1 Introduction
In 1988, Hilger introduced the theory of time scales in order to unify and extend the difference and differential calculus in a consistent way (see [1]). Since then, more and more researchers are getting involved in this fast-growing field, for example, [2‐10] and the references therein. Among various aspects of the theory, we notice that dynamic inequalities on time scales is an object of long standing interest [11‐29]. However, to the best of our knowledge, there are few results dealing with Volterra-Fredholm type integral inequalities on time scales. Recent results in this direction include the works of Gu and Meng [16] and Meng and Shao [20].
The purpose of this paper is to investigate some new retarded Volterra-Fredholm type integral inequalities on time scales, which not only generalize and extend the results of [16, 20] and some known integral inequalities but also provide a handy and effective tool for the study of qualitative properties of solutions of some complicated Volterra-Fredholm type dynamic equations.
Anzeige
The paper is organized as follows. In Section 2, some necessary definition and lemmas are presented. In Section 3, some new retarded Volterra-Fredholm type integral inequalities on time scales are investigated. Finally, Section 4 is devoted to applying our results to a retarded Volterra-Fredholm type dynamic integral equation on time scales.
2 Preliminaries
Throughout this paper, knowledge and understanding of time scales and time scale notation is assumed. For an excellent introduction to the calculus on time scales, we refer the reader to [30] and [31].
List of abbreviations. In what follows, we always assume that \(\mathbb{R}\) denotes the set of real numbers, \(\mathbb{R}_{+}=[0,\infty)\), \(\mathbb{Z}\) denotes the set of integers, \(\mathbb{T}\) is an arbitrary time scale (nonempty closed subset of \(\mathbb{R}\)), \(\mathcal{R}\) denotes the set of all regressive and rd-continuous functions, \(\mathcal{R}^{+}=\{p\in\mathcal{R}:1+\mu (t)p(t)>0\mbox{ for all }t\in\mathbb{T}\}\) and \(I=[t_{0},T]\cap\mathbb{T}^{\kappa}\), where \(t_{0}\in\mathbb{T}^{\kappa}\), \(T\in\mathbb{T}^{\kappa}\), \(T>t_{0}\). The set \(\mathbb{T}^{\kappa}\) is defined as follows: If \(\mathbb{T}\) has a maximum m and m is left-scattered, then \(\mathbb{T}^{\kappa}=\mathbb{T}-\{m\}\). Otherwise \(\mathbb{T}^{\kappa}=\mathbb{T}\). The graininess function \(\mu: \mathbb {T}\rightarrow[0,\infty)\) is defined by \(\mu(t):=\sigma(t)-t\), the forward jump operator \(\sigma: \mathbb{T}\rightarrow\mathbb{T}\) by \(\sigma(t):=\inf\{s\in\mathbb{T}: s>t\}\), and the “circle plus” addition ⊕ is defined by \((p\oplus q)(t):=p(t)+q(t)+\mu(t)p(t)q(t)\) for all \(t\in\mathbb {T}^{\kappa}\).
The following lemmas and definition are useful in the proof of the main results of this paper.
Anzeige
Lemma 2.1
([30, Theorem 1.16])
Assume that
\(f:\mathbb {T}\rightarrow\mathbb{R}\)
is a function and let
\(t\in\mathbb{T}\). If
f
is differentiable at
t, then
$$f\bigl(\sigma(t)\bigr)=f(t)+\mu(t)f^{\Delta}(t). $$
Lemma 2.2
([30, Theorem 1.98])
Assume that
\(\nu :\mathbb{T}\rightarrow\mathbb{R}\)
is a strictly increasing function and
\(\widetilde{\mathbb{T}}:=\nu(\mathbb{T})\)
is a time scale. If
\(f: \mathbb{T}\rightarrow\mathbb{R}\)
is an rd-continuous function and
ν
is differentiable with rd-continuous derivative, then for
\(a, b\in \mathbb{T}\),
$$\int_{a}^{b}f(t)\nu^{\Delta}(t)\Delta t= \int_{\nu(a)}^{\nu(b)}\bigl(f\circ\nu ^{-1}\bigr) (s) \widetilde{\Delta} s. $$
Lemma 2.3
Let
\(\alpha:I\rightarrow I\)
be a continuous and strictly increasing function such that
\(\alpha(t)\leq t\), and
