Abstract
This paper investigates some new retarded weakly singular integral inequalities with discontinuous functions for two independent variables. The inequalities given here can be used in the qualitative analysis of various problems for integral equations and differential equations. Some examples are also given to illustrate the application of the conclusion.
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1 Introduction
Integral inequalities are used as handy tools in the study of the qualitative properties of solutions to differential and integral equations, such as existence, uniqueness, boundedness, stability, and other properties. The literature on such inequalities and their applications is vast (see [1,2,3,4,5,6,7,8,9,10,11,12] and the references therein). With the development of theory for fractional differential equations, integral inequalities with weakly singular kernels have attracted great interest [13,14,15,16,17,18]. In 1981, Henry [8] proved global existence and exponential decay results for a parabolic Cauchy problem by using the following singular integral inequality:
Sano and Kunimatsu [9] gave a sufficient condition for stabilization of semilinear parabolic distributed systems by using a modification of Henry-type inequalities
Ye et al. [11] provided a generalization of inequality (1)
and used it to study the dependence of the solution and the initial condition to a certain fractional differential equation. All such inequalities are studied by an iteration argument, and the estimation formulas are expressed by a complicated power series which are sometimes not very convenient for applications. To avoid this, Medved̆’ [12] presented a new method for studying Henry-type inequalities and established explicit bounds with relatively simple formulas which are similar to the classic Gronwall–Bellman inequalities. Recently, by using a modification of Medved̆’s method, Ma and Pec̆arić [14] studied a certain class of nonlinear inequalities of Henry-type
The results were further generalized by Cheung et al. [15] to the following form:
In 2017, Xu [18] studied the following new generalization of weakly singular integral inequalities in two variables:
with the initial condition \(u(x,y)=\phi(x,y), (x,y)\in D'=[\mu,0]\times[\mu,0], \phi(\sigma(x),\sigma(y))\leq(a(x,y))^{1/p}\) for \((x,y)\in D\) with \(\sigma(x)\leq0,\sigma(y)\leq0\).
In recent ten years, a series of achievements have been made in the research of integral inequalities for discontinuous functions (see [19,20,21,22]). In 2007, Iovane [19] studied the following discontinuous function integral inequality:
In 2009, Gllo et al. [20] studied the impulsive integral inequality
In 2015, Mi et al. [21] studied the integral inequality of complex functions with unknown function
Very recently, Li et al. [22] studied the following weakly singular retarded integral inequality for discontinuous function:
In this paper, we establish some new weakly singular retarded integral inequalities with discontinuous functions in two variables
Finally, two examples are included to illustrate the usefulness of our results.
2 Preliminaries
In this paper, let \(\varOmega=\bigcup_{i,j\geq1}\varOmega_{ij}, \varOmega_{ij}=\{(x,y):x_{i-1}\leq x< x_{i},y_{j-1}\leq y< y_{j},i,j=1,2, \ldots,x_{0}>0,y_{0}>0\}\). Let \(\mathbb{R}\) denote the set of real numbers and \(\mathbb{R}^{+}=[0,\infty)\), \(C(M,S)\) denote the class of all continuous functions defined on the set M with range in the set S.
Lemma 1
([23])
Let \(a_{1},a_{2},\ldots,a_{n}\) be nonnegative real numbers, \(m>1\) is a real number, and n is a natural number. Then
Lemma 2
([13])
Let \(\beta,\gamma,\xi\), and p be positive constants. Then
where \(B[x,y]=\int_{0}^{1}s^{x-1}(1-s)^{y-1}\,\mathrm{d}s\ (x>0,y>0)\) is the well-known beta-function and \(\theta=p[\beta(\gamma-1)+\xi-1]+1\).
In addition, Li et al. [22] gave a generalization of equality (5), that is,
where \(\alpha(t)\) is a continuous, differentiable, and increasing function on \([t_{0},+\infty)\) with \(\alpha(t)\leq t\), \(\alpha(t_{0})=t_{0}\geq0\).
