In this section we prove the main results in the paper. We prove two versions of the FilippovPliss lemma which have many applications in optimal control (cf. [18]).
We need the following result, which is a particular case of Proposition 1.43 of [25].
Proposition 4.1
Let
\(F, G: \mathbb{I}\times \mathbb{R}^{n}\rightrightarrows \mathbb{R}^{n}\)
be
\(\Delta \times \mathcal{B}\)
measurable and at least one with compact values. Then the map
\(H(t,x)= F(t,x)\cap G(t,x)\)
is also
\(\Delta \times \mathcal{B}\)
measurable.
Now we will prove two variants of the FilippovPliss theorem for dynamical inclusion on a time scale. The first proof deals with Lipschitz righthand side.
Theorem 4.2
Let
\(F(\cdot ,\cdot )\)
satisfy
F1, F2, and let
\(F(t,\cdot )\)
be
LLipschitz. If
\(f(\cdot )\)
is a Δintegrable function on
\(\mathbb{I}\)
and if
\(y(\cdot )\)
is an AC function with
$$ \operatorname{dist} \bigl(y^{\Delta }(t), F \bigl(t, y(t) \bigr) \bigr) \leq f(t), $$
then there exists a solution
\(x(\cdot )\)
of (3.1) such that
\(\vert x(t)  y(t)\vert \leq r(t)\), where
\(r^{\Delta }(t) = L r(t) + f(t)\)
and
\(r(t_{0}) = \vert x_{0}  y_{0}\vert \). Furthermore, \(\vert x^{\Delta }(t) y^{ \Delta }(t)\vert \leq L r(t)+ f(t)\).
Proof
Define the map
$$ G(t, u) = \bigl\{ v \in F(t, u) : \bigl\vert y^{\Delta }(t)  v \bigr\vert \leq L \bigl\vert y(t)  u \bigr\vert + f(t) \bigr\} . $$
We claim that \(G(\cdot ,\cdot )\) satisfies F1 and F2.
Namely, \(G(t, u)\) admits nonempty values because \(F(t,\cdot )\) is Lipschitz and it is with nonempty convex compact values. We are going to prove that \(G(t, u)\) is closed and convex. Indeed if \(v_{i} \in F(t, u)\) and \(v_{i}\rightarrow v\), we know that \(\vert v  y^{\Delta }(t)\vert = \lim_{i\rightarrow \infty } \vert v_{i}  y^{\Delta }(t)\vert \). Therefore, if \(v_{1}, v_{2} \in G(t, u)\), then we have that \(\vert y^{\Delta }(t)  \lambda v_{1}  (1  \lambda )v_{2}\vert \leq \lambda \vert y^{\Delta }(t)  v _{1}\vert + (1  \lambda )\vert y^{\Delta }(t)  v_{2}\vert \leq L\vert y(t)  u\vert + f(t)\), \(\forall \lambda \in (0, 1)\). Also \(G(t, u)\) is USC. For this, it is enough to see that \(G(t,\cdot )\) has a closed graph. Let \(v_{i} \in G(t, u_{i})\), \(u_{i} \rightarrow u\) and \(v_{i} \rightarrow v\). Since \(F(t,\cdot )\) is USC, one has that \(\lim_{i \rightarrow \infty }v_{i} = v \in F(t,u)\). Furthermore, \(\vert v_{i}  y^{\Delta }(t)\vert \leq L\vert y(t)  u_{i}\vert + f(t)\), and hence \(\vert v  y^{\Delta }(t)\vert \leq L\vert y(t)  u\vert + f(t)\).
Now, we have to show that \(G(\cdot ,\cdot )\) is \(\Delta \times \mathcal{B}\) measurable. Let \(X\in \mathbb{R}^{n}\). Since \(y(\cdot)\) is AC, then \(y^{\Delta }(\cdot)\) is Δmeasurable, and hence the multimap \(H(t,X) = \{ (t,z) \in [t_{0},T)_{\mathbb{T}}\times \mathbb{R}^{n}:\vert z  y^{\Delta }(t)\vert \leq L\vert y(t)  X\vert + f(t) \}\) is \(\Delta \times \mathcal{B}\) measurable. Due to Proposition 4.1, the map \(G(t, X) = H(t,X) \cap F(t,X)\) is also \(\Delta \times \mathcal{B}\) measurable. The claim is therefore proved. It follows from Theorem 3.3 that \(x^{\Delta } \in G(t, x(t))\), \(x(t_{0}) = x_{0}\) admits a solution \(x(\cdot )\).
