An Extreme Problem for a Volterra Type Integral Inclusion

Abstract

In the work, we have studied the dependencies of the solutions to integral in-clusions from perturbation and investigated an extremal problem for integral inclusions. We obtained necessary and sufficient minimum conditions for ex-tremal problems of Volterra type convex inclusions. We also studied a non-convex extremal problem for the Volterra type inclusion. We obtained a high order necessary condition in the extremal problem for the Volterra type inclu-sion.

Share and Cite:

Sadygov, M. (2019) An Extreme Problem for a Volterra Type Integral Inclusion. Open Access Library Journal, 6, 1-8. doi: 10.4236/oalib.1105605.

1. Dependence of the Solution to the Integral Inclusion from Perturbation

Let ${R}^{n}$ be the n-dimensional Euclidean space. The set of all nonempty compact (convex compact) subsets in ${R}^{n}$ we will designate as $comp{R}^{n}\left(conv{R}^{n}\right)$ ; $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ is the continuous matrix function, wherewith ${M}_{n}$ being the set of all square $n×n$ matrices of real elements ( ${b}_{ij}$ ); $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function; $F:\left[{t}_{0},T\right]×{R}^{n}\to comp{R}^{n}$ the setvalued mapping.

Assume that if a vector is multiplied by a matrix, then the vector is a row vector, if a matrix is multiplied by a vector, then the vector is a column vector.

Let us consider a problem for inclusion

$u\left(t\right)\in F\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)$ (1)

The function $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ satisfying (1) we will call the solution to problem (1) (see [1] ).

Let $a=\underset{t,s\in \left[{t}_{0},T\right]}{\mathrm{max}}‖k\left(t,s\right)‖=\underset{t,s\in \left[{t}_{0},T\right]}{\mathrm{max}}{\sum }_{i=1}^{n}{\sum }_{j=1}^{n}|{k}_{i,j}\left(t,s\right)|$, if $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ is the continuous matrix function.

Theorem 1. Let $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ be the continuous matrix function, $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function, $F:\left[{t}_{0},T\right]×{R}^{n}\to comp{R}^{n}$ the multivalued mapping, $t\to F\left(t,x\right)$ is measurable on t, and there exists a summable function $M\left(t\right)>0$ such that ${\rho }_{x}\left(F\left(t,x\right),F\left(t,{x}_{1}\right)\right)\le M\left(t\right)|x-{x}_{1}|$ for

$x,{x}_{1}\in {R}^{n}$. Moreover, let $\rho \left(\cdot \right)\in {L}_{1}\left[{t}_{0},T\right]$ and $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ be such that

$d\left(\stackrel{¯}{u}\left(t\right),F\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)\right)\right)\le \rho \left(t\right)$ for $t\in \left[{t}_{0},T\right]$. Then there exists such a solution $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ to problem (1) that

$|{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s-{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s|\le a{\int }_{{t}_{0}}^{t}\text{ }\text{ }{\text{e}}^{m\left(t\right)-m\left(s\right)}\rho \left(s\right)\text{d}s,$

$|u\left(t\right)-\stackrel{¯}{u}\left(t\right)|\le \rho \left(t\right)+aM\left(t\right){\int }_{{t}_{0}}^{t}\text{ }\text{ }{\text{e}}^{m\left(t\right)-m\left(s\right)}\rho \left(s\right)\text{d}s$

for $t\in \left[{t}_{0},T\right]$, where $m\left(t\right)=a{\int }_{{t}_{0}}^{t}\text{ }M\left(s\right)\text{d}s$.

2. On Subdifferential of the Integral Functional

Let $f:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ is the normal convex integrant (see [2] ).

Let consider a subdifferential of the integral functional

$J\left(u\left(\cdot \right)\right)={\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t$

in ${L}_{1}^{n}\left[{t}_{0},T\right]$.