Lemma 3
([13])
Suppose that the positive constants \(\beta,\gamma,\xi ,p_{1}\), and \(p_{2}\) satisfy conditions:
-
(1)
if \(\beta\in(0,1],\gamma\in(\frac{1}{2},1)\) and \(\xi\geq\frac {3}{2}-\gamma,p_{1}=\frac{1}{\gamma}\);
-
(2)
if \(\beta\in(0,1],\gamma\in(0,\frac{1}{2})\) and \(\xi>\frac{1-2\gamma^{2}}{1-\gamma^{2}},p_{2}=\frac{1+4\gamma }{1+3\gamma}\), then
$$ B \biggl[\frac{p_{i}(\xi-1)+1}{\beta},p_{i}(\gamma -1)+1 \biggr]\in \mathbb{R}^{+},\qquad \theta_{i}=p_{i}\bigl[\beta( \gamma-1)+\xi-1\bigr]+1\geq0 $$are valid for \(i=1,2\).
Lemma 4
Let \(u(x,y),a(x,y),b(x,y),h(x,y)\in C(\mathbb{R}^{+}\times\mathbb {R}^{+},\mathbb{R}^{+}),\alpha(x),\beta(y)\) be continuous, differentiable, and increasing functions on \(\mathbb{R}^{+}\) with \(\alpha(x)\leq x,\beta(y)\leq y,\alpha(0)=0,\beta(0)=0\). If \(u(x,y)\) satisfied the following inequality
then
where
Proof
Define a function \(v(x,y)\) on \(\mathbb{R}^{+}\) by
we have \(v(x,0)=0,v(0,y)=0\). Differentiating \(v(x,y)\) with respect to \(x,y\), we have
Using (6) and \(\alpha(x)\leq x,\beta(y)\leq y\), we have
Integrating both sides of inequality (9), because \(v(x,0)=v(0,y)=0\), we have
Substituting inequality (11) into (6), we can get the required estimation (7). This completes the proof. □
Lemma 5
([24])
Let \(a\geq0,p\geq q\geq0\), and \(p\neq0\), then
We give two special cases of the above result:
-
(a)
If \(K=1\), we have
$$\begin{aligned} a^{\frac{q}{p}}\leq\frac{q}{p}a+\frac{p-q}{p},\quad a\geq0,p\geq q\geq0,p\neq0. \end{aligned}$$ -
(b)
If \(K=1,p=1\), we have
$$\begin{aligned} a^{q}\leq qa+(1-q),\quad a\geq0,q\geq0. \end{aligned}$$
3 Main results
Firstly, we study inequality (2) and assume that the following conditions hold:
- (\(H_{1}\)):
-
\(a(x,y)\in C(\mathbb{R}^{+}\times\mathbb{R}^{+},\mathbb{R} ^{+})\), and \(a(x,y)\) is a nondecreasing function;
- (\(H_{2}\)):
-
\(f_{i}(x,y)\ (i=1,2,3)\) are continuous and nonnegative on Ω;
- (\(H_{3}\)):
-
\(\alpha(x),\beta(y)\) are continuous, differentiable, and increasing functions on \(\mathbb{R}^{+}\) with \(\alpha(x)\leq x, \beta(y)\leq y,\alpha(0)=0,\beta(0)=0\);
- (\(H_{4}\)):
-
\(\zeta,\gamma,\xi\) are positive constants.