From Theorem 1.67 of [3] we know that \(\vert x(t)  y(t)\vert = r(t)\), where \(r^{\Delta }(t) \leq L r(t) + f(t)\) for Δa.e. t and \(r(t_{0}) =\vert x_{0}  y_{0}\vert \).
The definition of \(G(\cdot ,\cdot )\) then implies the last statement of the theorem. □
Now we will prove a FilippovPliss type theorem under much weaker condition, which gives the estimation only of the difference between \(x(\cdot )\) and \(y(\cdot )\) but not between their derivatives.
Definition 4.3
The multivalued map \(F: \mathbb{I}\times \mathbb{R}^{n}\rightrightarrows \mathbb{R}^{n}\) is said to be OSP (onesided Perron) (on the state variable) if there exists a Perron function \(w(\cdot ,\cdot )\) such that:
For every \(x, y\in \mathbb{R}^{n}\), almost every \(t\in \mathbb{I}\) and every \(f_{x}\in F(t,x)\), there exists \(f_{y}\in F(t,y)\) such that
$$\begin{aligned} & \langle xy,f_{x}f_{y}\rangle \leq \frac{1}{2}w \bigl(t,\vert xy\vert \bigr)\vert xy\vert , \quad \mbox{if } t\in \mathbb{T}_{\mathrm {rd}} \\ & \vert f_{x}f_{y}\vert \leq w \bigl(t,\vert xy \vert \bigr),\quad \mbox{if } t\in \mathbb{T}_{\mathrm {rs}} . \end{aligned}$$
Recall that the function \(v(\cdot ,\cdot )\) is said to be a Perron function if

\(v(\cdot ,\cdot )\) is \(\Delta \times \mathcal{B}\) measurable, \(v(t,\cdot )\) is continuous;

\(v(\cdot ,\cdot )\) is Δintegrally bounded on the bounded sets and \(v(t,0)= 0\);

the unique solution of \(r^{\Delta }(t)= v(t,r(t))\), \(r(t_{0})=0\) is \(r(t)= 0\).
\(v(\cdot ,\cdot )\) is called module if it satisfies only the first two conditions, but not necessarily the third one.
The definition of OSP condition on a time scale is different than in ordinary differential inclusions. Here it depends also on the point t.
Now we extend the previous theorem to the case of OSP multifunctions.
Theorem 4.4
Let
\(F(\cdot ,\cdot )\)
satisfy
F1, F2, and let
\(F(t,\cdot )\)
be OSP w.r.t. a Perron function
\(w(\cdot ,\cdot )\). If
\(f(\cdot )\)
is a Δintegrable function on
\(\mathbb{I}\)
and if
\(y(\cdot )\)
is an AC function with
\(\operatorname{dist}(y^{\Delta }(t), F(t, y(t))) \leq f(t)\), then there exists a solution
\(x(\cdot )\)
of (3.1) such that
\(\vert x(t)  y(t)\vert \leq r(t)\), where
\(r^{\Delta }(t) = w(t,r(t))+ f(t)\)
and
\(r(t_{0}) = \vert x_{0}  y_{0}\vert \).
Proof
Clearly, the setvalued map \(t\to y^{\Delta }(t)+f(t)\mathbb{B}\) is Δmeasurable. Therefore \(H(t)=F(t,y(t))\cap (y^{\Delta }(t)+f(t) \mathbb{B})\) is also Δmeasurable and hence \(t\to F(t,y(t))\) is Δmeasurable. Thus there exists a Δmeasurable selection \(h(t)\in H(t)\). Evidently \(h(t)\in F(t,y(t))\).
