M. A. Sadygov
Baku State University, Baku, Azerbaijan.
DOI: 10.4236/oalib.1105605   PDF    HTML   XML   312 Downloads   1,133 Views   Citations

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.

Keywords

An Extreme Problem for a Volterra Type Integral Inclusion

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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 $comp{R}^n\left(conv{R}^n\right)$ ; $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ is the continuous matrix function, wherewith $M_n$ being the set of all square $n\times n$ matrices of real elements ( $b_{ij}$ ); $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function; $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n$ 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

$u\left(t\right)\in F\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)$ (1)

The function $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ satisfying (1) we will call the solution to problem (1) (see [1] ).

Let $a=\underset{t,s\in \left[t_0,T\right]}{\max }\Vert k\left(t,s\right)\Vert =\underset{t,s\in \left[t_0,T\right]}{\max }{\displaystyle {\sum }_{i=1}^n}{\displaystyle {\sum }_{j=1}^n}\left|k_{i,j}\left(t,s\right)\right|$, if $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ is the continuous matrix function.

Theorem 1. Let $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ be the continuous matrix function, $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function, $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n$ the multivalued mapping, $t\to F\left(t\mathrm{,}x\right)$ is measurable on t, and there exists a summable function $M\left(t\right)>0$ such that ${\rho }_x\left(F\left(t\mathrm{,}x\right)\mathrm{,}F\left(t\mathrm{,}{x}_1\right)\right)\le M\left(t\right)\left|x-x_1\right|$ for

$x\mathrm{,}{x}_1\in R^n$. Moreover, let $\rho (\cdot )\in L_1\left[t_0\mathrm{,}T\right]$ and $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ be such that

$d\left(\bar{u}\left(t\right)\mathrm{,}F\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)\right)\right)\le \rho \left(t\right)$ for $t\in \left[t_0\mathrm{,}T\right]$. Then there exists such a solution $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ to problem (1) that

$\left|{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s-{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s\right|\le a{\displaystyle {\int }_{t_0}^t}{\text{e}}^{m\left(t\right)-m\left(s\right)}\rho \left(s\right)\text{d}s,$

$\left|u\left(t\right)-\bar{u}\left(t\right)\right|\le \rho \left(t\right)+aM\left(t\right){\displaystyle {\int }_{t_0}^t}{\text{e}}^{m\left(t\right)-m\left(s\right)}\rho \left(s\right)\text{d}s$

for $t\in \left[t_0\mathrm{,}T\right]$, where $m\left(t\right)=a{\displaystyle {\int }_{t_0}^t}M\left(s\right)\text{d}s$.

2. On Subdifferential of the Integral Functional

Let $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ is the normal convex integrant (see [2] ).

Let consider a subdifferential of the integral functional

$J\left(u(\cdot )\right)={\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t$

in $L_1^n\left[t_0\mathrm{,}T\right]$.

Theorem 2. If $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ be the continuous matrix function, $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function, $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ is the

normal convex integrant and function $f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)+x\right)$ is

summable for $x\in R^n$, $\left|x\right|\le \delta$, where $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$, then $\partial J\left(\bar{u}(\cdot )\right)$ is nonempty and ${\upsilon }^{\mathrm{<mo>\; *\; </mo>}}\in L_{\infty }^n\left[t_0\mathrm{,}T\right]$ belongs to $\partial J\left(\bar{u}(\cdot )\right)$ if and only if, there exist

$u^{\mathrm{<mo>\; *\; </mo>}}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$, $u^{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\in \partial f\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)\right)$, such, that ${\upsilon }^{*}\left(s\right)={\displaystyle {\int }_s^T}k{\left(t,s\right)}^t{u}^{*}\left(t\right)\text{d}t$, where $k{\mathrm{<mo\; stretchy="false">\; (\; </mo>}\tau \mathrm{,}t\mathrm{<mo\; stretchy="false">\; )\; </mo>}}^t$ is the transpose of the matrix $k\left(\tau \mathrm{,}t\right)$.

