left\{ u_n \ge k\right\} \subset \Omega , \end{aligned}
hence by the sub-additivity of the Lebesgue measure m we get
\begin{aligned}&m\left( \left\{ |(-\Delta )^{s/2} u_n|\ge t\right\} \right) \le m\left( \left\{ |(-\Delta )^{s/2} u_n|\ge t,u_n< k\right\} \right) \nonumber \\&\quad +\; m(\{u_n \ge k\}). \end{aligned}
(2.22)
Therefore from the Sobolev inequality
\begin{aligned} \left( \int \limits _\Omega |T_k(u_n)|^{2^*}\right) ^{\frac{2}{2^*}}\le \frac{1}{\lambda _1}\int \limits _{\Omega }|(-\Delta )^{s/2} T_k(u_n)|^2 \le Ck, \end{aligned}
where $$\lambda _1$$ is the first eigenvalue of the fractional Laplacian operator. Now if we restrict the integral on the left hand side on $$I_1=\left\{ x\in \Omega :u_n(x)\ge k \right\}$$, on which $$T_k(u_n)=k$$, we then obtain
\begin{aligned} k^2m(\{u_n\ge k\})^{\frac{2}{2^*}}\le Ck, \end{aligned}
so that
\begin{aligned} m(\{u_n\ge k\})\le \frac{C}{k^\frac{N}{N-2s}}, \,\, \forall ~k\ge 1. \end{aligned}
So $$(u_n)$$ is bounded in the Marcinkiewicz space $$M^{\frac{N}{N-2s}}(\Omega )$$. Similarly proceeding for $$I_2=\{|(-\Delta )^{s/2} u_n|\ge t,u_n< k\}$$, we have
\begin{aligned} m\left( \{|(-\Delta )^{s/2} u_n|\ge t,u_n< k\}\right) \le \frac{1}{t^2}\int \limits _\Omega |(-\Delta )^{s/2} T_k(u_n)|^2\le \frac{Ck}{t^2}, ~\forall \,k>1. \end{aligned}
From Eq. (2.22), we have
\begin{aligned} m\left( \left\{ |(-\Delta )^{s/2} u_n|\ge t\right\} \right)&\le m\left( \left\{ |(-\Delta )^{s/2} u_n|\ge t,u_n< k\right\} \right) + m(\{u_n \ge k\}) \\&\le \frac{Ck}{t^2} + \frac{C}{k^\frac{N}{N-2s}}, ~\forall \,k>1. \end{aligned}
We then choose $$k=t^{\frac{N-2s}{N-s}}$$ to get
\begin{aligned} m\left( \left\{ |(-\Delta )^{s/2} u_n|\ge t\right\} \right) \le \frac{C}{t^\frac{N}{N-s}},~\forall \,t\ge 1. \end{aligned}
We have proved that $$((-\Delta )^{s/2} u_n)$$ is bounded in the Marcinkiewicz space $$M^{\frac{N}{N-s}}(\Omega )$$. By the property in Eq. (1.3) the sequence $$(u_n)$$ is bounded in $$W_0^{s,q}(\Omega )$$ for every $$1\le q<\frac{N}{N-s}$$. $$\square$$

### Theorem 2.6

Let $$\gamma \le 1$$. Then there exists a weak solution u of the problem (2.1) in $$W_0^{s,q}(\Omega )$$ for every $$q< \frac{N}{N-s}$$.

### Proof

With the consideration of Lemma 2.5, it implies that there exists u such that the sequence $$(u_n)$$ converges weakly to u in $$W_0^{s,q}(\Omega )$$ for every $$q< \frac{N}{N-s}$$. This implies that for $$\varphi$$ in $$C_c^1(\Omega )$$
\begin{aligned} \lim _{n\rightarrow +\infty } \int \limits _{\Omega } (-\Delta )^{s/2} u_n \cdot (-\Delta )^{s/2}\varphi = \int \limits _{\Omega }(-\Delta )^{s/2} u \cdot (-\Delta )^{s/2}\varphi . \end{aligned}
In addition to this, by compact embeddings we can assume that $$u_n$$ converges to u both strongly in $$L^1(\Omega )$$ and a.e. in $$\Omega$$. Thus, taking $$\varphi$$ in $$C_c^1(\Omega )$$, we have
\begin{aligned} 0&\le |h_n\left( u_n+\frac{1}{n}\right) f_n\varphi | \\&\le { \left\{ \begin{array}{ll} \frac{C_1\parallel \varphi \parallel _{L^\infty (\Omega )}f}{C_K^\gamma }, &{} \text {if }\,\, \left( u_n+\frac{1}{n}\le \underline{K}\right) \\ {M\parallel \varphi \parallel _{L^\infty (\Omega )}f}, &{} \text {if }\,\, \left( \underline{K}\le u_n+\frac{1}{n}\le \overline{K}\right) \\ \frac{C_2\parallel \varphi \parallel _{L^\infty (\Omega )}f}{C_K^\theta }, &{} \text {if }\,\, \left( u_n+\frac{1}{n}\ge \overline{K}\right) \end{array} \right. } \end{aligned}
where $$M>0$$ and K is the set $$\{x\in \Omega : \varphi (x)\ne 0\}$$. This is sufficient to apply the dominated convergence theorem to obtain
\begin{aligned} \lim _{n\rightarrow +\infty } \int _{\Omega } h_n\left( u_n+\frac{1}{n}\right) f_n\varphi = \int \limits _{\Omega }h( u) f\varphi . \end{aligned}
Hence, we can pass the limit $$n\rightarrow \infty$$ in the last term of Eq. (2.8) involving $$\mu _n$$. This concludes the proof of the result as it is easy to pass to the limit in Eq. (2.8). Therefore, we obtain a weak solution of Eq. (2.1) in $$W_0^{s,q}(\Omega )$$ for every $$q< \frac{N}{N-s}$$. $$\square$$

