Existence of optimal solutions to Lagrange problems for ordinary control systems involving fractional Laplace operators

Abstract

In this paper, we study optimal control problems containing ordinary control systems, linear with respect to a control variable, described by fractional Dirichlet and Dirichlet–Neumann Laplace operators and a nonlinear integral performance index. The main result is a theorem on the existence of optimal solutions for such problems. In our approach we use a characterization of a weak lower semicontinuity of integral functionals.

1 Introduction

During the last few decades fractional calculus has been attracted the intersts of many scientists. Recent investigations shown that many phenomena can be accurately described by using differential operators of fractional orders. There exist a lot of types of such operators. In particular, fractional Laplace operators appear in many fields of science; for example in economics [2, 15], probability [2, 6, 7, 14], mechanics [5, 7], material science [4], fluid mechanics and hydrodynamics [8, 11,12,13, 28,29,30].

They can be defined in many ways (e.g. Fourier transform [16, 17], hypersingular integral [16], Riesz potential operator [23], Bochner’s definition [27], spectral decomposition [3, 15]).

Another field of research, in which fractional Laplacians appear, is optimal control theory. In [21], the following two optimal control problems are studied:

where $k=1,2$, $\beta >\frac{1}{4}$, $g:(0,\pi )\times {\mathbb {R}}^n\times M\rightarrow {\mathbb {R}}^n$ and $f_0:(0,\pi )\times {\mathbb {R}}^n\times M\rightarrow {\mathbb {R}}$. Here, the control system ($Eq_1$) is described by the one-dimensional Dirichlet Laplace operator $(-\varDelta _1)^\beta$ of order $\beta$, while ($Eq_2$) involves the Dirichlet–Neumann Laplace operator $(-\varDelta _2)^\beta$. These operators are defined through the spectral decomposition of the Laplace operator $-\varDelta$ in $(0,\pi )$ with zero Dirichlet and Dirichlet–Neumann boundary conditions, respectively (cf. Sect. 2). The main result, obtained in cited paper, are the necessary optimality conditions for the problems ($P_k$), $k=1,2$ (Pontryagin maximum principle).

In this paper we study the existence of optimal solutions of problems ($P_k$), $k=1,2$, where $\beta >\frac{1}{2}$ and control systems ($Eq_1$) and ($Eq_2$) are linear with respect to the control variable u. Precisely, we consider the following problems:

where $k=1,2$, $B:(0,\pi )\rightarrow {\mathbb {R}}^{n\times m}$, $f:(0,\pi )\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ and $f_0:(0,\pi )\times {\mathbb {R}}^n\times M\rightarrow {\mathbb {R}}$.

Our study is based on the $L^1$ weak lower-semicontinuity of integral functionals [26]. The existence of optimal solutions is also investigated in [9], where an optimal control problem with a fractional Dirichlet Laplacian, defined in ${\mathbb {R}}^n$ is considered. The control system, studied there, has a variational structure and the cost functional depends also on the fractional Laplacian.

The paper is organized as follows. In Sect. 2, we give necessary notions and facts concerning ordinary Dirichlet and Dirichlet–Neumann Laplace operators of fractional order. In Sect. 3, based on a some version of a global implicit function theorem [18], we formulate and prove a theorem on the existence of a unique solution of the control sysytems ($E_k$), $k=1,2$. In Sect. 4, we derive the main result of this paper, namely a theorem on the existence of optimal solutions for problems ($\hbox {OCP}_k$), $k=1,2$. Section 5 contains an illustrative, theoretical example. We finish with Sect. A containing some basics from the spectral theory of self-adjoint operators in a real Hilbert space.

2 Preliminaries

This part of the paper concerns fractional ordinary Dirichlet and mixed Dirichlet–Neumann Laplace operators. Definitions of these operators are based on the spectral integral representation theorem for a self-adjoint operator in a Hilbert space (cf. [19] and Appendix A).

$$\begin{aligned}&\mathbf{One-dimensional\,Dirichlet\, and\, Dirichlet--Neumann }\\&\quad \mathbf{\, Laplace\,operators\, of\, fractional\, order } \end{aligned}$$

Let us consider the one-dimensional Laplace operator $-\varDelta$ on the interval $(0,\pi )$ given by

$$\begin{aligned} -\varDelta u=-u''. \end{aligned}$$

(1)

We define the following spaces of functions:

$$\begin{aligned} H_D:=H^1_0\cap H^2\qquad and\qquad H_{DN}:=\{z\in H^2;\quad z(0)=z'(\pi )=0\}, \end{aligned}$$

where $H^1_0=H^1_0((0,\pi ),{\mathbb {R}}^n)$ and $H^2=H^2((0,\pi ),{\mathbb {R}}^n)$ are classical Sobolev spaces.

We recall that conditions $z(0)=z(\pi )=0$ (hidden in the definition of $H_D$) and $z(0)=z'(\pi )=0$ are called Dirichlet and Dirichlet–Neumann boundary conditions, respectively. Moreover, $H_D$ and $H_{DN}$ are dense subspeces of the space $L^2=L^2((0,\pi ),{\mathbb {R}}^n)$.

The operator $-\varDelta :H_D\subset L^2\rightarrow L^2$ given by (1) under Dirichlet boundary conditions is called the Dirichlet Laplace operator and denoted by $-\varDelta _D$. Similarly, by the Dirichlet–Neumann Laplace operator $-\varDelta _{DN}:H_{DN}\subset L^2\rightarrow L^2$ we mean the operator $-\varDelta$ under Dirichlet–Neumann boundary conditions.

In an elementary way one can show that operators $-\varDelta _D$ and $-\varDelta _{DN}$ are self-adjoint. Moreover, their spectrum is given by

$$\begin{aligned} \sigma (-\varDelta _D)&=\sigma _p(-\varDelta _D)=\{j^2;\quad j=1,2,\dots \},\\ \sigma (-\varDelta _{DN})&=\sigma _p(-\varDelta _{DN})=\left\{ \left( j-\tfrac{1}{2}\right) ^2;\quad j=1,2,\dots \right\} , \end{aligned}$$

respectively and the eigenspaces ${\mathrm{Eig}}_j(-\varDelta _D)$ (associated with the eigenvalues $\lambda _j=j^2$), ${\mathrm{Eig}}_j(-\varDelta _{DN})$ (associated with the eigenvalues $\lambda _j=\left( j-\frac{1}{2}\right) ^2$) are sets

$$\begin{aligned} {\mathrm{Eig}}_j(-\varDelta _D)&=\{c\sin jt;\quad c\in {\mathbb {R}}^n\},\\ {\mathrm{Eig}}_j(-\varDelta _{DN})&=\{d\sin \left( j-\tfrac{1}{2}\right) t;\quad d\in {\mathbb {R}}^n\}. \end{aligned}$$

It is well known that systems of functions

$$\begin{aligned} c_{i,j}&=\left( 0,\dots ,0,\underbrace{\sqrt{\frac{2}{\pi }}\sin jt}_{i},0,\dots ,0\right) ,\quad i=1,\dots ,n,\,\,j=1,2,\dots ,\\ d_{i,j}&=\left( 0,\dots ,0,\underbrace{\sqrt{\frac{2}{\pi }}\sin \left( j-\tfrac{1}{2}\right) t}_{i},0,\dots ,0\right) ,\quad i=1,\dots ,n,\,\,j=1,2,\dots \end{aligned}$$

are complete orthonormal systems in $L^2$.