Now we define the following multifunction:
$$\begin{aligned}& G(t,u)= \bigl\{ v\in F(t,u) \bigr\} \quad \mbox{such that} \\& \textstyle\begin{cases} \langle y(t)u,h(t)v\rangle \leq w(t,\vert xy\vert )\vert xy\vert ,& t \mbox{ is right dense} \\ \vert h(t) v\vert \leq w(t,\vert xy\vert ), &t \mbox{ is right scattered}. \end{cases}\displaystyle \end{aligned}$$
(4.1)
We claim that \(G(t,\cdot )\) is upper semicontinuous for every \(t\in \mathbb{I}\). Indeed we have to prove that the graph of \(G(t,\cdot )\) is compact. However, the graph is bounded, and hence it remains to show that it is closed.
Let \(u_{i}\to u\), \(v_{i}\in G(t,u_{i})\) and \(v_{i}\to v\). We have to show that \(v\in G(t,u)\). Clearly \(v\in F(t,u)\) because \(F(t,\cdot )\) is USC. If t is right dense, then \(\langle y(t)u_{i},h(t)v_{i}\rangle \to \langle y(t)u,h(t)v\rangle \), \(\vert y(t)u_{i}\vert \to \vert y(t)u\vert \) and \(w(\vert y(t)u_{i}\vert )\to w(t,\vert y(t)u\vert )\). Thus \(\langle y(t)u,h(t)v\rangle \leq w(t,\vert y(t)u\vert )\vert y(t)u\vert \), i.e., \(v\in G(t,u)\). If t is right scattered, then \(\vert h(t)v\vert \leq w(t,\vert y(t)u\vert )\) because \(\lim_{i\to \infty }\vert h(t)v_{i}\vert =\vert h(t)v\vert \).
We have to prove that \(G(\cdot ,\cdot )\) is \(\Delta \times \mathcal{B}\) measurable.
Consider first the case \(G: \mathbb{T}_{\mathrm {rs}}\times \mathbb{R}^{n} \rightrightarrows \mathbb{R}^{n}\). Since \(w(t,\cdot )\) and \(y(\cdot )\) are continuous as well as \(h(\cdot )\) is Δmeasurable, one has that \(S(t,u)=\{v\in \mathbb{R}^{n}:\vert h(t)v\vert \leq w(t, \vert y(t)u\vert )\}\) is \(\Delta \times \mathcal{B}\) measurable. Then \(\overline{G}(t,u)=F(t,u) \cap \overline{S}(t,u)\) is \(\Delta \times \mathcal{B}\) measurable.
Let \(G: \mathbb{T}_{\mathrm {rd}}\times \mathbb{R}^{n}\rightrightarrows \mathbb{R}^{n}\). It is easy to see that the map
$$ S(t,u)= \bigl\{ v\in \mathbb{R}^{n}: \bigl\langle y(t) u, h(t) v \bigr\rangle \leq w \bigl(t, \bigl\vert y(t)u \bigr\vert \bigr) \bigl\vert y(t)u \bigr\vert \bigr\} $$
is \(\Delta \times \mathcal{B}\) measurable.
Therefore \(\overline{G}(t,u)=F(t,u)\cap \overline{S}(t,u)\) is also Δmeasurable. Consequently, \(G(\cdot ,\cdot )\) is \(\Delta \times \mathcal{B}\) measurable.
Due to Theorem 3.3, the system
$$ \dot{x}\in G \bigl(t,x(t) \bigr),\quad x(t_{0})=x_{0} $$
has a solution \(x(\cdot )\). Therefore
$$\begin{aligned} \bigl\langle y(t)x(t),y^{\Delta }(t)x^{\Delta }(t) \bigr\rangle \leq \bigl\vert y(t)x(t) \bigr\vert \bigl\vert y ^{\Delta }(t) h(t) \bigr\vert +w \bigl(t, \bigl\vert y(t)x(t) \bigr\vert \bigr) \bigl\vert y(t)x(t) \bigr\vert , \end{aligned}$$
\(\forall t\in \mathbb{T}_{\mathrm {rd}}\), i.e., \((\vert y(t)x(t)\vert ^{2}) ^{\Delta } =2\vert x(t)y(t)\vert \cdot (\vert x(t)y(t)\vert )^{\Delta } \leq 2\vert y(t)x(t)\vert (f(t)+w(t,\vert y(t)x(t)\vert )\). Clearly \(t\to \vert x(t)y(t)\vert \) is ΔAC.