### 2.2 When $$\gamma >1$$

This case corresponds to a strongly singular case, which is why we can produce some local estimates on $$u_n$$ in the fractional Sobolev space. We shall give global estimates on $$T_k^{\frac{\gamma +1}{2}}(u_n)$$ in $$W_0^{s,2}(\Omega )$$ with the aim of giving sense, at least in a weak sense, to the boundary values of u.

### Lemma 2.7

Let $$u_n$$ be a weak solution of the problem in Eq. (2.7) with $$\gamma >1$$. Then $$T_k^{\frac{\gamma +1}{2}}(u_n)$$ is bounded in $$W_0^{s,2}(\Omega )$$ for every fixed $$k>0$$.

### Proof

We use a result due to Canino et al. (2017), Proposition 3.3. Consider
\begin{aligned}&\int \limits _\Omega |(-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)|^2=\int \limits _\Omega (-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)\cdot (-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)\nonumber \\&\quad \le \int \limits _{\Omega }\left[ h_n\left( u_n+\frac{1}{n}\right) f_n+\mu _n\right] \cdot \left( \frac{\gamma +1}{2}\right) T_k^{\frac{\gamma -1}{2}}(u_n)T_k^{\frac{\gamma +1}{2}}(u_n)\nonumber \\&\quad =\frac{\gamma +1}{2}\left( \int \limits _{\Omega } h_n\left( u_n+\frac{1}{n}\right) f_n T_k^{\gamma }(u_n)+\int \limits _{\Omega }T_k^{\gamma }(u_n)\mu _n\right) \end{aligned}
(2.23)
Since $$\frac{T_k^\gamma (u_n)}{(u_n+\frac{1}{n})^\gamma }\le \frac{u_n^\gamma }{(u_n+\frac{1}{n})^\gamma }\le 1$$, the term on the right hand side of Eq. (2.23) can be estimated as
\begin{aligned}&\int \limits _{\Omega } h_n\left( u_n+\frac{1}{n}\right) f_nT_k^\gamma (u_n) + \int \limits _{\Omega }T_k^\gamma (u_n)\mu _n\\&\quad \le { C_1 \int \limits _{\left( u_n +\frac{1}{n}\le \underline{K}\right) }\frac{f_n T_k^\gamma (u_n)}{\left( u_n +\frac{1}{n}\right) ^\gamma }} + C_2 \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) }\frac{f_n T_k^\gamma (u_n)}{\left( u_n+\frac{1}{n}\right) ^\theta } \\&\quad \quad + \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le (u_n+\frac{1}{n})\le \overline{K}\right) } f_n T_k^\gamma (u_n) + k^\gamma \int \limits _{\Omega } \mu _n \\&\quad \le C_1 \int \limits _{\left( u_n+\frac{1}{n}\le \underline{K}\right) }f + \frac{C_2 k^\gamma }{\overline{K}^{\theta }} \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) } f +k^\gamma \max _{[\underline{K},\overline{K}]} h(s) \int _{\left( \underline{K}\le \left( u_n+\frac{1}{n}\right) \le \overline{K}\right) }f\\&\quad \quad + k^\gamma \int \limits _\Omega \mu _n \\&\quad \le C(k,\gamma )k^\gamma \end{aligned}
From the inequality in Eq. (2.23) we get
\begin{aligned} \int \limits _{\Omega } |(-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)|^2 \le Ck^\gamma . \end{aligned}
(2.24)
Therefore, $$\left( T_k^{\frac{\gamma +1}{2}}(u_n)\right)$$ is bounded in $$W_0^{s,2}(\Omega )$$ for every fixed $$k>0$$. $$\square$$

Now, in order to pass to the limit $$n\rightarrow \infty$$ in the weak formulation (2.8), we require to prove some local estimates on $$u_n$$. We first prove the following.

### Lemma 2.8

Let $$u_n$$ be a weak solution of the problem (2.7) with $$\gamma >1$$. Then ($$u_n$$) is bounded in $$W_{loc}^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$.

### Proof

We follow Benilan et al. (1995) to prove this lemma. We prove the lemma in two steps.

$${\textit{Step 1}}$$ We claim that the sequence $$\left( G_1(u_n)\right)$$ is bounded in $$W_0^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$.

