This section states and proves our results. Throughout this paper, we always assume that \(t_{0}\in\mathbb{T}\) and \(\mathbb{T}_{0}=[t_{0},\infty )\cap\mathbb{T}\).
Theorem 2.1
Suppose that
\(u, a\in C_{\mathrm{rd}}(\mathbb{T}_{0},\mathbb{R^{+}})\), a
is nondecreasing, \(g(t,s)\), \(g^{\Delta_{t}}(t,s)\), \(h(t,s)\), \(h^{\Delta _{t}}(t,s)\in C_{\mathrm{rd}}(\mathbb{T}_{0}\times\mathbb{T}_{0},\mathbb {R^{+}})\), \(\omega_{1}\), \(\omega_{2}\in C(\mathbb{R^{+}},\mathbb{R^{+}})\)
are nondecreasing, \(\tau(t)\in(\mathbb{T}_{0},\mathbb{T})\)
with
\(\tau (t)< t\), \(-\infty<\alpha=\inf\{\tau(t), t\in\mathbb{T}_{0}\}\leq t_{0}\), and
\(\phi\in C_{\mathrm{rd}}([\alpha,t_{0}]\cap\mathbb{T},\mathbb {R^{+}})\). If for
\(t\in\mathbb{T}_{0}\), u
satisfies the inequality:
$$ \omega_{1}\bigl(u(t)\bigr)\leq a(t)+ \int_{t_{0}}^{t}g(t,s) \omega_{1}\bigl(u \bigl(\tau (s)\bigr)\bigr)\Delta s + \int_{t_{0}}^{t} h(t,s)\omega_{2}\bigl(u \bigl(\tau(s)\bigr)\bigr)\Delta s $$
(2.1)
with the initial condition
$$ \textstyle\begin{cases} \omega_{1}(u(t)) = \phi(t) \quad \textit{if } t\in[\alpha,t_{0}]\cap\mathbb{T}, \\ \phi(\tau(t)) \leq a(t) \quad \textit{if } \tau(t)\leq t_{0}, t\in\mathbb{T}_{0}, \end{cases} $$
(2.2)
then
$$ u(t)\leq\omega_{1}^{-1} \biggl\{ G^{-1} \biggl[H^{-1} \biggl\{ H \biggl(G\bigl(a(t)\bigr)+ \int_{t_{0}}^{t}h(t,s)\Delta s \biggr)+ \int _{t_{0}}^{t}g(t,s)\Delta s \biggr\} \biggr] \biggr\} $$
(2.3)
with
$$ H \biggl(G\bigl(a(t)\bigr)+ \int_{t_{0}}^{t}h(t,s)\Delta s \biggr)+ \int _{t_{0}}^{t}g(t,s)\Delta s\in \operatorname{Dom}\bigl(H^{-1}\bigr), $$
where
G, H
are increasing bijective functions defined by
$$\begin{aligned}& (G\circ\nu)^{\Delta}(t) =\frac{\nu^{\Delta}(t)}{\omega_{2}(\omega _{1}^{-1}(\nu(t)))}, \end{aligned}$$
(2.4)
$$\begin{aligned}& (H\circ z)^{\Delta}(t) =\frac{\omega_{2}(\omega _{1}^{-1}(G^{-1}(z(t))))z^{\Delta}(t)}{G^{-1}(z(t))}. \end{aligned}$$
(2.5)
Proof
Fixing an arbitrary number \(T^{*}\in\mathbb{T}_{0}\), for \(t\in [t_{0},T^{*}]\cap\mathbb{T}\), we define the function
$$ \nu(t)=a\bigl(T^{*}\bigr)+ \int_{t_{0}}^{t}g(t,s) \omega_{1}\bigl(u \bigl(\tau(s)\bigr)\bigr)\Delta s + \int_{t_{0}}^{t} h(t,s)\omega_{2}\bigl(u \bigl(\tau(s)\bigr)\bigr)\Delta s. $$
(2.6)
Clearly, ν is a nonnegative nondecreasing function, and we have
$$ u(t)\leq\omega_{1}^{-1}\bigl(\nu(t)\bigr), \quad t\in\bigl[t_{0},T^{*}\bigr]\cap\mathbb{T}. $$
(2.7)
Therefore, for \(t\in[t_{0},T^{*}]\cap\mathbb{T}\), if \(\tau(t)\geq t_{0}\), then \(\tau(t)\in[t_{0},T^{*}]\cap\mathbb{T}\), and from (2.7) we obtain
$$ u\bigl(\tau(t)\bigr)\leq\omega_{1}^{-1}\bigl( \nu\bigl(\tau(t)\bigr)\bigr)\leq\omega _{1}^{-1}\bigl(\nu(t) \bigr). $$
(2.8)
On the other hand, if \(\tau(t)\leq t_{0}\), then, using the initial condition (2.2), we get
$$ u\bigl(\tau(t)\bigr)=\phi\bigl(\tau(t)\bigr)\leq a(t)\leq \omega_{1}^{-1}\bigl(\nu(t)\bigr). $$
(2.9)
So, combining (2.8) and (2.9), we have
$$ u\bigl(\tau(t)\bigr)\leq\omega_{1}^{-1}\bigl( \nu(t)\bigr). $$
(2.10)
Moreover, from (2.6), Lemma 1.2, and (2.10) we get
$$\begin{aligned} \nu^{\Delta}(t) =& \int_{t_{0}}^{t}g^{\Delta_{t}}(t,s) \omega_{1}\bigl(u\bigl(\tau (s)\bigr)\bigr)\Delta s +g\bigl(\sigma(t),t \bigr) \omega_{1}\bigl(u\bigl(\tau(t)\bigr)\bigr) \\ &{}+ \int_{t_{0}}^{t} h^{\Delta_{t}}(t,s) \omega_{2}\bigl(u\bigl(\tau(s)\bigr)\bigr)\Delta s+h\bigl(\sigma(t),t \bigr)\omega_{2}\bigl(u\bigl(\tau(t)\bigr)\bigr) \\ \leq& \int_{t_{0}}^{t}g^{\Delta_{t}}(t,s)\nu(s)\Delta s +g \bigl(\sigma (t),t\bigr) \nu(t) \\ &{}+ \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s) \omega_{2}\bigl(\omega_{1}^{-1}\bigl(\nu (t)\bigr) \bigr)\Delta s+h\bigl(\sigma(t),t\bigr)\omega_{2}\bigl( \omega_{1}^{-1}\bigl(\nu (s)\bigr)\bigr) \\ \leq& \biggl\{ \int_{t_{0}}^{t}g^{\Delta_{t}}(t,s)\Delta s+g\bigl( \sigma (t),t\bigr) \biggr\} \nu(t) \\ &{}+ \biggl\{ \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s)\Delta s+h\bigl( \sigma (t),t\bigr) \biggr\} \omega_{2}\bigl(\omega_{1}^{-1} \bigl(\nu(t)\bigr)\bigr), \end{aligned}$$