Now, let us assume that $\beta >0$. We define the operator

$$\begin{aligned} (-\varDelta _D)^\beta :D((-\varDelta _D)^\beta )\subset L^2\rightarrow L^2 \end{aligned}$$

in the following way (cf. [19, Theorem 2.1])

$$\begin{aligned} (-\varDelta _D)^\beta x(t)=\left( \int \limits _{\sigma (-\varDelta _D)}\lambda ^\beta E(d\lambda )x\right) (t)=\sum \limits _{j=1}^\infty (j^2)^\beta a_j\sqrt{\frac{2}{\pi }}\sin jt \end{aligned}$$

for $x\in D((-\varDelta _D)^\beta )$, where

$$\begin{aligned} D((-\varDelta _D)^\beta )=\Bigg \{x(t)=&\left( \int \limits _{\sigma (-\varDelta _D)}1 E(d\lambda )x\right) (t)=\sum \limits _{j=1}^\infty a_j\sqrt{\frac{2}{\pi }}\sin jt\in L^2;\\&\int \limits _{\sigma (-\varDelta _D)}\left| \lambda ^\beta \right| ^2 \Vert E(d\lambda )x\Vert _{L^2}^2=\sum \limits _{j=1}^\infty \left( \left( j^2\right) ^\beta \right) ^2|a_j|^2<\infty \Bigg \} \end{aligned}$$

(here E is the spectral measure for the operator $-\varDelta _D$ and $a_j\sqrt{\frac{2}{\pi }}\sin jt$ is the projection of x on the n-dimensional eigenspace ${\mathrm{Eig}}_j(-\varDelta _D)$).

The operator $(-\varDelta _D)^\beta$ is called the fractional Dirichlet Laplace operator of order $\beta$ and the function $(-\varDelta _D)^\beta x$ - the fractional Dirichlet Laplacian of order $\beta$ of x.

Similarly, we define the fractional Dirichlet–Neumann Laplace operator of order $\beta$

$$\begin{aligned} (-\varDelta _{DN})^\beta :D((-\varDelta _{DN})^\beta )\subset L^2\rightarrow L^2. \end{aligned}$$

It is given by

$$\begin{aligned} (-\varDelta _{DN})^\beta x(t)=\left( \int \limits _{\sigma (-\varDelta _{DN})}\lambda ^\beta F(d\lambda )x\right) (t)=\sum \limits _{j=1}^\infty \left( \left( j-\tfrac{1}{2}\right) ^2\right) ^\beta b_j\sqrt{\frac{2}{\pi }}\sin \left( j-\tfrac{1}{2}\right) t \end{aligned}$$

for $x\in D((-\varDelta _{DN})^\beta )$, where

$$\begin{aligned} D((-\varDelta _{DN})^\beta )=\Bigg \{&x(t)=\left( \int \limits _{\sigma (-\varDelta _{DN})}1 F(d\lambda )x\right) (t)=\sum \limits _{j=1}^\infty b_j\sqrt{\frac{2}{\pi }}\sin \left( j-\tfrac{1}{2}\right) t\in L^2;\\&\int \limits _{\sigma (-\varDelta _{DN})}\left| \lambda ^\beta \right| ^2 \Vert F(d\lambda )x\Vert _{L^2}^2=\sum \limits _{j=1}^\infty \left( \left( \left( j-\tfrac{1}{2}\right) ^2\right) ^\beta \right) ^2|b_j|^2<\infty \Bigg \} \end{aligned}$$

(here F is the spectral measure for the operator $-\varDelta _{DN}$ and $b_j\sqrt{\frac{2}{\pi }}\sin \left( j-\tfrac{1}{2}\right) t$ is the projection of x on the n-dimensional eigenspace ${\mathrm{Eig}}_j(-\varDelta _{DN})$).

Remark 1

To shorten the notation, in the rest of this paper the fractional Dirichlet (Dirichlet–Neumann) Laplace operator of order $\beta$ is denoted by $(-\varDelta _{1})^\beta$ $((-\varDelta _{2})^\beta )$.

Now, we formulate some useful facts concerning mentioned operators and their domains (cf. [19]).

Lemma 1

The spaces $D((-\varDelta _k)^\beta )$, $k=1,2$ are complete with the scalar products

$$\begin{aligned} \langle x,y\rangle _{k_\beta }=\langle x,y\rangle _{L^2}+\left\langle (-\varDelta _k)^\beta x,(-\varDelta _k)^\beta y\right\rangle _{L^2},\quad k=1,2. \end{aligned}$$

The above result follows from the fact that operators $(-\varDelta _k)^\beta )$, $k=1,2$ are self-adjoint, so also closed.

In our paper we shall use a scalar products given by

$$\begin{aligned} \langle x,y\rangle _{k_{\sim \beta }}=\left\langle (-\varDelta _k)^\beta x,(-\varDelta _k)^\beta y\right\rangle _{L^2},\quad k=1,2, \end{aligned}$$

(2)

which generate equivalent norms $\Vert \cdot \Vert _{k_\beta }$ and $\Vert \cdot \Vert _{k_{\sim \beta }}$ in $D((-\varDelta _k)^\beta )$ due to the following Poincaré inequalities:

$$\begin{aligned}&\Vert x\Vert ^2_{L^2}\le \Vert x\Vert ^2_{1_{\sim \beta }},\quad x\in D((-\varDelta _1)^\beta ) \end{aligned}$$

(3)

$$\begin{aligned}&\Vert x\Vert ^2_{L^2}\le 16^\beta \Vert x\Vert ^2_{2_{\sim \beta }},\quad x\in D((-\varDelta _2)^\beta ). \end{aligned}$$

(4)

The proof of (3) can be found in [19, formula (11)]. Analogously, we prove inequality (4):

$$\begin{aligned} \Vert x\Vert ^2_{L^2}=&\sum \limits _{j=1}^\infty |b_j|^2\le \sum \limits _{j=1}^\infty (((2j-1)^2)^\beta )^2 |b_j|^2\\ =&16^\beta \sum \limits _{j=1}^\infty (((j-\tfrac{1}{2})^2)^\beta )^2 |b_j|^2= 16^\beta \Vert x\Vert ^2_{2_{\sim \beta }},\quad x\in D((-\varDelta _{2})^\beta ). \end{aligned}$$

Lemma 2

If $\beta >\frac{1}{4}$ then

$$\begin{aligned} \Vert x\Vert _{L^\infty }\le&\sqrt{\tfrac{2}{\pi }\zeta (4\beta )}\Vert x\Vert _{1_{\sim \beta }},\quad x\in D((-\varDelta _1)^\beta ), \end{aligned}$$

(5)

$$\begin{aligned} \Vert x\Vert _{L^\infty }\le&4^\beta \sqrt{\tfrac{2}{\pi }\zeta (4\beta )}\Vert x\Vert _{2_{\sim \beta }},\quad x\in D((-\varDelta _{2})^\beta ), \end{aligned}$$

(6)

so embeddings

$$\begin{aligned} D((-\varDelta _k)^\beta )\subset L^\infty ,\quad k=1,2 \end{aligned}$$

are continuous (here $\zeta$ is the Riemann zeta function given by $\zeta (\gamma )=\sum \limits _{k=1}^\infty \frac{1}{k^\gamma }$).

Proof

For the convenience of the reader, we recall the proof of the inequality (6) which can be found in [21] (the proof of (5) for $n=1$ can be found in [19]).

Let $x\in D((-\varDelta _{2})^\beta )$. Then

$$\begin{aligned} |x(t)|^2&=\left| \sum \limits _{j=1}^\infty b_j\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t\right| ^2=\sum \limits _{i=1}^n\left| \sum \limits _{j=1}^\infty b^i_j\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t\right| ^2\\&\le \frac{2}{\pi }\sum \limits _{i=1}^n\left( \sum \limits _{j=1}^\infty |b^i_j|\right) ^2=\frac{2}{\pi }\sum \limits _{i=1}^n\left( \sum \limits _{j=1}^\infty \frac{((j-\tfrac{1}{2})^2)^\beta |b^i_j|}{((j-\tfrac{1}{2})^2)^\beta }\right) ^2\\&\le \frac{2}{\pi }\left( \sum \limits _{j=1}^\infty \frac{1}{(((j-\tfrac{1}{2})^2)^\beta )^2}\right) \sum \limits _{i=1}^n \left( \sum \limits _{j=1}^\infty (((j-\tfrac{1}{2})^2)^\beta )^2 |b^i_j|^2\right) \\&\le \frac{2}{\pi }\Vert x\Vert ^2_{2_{\sim \beta }}\left( \sum \limits _{j=1}^\infty \frac{1}{(((j-\tfrac{1}{2}j)^2)^\beta )^2}\right) =\frac{2}{\pi }\Vert x\Vert ^2_{2_{\sim \beta }}\left( \sum \limits _{j=1}^\infty \frac{1}{(((\tfrac{1}{2}j)^2)^\beta )^2}\right) \\&=(4^{\beta })^2\frac{2}{\pi }\Vert x\Vert ^2_{2_{\sim \beta }}\zeta (4\beta )<\infty ,\quad t\in (0,\pi )\,\,a.e. \end{aligned}$$

Hence, we obtain inequality (6).