We can see that $$G_1(u_n)=0$$ when $$0\le u_n\le 1$$, $$G_1(u_n)=u_n-1$$, otherwise i.e when $$u_n>1$$. So $$(-\Delta )^{s/2} G_1(u_n)=(-\Delta )^{s/2} u_n$$ for $$u_n>1$$.

Now, we need to show that $$\left( (-\Delta )^{s/2} G_1(u_n)\right)$$ is bounded in the Marcinkiewicz space $$M^{\frac{N}{N-s}}(\Omega )$$. Since
\begin{aligned}&\left\{ |(-\Delta )^{s/2} u_n|> t, u_n>1\right\} \\&\quad = \left\{ |(-\Delta )^{s/2} u_n|> t,1<u_n\le k+1\right\} \cup \{|(-\Delta )^{s/2} u_n|> t,u_n> k+1\} \\&\quad \subset \left\{ |(-\Delta )^{s/2} u_n|> t,1<u_n\le k+1\right\} \cup \{u_n > k+1\}\subset \Omega . \end{aligned}
hence, by the sub-additivity of the Lebesgue measure m we have
\begin{aligned}&m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,u_n>1\right\} \right) \le m\left( \left\{ |(-\Delta )^{s/2} u_n|> t, ~1<u_n\le k+1\right\} \right) \nonumber \\&\qquad + m(\{u_n > k+1\}). \end{aligned}
(2.25)
In order to estimate Eq. (2.25), we take $$\varphi =T_k(G_1(u_n))$$, for $$k>1$$ as a test function in Eq. (2.8).
We observe that $$(-\Delta )^{s/2} T_k(G_1(u_n))= (-\Delta )^{s/2} u_n$$ only when $$1<u_n \le k+1$$, otherwise is zero, and $$T_k(G_1(u_n))=0$$ on $$\left\{ u_n\le 1\right\}$$. It follows that
\begin{aligned}&\int \limits _\Omega |(-\Delta )^{s/2} T_k(G_1(u_n))|^2 = \int \limits _\Omega h_n\left( u_n+\frac{1}{n}\right) f_nT_k(G_1(u_n)) + \int \limits _{\Omega }T_k(G_1(u_n))\mu _n \\&\quad \le { C_1 \int \limits _{\left( u_n +\frac{1}{n}\le \underline{K}\right) }\frac{f_n T_k(G_1(u_n))}{\left( u_n +\frac{1}{n}\right) ^\gamma }}+\max _{[\underline{K},\overline{K}]} h(s) \int \limits _{(\underline{K}\le \left( u_n+\frac{1}{n}\right) \le \overline{K})} f_n T_k(G_1(u_n)) \\&\quad \quad + C_2 \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) }\frac{f_n T_k(G_1(u_n))}{\left( u_n+\frac{1}{n}\right) ^\theta }+\int \limits _{\Omega }T_k(G_1(u_n))\mu _n \\&\quad \le { C_1k \int \limits _{\left( u_n+\frac{1}{n}\le \underline{K}\right) }\frac{f_n}{\left( 1+\frac{1}{n}\right) ^\gamma } + k \max _{[\underline{K},\overline{K}]} h(s) \int \limits _{\left( \underline{K}\le \left( u_n+\frac{1}{n}\right) \le \overline{K}\right) }f_n}\\&\quad \quad + {\frac{C_2 k}{\overline{K}^{\theta }} \int \limits _{\left( u_n+\frac{1}{n} \ge \overline{K}\right) } f_n }+\int \limits _{\Omega }T_k(G_1(u_n))\mu _n \\&\quad \le Ck. \end{aligned}
Since
\begin{aligned} \int \limits _{\Omega } |(-\Delta )^{s/2} T_k(G_1(u_n))|^2= & {} \int \limits _{\left\{ 1<u_n\le k+1 \right\} } |(-\Delta )^{s/2} u_n|^2\nonumber \\\ge & {} \int \limits _{\left\{ |(-\Delta )^{s/2} u_n|>t, 1<u_n\le k+1 \right\} } |(-\Delta )^{s/2} u_n|^2\nonumber \\\ge & {} t^2m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,1<u_n\le k+1\right\} \right) \quad \quad \quad \end{aligned}
(2.26)
so that
\begin{aligned} m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,1<u_n\le k+1\right\} \right) \le \frac{Ck}{t^2},~\forall \,k \ge 1. \end{aligned}
According to Eq. (2.24) in the proof of Lemma 2.7 one can see that
\begin{aligned} \int \limits _{\Omega } |(-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)|^2 \le Ck^\gamma , \quad \text {for any}\quad k>1, \end{aligned}
Therefore from the Sobolev inequality (Di Nezza et al. 2012) we get
\begin{aligned} \left( \int \limits _\Omega |T_k^{\frac{\gamma +1}{2}}(u_n)|^{2^*}\right) ^{\frac{2}{2^*}}\le \frac{1}{\lambda _1}\int \limits _{\Omega }|(-\Delta )^{s/2} T_k^{\frac{\gamma +1}{2}}(u_n)|^2 \le Ck^{\gamma }, \end{aligned}
where $$\lambda _1$$ is the first eigenvalue of the fractional Laplacian operator. Now if we restrict the integral on the left hand side on $$I_k=\left\{ u_n>k+1 \right\} _{x\in \Omega }$$, on which $$T_k(u_n)=k$$, we then obtain
\begin{aligned} k^{\gamma +1}m(\{u_n>k+1\})^{\frac{2}{2^*}}\le Ck^{\gamma }, \end{aligned}
so that
\begin{aligned} m(\{u_n>k+1\})\le \frac{C}{k^\frac{N}{N-2s}}, ~\forall \, k\ge 1. \end{aligned}
So $$(u_n)$$ is bounded in the Marcinkiewicz space $$M^{\frac{N}{N-2s}}(\Omega )$$, i.e. the sequence $$(G_1(u_n))$$ is also bounded in $$M^{\frac{N}{N-2s}}(\Omega )$$.
Now Eq. (2.25) becomes
\begin{aligned}&m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,u_n>1\right\} \right) \le m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,1<u_n\le k+1\right\} \right) \\&\qquad + m(\{u_n> k+1\}) \le \frac{Ck}{t^2} + \frac{C}{k^\frac{N}{N-2s}}, ~\forall \,k>1. \end{aligned}
On choosing $$k=t^{\frac{N-2s}{N-s}}$$, we get
\begin{aligned} m\left( \left\{ |(-\Delta )^{s/2} u_n|> t,u_n>1\right\} \right) \le \frac{C}{t^\frac{N}{N-s}}, ~\forall \,t\ge 1, \end{aligned}
We just proved that $$((-\Delta )^{s/2} u_n)=((-\Delta )^{s/2} G_1(u_n))$$ is bounded in $$M^{\frac{N}{N-s}}(\Omega )$$. This implies by the property (1.3) that $$(G_1(u_n))$$ is bounded in $$W_0^{s,q}$$ for every $$q<\frac{N}{N-s}$$.

