Some new weakly singular integral inequalities with discontinuous functions for two variables and their applications

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.

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:

$$\begin{aligned} u(t)\leq a+b \int_{0}^{t}(t-s)^{\beta-1}u(s)\,{ \mathrm{d}}s. \end{aligned}$$

Sano and Kunimatsu [9] gave a sufficient condition for stabilization of semilinear parabolic distributed systems by using a modification of Henry-type inequalities

$$\begin{aligned} 0\leq u(t)\leq c_{1}+c_{2}t^{\alpha-1}+c_{3} \int_{0}^{t}u(s)\,\mathrm{d}s+c_{4} \int_{0}^{t}(t-s)^{\beta-1}u(s)\,{ \mathrm{d}}s. \end{aligned}$$

(1)

Ye et al. [11] provided a generalization of inequality (1)

$$\begin{aligned} u(t)\leq a(t)+b(t) \int_{0}^{t}(t-s)^{\beta-1}u(s)\,{ \mathrm{d}}s, \end{aligned}$$

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

$$\begin{aligned} u^{p}(t)\leq a(t)+b(t) \int_{0}^{t}\bigl(t^{\alpha}-s^{\alpha} \bigr)^{\beta-1}s ^{\gamma-1}f(s)u^{q}(s)\,{\mathrm{d}}s,\quad t \in\mathbb{R}^{+}. \end{aligned}$$

The results were further generalized by Cheung et al. [15] to the following form:

$$\begin{aligned} &u^{p}(x,y)\leq a(x,y)+b(x,y) \int_{0}^{x} \int_{0}^{y}\bigl(x^{\alpha}-s ^{\alpha}\bigr)^{\beta-1}s^{\gamma-1}\bigl(y^{\alpha}-t^{\alpha} \bigr)^{\beta-1}t ^{\gamma-1}f(s,t)u^{q}(s,t)\,{ \mathrm{d}}t\,{\mathrm{d}}s, \\ &\quad (x,y)\in\mathbb{R}^{+}\times\mathbb{R}^{+}. \end{aligned}$$

In 2017, Xu [18] studied the following new generalization of weakly singular integral inequalities in two variables:

$$\begin{aligned} u^{p}(x,y)\leq{}& a(x,y)+b(x,y) \int_{0}^{x} \int_{0}^{y}\bigl(x^{\alpha}-s ^{\alpha}\bigr)^{\beta-1}s^{\gamma-1}\bigl(y^{\alpha}-t^{\alpha} \bigr)^{\beta-1}t ^{\gamma-1} \\ &{} \times\bigl[f(s,t)u^{q}(s,t)+h(s,t)u^{r}\bigl( \sigma(s),\sigma(t)\bigr)\bigr]\,{\mathrm{d}}t \,{\mathrm{d}}s, \end{aligned}$$

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:

$$\begin{aligned} u(t)\leq a(t)+ \int_{t_{0}}^{t}f(s)u\bigl(\tau(s)\bigr)\,{ \mathrm{d}}s+ \sum_{t_{0}< t_{i}< t}\beta_{i}u^{m}(t_{i}-0),\quad t\geq t_{0}. \end{aligned}$$

In 2009, Gllo et al. [20] studied the impulsive integral inequality

$$\begin{aligned} u(t)\leq a(t)+g(t) \int_{t_{0}}^{t}q(s)u^{n}\bigl(\tau(s) \bigr)\,{\mathrm{d}}s+p(t) \sum_{t_{0}< t_{i}< t} \beta_{i}u^{m}(t_{i}-0), \quad t\geq t_{0}. \end{aligned}$$

In 2015, Mi et al. [21] studied the integral inequality of complex functions with unknown function

$$\begin{aligned} u(t)\leq a(t)+ \int_{t_{0}}^{t}f(t,s) \int_{t_{0}}^{s}g(s,\tau)w\bigl(u( \tau)\bigr)\,{ \mathrm{d}}\tau\,{\mathrm{d}}s+q(t)\sum_{t_{0}< t_{i}< t}\beta _{i}u^{m}(t_{i}-0), \quad t\geq t_{0}. \end{aligned}$$

Very recently, Li et al. [22] studied the following weakly singular retarded integral inequality for discontinuous function:

$$\begin{aligned} u^{p}(t)\leq{}& a(t)+b(t) \int_{t_{0}}^{\alpha(t)}\bigl(\alpha^{\beta}(t)-s ^{\beta}\bigr)^{\gamma-1}s^{\xi-1}f(s) \biggl[u^{m}(s)+ \int_{t_{0}}^{s}g( \tau)u^{n}(\tau) \,{\mathrm{d}}\tau \biggr]^{q}\,{\mathrm{d}}s \\ &{}+\sum_{t_{0}< t_{i}< t}\beta_{i}u^{p}(t_{i}-0). \end{aligned}$$

In this paper, we establish some new weakly singular retarded integral inequalities with discontinuous functions in two variables

$$\begin{aligned} &u(x,y)\leq a(x,y) \\ &\phantom{u(x,y)\leq}{}+ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}\bigl( \alpha^{\zeta}(x)-t^{\zeta} \bigr)^{(\gamma-1)}t^{(\xi-1)}\bigl(\beta^{ \zeta}(y)-s^{\zeta} \bigr)^{(\gamma-1)}s^{(\xi-1)}f_{1}(t,s)u(t,s)\,{ \mathrm{d}}s \,{\mathrm{d}}t \\ &\phantom{u(x,y)\leq}{}+ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{ \zeta} \bigr)^{(\gamma-1)}t^{(\xi-1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{( \gamma-1)}s^{(\xi-1)} \\ &\phantom{u(x,y)\leq}{}\times f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)u( \tau, \eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{\mathrm{d}}t, \end{aligned}$$

(2)

$$\begin{aligned} &u(x,y)\leq a(x,y)+ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(x ^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f(t,s)u(t,s) \\ &\phantom{u(x,y)\leq }{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau, \eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p} \,{\mathrm{d}}s\,{ \mathrm{d}}t \\ &\phantom{u(x,y)\leq }{}+\sum_{(x_{0},y_{0})< (x_{i},y_{i})< (x,y)} \zeta_{i}u(x_{i}-0,y_{i}-0), \end{aligned}$$

(3)

$$\begin{aligned} &u^{p}(x,y) \leq a(x,y)+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}} ^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl( \beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}f(t,s) \\ &\phantom{u^{p}(x,y) \leq}{}\times \biggl[u^{m}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau, \eta)u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr] ^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &\phantom{u^{p}(x,y) \leq}{}+\sum_{(x_{0},y_{0})< (x_{i},y_{i})< (x,y)} \zeta_{i}u^{p}(x_{i}-0,y_{i}-0). \end{aligned}$$