The proof is completed. $\square$

Lemma 3

If $\beta >\frac{1}{2}$ then the operators

$$\begin{aligned}{}[(-\varDelta _{k})^\beta ]^{-1} : L^2 \ni g \rightarrow x_g \in L^2,\quad k=1,2 \end{aligned}$$

are compact.

Proof

The proof of this fact for $k=1$ (in the case of $n=1$) is given in [19, proof of Lemma 5.1]. It is analogous for vector valuable functions, so we present only the sketch of it in the case of $k=2$.

Let $F\in L^2((0,\pi ),{\mathbb {R}}^n)$ be any bounded (by a constant D) set in $L^2((0,\pi ),{\mathbb {R}}^n)$ and consider a function

$$\begin{aligned} f(t)=\sum \limits _{j=1}^\infty b_j^f\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t\in F. \end{aligned}$$

In the same way as in [19, Section 5.3] we can show that there exists a unique function

$$\begin{aligned} x_f(t)=\sum \limits _{j=1}^\infty c_j^f\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t \end{aligned}$$

such that

$$\begin{aligned} (-\varDelta _{2})^\beta x_f(t)=f(t). \end{aligned}$$

Consequently,

$$\begin{aligned} \sum \limits _{j=1}^\infty \left( (j-\tfrac{1}{2})^2\right) ^\beta c_j^f\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t=\sum \limits _{j=1}^\infty b_j^f\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t, \end{aligned}$$

so

$$\begin{aligned} c_j^f=\frac{b_j^f}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }. \end{aligned}$$

Let us consider the set of functions

$$\begin{aligned} {\mathcal {F}}:=\{{\tilde{x}}_f:\quad f\in F\}, \end{aligned}$$

where

$$\begin{aligned} {\tilde{x}}_f:(-\infty ,\infty )\ni t\rightarrow {\left\{ \begin{array}{ll} x_f(t);&{}t\in (0,\pi )\\ 0;&{} otherwise. \end{array}\right. } \end{aligned}$$

Then, for any fixed $h\in (0,\pi )$ we have

$$\begin{aligned} \int \limits _{-\infty }^\infty |{\tilde{x}}_f(t+h)-{\tilde{x}}_f(t)|^2dt=&\int \limits _0^h|x_f(t)|^2dt+\int \limits _{0}^{\pi -h}|x_f(t+h)-x_f(t)|^2dt+\int \limits _{\pi -h}^\pi |x_f(t)|^2dt\\ =&I_1+I_2+I_3. \end{aligned}$$

Using the Hölder inequality (for series) we obtain

$$\begin{aligned} I_1=&\int \limits _0^h\sum \limits _{i=1}^n\left| \sum \limits _{j=1}^\infty \tfrac{(b^i_j)^f}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }\sqrt{\frac{2}{\pi }}\sin (j-\tfrac{1}{2})t\right| ^2dt\le \frac{2}{\pi }\int \limits _0^h\sum \limits _{i=1}^n\left( \sum \limits _{j=1}^\infty \tfrac{|(b^i_j)^f|}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }\right) ^2dt\\ \le&\frac{2}{\pi }h\sum \limits _{j=1}^\infty |b_j^f|^2\sum \limits _{j=1}^\infty \tfrac{1}{\left( (j-\tfrac{1}{2})^2\right) ^{2\beta }}\le 16^\beta \frac{2}{\pi }h\Vert f\Vert ^2_{L^2}\zeta (4\beta )\le 16^\beta \frac{2}{\pi }D\zeta (4\beta )h. \end{aligned}$$

Similarly we estimate the term $I_3$. Now, we estimate the term $I_2$.

$$\begin{aligned} \int \limits _{0}^{\pi -h}|x_f(t+h)-&x_f(t)|^2dt=\int \limits _{0}^{\pi -h}\left| \sum \limits _{j=1}^\infty \tfrac{b_j^f}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }\sqrt{\tfrac{2}{\pi }}\left( \sin (j-\tfrac{1}{2})(t+h)-\sin (j-\tfrac{1}{2})t\right) \right| ^2dt\\ \le&\int \limits _{0}^{\pi -h}\sum \limits _{i=1}^n\left( \sum \limits _{j=1}^\infty \tfrac{|(b^i_j)^f|}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }\sqrt{\tfrac{2}{\pi }}\left| 2\sin \tfrac{(j-\tfrac{1}{2})h}{2}\cos \left( (j-\tfrac{1}{2})t+\tfrac{(j-\tfrac{1}{2})h}{2}\right) \right| \right) ^2dt\\\le&\frac{8}{\pi }\int \limits _{0}^{\pi -h}\sum \limits _{i=1}^n\left( \sum \limits _{j=1}^\infty \tfrac{|(b^i_j)^f|}{\left( (j-\tfrac{1}{2})^2\right) ^\beta }\left| \sin \tfrac{(j-\tfrac{1}{2})h}{2}\right| \right) ^2dt\\\le&\frac{8}{\pi }(\pi -h)\sum \limits _{j=1}^\infty |b_j^f|^2\sum \limits _{j=1}^\infty \tfrac{\sin ^2\tfrac{(j-\tfrac{1}{2})h}{2}}{\left( (j-\tfrac{1}{2})^2\right) ^{2\beta }}\le 8D\sum \limits _{j=1}^\infty \tfrac{(j-\tfrac{1}{2})h}{2\left( (j-\tfrac{1}{2})^2\right) ^{2\beta }}\\ {}&=4Dh\sum \limits _{j=1}^\infty \tfrac{1}{\left( j-\tfrac{1}{2}\right) ^{4\beta -1}}=2D16^\beta \zeta (4\beta -1)h. \end{aligned}$$

Analogously, we estimate terms $I_1, I_2, I_3$ for any fixed $h\in (-\pi ,0)$.

Finally,

$$\begin{aligned} \int \limits _{-\infty }^\infty |{\tilde{x}}_f(t+h)-{\tilde{x}}_f(t)|^2dt\le const|h|,\quad |h|\le \pi . \end{aligned}$$

Consequently,

$$\begin{aligned} \underset{|h|\rightarrow 0}{\lim }\Vert \tau _h{\tilde{x}}-{\tilde{x}}\Vert _{L^2({\mathbb {R}},{\mathbb {R}}^n)}=0\quad \mathrm {uniformly\,\,on}\,\,{\mathcal {F}}, \end{aligned}$$

where $\tau _h{\tilde{x}}(t)={\tilde{x}}(t+h)$, so the set

$$\begin{aligned} {\mathcal {F}}|_{(0,\pi )}=\{x_f:\quad f\in F\} \end{aligned}$$

is relatively compact in $L^2((0,\pi ),{\mathbb {R}}^n)$.

The proof is completed. $\square$

Remark 2

The relatively compactness of ${\mathcal {F}}$ follows from the following Kolmogorov-Fréchet-Riesz theorem (cf. [10, Theorem 4.26]):

Theorem 1

(Kolmogorov-Fréchet-Riesz) Let ${\mathcal {F}}$ be a bounded set in $L^p({\mathbb {R}}^N)$ with $1 \le p < \infty$. Assume that

$$\begin{aligned} \underset{|h|\rightarrow 0}{lim}\Vert \tau _hf-f\Vert _{L^p}=0\quad uniformly\,\,in\,\, f\in {\mathcal {F}}, \end{aligned}$$

i.e.

$$\begin{aligned} \forall _{\varepsilon>0}\,\, \exists _{\delta >0}\,\, such\,\,that\,\, \Vert \tau _hf-f\Vert _{L^p}<\varepsilon ,\quad \forall _{f\in {\mathcal {F}}},\,\, \forall _{h\in {\mathbb {R}}^N}\,\,with\,\, |h| < \delta . \end{aligned}$$

Then the closure of ${\mathcal {F}}_{|\varOmega }$ in $L^p({\mathbb {R}}^N)$ is compact for any measurable set $\varOmega \subset {\mathbb {R}}^N$ with finite measure.