Step 2 We claim that $$T_1(u_n)$$ is bounded in $$W_{loc}^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$.

We have to examine the behaviour of $$u_n$$ for small values of $$u_n$$ for each n. We want to show that for every $$K\subset \subset \Omega$$,
\begin{aligned} \int \limits _K |(-\Delta )^{s/2} T_1(u_n)|^2 \le C. \end{aligned}
(2.27)
We have already proved that $$u_n\ge C_K>0$$ on K in Lemma 2.4. We again use the Proposition 3.3 due to Canino et al. (2017) to obtain
\begin{aligned} \int \limits _{K}|(-\Delta )^{s/2} T_1(u_n)|^2\le \int \limits _\Omega h_n\left( u_n+\frac{1}{n}\right) f_nT_1(u_n) +\int \limits _\Omega T_1(u_n)\mu _n \le C.\quad \quad \end{aligned}
(2.28)
Thus we get the inequality (2.27).

The proof is complete since $$u_n=T_1(u_n)+G_1(u_n)$$ and hence the sequence ($$u_n$$) is bounded in $$W_{loc}^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$. $$\square$$

As a consequence we have the following existence result.

### Theorem 2.9

Let $$\gamma >1$$. Then there exists a weak solution u of Eq. (2.1) in $$W_{loc}^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$.

The proof of this theorem is a straightforward application of the Theorem 2.6 and using the results in Lemmas 2.7 and 2.8.

## 3 Further Analysis of the Case $$\gamma <1$$

In this section we will consider $$\Omega$$ to be a bounded open subset of $$\mathbb {R}^N (N\ge 2)$$, with boundary $$\partial \Omega$$ of class $$C^{2,\beta }$$ for some $$0<\beta <1$$. We consider the following semilinear elliptic problem
\begin{aligned} \begin{aligned} (-\Delta )^s u&= h(u)f+\mu ~\text {in}~\Omega ,\\ u&= 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega , \end{aligned} \end{aligned}
(3.1)
where $$0<\gamma <1$$, $$f\in C^{\beta } (\bar{\Omega })$$ such that $$f>0$$ in $$\bar{\Omega }$$ and $$\mu$$ is a nonnegative bounded Radon measure on $$\Omega$$ and guarantee the existence of a ‘very weak’ solution which is defined as follows.

### Definition 3.1

A very weak solution to the problem (3.1) is a function $$u\in L^1(\Omega )$$ such that $$u>0$$ a.e. in $$\Omega$$, $$fh(u) \in L^1(\Omega )$$, and
\begin{aligned} \int \limits _{\Omega }u(-\Delta )^s \varphi= & {} \int \limits _{\Omega }h(u)f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^1({\bar{\Omega }}) \nonumber \\&\text {such that}~ (-\Delta )^s\phi \in L^{\infty }(\Omega ). \end{aligned}
(3.2)

We will show existence of a nonnegative very weak solution to the problem (3.1).