Theorem 2. If $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ be the continuous matrix function, $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function, $f:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ is the

normal convex integrant and function $f\left(t,{\int }_{{t}_{0}}^{t}\text{ }\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)+x\right)$ is

summable for $x\in {R}^{n}$, $|x|\le \delta$, where $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$, then $\partial J\left(\stackrel{¯}{u}\left(\cdot \right)\right)$ is nonempty and ${\upsilon }^{*}\in {L}_{\infty }^{n}\left[{t}_{0},T\right]$ belongs to $\partial J\left(\stackrel{¯}{u}\left(\cdot \right)\right)$ if and only if, there exist

$\text{ }{u}^{*}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$, $\text{ }{u}^{*}\left(t\right)\in \partial f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)\right)$, such, that ${\upsilon }^{*}\left(s\right)={\int }_{s}^{T}\text{ }k{\left(t,s\right)}^{t}{u}^{*}\left(t\right)\text{d}t$, where $k{\left(\tau ,t\right)}^{t}$ is the transpose of the matrix $k\left(\tau ,t\right)$.

Theorem 3. If $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ be measurable bounded matrix function, $z:\left[{t}_{0},T\right]\to {R}^{n}$ be measurable bounded function, $f:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ is

the normal convex integrant and function $f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)+x\right)$ is

summable for $x\in {R}^{n}$, $|x|\le \delta$, where $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$, then $\partial J\left(\stackrel{¯}{u}\left(\cdot \right)\right)$ is nonempty and functional ${\upsilon }^{*}\in {L}_{\infty }^{n}\left[{t}_{0},T\right]$ belongs to $\partial J\left(\stackrel{¯}{u}\left(\cdot \right)\right)$ if and only if,

there exist $\text{ }{u}^{*}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$, $\text{ }{u}^{*}\left(t\right)\in \partial f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)\right)$, such, that ${\upsilon }^{*}\left(s\right)={\int }_{s}^{T}k{\left(t,s\right)}^{t}{u}^{*}\left(t\right)\text{d}t$.

3. On Subdifferential of the Terminal Functional

Let $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ continuous matrix function, $z:\left[{t}_{0},T\right]\to {R}^{n}$ continuous function, $\phi :{R}^{n}\to \left(-\infty ,+\infty \right]$ proper convex function in ${R}^{n}$. Consider a subdifferential of the terminal functional $F\left(u\left(\cdot \right)\right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)$ in ${L}_{1}^{n}\left[{t}_{0},T\right]$, where $z\left(\cdot \right)\in {C}^{n}\left[{t}_{0},T\right]$.

Theorem 4. If $\phi$ -proper convex function in ${R}^{n}$ and continuous in the

point ${\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(T\right)$, then

$\partial F\left(\stackrel{¯}{u}\left(\cdot \right)\right)=\left\{bk\left(T,s\right):b\in \partial \phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(T\right)\right)\right\}.$

4. Convex Extremal Problem for Integral Inclusions

Let $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ be the continuous matrix function, $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function. Hereafter we will assume that $f:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ is the normal convex integrant, $\phi :{R}^{n}\to \left(-\infty ,+\infty \right]$ the convex function. Let ${t}_{0}, $F:\left[{t}_{0},T\right]×{R}^{n}\to comp\text{ }{R}^{n}\cup \left\{\varnothing \right\}$ is the multivalued mapping.

The problem of minimization of the functional

$J\left(u\right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t$ (2)

is considered under the following constraints

$u\left(t\right)\in F\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right),$ (3)

where $t\in \left[{t}_{0},T\right]$, $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Introducing the notation $\omega \left(t,x,z\right)=\left\{\begin{array}{l}0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }z\in F\left(t,x\right)\\ +\infty ,\text{ }\text{\hspace{0.17em}}z\notin F\left(t,x\right)\end{array}$ we have that problem (2) and (3) is equivalent to the minimization of the functional

$\begin{array}{c}{J}_{1}\left(u\right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{{t}_{0}}^{T}\text{ }\omega \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\end{array}$

among all functions $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Let the mapping $t\to gr{F}_{t}=\left\{\left(x,y\right):y\in F\left(t,x\right)\right\}$ be measurable on $\left[{t}_{0},T\right]$, the set $gr{F}_{t}$ be closed and convex for almost all $t\in \left[{t}_{0},T\right]$ and $F\left(t,x\right)$ be compact for all $\left(t,x\right)$. From here it follows that $\omega \left(t,x,z\right)$ is a convex normal integrant on $\left[{t}_{0},T\right]×\left({R}^{n}×{R}^{n}\right)$.