Theorem 1
Suppose that \((H_{1})\)–\((H_{4})\) hold and \(u(x,y)\) satisfies inequality (2), then we have the following results:
- (¡):
-
If \(\zeta\in(0,1],\gamma\in(\frac{1}{2},1)\), and \(\xi\geq \frac{3}{2}-\gamma\), we have
$$\begin{aligned} & u(x,y)\leq \biggl(\tilde{a}_{1}(x,y)+ \frac{\tilde{b}_{1}(x,y)}{\tilde {e}_{1}(\alpha(x),\beta(y))} \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}\tilde{h}_{1}(t,s) \tilde{a}_{1}(t,s)\tilde{e}_{1}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{1-\gamma}, \\ &\quad(x,y)\in\varOmega, \end{aligned}$$(12)where
$$\begin{aligned} &\tilde{a}_{1}(x,y)=3^{\frac{\gamma}{1-\gamma}}a^{\frac{1}{1-\gamma }}(x,y), \\ &\tilde{b}_{1}(x,y)=\bigl(3M_{1}^{2}\times \bigl(\alpha(x)\beta (y)\bigr)^{\theta_{1}}\bigr)^{\frac{\gamma}{1-\gamma}}, \\ &\tilde{h}_{1}(x,y)=f_{1}^{\frac{1}{1-\gamma}}(x,y)+ \biggl(f_{2}(x,y) \int_{0}^{x} \int_{0}^{y}f_{3}(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr)^{\frac{1}{1-\gamma}}, \\ &\tilde{e}_{1}(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}\tilde {h}_{1}(t,s) \tilde{b}_{1}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr), \\ &M_{1}=\frac{1}{\zeta}B \biggl[\frac{\gamma+\xi-1}{\zeta\gamma }, \frac{2\gamma-1}{\gamma} \biggr], \\ &\theta_{1}=\frac{1}{\gamma}\bigl[\zeta(\gamma-1)+\xi-1 \bigr]+1. \end{aligned}$$ - (¡¡):
-
If \(\zeta\in(0,1],\gamma\in(0,\frac{1}{2}]\), and \(\xi>\frac {1-2\gamma^{2}}{1-\gamma^{2}}\), we have
$$\begin{aligned} &u(x,y)\leq \biggl(\tilde{a}_{2}(x,y)+ \frac{\tilde{b}_{2}(x,y)}{\tilde {e}_{2}(\alpha(x),\beta(y))} \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}\tilde{h}_{2}(t,s) \tilde{a}_{2}(t,s)\tilde{e}_{2}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{\frac{\gamma}{1+4\gamma}}, \\ &\quad (x,y)\in \varOmega, \end{aligned}$$(13)where
$$\begin{aligned} &\tilde{a}_{2}(x,y)=3^{\frac{1+3\gamma}{\gamma}}a^{\frac{1+4\gamma }{\gamma}}(x,y), \\ &\tilde{b}_{2}(x,y)=\bigl(3M_{2}^{2}\times \bigl(\alpha (x)\beta(y)\bigr)^{\theta_{2}}\bigr)^{\frac{1+3\gamma}{\gamma}}, \\ &\tilde{h}_{2}(x,y)=f_{1}^{\frac{1+4\gamma}{\gamma}}(x,y)+ \biggl(f_{2}(x,y) \int_{0}^{x} \int_{0}^{y}f_{3}(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr)^{\frac{1+4\gamma}{\gamma}}, \\ &\tilde{e}_{2}(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}\tilde {h}_{1}(t,s) \tilde{b}_{1}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr), \\ &M_{2}=\frac{1}{\zeta}B \biggl[\frac{\xi(1+4\gamma)-\gamma}{\zeta (1+3\gamma)}, \frac{4\gamma^{2}}{1+3\gamma} \biggr], \\ &\theta_{2}=\frac{1+4\gamma}{1+3\gamma}\bigl[\zeta(\gamma-1)+\xi-1 \bigr]+1. \end{aligned}$$
Proof
If \(\zeta\in(0,1],\gamma\in(\frac{1}{2},1)\), and \(\xi\geq\frac {3}{2}-\gamma\), let
If \(\zeta\in(0,1],\gamma\in(0,\frac{1}{2}]\), and \(\xi>\frac {1-2\gamma^{2}}{1-\gamma^{2}}\), let
then
Using Hölder’s inequality in (2), we have
Set \(z(x,y)\) as the right-hand side of the above inequality, and
that is,
Then \(z(x,y)\) is a nondecreasing function, and \(u(x,y)\leq z(x,y)\), we have
Using the discrete Jensen inequality in Lemma 1 with \(n=3,m=q_{i}\), we get
Using Lemma 2, we obtain
for \((x,y)\in\varOmega\), where
Set
we have
Applying Lemma 4 to (18), we obtain
Substituting \(p_{1}=\frac{1}{\gamma},q_{1}=\frac{1}{1-\gamma}\), and \(p_{2}=\frac{1+4\gamma}{1+3\gamma},q_{2}=\frac{1+4\gamma}{\gamma }\) to (19), respectively, we can get the desired estimations (12) and (13). This completes the proof. □
Secondly, we study inequality (3) and assume that the following conditions hold:
- (\(H_{5}\)):
-
\(a(x,y)\geq1\);
- (\(H_{6}\)):
-
\(f(x,y)\) is continuous and nonnegative on Ω;
- (\(H_{7}\)):
-
\(\alpha(x),\beta(y)\) are continuous, differentiable, and increasing functions on \([x_{0},+\infty), [y_{0},+\infty)\), respectively, and \(\alpha(x)\leq x,\beta(y)\leq y,\alpha (x_{i})=x_{i},\beta(y_{i})=y_{i},i=0,1,2,\ldots \) ;
- (\(H_{8}\)):
-
\(u(x,y)\) is nonnegative and continuous on Ω with the exception of the points \((x_{i},y_{i})\), where there is a finite jump: \(u(x_{i}-0,y_{i}-0)\neq u(x_{i}+0,y_{i}+0),i=1,2,\ldots \) ;
- (\(H_{9}\)):
-
\(p,\zeta,\gamma\) are positive constants;
- (\(H_{10}\)):
-
\(\zeta_{i}\) are nonnegative constants for any positive integer i.