(2.11)
which implies
$$ \frac{\nu^{\Delta}(t)}{\omega_{2}(\omega_{1}^{-1}(\nu(t)))}\leq \biggl\{ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr\} ^{\Delta_{t}} \frac{\nu (t)}{\omega_{2}(\omega_{1}^{-1}(\nu(t)))}+ \biggl\{ \int _{t_{0}}^{t}h(t,s)\Delta s \biggr\} ^{\Delta_{t}}. $$
(2.12)
Integrating (2.12) from \(t_{0}\) to t yields
$$ G\bigl(\nu(t)\bigr)-G\bigl(\nu(t_{0})\bigr)\leq \int_{t_{0}}^{t} \biggl\{ \int _{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr\} ^{\Delta_{s}} \frac{\nu(s)}{\omega_{2}(\omega_{1}^{-1}(\nu(s)))}\Delta s+ \int _{t_{0}}^{t}h(t,s)\Delta s. $$
(2.13)
Since \(\nu(t_{0})=a(T^{*})\) and G is increasing, from (2.13) we have
$$\begin{aligned} \nu(t) \leq& G^{-1} \biggl[G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{t} \biggl\{ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr\} ^{\Delta_{s}} \frac{\nu(s)}{\omega_{2}(\omega_{1}^{-1}(\nu(s)))}\Delta s+ \int _{t_{0}}^{t}h(t,s)\Delta s \biggr] \\ \leq&G^{-1} \biggl[G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{t} \biggl\{ \int _{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr\} ^{\Delta_{s}} \frac{\nu(s)}{\omega_{2}(\omega_{1}^{-1}(\nu(s)))}\Delta s \\ &{}+ \int _{t_{0}}^{T^{*}}h(t,s)\Delta s \biggr]. \end{aligned}$$
(2.14)
For brevity, let
$$ z(t)= G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{t} \biggl\{ \int_{t_{0}}^{s}g(s,\lambda )\Delta\lambda \biggr\} ^{\Delta_{s}} \frac{\nu(s)}{\omega_{2}(\omega_{1}^{-1}(\nu(s)))}\Delta s + \int _{t_{0}}^{T^{*}}h(t,s)\Delta s, $$
(2.15)
and hence
$$z(t_{0})=G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}h(t,s)\Delta s. $$
Then
$$ \nu(t)\leq G^{-1}\bigl(z(t)\bigr),\quad t\in \bigl[t_{0},T^{*}\bigr]\cap\mathbb{T}. $$
(2.16)
Using Lemma 1.2 and (2.15), we obtain
$$\begin{aligned} z^{\Delta}(t) =& \biggl\{ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr\} ^{\Delta_{t}} \frac{\nu(t)}{\omega_{2}(\omega_{1}^{-1}(\nu(t)))} \\ \leq& \biggl\{ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr\} ^{\Delta_{t}} \frac{G^{-1}(z(t))}{\omega_{2}(\omega_{1}^{-1}(G^{-1}(z(t))))}, \end{aligned}$$
(2.17)
which implies
$$ \frac{\omega_{2}(\omega_{1}^{-1}(G^{-1}z(t)))z^{\Delta}(t)}{G^{-1}(z(t))}\leq \biggl\{ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr\} ^{\Delta_{t}}. $$
(2.18)
Integrating (2.18) from \(t_{0}\) to t yields
$$ H\bigl(z(t)\bigr)-H\bigl(z(t_{0})\bigr)\leq \int_{t_{0}}^{t}g(t,s)\Delta s, $$
from which it follows that
$$\begin{aligned} z(t) \leq& H^{-1} \biggl\{ H\bigl(z(t_{0})\bigr)+ \int_{t_{0}}^{t}g(t,s) \Delta s \biggr\} \\ =& H^{-1} \biggl\{ H \biggl(G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}h(t,s)\Delta s \biggr)+ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr\} . \end{aligned}$$
(2.19)
From (2.9), (2.16), and (2.19) we obtain
$$ u(t)\leq\omega_{1}^{-1} \biggl\{ G^{-1} \biggl[H^{-1} \biggl\{ H \biggl(G\bigl(a\bigl(T^{*} \bigr)\bigr)+ \int_{t_{0}}^{T^{*}}h(t,s)\Delta s \biggr)+ \int _{t_{0}}^{t}g(t,s)\Delta s \biggr\} \biggr] \biggr\} . $$
(2.20)
Taking \(t=T^{*}\) in (2.20) yields
$$ u\bigl(T^{*}\bigr)\leq\omega_{1}^{-1} \biggl\{ G^{-1} \biggl[H^{-1} \biggl\{ H \biggl[G\bigl(a\bigl(T^{*} \bigr)\bigr)+ \int_{t_{0}}^{T^{*}}h(t,s)\Delta s \biggr]+ \int _{t_{0}}^{T^{*}}g(t,s)\Delta s \biggr\} \biggr] \biggr\} . $$
(2.21)
Since \(T^{*}\in\mathbb{T}_{0}\) is chosen arbitrarily, we obtain the desired inequality after substituting \(T^{*}\) with t into (2.21). This completes the proof. □
Remark 2.2
It is interesting to note that, in particular, if we put \(\omega _{1}(u)=u^{p}\) and \(g(t,s)=0\) in Theorem 2.1, then Theorem 2.1 reduces to [37, Thm. 3.1]. If also \(\omega_{1}(u)=u^{p}\), \(g(t,s)=0\), and \(h(t,s)=b(t)f(s)\) in Theorem 2.1, then Theorem 2.1 reduces to [37, Thm. 3.2].
Theorem 2.1 unifies some known continuous and discrete inequalities in the literature as shown in the following remarks.