\(\alpha^{\Delta}\)
is rd-continuous. Assume that
\(f: I\rightarrow\mathbb{R}\)
is an rd-continuous function, then
implies
$$ g(t)= \int_{\alpha(t_{0})}^{\alpha(t)}f(s)\Delta s,\quad t\in I, $$
(2.1)
$$ g^{\Delta}(t)=f\bigl(\alpha(t)\bigr)\alpha^{\Delta}(t),\quad t\in I. $$
(2.2)
Proof
From (2.1), we get, for any \(t\in I\), By Lemma 2.2, we obtain so we get □
$$g(t)= \int_{\alpha(t_{0})}^{\alpha(t)}f(s)\Delta s= \int_{\alpha (t_{0})}^{\alpha(t)}f\bigl(\alpha\bigl(\alpha^{-1}(s) \bigr)\bigr)\Delta s. $$
$$g(t)= \int_{t_{0}}^{t}f\bigl(\alpha(s)\bigr) \alpha^{\Delta}(s)\Delta s, \quad t\in I, $$
$$g^{\Delta}(t)=f\bigl(\alpha(t)\bigr)\alpha^{\Delta}(t). $$
Lemma 2.4
([30, Theorem 1.117])
Suppose that, for each
\(\varepsilon>0\), there exists a neighborhood
U
of
t, independent of
\(\tau\in[t_{0},\sigma(t)]\), such that
where
\(w: \mathbb{T}\times\mathbb{T}^{\kappa}\rightarrow\mathbb{R}_{+}\)
is continuous at
\((t,t)\), \(t\in\mathbb{T}^{\kappa}\)
with
\(t>t_{0}\), and
\(w^{\Delta}_{t}(t,\cdot)\)
are rd-continuous on
\([t_{0},\sigma(t)]\). Then
implies
$$ \bigl\vert w\bigl(\sigma(t),\tau\bigr)-w(s,\tau) -w^{\Delta}_{t}(t, \tau) \bigl(\sigma (t)-s\bigr) \bigr\vert \leq\varepsilon \bigl\vert \sigma(t)-s \bigr\vert ,\quad s\in U, $$
(2.3)
$$g(t):= \int_{t_{0}}^{t}w(t,\tau)\Delta\tau $$
$$ g^{\Delta}(t)= \int_{t_{0}}^{t}w^{\Delta}_{t}(t,\tau) \Delta\tau+w\bigl(\sigma (t),t\bigr),\quad t\in\mathbb{T}^{\kappa}. $$
(2.4)
Lemma 2.5
([30, Theorem 6.1])
Suppose that
y
and
f
are rd-continuous functions and
\(p\in\mathcal {R}^{+}\). Then
implies
$$y^{\Delta}(t)\leq p(t)y(t)+f(t) \quad \textit{for all } t \in\mathbb{T} $$
$$y(t)\leq y(t_{0})e_{p}(t,t_{0})+ \int_{t_{0}}^{t}e_{p}\bigl(t,\sigma(\tau) \bigr)f(\tau)\Delta \tau\quad \textit{for all } t \in\mathbb{T}. $$
Definition 2.1
A function \(x:I\rightarrow I\) is said to belong to the class ϒ if
(1)
x is continuous and strictly increasing, and
(2)
\(x(t)\leq t\) and \(x^{\Delta}\) is rd-continuous.
3 Main results
Theorem 3.1
Assume that
\(\alpha\in\Upsilon\)
and
\(u, a, b, f_{1}, f_{2}, g: I\rightarrow\mathbb{R}_{+}\)
are rd-continuous functions, a
is nondecreasing, \(\lambda\geq0\)
is a constant, \(\alpha^{\Delta}(t)\geq0\), \(b^{\Delta}(t)\geq0\), \(\mu(t)A(t)<1\). Suppose that
u
satisfies
If
then
where we use the convention that
\(\frac{1}{0}=+\infty\),
$$\begin{aligned} u(t) \leq& a(t)+b(t) \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{} +\lambda b(T) \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int _{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.1)
$$ \beta:=e_{B \oplus C}(T,t_{0})< 1+\frac{1}{\lambda}, $$
(3.2)
$$ u(t) \leq\frac{a(T)}{\lambda+1-\beta\lambda}e_{B \oplus C}(t,t_{0}), \quad t\in I, $$
(3.3)
$$\begin{aligned}& A(t) := b^{\Delta}(t) \int_{\alpha(t_{0})}^{\alpha(\sigma (t))}\biggl[f_{1}(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)\Delta\tau\biggr] \Delta s, \qquad B(t):=\frac{A(t)}{1-\mu(t)A(t)}, \end{aligned}$$
(3.4)
$$\begin{aligned}& C(t) := b(t)\biggl[f_{1}\bigl(\alpha(t)\bigr)+f_{2}\bigl( \alpha(t)\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)\Delta\tau\biggr] \alpha^{\Delta}(t). \end{aligned}$$
(3.5)
Proof
Denote Then z is nondecreasing on I, and From Lemma 2.3 and (3.4)-(3.7), we have Note that from (3.4) we get and from (3.9) we have which yields i.e., Note that z is rd-continuous and \(B\oplus C\in\mathcal{R}^{+}\), from Lemma 2.5 and (3.10), we obtain From (3.6) and (3.8), we get i.e., From (3.2), (3.11) and (3.12), we have In view of (3.2) and (3.13), we get Substituting (3.14) into (3.11), we obtain Noting \(u(t)\leq z(t)\), we get the desired inequality (3.3). This completes the proof. □