(4)

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

$$\begin{aligned} (a_{1}+a_{2}+\cdots+a_{n})^{m} \leq n^{m-1}\bigl(a_{1}^{m}+a_{2}^{m}+ \cdots+a_{n}^{m}\bigr). \end{aligned}$$

Lemma 2

([13])

Let $\beta,\gamma,\xi$, and p be positive constants. Then

$$\begin{aligned} \int_{0}^{t}\bigl(t^{\beta}-s^{\beta} \bigr)^{p(\gamma-1)}s^{p(\xi-1)}\,\mathrm{d}s=\frac {t^{\theta}}{\beta}B \biggl[\frac{p(\xi-1)+1}{\beta},p(\gamma -1)+1 \biggr],\quad t\in\mathbb{R}^{+}, \end{aligned}$$

(5)

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,

$$\begin{aligned} \int_{\alpha(t_{0})}^{\alpha(t)}\bigl(\alpha^{\beta}(t)-s^{\beta } \bigr)^{p(\gamma-1)}s^{p(\xi-1)}\,\mathrm{d}s\leq\frac{\alpha^{\theta}(t)}{\beta }B \biggl[\frac{p(\xi-1)+1}{\beta},p(\gamma-1)+1 \biggr],\quad t\in\mathbb{R}^{+}, \end{aligned}$$

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

$$\begin{aligned} u(x,y)\leq a(x,y)+b(x,y) \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \end{aligned}$$

(6)

then

$$\begin{aligned} u(x,y)\leq a(x,y)+\frac{b(x,y)}{e(\alpha(x),\beta(y))} \int _{0}^{\alpha(x)} \int_{0}^{\beta(y)}h(t,s)a(t,s)e(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t, \end{aligned}$$

(7)

where

$$\begin{aligned} e(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}h(t,s)b(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr). \end{aligned}$$

Proof

Define a function $v(x,y)$ on $\mathbb{R}^{+}$ by

$$\begin{aligned} v(x,y)=e\bigl(\alpha(x),\beta(y)\bigr) \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \end{aligned}$$

(8)

we have $v(x,0)=0,v(0,y)=0$. Differentiating $v(x,y)$ with respect to $x,y$, we have

$$\begin{aligned} v_{xy}={}&e\bigl(\alpha(x),\beta(y)\bigr)\alpha'(x) \beta'(y)h\bigl(\alpha(x),\beta (y)\bigr)u\bigl(\alpha(x),\beta(y) \bigr) \\ &{}-e\bigl(\alpha(x),\beta(y)\bigr)\alpha'(x)\beta'(y)h \bigl(\alpha(x),\beta (y)\bigr)b\bigl(\alpha(x),\beta(y)\bigr) \\ &{}\times\int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t. \end{aligned}$$

Using (6) and $\alpha(x)\leq x,\beta(y)\leq y$, we have

$$\begin{aligned} v_{xy}\leq{}&e\bigl(\alpha(x),\beta(y)\bigr)\alpha'(x) \beta'(y)h\bigl(\alpha (x),\beta(y)\bigr) \\ &{}\times\biggl[a\bigl(\alpha(x), \beta(y)\bigr)+ b\bigl(\alpha(x),\beta(y)\bigr) \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr] \\ &{}-e\bigl(\alpha(x),\beta(y)\bigr)\alpha'(x)\beta'(y)h \bigl(\alpha(x),\beta (y)\bigr)b\bigl(\alpha(x),\beta(y)\bigr) \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \\ ={}&e\bigl(\alpha(x),\beta(y)\bigr)\alpha'(x)\beta'(y)h \bigl(\alpha(x),\beta (y)\bigr)a\bigl(\alpha(x),\beta(y)\bigr). \end{aligned}$$

(9)

Integrating both sides of inequality (9), because $v(x,0)=v(0,y)=0$, we have

$$\begin{aligned} v(x,y)&\leq \int_{0}^{x} \int_{0}^{y}e\bigl(\alpha(x),\beta(y)\bigr) \alpha '(x)\beta'(y)h\bigl(\alpha(x),\beta(y)\bigr)a \bigl(\alpha(x),\beta(y)\bigr)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &= \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}e(t,s)h(t,s)a(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t. \end{aligned}$$

(10)

From (8) and (10), we have

$$\begin{aligned} \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}h(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t\leq\frac{1}{e(\alpha(x),\beta(y))} \int _{0}^{\alpha(x)} \int_{0}^{\beta(y)}h(t,s)a(t,s)e(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t. \end{aligned}$$

(11)

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

$$\begin{aligned} a^{\frac{q}{p}}\leq\frac{q}{p}K^{\frac{q-p}{p}}a+ \frac{p-q}{p}K^{ \frac{q}{p}},\quad K>0. \end{aligned}$$

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

$$\begin{aligned} p_{1}=\frac{1}{\gamma},\qquad q_{1}= \frac{1}{1-\gamma}. \end{aligned}$$

If $\zeta\in(0,1],\gamma\in(0,\frac{1}{2}]$, and $\xi>\frac {1-2\gamma^{2}}{1-\gamma^{2}}$, let

$$\begin{aligned} p_{2}=\frac{1+4\gamma}{1+3\gamma},\qquad q_{2}= \frac{1+4\gamma}{\gamma}, \end{aligned}$$

then

$$\begin{aligned} \frac{1}{p_{i}}+\frac{1}{q_{i}}=1,\quad i=1,2. \end{aligned}$$

Using Hölder’s inequality in (2), we have

$$\begin{aligned} u(x,y)\leq{}& a(x,y)\\ &{}+ \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/p_{i}} \\ &{}\times \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}f_{1}^{q_{i}}(t,s)u^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}} \\ &{}+ \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}\bigl(\alpha^{\zeta }(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi-1)}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi-1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/p_{i}} \\ &{}\times \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)} \biggl(f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)u( \tau,\eta )\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}}. \end{aligned}$$

Set $z(x,y)$ as the right-hand side of the above inequality, and

$$\begin{aligned} A(x,y)= \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi-1)}\bigl(\beta ^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi-1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/p_{i}}, \end{aligned}$$

that is,

$$\begin{aligned} z(x,y)={}&a(x,y)+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}f_{1}^{q_{i}}(t,s)u^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}} \\ &{}+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)} \biggl(f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)u( \tau,\eta )\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}}. \end{aligned}$$