Remark 2.3
If we put \(\mathbb{T}=\mathbb{R}\), \(\omega_{1}(u)=u^{p}\), \(\omega _{2}(u)=u^{q}\), \(g(t,s)=f(s)\), \(h(t,s)=h(s)\), \(t_{0}=0\), and \(a(t)=u_{0}\) (any constant) in Theorem 2.1, then Theorem 2.1 reduces to [7, Thm. 3.1].
Remark 2.4
If we put \(\mathbb{T}=\mathbb{R}\), \(\omega_{1}(u)=u\), \(g(t,s)=0\), \(t_{0}=0\) in Theorem 2.1, then Theorem 2.1 reduces to [39, Thm. 2.1]. If furthermore \(h(t,s)=f_{1}(t)f_{2}(s)\) in Theorem 2.1, then Theorem 2.1 reduces to [39, Thm. 2.2].
Remark 2.5
If we take \(\mathbb{T}=\mathbb{R}\), \(\omega_{1}(u)=u^{p}\), \(g(t,s)=0\), \(t_{0}=0\), \(\tau(t)=t\), and \(h(t,s)=b(t)f(s)\) in Theorem 2.1, then Theorem 2.1 reduces to [8, Thm. 2(\(b_{3}\))]. If \(\mathbb{T}=\mathbb{Z}\), \(\omega_{1}(u)=u^{p}\), \(g(t,s)=0\), \(t_{0}=0\), \(\tau (t)=t\), and \(h(t,s)=b(t)f(s)\) in Theorem 2.1, then Theorem 2.1 reduces to [8, Thm. 4(\(d_{3}\))].
Theorem 2.6
Let
u, a, τ, \(\omega_{1}\), \(\omega_{2}\), g
be the same as in Theorem 2.1, and let
\(f,p\in C_{\mathrm{rd}}(\mathbb{T}_{0},\mathbb {R^{+}})\). If
\(t\in\mathbb{T}_{0}\)
and
u
satisfies the inequality
$$\begin{aligned} \omega_{1}\bigl(u(t)\bigr) \leq& a(t)+ \int_{t_{0}}^{t}\omega_{2}\bigl(u\bigl( \tau(s)\bigr)\bigr) \bigl[ f(s)\omega_{1}\bigl(u\bigl(\tau(s)\bigr) \bigr)+p(s) \bigr]\Delta s \\ &{}+ \int_{t_{0}}^{t}\omega_{2}\bigl(u\bigl( \tau(s)\bigr)\bigr)f(s) \biggl[ \int_{t_{0}}^{s}g(s,\lambda)\omega_{1} \bigl(u\bigl(\tau\bigl(\tau(\lambda )\bigr)\bigr)\bigr)\Delta\lambda \biggr]\Delta s, \end{aligned}$$
(2.22)
then
$$\begin{aligned} u(t) \leq&\omega_{1}^{-1} \biggl\{ \tilde{G}^{-1} \biggl[\tilde {H}^{-1} \biggl\{ \tilde{H}\biggl[G\bigl(a(t)\bigr)+ \int_{t_{0}}^{t}p(s)\Delta s\biggr] \\ &{}+ \int_{t_{0}}^{t}f(s) \biggl[1+ \int _{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr] \Delta s \biggr\} \biggr] \biggr\} \end{aligned}$$
(2.23)
with
$$ \tilde{H} \biggl[G\bigl(a(t)\bigr)+ \int_{t_{0}}^{t}p(s)\Delta s \biggr]+ \int _{t_{0}}^{t}f(s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \in \operatorname{Dom}\bigl(\tilde{H}^{-1}\bigr), $$
where
G̃
and
H̃
are increasing bijective functions defined by
$$\begin{aligned}& (\tilde{G}\circ\nu)^{\Delta}(t) =\frac{\nu^{\Delta}(t)}{\omega _{2}(\omega_{1}^{-1}(\nu(t)))}, \end{aligned}$$
(2.24)
$$\begin{aligned}& (\tilde{H}\circ z)^{\Delta}(t) =\frac{z^{\Delta}(t)}{\tilde{G}^{-1}(z(t))}. \end{aligned}$$
(2.25)
Proof
Fixing an arbitrary number \(T^{*}\in\mathbb{T}_{0}\), for \(t\in [t_{0},T^{*}]\cap\mathbb{T}\), we define the function
$$\begin{aligned} \nu(t) =& a\bigl(T^{*}\bigr)+ \int_{t_{0}}^{t}\omega_{2}\bigl(u\bigl( \tau(s)\bigr)\bigr) \bigl[ f(s)\omega_{1}\bigl(u\bigl(\tau(s)\bigr) \bigr)+p(s) \bigr]\Delta s \\ &{}+ \int_{t_{0}}^{t}\omega_{2}\bigl(u\bigl( \tau(s)\bigr)\bigr)f(s) \biggl[ \int_{t_{0}}^{s}g(s,\lambda)\omega_{1} \bigl(u\bigl(\tau\bigl(\tau(\lambda)\bigr)\bigr)\bigr)\Delta \lambda \biggr]\Delta s. \end{aligned}$$
(2.26)
Clearly, ν is a nonnegative nondecreasing function with \(\nu (t_{0})=a(T^{*})\), and we have
$$ u(t)\leq\omega_{1}^{-1}\bigl(\nu(t)\bigr), \quad t\in\bigl[t_{0},T^{*}\bigr]\cap\mathbb{T}. $$
(2.27)
Similarly to (2.8)–(2.9), we get
$$ u\bigl(\tau(t)\bigr)\leq\omega_{1}^{-1}\bigl( \nu(t)\bigr). $$
(2.28)
Furthermore, from (2.26), by Lemma 1.2 and (2.28) we have
$$\begin{aligned} \nu^{\Delta}(t) =&\omega_{2}\bigl(u\bigl(\tau(t)\bigr)\bigr) \bigl[f(t)\omega_{1}\bigl(u\bigl(\tau (t)\bigr)\bigr)+p(t) \bigr] \\ &{}+\omega_{2}\bigl(u\bigl(\tau(t)\bigr)\bigr)f(t) \biggl[ \int_{t_{0}}^{t}g(t,s)\omega _{1}\bigl(u \bigl(\tau(s)\bigr)\bigr)\Delta s \biggr] \\ \leq&p(t)\omega_{2}\bigl(\omega_{1}^{-1}\bigl( \nu(t)\bigr)\bigr)+f(t)\omega_{2}\bigl(\omega _{1}^{-1} \bigl(\nu(t)\bigr)\bigr)\biggl[\nu(t)+ \int_{t_{0}}^{t}g(t,s)\nu(s)\Delta s \biggr], \end{aligned}$$
(2.29)
which implies
$$ \frac{\nu^{\Delta}(t)}{\omega_{2}(\omega_{1}^{-1}(\nu(t)))}\leq p(t)+f(t) \biggl[\nu(t)+ \int_{t_{0}}^{t}g(t,s)\nu(s)\Delta s \biggr]. $$
(2.30)
Integrating both sides of (2.30) from \(t_{0}\) to t yields