Let us consider the following functional

$\begin{array}{c}S\left(u,\upsilon \right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}f\left(t,{\int }_{{t}_{0}}^{t}\text{ }\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\int }_{{t}_{0}}^{T}\text{ }\omega \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)+\upsilon \left(t\right)\right)\text{d}t,\end{array}$

where $\upsilon \left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$. Let $h\left(\upsilon \right)=\underset{u\in {L}_{1}^{n}\left[{t}_{0},T\right]}{\mathrm{inf}}S\left(u,\upsilon \right)$. The problem (2) and (3) is called stable, if $h\left(0\right)$ is finite and function h is subdifferentiable at zero (see [3] ).

Lemma 1. Let $F:\left[{t}_{0},T\right]×{R}^{n}\to comp\text{ }{R}^{n}\cup \left\{\varnothing \right\}$ ; the mapping $t\to F\left(t,x\right)$ be measurable on $\left[{t}_{0},T\right]$ ; the mapping $x\to F\left(t,x\right)$ be closed and convex for almost all $t\in \left[{t}_{0},T\right]$, i.e. $gr{F}_{t}$ be closed and convex for almost all $t\in \left[{t}_{0},T\right]$ ; there exist such a summable function $\lambda \left(t\right)$ that $‖F\left(t,x\right)‖\le \lambda \left(t\right)\left(1+|x|\right)$ for $x\in {R}^{n}$ ; there exist a solution ${u}_{0}\left(t\right)$ to the problem

${u}_{0}\left(t\right)\in F\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right){u}_{0}\left(s\right)\text{d}s+z\left(t\right)\right)$ such that ${x}_{0}\left(t\right)={\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right){u}_{0}\left(s\right)\text{d}s+z\left(t\right)$ belongs to $dom\text{ }\text{ }{F}_{t}=\left\{x:F\left(t,x\right)\ne \varnothing \right\}$ coupled with some $\epsilon$ tube, i.e. $\left\{x:|{x}_{0}\left(t\right)-x|\le \epsilon \right\}\subset dom\text{ }{F}_{t}$ ; $f:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ the normal convex integrant; $\phi :{R}^{n}\to \left(-\infty ,+\infty \right]$ the convex function and $\underset{u\in {L}_{1}^{n}\left[{t}_{0},T\right]}{\mathrm{inf}}{J}_{1}\left(u\right)$ is finite; the function $f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right){u}_{0}\left(s\right)\text{d}s+z\left(t\right)+y\right)$ be summarized for $y\in {R}^{n}$, $|y|, where $r>0$, and function $\phi \left(\cdot \right)$ be continuous at the point ${\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right){u}_{0}\left(s\right)\text{d}s+z\left(T\right)$. Then the function h is subdifferentiable at zero, i.e. problem (2) and (3) is stable.

Let $\upsilon \in {R}^{n}$. Assume

${\omega }^{0}\left(t,x,\upsilon \right)=\underset{z\in {R}^{n}}{\mathrm{inf}}\left\{\left(z|\upsilon \right)+\omega \left(t,x,z\right)\right\}=\mathrm{inf}\left\{\left(z|\upsilon \right):z\in F\left(t,x\right)\right\}$,

where $\mathrm{inf}\varnothing =+\infty$.

Theorem 5. Let $F:\left[{t}_{0},T\right]×{R}^{n}\to comp\text{ }{R}^{n}\cup \left\{\varnothing \right\}$ ; the mapping $t\to F\left(t,x\right)$ be measurable on $\left[{t}_{0},T\right]$ ; the mapping $x\to F\left(t,x\right)$ be closed and convex for almost all $t\in \left[{t}_{0},T\right]$ ; f be the normal convex integrant on $\left[{t}_{0},T\right]×{R}^{n}$ ; $\phi$ the convex function on ${R}^{n}$ ; $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ the continuous matrix function; $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function. For the function $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ to minimize the functional (2) among all the solutions to the problem (3), it is sufficient that there exist ${u}_{1}^{*}\left(\cdot \right)$, ${u}_{2}^{*}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ and $\text{ }b\in {R}^{n}$ such that