Theorem 3. If $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ be measurable bounded matrix function, $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ be measurable bounded function, $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ is

the normal convex integrant and function $f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)+x\right)$ is

summable for $x\in R^n$, $\left|x\right|\le \delta$, where $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$, then $\partial J\left(\bar{u}(\cdot )\right)$ is nonempty and functional ${\upsilon }^{\mathrm{<mo>\; *\; </mo>}}\in L_{\infty }^n\left[t_0\mathrm{,}T\right]$ belongs to $\partial J\left(\bar{u}(\cdot )\right)$ if and only if,

there exist $u^{\mathrm{<mo>\; *\; </mo>}}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$, $u^{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\in \partial f\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)\right)$, such, that ${\upsilon }^{\mathrm{<mo>\; *\; </mo>}}\left(s\right)={\displaystyle {\int }_s^T}k{\left(t\mathrm{,}s\right)}^t{u}^{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\text{d}t$.

3. On Subdifferential of the Terminal Functional

Let $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ continuous matrix function, $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ continuous function, $\phi \mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$ proper convex function in $R^n$. Consider a subdifferential of the terminal functional $F\left(u(\cdot )\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)$ in $L_1^n\left[t_0\mathrm{,}T\right]$, where $z(\cdot )\in C^n\left[t_0\mathrm{,}T\right]$.

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

point ${\displaystyle {\int }_{t_0}^T}k\left(T,s\right)\bar{u}\left(s\right)\text{d}s+z\left(T\right)$, then

$\partial F\left(\bar{u}(\cdot )\right)=\left\{bk\left(T,s\right):b\in \partial \phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)\bar{u}\left(s\right)\text{d}s+z\left(T\right)\right)\right\}.$

4. Convex Extremal Problem for Integral Inclusions

Let $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ be the continuous matrix function, $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function. Hereafter we will assume that $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ is the normal convex integrant, $\phi \mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$ the convex function. Let $t_0<T$, $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n\cup \{\varnothing \}$ is the multivalued mapping.

The problem of minimization of the functional

$J\left(u\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t$ (2)

is considered under the following constraints

$u\left(t\right)\in F\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\mathrm{,}$ (3)

where $t\in \left[t_0\mathrm{,}T\right]$, $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

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

$\begin{array}{c}{J}_1\left(u\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_{t_0}^T}\omega \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\end{array}$

among all functions $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

Let the mapping $t\to gr{F}_t=\left\{\left(x\mathrm{,}y\right)\mathrm{<mo>\; :\; </mo>}y\in F\left(t\mathrm{,}x\right)\right\}$ be measurable on $\left[t_0\mathrm{,}T\right]$, the set $gr{F}_t$ be closed and convex for almost all $t\in \left[t_0\mathrm{,}T\right]$ and $F\left(t\mathrm{,}x\right)$ be compact for all $\left(t\mathrm{,}x\right)$. From here it follows that $\omega \left(t\mathrm{,}x\mathrm{,}z\right)$ is a convex normal integrant on $\left[t_0\mathrm{,}T\right]\times \left(R^n\times R^n\right)$.

Let us consider the following functional

$\begin{array}{c}S\left(u,\upsilon \right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ +{\displaystyle {\int }_{t_0}^T}\omega \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)+\upsilon \left(t\right)\right)\text{d}t,\end{array}$

where $\upsilon (\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$. Let $h\left(\upsilon \right)=\underset{u\in L_1^n\left[t_0\mathrm{,}T\right]}{\inf }S\left(u\mathrm{,}\upsilon \right)$. 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 $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n\cup \{\varnothing \}$ ; the mapping $t\to F\left(t\mathrm{,}x\right)$ be measurable on $\left[t_0\mathrm{,}T\right]$ ; the mapping $x\to F\left(t\mathrm{,}x\right)$ be closed and convex for almost all $t\in \left[t_0\mathrm{,}T\right]$, i.e. $gr{F}_t$ be closed and convex for almost all $t\in \left[t_0\mathrm{,}T\right]$ ; there exist such a summable function $\lambda \left(t\right)$ that $\Vert F\left(t\mathrm{,}x\right)\Vert \le \lambda \left(t\right)\left(1+\left|x\right|\right)$ for $x\in R^n$ ; there exist a solution $u_0\left(t\right)$ to the problem