$$ \tilde{G}\bigl(\nu(t)\bigr)-\tilde{G}\bigl(\nu(t_{0}) \bigr)\leq \int_{t_{0}}^{t} p(s)\Delta s+ \int_{t_{0}}^{t}f(s)\biggl[\nu(s)+ \int_{t_{0}}^{s}g(s,\lambda )\nu(\lambda)\Delta\lambda \biggr]\Delta s. $$
(2.31)
Since \(\nu(t_{0})=a(T^{*})\) and G̃ is an increasing function, from (2.31), we get
$$\begin{aligned} \nu(t) \leq& \tilde{G}^{-1} \biggl[G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int _{t_{0}}^{t} p(s)\Delta s+ \int_{t_{0}}^{t}f(s) \biggl[\nu(s)+ \int _{t_{0}}^{s}g(s,\lambda)\nu(\lambda)\Delta\lambda \biggr]\Delta s \biggr] \\ \leq& \tilde{G}^{-1} \biggl[G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}} p(s)\Delta s+ \int_{t_{0}}^{t}f(s) \biggl[\nu(s)+ \int_{t_{0}}^{s}g(s,\lambda)\nu (\lambda)\Delta\lambda \biggr]\Delta s \biggr]. \end{aligned}$$
(2.32)
For brevity, let
$$ z(t)= \tilde{G}\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}} p(s)\Delta s+ \int _{t_{0}}^{t}f(s) \biggl[\nu(s)+ \int_{t_{0}}^{s}g(s,\lambda)\nu(\lambda )\Delta\lambda \biggr]\Delta s, $$
(2.33)
and hence
$$z(t_{0})=\tilde{G}\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}p(s)\Delta s. $$
Then
$$ \nu(t)\leq\tilde{G}^{-1}\bigl(z(t)\bigr), \quad t\in \bigl[t_{0},T^{*}\bigr]\cap\mathbb{T}. $$
(2.34)
Using Lemma 1.2, (2.34), and (2.33), we have
$$\begin{aligned} z^{\Delta}(t) =&f(t) \biggl[\nu(t)+ \int_{t_{0}}^{t}g(s)\nu(s)\Delta s \biggr] \\ \leq&f(t) \biggl[1+ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr]\tilde {G}^{-1}\bigl(z(t)\bigr). \end{aligned}$$
(2.35)
From (2.35) we have
$$ \frac{z^{\Delta}(t)}{\tilde{G}^{-1}(z(t))}\leq f(t) \biggl[1+ \int _{t_{0}}^{t}g(t,s)\Delta s \biggr]. $$
(2.36)
Integrating both sides of (2.36) from \(t_{0}\) to t yields
$$ \tilde{H}\bigl(z(t)\bigr)-\tilde{H}\bigl(z(t)\bigr)\leq \int_{t_{0}}^{t}f(s) \biggl[1+ \int _{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr] \Delta s, $$
which implies
$$\begin{aligned} z(t) \leq& \tilde{H}^{-1} \biggl\{ \tilde{H} \biggl[\bigl(G\bigl(a \bigl(T^{*}\bigr)\bigr)\bigr)+ \int _{t_{0}}^{T^{*}}p(s)\Delta s \biggr] \\ &{} + \int_{t_{0}}^{t}f(s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta \lambda \biggr] \Delta s \biggr\} . \end{aligned}$$
(2.37)
From (2.27), (2.34), and (2.37) we obtain
$$\begin{aligned} u(t) \leq&\omega_{1}^{-1} \biggl\{ \tilde{G}^{-1} \biggl[\tilde {H}^{-1} \biggl\{ \tilde{H} \biggl[G\bigl(a\bigl(T^{*}\bigr) \bigr)+ \int_{t_{0}}^{T^{*}}p(s)\Delta s \biggr] \\ &{}+ \int_{t_{0}}^{T^{*}}f(s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \biggr\} \biggr] \biggr\} . \end{aligned}$$
(2.38)
Taking \(t=T^{*}\) in (2.38) yields
$$\begin{aligned} u\bigl(T^{*}\bigr) \leq&\omega_{1}^{-1} \biggl\{ \tilde{G}^{-1} \biggl[\tilde {H}^{-1} \biggl\{ \tilde{H} \biggl[G\bigl(a\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}p(s)\Delta s \biggr] \\ &{}+ \int_{t_{0}}^{T^{*}}f(s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \biggr\} \biggr] \biggr\} . \end{aligned}$$
(2.39)
Since \(T^{*}\in\mathbb{T}_{0}\) is chosen arbitrarily, we obtain the required result after substituting \(T^{*}\) with t into (2.39). This completes the proof. □
Remark 2.7
It is interesting to note that, as a particular case, if we put \(\omega _{1}(u)=u\), \(\omega_{2}(u)=1\), \(\tau(t)=t\), and \(a(t)=u_{0}\) (any constant) in Theorem 2.6, then the inequality given in Theorem 2.6 reduces to the inequality given in [31, Thm. 3.1].
Remark 2.8
In addition to the assumptions in Remark 2.7, if we put \(p(t)=0\) and \(g(t,s)=g(s)\) in Theorem 2.6, then we get [9, Thm. 1] as a particular case.
Theorem 2.6 unifies some known continuous and discrete inequalities in the literature as shown in the following remarks:
Remark 2.9
In addition to the assumptions in Remark 2.7, when \(p(t)=0\) in Theorem 2.6, if \(\mathbb{T}=\mathbb{R}\), then we can easily obtain the continuous version proved by Pachpatte [8, Thm. 2.1(\(a_{1}\))]. If \(\mathbb{T}=\mathbb{Z}\), then we can also get the discrete version proved by Pachpatte [8, Thm. 2.3(\(c_{1}\))].
