The following theorem is proved by utilizing the idea of a retract method, which is well known for ordinary differential equations and goes back to Ważewski [13]. In the next theorem, we assume that the function *f*, except for the indicated conditions, satisfies all the assumptions given in Section 2. Namely, we assume that the function *f* is bounded and Lipschitz continuous on an open set *S* and \overline{\mathrm{\Omega}}\subset S.

**Theorem 2** *Let* f:\mathbb{T}\times {\mathbb{R}}^{n}\to {\mathbb{R}}^{n}. *Let* {b}_{i},{c}_{i}:\mathbb{T}\to \mathbb{R}, i\in \{1,\dots ,n\} *be* Δ-*differentiable functions on* \mathbb{T} *such that* {b}_{i}(t)<{c}_{i}(t) *for each* t\in {[{\alpha}_{0},\mathrm{\infty})}_{\mathbb{T}}. *If*, *moreover*, *every point* M\in {\partial}_{y}\mathrm{\Omega} *is the point of strict egress for the set* Ω *with respect to system* (1), *then there exists an rd*-*continuous initial function* {\phi}^{\ast}:{[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}}\to {\mathbb{R}}^{n} *satisfying*

{b}_{i}(t)<{\phi}_{i}^{\ast}(t)<{c}_{i}(t)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}t\in {[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}},i=1,\dots ,n

(6)

*such that the initial problem*

y(t)={\phi}^{\ast}(t),\phantom{\rule{1em}{0ex}}t\in {[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}}

(7)

*defines a solution* *y* *of* (1) *on the interval* {[{\alpha}_{0},\mathrm{\infty})}_{\mathbb{T}} *satisfying*

(t,y(t))\in \mathrm{\Omega}\phantom{\rule{1em}{0ex}}\mathit{\text{for every}}t\in {[{\alpha}_{0},\mathrm{\infty})}_{\mathbb{T}}.

(8)

*Proof* The idea of the proof is the following. By contrary, we assume that a solution *y* satisfying (8) does not exist. Then we are able to prove (by a construction of a chain of auxiliary mappings) an existence of a retraction of an *n*-dimensional ball into its boundary. However, it is well known that the boundary of an *n*-dimensional ball cannot be its retract (see, *e.g.*, [12]) and we get a contradiction.

Without any special comment, throughout the proof, we use the fact that the initial value problem has a unique solution and this solution depends continuously on the initial data (this is guaranteed by Theorem 1). Suppose now that the initial function {\phi}^{\ast} satisfying (6) generates the solution y=y(t) which does not satisfy (8) for at least one t\in {({t}_{0},\mathrm{\infty})}_{\mathbb{T}}. This means, in general, that for any rd-continuous initial function {\phi}_{0} satisfying the inequality

{b}_{i}(t)<{\phi}_{0i}(t)<{c}_{i}(t)\phantom{\rule{1em}{0ex}}\text{for all}t\in {[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}},i=1,\dots ,n,

(9)

there exists {t}^{0}\in \mathbb{T}, {t}^{0}>{t}_{0} such that, for a corresponding solution y={y}^{0}(t) of the initial problem

{y}^{0}(t)={\phi}_{0}(t),\phantom{\rule{1em}{0ex}}t\in {[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}},

we have

({t}^{0},{y}^{0}\left({t}^{0}\right))\notin \mathrm{\Omega}

and

(t,{y}^{0}(t))\in \mathrm{\Omega}\phantom{\rule{1em}{0ex}}\text{for all}t\in {[{t}_{0},{t}^{0})}_{\mathbb{T}}.

Let us define auxiliary mappings {P}_{1}, {P}_{2} and {P}_{3}. Note that in this part of the proof, without loss of generality, we admit the eventuality that the function {\phi}_{0}(t) instead of (9) satisfies weaker restrictions

{b}_{i}(t)<{\phi}_{0i}(t)<{c}_{i}(t)\phantom{\rule{1em}{0ex}}\text{for all}t\in {[{\alpha}_{0},{t}_{0})}_{\mathbb{T}},i=1,\dots ,n,

and {\phi}_{0}({t}_{0})\in \overline{\omega}({t}_{0}).