$u_0\left(t\right)\in F\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u_0\left(s\right)\text{d}s+z\left(t\right)\right)$ such that $x_0\left(t\right)={\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u_0\left(s\right)\text{d}s+z\left(t\right)$ belongs to $dom{F}_t=\left\{x\mathrm{<mo>\; :\; </mo>}F\left(t\mathrm{,}x\right)\ne \varnothing \right\}$ coupled with some $\epsilon$ tube, i.e. $\left\{x\mathrm{<mo>\; :\; </mo>}\left|x_0\left(t\right)-x\right|\le \epsilon \right\}\subset dom{F}_t$ ; $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ the normal convex integrant; $\phi \mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$ the convex function and $\underset{u\in L_1^n\left[t_0\mathrm{,}T\right]}{\inf }{J}_1\left(u\right)$ is finite; the function $f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u_0\left(s\right)\text{d}s+z\left(t\right)+y\right)$ be summarized for $y\in R^n$, $\left|y\right|<r$, where $r>0$, and function $\phi (\cdot )$ be continuous at the point ${\displaystyle {\int }_{t_0}^T}k\left(T\mathrm{,}s\right)u_0\left(s\right)\text{d}s+z\left(T\right)$. Then the function h is subdifferentiable at zero, i.e. problem (2) and (3) is stable.

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

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

where $\inf \varnothing =+\infty$.

Theorem 5. Let $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n\cup \{\varnothing \}$ ; the mapping $t\to F\left(t\mathrm{,}x\right)$ be measurable on $\left[t_0\mathrm{,}T\right]$ ; the mapping $x\to F\left(t\mathrm{,}x\right)$ be closed and convex for almost all $t\in \left[t_0\mathrm{,}T\right]$ ; f be the normal convex integrant on $\left[t_0\mathrm{,}T\right]\times R^n$ ; $\phi$ the convex function on $R^n$ ; $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ the continuous matrix function; $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function. For the function $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ to minimize the functional (2) among all the solutions to the problem (3), it is sufficient that there exist $u_1^{\mathrm{<mo>\; *\; </mo>}}(\cdot )$, $u_2^{\mathrm{<mo>\; *\; </mo>}}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ and $b\in R^n$ such that

1) $u_1^{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\in \partial f\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)\right)$,

2) $b\in \partial \phi \left(z\left(T\right)+{\displaystyle {\int }_{t_0}^T}k\left(T\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s\right)$,

3) $u_2^{\mathrm{<mo>\; *\; </mo>}}\left(t\right)\in \partial {\omega }^0\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)\mathrm{,}{\displaystyle {\int }_t^T}k{\left(\tau \mathrm{,}t\right)}^t\left(u_1^{\mathrm{<mo>\; *\; </mo>}}\left(\tau \right)+u_2^{\mathrm{<mo>\; *\; </mo>}}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right)\mathrm{,}$

4) $\begin{array}{l}{\omega }^0\left(t,z\left(t\right)+{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s,{\displaystyle {\int }_t^T}k{\left(\tau ,t\right)}^t\left(u_1^{*}\left(\tau \right)+u_2^{*}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right)\\ =\left(\bar{u}\left(t\right)|{\displaystyle {\int }_t^T}k{\left(\tau ,t\right)}^t\left(u_1^{*}\left(\tau \right)+u_2^{*}\left(\tau \right)\right)\text{d}\tau +K{\left(T,t\right)}^{t}b\right)\\ +\omega \left(t,z\left(t\right)+{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s,\bar{u}\left(t\right)\right),\end{array}$

and if for $u_0\left(t\right)=\bar{u}\left(t\right)$ the condition of lemma 1 is satisfied, then conditions 1) - 4) become necessary.

5. Nonconvex Extremal Problem for Integral Inclusions

Let $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ be the continuous matrix function; $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function, i.e. $z(\cdot )\in C^n\left[t_0\mathrm{,}T\right]$. Hereafter we will assume that $f\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ is the normal integrant and $\phi \mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$ is the function. Let $t_0<T$, $F:\left[t_0,T\right]\times R^n\to comp{R}^n$ be the multivalued mapping.

We consider the following problem of minimization of the functional

$J\left(u\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t,$ (4)

under the following constraints

$u\left(t\right)\in F\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right),$ (5)

where $t\in \left[t_0\mathrm{,}T\right]$, $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

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

$\begin{array}{c}{J}_r\left(u\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\\ +r{\displaystyle {\int }_{t_0}^T}\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\end{array}$

among all the functions $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

Theorem 6. If $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ is the solution to the problem (4) and (5), $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n\cup \{\varnothing \}$ and $t\to F\left(t\mathrm{,}x\right)$ are measurable on t, $\bar{x}\left(t\right)={\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)$, there exist $k(\cdot )\in L_1\left[t_0\mathrm{,}T\right]$, $M(\cdot )\in L_1\left[t_0\mathrm{,}T\right]$,