Remark 2.10
It is interesting to note that if, in addition to the assumptions in Remark 2.7, we take \(g(t,s)=g(s)\) in Theorem 2.6 and \(\mathbb{T}=\mathbb{R}\), then we get the continuous version established by Pachpatte [8, Thm. 1.7.2(i)]. Furthermore, if \(\mathbb{T}=\mathbb{Z}\), then we can also get the discrete version inequality established by Pachpatte [8, Thm. 1.8.7].
Theorem 2.11
Let
u, a, τ, \(\omega(u)\), h, g
be defined as in Theorem 2.1. If
\(t\in\mathbb{T}_{0}\)
and
u
satisfies the retarded dynamic inequality
$$ u^{p}(t)\leq a(t)+ \int_{t_{0}}^{t}h(t,s)\omega\bigl(u\bigl(\tau(s)\bigr) \bigr) \biggl[ \omega\bigl(u\bigl(\tau(s)\bigr)\bigr)+ \int_{t_{0}}^{s}g(s,\lambda)\omega\bigl(u\bigl(\tau ( \lambda)\bigr)\bigr)\Delta\lambda \biggr]\Delta s $$
(2.40)
with the initial condition
$$ \textstyle\begin{cases} u(t) = \phi(t) \quad \textit{if } t\in [\alpha,t_{0}]\cap\mathbb{T}, \\ \phi(\tau(t)) \leq a^{\frac{1}{p}}(t) \quad \textit{if } \tau (t)\leq t_{0}, t\in\mathbb{T}_{0}, \end{cases} $$
(2.41)
then
$$\begin{aligned} u(t) \leq& \biggl\{ \varOmega^{-1} \biggl[\varUpsilon^{-1} \biggl\{ \varUpsilon \bigl(\varOmega\bigl(\nu\bigl(a(t)\bigr)\bigr)\bigr) \\ &{}+ \int_{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta \lambda \biggr] \Delta s \biggr\} \biggr] \biggr\} ^{\frac {1}{p}} \end{aligned}$$
(2.42)
with
$$ \varUpsilon\bigl(\varOmega\bigl(\nu\bigl(a(t)\bigr)\bigr)\bigr) + \int_{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta \lambda \biggr] \Delta s\in \operatorname{Dom}\bigl(\varUpsilon^{-1}\bigr), $$
where
Ω
and
ϒ
are increasing bijective functions defined by
$$\begin{aligned}& (\varOmega\circ\nu)^{\Delta}(t) =\frac{\nu^{\Delta}(t)}{\omega(\nu ^{\frac{1}{p}}(t))}, \end{aligned}$$
(2.43)
$$\begin{aligned}& (\varUpsilon\circ z)^{\Delta}(t) =\frac{z^{\Delta}(t)}{\omega([\varOmega ^{-1}(z(t))]^{\frac{1}{p}})}. \end{aligned}$$
(2.44)
Proof
Fixing an arbitrary number \(T^{*}\in\mathbb{T}_{0}\), for \(t\in [t_{0},T^{*}]\cap\mathbb{T}\), we define the function
$$\begin{aligned} \nu(t) =& a\bigl(T^{*}\bigr)+ \int_{t_{0}}^{t}h(t,s)\omega\bigl(u\bigl(\tau(s)\bigr) \bigr) \biggl[ \omega\bigl(u\bigl(\tau(s)\bigr)\bigr) \\ &{}+ \int_{t_{0}}^{s}\omega\bigl(u\bigl(\tau(\lambda)\bigr) \bigr)\Delta\lambda \biggr]\Delta s. \end{aligned}$$
(2.45)
Clearly, ν is a nonnegative nondecreasing function with \(\nu (t_{0}=a(T^{*}))\), and we have
$$ u(t)\leq\nu^{\frac{1}{p}}(t), \quad t\in\bigl[t_{0},T^{*} \bigr]\cap\mathbb{T}. $$
(2.46)
Therefore, for \(t\in[t_{0},T^{*}]\cap\mathbb{T}\), if \(\tau(t)\geq t_{0}\), then \(\tau(t)\in[t_{0},T^{*}]\cap\mathbb{T}\), and from (2.46) we obtain
$$ u\bigl(\tau(t)\bigr)\leq\nu^{\frac{1}{p}}\bigl(\tau(t)\bigr)\leq \nu^{\frac{1}{p}}(t). $$
(2.47)
On the other hand, if \(\tau(t)\leq t_{0}\), then, using the initial condition (2.41), we get
$$ u\bigl(\tau(t)\bigr)=\phi\bigl(\tau(t)\bigr)\leq a^{\frac{1}{p}}(t)\leq\nu^{\frac{1}{p}}(t). $$
(2.48)
So, combining (2.47) and (2.48), we have
$$ u\bigl(\tau(t)\bigr)\leq\nu^{\frac{1}{p}}(t). $$
(2.49)
Thus from (2.45), Lemma 1.2, and (2.49) we have
$$\begin{aligned} \nu^{\Delta}(t) =& \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s)\omega\bigl(u\bigl( \tau(s)\bigr)\bigr) \biggl[\omega\bigl(u\bigl(\tau(s)\bigr)\bigr)+ \int_{t}^{s}g(t,s)\omega\bigl(u\bigl(\tau(\lambda ) \bigr)\bigr)\Delta\lambda \biggr]\Delta s \\ &{}+h\bigl(\sigma(t),t\bigr)\omega\bigl(u\bigl(\tau(t)\bigr)\bigr) \biggl[ \omega\bigl(u\bigl(\tau(t)\bigr)\bigr)+ \int _{t}^{t}g(t,s)\omega\bigl(u\bigl(\tau(s)\bigr) \bigr)\Delta s \biggr] \\ \leq& \biggl\{ \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s) \biggl[\omega \bigl(\nu^{\frac{1}{p}}(s)\bigr)+ \int_{t}^{s}g(t,s)\omega\bigl(\nu ^{\frac{1}{p}}( \lambda)\bigr)\Delta\lambda \biggr]\Delta s \\ &{}+h\bigl(\sigma(t),t\bigr) \biggl[\omega\bigl(\nu^{\frac{1}{p}}(t)\bigr)+ \int _{t}^{t}g(t,s)\omega\bigl(\nu^{\frac{1}{p}}(s) \bigr)\Delta s \biggr] \biggr\} \omega\bigl(\nu^{\frac{1}{p}}(t) \bigr), \end{aligned}$$
(2.50)
which implies
