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We are pleased to announce the International Seminar on Time Scales Analysis.

Non-trivial solutions of a non-local elliptic equation with a critical Sobolev exponent and a singular term

Jan 24, 2026, 2:40 PM
30m

Speaker

Dr Abdelaziz Bennour (University of Oran 1)

Description

The paper deals with the following fractional Hardy-Sobolev equation with nonhomogeneous term
\begin{equation}
%\label{eq1}
\begin{cases}
{(-\Delta)}^{s}u-\mu \frac{u}{|x|^{2s}}=|u|^{2_{s}^{}-2}u+\lambda \frac{u}{|x|^{2s-\alpha}}+f(x),&x\in \Omega,\
u=0&x\in \partial\Omega,
\end{cases}
\end{equation}
being $02s)$ containing the origin $0$ in its interior, $0\leq \mu <\overline{\mu_{s}}:=2^{2s}\frac{\Gamma^{2}(\frac{N+2s}{4})}{\Gamma^{2}(\frac{N-2s}{4})}$, $\lambda$ is a positive parameter, $0<\alpha<2s$, $2_{s}^{*}=\frac{2N}{N-2s}$ is the fractional critical Hardy-Sobolev exponent. The fractional Laplacian $(-\Delta)^{s}$ is defined by
\begin{equation
}
-2(-\Delta)^{s}u(x)=C_{N,s}\underset{\mathbb{R}^{N}}{\int}\dfrac{u(x+y)+u(x-y)-2u(x)}{|x-y|^{N+2s}}dy
\end{equation*}
where
$$C_{N,s}=\dfrac{4^{s}\Gamma(N\setminus2+s)}{\pi^{N\setminus2}|\Gamma(-s)|}.$$ $\Gamma$ is the Gamma function, $f$ is a given bounded measurable function. by virtue of Ekeland’s Variational Principle and the Mountain Pass Lemma and for which we consider the following hypothesis \begin{equation*} \inf\left\lbrace \gamma_{N,s}(T(u))^{\frac{N+2s}{4s}}-\underset{\Omega}{\int}f u dx:\;u\in X,\underset{\Omega}{\int}|u|^{2_{s}^{*}} dx=1\right\rbrace>0.\;\;(\mathcal{F}) %\label{ast} \end{equation*} Where $X$ is a Hilbert space defined as $$X=\lbrace u\in H^{2s}(\mathbb{R}^{N}):u=0\;\text{in}\;\mathbb{R}^{N}\setminus\Omega\rbrace,$$ where $H^{2s}(\mathbb{R}^{N})$ the usual fractional Sobolev space, $$\gamma_{N,s}=\frac{4s}{N-2s}(\frac{N-2s}{N+2s})^{\frac{N+2s}{4s}}$$ and $$T(u)=C_{N,s}\underset{\mathbb{R}^{N}}{\int}\underset{\mathbb{R}^{N}}{\int}\dfrac{|u(x)-u(y)|^{2}}{|x-y|^{N+2s}}dxdy-\mu \underset{\Omega}{\int}\frac{u^{2}}{|x|^{2s}}dx-\lambda\underset{\Omega}{\int}\frac{u^{2}}{|x|^{2s-\alpha}}dx.$$\\ Moreover, the following eigenvalue problem with Hardy potentials and singular coefficient \begin{equation*} \begin{cases} {(-\Delta)}^{s}u-\mu \frac{u}{|x|^{2s}}=\lambda \frac{u}{|x|^{2s-\alpha}}& x\in\Omega, \\ u=0 & x\in\partial \Omega, \end{cases} \end{equation*} where $0 < \alpha <2s$, $\lambda \in \mathbb{R}$, has the first eigenvalue $\lambda_{1}$ given by: \begin{equation*} \lambda_{1}= \underset{u\in X\setminus\lbrace0\rbrace}{\inf}\dfrac{C_{N,s}\underset{\mathbb{R}^{N}}{\int}\underset{\mathbb{R}^{N}}{\int}\dfrac{|u(x)-u(y)|^{2}}{|x-y|^{N+2s}}dxdy-\mu \underset{\Omega}{\int}\frac{u^{2}}{|x|^{2s}}dx}{\underset{\Omega}{\int}\frac{u^{2}}{|x|^{2s-\alpha}}dx}. \end{equation*} We get the following results: \\ Let $0<\mu<\overline{\mu_{s}}$, $0<\lambda<\lambda_{1}$ and $f$ is a bounded measurable function satisfying the condition $(\mathcal{F})$, then the problem has at least two nontrivial solutions, if $0<\alpha<2\beta^{+}(\mu)+2s-N.$ \\ Where $\beta^{+}(\mu)$ comes through the processes and techniques of calculations.
%\label{th}

Authors

Dr Abdelaziz Bennour (University of Oran 1) Prof. sofiane messirdi (University of Oran 1)

Presentation materials

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