Theorem 2
Suppose that \((H_{1})\), \((H_{5})\)–\((H_{10})\) hold and \(u(x,y)\) satisfies inequality (3), then we have
where
Proof
Firstly, we consider the case \((x,y)\in\varOmega_{11}\). Denoting
then \(v(x,y)\) is a nonnegative and nondecreasing continuous function, and \(u(x,y)\leq v(x,y), v(x_{0},y_{0})=a(x_{0},y_{0})\).
Differentiating (21), we have
Set
then \(F(x,y)\leq G(x,y)\), \(G(x,y)\) is a nonnegative and nondecreasing continuous function, and \(G(x_{0},y_{0})=a^{2}(x_{0},y_{0})\). Since \(a(x,y)\geq1\), we have \(v(x,y)\geq1\), then \(v(x,y)\leq v^{2}(x,y)\leq G(x,y)\), that is, \(v(x,y)\leq G(x,y)\). Differentiating (24) with respect to x, from (22), we have
Set
then
From (26), we have
Let \(\eta(x,y)=G^{-(p+1)}(x,y)\), then \(\eta _{x}(x,y)=-(p+1)G^{-(p+2)}(x,y)G_{x}(x,y)\), (27) can be restated as
Multiplying by \(\exp ((p+1)\int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t )\) on both sides of (28), we have
Integrating both sides of (29) from \(x_{0}\) to x, we get
set
then
Since \(\eta(x_{0},y)=G^{-(p+1)}(x_{0},y)=a^{-2(p+1)}(x_{0},y)\), from (30) we have
By the relation \(\eta(x,y)=G^{-(p+1)}(x,y)\), from (31) we get
where \(1-(p+1)a^{2(p+1)}(x_{0},y)\int_{x_{0}}^{x}B(t,y)\Delta (t,y)\,\mathrm{d}t>0\). Setting
from (23), (24), (32), and (33), we obtain
Integrating both sides of (34), we get
where \(\tilde{h}(t,s)=(x^{\zeta}-t^{\zeta})^{\gamma-1}(y^{\zeta }-s^{\zeta})^{\gamma-1}f(t,s)\varPhi(\alpha^{-1}(t),\beta^{-1}(s))\). Inequality (35) has the same form as inequality (6) of Lemma 4. By using Lemma 4, we can obtain the estimate of \(u(x,y)\) as follows:
Set
then \(u(x,y)\leq\tilde{u}_{1}(x,y),(x,y)\in\varOmega_{11}\).
Next, if \((x,y)\in\varOmega_{22}\), (3) can be restated as
Setting
then \(\varPsi(x,y)\) is a nonnegative and nondecreasing function, and
Differentiating both sides of (37), we obtain
(38) has the same form as (23), and using the same procedure, we can get the desired estimations (20) for \((x,y)\in\varOmega_{22}\).
Consequently, by using a similar procedure, we can get the desired estimations (20) for \((x,y)\in\varOmega_{ii}\ (i=3,4,5,\ldots)\). Thus we complete the proof of Theorem 2. □
Finally, we study inequality (4) and assume that the following conditions hold:
\((H_{11})\) \(g(x,y)\) is continuous and nonnegative on Ω;
\((H_{12})\) \(p,q,m,n,\xi,\zeta,\gamma\) are positive constants with \(p\geq m,p\geq n,q\in[0,1]\).