Then $z(x,y)$ is a nondecreasing function, and $u(x,y)\leq z(x,y)$, we have

$$\begin{aligned} z(x,y)\leq{}& a(x,y)+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}f_{1}^{q_{i}}(t,s)z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}} \\ &{}+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)} \biggl(f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)z( \tau,\eta )\,{\mathrm{d}} \eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}} \\ \leq{}& a(x,y)+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)}f_{1}^{q_{i}}(t,s)z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}} \\ &{}+A(x,y) \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)} \biggl(f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)\,{ \mathrm{d}} \eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}}. \end{aligned}$$

(14)

Using the discrete Jensen inequality in Lemma 1 with $n=3,m=q_{i}$, we get

$$\begin{aligned} &z^{q_{i}}(x,y) \\ &\quad\leq 3^{q_{i}-1}a^{q_{i}}(x,y)+3^{q_{i}-1}A^{q_{i}}(x,y) \int_{0}^{\alpha (x)} \int_{0}^{\beta(y)}f_{1}^{q_{i}}(t,s)z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \\ &\qquad{}+3^{q_{i}-1}A^{q_{i}}(x,y) \int_{0}^{\alpha(x)} \int_{0}^{\beta (y)} \biggl(f_{2}(t,s) \int_{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta )\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

(15)

Using Lemma 2, we obtain

$$\begin{aligned} A(x,y)={}& \biggl[ \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi-1)} \bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/p_{i}} \\ \leq{}&\bigl(M_{i}^{2}\times\bigl(\alpha(x)\beta(y) \bigr)^{\theta _{i}}\bigr)^{1/p_{i}}, \end{aligned}$$

(16)

for $(x,y)\in\varOmega$, where

$$\begin{aligned} &M_{i}=\frac{1}{\zeta}B \biggl[\frac{p_{i}(\xi-1)+1}{\zeta },p_{i}( \gamma-1)+1 \biggr], \\ &\theta_{i}=p_{i}\bigl[\zeta(\gamma-1)+\xi-1\bigr]+1 \geq0, \quad i=1,2. \end{aligned}$$

From (15) and (16), we get

$$\begin{aligned} z^{q_{i}}(x,y)\leq{}& 3^{q_{i}-1}a^{q_{i}}(x,y)+3^{q_{i}-1} \bigl[M_{i}^{2}\times\bigl(\alpha(x)\beta (y) \bigr)^{\theta_{i}}\bigr]^{q_{i}/p_{i}} \\ &{}\times \int_{0}^{\alpha(x)} \int_{0}^{\beta(y)} \biggl[f_{1}^{q_{i}}(t,s)+ \biggl(f_{2}(t,s) \int_{0}^{t} \int _{0}^{s}f_{3}(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}} \biggr]z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

(17)

Set

$$\begin{aligned} &\tilde{a}_{i}(x,y)=3^{q_{i}-1}a^{q_{i}}(x,y), \\ &\tilde{b}_{i}(x,y)=3^{q_{i}-1}\bigl[M_{i}^{2} \times\bigl(\alpha(x)\beta (y)\bigr)^{\theta_{i}}\bigr]^{q_{i}/p_{i}}, \\ &\tilde{h}_{i}(t,s)=f_{1}^{q_{i}}(t,s)+ \biggl(f_{2}(t,s) \int _{0}^{t} \int_{0}^{s}f_{3}(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}, \\ &\tilde{e}_{i}(x,y)=\exp \biggl(- \int_{0}^{x} \int_{0}^{y}\tilde {h}_{i}(t,s) \tilde{b}_{i}(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr),\quad i=1,2, \end{aligned}$$

we have

$$\begin{aligned} z^{q_{i}}(x,y)\leq\tilde{a}_{i}(x,y)+\tilde{b}_{i}(x,y) \int _{0}^{\alpha(x)} \int_{0}^{\beta(y)} \tilde{h}_{i}(t,s)z^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t,\quad i=1,2, (x,y)\in\varOmega. \end{aligned}$$

(18)

Applying Lemma 4 to (18), we obtain

$$\begin{aligned} &u^{q_{i}}(x,y)\leq z^{q_{i}}(x,y)\leq\tilde{a}_{i}(x,y)+ \frac{\tilde {b}_{i}(x,y)}{\tilde{e}_{i}(\alpha(x),\beta(y))} \int_{0}^{\alpha (x)} \int_{0}^{\beta(y)}\tilde{h}_{i}(t,s) \tilde{a}_{i}(t,s)\tilde{e}_{i}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \\ &\quad i=1,2, (x,y)\in\varOmega. \end{aligned}$$

(19)

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

$$\begin{aligned} u(x,y)\leq\tilde{a}_{i}(x,y)+\frac{1}{\tilde{e}_{i}(\alpha(x),\beta (y))} \int_{x_{i}}^{\alpha(x)} \int_{y_{i}}^{\beta(y)}\tilde {h}(t,s) \tilde{a}_{i}(t,s)\tilde{e}_{i}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \quad (x,y)\in\varOmega, \end{aligned}$$

(20)

where

$$\begin{aligned} &\tilde{a}_{i}(x,y)=A_{i}^{\frac{1}{1-\gamma}}(x,y),\quad i=0,1,2,\ldots, \\ &A_{i}(x,y)=a(x,y)+\sum_{j=1}^{i} \int_{x_{j-1}}^{\alpha(x_{j})} \int _{y_{j-1}}^{\beta(y_{j})}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta }-s^{\zeta} \bigr)^{\gamma-1}f(t,s)\tilde{u}_{j}(t,s) \\ &\phantom{A_{i}(x,y)=}{} \times \biggl[\tilde{u}_{j}^{2}(t,s)+ \int_{x_{j-1}}^{t} \int _{y_{j-1}}^{s}g(\tau,\eta)\tilde{u}_{j}( \tau,\eta)\,{\mathrm{d}}\eta \,{\mathrm{d}}\tau \biggr]^{p}\,{ \mathrm{d}}s\,{\mathrm{d}}t+\sum_{j=1}^{i} \zeta_{j}\tilde{u}_{j}(x_{j}-0,y_{j}-0), \\ &\quad i=0,1,2,\ldots , \\ &\tilde{u}_{j}(x,y)= \tilde{a}_{j-1}(x,y)+ \frac{1}{\tilde {e}_{j-1}(\alpha(x),\beta(y))} \int_{x_{j-1}}^{\alpha(x)} \int _{y_{j-1}}^{\beta(y)}\tilde{h}(t,s) \tilde{a}_{j-1}(t,s)\tilde {e}_{j-1}(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t,\\ &\quad j=1,2,3,\ldots, \\ &\tilde{h}(t,s)=\bigl(x^{\zeta}-t^{\zeta}\bigr)^{\gamma-1} \bigl(y^{\zeta}-s^{\zeta }\bigr)^{\gamma-1}f(t,s)\varPhi\bigl( \alpha^{-1}(t),\beta^{-1}(s)\bigr), \\ &\tilde{e}_{i}(x,y)=\exp \biggl(- \int_{x_{i}}^{x} \int _{y_{i}}^{y}\tilde{h}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr),\quad i=0,1,2,\ldots. \end{aligned}$$