$$\begin{aligned} z(t) :=& a(T)+b(t) \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{}+\lambda b(T) \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int _{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.6)
$$ u(t)\leq z(t),\quad t\in I, $$
(3.7)
$$ z(t_{0})=a(T)+\lambda b(T) \int_{\alpha(t_{0})}^{\alpha (T)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s. $$
(3.8)
$$\begin{aligned} z^{\Delta}(t) =&b^{\Delta}(t) \int_{\alpha(t_{0})}^{\alpha (\sigma(t))}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau )\Delta\tau\biggr] \Delta s \\ &{}+b(t)\biggl[f_{1}\bigl(\alpha(t)\bigr)u\bigl(\alpha(t) \bigr)+f_{2}\bigl(\alpha(t)\bigr) \int_{\alpha (t_{0})}^{\alpha(t)}g(\tau)u(\tau)\Delta\tau\biggr] \alpha^{\Delta}(t) \\ \leq&b^{\Delta}(t) \int_{\alpha(t_{0})}^{\alpha(\sigma (t))}\biggl[f_{1}(s)z(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)z(\tau)\Delta\tau\biggr] \Delta s \\ &{}+b(t)\biggl[f_{1}\bigl(\alpha(t)\bigr)z\bigl(\alpha(t) \bigr)+f_{2}\bigl(\alpha(t)\bigr) \int_{\alpha (t_{0})}^{\alpha(t)}g(\tau)z(\tau)\Delta\tau\biggr] \alpha^{\Delta}(t) \\ \leq&z\bigl(\sigma(t)\bigr)b^{\Delta}(t) \int_{\alpha(t_{0})}^{\alpha(\sigma (t))}\biggl[f_{1}(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)\Delta\tau\biggr] \Delta s \\ &{}+z(t)b(t)\biggl[f_{1}\bigl(\alpha(t)\bigr)+f_{2}\bigl( \alpha(t)\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)\Delta\tau\biggr] \alpha^{\Delta}(t) \\ =&A(t)z\bigl(\sigma(t)\bigr)+C(t)z(t), \quad t\in I. \end{aligned}$$
(3.9)
$$A(t)=\frac{B(t)}{1+\mu(t)B(t)}, $$
$$\begin{aligned} z^{\Delta}(t) &\leq\frac{B(t)}{1+\mu(t)B(t)}z\bigl(\sigma(t)\bigr)+C(t)z(t) \\ &=\frac{B(t)}{1+\mu(t)B(t)}\bigl[z(t)+\mu(t)z^{\Delta}(t)\bigr]+C(t)z(t), \end{aligned} $$
$$\frac{1}{1+\mu(t)B(t)}z^{\Delta}(t)\leq\biggl[\frac{B(t)}{1+\mu(t)B(t)}+C(t) \biggr]z(t), $$
$$\begin{aligned} z^{\Delta}(t) \leq&\bigl[B(t)+\bigl(1+\mu(t)B(t)\bigr)C(t)\bigr]z(t) \\ =&(B\oplus C) (t)z(t). \end{aligned}$$
(3.10)
$$ z(t)\leq z(t_{0})e_{B \oplus C}(t,t_{0}), \quad t\in I. $$
(3.11)
$$\bigl(z(t_{0})-a(T)\bigr)\frac{\lambda+1}{\lambda}+a(T)=z(T), $$
$$ \frac{\lambda+1}{\lambda}z(t_{0})-\frac{1}{\lambda}a(T)=z(T). $$
(3.12)
$$ \frac{\lambda+1}{\lambda}z(t_{0})-\frac{1}{\lambda}a(T)=z(T)\leq z(t_{0})e_{B \oplus C}(T,t_{0})=z(t_{0}) \beta. $$
(3.13)
$$ z(t_{0})\leq\frac{a(T)}{\lambda+1-\beta\lambda}. $$
(3.14)
$$z(t)\leq\frac{a(T)}{\lambda+1-\beta\lambda}e_{B \oplus C}(t,t_{0}),\quad t\in I. $$
If we let \(\lambda=\frac{1}{b(T)}\) in Theorem 3.1, then we obtain the following corollary.
Corollary 3.1
Assume that
α, u, a, b, \(f_{1}\), \(f_{2}\), g, A, B, C
are the same as in Theorem
3.1
and
\(b(T)\neq0\). Suppose that
u
satisfies
If
then
$$\begin{aligned} u(t) \leq& a(t)+b(t) \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ & {}+ \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha (t_{0})}^{s}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
$$\beta:=e_{B \oplus C}(T,t_{0})< 1+b(T), $$
$$u(t) \leq\frac{a(T)b(T)}{1+b(T)-\beta}e_{B \oplus C}(t,t_{0}), \quad t\in I. $$
Remark 3.1
Theorem 3.2
Assume that
α, u, a, b, λ
are the same as in Theorem
3.1
and
\(\mu(t)\widetilde{A}(t)<1\). Let
\(v(t,s)\)
and
\(w(t,s)\)
be defined as in Lemma
2.3
such that
\(v^{\Delta}_{t}(t,s)\geq0\), \(w^{\Delta}_{t}(t,s)\geq0\)
for
\(t\geq s\)
and (2.3) holds. Suppose that
u
satisfies
If
then
where
$$\begin{aligned} u(t) \leq& a(t)+b(t) \int _{t_{0}}^{t}\biggl[v(t,s)u(s)+w(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)u(\tau )\Delta\tau\biggr] \Delta s \\ &{}+\lambda b(T) \int_{t_{0}}^{T}\biggl[v(T,s)u(s)+w(T,s) \int_{\alpha (t_{0})}^{\alpha(s)}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.15)
