Proof of Theorem 1. The theorem is proved
by a straightforward substitution of the Cauchy formula for system (2.1),
$$ x(t)=F(t,t^{1} )x^{1} +\int \limits _{t^{1} }^{t}F(t,\tau
) A_{2} (\tau )u(\tau )d\tau ,$$
(A.1)
into conditions
(
2.2). After easy transformations, by arranging the
terms, we obtain the algebraic system
$$ \begin {gathered} Lx^{1} =D, \\ L=\sum _{i=1}^{l_{1}
}\alpha _{i} F(\tilde {t}^{i} ,t^{1} )+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat
{t}^{2j-1} }^{\hat {t}^{2j} }\beta _{j} (t )F(t,t^{1} )dt , \\ {D=\vartheta -\sum
_{i=1}^{l_{1} }\alpha _{i}\; \int \limits _{t^{1} }^{\tilde {t}^{i} }F(\tilde
{t}^{i} ,\tau )A_{2} (\tau )u(\tau )d\tau -\sum _{j=1}^{l_{2} }\;\int \limits
_{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta _{j} (t )\int \limits _{t^{1} }^{t}A_{2}
(\tau )u(\tau )d\tau dt} \end {gathered}$$
(A.2)
for
\(x^{1}=x(t^{1}) \).
It is well known that system (A.2)
has a unique solution provided that the matrix \(L\) is invertible, i.e.,
under condition (3.1). It is clear that
\(\mathrm {rank}\,L\) is independent of the values of the
vector \(D \) and hence of the vector function \(u(t) \) and the vector \(\vartheta \). At the same time, owing to the uniqueness of the
representation (A.1) for the solution of the
Cauchy problem for system (2.1),
problem (2.1), (2.2) has a unique solution under condition (3.1). This completes the proof of Theorem 1.\(\quad \blacksquare \)
Proof of Theorem 2. Let \((u^{1} ,\vartheta ^{1} ) \) and \((u^{2} ,\vartheta ^{2}) \) be two pairs of arbitrary feasible control and
parameters, and let \(x^{1} (t)\) and
\(x^{2} (t) \) be the respective solutions of the boundary value
problem (2.1), (2.2). Then
$$ \dot {x}^{1} (t)=A_{1} (t)x^{1} (t)+A_{2} (t)u^{1}
(t),\quad t\in [t^{1} ,t^{f} ],$$
(A.3)
$$ \sum _{i=1}^{l_{1} }\alpha _{i} x^{1} (\tilde {t}_{i}
)+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta
_{j} (t )x^{1} (t)dt =\vartheta ^{1} ,$$
(A.4)
$$ \dot {x}^{2} (t)=A_{1} (t)x^{2} (t)+A_{2} (t)u^{2}
(t),\quad t\in [t^{1} ,t^{f} ],$$
(A.5)
$$ \sum _{i=1}^{l_{1} }\alpha _{i} x^{2} (\tilde {t}_{i}
)+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta
_{j} (t )x^{2} (t)dt =\vartheta ^{2} .$$
(A.6)
By virtue of the convexity of the feasible sets \(U \) and \(V \), for an arbitrary \(\sigma \in [0; 1] \) one has
$$ u(t)=\sigma u^{1} (t)+(1-\sigma )u^{2} (t)\in U,\quad
\vartheta =\sigma \vartheta ^{1} +(1-\sigma )\vartheta ^{2} \in V.
$$
(A.7)
Set \(x(t)=\sigma x^{1} (t)+(1-\sigma )x^{2} (t) \).
We multiply both sides of (A.3)
by \(\sigma \) and of (A.5) by \((1-\sigma )\), add the
resulting equalities term by term, and arrange the terms to obtain
$$ \sigma \dot {x}^{1}
(t)+(1-\sigma ) \dot {x}^{2} (t) =A_{1} (t)\left [\sigma x^{1} (t)+(1-\sigma
)x^{2} (t)\right ]+A_{2} (t)\left [\sigma u^{1} (t)+(1-\sigma )u^{2} (t)\right ].
$$
It follows that
\(x(t) \) satisfies the system of differential equations
(
2.1).
We multiply both sides of (A.4)
by \(\sigma \) and of (A.6) by \((1-\sigma )\), add the
resulting equalities, and arrange terms to obtain
$$ \sum _{i=1}^{l_{1} }\alpha _{i} \left [\sigma x^{1}
(\tilde {t}^{i} )+(1-\sigma )x^{2} (\tilde {t}^{i} )\right ]+\sum _{j=1}^{l_{2}
}\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta _{j} (t)\left [\sigma
x^{1} (t)+(1-\sigma )x^{2} (t)\right ] dt =\sigma \vartheta ^{1} +(1-\sigma )
\vartheta ^{2} .$$
In view of notation (
A.7), this implies that the triple
\((x(t),u(t),\vartheta ) \) satisfies conditions (
2.2).