Proof

Firstly, we consider the case $(x,y)\in\varOmega_{11}$. Denoting

$$\begin{aligned} v(x,y)={}&a(x,y)+ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma -1}f(t,s)u(t,s) \\ &{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t, \end{aligned}$$

(21)

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

$$\begin{aligned} v_{x}(x,y)={}& a_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta }- \alpha^{\zeta}(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma -1}f\bigl(\alpha(x),s\bigr)u\bigl(\alpha(x),s\bigr) \\ &{}\times \biggl[u^{2}\bigl(\alpha(x),s\bigr)+ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{s}g(\tau,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s \\ \leq{}& a_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta}-\alpha ^{\zeta}(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f\bigl(\alpha (x),s\bigr)v\bigl(\alpha(x),s\bigr) \\ &{}\times \biggl[v^{2}\bigl(\alpha(x),s\bigr)+ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{s}g(\tau,\eta)v(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s. \end{aligned}$$

(22)

$$\begin{aligned} v_{xy}(x,y)\leq{}& a_{xy}(x,y)+\alpha'(x) \beta'(y) \bigl(x^{\zeta}-\alpha ^{\zeta}(x) \bigr)^{\gamma-1}\bigl(y^{\zeta}-\beta^{\zeta}(y) \bigr)^{\gamma -1} \\ &{}\times f\bigl(\alpha(x),\beta(y)\bigr)v\bigl(\alpha(x),\beta(y) \bigr) \\ &{}\times \biggl[v^{2}\bigl(\alpha(x),\beta(y)\bigr)+ \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}g(t,s)v(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr]^{p} . \end{aligned}$$

(23)

Set

$$\begin{aligned} \begin{aligned} &F(x,y)=v^{2}\bigl(\alpha(x),\beta(y)\bigr)+ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}g(t,s)v(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \\ &G(x,y)=v^{2}(x,y)+ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}g(t,s)v(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \end{aligned} \end{aligned}$$

(24)

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

$$\begin{aligned} G_{x}(x,y)={}&2v(x,y)v_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta (y)}g\bigl(\alpha(x),s\bigr)v\bigl( \alpha(x),s\bigr)\,{\mathrm{d}}s \\ \leq{}&2G(x,y)\left [a_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta (y)}\bigl(x^{\zeta}- \alpha^{\zeta}(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta } \bigr)^{\gamma-1}f\bigl(\alpha(x),s\bigr)v\bigl(\alpha(x),s\bigr)\right . \\ &{}\times \left . \biggl[v^{2}\bigl(\alpha(x),s\bigr)+ \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{s}g(\tau,\eta)v(\tau,\eta)\,{ \mathrm{d}}\eta \,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\right ] \\ &{}+\alpha'(x) \int _{y_{0}}^{\beta(y)}g\bigl(\alpha(x),s\bigr)v\bigl( \alpha(x),s\bigr)\,{\mathrm{d}}s \\ \leq{}&2G(x,y) \biggl[a_{x}(x,y)+G^{p+1}(x,y) \alpha'(x) \\ &{}\times \int _{y_{0}}^{\beta(y)}\bigl(x^{\zeta}- \alpha^{\zeta}(x)\bigr)^{\gamma -1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f\bigl(\alpha(x),s\bigr)\,{\mathrm{d}}s \biggr] \\ &{}+\alpha'(x)G(x,y) \int_{y_{0}}^{\beta(y)}g\bigl(\alpha(x),s\bigr)\,{ \mathrm{d}}s \\ ={}& \biggl[2\alpha'(x) \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta}-\alpha ^{\zeta}(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f\bigl(\alpha (x),s\bigr)\,{\mathrm{d}}s \biggr]G^{p+2}(x,y) \\ &{}+ \biggl[2a_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta(y)}g\bigl(\alpha (x),s\bigr)\,{ \mathrm{d}}s \biggr]G(x,y). \end{aligned}$$

(25)

Set

$$\begin{aligned} &A(x,y)=2a_{x}(x,y)+\alpha'(x) \int_{y_{0}}^{\beta(y)}g\bigl(\alpha (x),s\bigr)\,{ \mathrm{d}}s, \\ &B(x,y)=2\alpha'(x) \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta}- \alpha^{\zeta }(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f\bigl(\alpha (x),s\bigr)\,{\mathrm{d}}s, \end{aligned}$$

then

$$\begin{aligned} G_{x}(x,y)\leq B(x,y)G^{p+2}(x,y)+A(x,y)G(x,y). \end{aligned}$$

(26)

From (26), we have

$$\begin{aligned} G^{-(p+2)}(x,y)G_{x}(x,y)\leq B(x,y)+A(x,y)G^{-(p+1)}(x,y). \end{aligned}$$

(27)

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

$$\begin{aligned} \eta_{x}(x,y)+(p+1)A(x,y)\eta(x,y)\geq(p+1)B(x,y). \end{aligned}$$

(28)

Multiplying by $\exp ((p+1)\int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t )$ on both sides of (28), we have

$$\begin{aligned} &\frac{\partial}{\partial x} \biggl[\eta(x,y)\times\exp \biggl((p+1) \int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t \biggr) \biggr] \\ &\quad \geq -(p+1)B(x,y)\times\exp \biggl((p+1) \int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t \biggr). \end{aligned}$$

(29)

Integrating both sides of (29) from $x_{0}$ to x, we get

$$\begin{aligned} &\eta(x,y)\times\exp \biggl((p+1) \int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t \biggr)- \eta(x_{0},y)\\ &\quad\geq \int_{x_{0}}^{x}-(p+1)B(t,y)\times \exp \biggl((p+1) \int_{x_{0}}^{t}A(\tau,y)\,{\mathrm{d}}\tau\,{ \mathrm{d}}t \biggr), \end{aligned}$$

set

$$\begin{aligned} \Delta(x,y)=\exp \biggl((p+1) \int_{x_{0}}^{x}A(t,y)\,{\mathrm{d}}t \biggr), \end{aligned}$$

then

$$\begin{aligned} \eta(x,y)\geq\frac{\eta(x_{0},y)-\int_{x_{0}}^{x}(p+1)B(t,y)\Delta (t,y)\,{\mathrm{d}}t}{\Delta(x,y)} . \end{aligned}$$