First, define a mapping {P}_{1}. For ({t}_{0},{\phi}_{0}({t}_{0})) with {\phi}_{0}({t}_{0})\in \overline{\omega}({t}_{0}),

{P}_{1}({t}_{0},{\phi}_{0}({t}_{0})):=\{\begin{array}{ll}({t}^{0},{y}^{0}({t}^{0}))& \text{if}{\phi}_{0}({t}_{0})\in \omega ({t}_{0}),\\ ({t}^{0},{y}^{0}({t}^{0}))\equiv ({t}_{0},{y}^{0}({t}_{0}))& \text{if}{\phi}_{0}({t}_{0})\in \partial \omega ({t}_{0})\end{array}

with {t}^{0} having the value defined above except for the case {\phi}_{0}({t}_{0})\in \partial \omega ({t}_{0}). In the latter case, we put {t}^{0}={t}_{0}.

Second, we define a mapping {P}_{2} for every ({t}^{0},{y}^{0}({t}^{0})). For this we will need a set

{\mathrm{\Omega}}_{\mathbb{R}}:=\{(t,y):t\in [{\alpha}_{0},\mathrm{\infty}),y\in {\omega}_{\mathbb{R}}(t)\}

with {\omega}_{\mathbb{R}}(t)\equiv \omega (t) if t\in \mathbb{T} and

\begin{array}{rcl}{\omega}_{\mathbb{R}}(t)& :=& \{y\in {\mathbb{R}}^{n}:{b}_{i}({t}_{a})+[{b}_{i}({t}_{b})-{b}_{i}({t}_{a})]\cdot \frac{t-{t}_{a}}{{t}_{b}-{t}_{a}}\\ <{y}_{i}<{c}_{i}({t}_{a})+[{c}_{i}({t}_{b})-{c}_{i}({t}_{a})]\cdot \frac{t-{t}_{a}}{{t}_{b}-{t}_{a}},i=1,\dots ,n\}\end{array}

if t\notin \mathbb{T} and {t}_{a}\in \mathbb{T}, {t}_{b}\in \mathbb{T} are such that {t}_{a}<{t}_{b}, t\in ({t}_{a},{t}_{b}) and ({t}_{a},{t}_{b})\cap \mathbb{T}=\mathrm{\varnothing}. It is clear that \mathrm{\Omega}\subseteq {\mathrm{\Omega}}_{\mathbb{R}}. Further,

V({t}_{a},{t}_{b}):=\{(t,y):{t}_{a}\le t\le {t}_{b},y\in {\overline{\omega}}_{\mathbb{R}}(t)\}

is obviously convex. We define, moreover, the *y*-part of the boundary of {\mathrm{\Omega}}_{\mathbb{R}} as

{\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}:=\{(t,y):t\in [{\alpha}_{0},\mathrm{\infty}),y\in \partial {\omega}_{\mathbb{R}}(t)\},

where \partial {\omega}_{\mathbb{R}}(t)\equiv \partial \omega (t) if t\in \mathbb{T} and

\begin{array}{rcl}\partial {\omega}_{\mathbb{R}}(t)& :=& \{y\in {\overline{\omega}}_{\mathbb{R}}(t):\prod _{i=1}^{n}({b}_{i}({t}_{a})+[{b}_{i}({t}_{b})-{b}_{i}({t}_{a})]\cdot \frac{t-{t}_{a}}{{t}_{b}-{t}_{a}}-{y}_{i})\\ \times ({y}_{i}-{c}_{i}({t}_{a})-[{c}_{i}({t}_{b})-{c}_{i}({t}_{a})]\cdot \frac{t-{t}_{a}}{{t}_{b}-{t}_{a}})=0\}\end{array}

if t\notin \mathbb{T}, where {t}_{a}\in \mathbb{T} and {t}_{b}\in \mathbb{T} are as above.

Now we are ready to consider an auxiliary mapping {P}_{2}. Let ({t}^{0},{y}^{0}({t}^{0}))\notin \mathrm{\Omega}. Then, due to the convexity of V({t}_{a},{t}_{b}), there exists a unique intersection of the segment connecting the points (\rho ({t}^{0}),{y}^{0}(\rho ({t}^{0}))) and ({t}^{0},{y}^{0}({t}^{0})) with {\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}. We denote this point as {M}^{0} and define

{P}_{2}({t}^{0},{y}^{0}\left({t}^{0}\right)):={M}^{0}.

Note that if ({t}^{0},{y}^{0}({t}^{0}))\in {\partial}_{y}\mathrm{\Omega}, we get the particular case {M}^{0}=({t}^{0},{y}^{0}({t}^{0})) and {P}_{2}({t}^{0},{y}^{0}({t}^{0}))=({t}^{0},{y}^{0}({t}^{0})).