By virtue of the convexity of the functions \(f^{0} (x(t),u(t),\vartheta ,t) \) and \(\Phi (\tilde {x},\hat {x},\vartheta )\) in the arguments
\(x \), \(\tilde {x} \), \(\hat {x} \), \(u \), and \(\vartheta \), we have
$$ \begin {aligned} J(u,\vartheta )&=J\Big (\sigma u^{1}
(t)+(1-\sigma )u^{2} (t), \sigma \vartheta ^{1} +(1-\sigma )\vartheta ^{2} \Big )
\\ &=\int \limits _{t^{1} }^{t^{f} }f^{0} \Big (\sigma x^{1} (t)+(1-\sigma )x^{2}
(t), \sigma u^{1} (t)+(1-\sigma )u^{2} (t), \sigma \vartheta ^{1} +(1-\sigma
)\vartheta ^{2} \Big )dt \\ &\quad {}+\Phi \Big (\sigma \tilde {x}^{1} +(1-\lambda
\sigma )\tilde {x}^{2} ,\sigma \hat {x}^{1} +(1-\lambda \sigma )\hat {x}^{2}
,\sigma \vartheta ^{1} +(1-\sigma )\vartheta ^{2} \Big ) \\ &{}\le \sigma \int
\limits _{t^{1} }^{t^{f} }f^{0} \Big (x^{1} (t),u^{1} (t),\vartheta ^{1} \Big
)dt+(1-\sigma )\int \limits _{t^{1} }^{t^{f} }f^{0} \Big (x^{2} (t),u^{2}
(t),\vartheta ^{2} \Big )dt \cr &\quad {}+\sigma \Phi \Big (\tilde {x}^{1} ,\hat
{x}^{1} ,\vartheta ^{1} \Big )+(1-\sigma )\Phi \Big (\tilde {x}^{2} ,\hat {x}^{2}
,\vartheta ^{2} \Big )=\sigma J\Big (u^{1} ,\vartheta ^{1} \Big )+(1-\sigma )J\Big
(u^{2} ,\vartheta ^{2} \Big ) . \end {aligned}$$
(A.8)
This implies the convexity of the functional \(J(u,\vartheta ) \). It is clear that if one of the functions
\(\Phi (\tilde {x},\hat {x},\vartheta )\) and \(f^{0} (x(t),u(t),\vartheta ,t) \) is strictly convex, then the inequality in
(A.8) will be strict. Consequently, the functional
of problem (2.1)–(2.3) will be strictly convex. This implies the assertion in
Theorem 2.\(\quad \blacksquare \)
Proof of Theorem 3. Let \(x(t)\in \mathrm {R}^{n} \) be a solution of the boundary value
problem (2.1), (2.2) for some feasible control \(u(t)\in U \) and parameter vector \(\vartheta \in V \), and let \(x^{1} (t)=x(t)+\Delta x(t) \) be the solution of problem (2.1), (2.2)
corresponding to incremented feasible control \({ u^{1} (t)}{=u(t)+\Delta u(t)\in U} \) and vector \(\vartheta ^{1} =\vartheta +\Delta \vartheta \in V\),
$$ \dot {x}^{1} (t)=A_{1} (t)x^{1} (t)+A_{2} (t)u^{1}
(t),\quad t\in [t^{1} ,t^{f} ],$$
(A.9)
$$ \sum _{i=1}^{l_{1} }\alpha _{i} x^{1} (\tilde {t}^{i}
)+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta
_{j} (t )x^{1} (t)dt =\vartheta ^{1} .$$
(A.10)
It follows from (2.1),
(2.2) and (A.9), (A.10)
that
$$ \Delta \dot {x}(t)=A_{1} (t)\Delta x(t)+A_{2} (t)\Delta
u(t),\quad t\in [t^{1} ,t^{f} ],$$
(A.11)
$$ \sum _{i=1}^{l_{1} }\alpha _{i} \Delta x(\tilde {t}^{i}
)+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta
_{j} (t )\Delta x(t)dt =\Delta \vartheta .$$
(A.12)
Then for the increment of the functional (2.3) we have
$$ \begin {aligned} &\Delta J(u,\vartheta )=J(u^{1}
,\vartheta ^{1} )-J(u,\vartheta ) \\ &=\int \limits _{t^{1} }^{t^{f} }\Big [f^{0}
(x^{1} (t),u^{1} (t),\vartheta ^{1},t)-f^{0} (x(t),u(t),\vartheta ,t)\Big ]
dt+\Phi (\tilde {x}^{1},\hat {x}^{1},\vartheta ^{1})-\Phi (\tilde {x},\hat
{x},\vartheta ) \\ &=\int \limits _{t^{1}}^{t^{f} }\left [\frac {\partial
f^{0}(x(t),u(t),\vartheta ,t)}{\partial x}\Delta x(t) + \frac {\partial
f^{0}(x(t),u(t),\vartheta ,t)}{\partial u}\Delta u(t) + \frac {\partial f^{0}
(x(t),u(t),\vartheta , t)}{\partial \vartheta } \Delta \vartheta \right ] dt \\
&\qquad {}+\sum _{i=1}^{l_{1} }\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \tilde {x}^{i}}\Delta x(\tilde {t}^{i})+\sum _{j=1}^{2l_{2} }\frac
{\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \hat {x}^{j}}\Delta
x(\hat {t}^{j} )+\frac {\partial \Phi (\tilde {x}, \hat {x},\vartheta )}{\partial
\vartheta }\Delta \vartheta +R, \\ &\qquad \qquad \qquad \qquad \qquad R=o\left
(\left \|\Delta x(t)\right \|_{C^{1,n} [t^{1},t^{f} ]},\left \|\Delta u(t)\right
\|_{L_{2}^{r} [t^{1},t^{f}]},\left \|\Delta \vartheta \right \|_{\mathrm {R}^{n}
}\right ). \end {aligned}$$
(A.13)
Here \(R\) is a remainder term.