$$ \beta:=e_{\widetilde{B} \oplus\widetilde{C}}(T,t_{0})< 1+\frac{1}{\lambda }, $$
(3.16)
$$ u(t) \leq\frac{a(T)}{\lambda+1-\beta\lambda}e_{\widetilde{B} \oplus \widetilde{C}}(t,t_{0}),\quad t\in I, $$
(3.17)
$$\begin{aligned}& \begin{gathered} \widetilde{A}(t):=b^{\Delta}(t) \int_{t_{0}}^{\sigma(t)}\biggl[v\bigl(\sigma (t),s\bigr)+w \bigl(\sigma(t),s\bigr) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)\Delta\tau\biggr] \Delta s, \\ \widetilde{B}(t):=\frac{\widetilde{A}(t)}{1-\mu (t)\widetilde{A}(t)},\end{gathered} \end{aligned}$$
(3.18)
$$\begin{aligned}& \begin{aligned} \widetilde{C}(t)&:=b(t)\biggl[v\bigl(\sigma(t),t\bigr)+w\bigl(\sigma(t),t\bigr) \int_{\alpha (t_{0})}^{\alpha(t)}g(\tau)\Delta\tau\\ &\quad {}+ \int_{t_{0}}^{t} \biggl[v^{\Delta}_{t}(t,s)+w^{\Delta}_{t}(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau )\Delta\tau\biggr]\Delta s \biggr].\end{aligned} \end{aligned}$$
(3.19)
Proof
Denote Then z is nondecreasing on I, and From Lemma 2.3 and (3.15)-(3.21), we have Similar to the proof of Theorem 3.1, we get (3.17). This completes the proof. □
$$\begin{aligned} z(t) :=& a(T)+b(t) \int _{t_{0}}^{t}\biggl[v(t,s)u(s)+w(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)u(\tau )\Delta\tau\biggr] \Delta s \\ &{}+\lambda b(T) \int_{t_{0}}^{T}\biggl[v(t,T)u(s)+w(T,s) \int_{\alpha (t_{0})}^{\alpha(s)}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.20)
$$ u(t)\leq z(t),\quad t\in I, $$
(3.21)
$$ z(t_{0})=a(T)+\lambda b(T) \int_{t_{0}}^{T}\biggl[v(T,s)u(s)+w(T,s) \int_{\alpha (t_{0})}^{\alpha(s)}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s. $$
(3.22)
$$\begin{aligned} z^{\Delta}(t) =&b^{\Delta}(t) \int_{t_{0}}^{\sigma (t)}\biggl[v\bigl(\sigma(t),s\bigr)u(s)+w \bigl(\sigma(t),s\bigr) \int_{\alpha(t_{0})}^{\alpha (s)}g(\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{}+b(t)\biggl[v\bigl(\sigma(t),t\bigr)u(t)+w\bigl(\sigma(t),t\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)u(\tau)\Delta\tau \\ &{}+ \int_{t_{0}}^{t} \biggl[v^{\Delta}_{t}(t,s)u(s)+w^{\Delta}_{t}(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau )u(\tau)\Delta\tau\biggr]\Delta s\biggr] \\ \leq&b^{\Delta}(t) \int_{t_{0}}^{\sigma(t)}\biggl[v\bigl(\sigma(t),s\bigr)z(s)+w \bigl(\sigma (t),s\bigr) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)z(\tau)\Delta\tau\biggr] \Delta s \\ &{}+b(t)\biggl[v\bigl(\sigma(t),t\bigr)z(t)+w\bigl(\sigma(t),t\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)z(\tau)\Delta\tau \\ &{}+ \int_{t_{0}}^{t} \biggl[v^{\Delta}_{t}(t,s)z(s)+w^{\Delta}_{t}(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau )z(\tau)\Delta\tau\biggr]\Delta s\biggr] \\ \leq&z\bigl(\sigma(t)\bigr)b^{\Delta}(t) \int_{t_{0}}^{\sigma(t)}\biggl[v\bigl(\sigma (t),s\bigr)+w \bigl(\sigma(t),s\bigr) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)\Delta\tau\biggr] \Delta s \\ &{}+z(t)b(t)\biggl[v\bigl(\sigma(t),t\bigr)+w\bigl(\sigma(t),t\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)\Delta\tau \\ &{}+ \int_{t_{0}}^{t} \biggl[v^{\Delta}_{t}(t,s)+w^{\Delta}_{t}(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau )\Delta\tau\biggr]\Delta s \biggr] \\ =&\widetilde{A}(t)z\bigl(\sigma(t)\bigr)+\widetilde{C}(t)z(t),\quad t\in I. \end{aligned}$$
(3.23)
If we let \(\lambda=\frac{1}{b(T)}\) in Theorem 3.2, then we obtain the following corollary.
Corollary 3.2
Assume that
u, a, b, v, w, Ã, B̃, C̃
are the same as in Theorem
3.2
and
\(b(T)\neq0\). Suppose that
u
satisfies
If
then
$$\begin{aligned} u(t) \leq& a(t)+b(t) \int _{t_{0}}^{t}\biggl[v(t,s)u(s)+w(t,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau)u(\tau )\Delta\tau\biggr] \Delta s \\ &{}+ \int_{t_{0}}^{T}\biggl[v(T,s)u(s)+w(T,s) \int_{\alpha(t_{0})}^{\alpha(s)}g(\tau )u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