$$\begin{aligned} \frac{\nu^{\Delta}(t)}{\omega(\nu^{\frac{1}{p}}(t))} \leq& \int _{t_{0}}^{t}h^{\Delta_{t}}(t,s) \biggl[\omega \bigl(\nu^{\frac{1}{p}}(t)\bigr)+ \int_{t}^{s}g(t,s)\omega\bigl(\nu ^{\frac{1}{p}}( \lambda)\bigr)\Delta\lambda \biggr]\Delta s \\ &{}+h\bigl(\sigma(t),t\bigr) \biggl[\omega\bigl(\nu^{\frac{1}{p}}(t)\bigr)+ \int _{t}^{t}g(t,s)\omega\bigl(\nu^{\frac{1}{p}}(s) \bigr)\Delta s \biggr] \\ =& \biggl\{ \int_{t_{0}}^{t}h\bigl(\sigma(t),t\bigr)\biggl[\omega \bigl(\nu^{\frac {1}{p}}(t)\bigr)+ \int_{t}^{s}g(s,\lambda)\omega\bigl( \nu^{\frac {1}{p}}(\lambda)\bigr)\Delta\lambda\biggr]\Delta s \biggr\} ^{\Delta _{t}}. \end{aligned}$$
(2.51)
Integrating both sides of (2.51) from \(t_{0}\) to t yields
$$ \varOmega\bigl(\nu(t)\bigr)-\varOmega\bigl(\nu(t_{0})\bigr) \leq \int_{t_{0}}^{t}h(t,s) \biggl[\omega\bigl( \nu^{\frac{1}{p}}(s)\bigr)+ \int_{t_{0}}^{s}g(s,\lambda )\omega\bigl( \nu^{\frac{1}{p}}(\lambda)\bigr)\Delta\lambda \biggr]\Delta s. $$
(2.52)
Since Ω is increasing, (2.52) implies
$$\begin{aligned} \nu(t) \leq& \varOmega^{-1} \biggl[ \varOmega\bigl(\nu\bigl(a\bigl(T^{*}\bigr) \bigr)\bigr) + \int _{t_{0}}^{t}h(t,s) \biggl[\omega\bigl( \nu^{\frac{1}{p}}(s)\bigr) \\ &{}+ \int_{t_{0}}^{s}g(s,\lambda)\omega\bigl( \nu^{\frac{1}{p}}(\lambda )\bigr)\Delta\lambda \biggr]\Delta s \biggr]. \end{aligned}$$
(2.53)
For brevity, let
$$ z(t)=\varOmega\bigl(\nu\bigl(a\bigl(T^{*}\bigr)\bigr)\bigr) + \int_{t_{0}}^{t}h(t,s) \biggl[\omega\bigl( \nu^{\frac{1}{p}}(s)\bigr)+ \int_{t_{0}}^{s}g(s,\lambda)\omega \bigl( \nu^{\frac{1}{p}}(\lambda)\bigr)\Delta\lambda \biggr]\Delta s, $$
(2.54)
and hence \(z(t_{0})=\varOmega(\nu(T^{*}))\).
Then
$$ \nu(t)\leq\varOmega^{-1}\bigl(z(t)\bigr),\quad t\in \bigl[t_{0},T^{*}\bigr]\cap\mathbb{T}. $$
(2.55)
By Lemma 1.2, (2.54), and (2.55) we get
$$\begin{aligned} z^{\Delta}(t) =& \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s)\biggl[\omega\bigl( \nu^{\frac {1}{p}}(s)\bigr)+ \int_{t_{0}}^{t}g(s,\lambda)\omega\bigl( \nu^{\frac {1}{p}}(\lambda)\bigr)\Delta\lambda\biggr]\Delta s \\ &{}+ h\bigl(\sigma(t),t\bigr) \biggl[\omega\bigl(\nu^{\frac{1}{p}}(t)\bigr)+ \int_{t_{0}}^{t}g(t,s)\omega\bigl(\nu ^{\frac{1}{p}}(s) \bigr)\Delta s \biggr] \\ \leq& \biggl\{ \int_{t_{0}}^{t}h^{\Delta_{t}}(t,s) \biggl[1+ \int _{t_{0}}^{t}g(s,\lambda)\Delta\lambda \biggr]\Delta s+ h\bigl(\sigma (t),s\bigr) \biggl[1+ \int_{t_{0}}^{t}g(t,s)\Delta s \biggr] \biggr\} \\ &{}\times\omega\bigl(\bigl[\varOmega^{-1}\bigl(z(t)\bigr) \bigr]^{\frac{1}{p}}\bigr), \end{aligned}$$
which implies
$$ \frac{z^{\Delta}(t)}{\omega([\varOmega^{-1}(z(t))]^{\frac{1}{p}})}\leq \biggl\{ \int_{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda )\Delta\lambda \biggr] \Delta s \biggr\} ^{\Delta_{t}}. $$
(2.56)
Integrating both sides of (2.56) from \(t_{0}\) to t yields
$$ \varUpsilon\bigl(z(t)\bigr)-\varUpsilon\bigl(z(t_{0})\bigr) \leq \int_{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s. $$
(2.57)
Since ϒ is increasing, (2.57) implies
$$ z(t)\leq\varUpsilon^{-1} \biggl[\varUpsilon\bigl(\varOmega\bigl( \nu\bigl(a\bigl(T^{*}\bigr)\bigr)\bigr)\bigr)+ \int _{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \biggr]. $$
(2.58)
Combining (2.46), (2.55), and (2.58), we obtain
$$\begin{aligned} u(t) \leq& \biggl[\varOmega^{-1} \biggl\{ \varUpsilon^{-1} \biggl[\varUpsilon \bigl(\varOmega\bigl(\nu\bigl(a\bigl(T^{*} \bigr)\bigr)\bigr)\bigr) \\ &{}+ \int_{t_{0}}^{t}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \biggr] \biggr\} \biggr]^{\frac{1}{p}}. \end{aligned}$$
(2.59)
Taking \(t=T^{*}\) in (2.38) yields
$$\begin{aligned} u\bigl(T^{*}\bigr) \leq& \biggl[\varOmega^{-1} \biggl\{ \varUpsilon^{-1} \biggl[\varUpsilon \bigl(\varOmega\bigl(\nu\bigl(a\bigl(T^{*} \bigr)\bigr)\bigr)\bigr) \\ &{}+ \int_{t_{0}}^{T^{*}}h(t,s) \biggl[1+ \int_{t_{0}}^{s}g(s,\lambda)\Delta\lambda \biggr]\Delta s \biggr] \biggr\} \biggr]^{\frac{1}{p}}. \end{aligned}$$
(2.60)
Since \(T^{*}\in\mathbb{T}_{0}\) is chosen arbitrarily, we obtain the required inequality after substituting \(T^{*}\) with t into (2.60). This completes the proof. □
Remark 2.12
If we take \(h(t,s)=f(s)\), \(g(s,\lambda)=g(\lambda)\), \(p=1\), \(a(t)=u_{0}\) (any constant), and \(\mathbb{T}=\mathbb{R}\) in Theorem 2.11, then the inequality given in Theorem 2.11 reduces to the inequality given in [10, Thm. 2.1].