Under the adopted assumptions about the data of problem (2.1), (2.2), using
the well-known technique in [24], we can
produce an estimate of the form
$$ \left \| \Delta x(t)\right \| _{C^{1,n} [t^{1} ,t^{f} ]}
\le c_{1} \left \| \Delta u(t)\right \| _{L_{2}^{r} [t^{1} ,t^{f} ]} +c_{2} \left
\| \Delta \vartheta \right \| _{\mathrm {R}^{n} } ,
$$
where the positive constants
\(c_{1} \) and
\(c_{2} \) are independent of
\(x(t) \). In view of (
A.13), this implies the differentiability of the functional
\(J(u,\vartheta )\) with respect to both
\(u(t) \) and
\(\vartheta \). Let us merge and order the sets of points
\(\tilde {t}_{i} \),
\(i=1,2,\ldots ,l_{1} \), and
\(\hat {t}_{j} \),
\(j=1,2,\ldots ,2l_{2} \), denoting the resulting tuple of points by
\(\bar {t}_{s} \),
\(s=1,2,\ldots ,(l_{1}+2l_{2}) \).
We transpose the right-hand side of (A.11) to the left and take the inner product of both sides of the
resulting relation by as yet arbitrary \(n\)-dimensional vector
function \(\psi (t)\) continuously differentiable on the intervals
\((\bar {t}_{i},\bar {t}_{i+1})\), \(i=1,2,\ldots ,l_{1} +2l_{2} -1 \). Integrating by parts in the resulting relation and
using the notation
$$
\eqalign { \psi (t_{+}^{i} )&={\mathop {\lim }\limits _{\varepsilon \to +0}} \psi
(t_{i} +\varepsilon ), \cr \psi (t_{-}^{i} )&={\mathop {\lim }\limits
_{\varepsilon \to +0}} \psi (t_{i}-\varepsilon ),}
$$
we obtain
$$ \begin {aligned} 0&=\int \limits _{t^{1}}^{t^{f} }\psi
^{\mathrm {T}} (t)\Big [\Delta \dot {x}(t)-A_{1} (t)\Delta x(t)-A_{2} (t)\Delta
u(t)\Big ]dt \\ &=\sum _{i=1}^{(l_{1} +2l_{2} -1)}\int \limits _{\bar
{t}^{i}}^{\bar {t}^{i+1} } \left [\psi ^{\mathrm {T}} (t)\Delta \dot {x}(t)-\psi
^{\mathrm {T}} (t)A_{1} (t)\Delta x(t)-\psi ^{\mathrm {T}} (t)A_{2}(t)\Delta
u(t)\right ]dt \\ &=\psi ^{\mathrm {T}} (t^{f} )\Delta x(t^{f} )-\psi ^{\mathrm
{T}}(t^{1} )\Delta x(t^{1} )+\int \limits _{t^{1} }^{t^{f}}\left [-\dot {\psi
}^{\mathrm {T}} (t)-\psi ^{\mathrm {T}} (t)A_{1}(t)\right ] \Delta x(t)dt \\
&\qquad {}+\int \limits _{t^{1} }^{t^{f} }\Big [-\psi ^{\mathrm {T}}(t)A_{2}
(t)\Big ] \Delta u(t)dt+\sum _{i=2}^{l_{1} -1} \Big [\psi (\tilde {t}_{-}^{i}
)-\psi (\tilde {t}_{+}^{i} )\Big ]^{\mathrm {T}} \Delta x(\tilde {t}^{i} )+\sum
_{j=1}^{2l_{2} } \left [\psi (\hat {t}_{-}^{j} )-\psi (\hat {t}_{+}^{j} )\right ]
^{\mathrm {T}} \Delta x(\hat {t}^{j} ). \end {aligned}
$$
Adding the resulting expression, which is zero, to (A.13), after simple transformations we obtain
$$ \begin {aligned} \Delta J(u,\vartheta )&=\int \limits