(Here ${\mathcal {F}}_{|\varOmega }$ denotes the restrictions to of the functions in ${\mathcal {F}}$).

Using the above lemma and analogous arguments as in the proof of [19, Lemma 5.2] we obtain

Corollary 1

Let $k=1,2$ and $\beta >\frac{1}{2}$. If $x_n \rightharpoonup x_0$ weakly in $D((-\varDelta _{k})^\beta )$ then $x_n \rightarrow x_0$ strongly in $L^2$ and $(-\varDelta _{k})^\beta x_n \rightharpoonup (-\varDelta _{k})^\beta x_0$ weakly in $L^2$.

3 Existence and uniqueness of a solution to the control systems ($\hbox {E}_1$) and ($\hbox {E}_2$)

The main result of this section is a theorem on the existence of a unique solution to the control systems ($\hbox {E}_k$), $k=1,2.$ In the proof of this fact we use the following result.

Theorem 2

(Corollary 3.3, [18]) Let X be a real Banach space, Y a non-empty set, and H a real Hilbert space. If $F : X \times Y \rightarrow H$ is continuously differentiable with respect to $x\in X$ and

• for any $y\in Y$ the functional

  $$\begin{aligned} \phi _y:X\ni x\rightarrow \frac{1}{2}\Vert F(x,y)\Vert ^2\in {\mathbb {R}} \end{aligned}$$

  (7)

  satisfies the Palais-Smale (PS) condition¹,

• $F'_x(x,y):X\rightarrow Y$ is bijective for any $(x,y)\in X \times Y$ such that $F(x,y) = 0$ and

  $$\begin{aligned} F(x,y) \notin (Im F'_x(x, y))^{\perp } \end{aligned}$$

  (8)

  for any $(x,y) \in X \times Y$ such that $F(x,y)\ne 0$

then, for any $y\in Y$, there exists a unique $x_y\in X$ such that $F(x_y,y) = 0$.

In the rest of this paper we assume that $\beta >\frac{1}{2}$. Let us define the following set of controls:

$$\begin{aligned} {\mathcal {U}}_M:=\{u\in L^2((0,\pi ),{\mathbb {R}}^m);\quad u(t)\in M,\quad t\in (0,\pi )\}. \end{aligned}$$

We have

Theorem 3

Let us fix $k=1,2$. If

• (A1) f is measurable in $t\in (0,\pi )$, continuously differentiable in $x\in {\mathbb {R}}^n$ and

  $$\begin{aligned}&|f(t,x)|\le a(t)|x|+b(t),\quad t\in (0,\pi )\,\,a.e., x\in {\mathbb {R}}^n, \end{aligned}$$

  (9)
  $$\begin{aligned}&|f_x(t,x)|\le a(t)\delta (|x|),\quad t\in (0,\pi )\,\,a.e., x\in {\mathbb {R}}^n, \end{aligned}$$

  (10)

  where $\delta \in C({\mathbb {R}}_0^+,{\mathbb {R}}_0^+)$ and $a,b\in L^2((0,\pi ),{\mathbb {R}}^+)$ is such that

  $$\begin{aligned}&\sqrt{\frac{2}{\pi }\zeta (4\beta )}\Vert a\Vert _{L^2}<1 \qquad if\quad k=1, \end{aligned}$$

  (11)
  $$\begin{aligned}&\sqrt{\frac{2}{\pi }\zeta (4\beta )}\Vert a\Vert _{L^2}<\frac{1}{4^\beta } \qquad if\quad k=2, \end{aligned}$$

  (12)

• (A2) $B\in L^\infty ((0,\pi ),{\mathbb {R}}^{n\times m})$

• (A3) for any pair $(x,u)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ one of the following three conditions are satisfied

  • (a)

    $$\begin{aligned}&\Vert \varLambda \Vert _{L^1}\le \frac{\pi }{2\zeta (2\beta )} \qquad if\quad k=1, \end{aligned}$$

    (13)
    $$\begin{aligned}&\Vert \varLambda \Vert _{L^1}\le \frac{\pi }{2\zeta (2\beta )4^\beta } \qquad if\quad k=2, \end{aligned}$$

    (14)

  • (b) $\varLambda (t)\le 0,\quad t\in (0,\pi )\,\,a.e.,$

  • (c) $\varLambda \in L^\infty ((0,\pi ),{\mathbb {R}}^{n\times n})$ and $\Vert \varLambda \Vert _{L^\infty }<1$,

  where $\varLambda (\cdot )=f_x(\cdot ,x(\cdot ))$

then for any fixed control $u\in {\mathcal {U}}_M$ there exists a unique solution $x_u\in D((-\varDelta _k)^\beta )$ of the control system ($\hbox {E}_k$).

Proof

Let us fix $k=1,2$ and define the operator

$$\begin{aligned} F_k:D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M\ni (x,u)\rightarrow (-\varDelta _k)^\beta x(t)-f(t,x(t))-B(t)u(t)\in L^2. \end{aligned}$$

It is sufficient to show that $F_k$ satisfies all assumptions of Theorem 2.

• Using assumptions (A1), (A2) and analogous arguments as in [19, Proposition 5.1], we check that the mapping $F_k$ is continuously differentiable with respect to $x\in D((-\varDelta _k)^\beta )$ and the differential $(F_k)_x: D((-\varDelta _k)^\beta )\rightarrow L^2$ of $F_k$ at the point (x, u) is given by

  $$\begin{aligned} (F_k)_x(x,u)h=(-\varDelta _k)^\beta h(t)-f_x(t,x(t))h(t) \end{aligned}$$

  for any fixed $u\in {\mathcal {U}}_M$.

• (c) Now, we show that for any $u\in {\mathcal {U}}_M$ the functional

  $$\begin{aligned} \phi ^k_u:D((-\varDelta _k)^\beta )\ni x\rightarrow \frac{1}{2}\Vert F_k(x,u)\Vert ^2\in {\mathbb {R}} \end{aligned}$$

  satisfies the Palais-Smale condition. First, let us observe that the growth condition (9) and conditions (11), (12) guarantee coercivity of $\phi ^k_u$ for any $u\in {\mathcal {U}}_M$ (it is sufficient to use the same arguments as in the proof of [19, Lemma 5.3]). Moreover, it is continuously differentiable with respect to x and its differential $(\phi ^k_u)':D((-\varDelta _k)^\beta )\rightarrow {\mathbb {R}}$ is given by

  $$\begin{aligned} (\phi ^k_u)'(x)h=\int \limits _0^\pi \left\langle (-\varDelta _k)^\beta x(t)-f(t,x(t))-B(t)u(t), (-\varDelta _k)^\beta h(t)-f_x(t,x(t))h(t)\right\rangle dt \end{aligned}$$

  for any $h\in D((-\varDelta _k)^\beta )$. Let $x_0\in D((-\varDelta _k)^\beta )$ and $(x_l)_{l\in {\mathbb {N}}}\subset D((-\varDelta _k)^\beta )$. Then

  $$\begin{aligned} (\phi ^k_u)'(x_l)-(\phi ^k_u)'(x_0)(x_l-x_0)=\Vert x_l-x_0\Vert ^2_{\sim \frac{\beta }{2}}+\sum \limits _{i=1}^5\psi ^k_i(x_l), \end{aligned}$$

  where

  $$\begin{aligned} \psi ^k_1(x_l)= & {} \int \limits _0^\pi \left\langle (-\varDelta _k)^\beta x_l(t),f_x(t,x_l(t))(x_0(t)-x_l(t))\right\rangle dt,\\ \psi ^k_2(x_l)= & {} \int \limits _0^\pi \left\langle (-\varDelta _k)^\beta x_0(t),f_x(t,x_0(t))(x_l(t)-x_0(t))\right\rangle dt,\\ \psi ^k_3(x_l)= & {} \int \limits _0^\pi \left\langle f(t,x_l(t)),f_x(t,x_l(t))(x_l(t)-x_0(t))\right\rangle dt,\\ \psi ^k_4(x_l)= & {} \int \limits _0^\pi \left\langle f(t,x_0(t)),f_x(t,x_0(t))(x_0(t)-x_l(t))\right\rangle dt,\\ \psi ^k_5(x_l)= & {} \int \limits _0^\pi \left\langle f(t,x_0(t))-f(t,x_l(t)),(-\varDelta _k)^\beta x_l(t)-(-\varDelta _k)^\beta x_0(t)\right\rangle dt. \end{aligned}$$