### Definition 3.2

A function $$\underline{u}$$ is said to be a subsolution for Eq. (3.1) if $$\underline{u}\in L^1(\Omega ), \,\underline{u}>0$$ in $$\Omega$$, $$fh(\underline{u})\in L^1(\Omega )$$ and
\begin{aligned} \int \limits _{\Omega }\underline{u}(-\Delta )^s \varphi \le \int \limits _{\Omega }h(\underline{u})f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^1({\bar{\Omega }}) \end{aligned}
(3.3)
such that $$(-\Delta )^s\phi \in L^{\infty }(\Omega ),~\varphi \ge 0.$$
Equivalently, $$\bar{u}$$ is said to be a supersolution for the problem (3.1) if $$\bar{u}\in L^1(\Omega )$$, $$\bar{u}>0$$ in $$\Omega$$, $$fh(\bar{u})\in L^1(\Omega )$$ and
\begin{aligned} \int \limits _{\Omega }\bar{u}(-\Delta )^s \varphi \ge \int \limits _{\Omega }h(\bar{u})f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^1({\bar{\Omega }}) \end{aligned}
(3.4)
such that $$(-\Delta )^s\phi \in L^{\infty }(\Omega ),~\varphi \ge 0.$$

### Theorem 3.3

Let $$\underline{u}$$ is a subsolution and $$\bar{u}$$ is a supersolution to the problem (3.1) with $$\underline{u}\le \bar{u}$$ in $$\Omega$$, then there exists a solution u to Eq. (3.1) according to the Definition 3.1 such that $$\underline{u}\le u\le \bar{u}$$.

### Proof

The proof is a re-adaptation of an argument in Montenegro and Ponce (2008). First of all we will try to modify the singular nonlinearity by defining $$\bar{g}:\Omega \times \mathbb {R}\rightarrow \mathbb {R}$$ as
\begin{aligned} \bar{g}(x,t)= \left\{ \begin{array}{ll} f(x)h(\underline{u}(x)), &{} \quad \text {if}\,\,\, t<\underline{u}(x), \\ f(x)h(t), &{} \quad \text {if}\,\,\, \underline{u}(x)\le t\le \bar{u}(x),\\ f(x)h(\bar{u}(x)), &{} \quad \text {if}\,\,\, t>\bar{u}(x). \end{array} \right. \end{aligned}
Moreover, $$\underline{u}>0$$ so that, $$\bar{g}$$ is well defined a.e. in $$\Omega$$ and for every fixed $$v\in L^1(\Omega )$$ we have that $$\bar{g}(x,v(x))\in L^1(\Omega )$$. We divide the proof of the theorem into two steps.
$${\textit{Step 1}}$$ We claim that if u satisfies
\begin{aligned} (-\Delta )^s u= & {} \bar{g}(x,u)+\mu ~\text {in}~\Omega ,\nonumber \\ u= & {} 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega , \end{aligned}
(3.5)
in the very weak sense as defined in Definition 3.1, then $$\underline{u}\le u\le \bar{u}$$. Thus $$\bar{g}(.,u)=fh(u)\in L^1(\Omega )$$, and u is a very weak solution to Eq. (3.1).
The very weak formulation of Eq. (3.5) is given by
\begin{aligned} \int \limits _{\Omega }u(-\Delta )^s \varphi = \int \limits _{\Omega }\bar{g}(x,u)\varphi +\int \limits _\Omega \varphi d\mu ,\,\,\,\forall \varphi \in C_0^1(\bar{\Omega }) \end{aligned}
(3.6)
such that $$(-\Delta )^s\phi \in L^{\infty }(\Omega ).$$

We only show that $$u\le \bar{u}$$ in $$\Omega$$. The proof of the other side of the inequality, $$\underline{u}\le u$$, follows similarly.