1) ${u}_{1}^{*}\left(t\right)\in \partial f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)\right)$,

2) $b\in \partial \phi \left(z\left(T\right)+{\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s\right)$,

3) ${u}_{2}^{*}\left(t\right)\in \partial {\omega }^{0}\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right),{\int }_{t}^{T}\text{ }k{\left(\tau ,t\right)}^{t}\left({u}_{1}^{*}\left(\tau \right)+{u}_{2}^{*}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right),$

4) $\begin{array}{l}{\omega }^{0}\left(t,z\left(t\right)+{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s,{\int }_{t}^{T}\text{ }k{\left(\tau ,t\right)}^{t}\left({u}_{1}^{*}\left(\tau \right)+{u}_{2}^{*}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right)\\ =\left(\stackrel{¯}{u}\left(t\right)|{\int }_{t}^{T}k{\left(\tau ,t\right)}^{t}\left({u}_{1}^{*}\left(\tau \right)+{u}_{2}^{*}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+\omega \left(t,z\left(t\right)+{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s,\stackrel{¯}{u}\left(t\right)\right),\end{array}$

and if for ${u}_{0}\left(t\right)=\stackrel{¯}{u}\left(t\right)$ the condition of lemma 1 is satisfied, then conditions 1) - 4) become necessary.

5. Nonconvex Extremal Problem for Integral Inclusions

Let $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ be the continuous matrix function; $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function, i.e. $z\left(\cdot \right)\in {C}^{n}\left[{t}_{0},T\right]$. Hereafter we will assume that $f:\left[{t}_{0},T\right]×{R}^{n}×{R}^{n}\to \left(-\infty ,+\infty \right]$ is the normal integrant and $\phi :{R}^{n}\to \left(-\infty ,+\infty \right]$ is the function. Let ${t}_{0}, $F:\left[{t}_{0},T\right]×{R}^{n}\to comp\text{ }{R}^{n}$ be the multivalued mapping.

We consider the following problem of minimization of the functional

$J\left(u\right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t,$ (4)

under the following constraints

$u\left(t\right)\in F\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right),$ (5)

where $t\in \left[{t}_{0},T\right]$, $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Let $\psi \left(s,x,y\right)=\mathrm{inf}\left\{|z-y|:z\in F\left(s,x\right)\right\}$ and consider the minimization of the functional

$\begin{array}{c}{J}_{r}\left(u\right)=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+r{\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\end{array}$

among all the functions $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Theorem 6. If $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ is the solution to the problem (4) and (5), $F:\left[{t}_{0},T\right]×{R}^{n}\to comp\text{ }{R}^{n}\cup \left\{\varnothing \right\}$ and $t\to \text{ }F\left(t,x\right)$ are measurable on t, $\stackrel{¯}{x}\left(t\right)={\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)$, there exist $k\left(\cdot \right)\in {L}_{1}\left[{t}_{0},T\right]$, $M\left(\cdot \right)\in {L}_{1}\left[{t}_{0},T\right]$,

${k}_{1}>0$, ${k}_{2}>0$ and $\alpha >0$ such that $B\left(\stackrel{¯}{x}\left(t\right),\alpha \right)\subset dom\text{ }{F}_{t}=\left\{x\in {R}^{n}:F\left(t,x\right)\ne \varnothing \right\}$ at $t\in \left[{t}_{0},T\right]$ and

$|\phi \left(z\right)-\phi \left(u\right)|\le {k}_{2}|z-u|\text{ },$

$|f\left(t,{x}_{1},{y}_{1}\right)-f\left(t,{x}_{2},{y}_{2}\right)|\le k\left(t\right)|{x}_{1}-{x}_{2}|+{k}_{1}|{y}_{1}-{y}_{2}|,$

