Let $I=[0,1]$. We deal with the existence of solutions $u\in L^p(I)$ of the implicit functional-integral equation $$ \psi(u(t))=f\Big(t,\int_I k(t,s)\,u(\varphi(s))\,ds\Big)\quad\hbox{for a.e.}\quad t\in I, $$ where $Y\subseteq{\bf R}$ is a closed interval, $p\in]1,+\infty[\,$, and $\psi:Y\to{\bf R}$, $f:I\times {\bf R}\to{\bf R}$, $k:I\times I\to [0,+\infty[$ and $\varphi:I\to I$ are given functions. We prove an existence result where the function $f$ can be discontinuous, with respect to the second variable, even at all points $x\in{\bf R}$. Our result improve in several aspects a very recent result in the field. In particular, we impose a linear growth condition for the function $\psi^{-1}(f(t,\cdot))$, meaningfully weaker than boundedness condition which was previously imposed. As regards the function $\psi$, we only require that it is continuous and non-constant on intervals.
Implicit functional-integral equations associated with unbounded discontinuous functions
P. Cubiotti;
2026-01-01
Abstract
Let $I=[0,1]$. We deal with the existence of solutions $u\in L^p(I)$ of the implicit functional-integral equation $$ \psi(u(t))=f\Big(t,\int_I k(t,s)\,u(\varphi(s))\,ds\Big)\quad\hbox{for a.e.}\quad t\in I, $$ where $Y\subseteq{\bf R}$ is a closed interval, $p\in]1,+\infty[\,$, and $\psi:Y\to{\bf R}$, $f:I\times {\bf R}\to{\bf R}$, $k:I\times I\to [0,+\infty[$ and $\varphi:I\to I$ are given functions. We prove an existence result where the function $f$ can be discontinuous, with respect to the second variable, even at all points $x\in{\bf R}$. Our result improve in several aspects a very recent result in the field. In particular, we impose a linear growth condition for the function $\psi^{-1}(f(t,\cdot))$, meaningfully weaker than boundedness condition which was previously imposed. As regards the function $\psi$, we only require that it is continuous and non-constant on intervals.Pubblicazioni consigliate
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