_{t^{1} }^{t^{f} }\left [-\dot {\psi }^{\mathrm {T}} (t)-\psi ^{\mathrm {T}}
(t)A_{1} (t)+\frac {\partial f^{0}(x(t), u(t),\vartheta , t)}{\partial x} \right ]
\Delta x(t)dt \\ &\quad {}+ \int \limits _{t^{1} }^{t^{f} }\left [-\psi ^{\mathrm
{T}}(t)A_{2} (t) + \frac {\partial f^{0} (x(t), u(t),\vartheta ,t)}{\partial u}
\right ] \Delta u(t)dt\\ &\quad {}+ \left [\int \limits _{t^{1} }^{t^{f} }\frac
{\partial f^{0} (x(t),u(t),\vartheta , t)}{\partial \vartheta } dt + \frac
{\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \vartheta } \right ]
\Delta \vartheta \\ &\quad {}+\left \{\sum _{i=1}^{l_{1} }\frac {\partial \Phi
(\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}^{i}}\Delta x(\tilde {t}^{i}
)+\sum _{j=1}^{2l_{2} }\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \hat {x}^{j}}\Delta x(\hat {t}^{j} )+\psi ^{\mathrm {T}} (t^{f}
)\Delta x(t^{f} )\right . \\ &\qquad \qquad \qquad \qquad {}-\left .\psi ^{\mathrm
{T}} (t^{1} )\Delta x(t^{1} )+\sum _{i=2}^{l_{1} -1}\left [ \psi (\tilde
{t}_{-}^{i} )-\psi (\tilde {t}_{+}^{i} )\right ]^{\mathrm {T}} \Delta x(\tilde
{t}^{i})\right . \\ &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
{}+\left .\sum _{j=1}^{2l_{2} }\left [\psi (\hat {t}_{-}^{j} )-\psi (\hat
{t}_{+}^{j} )\right ]^{\mathrm {T}} \Delta x(\hat {t}^{j} ) \right \}+R. \end
{aligned}$$
(A.14)
Now let us deal with the terms in curly braces.
In (A.14), we use conditions
(A.12) to express some \(n \) components of the \(nl_{1} \)-dimensional vector
$$ \Delta x(\tilde {t})=\Delta
\tilde {x}=\left (\Delta x_{1} (\tilde {t}^{1} ), \Delta x_{2} (\tilde {t}^{1}
),\ldots , \Delta x_{n} (\tilde {t}^{1} ), \ldots , \Delta x_{i} (\tilde {t}^{j}
), \ldots , \Delta x_{n} (\tilde {t}^{l_{1} } )\right ),
$$
via the other
\(nl_{1}-1 \) components.
In what follows, to simplify the exposition of technical details, we use
componentwise notation of formulas jointly with the matrix operations.
Then relation (A.12) can be
written as
$$ {\overset{{}_\frown}{\alpha}{}}
\Delta \, {\overset{{}_\frown}{x}{}} +\breve {\alpha }\Delta \breve {x}+\sum _{j=1}^{l_{2} }\;\int
\limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta _{j} (t )\Delta x(t)dt =\Delta
\vartheta .$$
Based on this, in view of (
3.2), we have
$$ \Delta \, {\overset{{}_\frown}{x}{}} = {\overset{{}_\frown}{\alpha}{}} ^{-1} \Delta \vartheta -
{\overset{{}_\frown}{\alpha}{}} ^{-1} \breve {\alpha }\Delta \breve {x}-\sum _{j=1}^{l_{2} }\;\int \limits
_{\hat {t}^{2j-1} }^{\hat {t}^{2j} } {\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{j} (t )\Delta x(t)dt .