We will show that u is a solution to Eq. (3.5) and $$\bar{u}$$ is a supersolution to Eq. (3.1). Subtracting Eq. (3.6) from Eq. (3.4) we have, for every $$\varphi \in C_0^1(\bar{\Omega })$$ such that $$\varphi \ge 0$$ and $$(-\Delta )^s\phi \in L^{\infty }(\Omega )$$,
\begin{aligned} \int \limits _{\Omega }(u-\bar{u})(-\Delta )^s \varphi \le \int \limits _{\Omega }\left( \bar{g}(x,u)-fh(\bar{u})\right) \varphi =\int \limits _{\Omega }\chi _{\{u\le \bar{u}\}}\left( \bar{g}(x,u)-fh(\bar{u})\right) \varphi . \end{aligned}
Now applying Kato type inequality (4.2) from the Appendix we get,
\begin{aligned} \int \limits _{\Omega }(u-\bar{u})^+ \le \int \limits _{\Omega }\chi _{\{u\le \bar{u}\}} \left( \bar{g}(x,u)-fh(\bar{u})\right) (sign_+(u-\bar{u}))\varphi =0, \end{aligned}
which further implies that
\begin{aligned} \int \limits _{\Omega }(u-\bar{u})^+\le 0. \end{aligned}
Thus we get $$u\le \bar{u}$$ a.e. in $$\Omega$$, and the proof of the claim is complete.
$${\textit{Step 2}}$$ We now show that a solution to problem (3.5) does exist. Let us define a map as follows.
\begin{aligned} G:L^1(\Omega )&\rightarrow L^1(\Omega ) \\ v&\mapsto u. \end{aligned}
This map assigns to every $$v\in L^1(\Omega )$$ the very weak solution u to the following linear problem
\begin{aligned} (-\Delta )^s u= & {} \bar{g}(x,v)+\mu ~\text {in}~\Omega ,\nonumber \\ u= & {} 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned}
(3.7)
The map G is well-defined since a unique solution exists to the problem in Eq. (3.7) due to Petitta (2016). We need to show that this map is continuous in $$L^1(\Omega )$$. Let us choose a sequence $$(v_n)$$ converging to some function v in $$L^1(\Omega )$$, then as h is a non-increasing continuous function we get
\begin{aligned} |\bar{g}(x,v_n(x))|\le fh(\underline{u}). \end{aligned}
Hence, using the dominated convergence theorem, we conclude that
\begin{aligned} \Vert \bar{g}(x, v_n)-\bar{g}(x, v)\Vert _{L^1(\Omega )}\rightarrow 0. \end{aligned}
Thus, by Petitta (2016) the linear problem (3.5) has a unique solution and we can say that $$u_n= G(v_n)$$, $$u = G(v)$$ and $$\Vert u_n-u\Vert _{L^1(\Omega )}\rightarrow 0$$. Therefore, we proved that G is continuous.
What is left to be proved is that the set $$G(L^1(\Omega ))$$ is bounded and relatively compact in $$L^1(\Omega )$$. For every $$v\in L^1(\Omega )$$ we have
\begin{aligned} \Vert \bar{g}(x,v)+\mu \Vert _{\mathcal {M}(\Omega )} \le \,\, \Vert \bar{g}(x,v)\parallel _{\mathcal {M}(\Omega )}+\Vert \mu \Vert _{\mathcal {M}(\Omega )}\le \,\, \Vert f h(\underline{u})\Vert _{L^1(\Omega )}+\Vert \mu \Vert _{\mathcal {M}(\Omega )}. \end{aligned}
Again, by the linear theory in Petitta (2016) , we see that G(v) is bounded in $$W_0^{s,q}(\Omega )$$ for every $$q<\frac{N}{N-s}$$ and therefore, by Rellich-Kondrachov theorem we get $$G(L^1(\Omega ))$$ is bounded and hence relatively compact in $$L^1(\Omega )$$.

We can now apply the Schauder fixed point theorem to see that G has a fixed point $$u\in L^1(\Omega )$$. According to the result from Step 1, we conclude that u is a very weak solution to Eq. (3.1) such that $$\underline{u}\le u\le \bar{u}$$. $$\square$$

### Theorem 3.4

There exists a solution to the problem (3.1) in the sense of Definition 3.1.

### Proof

We want to find both a subsolution and a supersolution to the problem (3.1) in the sense of Definition 3.2. Then we will use the result of Theorem 3.3 to prove the existence of a solution to the problem (3.1) in the sense of Definition 3.1.