Theorem 2.13
Letu, a, \(\omega_{1}\), \(\omega_{2}\), τ
be defined as in Theorem 2.1, and let
\(f, g \in C_{\mathrm{rd}}(\mathbb{T}_{0},\mathbb{R^{+}})\). If
u
satisfies the inequality
$$ u(t)\leq a(t) + \int_{0}^{t}f(s)\psi_{1}\bigl(u\bigl( \tau(s)\bigr)\bigr){ \biggl[u^{q}\bigl(\tau(s)\bigr)+ \int_{0}^{s}g(\lambda)\psi_{2}\bigl(u \bigl(\tau(\lambda)\bigr)\bigr)\Delta\lambda \biggr]}^{p}\Delta s $$
(2.61)
with the initial condition
$$ \textstyle\begin{cases} u(t) = \phi(t) \quad \textit{if } t\in [\alpha,t_{0}]\cap\mathbb{T}, \\ \phi(\tau(t)) \leq a(t) \quad \textit{if } \tau(t)\leq t_{0}, t\in\mathbb{T}_{0}, \end{cases} $$
(2.62)
where
\(p>0\)
and
\(q\geq1\)
are constants such that
\(p+q>1\), then
$$ u(t)\leq\varPsi_{1}^{-1} \biggl[ \varPsi_{2}^{-1} \biggl(\varPsi_{2} \biggl(\varPsi _{1}\bigl(u_{0}^{q}\bigr)+ \int_{0}^{t}g(s)\Delta s \biggr)+ \int_{0}^{t}f(s)\Delta s \biggr) \biggr] $$
(2.63)
with
$$ \varPsi_{2} \biggl(\varPsi_{1}\bigl(u_{0}^{q} \bigr)+ \int_{0}^{t}g(s)\Delta s \biggr)+ \int_{0}^{t}f(s)\Delta s\in \operatorname{Dom}{ \varPsi_{2}^{-1}}, $$
where
$$\begin{aligned}& (\varPsi_{1}\circ Y)^{\Delta}(t)= \frac{Y^{\Delta}(t)}{\psi_{2}(Y(t))}, \end{aligned}$$
(2.64)
$$\begin{aligned}& (\varPsi_{2}\circ Y_{1})^{\Delta}(t)= \frac{\psi_{2}(\varPsi _{1}^{-1}(Y_{1}(t)))Y^{\Delta}_{1}(t)}{ \psi_{1}(\varPsi_{1}^{-1}(Y_{1}(t)))[\varPsi_{1}^{-1}(Y_{1}(\sigma(t)))]^{\frac {q+pq-1}{q}}}. \end{aligned}$$
(2.65)
Proof
Fixing an arbitrary number \(T^{*}\in\mathbb{T}_{0}\), for \(t\in [t_{0},T^{*}]\cap\mathbb{T}\), we define the function
$$ \nu(t)= a\bigl(T^{*}\bigr) + \int_{t_{0}}^{t}f(s)\omega_{1}\bigl(u\bigl( \tau(s)\bigr)\bigr){ \biggl[u^{q}\bigl(\tau(s)\bigr)+ \int_{0}^{s}g(\lambda)\omega_{2}\bigl(u \bigl(\tau(\lambda)\bigr)\bigr)\Delta\lambda \biggr]}^{p}\Delta s, $$
(2.66)
which is a positive nondecreasing function with \(\nu(t_{0})=a(T^{*})\). Then we have
$$ u(t)\leq\nu(t), \quad t\in\bigl[t_{0},T^{*} \bigr]\cap\mathbb{T}. $$
(2.67)
Therefore, for \(t\in[t_{0},T^{*}]\cap\mathbb{T}\), if \(\tau(t)\geq t_{0}\), then \(\tau(t)\in[t_{0},T^{*}]\cap\mathbb{T}\), and from (2.46) we obtain
$$ u\bigl(\tau(t)\bigr)\leq\nu(t). $$
(2.68)
On the other hand, if \(\tau(t)\leq t_{0}\), then, using the initial condition (2.62), we get
$$ u\bigl(\tau(t)\bigr)=\phi\bigl(\tau(t)\bigr)\leq a(t)\leq\nu(t). $$
(2.69)
So, combining (2.68) and (2.69), we have
$$ u\bigl(\tau(t)\bigr)\leq\nu(t), \quad t\in\bigl[t_{0},T^{*} \bigr]\cap\mathbb{T}. $$
(2.70)
Thus, by Lemma 1.2 from (2.66) and (2.70) we get
$$ \nu^{\Delta}(t)= f(t)\omega_{1}\bigl(\nu(t) \bigr)Y^{p}(t), $$
(2.71)
where \(Y(t)=\nu^{q}(t)+\int_{t_{0}}^{t}g(s)\omega_{2}(\nu(s))\Delta s\) is nondecreasing. Hence \(Y(t_{0})=\nu^{q}(t_{0})= a^{q}(T^{*})\) and \(\nu ^{q}(t)\leq Y(t)\), but \(q\geq1\), and thus
$$ \begin{aligned} &\nu(t) \le Y^{\frac{1}{q}}(t), \\ &\nu^{\sigma}(t) \le Y^{\frac{1}{q}}\bigl(\sigma(t)\bigr) \end{aligned} $$
(2.72)
for all \(t\in[t_{0},T^{*}]\cap\mathbb{T}\).