(30)

Since $\eta(x_{0},y)=G^{-(p+1)}(x_{0},y)=a^{-2(p+1)}(x_{0},y)$, from (30) we have

$$\begin{aligned} \eta(x,y)\geq\frac{1-(p+1)a^{2(p+1)}(x_{0},y)\int _{x_{0}}^{x}B(t,y)\Delta(t,y)\,{\mathrm{d}}t}{a^{2(p+1)}(x_{0},y)\Delta (x,y)} . \end{aligned}$$

(31)

By the relation $\eta(x,y)=G^{-(p+1)}(x,y)$, from (31) we get

$$\begin{aligned} G^{p}(x,y)\leq \biggl[\frac{a^{2(p+1)}(x_{0},y)\Delta (x,y)}{1-(p+1)a^{2(p+1)}(x_{0},y)\int_{x_{0}}^{x}B(t,y)\Delta (t,y)\,{\mathrm{d}}t} \biggr]^{\frac{p}{p+1}} , \end{aligned}$$

(32)

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

$$\begin{aligned} \varPhi(x,y)= \biggl[\frac{a^{2(p+1)}(x_{0},y)\Delta (x,y)}{1-(p+1)a^{2(p+1)}(x_{0},y)\int_{x_{0}}^{x}B(t,y)\Delta (t,y)\,{\mathrm{d}}t} \biggr]^{\frac{p}{p+1}} , \end{aligned}$$

(33)

from (23), (24), (32), and (33), we obtain

$$\begin{aligned} v_{xy}(x,y)\leq{}&a_{xy}(x,y)+\alpha'(x) \beta'(y) \bigl(x^{\zeta}-\alpha ^{\zeta}(x) \bigr)^{\gamma-1}\bigl(y^{\zeta}-\beta^{\zeta}(y) \bigr)^{\gamma-1} \\ &{}\times f\bigl(\alpha(x),\beta(y)\bigr)v\bigl(\alpha(x),\beta(y)\bigr) \varPhi(x,y). \end{aligned}$$

(34)

Integrating both sides of (34), we get

$$\begin{aligned} v(x,y)\leq{}&a(x,y)+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\alpha'(t)\beta '(s) \bigl(x^{\zeta}-\alpha^{\zeta}(t) \bigr)^{\gamma-1}\bigl(y^{\zeta}-\beta ^{\zeta}(s) \bigr)^{\gamma-1} \\ &{}\times f\bigl(\alpha(t),\beta(s)\bigr)v\bigl(\alpha(t),\beta (s) \bigr)\varPhi(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ ={}&a(x,y)+ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta }-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f(t,s)\varPhi \bigl(\alpha^{-1}(t), \beta^{-1}(s)\bigr)v(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ ={}&a(x,y)+ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\tilde {h}(t,s)v(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t, \end{aligned}$$

(35)

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:

$$\begin{aligned} u(x,y)\leq\tilde{a}_{0}(x,y)+\frac{1}{\tilde{e}_{0}(\alpha(x),\beta (y))} \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\tilde {h}(t,s) \tilde{a}_{0}(t,s)\tilde{e}_{0}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t, \quad(x,y)\in\varOmega_{11}. \end{aligned}$$

Set

$$\begin{aligned} \tilde{u}_{1}(x,y)=\tilde{a}_{0}(x,y)+ \frac{1}{\tilde{e}_{0}(\alpha (x),\beta(y))} \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\tilde {h}(t,s) \tilde{a}_{0}(t,s)\tilde{e}_{0}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t,\quad (x,y)\in\varOmega_{11}, \end{aligned}$$

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

$$\begin{aligned} u(x,y)\leq{}&a(x,y)+ \int_{x_{0}}^{\alpha(x_{1})} \int_{y_{0}}^{\beta (y_{1})}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta } \bigr)^{\gamma-1}f(t,s)u(t,s) \\ &{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t \\ &{}+ \int_{x_{1}}^{\alpha(x)} \int _{y_{1}}^{\beta(y)}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta }-s^{\zeta} \bigr)^{\gamma-1}f(t,s)u(t,s) \\ &{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t+\zeta_{1}u(x_{1}-0,y_{1}-0) \\ \leq{} &a(x,y)+ \int_{x_{0}}^{\alpha(x_{1})} \int_{y_{0}}^{\beta (y_{1})}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta } \bigr)^{\gamma-1}f(t,s)\tilde{u}_{1}(t,s) \\ &{}\times \biggl[\tilde{u}_{1}^{2}(t,s)+ \int_{x_{0}}^{t} \int _{y_{0}}^{s}g(\tau,\eta)\tilde{u}_{1}( \tau,\eta)\,{\mathrm{d}}\eta \,{\mathrm{d}}\tau \biggr]^{p}\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &{}+ \int _{x_{1}}^{\alpha(x)} \int_{y_{1}}^{\beta(y)}\bigl(x^{\zeta}-t^{\zeta } \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f(t,s)u(t,s) \\ &{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t+\zeta_{1}\tilde {u}_{1}(x_{1}-0,y_{1}-0). \end{aligned}$$

(36)

Setting

$$\begin{aligned} \begin{aligned} &A_{1}(x,y)= a(x,y)+ \int_{x_{0}}^{\alpha(x_{1})} \int_{y_{0}}^{\beta (y_{1})}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta } \bigr)^{\gamma-1}f(t,s)\tilde{u}_{1}(t,s) \\ &\phantom{A_{1}(x,y)=}{}\times \biggl[\tilde{u}_{1}^{2}(t,s)+ \int_{x_{0}}^{t} \int _{y_{0}}^{s}g(\tau,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t+\zeta_{1}\tilde {u}_{1}(x_{1}-0,y_{1}-0),\\ &\quad (x,y)\in\varOmega_{22}, \\ &\varPsi(x,y)=A_{1}(x,y)+ \int_{x_{1}}^{\alpha(x)} \int_{y_{1}}^{\beta (y)}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma -1}f(t,s)u(t,s) \\ &\phantom{\varPsi(x,y)=}{}\times \biggl[u^{2}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{p}\,{\mathrm{d}}s\,{ \mathrm{d}}t, \quad (x,y)\in\varOmega_{22}, \end{aligned} \end{aligned}$$