  Using analogous arguments as in the proof of [19, Proposition 5.3] (including coercivity of $\phi ^k_u$, Corollary 1 and the Lebesque dominated convergence theorem) we conclude that there exists a subsequence $(x_{l_j})_{j\in {\mathbb {N}}}$ such that

  $$\begin{aligned} \psi ^k_i(x_{l_j})\underset{j\rightarrow \infty }{\longrightarrow } 0,\quad i=1,\dots ,5,\quad k=1,2. \end{aligned}$$

  This means that $x_{l_j}\longrightarrow x_0$ in $D((-\varDelta _k)^\beta )$, so for any $u\in {\mathcal {U}}_M$ the functional $\phi ^k_u$ satisfies the Palais-Smale condition.

• Analogously as in [19, Proposition 5.2] (using the assumption (A3)) we show that for any pair $(x,u)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ the differential $(F_k)_x$ is bijective².

The proof is completed. $\square$

We also have the following two results

Proposition 1

If M is a bounded set and assumptions (A1), (A2), (A3) of Theorem 3 are satisfied then there exists constants $C_1, C_2>0$ (inpedendent on u) such that for any control $u\in {\mathcal {U}}_M$

$$\begin{aligned} \Vert x_u\Vert _{k_{\sim \beta }}\le C_k,\quad k=1,2. \end{aligned}$$

(15)

Proof

Let us fix $k=1,2$ and any control $u\in {\mathcal {U}}_M$. Let C be a constant such that $|u(t)|\le C$ for a.e. $t\in (0,\pi )$. Assume that $x_u\in D((-\varDelta _k)^\beta )$ is a solution of the control system ($\hbox {E}_k$), corresponding to u. Then, using (9), we obtain

$$\begin{aligned} \Vert x_u\Vert _{k_{\sim \beta }}=&\Vert (-\varDelta _k)^\beta x_u\Vert _{L^2}=\left( \int \limits _0^\pi \left| f(t,x_u(t)) +B(t)u(t)\right| ^2dt\right) ^{\frac{1}{2}}\\ \le&\left( \int \limits _0^\pi |a(t)|^2|x_u(t)|^2dt\right) ^{\frac{1}{2}}+\left( \int \limits _0^\pi |B(t)|^2|u(t)|^2dt\right) ^{\frac{1}{2}}+\Vert b\Vert _{L^2}\\ \le&\Vert a\Vert _{L^2}\Vert x_u\Vert _{L^\infty }+C\sqrt{\pi }\Vert B\Vert _{L^\infty }+\Vert b\Vert _{L^2}. \end{aligned}$$

Thus and from Lemma 2 we have

$$\begin{aligned} \Vert x_u\Vert _{k_{\sim \beta }}\le {\left\{ \begin{array}{ll} \sqrt{\frac{2}{\pi }\zeta (4\beta )}\Vert a\Vert _{L^2}\Vert x_u\Vert _{k_{\sim \beta }}+C\sqrt{\pi }\Vert B\Vert _{L^\infty }+\Vert b\Vert _{L^2}&{} if\quad k=1\\ 4^\beta \sqrt{\frac{2}{\pi }\zeta (4\beta )}\Vert a\Vert _{L^2}\Vert x_u\Vert _{k_{\sim \beta }}+C\sqrt{\pi }\Vert B\Vert _{L^\infty }+\Vert b\Vert _{L^2}&{} if\quad k=2. \end{array}\right. } \end{aligned}$$

This means that

$$\begin{aligned} \Vert x_u\Vert _{k_{\sim \beta }}\le C_k, \end{aligned}$$

where

$$\begin{aligned} C_k={\left\{ \begin{array}{ll} \frac{\sqrt{\pi }C\Vert B\Vert _{L^\infty +\Vert b\Vert _{L^2}}}{1-\sqrt{ \frac{2}{\pi }\zeta (4\beta )\Vert a\Vert _{L^2}}}&{} if\quad k=1\\ \frac{\sqrt{\pi }C\Vert B\Vert _{L^\infty }+\Vert b\Vert _{L^2}}{1-4^\beta \sqrt{ \frac{2}{\pi }\zeta (4\beta )\Vert a\Vert _{L^2}}}&{} if\quad k=2. \end{array}\right. } \end{aligned}$$

The proof is completed. $\square$

Proposition 2

Let us fix $k=1,2$. Assume that all assumptions of Theorem 3 are satisfied and the set M is convex and compact. If $(u_l)_{l\in {\mathbb {N}}}\subset {\mathcal {U}}_M$ is a sequence of controls and $(x_l)_{l\in {\mathbb {N}}}\subset D((-\varDelta _k)^\beta )$ is a sequence of corresponding solutions of the control system $(E_k)$ then there exist a control $u_0\in {\mathcal {U}}_M$, a function $x_0\in D((-\varDelta _k)^\beta )$ and a subsequence $(l_i)_{i\in {\mathbb {N}}}$ such that the pair $(x_0,u_0)$ satisfies $(E_k)$ and

• (Z1) $x_{l_i}\underset{i\rightarrow \infty }{\longrightarrow }x_0$ strongly in $L^2$,

• (Z2) $(-\varDelta _{k})^\beta x_{l_i} \rightharpoonup (-\varDelta _{k})^\beta x_0$ weakly in $L^2$,

• (Z3) $u_{l_i}\underset{i\rightarrow \infty }{\rightharpoonup }u_0$ weakly in $L^2((0,\pi ),{\mathbb {R}}^m)$.

Proof

Let us fix $k=1,2$ and consider a sequence of controls $(u_l)_{l\in {\mathbb {N}}}\in {\mathcal {U}}_M$ and a sequence of corresponding solutions $(x_l)_{l\in {\mathbb {N}}}\subset D((-\varDelta _k)^\beta )$ of the system $(E_k)$. Using the standard arguments we check that compactness and convexity of the set M imply a convexity, boundedness and closure of the set ${\mathcal {U}}_M$ in $L^2((0,\pi ),{\mathbb {R}}^m)$. This means that ${\mathcal {U}}_M$ is sequentially weakly compact, while $L^2((0,\pi ),{\mathbb {R}}^m)$ is a reflexive space. Consequently, there exist a subsequence $(u_{l_i})_{i\in {\mathbb {N}}}$ and $u_0\in {\mathcal {U}}_M$ such that

$$\begin{aligned} u_{l_i}\underset{i\rightarrow \infty }{\rightharpoonup }u_0\quad {\mathrm{weakly\,\, in}}\,\, L^2((0,\pi ),{\mathbb {R}}^m), \end{aligned}$$

so the condition (Z3) of this proposition is satisfied.

From Proposition 1 it follows that the sequence of norms $\Vert x_l\Vert _{k_{\sim \beta }}$ is bounded, so, there exist a subsequence $(x_{l_i})_{i\in {\mathbb {N}}}$ and a function $x_0\in D((-\varDelta _k)^\beta )$ such that

$$\begin{aligned} x_{l_i}\underset{i\rightarrow \infty }{\rightharpoonup }x_0\quad {\mathrm{weakly\,\, in}}\,\, D((-\varDelta _k)^\beta ). \end{aligned}$$

Consequently, Corollary 1 implies convergences (Z1) and (Z2).