Since ν is nondecreasing, from \(Y(t)\), by Lemma 1.7 and Lemma 1.2 we see that
$$ Y^{\Delta}(t)\leq q\bigl[\nu^{\sigma}(t) \bigr]^{q-1}\nu^{\Delta }(t)+g(t)\omega_{2}\bigl(\nu(t) \bigr). $$
(2.73)
From (2.73), using (2.71), (2.72), and the inequality \(Y(t)\leq Y^{\sigma}(t)\), we get
$$\begin{aligned} Y^{\Delta}(t) \leq& q\bigl[Y^{\sigma}(t)\bigr]^{\frac{q-1}{q}}f(t) \omega _{1}\bigl(Y(t)\bigr)Y^{p}(t)+g(t)\omega_{2} \bigl(Y(t)\bigr), \\ \leq& qf(t)\omega_{1}\bigl(Y(t)\bigr)\bigl[Y^{\sigma}(t) \bigr]^{\frac{q-1}{q}}\bigl[Y^{\sigma }(t)\bigr]^{p}+g(t) \omega_{2}\bigl(Y(t)\bigr), \\ =& qf(t)\omega_{1}\bigl(Y(t)\bigr)\bigl[Y^{\sigma}(t) \bigr]^{\frac{q+pq-1}{q}}+g(t)\omega _{2}\bigl(Y(t)\bigr), \end{aligned}$$
(2.74)
\(t\in[t_{0},T^{*}]\cap\mathbb{T}\).
Since \(\omega_{2}(Y(t))>0\) for \(t>0\), from (2.74) we get
$$\frac{Y^{\Delta}(t)}{\omega_{2}(Y(t))}\leq qf(t)\frac{\omega _{1}(Y(t))[Y^{\sigma}(t)]^{\frac{q+pq-1}{q}}}{\omega_{2}(Y(t))}+g(t). $$
Taking \(t = s\) in the last inequality, integrating both sides from \(t_{0}\) to t, and using (2.64), we have
$$ \varPsi_{1}\bigl(Y(t)\bigr)\leq\varPsi_{1}\bigl(a^{q} \bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{t}qf(s)\frac{\omega _{1}(Y(s))[Y^{\sigma}(s)]^{\frac{q+pq-1}{q}}}{\omega_{2}(Y(s))}\Delta s+ \int_{t_{0}}^{t}g(s)\Delta s, $$
where \(\varPsi_{1}\) is defined by (2.64). From this inequality we have
$$\begin{aligned} \varPsi_{1}\bigl(Y(t)\bigr) \leq&\varPsi_{1} \bigl(a^{q}\bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}g(s)\Delta s \\ &{}+ \int_{t_{0}}^{t}qf(s) \frac{\omega_{1}(Y(s))[Y^{\sigma}(s)]^{\frac {q+pq-1}{q}}}{\omega_{2}(Y(s))}\Delta s. \end{aligned}$$
Let \(Y_{1}(t)\) denote the function on the right-hand side of this inequality. Then
$$\begin{aligned} Y_{1}(t) =& \varPsi_{1}\bigl(a^{q} \bigl(T^{*}\bigr)\bigr)+ \int_{t_{0}}^{T^{*}}g(s)\Delta s \\ &{}+ \int _{t_{0}}^{t}qf(s) \frac{\omega_{1}(Y(s))Y^{\frac{q+pq-1}{q}}(s)}{\omega _{2}(Y(s))}\Delta s, \end{aligned}$$
(2.75)
which is a positive nondecreasing function with \(Y_{1}(T^{*})=\varPsi _{2}(a^{q}(T^{*}))+\int_{t_{0}}^{T^{*}}g(s)\Delta s\), and
$$ \begin{aligned} &Y(t)\leq \varPsi_{1}^{-1} \bigl(Y_{1}(t)\bigr), \\ &Y^{\sigma}(t)\leq \varPsi_{1}^{-1}\bigl(Y_{1} \bigl(\sigma(t)\bigr)\bigr). \end{aligned} $$
(2.76)
From (2.75), (2.76), and Lemma 1.2 we obtain
$$\begin{aligned} Y^{\Delta}_{1}(t) \leq& qf(t) \frac{\omega_{1}(Y(t))Y^{\frac{q+pq-1}{q}}(t)}{\omega_{2}(Y(t))} \\ \leq& qf(t) \frac{\omega_{1}(\varPsi^{-1}_{1}(Y_{1}(t)))[\varPsi_{1}^{-1}(Y_{1}(\sigma (t)))]^{\frac{q+pq-1}{q}}(t)}{ \omega_{2}(\varPsi_{1}^{-1}(Y_{1}(\sigma(t))))}. \end{aligned}$$
From this inequality, by the definition of \(\varPsi_{2}\) in (2.65), letting \(t=T^{*}\), we have
$$ \varPsi_{2}\bigl(Y_{1}\bigl(T^{*} \bigr)\bigr) \leq\varPsi_{2} \biggl(\varPsi_{1} \bigl(a^{q}\bigl(T^{*}\bigr)\bigr)+ \int _{t_{0}}^{T^{*}}g(s)\Delta s \biggr)+ \int_{t_{0}}^{T^{*}}qf(s)\Delta s. $$
(2.77)
Since \(T^{*}\in\mathbb{T}_{0}\) is chosen, from (2.70), (2.72), (2.76), and (2.77) we get the required inequality in (2.63). This completes the proof. □
Remark 2.14
By taking \(\mathbb{T}=\mathbb{R}\) and \(a(t)=u_{0}\) (any constant) in Theorem 2.13 it is easy to observe that the inequality obtained in Theorem 2.13 reduces to the inequality obtained by Abdeldaim and El-Deeb in [13, Thm. 2.5].
Remark 2.15
If we put \(\omega_{1}(u)=\omega_{2}(u)=u\), \(a(t)=u_{0}\) (any constant), and \(\mathbb{T}=\mathbb{R}\) in Theorem 2.13, then we get the inequality obtained by Abdeldaim and El-Deeb in [13, Thm. 2.3].
Remark 2.16
If we take \(q=1\), \(a(t)=u_{0}\) (any constant), and \(\mathbb{T}=\mathbb {R}\) in Theorem 2.13, then the inequality given in Theorem 2.13 reduces to the inequality given in [40, Thm. 4].