$$
(A.15)
With the adopted notation \(C=-{\overset{{}_\frown}{\alpha}{}} ^{-1}\) and
\(B=-{\overset{{}_\frown}{\alpha}{}} ^{-1}\breve {\alpha }\), relation (A.15) acquires the form
$$ \Delta \, {\overset{{}_\frown}{x}{}} =C \Delta \vartheta +B\Delta \breve
{x}-\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }
{\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{j} (t)\Delta x(t)dt,$$
(A.16)
or, in
componentwise form,
$$ \eqalign { \Delta {\overset{{}_\frown}{x}{}} _{i} &=\Delta x_{k_{i} } (\tilde
{t}^{s_{i} } )=\sum _{k=1}^{n}c_{ik} \Delta \vartheta _{k} +\sum _{\nu =1}^{l_{1}
n}b_{i\nu } \Delta x_{g_{\nu } } (\tilde {t}^{q_{\nu } } ) \cr &\qquad {}-\sum
_{j=1}^{l_{2} }\sum _{k=1}^{n}\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }
{\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{ik}^{j} (t) \Delta x_{k} (t)dt , \quad i=1,2,\ldots ,n
,\quad 1\le g_{\nu } \le n.}$$
(A.17)
We write the last 4–7
terms in (
A.14) as
$$ \eqalign { \psi ^{\mathrm
{T}} (t^{f} )\Delta x(t^{f} )&=\sum _{j=1}^{n}\psi _{j} (t^{f} )\Delta x_{j}
(t^{f} ) , \cr \psi ^{\mathrm {T}} (t^{1} )\Delta x(t^{1} )&=\sum _{j=1}^{n}\psi
_{j} (t^{1} )\Delta x_{j} (t^{1} ) .}$$
Combining the fourth and eighth terms in (A.14) and taking into account (A.17), we obtain
$$ \eqalign { \sum _{i=1}^{l_{1} }\sum _{j=1}^{n}&{}\left
[\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde
{x}_{i}^{j} } + \Delta \psi _{j} (\tilde {t}^{i} )\right ] \Delta x_{j} (\tilde
{t}^{i} )\cr &\qquad {}=\sum _{i=1}^{n}\left [\frac {\partial \Phi (\tilde
{x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } + \Delta \psi
_{k_{i} } (\tilde {t}^{s_{i} } )\right ]\Delta x_{k_{i} } (\tilde {t}^{s_{i} }
)\cr &\qquad \qquad {}+\sum _{\nu =1}^{l_{1} n} \left [\frac {\partial \Phi
(\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{g_{\nu } }^{q_{\nu } } }
+\Delta \psi _{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right ] \Delta x_{g_{\nu } }
(\tilde {t}^{q_{\nu } } ) \cr &\qquad {}=\sum _{i=1}^{n}\left [\frac {\partial
\Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +
\Delta \psi _{k_{i} } (\tilde {t}^{s_{i} } )\right ] \cr &\qquad \qquad \qquad
{}\times \left [\sum _{k=1}^{n}c_{ik} \Delta \vartheta _{k} + \sum _{\nu
=1}^{l_{1} n}b_{i\nu } \Delta x_{g_{\nu } } (\tilde {t}^{q_{\nu } } ) - \sum
_{j=1}^{l_{2} }\sum _{k=1}^{n}\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }
{\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{ik}^{j} (t) \Delta x_{k} (t)dt \right ] \cr &\qquad \qquad
{}+\sum _{\nu =1}^{l_{1} n} \left [\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{g_{\nu } }^{q_{\nu } } } +\Delta \psi
_{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right ] \Delta x_{g_{\nu } } (\tilde
{t}^{q_{\nu } } ).}$$
From (A.14), considering the
resulting relation, we have
$$ \begin {aligned} \Delta J(u,\vartheta )&=\int \limits
_{t^{1} }^{t^{f} }\left [\vphantom {\sum _{j=1}^{l_{2} }}-\dot {\psi }^{\mathrm
{T}} (t)-\psi ^{\mathrm {T}} (t)A_{1} (t)+\frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial x} \right . \\ &\qquad \qquad {}-\left .\sum
_{i=1}^{n}\left (\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial
\tilde {x}_{k_{i} }^{s_{i} } } +\Delta \psi _{k_{i} } (\tilde {t}^{s_{i} } )\right
) \sum _{j=1}^{l_{2} }\chi _{[\hat {t}^{2j-1} , \hat {t}^{2j}]} (t){\overset{{}_\frown}{\alpha}{}} ^{-1}
\beta _{j} (t) \right ]\Delta x(t)dt \\ &\qquad {}+\int \limits _{t^{1} }^{t^{f}
}\left [-\psi ^{\mathrm {T}} (t)A_{2} (t)+\frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial u} \right ]\Delta u(t)dt \\ &\qquad {}+\sum
_{k=1}^{n}\left \{\vphantom {\int \limits _{t^{1} }^{t^{f} }}\sum _{i=1}^{n}\left
(\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i}
}^{s_{i} } } +\Delta \psi _{k_{i} } (\tilde {t}^{s_{i} } )\right ) c_{ik}\right .