(37)

then $\varPsi(x,y)$ is a nonnegative and nondecreasing function, and

$$\begin{aligned} u(x,y)\leq\varPsi(x,y),\qquad u(x_{1},y_{1})\leq \varPsi(x_{1},y_{1})=A_{1}(x_{1},y_{1}). \end{aligned}$$

Differentiating both sides of (37), we obtain

$$\begin{aligned} \varPsi_{xy}(x,y)={}&\bigl(A_{1}(x,y)\bigr)_{xy}+ \alpha'(x)\beta'(y) \bigl(x^{\zeta }- \alpha^{\zeta}(x)\bigr)^{\gamma-1}\bigl(y^{\zeta}- \beta^{\zeta }(y)\bigr)^{\gamma-1} \\ &{}\times f\bigl(\alpha(x),\beta(y)\bigr)u \bigl(\alpha(x),\beta(y)\bigr) \\ &{}\times \biggl[u^{2}\bigl(\alpha(x),\beta(y)\bigr)+ \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}g(t,s)u(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr]^{p} \\ \leq{}&\bigl(A_{1}(x,y)\bigr)_{xy}+\alpha'(x) \beta'(y) \bigl(x^{\zeta}-\alpha^{\zeta }(x) \bigr)^{\gamma-1}\bigl(y^{\zeta}-\beta^{\zeta}(y) \bigr)^{\gamma-1} \\ &{}\times f\bigl(\alpha (x),\beta(y)\bigr)\varPsi\bigl(\alpha(x),\beta(y) \bigr) \\ &{}\times \biggl[\varPsi^{2}\bigl(\alpha(x),\beta(y)\bigr)+ \int_{x_{0}}^{\alpha (x)} \int_{y_{0}}^{\beta(y)}g(t,s)\varPsi(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr]^{p}. \end{aligned}$$

(38)

(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

$$\begin{aligned} u^{p}(x,y)\leq{}& a(x,y)+b(x,y) \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi -1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}f(t,s) \\ &{}\times \biggl[u^{m}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau ,\eta)u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

(41)

By Lemma 5, we obtain

$$\begin{aligned} &\biggl[u^{m}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta )u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q} \\ &\quad \leq q \biggl[u^{m}(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta )u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]+(1-q). \end{aligned}$$

(42)

Substituting (42) into (41), we get

$$\begin{aligned} u^{p}(x,y)\leq{}& a(x,y)+b(x,y) \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi -1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}f(t,s) \\ &{}\times \biggl[q \biggl(u^{m}(t,s)+ \int_{x_{0}}^{t} \int _{y_{0}}^{s}g(\tau,\eta)u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)+(1-q) \biggr]\,{ \mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

(43)

Define a function $w(x,y)$ as the second items of the right-hand side of (43), i.e.,

$$\begin{aligned} w(x,y)={}&b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta ^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}(1-q)f(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s)u^{m}(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta)u^{n}( \tau ,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{ \mathrm{d}}t. \end{aligned}$$

(44)

From (43) and (44), we have

$$\begin{aligned} u^{p}(x,y)\leq a(x,y)+w(x,y)\quad \text{or} \quad u(x,y)\leq\bigl(a(x,y)+w(x,y) \bigr)^{1/p}. \end{aligned}$$

(45)

By Lemma 5 and (45), we obtain

$$\begin{aligned} &u^{m}(x,y)\leq\bigl(a(x,y)+w(x,y)\bigr)^{m/p}\leq \frac {m}{p}\bigl(a(x,y)+w(x,y)\bigr)+\frac{p-m}{p}, \end{aligned}$$

(46)

$$\begin{aligned} &u^{n}(x,y)\leq\bigl(a(x,y)+w(x,y)\bigr)^{n/p}\leq \frac {n}{p}\bigl(a(x,y)+w(x,y)\bigr)+\frac{p-n}{p}. \end{aligned}$$

(47)

Substituting inequalities (46) and (47) into (44), we have

$$\begin{aligned} w(x,y)\leq{}& b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta ^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}(1-q)f(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s) \\ &{}\times \biggl[\frac{m}{p}\bigl(a(t,s)+w(t,s)\bigr)+ \frac{p-m}{p} \biggr]\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta) \biggl[ \frac {n}{p}\bigl(a(\tau,\eta)+w(\tau,\eta)\bigr)+\frac{p-n}{p} \biggr]\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{\mathrm{d}}t \\ ={}&b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\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) \\ &{}\times \biggl[(1-q)+q \biggl(\frac{m}{p}a(x,y)+\frac{p-m}{p} \biggr) \biggr]\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta) \biggl[ \frac {n}{p}a(\tau,\eta)+\frac{p-n}{p} \biggr]\,{\mathrm{d}}\eta \,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}\frac{mq}{p}f(t,s)w(t,s) \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}qf(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}\frac{n}{p}g(\tau,\eta )w(\tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{\mathrm{d}}t \\ = {}&b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(x^{\zeta }-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma -1}s^{\xi-1}B_{0}(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}g_{1}(t,s)w(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}g_{2}(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}g_{3}(\tau,\eta)w( \tau ,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{\mathrm{d}}t, \end{aligned}$$

that is,

$$\begin{aligned} w(x,y)\leq{}& b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(y^{\zeta}-s^{\zeta } \bigr)^{\gamma-1}s^{\xi-1}B_{0}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}g_{1}(t,s)w(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &{}+b(x,y) \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}g_{2}(t,s) \\ &{}\times \int_{x_{0}}^{t} \int_{y_{0}}^{s}g_{3}(\tau,\eta)w( \tau ,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau\,{\mathrm{d}}s\,{ \mathrm{d}}t, \end{aligned}$$

(48)

where

$$\begin{aligned} &B_{0}(x,y)=f(x,y) \biggl[(1-q)+q \biggl(\frac{m}{p}a(x,y)+ \frac {p-m}{p} \biggr) \biggr] \\ &\phantom{B_{0}(x,y)=}{} +qf(x,y) \int_{x_{i}}^{x} \int _{y_{i}}^{y}g(\tau,\eta) \biggl[ \frac{n}{p}a(\tau,\eta)+\frac {p-n}{p} \biggr]\,{\mathrm{d}}\eta \,{\mathrm{d}}\tau,\quad i=0,1,2,\ldots, \\ &g_{1}(x,y)=\frac{mq}{p}f(x,y),\qquad g_{2}(x,y)=qf(x,y),\qquad g_{3}(x,y)=\frac {n}{p}g(x,y). \end{aligned}$$

