The proof of the following theorem is based on the retract method, which is well known for ordinary differential equations and goes back to Ważewski [18]. Below we will assume that the function f, except for the indicated conditions, satisfies all the assumptions given in Section 2.
Theorem 2 Let f:\mathbb{T}\times \mathbb{R}\to \mathbb{R}. Let b,c:\mathbb{T}\to \mathbb{R} be delta differentiable functions such that b(t)<c(t) for each t\in {[\tau ({t}_{0}),\mathrm{\infty})}_{\mathbb{T}}. If, moreover, every point M\in \partial {\mathrm{\Omega}}_{B}\cup \partial {\mathrm{\Omega}}_{C} is the point of strict egress for the set Ω with respect to equation (1), then there exists an rdcontinuous initial function {\phi}^{\ast}:{[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}}\to \mathbb{R} satisfying
b(t)<{\phi}^{\ast}(t)<c(t)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}t\in {[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}},
such that the initial problem
y(t)={\phi}^{\ast}(t),\phantom{\rule{1em}{0ex}}t\in {[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}}
(6)
defines a solution y of (1) on the interval {[\tau ({t}_{0}),\mathrm{\infty})}_{\mathbb{T}} satisfying
b(t)<y(t)<c(t)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}t\in {[\tau ({t}_{0}),\mathrm{\infty})}_{\mathbb{T}}.
(7)
Proof The idea of the proof is simple. We suppose that the statement of the theorem is not valid. Then it is possible to prove that there exists a retraction of a segment B:=[\alpha ,\beta ] with \alpha <\beta onto a twopoint set A:=\{\alpha ,\beta \}. But it is well known that the boundary of a nonempty (closed) interval cannot be its retract (see [19]). So, in our case, such a retractive mapping cannot exist because it is incompatible with continuity.
Without any special comment, throughout the proof, we use the property that the initial value problem in question has a unique solution and the property of continuous dependence of solutions on their initial data.
Suppose now that {\phi}^{\ast} satisfying the inequality
b(t)<{\phi}^{\ast}(t)<c(t)\phantom{\rule{1em}{0ex}}\text{for all}t\in {[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}}
and generating the solution y=y(t) which satisfies (7) for any t\in {[\tau ({t}_{0}),\mathrm{\infty})}_{\mathbb{T}} does not exist. This means that for any rdcontinuous initial function {\phi}_{0} satisfying the inequality
b(t)<{\phi}_{0}(t)<c(t)\phantom{\rule{1em}{0ex}}\text{for all}t\in {[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}},
(8)
there exists a {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 {[\tau ({t}_{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}.
First, define the mapping {P}_{1}:{B}_{1}\to \mathbb{T}\times \mathbb{R}, where
{B}_{1}=\{({t}_{0},{\phi}_{0}({t}_{0}))\in \mathbb{T}\times \mathbb{R}:b({t}_{0})\le {\phi}_{0}({t}_{0})\le c({t}_{0})\},
such that

(i)
for {\phi}_{0}(t), b(t)<{\phi}_{0}(t)<c(t), t\in {[\tau ({t}_{0}),{t}_{0}]}_{\mathbb{T}}, we define
{P}_{1}:({t}_{0},{\phi}_{0}({t}_{0}))\to ({t}^{0},{y}^{0}\left({t}^{0}\right));

(ii)
for {\phi}_{0}(t) satisfying b(t)<{\phi}_{0}(t)<c(t), t\in {[\tau ({t}_{0}),{t}_{0})}_{\mathbb{T}} and {\phi}_{0}({t}_{0})=b({t}_{0}), we put {t}^{0}={t}_{0} and define
{P}_{1}:({t}_{0},{\phi}_{0}({t}_{0}))\to ({t}^{0},{y}^{0}\left({t}^{0}\right))=({t}_{0},b({t}_{0}));

(iii)
for {\phi}_{0}(t), b(t)<{\phi}_{0}(t)<c(t), t\in {[\tau ({t}_{0}),{t}_{0})}_{\mathbb{T}} and {\phi}_{0}({t}_{0})=c({t}_{0}), we put {t}^{0}={t}_{0} and define
{P}_{1}:({t}_{0},{\phi}_{0}({t}_{0}))\to ({t}^{0},{y}^{0}\left({t}^{0}\right))=({t}_{0},c({t}_{0})).
Second, we define the mapping {P}_{2}:{B}_{2}\to \mathbb{T}\times \mathbb{R}, where
{B}_{2}=P({B}_{1})=\{({t}^{0},{y}^{0}({t}_{0}))\in \mathbb{T}\times \mathbb{R}:{y}^{0}({t}_{0})\le b\left({t}^{0}\right)\text{or}{y}^{0}({t}_{0})\ge c\left({t}^{0}\right)\},
as
{P}_{2}:({t}^{0},{y}^{0}\left({t}^{0}\right))\to \{\begin{array}{cc}({t}^{0},c({t}^{0}))\hfill & \text{if}{y}^{0}({t}^{0})\ge c({t}^{0}),\hfill \\ ({t}^{0},b({t}^{0}))\hfill & \text{if}{y}^{0}({t}^{0})\le b({t}^{0}).\hfill \end{array}
Third, we define the mapping {P}_{3}:{B}_{3}\to \mathbb{T}\times \mathbb{R}, where
{B}_{3}=P({B}_{2})=\{({t}^{0},\tilde{y})\in \mathbb{T}\times \mathbb{R}:\tilde{y}=b\left({t}^{0}\right)\text{or}\tilde{y}=c\left({t}^{0}\right)\},
as
{P}_{3}:({t}^{0},\tilde{y})\to \{\begin{array}{cc}({t}_{0},c({t}_{0}))\hfill & \text{if}\tilde{y}=c({t}^{0}),\hfill \\ ({t}_{0},b({t}_{0}))\hfill & \text{if}\tilde{y}=b({t}^{0}).\hfill \end{array}
We will show that the composite mapping
P:={P}_{3}\circ {P}_{2}\circ {P}_{1},\phantom{\rule{2em}{0ex}}P:{B}_{1}\to {A}_{1},
where
{A}_{1}=\{({t}_{0},b({t}_{0})),({t}_{0},c({t}_{0}))\},
is continuous with respect to the second coordinate {\phi}_{0}({t}_{0}) of the point ({t}_{0},{\phi}_{0}({t}_{0}))\in {B}_{1}. The definition of the mapping P implies that only two resulting points are possible, namely either P({B}_{1})=({t}_{0},c({t}_{0})) or P({B}_{1})=({t}_{0},b({t}_{0})).