From Proposition 2.4 we know that the intersection of the sets \(\{t\in \mathbb{T}:\vert x(t)y(t)\vert =0\}\) and \(\{t\in \mathbb{T}: \vert x^{\Delta }(t)y^{\Delta }(t)\vert \neq 0\}\) has Δmeasure zero. Thus \(\vert x(t)y(t)\vert ^{\Delta }\leq w(t,\vert y(t)x(t)\vert ) +f(t)\) for Δ almost every \(t\in \mathbb{T}_{\mathrm {rd}}\).
If \(t\in \mathbb{T}_{\mathrm {rs}}\), then \(\vert y^{\Delta }(t)x^{\Delta }(t)\vert \leq \vert y^{\Delta }(t)h(t)\vert + \vert h(t)x^{\Delta }(t)\vert \leq w(t,\vert y(t)x(t)\vert )\). However, \(\vert x(t)y(t)\vert ^{\Delta }\leq \vert x^{\Delta }(t) y^{\Delta }(t)\vert \) and hence \(\vert x(t)y(t)\vert ^{\Delta }\leq w(t,\vert x(t)y(t)\vert )+f(t)\).
Consequently,
$$ \bigl\vert x(t)y(t) \bigr\vert \leq r(t),\mbox{ where } r^{\Delta }(t)= w \bigl(t,r(t) \bigr)+ f(t), r(t_{0})=\vert x_{0}y_{0} \vert . $$
The proof is therefore complete. □
Remark 4.5
It is easy to see that Theorem 4.2 remains true also when \(w(\cdot ,\cdot )\) is only module.
As it is well known,
$$\begin{aligned} \bigl({ \bigl\Vert x(t) \bigr\Vert }^{2} \bigr)^{\Delta }&={ \bigl\langle x(t),x(t) \bigr\rangle }^{\Delta }= \bigl\langle x(t)+x \bigl( \sigma (t) \bigr),x^{\Delta }(t) \bigr\rangle \\ & = \bigl\langle x(t)+x(t)+\mu (t)x^{\Delta }(t),x^{\Delta }(t) \bigr\rangle = 2 \bigl\langle x(t),x^{\Delta }(t) \bigr\rangle +\mu (t) \bigl\Vert x^{\Delta }(t) \bigr\Vert ^{2}. \end{aligned}$$
The above conclusion leads us to another definition of OSP on a time scale. Namely, \(F(t,\cdot )\) is said to be OSP if there exists a Perron function \(V(\cdot ,\cdot )\) such that, for every \(f_{x}\in F(t,x)\), there exists \(f_{y}\in F(t,y)\) such that
$$ 2\langle xy,f_{x}f_{y}\rangle +\mu (t){\vert f_{x}f_{y}\vert }^{2}\leq V \bigl(t, {\vert xy\vert }^{2} \bigr). $$
In this case, however, it is difficult to prove a meaningful version of the FilippovPliss lemma, although the following theorem is true.
Theorem 4.6
Let
\(x(\cdot )\)
be a solution of (3.1) with
\(x(t_{0})=x_{0}\). Then, for any
\(y_{0}\), there exists a solution
\(y(\cdot )\)
of (3.1) with
\(y(t_{0})=y_{0}\)
such that
$$ { \bigl\vert x(t)y(t) \bigr\vert }^{2}\leq r(t),\textit{ where } r(t_{0})={\vert x_{0}y_{0}\vert } ^{2} \quad \textit{and} \quad r^{\Delta }(t)=V \bigl(t,r(t) \bigr). $$
Example 4.7
Let \(\mathbb{T}\) be a time scale on \([0,1]\). Let \(\{ x_{i}\}_{i}\) be a dense subset of \(\mathbb{B}\). Define the multifunction \(H(x)= \overline {\operatorname {\mathit {co}}}\{f _{i}(t,x)\}_{i=1}^{k}+ \sum_{i=k}^{\infty }\frac{f_{i}(t,x)}{2^{i}}\), where \(k\geq 5\), \(f_{i}(t,x)= c(t)g_{i}(x)\) and
$$ g_{i}(x)= \textstyle\begin{cases} \frac{xx_{i}}{\sqrt{\vert xx_{i}\vert }}, & x\neq x_{i}, \\ 0, &x= x_{i}. \end{cases} $$
While \(c = \max \{ \tau  t: [t,\tau ]\subset \mathbb{T}\}\). Clearly every \(g_{i}(\cdot )\) is onesided Lipschitz with a constant 0. Therefore \(H(t,\cdot )\) is OSL with a constant 0 on any point \((t,x)\) with \(t\in \mathbb{T}_{\mathrm {rd}}\) and it is 0 on any point \((t,x)\) with \(t\in \mathbb{T}_{\mathrm {rs}}\).