Using Hölder’s inequality in (48), we have

$$\begin{aligned} &w(x,y)\\ &\quad\leq b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}} \\ &\qquad{}\times \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}B_{0}^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}} \\ &\qquad{}+b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}} \\ &\qquad{}\times \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}g_{1}^{q_{i}}(t,s)w^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}} \\ &\qquad{}+b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}} \\ &\qquad{}\times \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)} \biggl(g_{2}(t,s) \int_{x_{0}}^{t} \int_{y_{0}}^{s}g_{3}(\tau,\eta)w( \tau ,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}}. \end{aligned}$$

Using Lemma 2 to the first items of the right-hand side above, we have

$$\begin{aligned} &w(x,y) \\ &\quad \leq b(x,y) \bigl(M_{i}^{2}\times\bigl(\alpha(x) \beta(y)\bigr)^{\theta _{i}}\bigr)^{{1/p_{i}}} \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}B_{0}^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}} \\ &\qquad{}+b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}} \\ &\qquad{}\times \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}g_{1}^{q_{i}}(t,s)w^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}} \\ &\qquad{}+b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma-1)}t^{p_{i}(\xi -1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma-1)}s^{p_{i}(\xi -1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}} \\ &\qquad{}\times \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)} \biggl(g_{2}(t,s) \int_{x_{0}}^{t} \int_{y_{0}}^{s}g_{3}(\tau,\eta)w( \tau ,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\, \mathrm{d}s\,\mathrm{d}t \biggr]^{{1/q_{i}}} \\ &\quad =A_{0}(x,y)+\varGamma(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}g_{1}^{q_{i}}(t,s)w^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}} \\ &\qquad{}+\varGamma(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta (y)} \biggl(g_{2}(t,s) \int_{x_{0}}^{t} \int_{y_{0}}^{s}g_{3}(\tau,\eta )w( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr)^{q_{i}}\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/q_{i}}}, \end{aligned}$$

(49)

where

$$\begin{aligned} &A_{0}(x,y)=b(x,y) \bigl(M_{i}^{2}\times \bigl(\alpha(x)\beta(y)\bigr)^{\theta _{i}}\bigr)^{\frac{1}{p_{i}}} \biggl[ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}B_{0}^{q_{i}}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1/q_{i}}, \\ &\varGamma(x,y)=b(x,y) \biggl[ \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{p_{i}(\gamma -1)}t^{p_{i}(\xi-1)}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{p_{i}(\gamma -1)}s^{p_{i}(\xi-1)}\,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{{1/p_{i}}}\hspace{-1pt}. \end{aligned}$$

(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

$$\begin{aligned} u^{p}(x,y)\leq{}& a(x,y)+b(x,y) \int_{x_{0}}^{\alpha(x_{1})} \int _{y_{0}}^{\beta(y_{1})}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma -1}t^{\xi-1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi -1}f(t,s) \\ &{}\times \biggl[\tilde{u}_{1}^{m}(t,s)+ \int_{x_{0}}^{t} \int _{y_{0}}^{s}g(\tau,\eta)\tilde{u}_{1}^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t+\zeta _{1}\tilde{u}_{1}^{p}(x_{1}-0,y_{1}-0) \\ &{}+b(x,y) \int_{x_{1}}^{\alpha(x)} \int_{y_{1}}^{\beta(y)}\bigl(\alpha ^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi-1}\bigl(\beta^{\zeta }(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}f(t,s) \\ &{}\times \biggl[u^{m}(t,s) + \int_{x_{1}}^{t} \int_{y_{1}}^{s}g(\tau ,\eta)u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

Let

$$\begin{aligned} E_{1}(x,y)={}& a(x,y)+b(x,y) \int_{x_{0}}^{\alpha(x_{1})} \int _{y_{0}}^{\beta(y_{1})}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma -1}t^{\xi-1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi -1}f(t,s) \\ &{}\times \biggl[\tilde{u}_{1}^{m}(t,s)+ \int_{x_{0}}^{t} \int _{y_{0}}^{s}g(\tau,\eta)\tilde{u}_{1}^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ &{}+\zeta _{1}\tilde{u}_{1}^{p}(x_{1}-0,y_{1}-0),\quad (x,y)\in\varOmega_{22}, \end{aligned}$$

then we get

$$\begin{aligned} u^{p}(x,y)\leq{}& E_{1}(x,y)+b(x,y) \int_{x_{1}}^{\alpha(x)} \int _{y_{1}}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma-1}t^{\xi -1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma-1}s^{\xi-1}f(t,s) \\ &{}\times \biggl[u^{m}(t,s) + \int_{x_{1}}^{t} \int_{y_{1}}^{s}g(\tau ,\eta)u^{n}( \tau,\eta)\,{\mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{ \mathrm{d}}s\,{\mathrm{d}}t,\quad (x,y)\in\varOmega_{22}. \end{aligned}$$

(50)

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:

$$\begin{aligned} &u^{p}(x,y)- \int_{x_{0}}^{\alpha(x)} \int_{y_{0}}^{\beta(y)}\bigl(\alpha ^{\beta}(x)-t^{\beta} \bigr)^{\gamma-1}t^{\beta(1+\delta)-1}\bigl(\beta ^{\beta}(y)-s^{\beta} \bigr)^{\gamma-1}s^{\beta(1+\delta)-1} \\ &\quad{}\times \biggl[u(t,s)+ \int_{x_{0}}^{t} \int_{y_{0}}^{s}g(\tau,\eta )u(\tau,\eta)\,{ \mathrm{d}}\eta\,{\mathrm{d}}\tau \biggr]^{q}\,{\mathrm{d}}s\,{ \mathrm{d}}t=h(x,y),\quad (x,y)\in\varOmega, \end{aligned}$$

(51)

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

$$\begin{aligned} \bigl\vert u(x,y) \bigr\vert ^{p}\leq{}& \bigl\vert h(x,y) \bigr\vert + \int_{x_{0}}^{\alpha(x)} \int _{y_{0}}^{\beta(y)}\bigl(\alpha^{\zeta}(x)-t^{\zeta} \bigr)^{\gamma -1}t^{\zeta(1+\delta)-1}\bigl(\beta^{\zeta}(y)-s^{\zeta} \bigr)^{\gamma -1}s^{\beta(1+\delta)-1} \\ &{}\times \biggl[ \bigl\vert u(t,s) \bigr\vert + \int_{x_{0}}^{t} \int_{y_{0}}^{s} \bigl\vert g(\tau,\eta ) \bigr\vert \bigl\vert u(\tau,\eta) \bigr\vert \,{\mathrm{d}}\eta\,{\mathrm{d}} \tau \biggr]^{q}\,{\mathrm{d}}s\,{\mathrm{d}}t. \end{aligned}$$