$k_1>0$, $k_2>0$ and $\alpha >0$ such that $B\left(\bar{x}\left(t\right)\mathrm{,}\alpha \right)\subset dom{F}_t=\left\{x\in R^n\mathrm{<mo>\; :\; </mo>}F\left(t\mathrm{,}x\right)\ne \varnothing \right\}$ at $t\in \left[t_0\mathrm{,}T\right]$ and

$\left|\phi \left(z\right)-\phi \left(u\right)\right|\le k_2\left|z-u\right|\mathrm{,}$

$\left|f\left(t\mathrm{,}{x}_1\mathrm{,}{y}_1\right)-f\left(t\mathrm{,}{x}_2\mathrm{,}{y}_2\right)\right|\le k\left(t\right)\left|x_1-x_2\right|+k_1\left|y_1-y_2\right|\mathrm{,}$

${\rho }_X\left(F\left(t\mathrm{,}{x}_1\right)\mathrm{,}F\left(t\mathrm{,}{x}_2\right)\right)\le M\left(t\right)\left|x_1-x_2\right|$

for $z\mathrm{,}u\in B\left(\bar{x}\left(T\right)\mathrm{,}\alpha \right)$, $x_1\mathrm{,}{x}_2\in B\left(\bar{x}\left(t\right)\mathrm{,}\alpha \right)$, $y_1\mathrm{,}{y}_2\in R^n$. Then there exist a number $r_0>0$ such that $\bar{u}\left(t\right)$ minimizes the functional $J_r\left(u\right)$ in D for

$r\ge r_0$, where $D=\left\{u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]\mathrm{<mo>\; :\; </mo>}{\Vert u(\cdot )-\bar{u}(\cdot )\Vert }_{L_1^n\left[t_0\mathrm{,}T\right]}\le \frac{\alpha }{\beta }\right\}$, $\beta >\left(1+a\left({\text{e}}^{m\left(T\right)}+a{\text{e}}^{m\left(T\right)}{\displaystyle {\int }_{t_0}^T}M\left(t\right)\text{d}t\right)\right)\left(a{\displaystyle {\int }_{t_0}^T}M\left(t\right)\text{d}t+1\right)+a$, $m\left(t\right)=a{\displaystyle {\int }_{t_0}^t}M\left(s\right)\text{d}s$.

Theorem 7. Let the condition of the theorem 6 be satisfied and the function $\bar{u}\left(t\right)$ among all solutions to the problem (5) minimizes the functional (4). Then there exists $u^{\mathrm{<mo>\; *\; </mo>}}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ and $b\in R^n$ such that

1) $\begin{array}{l}\left(u^{*}\left(t\right),-{\displaystyle {\int }_t^T}k{\left(\tau ,t\right)}^t{u}^{*}\left(\tau \right)\text{d}\tau -K{\left(T,t\right)}^{t}b\right)\\ \in {\partial }_C(f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right),\bar{u}\left(t\right)\right)\\ +r\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right),\bar{u}\left(t\right)\right)),\end{array}$

2) $b\in {\partial }_C\phi \left(z\left(T\right)+{\displaystyle {\int }_{t_0}^T}k\left(T\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s\right),$

where ${\partial }_{C}g\left(\bar{x}\right)$ is Clarke subdifferential of the function g at the point $\bar{x}$ (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

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

We consider the following problem of minimization of the function

$\begin{array}{c}{J}_r\left(u\right)=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}f\left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ +r({\left({\displaystyle {\int }_{t_0}^T}\psi \left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(t\right)\mathrm{,}u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ +{\Vert \bar{u}(\cdot )-u(\cdot )\Vert }_{L_1^n}^{\beta -\nu }{\left({\displaystyle {\int }_{t_0}^T}\psi \left(t\mathrm{,}{\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)u\left(s\right)\text{d}s+z\left(t\right)\mathrm{,}u\left(t\right)\right)\text{d}t\right)}^{\nu })\end{array}$

among all functions $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

Let $\bar{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ be the solution to the problem (4) and (5). Let $\bar{x}\left(t\right)={\displaystyle {\int }_{t_0}^t}k\left(t\mathrm{,}s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right)$.