Third, we define a mapping {P}_{3}. For this we need a subset of the *y*-boundary {\partial}_{y}\mathrm{\Omega}. For s\in \mathbb{T}, let

{\partial}_{y}\mathrm{\Omega}{|}_{t=s}:={\partial}_{y}\mathrm{\Omega}\cap \{(s,y):y\in {\mathbb{R}}^{n}\}.

Now we are ready to consider an auxiliary continuous mapping {P}_{3},

{P}_{3}:{\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}\to {\partial}_{y}\mathrm{\Omega}{|}_{t={t}_{0}},

defined for {M}^{0}=(t,y)\in {\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}} as a point {M}^{\ast}=({t}_{0},{y}^{\ast}), where {y}^{\ast}\in \partial \omega ({t}_{0}) with

{y}_{i}^{\ast}={b}_{i}({t}_{0})+\frac{{c}_{i}({t}_{0})-{b}_{i}({t}_{0})}{{c}_{i}(t)-{b}_{i}(t)}\cdot ({y}_{i}-{b}_{i}(t)),\phantom{\rule{1em}{0ex}}i=1,\dots ,n.

(10)

It is easy to see that the {\partial}_{y}\mathrm{\Omega}{|}_{t={t}_{0}} is a retract of {\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}} and satisfies all the assumptions from Definition 3 (with A:={\partial}_{y}\mathrm{\Omega}{|}_{t={t}_{0}} and B:={\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}).

We show that the composite mapping

{P}_{3}\circ {P}_{2}\circ {P}_{1}:\{({t}_{0},y):y\in \overline{\omega}({t}_{0})\}\to {\partial}_{y}\mathrm{\Omega}{|}_{t={t}_{0}}

is continuous due to the continuous dependence of the solutions on the initial data, the convexity of sets of the type V({t}_{a},{t}_{b}) defined above and the continuity of the mapping {P}_{3}. Let \delta >0 be sufficiently small and {\phi}_{0},{\phi}_{0,\delta}:{[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}}\to {\mathbb{R}}^{n} be the initial functions such that

\parallel {\phi}_{0}(t)-{\phi}_{0,\delta}(t)\parallel <\delta \phantom{\rule{1em}{0ex}}\text{for all}t\in {[{\alpha}_{0},{t}_{0}]}_{\mathbb{T}}

and

\underset{\delta \to 0}{lim}{\phi}_{0,\delta}={\phi}_{0}.

(11)

Further, let

{P}_{1}({t}_{0},{\phi}_{0}({t}_{0}))=({t}^{0},{y}^{0}\left({t}^{0}\right)),

(12)

{P}_{1}({t}_{0},{\phi}_{0,\delta}({t}_{0}))=({t}^{0,\delta},{y}^{0,\delta}\left({t}^{0,\delta}\right)).

(13)

Suppose now that {\phi}_{0}({t}_{0}),{\phi}_{0,\delta}({t}_{0})\in \omega ({t}_{0}) (which is equivalent to {t}^{0}>{t}_{0}). We consider all the possible settings of ({t}^{0},{y}^{0}({t}^{0})) and characters of {t}^{0}.

(I)Point ({t}^{0},{y}^{0}({t}^{0}))\in {\partial}_{y}\mathrm{\Omega}.

First, assume that the point {t}^{0} is dense, *i.e.*, \rho ({t}^{0})={t}^{0}=\sigma ({t}^{0}). Then, due to the solutions depending continuously on initial data, we have {lim}_{\delta \to 0}{t}^{0,\delta}={t}^{0} and, consequently,

\underset{\delta \to 0}{lim}{P}_{1}({t}_{0},{\phi}_{0,\delta}({t}_{0}))={P}_{1}({t}_{0},{\phi}_{0}({t}_{0}))=({t}^{0},{y}^{0}\left({t}^{0}\right)).

In this case the mapping {P}_{1} is continuous and the composite mapping {P}_{2}\circ {P}_{1} is continuous as well because

({P}_{2}\circ {P}_{1})({t}_{0},{\phi}_{0}({t}_{0}))=({t}^{0},{y}^{0}\left({t}^{0}\right))

and

\underset{\delta \to 0}{lim}({P}_{2}\circ {P}_{1})({t}_{0},{\phi}_{0,\delta}({t}_{0}))=\underset{\delta \to 0}{lim}{P}_{2}({t}^{0,\delta},{y}^{0,\delta}\left({t}^{0,\delta}\right))=({t}^{0},{y}^{0}\left({t}^{0}\right)).