We first find a subsolution. Let us consider the problem
\begin{aligned} \begin{aligned} (-\Delta )^s v&= h(v)f ~\text {in}~\Omega ,\\ v&= 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned} \end{aligned}
(3.8)
The existence of a very weak solution in $$L^1(\Omega )$$ to the problem in Eq. (3.8) can be proved as in the argument in Theorem 3.3 using the Schauder fixed point theorem. Consider the eigenfunction $$\phi >0$$ of $$(-\Delta )^s$$ corresponding to an eigenvalue $$\lambda$$ with $$\phi |_{\partial \Omega }=0$$. Indeed, if $$\psi =\phi ^{-}$$ is chosen as a test function from $$W_0^{s,2}(\Omega )$$ to the weak formulation
\begin{aligned} \int \limits _{\Omega }(-\Delta )^s \phi \psi =\lambda \int \limits _{\Omega }\phi \psi , \end{aligned}
we obtain $$\phi ^{-}=0$$ a.e. in $$\Omega$$. Observe that
\begin{aligned} (-\Delta )^s\epsilon \phi -h(\epsilon \phi )f&<0 \\&=(-\Delta )^s v-h(v)f \end{aligned}
due to the facts (1) that $$\phi >0$$ and a choice of sufficiently small $$\epsilon >0$$, (2) the non-increasing nature of h and (3) v being a solution to Eq. (3.8). Hence we have $$v > 0$$ in $$\Omega$$.
Similarly, due to the assumption that $$\partial \Omega$$ is of class $$C^{2,\beta }$$ and the conditions over h given in Eqs. (2.3) and (2.4), we can see that $$h(\varphi _1)\in L^1(\Omega )$$ if and only if $$\gamma <1$$ by using arguments as in Lazer and McKenna (1991). We see that the first eigenfunction of the Laplacian, denoted by $$\phi _1$$, is a subsolution to the problem (3.8). It is proved in Lazer and McKenna (1991) that there exists a solution $$v\in L^1(\Omega )$$ to the problem (3.8) for $$h(s)=\frac{1}{s^{\gamma }}$$. On using the conditions over h in Eqs. (2.3) and (2.4) and the result in Lazer and McKenna (1991), we guarantee the existence of a solution to Eq. (3.8). Hence we have $$v > 0$$ in $$\Omega$$, $$h(v)f \in L^1(\Omega )$$, and
\begin{aligned} \int \limits _\Omega v(-\Delta )^s \varphi =\int \limits _{\Omega }h(v)f\varphi , ~\forall \,\varphi \in C_0^2(\bar{\Omega })~\text {such that}~ (-\Delta )^s\phi \in L^{\infty }(\Omega ). \end{aligned}
Since $$\mu$$ is a nonnegative, Radon measure we get the following inequality,
\begin{aligned} \int \limits _\Omega v(-\Delta )^s \varphi \le \int \limits _{\Omega }h(v)f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^2(\bar{\Omega }), \varphi \ge 0, \end{aligned}
such that $$(-\Delta )^s\phi \in L^{\infty }(\Omega ),$$ then we can say that v is a subsolution to the problem (3.1).
We now look for a supersolution of the problem (3.1). Let w be the solution of
\begin{aligned} (-\Delta )^s w= & {} \mu ~\text {in}~\Omega ,\nonumber \\ w= & {} 0 ~\text {on}~ \mathbb {R}^N{\setminus }\Omega . \end{aligned}
(3.9)
There exists a positive solution to the problem (3.9) in classical sense Petitta (2016). Let us denote $$Z=w + v$$, where v is a very weak solution to Eq. (3.8). Then we get
\begin{aligned} \int \limits _\Omega Z(-\Delta )^s \varphi= & {} \int \limits _\Omega w(-\Delta )^s \varphi +\int \limits _\Omega v(-\Delta )^s \varphi \\= & {} \int \limits _{\Omega }h(v)f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^1(\bar{\Omega }). \end{aligned}
We know that w is nonnegative, then we have $$0<h(Z)\le h(v)$$. Thus, we can say
\begin{aligned} \int \limits _{\Omega }h(Z)f\varphi +\int \limits _{\Omega }\varphi d\mu \le \int \limits _{\Omega }h(v)f\varphi +\int \limits _{\Omega }\varphi d\mu ~\forall \,\varphi \in C_0^1(\bar{\Omega }), ~\varphi \ge 0, \end{aligned}
such that $$(-\Delta )^s\phi \in L^{\infty }(\Omega )$$ i.e. Z is a positive function in $$L^1(\Omega )$$ such that $$h(Z)\le h(v)\in L^1(\Omega )$$ and
\begin{aligned}&\int \limits _\Omega Z(-\Delta )^s \varphi \ge \int \limits _{\Omega }h(Z)f\varphi +\int \limits _{\Omega }\varphi d\mu , ~\forall \,\varphi \in C_0^1(\bar{\Omega }), \varphi \ge 0 \\&\quad \text {such that}~ (-\Delta )^s\phi \in L^{\infty }(\Omega ). \end{aligned}
Therefore, Z is a supersolution to Eq. (3.1). We can now apply Theorem 3.3 to get the conclusion that there exists a solution u to problem (3.1) in the sense of Definition 3.1. $$\square$$

### 3.1 When $$f\in L^1(\Omega )\cap L^{\infty }(\Omega _\delta )$$.

We proved Theorem 3.4 by assuming a strong regularity on f i.e. f belongs to $$C^{\beta }(\bar{\Omega })$$ for some $$0<\beta < 1$$. In this section we do some relaxation on our assumption on f in order to prove the existence of solution.

For a fix $$\delta >0$$, let us define $$\Omega _\delta =\{x\in \Omega :\text {dist}(x,\partial \Omega )<\delta \}$$, and let f be an a.e. positive function in $$L^1(\Omega )\cap L^\infty (\Omega _\delta )$$.

### Theorem 3.5

Let $$f\in L^1(\Omega )\cap L^{\infty }(\Omega _\delta )$$ such that $$f>0$$ a.e. in $$\Omega$$ for some fixed $$\delta >0$$. Then there exists a solution for problem (3.1) in the sense of Definition 3.1.

### Proof

We consider the following approximating problems
\begin{aligned} \begin{aligned} (-\Delta )^s v_n&= h\left( v_n+\frac{1}{n}\right) f_n ~\text {in}~\Omega ,\\ v_n&= 0 ~\text {on}~ \partial \Omega , \end{aligned} \end{aligned}
(3.10)
where, $$f_n = T_n(f)$$. Boccardo and Orsina proved in (2010) that the nondecreasing sequence $$(v_n)$$ converges to a weak solution of the problem (3.8). Using the linear theory in Petitta (2016), the sequence $$(v_n)$$ belongs to $$L^{\infty }(\Omega )$$. So we can observe that the $$h(v_1 + 1)f_1$$ also belongs to $$L^{\infty }(\Omega )$$. We can now apply Lemma 3.2 in Brezis and Cabré (1998) so as to obtain that for a.e. x in $$\Omega$$
\begin{aligned} \frac{v_1(x)}{d(x)}\ge C\int _\Omega d(y)f_1(y) h\left( \parallel v_1\parallel _{L^{\infty }(\Omega )}+1\right) dy \ge C>0. \end{aligned}
where $$d(x) = d(x, \partial \Omega )$$ is the distance function of x from $$\partial \Omega$$. Thus, we have
\begin{aligned} v(x)\ge v_1(x)\ge Cd(x)~\text {a.e. on} \,\,\,\Omega . \end{aligned}
Therefore, as $$f\in L^{\infty }(\Omega _\delta )$$, then $$h(v)f\le Ch(d(x))f$$ is integrable in $$\Omega$$ for every $$\gamma <1$$. Hence the subsolution is bounded from below and this allows us to proceed as in the proof of Theorem 3.4. $$\square$$