Now, we show that the $x_0$ is a solution of $(E_k)$, corresponding to $u_0$. Indeed, first we note that since the matrix B is essentially bounded on $(0,\pi )$, therefore

$$\begin{aligned} Bu_{l_i}\underset{i\rightarrow \infty }{\rightharpoonup }Bu_0 \quad {\mathrm{weakly \,\,in}}\,\, L^2. \end{aligned}$$

Moreover, using condition (9) and Lemma 2 we have

$$\begin{aligned} \left| f(t, x_{l_i}(t))-f(t, x_{0}(t))\right| ^2\le&2\left( |f(t, x_{l_i}(t))|^2+|f(t, x_{0}(t))|^2\right) \\ \le&2a^2(t)(\Vert x_{l_i}\Vert _{L^\infty }^2+|x_0(t)|^2)\\ \le&{\left\{ \begin{array}{ll} 2a^2(t)\left( \tfrac{2}{\pi }\zeta (4\beta )\Vert x_{l_i}\Vert ^2_{k_{\sim \beta }}+|x_0(t)|^2\right) &{}if \,\,k=1\\ 2a^2(t)\left( 16^\beta \tfrac{2}{\pi }\zeta (4\beta )\Vert x_{l_i}\Vert ^2_{k_{\sim \beta }}+|x_0(t)|^2\right) &{}if \,\,k=2 \end{array}\right. }\\ \le&{\left\{ \begin{array}{ll} 2a^2(t)\left( \tfrac{2}{\pi }\zeta (4\beta )C_k^2+|x_0(t)|^2\right) &{}if \,\,k=1\\ 2a^2(t)\left( 16^\beta \tfrac{2}{\pi }\zeta (4\beta )C_k^2+|x_0(t)|^2\right) &{}if \,\,k=2, \end{array}\right. } \end{aligned}$$

where $C_k$, $k=1,2$ are constants from Proposition 1. Consequently, from the Lebesque dominated convergence theorem it follows that

$$\begin{aligned} f(\cdot ,x_{l_i}(\cdot ))\underset{i\rightarrow \infty }{\longrightarrow }f(\cdot ,x_0(\cdot ))\quad {\mathrm{strongly \,\,in}}\,\, L^2. \end{aligned}$$

Then, of course

$$\begin{aligned} f(\cdot ,x_{l_i}(\cdot ))\underset{i\rightarrow \infty }{\rightharpoonup }f(\cdot ,x_0(\cdot ))\quad {\mathrm{weakly \,\,in}}\,\, L^2. \end{aligned}$$

Thus, using (Z2) we get

$$\begin{aligned} (-\varDelta _k)^\beta x_{l_i}(\cdot )-f(\cdot , x_{l_i}(\cdot ))-B(\cdot )u_{l_i}(\cdot ) \underset{k\rightarrow \infty }{\rightharpoonup }(-\varDelta _k)^\beta x_{0}(\cdot )-f(\cdot , x_{0}(\cdot ))-B(\cdot )u_{0}(\cdot ) \end{aligned}$$

weakly in $L^2((0,\pi ),{\mathbb {R}}^n)$. On the other hand, $(x_{l_i})$ is a solution of $(E_k)$, corresponding to $(u_{l_i})$, so we have

$$\begin{aligned} (-\varDelta _k)^\beta x_{l_i}(t)-f(t, x_{l_i}(t))-B(t)u_{l_i}(t)=0,\quad t\in (0,\pi )\,\,a.e. \end{aligned}$$

This means that

$$\begin{aligned} (-\varDelta _k)^\beta x_{0}(t)-f(t, x_{0}(t))-B(t)u_{0}(t)=0,\quad t\in (0,\pi )\,\,a.e. \end{aligned}$$

The proof is completed. $\square$

4 Existence of optimal solutions

In this section we shall prove the main result of this paper, namely a theorem on the existence of optimal solutions of the problems ($\hbox {OCP}_k$), $k=1,2$.

Let us fix $k=1,2$. We shall say, that a pair $(x_*,u_*)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ is a globally optimal solution of the problem ($\hbox {OCP}_k$), if $x_*$ is the solution of the control system $(E_k)$, corresponding to the control $u_*$ and

$$\begin{aligned} J(x_*,u_*)\leqslant J(x,u) \end{aligned}$$

for every pair $(x,u)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ satisfying $(E_k)$.

We have

Theorem 4

Let us fix $k=1,2$ and assume that

• 1. M is convex and compact,

• 2. hypothesis (A1), (A2) and (A3) of Theorem 3 are satisfied,

• 3. $f_0(\cdot ,x,u)$ is measurable on $(0,\pi )$ for all $x\in {\mathbb {R}}^n$ and $u\in M$,

• 4. $f_0(t,\cdot ,\cdot )$ is continuous on ${\mathbb {R}}^n\times M$ for a.e. $t\in (0,\pi )$,

• 5. $f_0(t,x,\cdot )$ is convex on M for a.e. $t\in (0,\pi )$ and all $x\in {\mathbb {R}}^n$,

• 6. there exist a summable function $\psi :(0,\pi )\rightarrow {\mathbb {R}}^+_0$ and a constant $c\geqslant 0$ such that

  $$\begin{aligned} f_0(t,x,u)\geqslant -\psi (t)-c|x| \end{aligned}$$

  (16)

  for a.e. $t\in (0,\pi )$ and all $x\in {\mathbb {R}}^n$, $u\in M$.

Then the problem ($\hbox {OCP}_k$) possesses an optimal solution $(x_0,u_0)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_{M}$.

Proof

Let us fix $k=1,2$ and denote

$$\begin{aligned} \mu =\inf \big \{J(x_u,u),\quad u\in {\mathcal {U}}_M\big \}. \end{aligned}$$

(17)

It is clear that $\mu \leqslant J(x_u,u)$ for any pair $(x_u,u)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_{M}$. The condition (16), the Hölder inequality, Poincaré inequalities (3), (4) and Proposition 1 imply

$$\begin{aligned} J(x_u,u)&=\int \limits _0^\pi f_0(t,x_u(t),u(t))dt\geqslant -\int \limits _0^\pi \psi (t)dt-c\int \limits _0^\pi |x_u(t)|dt\\&\geqslant -\int \limits _0^\pi \psi (t)dt-c\sqrt{\pi }\Vert x_u\Vert _{L^2}\\&\geqslant -\int \limits _0^\pi \psi (t)dt- {\left\{ \begin{array}{ll} c\sqrt{\pi }\Vert x_u\Vert _{k_{\sim \beta }}&{}\quad if\,\,k=1\\ c\sqrt{\pi }4^\beta \Vert x_u\Vert _{k_{\sim \beta }}&{}\quad if\,\,k=2 \end{array}\right. }\\ {}&\geqslant -\int \limits _0^\pi \psi (t)dt- {\left\{ \begin{array}{ll} c\sqrt{\pi }C_k&{}\quad if\,\,k=1\\ c\sqrt{\pi }4^\beta C_k&{}\quad if\,\,k=2 \end{array}\right. } \,\, >-\infty , \end{aligned}$$

where $C_1, C_2$ are constants from Proposition 1. This means that $-\infty <\mu \leqslant +\infty$.

If $\mu =+\infty$ then the existence of optimal solutions is obvious.

So, let us assume that $-\infty<\mu <+\infty$ and $(x_l,u_l)_{l\in {\mathbb {N}}}\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ be a minimizing sequence of the functional J. This means that

$$\begin{aligned} \underset{l\rightarrow \infty }{\lim }J(x_l,u_l)=\mu . \end{aligned}$$

From Proposition 2 it follows that there exist a pair $(x_0,u_0)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_M$ and a subsequence $(l_i)_{i\in {\mathbb {N}}}$ such that the pair $(x_0,u_0)$ satisfies $(E_k)$ and

$$\begin{aligned}&u_{l_i}\underset{i\rightarrow \infty }{\rightharpoonup }u_0\,\, {\mathrm{weakly \,\,in}}\,\, L^2((0,\pi ),{\mathbb {R}}^m),\\&x_{l_i}\underset{i\rightarrow \infty }{\rightarrow }x_0\,\, {\mathrm{strongly \,\,in}}\,\, L^2. \end{aligned}$$

Thus, we obtain respective convergences in $L^1((0,\pi ),{\mathbb {R}}^m)$ and $L^1((0,\pi ),{\mathbb {R}}^n)$, respectively.