$$\beta:=e_{\widetilde{B} \oplus\widetilde{C}}(T,t_{0})< 1+b(T), $$
$$u(t) \leq\frac{a(T)}{\lambda+1-\beta\lambda}e_{\widetilde{B} \oplus \widetilde{C}}(t,t_{0}),\quad t\in I. $$
Theorem 3.3
Assume that
u, a, f, λ
are the same as in Theorem
3.1. Let
\(g(t,s)\)
be defined as in Lemma
2.4
such that
\(g^{\Delta}_{t}(t,s)\geq0\)
for
\(t\geq s\)
and (2.3) holds. Suppose that
u
satisfies
If
then
where
$$\begin{aligned} u(t) \leq& a(t)+ \int_{t_{0}}^{t}f(s)\biggl[u(s)+ \int _{t_{0}}^{s}g(s,\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{}+\lambda \int_{t_{0}}^{T}f(s)\biggl[u(s)+ \int_{t_{0}}^{s}g(s,\tau)u(\tau)\Delta \tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.24)
$$ \beta:= \int_{t_{0}}^{T}f(s)e_{A}(s,t_{0}) \Delta s< \frac{1}{\lambda}, $$
(3.25)
$$ u(t) \leq\frac{a(T)}{1-\beta\lambda}\biggl[1+ \int _{t_{0}}^{t}f(s)e_{A}(s,t_{0}) \Delta s\biggr],\quad t\in I, $$
(3.26)
$$ A(t):=f(t)+g\bigl(\sigma(t),t\bigr)+ \int_{t_{0}}^{t}g^{\Delta}_{t}(t,\tau) \Delta\tau . $$
(3.27)
Proof
Denote Then z is nondecreasing on I, and From Lemma 2.4 and (3.26)-(3.29), we get Let Obviously, From Lemma 2.4, (3.27) and (3.32), we obtain It is easy to see that \(A\in\mathcal{R}^{+}\). Therefore, from Lemma 2.5 and the above inequality, we have Combining (3.33) and (3.34), we get Setting \(t=\tau\) in (3.35), integrating it from \(t_{0}\) to t, we easily obtain From (3.28) and (3.30), we have i.e., From (3.25), (3.36) and (3.37), we have In view of (3.25), we get Substituting (3.38) into (3.36), we have Noting \(u(t)\leq z(t)\), we get the desired inequality (3.26). This completes the proof. □
$$\begin{aligned} z(t) :=& a(T)+ \int_{t_{0}}^{t}f(s)\biggl[u(s)+ \int _{t_{0}}^{s}g(s,\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{}+\lambda \int_{t_{0}}^{T}f(s)\biggl[u(s)+ \int_{t_{0}}^{s}g(s,\tau)u(\tau)\Delta \tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
(3.28)
$$ u(t)\leq z(t),\quad t\in I, $$
(3.29)
$$ z(t_{0})=a(T)+\lambda \int_{t_{0}}^{T}f(s)\biggl[u(s)+ \int_{t_{0}}^{s}g(s,\tau )u(\tau)\Delta\tau\biggr] \Delta s. $$
(3.30)
$$\begin{aligned} z^{\Delta}(t) =&f(t)u(t)+f(t) \int_{t_{0}}^{t}g(t,\tau )u(\tau)\Delta\tau \\ \leq&f(t)\biggl[z(t)+ \int_{t_{0}}^{t}g(t,\tau)z(\tau)\Delta\tau\biggr],\quad t\in I. \end{aligned}$$
(3.31)
$$ V(t)=z(t)+ \int_{t_{0}}^{t}g(t,\tau)z(\tau)\Delta\tau,\quad t\in I. $$
(3.32)
$$ V(t_{0})=z(t_{0}), \qquad z(t)\leq V(t),\qquad z^{\Delta}(t)\leq f(t)V(t). $$
(3.33)
$$\begin{aligned} V^{\Delta}(t) =&z^{\Delta}(t)+g\bigl(\sigma(t),t\bigr)z(t)+ \int _{t_{0}}^{t}g^{\Delta}_{t}(t, \tau)z(\tau)\Delta\tau \\ \leq&\biggl[f(t)+g\bigl(\sigma(t),t\bigr)+ \int_{t_{0}}^{t}g^{\Delta}_{t}(t,\tau) \Delta\tau \biggr]V(t) \\ =&A(t)V(t),\quad t\in I. \end{aligned}$$
$$ V(t)\leq V(t_{0})e_{A}(t,t_{0})=z(t_{0})e_{A}(t,t_{0}), \quad t\in I. $$
(3.34)
$$ z^{\Delta}(t)\leq f(t)z(t_{0})e_{A}(t,t_{0}). $$
(3.35)
$$ z(t)\leq z(t_{0})+z(t_{0}) \int_{t_{0}}^{t}f(s)e_{A}(s,t_{0}) \Delta s. $$
(3.36)
$$\bigl(z(t_{0})-a(T)\bigr)\frac{\lambda+1}{\lambda}+a(T)=z(T), $$
$$ \frac{\lambda+1}{\lambda}z(t_{0})-\frac{1}{\lambda}a(T)=z(T). $$
(3.37)
$$\frac{\lambda+1}{\lambda}z(t_{0})-\frac{1}{\lambda}a(T)=z(T)\leq z(t_{0})+z(t_{0}) \int_{t_{0}}^{T}f(s)e_{A}(s,t_{0}) \Delta s=z(t_{0})+z(t_{0})\beta. $$
$$ z(t_{0})\leq\frac{a(T)}{1-\beta\lambda}. $$
(3.38)
$$ z(t)\leq\frac{a(T)}{1-\beta\lambda}\biggl[1+ \int_{t_{0}}^{t}f(s)e_{A}(s,t_{0}) \Delta s\biggr]. $$
(3.39)
If we let \(\lambda=1\) in Theorem 3.3, then we obtain the following corollary.