\\ &\qquad \qquad \qquad \qquad {}+\left . \frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \vartheta _{k} } \right . +\left .\int \limits _{t^{1}
}^{t^{f} }\frac {\partial f^{0} (x(t),u(t),\vartheta ,t)}{\partial \vartheta _{k}
} dt\right \} \Delta \vartheta _{k} \\ &\qquad {}+\sum _{\nu =1}^{l_{1} n} \Bigg
[\sum _{i=1}^{n}b_{i\nu } \left (\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +\Delta \psi _{k_{i} }
(\tilde {t}^{s_{i} } )\right ) \\ &\qquad \qquad \qquad \qquad {}+\left (\frac
{\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{g_{\nu }
}^{q_{\nu } } } +\Delta \psi _{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right )\Bigg ]
\Delta x_{g_{\nu } } (\tilde {t}^{q_{\nu } } ) \\ &\qquad {}+\sum _{j=1}^{2l_{2}
}\sum _{i=1}^{n}\left [\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \hat {x}_{i}^{j} } +\Delta \psi _{i} (\hat {t}^{j} )\right ] \Delta
x_{i} (\hat {t}^{j} )+R. \end {aligned}$$
(A.18)
By virtue of the arbitrariness of the vector function \(\psi (t) \), we require that the expressions in the first and in
the last two brackets in (A.18) vanish. From the
first requirement, we obtain the adjoint system of differential equations (3.7), and from the other two requirements we obtain the
expressions
$$ \begin
{gathered} \sum _{i=1}^{n}b_{i\nu } \left [\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +\Delta \psi _{k_{i} }
(\tilde {t}^{s_{i} } )\right ]+ \left [\vphantom {\frac {\partial \Phi (\tilde
{x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } }}\frac {\partial
\Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{g_{\nu } }^{q_{\nu } }
} +\Delta \psi _{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right ]=0,\quad \nu
=1,2,\ldots ,l_{1} n, \\ \frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \hat {x}_{i}^{j} } +\Delta \psi _{i} (\hat {t}^{j} )=0,\quad
i=1,2,\ldots ,n, \quad j=1,2,\ldots ,2l_{2} . \end {gathered}
$$
This implies conditions (
3.8) and (
3.9).
Then the desired components of the gradient of the functional with respect to
\(u(t) \) and \(\vartheta \) will be determined from (A.18) as the linear parts of the increment of the functional for
\(\Delta u(t) \) and \(\Delta \vartheta \) by formulas (3.5) and (3.6).
This completes the proof of Theorem 3.\(\quad \blacksquare \)
Proof of Theorem 4. Let us proceed to the
optimality conditions for the pair \(\left (u,\vartheta \right ) \) with conditions (2.2) replaced by conditions (3.3) and (3.4).
Unlike the above manipulations for the case of \(\bar {n}=n\), in the
sequel, to take into account conditions (3.4), we
use the Lagrange method and introduce an \((n-\bar {n})\)
-dimensional additional vector of parameters—the Lagrange multipliers.
Once again using the method of increment in the parameters \(\left (u,\vartheta \right ) \) to be optimized, for the increment of the functional
we obtain formula (A.6). Condition
(3.4) in terms of increments now acquires the form
$$ \sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1}
}^{\hat {t}^{2j} }\beta _{j}^{2} (t )\Delta x(t)dt =\Delta \vartheta ^{(2)} .
$$
(A.19)
We transpose all terms in (A.19)
to the left, multiply the resulting expression by as yet arbitrary vector \(\lambda \in \mathrm {R}^{n-\bar {n}}\), and add to (A.14). For the increment of the functional, we obtain
$$ \begin {aligned} \Delta J(u,\vartheta )&=\int \limits
_{t^{1} }^{t^{f} }\left [\vphantom { \sum _{j=1}^{l_{2} }}-\dot {\psi }^{\mathrm
{T}} (t)-\psi ^{\mathrm {T}} (t)A_{1} (t)+\frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial x}\right . \\ &\qquad \qquad \qquad \qquad {}-\left
.\lambda ^{\mathrm {T}} \sum _{j=1}^{l_{2}}\chi _{[\hat {t}^{2j-1},\hat {t}^{2j}}
(t) \beta _{j}^{2} (t) \right ]\Delta x(t)dt \\ &\qquad {}+ \int \limits _{t^{1}
}^{t^{f} } \left [-\psi ^{\mathrm {T}} (t)A_{2} (t) + \frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial u} \right ] \Delta u(t)dt \\ &\qquad {}+\left [\int