Further, {P}_{3}\circ {P}_{2}\circ {P}_{1} is also continuous because of the continuity of mappings {P}_{3} and {P}_{2}\circ {P}_{1}.

Second, if the point {t}^{0} is left-scattered and right-dense, *i.e.*, \rho ({t}^{0})<{t}^{0}=\sigma ({t}^{0}), we can proceed analogously as in the case before. In this case, for fixed *δ*, either {t}^{0,\delta}={t}^{0} (if {y}^{0,\delta}({t}^{0})\notin \omega ({t}^{0})) or {t}^{0,\delta}>{t}^{0} (if {y}^{0,\delta}({t}^{0})\in \omega ({t}^{0})). In the alternative {t}^{0,\delta}={t}^{0}, it is obvious that

\underset{\delta \to 0}{lim}{y}^{0,\delta}\left({t}^{0,\delta}\right)={y}^{0}\left({t}^{0}\right)

and thus

\underset{\delta \to 0}{lim}({P}_{2}\circ {P}_{1})({t}_{0},{\phi}_{0,\delta}({t}_{0}))=\underset{\delta \to 0}{lim}{P}_{2}({t}^{0,\delta},{y}^{0,\delta}\left({t}^{0,\delta}\right))=({t}^{0},{y}^{0}\left({t}^{0}\right)).

Hence the composite mapping {P}_{2}\circ {P}_{1} is continuous. Further, {P}_{3}\circ {P}_{2}\circ {P}_{1} is also continuous because of the continuity of mappings {P}_{3} and {P}_{2}\circ {P}_{1}. The alternative {t}^{0,\delta}>{t}^{0} with {lim}_{\delta \to 0}{t}^{0,\delta}={t}^{0} can be proved by the same limit process as in the first case, where the point {t}^{0} is dense.

Third, let the point {t}^{0} be left-dense and right-scattered, *i.e.*, \rho ({t}^{0})={t}^{0}<\sigma ({t}^{0}). Then the approach used in previous cases can be modified as follows. Let, as before, (11), (12) and (13) hold. Then, for fixed *δ*, either {t}^{0,\delta}\le {t}^{0} (if {y}^{0,\delta}({t}^{0})\notin \omega ({t}^{0})) or {t}^{0,\delta}=\sigma ({t}^{0}) (if {y}^{0,\delta}({t}^{0})\in \omega ({t}^{0})). The alternative {t}^{0,\delta}\le {t}^{0} with {lim}_{\delta \to 0}{t}^{0,\delta}={t}^{0} can be proved by the same limit process as in the first case, where the point {t}^{0} is dense. However, the alternative {t}^{0,\delta}=\sigma ({t}^{0}) takes into account a possibility that the mapping {P}_{1} cannot be continuous. If {t}^{0,\delta}=\sigma ({t}^{0}) is valid for \delta \to 0, then, due to the convexity of V({t}^{0},\sigma ({t}^{0})), there exists a unique intersection of the segment connecting the points ({t}^{0},{y}^{0,\delta}({t}^{0})) and (\sigma ({t}^{0}),{y}^{0,\delta}(\sigma ({t}^{0}))) with {\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}. We denote this point as {M}^{0,\delta} and, in accordance with the above definition, {P}_{2}(\sigma ({t}^{0}),{y}^{0,\delta}(\sigma ({t}^{0})))={M}^{0,\delta}. We wish to show that {M}^{0,\delta}\to {M}^{0}=({t}^{0},{y}^{0}({t}^{0})) if \delta \to 0. However, in view of the definition of the mapping {P}_{2} and its geometric meaning,

\underset{\delta \to 0}{lim}{P}_{2}(\sigma \left({t}^{0}\right),{y}^{0,\delta}\left(\sigma \left({t}^{0}\right)\right))=\underset{\delta \to 0}{lim}{M}^{0,\delta}=({t}^{0},{y}^{0}\left({t}^{0}\right))

and

\underset{\delta \to 0}{lim}({P}_{2}\circ {P}_{1})({t}_{0},{\phi}_{0,\delta}({t}_{0}))=({t}^{0},{y}^{0}\left({t}^{0}\right)).

