A Method for Studying Integral Equations by Using a Covering Set of the Nemytskii Operator in Spaces of Measurable Functions

Abstract

Sufficient conditions are found for the existence of solutions $x$ of the scalar integral equations

$$ f\left (t,\int _0^1\mathcal {K}(t,s)x(s)\thinspace ds, x(t)\right )=z(t)\quad \text {and}\quad f\left (t,\int _0^t\mathcal {K}(t,s)x(s)\thinspace ds, x(t)\right )=z(t),\quad t\in [0,1], $$

in the space of measurable functions, where the functions $\mathcal {K}$, $z$, and $f$ are given; the kernel $\mathcal {K}\colon [0,1]\times [0,1]\to \mathbb {R}$ is measurable, and the integral of its modulus over the second argument is essentially bounded, the free term $z\colon [0,1]\to \overline {\mathbb {R}}$ is measurable, and the function $f\colon [0,1]\times \overline {\mathbb {R}}\times \mathbb {R}\to \overline {\mathbb {R}}$ is measurable in the first argument and jointly continuous in the second and third arguments ($\overline {\mathbb {R}}=\mathbb {R}\cup \{\infty \}$ is the real line supplemented with the point at infinity and equipped with the “usual” metric: $\rho _{\overline {\mathbb {R}}}(u,v)=|u-v|$, $\rho _{\overline {\mathbb {R}}}(\infty ,v) =\rho _{\overline {\mathbb {R}}}(u,\infty )=+\infty$, and $\rho _{\overline {\mathbb {R}}}(\infty ,\infty )=0$, where $u,v\in \mathbb {R}$). The conditions are obtained on the basis of a generalization of the concepts and results of the theory of covering mappings to spaces more general than metric ones.