(I)
We consider the first possibility, i.e., P({B}_{1})=({t}_{0},c({t}_{0})). Let b(t)<{\phi}_{0}(t)<c(t) for all t\in {[\tau ({t}_{0}),{t}_{0})}_{\mathbb{T}}. Then
{P}_{1}({t}_{0},{\phi}_{0}({t}_{0}))=({t}^{0},{y}^{0}\left({t}^{0}\right)),({t}^{0},{y}^{0}\left({t}^{0}\right))\notin \mathrm{\Omega}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{y}^{0}\left({t}^{0}\right)\ge c\left({t}^{0}\right).
Let {y}^{0}({t}^{0})>c({t}^{0}). Then \rho ({t}^{0})<{t}^{0} and the continuity of the mapping {P}_{1} is obvious. Indeed, if
{\phi}^{0,\epsilon}(t),\phantom{\rule{1em}{0ex}}t\in [\tau ({t}_{0}),{t}_{0}],\phantom{\rule{2em}{0ex}}b(t)<{\phi}^{0,\epsilon}(t)<c(t),\phantom{\rule{1em}{0ex}}t\in [\tau ({t}_{0}),{t}_{0})
(9)
is the initial problem defining the solution {y}^{\epsilon}(t), ε is a sufficiently small number and
{\phi}^{0,\epsilon}(t){\phi}_{0}(t)<\epsilon ,\phantom{\rule{1em}{0ex}}t\in [\tau ({t}_{0}),{t}_{0}],
then due to the property of continuous dependence of solutions on their initial data, {t}^{0,\epsilon}={t}^{0} and
{P}_{1}({t}_{0},{\phi}^{0,\epsilon}({t}_{0}))=({t}^{0,\epsilon},{y}^{\epsilon}\left({t}^{0,\epsilon}\right))=({t}^{0},{y}^{\epsilon}\left({t}^{0}\right))\phantom{\rule{1em}{0ex}}\text{with}\phantom{\rule{1em}{0ex}}{y}^{\epsilon}\left({t}^{0}\right)>c\left({t}^{0}\right).
Consequently, P({t}_{0},{\phi}^{0,\epsilon}({t}_{0}))=({t}_{0},c({t}_{0})).
Let {y}^{0}({t}^{0})=c({t}^{0}). By the assumption of the theorem, every boundary point of ∂ Ω is the point of strict egress for the set Ω with respect to equation (1). Then for the solution {y}^{\epsilon}(t) defined by (9), we have
{P}_{1}({t}_{0},{\phi}^{0,\epsilon}({t}_{0}))=({t}^{0,\epsilon},{y}^{\epsilon}\left({t}^{0,\epsilon}\right))
either with {y}^{\epsilon}({t}^{0,\epsilon})>c({t}^{0,\epsilon}) or with {t}^{0,\epsilon}={t}^{0}, {y}^{\epsilon}({t}^{0,\epsilon})=c({t}_{0}). (We do not describe all the possibilities for the occurrence of the first or of the second alternative.) In both alternatives we get P({t}_{0},{\phi}^{0,\epsilon}({t}_{0}))=({t}_{0},c({t}_{0})) again. Hence, the mapping P is continuous in the considered case.

(II)
We proceed analogously with the case P({B}_{1})=({t}_{0},b({t}_{0})).
The continuity of the mapping P was proved for initial functions {\phi}_{0} satisfying b(t)<{\phi}_{0}(t)<c(t) for all t\in {[\tau ({t}_{0}),{t}_{0})}_{\mathbb{T}} and b({t}_{0})<{\phi}_{0}({t}_{0})<c({t}_{0}). The desired retraction {P}_{r} can be defined as a mapping of the second coordinates realized by P. Then the mapping
[b({t}_{0}),c({t}_{0})]\stackrel{{P}_{r}}{\u27f6}\{b({t}_{0}),c({t}_{0})\}
is continuous and
\{b({t}_{0})\}\stackrel{{P}_{r}}{\u27f6}\{b({t}_{0})\},\phantom{\rule{2em}{0ex}}\{c({t}_{0})\}\stackrel{{P}_{r}}{\u27f6}\{c({t}_{0})\},
i.e., the points \{b({t}_{0})\}, \{c({t}_{0})\} are stationary.
In this situation we proved that there exists a retraction {P}_{r} of the set B:=[b({t}_{0}),c({t}_{0})] onto the twopoint set A:=\{b({t}_{0}),c({t}_{0})\} (see Definition 3). In regard to the above mentioned fact, this is impossible. Our supposition is false, and there exists the initial problem (6) such that the corresponding solution y={y}^{\ast}(t) satisfies the inequalities (7) for every t\in {[\tau ({t}_{0}),\mathrm{\infty}]}_{\mathbb{T}}. The theorem is proved. □