## 4 Appendix

We prove the Kato type inequality for the problem
\begin{aligned} (-\Delta )^s u&= h(u)f+\mu ~\text {in}~\Omega ,\nonumber \\ u&= 0 ~\text {on}~ \partial \Omega , \nonumber \\ u&> 0 ~\text {in}~ \Omega , \end{aligned}
(4.1)
where $$f>0$$ and $$u\in L^1(\Omega )$$ is a very weak solution with $$u>0$$ a.e. in $$\Omega$$ and $$fh(u)\in L^1(\Omega )$$.
Let $$u_1$$ and $$u_2$$ are two very weak solutions to the problem (4.1) with measure sourses $$\mu _1$$ and $$\mu _2$$, respectively. Hence, $$u_1,u_2\in L^1(\Omega )$$ and $$h(u_1)f,~h(u_2)f\in L^1(\Omega )$$. Then for every $$\phi \in C_0^1(\bar{\Omega })$$, the very weak formulations corresponding to the problem (4.1) are
\begin{aligned} \int \limits _{\Omega }u_1(-\Delta )^s\phi = \int \limits _{\Omega }h(u_1)f\phi +\int \limits _{\Omega }\phi d\mu _1, \end{aligned}
and
\begin{aligned} \int \limits _{\Omega }u_2(-\Delta )^s\phi = \int \limits _{\Omega }h(u_2)f\phi +\int \limits _{\Omega }\phi d\mu _2. \end{aligned}
Taking the difference between two formulations we get
\begin{aligned} \int \limits _{\Omega }(u_1-u_2)(-\Delta )^s\phi = \int \limits _{\Omega }(h(u_1)-h(u_2))f\phi +\int _{\Omega }\phi (d\mu _1-d\mu _2), \end{aligned}
and this is a very weak formulation of the problem
\begin{aligned} (-\Delta )^s(u_1-u_2)&= (h(u_1)-h(u_2))f + (\mu _1-\mu _2) ~\text {in}~\Omega , \\ u_1-u_2&= 0 ~\text {on}~ \partial \Omega , \end{aligned}
Now referring to the Proposition 1.5.4 (Kato’s inequality) of Marcus and Véron (2013), we can observe that
\begin{aligned} \int \limits _{\Omega }(u_1-u_2)^+(-\Delta )^s\phi\le & {} \int \limits _{\Omega }(h(u_1)-h(u_2)) (sign_+ (u_1-u_2))f\phi \\&+\int \limits _{\Omega }\phi (d\mu _1-d\mu _2), \end{aligned}
where, $${\text {sign}}_+ (u_1-u_2)={\chi _{\{x\in \Omega :u_1(x)\ge u_2(x)\}}}$$. Let us consider a standard choice $$\phi _0$$ such that $$(-\Delta )^s\phi _0=1$$ in $$\Omega$$ and $$\phi _0=0$$ on $$\partial \Omega$$. Now the above inequality becomes
\begin{aligned} \int \limits _{\Omega }(u_1-u_2)^+\le & {} \int \limits _{\Omega }(h(u_1)-h(u_2))({\text {sign}}_+(u_1-u_2))f\phi _0 \nonumber \\&+\int \limits _{\Omega }\phi _0 d(\mu _1-\mu _2) \end{aligned}
(4.2)
Equation (4.2) is our required Kato type inequality.
Now as $$f>0$$ and $$\phi _0\ge 0$$ we get
\begin{aligned} \int \limits _{\Omega }(u_1-u_2)^+ + \int \limits _{\Omega }f(h(u_2)-h(u_1))\phi _0 \le \int \limits _{\Omega }\phi _0 d(\mu _1-\mu _2). \end{aligned}
So if $$\mu _1$$ becomes equal to $$\mu _2$$, then
\begin{aligned} \int \limits _{\Omega }(h(u_2)-h(u_1))f\phi _0 \le 0. \end{aligned}
As $$h(u_2)-h(u_1)\ge 0$$ for $$u_1 \ge u_2$$. So we reach at the conclusion that $$h(u_1)=h(u_2)$$.

## Notes

### Acknowledgements

Two of the authors, Sekhar Ghosh and Ratan Kr. Giri, thanks the financial assistantship received from the Council of Scientic and Industrial Research (CSIR), India and the Ministry of Human Resource Development (MHRD), Govt. of India respectively.

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