Theorem 8. Let $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n$ be the multivalued mapping, the mapping $t\to F\left(t\mathrm{,}x\right)$ be measurable on $\left[t_0\mathrm{,}T\right]$ ; f be the normal integrant on $\left[t_0\mathrm{,}T\right]\times R^n$ ; $\phi$ the function in $R^n$ ; $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ the continuous matrix function; $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function, and there exists a summable function $M\left(t\right)>0$ such that ${\rho }_x\left(F\left(t\mathrm{,}x\right)\mathrm{,}F\left(t\mathrm{,}{x}_1\right)\right)\le M\left(t\right)\left|x-x_1\right|$ for $x\mathrm{,}{x}_1\in R^n$ ; there exist $k_1(\cdot )\in L_1\left[t_0\mathrm{,}T\right]$, $k_1\left(t\right)>0$ and number $k_2>0$ such that

$\left|f\left(t\mathrm{,}{x}_1\right)-f\left(t\mathrm{,}{x}_2\right)\right|\le k_1\left(t\right){\left|x_1-x_2\right|}^{\nu }\left({\left|x_2-\bar{x}\left(t\right)\right|}^{\beta -\nu }+{\left|x_1-x_2\right|}^{\beta -\nu }\right)$

for $x_1\mathrm{,}{x}_2\in R^n$,

$\left|\phi \left(x\right)-\phi \left(y\right)\right|\le k_2{\left|x-y\right|}^{\nu }\left({\left|y-\bar{x}\left(T\right)\right|}^{\beta -\nu }+{\left|x-y\right|}^{\beta -\nu }\right)$

for $x\mathrm{,}y\in R^n$ (see [5] ). If the function $\bar{u}\left(t\right)$ among all solutions of the problem (5) minimizes the functional (4), then there exists a number $r_0>0$ such that $\bar{u}\left(t\right)$ minimizes the functional $J_r\left(u\right)$ in $L_1^n\left[t_0\mathrm{,}T\right]$ for $r\ge r_0$.

Let $g\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$ be the normal integrant on $\left[t_0\mathrm{,}T\right]\times R^n$ ; $e\mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$ the function.

Let assume

$S\left(u\right)=e\left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)+{\displaystyle {\int }_{t_0}^T}g\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t,$

$\begin{array}{l}{H}_r\left(u\right)=J\left(u\right)-S\left(u\right)+r({\left({\displaystyle {\int }_{t_0}^T}\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ +{\Vert \bar{u}(\cdot )-u(\cdot )\Vert }_{L_1^n}^{\beta -\nu }{\left({\displaystyle {\int }_{t_0}^T}\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\nu })\end{array}$

$\begin{array}{l}=\phi \left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)-e\left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s+z\left(T\right)\right)\\ +{\displaystyle {\int }_{t_0}^T}f\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t-{\displaystyle {\int }_{t_0}^T}g\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right)\right)\text{d}t\\ +r({\left({\displaystyle {\int }_{t_0}^T}\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ +{\Vert \bar{u}(\cdot )-u(\cdot )\Vert }_{L_1^n}^{\beta -\nu }{\left({\displaystyle {\int }_{t_0}^T}\psi \left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s+z\left(t\right),u\left(t\right)\right)\text{d}t\right)}^{\nu }).\end{array}$

Theorem 9. Let $F\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to comp{R}^n$ the multivalued mapping, the mapping $t\to F\left(t\mathrm{,}x\right)$ be measurable on $\left[t_0\mathrm{,}T\right]$ ; f be the normal integrant on $\left[t_0\mathrm{,}T\right]\times R^n$ ; $\phi$ the function in $R^n$ ; $k\mathrm{<mo>\; :\; </mo>}{\left[t_0\mathrm{,}T\right]}^2\to M_n$ the continuous matrix function; $z\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\to R^n$ the continuous function and there exist a summable function $M\left(t\right)\mathrm{<mo>\; >\; </mo>0}$ such that ${\rho }_x\left(F\left(t\mathrm{,}x\right)\mathrm{,}F\left(t\mathrm{,}{x}_1\right)\right)\le M\left(t\right)\left|x-x_1\right|$ for $x\mathrm{,}{x}_1\in R^n$ ; there exist the normal integrant $g\mathrm{<mo>\; :\; </mo>}\left[t_0\mathrm{,}T\right]\times R^n\to \left(-\infty \mathrm{,}+\infty \right]$, the functions $e\mathrm{<mo>\; :\; </mo>}{R}^n\to \left(-\infty \mathrm{,}+\infty \right]$, $k_1(\cdot )\in L_1\left[t_0\mathrm{,}T\right]$, $k_1\left(t\right)>0$ and number $k_2>0$ such that