Theorem 3
Suppose \((H_{1})\), \((H_{5})\)–\((H_{8})\), \((H_{10})\)–\((H_{12})\) hold and \(u(x,y)\) satisfies inequality (4). Then we have the following results:
- (¡):
-
If \(\zeta\in(0,1],\gamma\in(\frac{1}{2},1)\), and \(\xi\geq \frac{3}{2}-\gamma\), we have
$$\begin{aligned} &u(x,y)\leq \biggl[ E_{i}(x,y)+ \biggl(\tilde{a}_{i}(x,y)+ \frac{\tilde {b}_{1}(x,y)}{\tilde{e}_{i}(\alpha(x),\beta(y))} \\ &\phantom{u(x,y)\leq}{}\times \int_{x_{i}}^{\alpha(x)} \int_{y_{i}}^{\beta(y)}\tilde{h}(t,s)\tilde {a}_{i}(t,s)\tilde{e}_{i}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{1-\gamma} \biggr]^{1/p}, \\ &\quad (x,y)\in\varOmega, \end{aligned}$$(39)where \(M_{1},\theta_{1}\) are the same as in Theorem 1, and
$$\begin{aligned} &E_{0}(x,y)=a(x,y),\quad (x,y)\in\varOmega_{11}, \\ &E_{i}(x,y)\\ &\quad =a(x,y)+b(x,y)\sum_{j=1}^{i} \int_{x_{j-1}}^{\alpha (x_{j})} \int_{y_{j-1}}^{\beta(y_{j})}\bigl(\alpha^{\zeta}(x)-t^{\zeta } \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma -1}s^{\xi-1}f(t,s) \\ &\qquad{}\times \biggl[\tilde{u}_{j}^{m}(t,s)+ \int_{x_{j-1}}^{t} \int _{y_{j-1}}^{s}g(\tau,\eta)\tilde{u}_{j}^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &\qquad{}+\sum_{j=1}^{i} \zeta_{j}\tilde{u}_{j}^{p}(x_{j}-0,y_{j}-0),\quad (x,y)\in\varOmega _{ii}, i=1,2,3,\ldots, \\ &\tilde{u}_{j}(x,y)= \biggl[ \biggl(\tilde{a}_{j-1}(x,y)+ \frac{\tilde {b}_{1}(x,y)}{\tilde{e}_{j-1}(\alpha(x),\beta(y))}\\ &\phantom{\tilde{u}_{j}(x,y)=}{}\times \int _{x_{j-1}}^{\alpha(x)} \int_{y_{j-1}}^{\beta(y)}\tilde {h}_{j-1}(t,s) \tilde{a}_{j-1}(t,s)\tilde{e}_{j-1}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{1-\gamma} \biggr]^{1/p},\quad j=1,2,3,\ldots , \\ &\tilde{a}_{i}(x,y)=3^{\frac{\gamma}{1-\gamma}}A_{i}^{\frac {1}{1-\gamma}}(x,y),\quad i=0,1,2,\ldots, \\ &A_{i}(x,y)=b(x,y) \bigl(M_{1}^{2}\times \bigl(\alpha(x)\beta(y)\bigr)^{\theta _{1}}\bigr)^{\gamma} \biggl[ \int_{x_{i}}^{\alpha(x)} \int_{y_{i}}^{\beta (y)}B_{i}^{\frac{1}{1-\gamma}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1-\gamma}, \\ &\quad i=0,1,2,\ldots, \\ &B_{i}(x,y)=f(x,y) \biggl[(1-q)+q \biggl(\frac{m}{p}E_{i}(x,y)+ \frac {p-m}{p} \biggr) \biggr] \\ &\phantom{B_{i}(x,y)=}{} +qf(x,y) \int_{x_{i}}^{x} \int _{y_{i}}^{y}g(\tau,\eta) \biggl[ \frac{n}{p}E_{i}(\tau,\eta)+\frac {p-n}{p} \biggr] \,{\mathrm{d}}\eta\,{\mathrm{d}}\tau,\quad i=0,1,2,\ldots, \\ &\tilde{b}_{1}(x,y)=\bigl(3M_{1}^{2} \times\bigl(\alpha(x)\beta(y)\bigr)^{\theta _{1}}\bigr)^{\frac{\gamma}{1-\gamma}} b^{\frac{1}{1-\gamma}}(x,y), \\ &\tilde{e}_{i}(x,y)=\exp \biggl(- \int_{x_{i}}^{x} \int _{y_{i}}^{y}\tilde{h}_{i}(t,s) \tilde{b}_{1}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr),\quad i=0,1,2, \ldots, \\ &\tilde{h}_{i}(x,y)=g_{1}^{\frac{1}{1-\gamma}}(x,y)+ \biggl(g_{2}(x,y) \int_{x_{i}}^{x} \int_{y_{i}}^{y}g_{3}(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{\frac{1}{1-\gamma}},\quad i=0,1,2,\ldots, \\ &g_{1}(x,y)=\frac{mq}{p}f(x,y),\qquad g_{2}(x,y)=qf(x,y),\qquad g_{3}(x,y)=\frac {n}{p}g(x,y). \end{aligned}$$ - (¡¡):