Corollary 3.3
Assume that
u, a, f, g
and
A
are the same as in Theorem
3.3. Suppose that
u
satisfies
If
then
$$\begin{aligned} u(t) \leq& a(t)+ \int_{t_{0}}^{t}f(s)\biggl[u(s)+ \int _{t_{0}}^{s}g(s,\tau)u(\tau)\Delta\tau\biggr] \Delta s \\ &{}+ \int_{t_{0}}^{T}f(s)\biggl[u(s)+ \int_{t_{0}}^{s}g(s,\tau)u(\tau)\Delta\tau\biggr] \Delta s,\quad t\in I. \end{aligned}$$
$$\beta:= \int_{t_{0}}^{T}f(s)e_{A}(s,t_{0}) \Delta s< 1, $$
$$u(t) \leq\frac{a(T)}{1-\beta}\biggl[1+ \int_{t_{0}}^{t}f(s)e_{A}(s,t_{0}) \Delta s\biggr],\quad t\in I. $$
Remark 3.2
If we take \(a(t)\equiv u_{0}\), \(g(s,t)=g(t)\) and \(\lambda=0\), then Theorem 3.3 reduces to [27, Theorem 1]. If we take \(\mathbb{T}=\mathbb{R}\), \(a(t)\equiv u_{0}\) and \(\lambda=0\), then Theorem 3.3 reduces to [32, Theorem 2.1 \((a_{1})\)]. If we take \(\mathbb{T}=\mathbb{Z}\), \(a(t)\equiv u_{0}\) and \(\lambda=0\), then Theorem 3.3 reduces to [32, Theorem 2.3 \((c_{1})\)].
4 Applications
In this section, we will present some simple applications for our results.
Example 4.1
Consider the following retarded Volterra-Fredholm type dynamic integral equation on time scales: where \(u, a, b: I\rightarrow\mathbb{R}\) are rd-continuous functions, \(\vert a \vert \) is nondecreasing, \(b(T)\neq0\), \(\alpha\in \Upsilon\), \(F, \widetilde{F}:I\times\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}\) and \(H, \widetilde{H}:I\times\mathbb{R}\rightarrow\mathbb{R}\) are continuous functions.
$$ \begin{aligned}[b] u(t)&=a(t)+b(t) \int_{\alpha(t_{0})}^{\alpha (t)}F\biggl(s,u(s), \int_{\alpha(t_{0})}^{s}H\bigl(\tau,u(\tau)\bigr)\Delta\tau \biggr)\Delta s \\ &\quad {}+ \int_{\alpha(t_{0})}^{\alpha(T)}\widetilde{F}\biggl(s,u(s), \int_{\alpha (t_{0})}^{s}\widetilde{H}\bigl(\tau,u(\tau)\bigr) \Delta\tau\biggr)\Delta s,\quad t\in I, \end{aligned} $$
(4.1)
The following theorem gives an estimate for the solutions of Eq. (4.1).
Theorem 4.1
Suppose that the functions
\(F, H, \widetilde{F}\)
and
H̃
in (4.1) satisfy the conditions
where
\(f_{1}, f_{2}, g: I\rightarrow\mathbb{R}_{+}\)
are rd-continuous functions. If
then all solutions of Eq. (4.1) satisfy
where
\(B(t)\)
and
\(C(t)\)
are defined as in Theorem
3.1.
$$\begin{aligned}& \bigl\vert F(t,u,v) \bigr\vert \leq f_{1}(t) \vert u \vert +f_{2}(t) \vert v \vert ,\quad t \in I, u, v\in \mathbb{R}, \end{aligned}$$
(4.2)
$$\begin{aligned}& \bigl\vert H(t,u) \bigr\vert \leq g(t) \vert u \vert ,\quad t\in I, u\in \mathbb{R}, \end{aligned}$$
(4.3)
$$\begin{aligned}& \bigl\vert \widetilde{F}(t,u,v) \bigr\vert \leq f_{1}(t) \vert u \vert +f_{2}(t) \vert v \vert ,\quad t \in I, u, v\in \mathbb{R}, \end{aligned}$$
(4.4)
$$\begin{aligned}& \bigl\vert \widetilde{H}(t,u) \bigr\vert \leq g(t) \vert u \vert ,\quad t\in I, u\in\mathbb{R}, \end{aligned}$$
(4.5)
$$\beta:=e_{B \oplus C}(T,t_{0})< 1+ \bigl\vert b(T) \bigr\vert , $$
$$ \bigl\vert u(t) \bigr\vert \leq\frac{ \vert a(T)b(T) \vert }{1+ \vert b(T) \vert -\beta}e_{B \oplus C}(t,t_{0}), \quad t\in I, $$
(4.6)
Proof
By (4.1)-(4.3), we get Using Corollary 3.1, we obtain the desired inequality (4.4). □
$$\begin{aligned} \bigl\vert u(t) \bigr\vert \leq& \bigl\vert a(t) \bigr\vert + \bigl\vert b(t) \bigr\vert \int_{\alpha(t_{0})}^{\alpha(t)} \biggl\vert F\biggl(s,u(s), \int_{\alpha(t_{0})}^{s}H\bigl(\tau,u(\tau)\bigr)\Delta\tau \biggr) \biggr\vert \Delta s \\ &{}+ \int_{\alpha(t_{0})}^{\alpha(T)} \biggl\vert \widetilde{F} \biggl(s,u(s), \int _{\alpha(t_{0})}^{s}\widetilde{H}\bigl(\tau,u(\tau)\bigr) \Delta\tau\biggr) \biggr\vert \Delta s \\ \leq& \bigl\vert a(t) \bigr\vert + \bigl\vert b(t) \bigr\vert \int _{\alpha(t_{0})}^{\alpha(t)}\biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s) \int _{\alpha(t_{0})}^{s}g(\tau) \bigl\vert u(\tau) \bigr\vert \Delta\tau\biggr]\Delta s \\ &{}+ \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau) \bigl\vert u(\tau) \bigr\vert \Delta\tau\biggr]\Delta s , \quad t\in I. \end{aligned}$$
The next result deals with the uniqueness of solutions of Eq. (4.1).