Now, let us consider a function ${\hat{f}}_0:(0,\pi )\times {\mathbb {R}}^n\times M\rightarrow {\mathbb {R}}$ given by

$$\begin{aligned} {\hat{f}}_0(t,x,u):= {\left\{ \begin{array}{ll} f_0(t,x,u);\qquad \,\,\, t\in T\\ -\psi (t)-c|x|;\quad t\notin T, \end{array}\right. } \end{aligned}$$

where $T\subset (0,\pi )$ is a set of the full measure consist of points, for which conditions 4, 5, 6 are satisfied. Then the function ${\hat{f}}_0$ satisfies mentioned conditions for all $t\in (0,\pi )$. From [22, Proposition 3.2] and [24, section IV § 3 Theorem 6] it follows that ${\hat{f}}_0$ is ${\mathcal {L}}((0,\pi ))\times {\mathcal {B}}({\mathbb {R}}^n\times M)$ measurable. Moreover, it can be extended to the function ${\tilde{f}}_0:(0,\pi )\times {\mathbb {R}}^n\times {\mathbb {R}}^m\rightarrow {\mathbb {R}}$ given by

$$\begin{aligned} {\tilde{f}}_0(t,x,u):= {\left\{ \begin{array}{ll} {\hat{f}}_0(t,x,u);\quad \, (t,x,y)\in (0,\pi )\times {\mathbb {R}}^n\times M\\ +\infty ;\qquad \quad (t,x,y)\notin (0,\pi )\times {\mathbb {R}}^n\times M. \end{array}\right. } \end{aligned}$$

[20, Lemma 16] guarantees ${\mathcal {L}}((0,\pi ))\times {\mathcal {B}}({\mathbb {R}}^n\times {\mathbb {R}}^m)$ - measurability of ${\tilde{f}}_0$. This function is also lower semicontinuous with respect to $(x,u)\in {\mathbb {R}}^n\times {\mathbb {R}}^m$ for any fixed $t\in (0,\pi )$, convex with respect to $u\in {\mathbb {R}}^m$ for any fixed $(t,x)\in (0,\pi )\times {\mathbb {R}}^n$ and satisfies inequality (16) for all points $(t,x,u)\in (0,\pi )\times {\mathbb {R}}^n\times {\mathbb {R}}^m$. Consequently, using a theorem on the $L^1$-weak lower semicontinuity of integral functionals (cf. [26]) we assert that

$$\begin{aligned} {\tilde{J}}(x_0,u_0)\leqslant \underset{i\rightarrow \infty }{\liminf }{\tilde{J}}(x_{l_i},u_{l_i}), \end{aligned}$$

where

$$\begin{aligned} {\tilde{J}}(x,u)=\int \limits _0^\pi {\tilde{f}}_0(t,x(t),u(t))dt. \end{aligned}$$

Thus, since

$$\begin{aligned} J(x_0,u_0)={\hat{J}}(x_0,u_0)={\tilde{J}}(x_0,u_0) \quad and\quad J(x_{l_i},u_{l_i})={\hat{J}}(x_{l_i},u_{l_i})={\tilde{J}}(x_{l_i},u_{l_i}), \end{aligned}$$

where

$$\begin{aligned} {\hat{J}}(x,u)=\int \limits _a^b{\hat{f}}_0(t,x(t),u(t))dt, \end{aligned}$$

therefore

$$\begin{aligned} J(x_0,u_0)\leqslant \underset{i\rightarrow \infty }{\liminf }J(x_{l_i},u_{l_i}). \end{aligned}$$

Hence

$$\begin{aligned} \mu \leqslant J(x_0,u_0)\leqslant \underset{i\rightarrow \infty }{\liminf }J(x_{l_i},u_{l_i})=\underset{i\rightarrow \infty }{\lim }J(x_{l_i},u_{l_i})=\mu , \end{aligned}$$

so

$$\begin{aligned} J(x_0,u_0)=\mu =\inf \big \{J(x_u,u),\quad u\in {\mathcal {U}}_M\big \}. \end{aligned}$$

The proof is completed. $\square$

5 Illustrative example

In this section we present the the following theoretical problems

$$\begin{aligned} {\left\{ \begin{array}{ll} {\left\{ \begin{array}{ll} (-\varDelta _k)^\beta x_1(t)=a\sin t(\sin x_1(t)+\sin x_2(t))+{\mathrm{e}}^tu(t),&{} \\ (-\varDelta _k)^\beta x_2(t)=a\sin t(\sin x_1(t)-\sin x_2(t))+t^2u(t), \end{array}\right. } t\in (0,\pi )\,\, {\mathrm{a.e.}\quad ({\mathrm{Equ_k}})}\\ u(t)\in [-1,1],\quad t\in (0,\pi ),\\ J(x,u)=\int \limits _0^\pi (\sin t+\cos t|\sin x_1(t)|+\sin t|x_2(t)|+u^2(t))dt\rightarrow \min , \end{array}\right. } \end{aligned}$$

(18)

where $k=1,2$, $\beta >\frac{1}{2}$ and $a>0$.

We see that $B:(0,\pi )\rightarrow {\mathbb {R}}^{2\times 1}$ and

$$\begin{aligned} B(t)= \left[ \begin{array}{c} {\mathrm{e}}^t\\ t^2 \end{array} \right] , \end{aligned}$$

$f:(0,\pi )\times {\mathbb {R}}^2\rightarrow {\mathbb {R}}^2$ and

$$\begin{aligned} f(t,x_1,x_2)=\left( a\sin t(\sin x_1+\sin x_2), a\sin t(\sin x_1-\sin x_2)\right) , \end{aligned}$$

$f_0:(0,\pi )\times {\mathbb {R}}^2\times [-1,1]\rightarrow {\mathbb {R}}$ and

$$\begin{aligned} f_0(t,x_1,x_2,u)=\sin t+\cos t|\sin x_1|+\sin t|x_2|+u^2. \end{aligned}$$

It is clear that f is measurable with respect to t, continuously differentiable on ${\mathbb {R}}^2$ and³,

$$\begin{aligned}&|f(t,x)|=|f(t,(x_1,x_2))|=\sqrt{2}a|\sin t|\sqrt{\sin ^2 x_1+\sin ^2 x_2}\le 2a+a|x|,\\&|f_x(t,x)|_{2\times 2}=|f_x(t,(x_1,x_2))|_{2\times 2}=\sqrt{2}a|\sin t|\sqrt{\cos ^2 x_1+\cos ^2 x_2}\le 2a \end{aligned}$$

for a.e. $t\in (0,\pi )$ and all $x\in {\mathbb {R}}^2$.

Consequently, conditions (9), (10) are satisfied with $a(t):= a$, $b(t):=2a$ and $\delta (s):=2$. Let us note that conditions (11), (12) hold if

$$\begin{aligned} a< {\left\{ \begin{array}{ll} \frac{1}{\sqrt{2\zeta (4\beta )}},&{}if\quad k=1\\ \frac{1}{4^\beta \sqrt{2\zeta (4\beta )}},&{}if\quad k=2. \end{array}\right. } \end{aligned}$$

Moreover, if

$$\begin{aligned} a\le {\left\{ \begin{array}{ll} \frac{1}{4\zeta (2\beta )},&{}if\quad k=1\\ \frac{1}{4^{\beta +1} \zeta (2\beta )},&{}if\quad k=2 \end{array}\right. } \end{aligned}$$

then conditions (13), (14) are satisfied.

Of course, the function $f_0$ satisfies assumptions 3,4,5 of Theorem 4. The assumption 6 also holds because

$$\begin{aligned} f_0(t,x,u)=f_0(t,(x_1,x_2),u)\ge -1-|x_1|-|x_2|\ge -1-2|x| \end{aligned}$$

for a.e. $t\in (0,\pi )$ and all $x\in {\mathbb {R}}^2$, $u\in [-1,1]$.