-
If \(\zeta\in(0,1],\gamma\in(0,\frac{1}{2})\), and \(\xi>\frac {1-2\gamma^{2}}{1-\gamma^{2}}\), we have
$$\begin{aligned} &u(x,y)\leq \biggl[ E_{i}(x,y)+ \biggl(\tilde{a}_{i}(x,y)+ \frac{\tilde {b}_{1}(x,y)}{\tilde{e}_{i}(\alpha(x),\beta(y))} \\ &\phantom{u(x,y)\leq}{}\times \int_{x_{i}}^{\alpha (x)} \int_{y_{i}}^{\beta(y)}\tilde{h}(t,s) \tilde{a}_{i}(t,s)\tilde {e}_{i}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{\frac{\gamma }{1+4\gamma}} \biggr]^{1/p}, \\ &\quad(x,y)\in\varOmega, \end{aligned}$$(40)where \(M_{2},\theta_{2}\) are the same as in Theorem 1, \(E_{i},B_{i},h_{i}\ (i=0,1,2,\ldots)\) are the same as in (2) of Theorem 3, and
$$\begin{aligned} &\tilde{a}_{i}(x,y)=3^{\frac{1+3\gamma}{\gamma}}A_{i}^{\frac {1+4\gamma}{\gamma}}(x,y),\quad i=0,1,2,\ldots, \\ &\tilde{b}_{2}(x,y)=\bigl(3M_{2}^{2} \times\bigl(\alpha(x)\beta(y)\bigr)^{\theta _{2}}\bigr)^{\frac{1+3\gamma}{\gamma}}b^{\frac{1+4\gamma}{\gamma }}(x,y), \\ &\tilde{e}_{i}(x,y)=\exp \biggl(- \int_{x_{i}}^{x} \int _{y_{i}}^{y}\tilde{h}_{i}(t,s) \tilde{b}_{2}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr),\quad i=0,1,2, \ldots, \\ &A_{i}(x,y)=b(x,y) \bigl(M_{2}^{2}\cdot \bigl(\alpha(x)\beta(y)\bigr)^{\theta _{2}}\bigr)^{\frac{1+3\gamma}{1+4\gamma}} \biggl[ \int_{x_{i}}^{\alpha (x)} \int_{y_{i}}^{\beta(y)}B_{i}^{\frac{1+4\gamma}{\gamma }}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{\frac{\gamma}{1+4\gamma}},\\ &\quad i=0,1,2,\ldots, \\ &\tilde{u}_{j}(x,y)= \biggl[ \biggl(\tilde{a}_{j-1}(x,y)+ \frac{\tilde {b}_{2}(x,y)}{\tilde{e}_{j-1}(\alpha(x),\beta(y))} \\ &\phantom{\tilde{u}_{j}(x,y)= }{}\times\int _{x_{j-1}}^{\alpha(x)} \int_{y_{j-1}}^{\beta(y)}\tilde {h}_{j-1}(t,s) \tilde{a}_{j-1}(t,s)\tilde{e}_{j-1}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{\frac{\gamma}{1+4\gamma}} \biggr]^{1/p}, \\ &\quad j=1,2,3,\ldots. \end{aligned}$$
Proof
If \((x,y)\in\varOmega_{11}\), (4) can be restated as
By Lemma 5, we obtain
Substituting (42) into (41), we get
Define a function \(w(x,y)\) as the second items of the right-hand side of (43), i.e.,
By Lemma 5 and (45), we obtain
Substituting inequalities (46) and (47) into (44), we have
that is,
where
Using Hölder’s inequality in (48), we have
Using Lemma 2 to the first items of the right-hand side above, we have
where
(49) has the same form as (14) of Theorem 1. Using the same procedure as that in Theorem 1, considering inequality (45), we can get the desired estimations (39) and (40) for \((x,y)\in\varOmega_{11}\).