Theorem 4.2
Suppose that the functions
F, H, F̃
and
H̃
in (4.1) satisfy the conditions
where
\(f_{1}, f_{2}, g: I\rightarrow\mathbb{R}_{+}\)
are rd-continuous functions. If
then Eq. (4.1) has at most one solution on
I, where
\(B(t)\)
and
\(C(t)\)
are defined as in Theorem
3.1.
$$\begin{aligned}& \bigl\vert F(t,u_{1},v_{1})-F(t,u_{2},v_{2}) \bigr\vert \leq f_{1}(t) \vert u_{1}-u_{2} \vert +f_{2}(t) \vert v_{1}-v_{2} \vert , \quad t \in I, u, v\in\mathbb{R}, \end{aligned}$$
(4.7)
$$\begin{aligned}& \bigl\vert H(t,u_{1})-H(t,u_{2}) \bigr\vert \leq g(t) \vert u_{1}-u_{2} \vert ,\quad t\in I, u_{1}, u_{2}\in\mathbb{R}, \end{aligned}$$
(4.8)
$$\begin{aligned}& \bigl\vert \widetilde{F}(t,u_{1},v_{1})- \widetilde{F}(t,u_{2},v_{2}) \bigr\vert \leq f_{1}(t) \vert u_{1}-u_{2} \vert +f_{2}(t) \vert v_{1}-v_{2} \vert ,\quad t \in I, u, v\in\mathbb{R}, \end{aligned}$$
(4.9)
$$\begin{aligned}& \bigl\vert \widetilde{H}(t,u_{1})-\widetilde{H}(t,u_{2}) \bigr\vert \leq g(t) \vert u_{1}-u_{2} \vert ,\quad t \in I, u_{1}, u_{2}\in\mathbb{R}, \end{aligned}$$
(4.10)
$$\beta:=e_{B \oplus C}(T,t_{0})< 1+ \bigl\vert b(T) \bigr\vert , $$
Proof
Let \(u(t)\) and \(v(t)\) be two solutions of Eq. (4.1) on I. From (4.1) and (4.7)-(4.10), we have Applying Corollary 3.1 to (4.11), we get \(\vert u(t)-v(t) \vert \leq0\), \(t\in I\). Therefore \(u(t)\equiv v(t)\), that is, there is at most one solution to Eq. (4.1). □
$$ \begin{gathered}[b] \bigl\vert u(t)-v(t) \bigr\vert \\ \quad \leq \bigl\vert b(t) \bigr\vert \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s) \bigl\vert u(s)-v(s) \bigr\vert +f_{2}(s) \int_{\alpha (t_{0})}^{s}g(\tau) \bigl\vert u(\tau)-v(\tau) \bigr\vert \Delta\tau\biggr]\Delta s \\ \qquad {}+ \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s) \bigl\vert u(s)-v(s) \bigr\vert +f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau) \bigl\vert u(\tau)-v(\tau ) \bigr\vert \Delta\tau\biggr]\Delta s , \quad t\in I. \end{gathered} $$
(4.11)
Example 4.2
Assume \(\mathbb{T}=q^{\mathbb{N}_{0}}=\{ 1,q,q^{2},\ldots\}\) with \(q>1\) and consider the following Volterra-Fredholm type q-difference equation: where \(u, a, b, f_{1}, f_{2}, g: I\rightarrow\mathbb{R}_{+}\) are functions, \(t_{0}, T\in q^{\mathbb{N}_{0}}\), \(T>t_{0}\) and \(I=[t_{0},T]\cap q^{\mathbb{N}_{0}}\).
$$\begin{aligned} u(t) =& a(t)+b(t)\sum_{s=t_{0}}^{t} \Biggl[f_{1}(s)u(s)+f_{2}(s)\sum _{\tau=t_{0}}^{s}g(\tau)u(\tau)\Biggr] \\ &{}+\sum_{s=t_{0}}^{T}\Biggl[f_{1}(s)u(s)+f_{2}(s) \sum_{\tau=t_{0}}^{s}g(\tau)u(\tau )\Biggr],\quad t \in I, \end{aligned}$$
(4.12)
Theorem 4.3
Assume that
a
is nondecreasing and
\((q-1)tA(t)<1\). If
then the solution
\(u(t)\)
of Eq. (4.12) satisfies the following inequality:
where
$$\beta:=\prod_{\tau=t_{0}}^{T/q}\bigl[1+(q-1)\tau \bigl(B(\tau)+C(\tau)+(q-1)\tau B(\tau)C(\tau)\bigr)\bigr]< 1+b(T), $$
$$ u(t) \leq\frac{a(T)b(T)}{1+b(T)-\beta}\prod_{\tau=t_{0}}^{t/q} \bigl[1+(q-1)\tau\bigl(B(\tau)+C(\tau)+(q-1)\tau B(\tau)C(\tau)\bigr)\bigr],\quad t\in I, $$
(4.13)
$$\begin{aligned}& A(t) := \frac{b(qt)-b(t)}{(q-1)t}\sum_{s=t_{0}}^{qt} \Biggl[f_{1}(s)+f_{2}(s)\sum_{\tau=t_{0}}^{s}g( \tau)\Biggr] ,\qquad B(t):=\frac{A(t)}{1-(q-1)tA(t)}, \\& C(t) := b(t)\Biggl[f_{1}(t)+f_{2}(t)\sum _{\tau=t_{0}}^{t}g(\tau)\Biggr]. \end{aligned}$$
Proof
Note that \(\sigma(t)=qt\), \(\mu(t)=(q-1)t\) for any \(t\in q^{\mathbb{N}_{0}}\) and for \(t>t_{0}\), where \(t, t_{0}, \tau\in q^{\mathbb{N}_{0}}\). Let \(u(t)\) be a solution of Eq. (4.12), we get Then a suitable application of Corollary 3.1 to (4.14) yields the desired result (4.13). □
$$e_{p}(t,t_{0})=\prod_{\tau=t_{0}}^{t/q} \bigl[1+(q-1)\tau p(\tau)\bigr] $$
$$\begin{aligned} \bigl\vert u(t) \bigr\vert \leq& a(t)+b(t)\sum _{s=t_{0}}^{t}\Biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s)\sum_{\tau =t_{0}}^{s}g( \tau) \bigl\vert u(\tau) \bigr\vert \Biggr] \\ &{}+\sum_{s=t_{0}}^{T}\Biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s)\sum _{\tau =t_{0}}^{s}g(\tau) \bigl\vert u(\tau) \bigr\vert \Biggr],\quad t\in I. \end{aligned}$$
(4.14)
Example 4.3
Assume \(\mathbb{T}=\mathbb{R}\) and consider the following retarded Volterra-Fredholm type integral equation: where \(u, a, b,\alpha, f_{1}, f_{2}, g: I\rightarrow\mathbb{R}_{+}\) are continuous functions, \(t_{0}, T\in\mathbb{R}\), \(T>t_{0}\) and \(I=[t_{0},T]\).