Hence the continuity of {P}_{2}\circ {P}_{1} is proved. Further, {P}_{3}\circ {P}_{2}\circ {P}_{1} is also continuous (for the same reason as before).

Fourth, suppose now that {t}^{0} is an isolated point, *i.e.*, \rho ({t}^{0})<{t}^{0}<\sigma ({t}^{0}). Let, as before, (11), (12) and (13) hold. Then, for sufficiently small *δ*, we have either {t}^{0,\delta}={t}^{0} (if {y}^{0,\delta}({t}^{0})\notin \omega ({t}^{0})) or {t}^{0,\delta}=\sigma ({t}^{0}) (if {y}^{0,\delta}({t}^{0})\in \omega ({t}^{0})). Without any special comment, in the alternative {t}^{0,\delta}={t}^{0}, we proceed in the same way as in the second case, where {t}^{0} is left-scattered and right-dense. Furthermore, in the alternative {t}^{0,\delta}=\sigma ({t}^{0}), we proceed in the same way as in the third case, where {t}^{0} is left-dense and right-scattered.

(II)Point ({t}^{0},{y}^{0}({t}^{0}))\notin {\partial}_{y}\mathrm{\Omega}. Then the point {t}^{0} is left-scattered and only two cases are possible (either {t}^{0} is left-scattered and right-dense or {t}^{0} is isolated). In both mentioned cases, we can proceed in the same way. Let, as before, {\phi}_{0}({t}_{0}),{\phi}_{0,\delta}({t}_{0})\in \omega ({t}_{0}) and (11), (12) and (13) hold. Then, for sufficiently small *δ*, we have only {t}^{0,\delta}={t}^{0} and, of course, {y}^{0,\delta}({t}^{0})\notin \overline{\omega}({t}^{0}). Further,

\underset{\delta \to 0}{lim}{P}_{1}({t}_{0},{\phi}_{0,\delta}({t}_{0}))=\underset{\delta \to 0}{lim}({t}^{0,\delta},{y}^{0,\delta}\left({t}^{0,\delta}\right))=({t}^{0},{y}^{0}\left({t}^{0}\right))\notin \overline{\mathrm{\Omega}}.

Due to the convexity of V(\rho ({t}^{0}),{t}^{0}), there exists a unique intersection of the segment connecting the points (\rho ({t}^{0}),{y}^{0,\delta}(\rho ({t}^{0}))) and ({t}^{0},{y}^{0,\delta}({t}^{0})) with {\partial}_{y}{\mathrm{\Omega}}_{\mathbb{R}}. We denote this point by {M}^{0,\delta}. In view of the definition of the mapping {P}_{2} and its geometric meaning, we can observe that

\underset{\delta \to 0}{lim}({P}_{2}\circ {P}_{1})({t}_{0},{\phi}_{0,\delta}({t}_{0}))=\underset{\delta \to 0}{lim}{P}_{2}({t}^{0},{y}^{0,\delta}\left({t}^{0}\right))=\underset{\delta \to 0}{lim}{M}^{0,\delta}={M}^{0}.

Hence the composite mapping {P}_{2}\circ {P}_{1} is continuous. Moreover, {P}_{3}\circ {P}_{2}\circ {P}_{1} is also continuous because of the continuity of mappings {P}_{3} and {P}_{2}\circ {P}_{1}.

We proved that the composite mapping {P}_{3}\circ {P}_{2}\circ {P}_{1} is continuous. Note that we omitted the special case {\phi}_{0}({t}_{0})\in \partial \omega ({t}_{0}) (which is equivalent to {t}^{0}={t}_{0}). However, this part can be shown in an analogous and simpler way to the one used above.

Now we are able to finish the proof. We proved that

P:={P}_{3}\circ {P}_{2}\circ {P}_{1}:B\to A,

where A:={\partial}_{y}\mathrm{\Omega}{|}_{t={t}_{0}}, B:=\{({t}_{0},y):y\in \overline{\omega}({t}_{0})\} is continuous. Moreover,

is an identity mapping. In this situation, we have proved that there exists a retraction of the set *B* onto the set *A* (see Definition 3). In view of the above-mentioned fact, this is impossible. Our assumption is false and there exists initial problem (7) such that the corresponding solution y={y}^{\ast}(t) satisfies (8) for every t\in {[{\alpha}_{0},\mathrm{\infty})}_{\mathbb{T}}. The theorem is proved. □