${\rho }_{X}\left(F\left(t,{x}_{1}\right),F\left(t,{x}_{2}\right)\right)\le M\left(t\right)|{x}_{1}-{x}_{2}|$

for $z,u\in B\left(\stackrel{¯}{x}\left(T\right),\alpha \right)$, ${x}_{1},{x}_{2}\in B\left(\stackrel{¯}{x}\left(t\right),\alpha \right)$, ${y}_{1},{y}_{2}\in {R}^{n}$. Then there exist a number ${r}_{0}>0$ such that $\stackrel{¯}{u}\left(t\right)$ minimizes the functional ${J}_{r}\left(u\right)$ in D for

$r\ge {r}_{0}$, where $D=\left\{u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]:{‖u\left(\cdot \right)-\stackrel{¯}{u}\left(\cdot \right)‖}_{{L}_{1}^{n}\left[{t}_{0},T\right]}\le \frac{\alpha }{\beta }\right\}$, $\beta >\left(1+a\left({\text{e}}^{m\left(T\right)}+a{\text{e}}^{m\left(T\right)}{\int }_{{t}_{0}}^{T}\text{ }M\left(t\right)\text{d}t\right)\right)\left(a{\int }_{{t}_{0}}^{T}\text{ }M\left(t\right)\text{d}t+1\right)+a$, $m\left(t\right)=a{\int }_{{t}_{0}}^{t}M\left(s\right)\text{d}s$.

Theorem 7. Let the condition of the theorem 6 be satisfied and the function $\stackrel{¯}{u}\left(t\right)$ among all solutions to the problem (5) minimizes the functional (4). Then there exists ${u}^{*}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ and $b\in {R}^{n}$ such that

1) $\begin{array}{l}\left({u}^{*}\left(t\right),-{\int }_{t}^{T}\text{ }k{\left(\tau ,t\right)}^{t}{u}^{*}\left(\tau \right)\text{d}\tau -K{\left(T,t\right)}^{t}b\right)\\ \in {\partial }_{C}\left(f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right),\stackrel{¯}{u}\left(t\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+r\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right),\stackrel{¯}{u}\left(t\right)\right)\right),\end{array}$

2) $b\in {\partial }_{C}\phi \left(z\left(T\right)+{\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s\right),$

where ${\partial }_{C}g\left(\stackrel{¯}{x}\right)$ is Clarke subdifferential of the function g at the point $\stackrel{¯}{x}$ (see [4] ).

6. A Higher Order Necessary Condition in the Extremal Problem for the Volterra Type Inclusion

Consider the problem (4) and (5), where $f\left(t,x,y\right)=f\left(t,x\right)$. Assume

$\psi \left(s,x,y\right)=\mathrm{inf}\left\{|z-y|:z\in F\left(s,x\right)\right\}.$

We consider the following problem of minimization of the function

$\begin{array}{c}{J}_{r}\left(u\right)=\phi \left({\int }_{{t}_{0}}^{T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+r\left({\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{‖\stackrel{¯}{u}\left(\cdot \right)-u\left(\cdot \right)‖}_{{L}_{1}^{n}}^{\beta -\nu }{\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\nu }\right)\end{array}$

among all functions $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Let $\stackrel{¯}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ be the solution to the problem (4) and (5). Let $\stackrel{¯}{x}\left(t\right)={\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right)$.

Theorem 8. Let $F:\left[{t}_{0},T\right]×{R}^{n}\to comp{R}^{n}$ be the multivalued mapping, the mapping $t\to F\left(t,x\right)$ be measurable on $\left[{t}_{0},T\right]$ ; f be the normal integrant on $\left[{t}_{0},T\right]×{R}^{n}$ ; $\phi$ the function in ${R}^{n}$ ; $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ the continuous matrix function; $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function, and there exists a summable function $M\left(t\right)>0$ such that ${\rho }_{x}\left(F\left(t,x\right),F\left(t,{x}_{1}\right)\right)\le M\left(t\right)|x-{x}_{1}|$ for $x,{x}_{1}\in {R}^{n}$ ; there exist ${k}_{1}\left(\cdot \right)\in {L}_{1}\left[{t}_{0},T\right]$, ${k}_{1}\left(t\right)>0$ and number ${k}_{2}>0$ such that