Let \(G: \mathbb{T}\times \mathbb{R}^{n} \rightrightarrows \mathbb{R} ^{n}\) be bounded full Perron. Define \(F(t,x)= H(x)+ G(t,x)\). Then \(F(\cdot ,\cdot )\) satisfies all the assumptions of Theorem 4.2.
Then clearly system (3.1) with \(t_{0}=0\), \(T=1\) and \(x_{0}=0\) satisfies the conditions of Theorem 4.4.
Now we discuss the closure of the solution set for system (3.1) when \(F(\cdot ,\cdot )\) is almost continuous and not necessarily convexvalued. Notice that we only show some further studying directions.
Clearly the closure of the solution set of (3.1) is not the solution set of
$$ x^{\Delta }(t)\in \overline {\operatorname {\mathit {co}}}F(t,x), \quad x(t_{0})= x_{0}. $$
The following theorem holds true.
Theorem 4.8
Let
\(F(\cdot ,\cdot )\)
be almost continuous with compact values. Suppose that
F1
holds. The closure of the solution set of (3.1) is a subset of the solution set of
$$ x^{\Delta }(t)\in H \bigl(t,x(t) \bigr), \quad x(t_{0})= x_{0}, $$
(4.2)
where
$$ H(t,x)= \textstyle\begin{cases} \overline {\operatorname {\mathit {co}}}F(t,x), & t\in \mathbb{I}_{\mathrm {rd}}, \\ F(t,x), & t\in \mathbb{I}_{\mathrm {rs}}. \end{cases} $$
Proof
Let \(x_{m}(\cdot )\) be a sequence of solutions of (3.1) such that \(x_{m}(t)\to x(t)\) uniformly on \(\mathbb{I}\). As in [9] we can extend every \(x_{k}^{\Delta }(\cdot )\) on I as a Lebesgue integrable function \(g_{k}(\cdot )\). Due to Diestel criterion (see, e.g., [27]), the sequence \(\{ g_{k}(\cdot )\}_{k}\) is weakly \(L_{1}\) precompact and passing to subsequences if necessary \(g_{k}(t)\to g(t)\)
\(L_{1}\) weakly. Due to Mazur’s lemma, there exists a convex combination \(\sum_{i=k}^{k_{i}} \alpha_{i} g_{i}(t)\) converging to \(g(t)\)
\(L_{1}\) strongly and passing to subsequences for a.a. \(t\in I\). Clearly its restriction to \(\mathbb{I}\)
\(g_{\mathbb{I}}(t) \in \overline {\operatorname {\mathit {co}}}F(t,x(t))\) for Δ a.e. t. Notice that every \(g_{(}\cdot )\) is constant on \([t,\sigma (t)]\), the latter is not a single point in the case \(t\in \mathbb{I}_{\mathrm {rs}}\). Since \(\mathbb{I} _{\mathrm {rs}}\) is countable, then it is easy to show that in \(g(t)\in F(t,x(t))\) for Δ a.a. \(t\in \mathbb{I}_{\mathrm {rs}}\). Consequently, \(g(t)\in H(t,x(t))\). □
It will be interesting to prove or disprove the following conjecture, which is an analogue of the very important in the optimal control relaxation theorem.
Conjecture 1
Let
\(F(\cdot ,\cdot )\)
be almost continuous with compact values. Suppose that
F1
holds and
\(F(t,\cdot )\)
is OSP. Then the closure of the solution set of (3.1) is the solution set of (4.2).
We hope that the reader will be able to prove this conjecture.