\limits _{t^{1} }^{t^{f} }\frac {\partial f^{0} (x(t), u(t),\vartheta ,
t)}{\partial \vartheta ^{(1)} } dt + \frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \vartheta ^{(1)} } \right ] \Delta \vartheta ^{(1)} \\
&\qquad {}+\left [\int \limits _{t^{1} }^{t^{f} }\frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial \vartheta ^{(2)} } dt+\frac {\partial \Phi (\tilde
{x},\hat {x},\vartheta )}{\partial \vartheta ^{(2)} } -\lambda ^{\mathrm {T}}
\right ]\Delta \vartheta ^{(2)} \\ &\qquad {}+\left \{\vphantom {\sum
_{j=1}^{2l_{2} }}\sum _{i=1}^{l_{1} }\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}^{i} } \Delta x(\tilde {t}^{i} )+\sum
_{j=1}^{2l_{2} }\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial
\hat {x}^{j} } \Delta x(\hat {t}^{j} )+\psi ^{\mathrm {T}} (t^{f} )\Delta x(t^{f}
)\right . \\ &\quad \quad \qquad \qquad \qquad {}-\psi ^{\mathrm {T}} (t^{1}
)\Delta x(t^{1} )+\sum _{i=2}^{l_{1} -1} \left [\psi (\tilde {t}_{-}^{i} )-\psi
(\tilde {t}_{+}^{i} )\right ] ^{\mathrm {T}} \Delta x(\tilde {t}^{i} ) \\ &\qquad
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad {}+\left .\sum
_{j=1}^{2l_{2} } \left [\psi (\hat {t}_{-}^{j} )-\psi (\hat {t}_{+}^{j} )\right ]
^{\mathrm {T}} \Delta x(\hat {t}^{j} ) \right \}+R. \end {aligned}
$$
(A.20)
Now let us deal with the terms in curly braces having in mind that the rank of the
augmented matrix \(\alpha =\left [\alpha _{1} ,\alpha _{2} ,\ldots ,\alpha _{l_{1} } \right ]\) in conditions (3.3) is \(\bar {n}\). From the
expression in increments
$$ \sum _{i=1}^{l_{1} }\alpha _{i} \Delta x(\tilde {t}^{i}
)+\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }\beta
_{j}^{1} (t )\Delta x(t)dt =\Delta \vartheta ^{(1)} ,
$$
obtained from (
3.3), we express some
\(\bar {n} \) components of the
\(nl_{1} \)-dimensional vector
\(\Delta \tilde {x}(\tilde {t}) \) via the remaining
\(nl_{1}-\bar {n} \) components. As was done above, from the matrix
\(\alpha \) we isolate a
\(\bar {n}\times \bar {n} \) submatrix (minor)
\({\overset{{}_\frown}{\alpha}{}} \). Let the remaining
\(nl_{1} -\bar {n} \) columns of the matrix
\(\alpha \) form a matrix
\(\breve {\alpha } \). We denote the components of the vector
\(\Delta \tilde {x}(\tilde {t})\) corresponding to the matrix
\({\overset{{}_\frown}{\alpha}{}} \) by
\(\Delta {\overset{{}_\frown}{x}{}} (\tilde {t}) \); the remaining components form the vector
\(\Delta \breve {x}(\tilde {t})\). Then
$$ \Delta \, {\overset{{}_\frown}{x}{}} (\tilde {t})=
{\overset{{}_\frown}{\alpha}{}} ^{-1} \Delta \vartheta ^{(1)} - {\overset{{}_\frown}{\alpha}{}} ^{-1} \breve {\alpha }\Delta
\breve {x}(\tilde {t})- {\overset{{}_\frown}{\alpha}{}} ^{-1} \sum _{j=1}^{l_{2} }\;\int \limits _{\hat
{t}^{2j-1} }^{\hat {t}^{2j} }\beta _{j}^{1} (t )\Delta x(t)dt ,
$$
$$ \Delta \, {\overset{{}_\frown}{x}{}} =C \Delta \vartheta +B\Delta \breve
{x}-\sum _{j=1}^{l_{2} }\;\int \limits _{\hat {t}^{2j-1} }^{\hat {t}^{2j} }
{\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{j}^{(1)} (t)\Delta x(t)dt,$$
or, in componentwise form,
$$ \begin {aligned} \Delta \,{\overset{{}_\frown}{x}{}} _{i}=\Delta x_{k_{i} }
(\tilde {t}^{s_{i} } )&=\sum _{k=1}^{n}c_{ik} \Delta \vartheta _{k} +\sum _{\nu
=1}^{l_{1} n-\bar {n}}b_{i\nu } \Delta x_{g_{\nu } } (\tilde {t}^{q_{\nu } } ) \\
&\qquad {}-\sum _{j=1}^{l_{2} }\sum _{k=1}^{n}\;\int \limits _{\hat {t}^{2j-1}
}^{\hat {t}^{2j} } {\overset{{}_\frown}{\alpha}{}} ^{-1} \beta _{jik}^{(1)} (t) \Delta x_{k} (t)dt , \quad
i=1,2,\ldots ,\bar {n} , \quad 1\le g_{\nu } \le n. \end {aligned}
$$
(A.21)
Taking (
A.21) into account in (
A.20), we obtain
$$ \begin {aligned} \Delta J(u,\vartheta )&=\int \limits
_{t^{1} }^{t^{f}} \left [\vphantom {\sum _{j=1}^{l_{2} }} -\dot {\psi }^{\mathrm
{T}}(t)-\psi ^{\mathrm {T}}(t)A_{1}(t)+ \frac {\partial f^{0} (x(t),u(t),\vartheta