$\begin{array}{l}\left|f\left(t\mathrm{,}{x}_1\right)-g\left(t\mathrm{,}{x}_1\right)-f\left(t\mathrm{,}{x}_2\right)+g\left(t\mathrm{,}{x}_2\right)\right|\\ \le k_1\left(t\right){\left|x_1-x_2\right|}^{\nu }\left({\left|x_2-\bar{x}\left(t\right)\right|}^{\beta -\nu }+{\left|x_1-x_2\right|}^{\beta -\nu }\right)\end{array}$

for $x_1\mathrm{,}{x}_2\in R^n$, where $\bar{x}\left(t\right)={\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\tilde{u}\left(s\right)\text{d}s+z\left(t\right)$,

$\left|\phi \left(x\right)-e\left(x\right)-\phi \left(y\right)+e\left(y\right)\right|\le k_2{\left|x-y\right|}^{\nu }\left({\left|y-\bar{x}\left(T\right)\right|}^{\beta -\nu }+{\left|x-y\right|}^{\beta -\nu }\right)$

for $x\mathrm{,}y\in R^n$ and let $\tilde{u}(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$ solutions of the problem (4)-(5). Then there exist a number $r_0>0$ such that $\tilde{u}\left(t\right)$ minimizes the functional $H_r\left(u\right)$ in $u\in \left\{\upsilon \in L_1^n\left[t_0\mathrm{,}T\right]\mathrm{<mo>\; :\; </mo>}S\left(w_{\upsilon }\right)\le S\left(\tilde{u}\right)\right\}$ for $r\ge r_0$, where $w_{\upsilon }$ solutions to the problem (5), which satisfy the main theorem 1 for $\bar{u}(\cdot )=\upsilon \left(t\right)$.

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

Let assume

$J_r^{\left\{\beta \right\}+}\left(\bar{u}\mathrm{<mo>\; ;\; </mo>}u\right)=\overline{\underset{\lambda \downarrow 0}{\lim }}\frac{1}{{\lambda }^{\beta }}\left(J_r\left(\bar{u}+\lambda u\right)-J_r\left(\bar{u}\right)\right)\mathrm{,}$

$J_r^{\left\{\beta \right\}-}\left(\bar{u}\mathrm{<mo>\; ;\; </mo>}u\right)=\underset{\lambda \downarrow 0}{\underset{\_}{\lim }}\frac{1}{{\lambda }^{\beta }}\left(J_r\left(\bar{u}+\lambda u\right)-J_r\left(\bar{u}\right)\right)$

for $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

Corollary 1. If the condition of theorem 8 is satisfied, then there exist a number $r_0>0$ such that $J_r^{\left\{\beta \right\}+}\left(\bar{u}\mathrm{<mo>\; ;\; </mo>}u\right)\ge J_r^{\left\{\beta \right\}-}\left(\bar{u}\mathrm{<mo>\; ;\; </mo>}u\right)\ge 0$ for $r\ge r_0$ and $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.

Let assume

$\begin{array}{l}{E}_r\left(u\right)={\phi }^{\left\{\beta \right\}-}\left({\displaystyle {\int }_{t_0}^T}k\left(T,s\right)\bar{u}\left(s\right)\text{d}s+z\left(T\right);{\displaystyle {\int }_{t_0}^T}k\left(T,s\right)u\left(s\right)\text{d}s\right)\\ +{\displaystyle {\int }_{t_0}^T}{f}^{\left\{\beta \right\}+}\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right);{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s\right)\text{d}t\\ +r({\left({\displaystyle {\int }_{t_0}^T}{\psi }^{\left\{1\right\}+}\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right),\bar{u}\left(t\right);{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s,u\left(t\right)\right)\text{d}t\right)}^{\beta }\\ +{\Vert u(\cdot )\Vert }_{L_1^n}^{\beta -\nu }{\left({\displaystyle {\int }_{t_0}^T}{\psi }^{\left\{1\right\}+}\left(t,{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)\bar{u}\left(s\right)\text{d}s+z\left(t\right),\bar{u}\left(t\right);{\displaystyle {\int }_{t_0}^t}k\left(t,s\right)u\left(s\right)\text{d}s,u\left(t\right)\right)\text{d}t\right)}^{\nu }).\end{array}$

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 $u(\cdot )\in L_1^n\left[t_0\mathrm{,}T\right]$.