If \((x,y)\in\varOmega_{22}\), (4) can be restated as
Let
then we get
Since (50) has the same form as (41), we can conclude that estimates (39) and (40) are valid for \((x,y)\in\varOmega_{22}\). Consequently, by using a similar procedure for \((x,y)\in\varOmega_{ii}\ (i=3,4,5,\ldots)\), we complete the proof. □
4 Applications
In this section, let \(\varOmega,\varOmega_{ij}\ (i,j=1,2,3,\ldots)\) be the as in the previous section.
• Consider the following Volterra-type retarded weakly singular integral equations:
which arise very often in various problems, especially in describing physical processes with after effect.
Example 1
Let \(u(x,y),g(x,y)\), and \(h(x,y)\) be continuous functions on Ω, and let \(\alpha(x),\beta(y)\) be continuous, differentiable, and increasing functions on \(\mathbb{R}^{+}\) with \(\alpha(x)\leq x,\beta(y)\leq y,\alpha (x_{0})=x_{0},\beta(y_{0})=y_{0}\). Let \(p,q,\zeta,\gamma,\delta\) be positive constants with \(p\geq q\). Suppose that \(u(x,y)\) satisfies equation (51). Then we have the estimate for \(u(x,y)\).
- (¡):
-
If \(\zeta\in(0,1],\gamma\in(\frac{1}{2},1)\), and \(\beta (1+\delta)\geq\frac{3}{2}-\gamma\), we have
$$\begin{aligned} & \bigl\vert u(x,y) \bigr\vert \leq \biggl[ \bigl\vert h(x,y) \bigr\vert + \biggl(\tilde{a}_{1}(x,y)+\frac{\tilde {b}_{1}(x,y)}{\tilde{e}_{1}(\alpha(x),\beta(y))} \\ &\phantom{\bigl\vert u(x,y) \bigr\vert \leq}{}\times\int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}\tilde{h}_{1}(t,s) \tilde{a}_{1}(t,s)\tilde {e}_{1}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{1-\gamma} \biggr]^{1/p}, \\ &\quad (x,y)\in\varOmega, \end{aligned}$$(52)where \(M_{1},\theta_{1}\) are the same as in Theorem 1, and
$$\begin{aligned} &\tilde{a}_{1}(x,y)=3^{\frac{\gamma}{1-\gamma}} \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}A_{1}^{\frac{1}{1-\gamma}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t, \\ &\tilde{b}_{1}(x,y)=\bigl(3M_{1}^{2} \times\bigl(\alpha(x)\beta(y)\bigr)^{\theta _{1}}\bigr)^{\frac{\gamma}{1-\gamma}}, \\ &\tilde{h}_{1}(x,y)=A_{2}^{\frac{1}{1-\gamma}}(x,y)+ \biggl(A_{3}(x,y) \int_{x_{0}}^{x} \int_{y_{0}}^{y}A_{4}(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{\frac{1}{1-\gamma}}, \\ &\tilde{e}_{1}(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}\tilde {h}_{1}(t,s) \tilde{b}_{1}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr), \\ &A_{1}(x,y)=(1-q)+q \biggl(\frac{1}{p} \bigl\vert h(x,y) \bigr\vert +\frac{p-1}{p} \biggr) \\ & \phantom{A_{1}(x,y)=}{}+ q \int_{0}^{x} \int_{0}^{y} \bigl\vert g(\tau,\eta) \bigr\vert \biggl(\frac{1}{p} \bigl\vert h(\tau ,\eta) \bigr\vert + \frac{p-1}{p} \biggr)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau, \\ &A_{2}(x,y)=\frac{q}{p},\qquad A_{3}(x,y)=q,\qquad A_{4}(x,y)=\frac{1}{p} \bigl\vert g(x,y) \bigr\vert . \end{aligned}$$ - (¡¡):
-
If \(\zeta\in(0,1],\gamma\in(0,\frac{1}{2})\), and \(\xi>\frac {1-2\gamma^{2}}{1-\gamma^{2}}\), we have