$$\begin{aligned} u(t) =& a(t)+b(t) \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)u(\tau)\,\mathrm{d}\tau\biggr] \,\mathrm{d} s \\ &{}+\lambda \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s)u(s)+f_{2}(s) \int_{\alpha (t_{0})}^{s}g(\tau)u(\tau)\,\mathrm{d}\tau\biggr] \,\mathrm{d} s,\quad t\in I, \end{aligned}$$
(4.15)
Theorem 4.4
Assume that
a
is nondecreasing, \(\lambda\geq0\)
is a constant, \(\alpha'(t)>0\), \(\alpha(t)\leq t\)
and
\(b'(t)\geq0\). If
then the solution
\(u(t)\)
of Eq. (4.15) satisfies the following inequality:
where we use the convention that
\(\frac{1}{0}=+\infty\),
$$\beta:=\exp\biggl( \int_{t_{0}}^{T}(A+C) (\tau)\,\mathrm{d}\tau\biggr)< 1+ \frac{b(T)}{\lambda}, $$
$$ u(t) \leq\frac{a(T)b(T)}{\lambda+b(T)-\beta\lambda}\exp\biggl( \int _{t_{0}}^{t}(A+C) (\tau)\,\mathrm{d}\tau\biggr), \quad t\in I, $$
(4.16)
$$\begin{gathered} A(t) := b'(t) \int_{\alpha(t_{0})}^{\alpha (t)}\biggl[f_{1}(s)+f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau)\,\mathrm{d} \tau\biggr] \,\mathrm{d} s\quad\textit{and} \\ C(t) := b(t)\biggl[f_{1}\bigl(\alpha(t)\bigr)+f_{2}\bigl( \alpha(t)\bigr) \int_{\alpha(t_{0})}^{\alpha (t)}g(\tau)\,\mathrm{d} \tau\biggr] \alpha'(t). \end{gathered} $$
Proof
Note that \(\sigma(t)=t\), \(\mu(t)=0\) for any \(t\in\mathbb{R}\), and for \(t>t_{0}\). Let \(u(t)\) be a solution of Eq. (4.15), we have Then a suitable application of Theorem 3.1 to (4.17) yields the desired result (4.16). □
$$e_{p}(t,t_{0})=\exp\biggl( \int_{t_{0}}^{t}p(\tau)\,\mathrm{d}\tau\biggr) $$
$$\begin{aligned} \bigl\vert u(t) \bigr\vert \leq& a(t)+b(t) \int_{\alpha (t_{0})}^{\alpha(t)}\biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s) \int_{\alpha (t_{0})}^{s}g(\tau) \bigl\vert u(\tau) \bigr\vert \,\mathrm{d}\tau\biggr] \,\mathrm{d} s \\ &{}+\lambda \int_{\alpha(t_{0})}^{\alpha(T)}\biggl[f_{1}(s) \bigl\vert u(s) \bigr\vert +f_{2}(s) \int_{\alpha(t_{0})}^{s}g(\tau) \bigl\vert u(\tau) \bigr\vert \,\mathrm{d}\tau\biggr] \,\mathrm{d} s,\quad t\in I. \end{aligned}$$
(4.17)
5 Conclusions
In this paper we have established some new retarded Volterra-Fredholm type integral inequalities on time scales. Unlike some existing results in the literature (e.g., [16, 20]), the integral inequalities considered in this paper involve the retarded term, which results in difficulties in the estimation on the explicit bounds of unknown functions \(u(t)\). These inequalities generalize and extend some known inequalities and can be used as tools in the qualitative theory of certain classes of retarded dynamic equations on time scales.
Acknowledgements
The author thanks the reviewers for their helpful and valuable suggestions and comments on this paper. This research was supported by the National Natural Science Foundation of China (No. 11671227) and the Project of Shandong Province Higher Educational Science and Technology Program (China) (No. J14LI09).
Competing interests
The author declares that there is no conflict of interests regarding the publication of this paper.
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