Nonlocal semilinear elliptic problems with singular nonlinearity

Abstract

We consider a Lazer-Mckenna-type problem involving the fractional Laplacian and singular nonlinearity. We investigate existence, regularity and uniqueness of solutions in light of the interplay between the nonlinearities and the summability of the datum.

Appendix

We start by proving the following lemma which we have used in the proof of Lemma 4.4.

Lemma 5.3

Let $F(x)=x^r$, $0<r<1$, for every $x>0$. Then for every function $v :\mathbb {R}^N\rightarrow ]0,+\infty [$ that satisfies

$$\begin{aligned} \int _{\mathbb {R}^{N}}\int _{\mathbb {R}^{N}}\frac{|v(x)-v(y)|^{2}}{|x-y|^{N+2s}}dydx<\infty , \end{aligned}$$

we have

$$\begin{aligned} \displaystyle \begin{array}{lll} (-\Delta )^{s}(F\circ v)(x)\le \\[2mm] \displaystyle F^{\prime }(v(x))(-\Delta )^{s}v(x)-\frac{F^{\prime \prime }(v(x))}{r}a(N,s)\int _{\mathbb {R}^{N}}\frac{\big (v(x)-v(y)\big )^{2}}{|x-y|^{N+2s}}dy. \end{array} \end{aligned}$$

(5.1)

Proof

Following [20, Lemma 2.3.], we can use Taylor’s formula obtaining for every $(x,y)\in \mathbb {R}^{N}\times \mathbb {R}^{N}$

$$\begin{aligned} F(v(y))-F(v(x))=F^{\prime }(v(x))(v(y)-v(x))+R(F), \end{aligned}$$

(5.2)

where

$$\begin{aligned} \begin{array}{lll}\displaystyle R(F)&{}=\displaystyle \int _{v(x)}^{v(y)}(v(y)-t)F^{\prime \prime }(t)dt\\ &{}=\displaystyle (v(y)-v(x))^{2}\int _{0}^{1}(1-s)F^{\prime \prime }(v(x)+s(v(y)-v(x)))ds. \end{array} \end{aligned}$$

On other hand, since the function $F^{\prime \prime }$ is increasing we have

$$\begin{aligned} \begin{array}{lll}\displaystyle (1-s)v(x)&{}\le v(x)+s(v(y)-v(x)) \\[2mm] &{}\Rightarrow F^{\prime \prime }((1-s)v(x))\le F^{\prime \prime }(v(x)+s(v(y)-v(x))). \end{array} \end{aligned}$$

Hence, it follows

$$\begin{aligned} \begin{array}{lll} -R(F)&{}\le -(v(y)-v(x))^{2}\int _{0}^{1}(1-s)F^{\prime \prime }((1-s)v(x))ds\\[4mm] &{}=-(v(y)-v(x))^{2}F^{\prime \prime }(v(x))\int _{0}^{1}(1-s)^{r-1}ds. \end{array} \end{aligned}$$

Then, from (5.2) we obtain

$$\begin{aligned} F(v(x))-F(v(y))\le F^{\prime }(v(x))(v(x)-v(y))-\frac{F^{\prime \prime }(v(x))}{r}(v(y)-v(x))^{2}. \end{aligned}$$

Dividing both sides of this inequality by $|x-y|^{N+2s}$ and then integrating with respect to the variable y we arrive at

$$\begin{aligned} \begin{array}{lll} a(N,s)P.V.\int _{\mathbb {R}^{N}} \frac{F(v(x))-F(v(y))}{|x-y|^{N+2s}}dy &{}\le F^{\prime }(v(x))a(N,s)P.V.\int _{\mathbb {R}^{N}} \frac{(v(x)-v(y))}{|x-y|^{N+2s}}dy\\[4mm] &{}-\frac{F^{\prime \prime }(v(x))}{r}a(N,s)P.V.\int _{\mathbb {R}^{N}} \frac{(v(y)-v(x))^{2}}{|x-y|^{N+2s}}dy, \end{array} \end{aligned}$$

which proves (5.1). $\square$

In the following result we extend the space of admissible test functions in (2.4).

Lemma 5.4

Let $u\in X_{0}^{s}(\Omega )$ be a solution of the problem (1.1) taken in the sense of Definition 2.1 with $f\in L^1(\Omega )$. Then for every $\phi \in X_{0}^{s}(\Omega )$ we get $\frac{f\phi }{u^\gamma }\in L^1(\Omega )$ and

$$\begin{aligned} \frac{a(N,s)}{2}\int _{{Q}}\frac{(u(x)-u(y))(\phi (x)-\phi (y))}{|x-y|^{N+2s}}dydx=\int _{\Omega }\frac{f\phi }{u^{\gamma }}dx. \end{aligned}$$

(5.3)

Proof

Take an arbitrary $\phi \in X_{0}^{s}(\Omega )$. By [29, Theorem 6] there exists a sequence $\{\varphi _n\}_n\subset \mathcal {C}_{0}^{\infty }(\Omega )$ such that $\varphi _n\rightarrow \phi$ in norm in $H^{s}(\mathbb {R}^N)$. Writing (2.4) with $\varphi _n\in \mathcal {C}_{0}^{\infty }(\Omega )$ we obtain

$$\begin{aligned} \frac{a(N,s)}{2}\int _{Q}\frac{(u(x)-u(y))(\varphi _{n}(x)-\varphi _{n}(y))}{|x-y|^{N+2s}}dydx=\int _{\Omega }\frac{f\varphi _n}{u^{\gamma }}dx, \end{aligned}$$