Consequently, we proved the following

Theorem 5

If

$$\begin{aligned} a< {\left\{ \begin{array}{ll} \min \left\{ \frac{1}{\sqrt{2\zeta (4\beta )}},\frac{1}{4\zeta (2\beta )}\right\} ,&{}if\quad k=1\\ \min \left\{ \frac{1}{4^\beta \sqrt{2\zeta (4\beta )}},\frac{1}{4^{\beta +1}\zeta (2\beta )}\right\} ,&{}if\quad k=2 \end{array}\right. }={\left\{ \begin{array}{ll} \frac{1}{4\zeta (2\beta )},&{}if\quad k=1\\ \frac{1}{4^{\beta +1} \zeta (2\beta )},&{}if\quad k=2 \end{array}\right. } \end{aligned}$$

then problems (18) have optimal solutions $((x_1,x_2),u)\in D((-\varDelta _k)^\beta )\times {\mathcal {U}}_{[-1,1]}$, $k=1,2$.

Basics of self-adjoint operators in a Hilbert space

In this section we give the necessary notions and facts concerning a theory of unbounded self-adjoint operators in a real Hilbert space (cf. [19]). More details can be found in [1, 25], where all results are obtained in the case of a complex Hilbert space. Nevertheless, their proofs can be reproduced (if required, with small changes) in the case of a real Hilbert space.

So, in this section we shall assume that H is a real Hilbert space with a scalar product $\langle \cdot ,\cdot \rangle _H$.

1.1 Self-adjoint operator

Let $T:D(T)\subset H\rightarrow H$ be a densely defined linear operator ($\overline{D(T)}=H$) with the domain D(T). We define

$$\begin{aligned} D(T^*):=\{x\in H; \quad \exists _{z\in H}\quad \langle x,Ty\rangle _H=\langle z,y\rangle _H,\,\,for\,\,all\,\,y\in D(T)\}. \end{aligned}$$

(19)

For $x\in D(T^*)$ we denote $T^*x=z$ (this element is uniquely determined due to the density of D(T)). The operator $T^*:D(T^*)\subset H\rightarrow H$ is called the adjoint operator to T. If $T=T^*$ and $D(T)=D(T^*)$, then T is called self-adjoint. We note that whenever T is self-adjoint operator one has

$$\begin{aligned} \langle Tx,y\rangle _H=\langle x,Ty\rangle _H,\,\,for\,\,all\,\,x,y\in D(T). \end{aligned}$$

1.2 Spectral integral and decomposition theorem

Let ${\mathcal {B}}$ be a $\sigma$ - algebra of Borel subsets of ${\mathbb {R}}$ and ${\mathcal {P}}(H)$ denotes the set of all orthogonal projection operators onto closed linear subspaces of H. A set function $E:{\mathcal {B}}\rightarrow {\mathcal {P}}(H)$ is called a spectral measure (or a decomposition of the identity) if

• 1. for any $x\in H$ the set function ${\mathcal {B}}\ni P\rightarrow E(P)x$ is $\sigma$ - additive,

• 2. $E({\mathbb {R}})=I$ (here I denotes the identity operator on H),

• 3. $E(P\cap Q)=E(P)\circ E(Q)$, for $P,Q\in {\mathcal {B}}$.

Let W be the union of all open sets $V\subset {\mathbb {R}}$ such that $E(V)=0$. Then the complement ${\mathbb {R}}{\setminus } W$ is called a support of a spectral measure E and denoted by supp(E).

Let assume that $u:{\mathbb {R}}\rightarrow {\mathbb {R}}$ defined a.e. E is a bounded Borel measurable function. Then in the usual way one can show (via a sequence of simple functions) that for any $x\in H$ there exists the integral (with respect to the vector measure $E(\cdot )x$)

$$\begin{aligned} \int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )x. \end{aligned}$$

We define the integral with respect to the spectral measure E

$$\begin{aligned} \int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda ):H\rightarrow H \end{aligned}$$

in the following way

$$\begin{aligned} \left( \int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )\right) x=\int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )x. \end{aligned}$$

One proves that the above operator is linear, continuous and Hermitian.

Now, let $u:{\mathbb {R}}\rightarrow {\mathbb {R}}$ defined a.e. E be an unbounded Borel measurable function. Let us define the sequence of functions $u_n$:

$$\begin{aligned} u_n(\lambda )= {\left\{ \begin{array}{ll} u(\lambda ),&{} {if} \quad |u(\lambda )|\le n\\ 0, &{}{if}\quad |u(\lambda )|> n. \end{array}\right. } \end{aligned}$$

Functions $u_n$ are Borel measurable and bounded. Consequently, there exist integrals

$$\begin{aligned} \int \limits _{-\infty }^{+\infty }f_n(\lambda )E(d\lambda ),\quad n\in {\mathbb {N}}. \end{aligned}$$

Let us consider the set

$$\begin{aligned} D=\left\{ x\in H:\quad \int \limits _{-\infty }^{+\infty }|u(\lambda )|^2\Vert E(d\lambda )x\Vert _H^2<\infty \right\} . \end{aligned}$$

(20)

One can show that D is a dense linear subspace of H and for $x\in D$ there exists the limit

$$\begin{aligned} \underset{n\rightarrow \infty }{\lim }\int \limits _{-\infty }^{+\infty }u_n(\lambda )E(d\lambda )x. \end{aligned}$$

So, we can define the operator $\int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda ):D\subset H\rightarrow H$ in the following way:

$$\begin{aligned} \left( \int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )\right) x=\underset{n\rightarrow \infty }{\lim }\int \limits _{-\infty }^{+\infty }u_n(\lambda )E(d\lambda )x. \end{aligned}$$

One can prove that

$$\begin{aligned} \left\| \int \limits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )x\right\| _H^2=\int \limits _{-\infty }^{+\infty }|u(\lambda )|^2\Vert E(d\lambda )x\Vert _H^2 \end{aligned}$$

(21)

and the operator $\int \nolimits _{-\infty }^{+\infty }u(\lambda )E(d\lambda )$ is self-adjoint.

If $u:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a Borel measurable function and $\omega \in {\mathcal {B}}$ then

$$\begin{aligned} \int \limits _\omega u(\lambda )E(d\lambda ):=\int \limits _{-\infty }^{+\infty }\chi _\omega (\lambda )u(\lambda )E(d\lambda ), \end{aligned}$$

where $\chi _\omega$ is a characteristic function of the set $\omega$.

In order to define the spectral integral in the case of a Borel measurable function $u:P\rightarrow {\mathbb {R}}$, where $P\in {\mathcal {B}}$ contains the support supp(E), it is sufficient to extend u on ${\mathbb {R}}$ to any Borel measurable function.

Now, we formulate a spectral decomposition theorem which plays a crucial role in the spectral theory of self-adjoint operators.

Theorem 6

(Spectral decomposition theorem for self-adjoint operators) Let $T:D(T)\subset H\rightarrow H$ be a self-adjoint operator such that the resolvent set $\rho (T)$ is non-empty. Then there exists a unique spectral measure E with the closed support $supp(E)=\sigma (T)$ such that

$$\begin{aligned} T=\int \limits _{-\infty }^{+\infty }\lambda E(d\lambda )=\int \limits _{\sigma (T)}\lambda E(d\lambda ). \end{aligned}$$

(22)

In conclusion of this section we shall define a function of a self-adjoint operator. Let $T:D(T)\subset H\rightarrow H$ be a self-adjoint operator with $\rho (T)\ne \emptyset$. From Theorem 6 it follows that T has the integral representation given by (22). For a Borel measurable function $u:{\mathbb {R}}\rightarrow {\mathbb {R}}$ defined a.e. E we define the operator u(T) as follows

$$\begin{aligned} u(T)=\int \limits _{-\infty }^{+\infty }u(\lambda ) E(d\lambda )=\int \limits _{\sigma (T)}u(\lambda ) E(d\lambda ). \end{aligned}$$

(23)

According to general properties of the spectral integrals presented above, the domain D(u(T)) is given by (20), the equality (21) holds and u(T) is self-adjoint. Moreover, its spectrum is given by

$$\begin{aligned} \sigma (u(T))=\overline{u(\sigma (T))}, \end{aligned}$$

provided that u is continuous on $\sigma (T)$.