, t)}{\partial x}\right . \\ &\qquad \qquad \quad {}-\lambda ^{\mathrm {T}} \sum
_{j=1}^{l_{2} }\chi _{[\hat {t}^{2j-1} , \hat {t}^{2j}]} (t) \beta _{j}^{2} (t) \\
&\qquad \qquad \qquad {}+ \left ( \sum _{i=1}^{n} \left ( \frac {\partial \Phi
(\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +(\psi
_{k_{i} } (\tilde {t}_{-}^{s_{i} } )-\psi _{k_{i} } (\tilde {t}_{+}^{s_{i} }
))^{\mathrm {T}} \right ) \right ) \\ &\qquad \qquad \qquad \qquad \qquad {}\times
\left . \sum _{j=1}^{l_{2} }\chi _{[\hat {t}^{2j-1} , \hat {t}^{2j}]} (t) {\overset{{}_\frown}{\alpha}{}}
^{-1} \beta _{j}^{1} (t) \right ]\Delta x(t)dt \\ &\qquad {}+\int \limits _{t^{1}
}^{t^{f} }\left [-\psi ^{\mathrm {T}} (t)A_{2} (t)+\frac {\partial f^{0} (x(t),
u(t),\vartheta , t)}{\partial u} \right ]\Delta u(t)dt \\ &\qquad {}+\sum
_{k=1}^{\bar {n}}\left [\vphantom {\int \limits _{t^{1} }^{t^{f} }}\sum
_{i=1}^{\bar {n}}\left [\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +\Delta \psi _{k_{i} } (\tilde
{t}^{s_{i} } )\right ] c_{ik}\right . \\ &\qquad \qquad \qquad {}+\left .\frac
{\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \vartheta _{k}^{(1)} }
+\int \limits _{t^{1} }^{t^{f} }\frac {\partial f^{0} (x(t),u(t),\vartheta
,t)}{\partial \vartheta _{k}^{(1)} } dt\right ] \Delta \vartheta _{k}^{(1)} \\
&\qquad {}+\sum _{k=1}^{n-\bar {n}} \left [-\lambda _{k} +\frac {\partial \Phi
(\tilde {x},\hat {x},\vartheta )}{\partial \vartheta _{k}^{(2)} } +\int \limits
_{t^{1} }^{t^{f} }\frac {\partial f^{0} (x(t),u(t),\vartheta ,t)}{\partial
\vartheta _{k}^{(2)} } dt\right ]\Delta \vartheta _{k}^{(2)} \\ &\qquad {}+\sum
_{\nu =1}^{l_{1} n-\bar {n}} \left [\sum _{i=1}^{n}b_{i\nu } \left (\frac
{\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial \tilde {x}_{k_{i}
}^{s_{i} } } +\Delta \psi _{k_{i} } (\tilde {t}^{s_{i} } )\right )\right . \\
&\qquad \qquad \qquad {}+\left .\left (\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{g_{\nu } }^{q_{\nu } } } +\Delta \psi
_{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right )\vphantom {\sum _{i=1}^{n}}\right ]
\Delta x_{g_{\nu } } (\tilde {t}^{q_{\nu } } ) \\ &\qquad {}+\sum _{j=1}^{2l_{2}
}\sum _{i=1}^{n}\left [\frac {\partial \Phi (\tilde {x},\hat {x},\vartheta
)}{\partial \hat {x}_{i}^{j} } +\Delta \psi _{i} (\hat {t}^{j} )\right ] \Delta
x_{i} (\hat {t}^{j} )+R. \end {aligned}$$
(A.22)
By virtue of the arbitrariness of the vector function \(\psi (t) \) and the vector \(\lambda \), we require that they be such that the expression in
the first bracket in (A.22) vanishes, and, based on
this, produce the adjoint differential equation (3.28). At the same time, by virtue of the arbitrariness of the
components of the remainder vector \(\Delta x(\tilde {t}) \) and increments \(\Delta x(\hat {t}^{i} ) \), \(j=1,2,\ldots ,2l_{2} \), we require that the expressions in the last two
brackets in (A.22) be zero,
$$ \begin {gathered} \sum
_{i=1}^{\bar {n}}b_{i\nu } \left [\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{k_{i} }^{s_{i} } } +\Delta \psi _{k_{i} }
(\tilde {t}^{s_{i} } )\right ]+\left [\frac {\partial \Phi (\tilde {x},\hat
{x},\vartheta )}{\partial \tilde {x}_{g_{\nu } }^{q_{\nu } } } +\Delta \psi
_{g_{\nu } } (\tilde {t}^{q_{\nu } } )\right ]=0 , \quad \nu =1, 2, \ldots , l_{1}
n-\bar {n}, \\ \frac {\partial \Phi (\tilde {x},\hat {x},\vartheta )}{\partial
\hat {x}_{i}^{j} } +\Delta \psi _{i} (\hat {t}^{j} )=0, \quad i=1, 2, \ldots ,n,
\quad j=1, 2, \ldots , 2l_{2} . \end {gathered}$$
Based on this, we obtain the boundary conditions (
3.29)–(
3.30) for
the adjoint equation (
3.28).
It is clear from (A.22) that the
formula for the gradient of the functional with respect to \(u(t) \) will be the same as in (3.5), and the components of the gradient with respect to
\(\vartheta \) are given by formulas (3.26) aand (3.27). Thus, the proof of Theorem 4 is complete.\(\quad \blacksquare \)