$$\begin{aligned} & \bigl\vert u(x,y) \bigr\vert \leq \biggl[ \bigl\vert h(x,y) \bigr\vert + \biggl(\tilde{a}_{2}(x,y)+\frac{\tilde {b}_{2}(x,y)}{\tilde{e}_{2}(\alpha(x),\beta(y))} \\ &\phantom{\bigl\vert u(x,y) \bigr\vert \leq}{}\times\int_{x_{0}}^{\alpha (x)} \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}\tilde{h}_{2}(t,s) \tilde{a}_{2}(t,s)\tilde {e}_{2}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{\frac{\gamma }{1+4\gamma}} \biggr]^{1/p}, \\ &\quad(x,y)\in\varOmega, \end{aligned}$$(53)where \(M_{2},\theta_{2}\) are the same as in Theorem 1 and \(A_{1},A_{2},A_{3},A_{4}\) are the same as in (¡) of Example 1
$$\begin{aligned} &\tilde{a}_{2}(x,y)=3^{\frac{1+3\gamma}{\gamma}} \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}A_{1}^{\frac{1+4\gamma}{\gamma }}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t, \\ &\tilde{b}_{2}(x,y)=\bigl(3M_{2}^{2} \times\bigl(\alpha(x)\beta(y)\bigr)^{\theta _{2}}\bigr)^{\frac{1+3\gamma}{\gamma}}, \\ &\tilde{h}_{2}(x,y)=A_{2}^{\frac{1+4\gamma}{\gamma}}(x,y)+ \biggl(A_{3}(x,y) \int_{x_{0}}^{x} \int_{y_{0}}^{y}A_{4}(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}} \tau \biggr)^{\frac{1+ 4\gamma}{\gamma}}, \\ &\tilde{e}_{2}(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}\tilde {h}_{2}(t,s) \tilde{b}_{2}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr). \end{aligned}$$
Proof
From (51), we have
Applying Theorem 3 for \((x,y)\in\varOmega_{11}\) (with \(m=n=1,\xi=\zeta (1+\delta),a(x,y)=|h(x,y)|,b(x,y)=1\)) to (54), we get the desired estimations (52) and (53). □
• Consider the following impulsive differential system:
where \((x_{i},y_{i})<(x_{i+1},y_{i+1})\), \(\lim_{i\rightarrow\infty }x_{i}=\infty,\lim_{i\rightarrow\infty}y_{i}=\infty\), \(v_{0}>0\) is a constant, \(H(x,y,v)\) is nonnegative and continuous on Ω.
Example 2
Suppose that \(H(x,y,v)\) satisfies
and \(f(x,y)\in C(\varOmega,\mathbb{R}^{+}),\zeta\in(0,1],\gamma\in (\frac{1}{2},1)\), then we have
where
Proof
The impulsive differential system (55) and (56) is equivalent to the integral equation
By using condition (57), from (58) we have
Let \(u(x,y)=|v(x,y)|\), from (59) we get
By Lemma 5, we have
Substituting (61) to (60), we get
that is,
where \(a(x,y)=v_{0}+\int_{x_{0}}^{x}\int_{y_{0}}^{y}(x^{\zeta }-t^{\zeta})^{\gamma-1}(y^{\zeta}-s^{\zeta})^{\gamma-1}\frac {f(t,s)}{2}\,{\mathrm{d}}s\,{\mathrm{d}}t\). We see that (62) is the particular form of (4), and the functions of (55) satisfy the conditions of Theorem 3. Using the result of Theorem 3, we complete the proof. □
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Luo, Y., Xu, R. Some new weakly singular integral inequalities with discontinuous functions for two variables and their applications. Adv Differ Equ 2019, 387 (2019). https://doi.org/10.1186/s13662-019-2288-9
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DOI: https://doi.org/10.1186/s13662-019-2288-9