$|f\left(t,{x}_{1}\right)-f\left(t,{x}_{2}\right)|\le {k}_{1}\left(t\right){|{x}_{1}-{x}_{2}|}^{\nu }\left({|{x}_{2}-\stackrel{¯}{x}\left(t\right)|}^{\beta -\nu }+{|{x}_{1}-{x}_{2}|}^{\beta -\nu }\right)$

for ${x}_{1},{x}_{2}\in {R}^{n}$,

$|\phi \left(x\right)-\phi \left(y\right)|\le {k}_{2}{|x-y|}^{\nu }\left({|y-\stackrel{¯}{x}\left(T\right)|}^{\beta -\nu }+{|x-y|}^{\beta -\nu }\right)$

for $x,y\in {R}^{n}$ (see [5] ). If the function $\stackrel{¯}{u}\left(t\right)$ among all solutions of the problem (5) minimizes the functional (4), then there exists a number ${r}_{0}>0$ such that $\stackrel{¯}{u}\left(t\right)$ minimizes the functional ${J}_{r}\left(u\right)$ in ${L}_{1}^{n}\left[{t}_{0},T\right]$ for $r\ge {r}_{0}$.

Let $g:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$ be the normal integrant on $\left[{t}_{0},T\right]×{R}^{n}$ ; $e:{R}^{n}\to \left(-\infty ,+\infty \right]$ the function.

Let assume

$S\left(u\right)=e\left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\int }_{{t}_{0}}^{T}\text{ }g\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t,$

$\begin{array}{l}{H}_{r}\left(u\right)=J\left(u\right)-S\left(u\right)+r\left({\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{‖\stackrel{¯}{u}\left(\cdot \right)-u\left(\cdot \right)‖}_{{L}_{1}^{n}}^{\beta -\nu }{\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\nu }\right)\end{array}$

$\begin{array}{l}=\phi \left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)-e\left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{\int }_{{t}_{0}}^{T}\text{ }f\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t-{\int }_{{t}_{0}}^{T}\text{ }g\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+r\left({\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }+{‖\stackrel{¯}{u}\left(\cdot \right)-u\left(\cdot \right)‖}_{{L}_{1}^{n}}^{\beta -\nu }{\left({\int }_{{t}_{0}}^{T}\text{ }\psi \left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\nu }\right).\end{array}$

Theorem 9. Let $F:\left[{t}_{0},T\right]×{R}^{n}\to comp{R}^{n}$ the multivalued mapping, the mapping $t\to F\left(t,x\right)$ be measurable on $\left[{t}_{0},T\right]$ ; f be the normal integrant on $\left[{t}_{0},T\right]×{R}^{n}$ ; $\phi$ the function in ${R}^{n}$ ; $k:{\left[{t}_{0},T\right]}^{2}\to {M}_{n}$ the continuous matrix function; $z:\left[{t}_{0},T\right]\to {R}^{n}$ the continuous function and there exist a summable function $M\left(t\right)>0$ such that ${\rho }_{x}\left(F\left(t,x\right),F\left(t,{x}_{1}\right)\right)\le M\left(t\right)|x-{x}_{1}|$ for $x,{x}_{1}\in {R}^{n}$ ; there exist the normal integrant $g:\left[{t}_{0},T\right]×{R}^{n}\to \left(-\infty ,+\infty \right]$, the functions $e:{R}^{n}\to \left(-\infty ,+\infty \right]$, ${k}_{1}\left(\cdot \right)\in {L}_{1}\left[{t}_{0},T\right]$, ${k}_{1}\left(t\right)>0$ and number ${k}_{2}>0$ such that

$\begin{array}{l}|f\left(t,{x}_{1}\right)-g\left(t,{x}_{1}\right)-f\left(t,{x}_{2}\right)+g\left(t,{x}_{2}\right)|\\ \le {k}_{1}\left(t\right){|{x}_{1}-{x}_{2}|}^{\nu }\left({|{x}_{2}-\stackrel{¯}{x}\left(t\right)|}^{\beta -\nu }+{|{x}_{1}-{x}_{2}|}^{\beta -\nu }\right)\end{array}$

for ${x}_{1},{x}_{2}\in {R}^{n}$, where $\stackrel{¯}{x}\left(t\right)={\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{˜}{u}\left(s\right)\text{d}s+z\left(t\right)$,