(5.4)

in which we shall pass to the limit as n tends to $+\infty$. Starting with the left-hand side of (5.4), we consider the following two functions

$$\begin{aligned} F_n(x,y)=\frac{(\varphi _{n}(x)-\varphi _{n}(y))}{|x-y|^{\frac{N+2s}{2}}} \text{ and } F(x,y)=\frac{(\phi (x)-\phi (y))}{|x-y|^{\frac{N+2s}{2}}}. \end{aligned}$$

Notice that the convergence $\varphi _n\rightarrow \phi$ in norm in $H^{s}(\mathbb {R}^N)$ implies that the sequence $\{F_n(x,y)\}_n$ converges to F(x, y) in $L^{2}(\mathbb {R}^{2N})$ and, up to a subsequence if necessary, we can assume that $\{F_n(x,y)\}_n$ converges almost everywhere in $\mathbb {R}^{2N}$.

As $u\in X_{0}^{s}(\Omega )$ we have $\frac{(u(x)-u(y))}{|x-y|^{\frac{N+2s}{2}}}\in L^{2}(\mathbb {R}^{2N})$ implying

$$\begin{aligned} \begin{array}{lll}\displaystyle \lim _{n\rightarrow \infty }\int _{Q}\frac{(u(x)-u(y))(\varphi _{n}(x)-\varphi _{n}(y))}{|x-y|^{N+2s}}dydx\\[2mm] = \displaystyle \int _{Q}\frac{(u(x)-u(y))(\phi (x)-\phi (y))}{|x-y|^{N+2s}}dydx. \end{array} \end{aligned}$$

For the term in the right-hand side of (5.4), we first note that thanks to [38, Proposition 3.] the two functions $(\varphi _n-\varphi _k)^+$ and $(\varphi _n-\varphi _k)^-$ are both admissible test functions in (2.4). Taking them so we obtain

$$\begin{aligned} \begin{array}{lll}\displaystyle \int _{\Omega }\frac{f}{u^{\gamma }}(\varphi _n-\varphi _k)^+(x)dx\\ =\displaystyle \frac{a(N,s)}{2}\int _{{Q}}\frac{(u(x)-u(y))\big ((\varphi _n-\varphi _k)^+(x)-(\varphi _n-\varphi _k)^+(y)\big )}{|x-y|^{N+2s}}dydx \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{lll}\displaystyle \int _{\Omega }\frac{f}{u^{\gamma }}(\varphi _n-\varphi _k)^-(x)dx\\ =\displaystyle \frac{a(N,s)}{2}\int _{{Q}}\frac{(u(x)-u(y))\big ((\varphi _n-\varphi _k)^-(x)-(\varphi _n-\varphi _k)^-(y)\big )}{|x-y|^{N+2s}}dydx. \end{array} \end{aligned}$$

Then, summing up both the two equalities we have

$$\begin{aligned} \begin{array}{lll}\displaystyle \int _{\Omega }\frac{f}{u^{\gamma }}\Big |\varphi _n-\varphi _k\Big |dx\\ =\displaystyle \frac{a(N,s)}{2}\int _{{Q}}\frac{(u(x)-u(y))\Big (|\varphi _n(x)-\varphi _k(x)|-|\varphi _n(y)-\varphi _k(y)|\Big )}{|x-y|^{N+2s}}dydx\\[3mm] \le \displaystyle \frac{a(N,s)}{2}\int _{{Q}}\frac{|u(x)-u(y)|\Big |(\varphi _n(x)-\varphi _k(x))-(\varphi _n(y)-\varphi _k(y))\Big |}{|x-y|^{N+2s}}dydx \end{array} \end{aligned}$$

and then the Hölder inequality implies

$$\begin{aligned} \int _{\Omega }\Big |\frac{f\varphi _n}{u^{\gamma }}-\frac{f\varphi _k}{u^{\gamma }}\Big |dx\le \frac{a(N,s)}{2}\Vert u\Vert _{X_{0}^{s}(\Omega )}\Vert \varphi _n-\varphi _k\Vert _{X_{0}^{s}(\Omega )}. \end{aligned}$$

Thus, we deduce that $\Big \{\frac{f\varphi _n}{u^{\gamma }}\Big \}_n$ is a Cauchy sequence in $L^1(\Omega )$. Since $\varphi _n$ converges to $\varphi$ a.e. in $\Omega$, the sequence $\Big \{\frac{f\varphi _n}{u^{\gamma }}\Big \}_n$ converges to $\frac{f\phi }{u^{\gamma }}\in L^1(\Omega )$ in norm in $L^1(\Omega )$. So that the passage to the limit as n tends to infinity in (5.4) yields

$$\begin{aligned} \frac{a(N,s)}{2}\int _{{Q}}\frac{(u(x)-u(y))(\phi (x)-\phi (y))}{|x-y|^{N+2s}}dydx=\int _{\Omega }\frac{f\phi }{u^{\gamma }}dx, \end{aligned}$$

for every $\phi \in X_{0}^{s}(\Omega )$. $\square$