$|\phi \left(x\right)-e\left(x\right)-\phi \left(y\right)+e\left(y\right)|\le {k}_{2}{|x-y|}^{\nu }\left({|y-\stackrel{¯}{x}\left(T\right)|}^{\beta -\nu }+{|x-y|}^{\beta -\nu }\right)$

for $x,y\in {R}^{n}$ and let $\stackrel{˜}{u}\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$ solutions of the problem (4)-(5). Then there exist a number ${r}_{0}>0$ such that $\stackrel{˜}{u}\left(t\right)$ minimizes the functional ${H}_{r}\left(u\right)$ in $u\in \left\{\upsilon \in {L}_{1}^{n}\left[{t}_{0},T\right]:S\left({w}_{\upsilon }\right)\le S\left(\stackrel{˜}{u}\right)\right\}$ for $r\ge {r}_{0}$, where ${w}_{\upsilon }$ solutions to the problem (5), which satisfy the main theorem 1 for $\stackrel{¯}{u}\left(\cdot \right)=\upsilon \left(t\right)$.

It’s possible to get the local variant of theorems 8 and 9 analogical to theorem 6.

Let assume

${J}_{r}^{\left\{\beta \right\}+}\left(\stackrel{¯}{u};u\right)=\stackrel{¯}{\underset{\lambda ↓0}{\mathrm{lim}}}\frac{1}{{\lambda }^{\beta }}\left({J}_{r}\left(\stackrel{¯}{u}+\lambda u\right)-{J}_{r}\left(\stackrel{¯}{u}\right)\right),$

${J}_{r}^{\left\{\beta \right\}-}\left(\stackrel{¯}{u};u\right)=\underset{\lambda ↓0}{\underset{_}{\mathrm{lim}}}\frac{1}{{\lambda }^{\beta }}\left({J}_{r}\left(\stackrel{¯}{u}+\lambda u\right)-{J}_{r}\left(\stackrel{¯}{u}\right)\right)$

for $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Corollary 1. If the condition of theorem 8 is satisfied, then there exist a number ${r}_{0}>0$ such that ${J}_{r}^{\left\{\beta \right\}+}\left(\stackrel{¯}{u};u\right)\ge {J}_{r}^{\left\{\beta \right\}-}\left(\stackrel{¯}{u};u\right)\ge 0$ for $r\ge {r}_{0}$ and $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Let assume

$\begin{array}{l}{E}_{r}\left(u\right)={\phi }^{\left\{\beta \right\}-}\left({\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(T\right);{\int }_{{t}_{0}}^{T}\text{ }k\left(T,s\right)u\left(s\right)\text{d}s\right)\\ \text{ }+{\int }_{{t}_{0}}^{T}\text{ }{f}^{\left\{\beta \right\}+}\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right);{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s\right)\text{d}t\\ \text{ }+r\left({\left({\int }_{{t}_{0}}^{T}\text{ }{\psi }^{\left\{1\right\}+}\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right),\stackrel{¯}{u}\left(t\right);{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s,u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ \text{ }+{‖u\left(\cdot \right)‖}_{{L}_{1}^{n}}^{\beta -\nu }{\left({\int }_{{t}_{0}}^{T}\text{ }{\psi }^{\left\{1\right\}+}\left(t,{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)\stackrel{¯}{u}\left(s\right)\text{d}s+z\left(t\right),\stackrel{¯}{u}\left(t\right);{\int }_{{t}_{0}}^{t}\text{ }k\left(t,s\right)u\left(s\right)\text{d}s,u\left(t\right)\right)\text{d}t\right)}^{\nu }\right).\end{array}$

Theorem 10. If the condition of theorem 8 is satisfied, then there exist a number ${r}_{0}>0$ such that ${E}_{r}\left(u\right)\ge 0$ for $r\ge {r}_{0}$ and $u\left(\cdot \right)\in {L}_{1}^{n}\left[{t}_{0},T\right]$.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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