(54)

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:

$$\begin{aligned} &\frac{\partial^{2}v(x,y)}{\partial x\,\partial y}=H\bigl(x,y,v(x,y)\bigr),\quad (x,y)\in\varOmega_{ii},x \neq x_{i},y\neq y_{i}, \end{aligned}$$

(55)

$$\begin{aligned} &\Delta v|_{x= x_{i},y=y_{i}}=\beta_{i}v(x_{i}-0,y_{i}-0), \\ &v(x_{0},y_{0})=v_{0}, \end{aligned}$$

(56)

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

$$\begin{aligned} H(x,y,v)\leq\bigl(x^{\zeta}-t^{\zeta}\bigr)^{\gamma-1} \bigl(y^{\zeta}-s^{\zeta }\bigr)^{\gamma-1}f(x,y)\sqrt{ \vert v \vert }, \end{aligned}$$

(57)

and $f(x,y)\in C(\varOmega,\mathbb{R}^{+}),\zeta\in(0,1],\gamma\in (\frac{1}{2},1)$, then we have

$$\begin{aligned} & \bigl\vert v(x,y) \bigr\vert \leq E_{i}(x,y)+ \biggl( \tilde{a}_{i}(x,y)+\frac{\tilde {b}(x,y)}{\tilde{e}_{i}(x,y)} \int_{x_{i}}^{x} \int_{y_{i}}^{y}\tilde {h}(t,s) \tilde{a}_{i}(t,s)\tilde{e}_{i}(t,s)\,{\mathrm{d}}s\,{ \mathrm{d}}t \biggr)^{1-\gamma},\\ &\quad(x,y)\in\varOmega, \end{aligned}$$

where

$$\begin{aligned} &E_{0}(x,y)=a(x,y),\quad (x,y)\in\varOmega_{11}, \\ &E_{i}(x,y)=a(x,y)+\sum_{j=1}^{i} \int_{x_{j-1}}^{x_{j}} \int _{y_{j-1}}^{y_{j}}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta }-s^{\zeta} \bigr)^{\gamma-1}f(t,s) \sqrt{\tilde{u}_{j}(t,s)} \\ &\phantom{E_{i}(x,y)=}{} +\sum_{j=1}^{i} \zeta_{j}u_{j}(x_{j}-0,y_{j}-0),\quad(x,y) \in\varOmega _{ii}, i=1,2,3,\ldots, \\ &\tilde{u}_{j}(x,y)= \biggl(\tilde{a}_{j-1}(x,y)+ \frac{\tilde {b}(x,y)}{\tilde{e}_{j-1}(x,y)} \int_{x_{j-1}}^{x} \int _{y_{j-1}}^{y}\tilde{h}(t,s) \tilde{a}_{j-1}(t,s)\tilde {e}_{j-1}(t,s)\,{\mathrm{d}}s \,{\mathrm{d}}t \biggr)^{1-\gamma}, \\ &\quad j=1,2,3,\ldots, \\ &\tilde{a}_{i}(x,y)=2^{\frac{\gamma}{1-\gamma}}A_{i}^{\frac {1}{1-\gamma}}(x,y),\quad i=0,1,2,\ldots, \\ &A_{i}(x,y)=b(x,y) \bigl(M_{1}^{2} \times(xy)^{\theta_{1}}\bigr)^{\gamma} \biggl[ \int_{x_{i}}^{x} \int_{y_{i}}^{y}B_{i}^{\frac{1}{1-\gamma }}(t,s) \,{\mathrm{d}}s\,{\mathrm{d}}t \biggr]^{1-\gamma},\quad i=0,1,2,\ldots , \\ &B_{i}(x,y)=f(x,y) \biggl(\frac{1}{2}+ \frac{1}{2}E_{i}(x,y) \biggr),\quad i=0,1,2,\ldots, \\ &\tilde{b}(x,y)=\bigl(2M_{1}^{2}\times(xy)^{\theta_{1}} \bigr)^{\frac{\gamma }{1-\gamma}}, \\ &\tilde{e}_{i}(x,y)=\exp \biggl(- \int_{x_{i}}^{x} \int _{y_{i}}^{y}\tilde{h}(t,s)\tilde{b}(t,s)\,{ \mathrm{d}}s\,{\mathrm{d}}t \biggr),\quad i=0,1,2,\ldots, \\ &\tilde{h}(x,y)=g_{1}^{\frac{1}{1-\gamma}}(x,y),\qquad g_{1}(x,y)= \frac {1}{2}f(x,y), \\ &M_{1}=\frac{1}{\zeta}B \biggl[\frac{1}{\zeta}, \frac{2\gamma -1}{\gamma} \biggr], \qquad\theta_{1}=\frac{1}{\gamma}\bigl[ \zeta(\gamma-1)\bigr]+1. \end{aligned}$$

Proof

The impulsive differential system (55) and (56) is equivalent to the integral equation

$$\begin{aligned} v(x,y)=v_{0}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}H\bigl(t,s,v(t,s)\bigr)\,{ \mathrm{d}}s\,{\mathrm{d}}t+\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta _{i}v(x_{i}-0,y_{i}-0). \end{aligned}$$

(58)

By using condition (57), from (58) we have

$$\begin{aligned} \bigl\vert v(x,y) \bigr\vert \leq{}& v_{0}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta }-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f(t,s)\sqrt { \bigl\vert v(t,s) \bigr\vert }\,{ \mathrm{d}}s\,{\mathrm{d}}t \\ & {}+\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta _{i} \bigl\vert v(x_{i}-0,y_{i}-0) \bigr\vert . \end{aligned}$$

(59)

Let $u(x,y)=|v(x,y)|$, from (59) we get

$$\begin{aligned} u(x,y)\leq{}& v_{0}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\beta }-t^{\beta} \bigr)^{\gamma-1}\bigl(y^{\beta}-s^{\beta} \bigr)^{\gamma-1}f(t,s)\sqrt {u(t,s)}\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{} +\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta _{i}u(x_{i}-0,y_{i}-0). \end{aligned}$$

(60)

By Lemma 5, we have

$$\begin{aligned} u^{\frac{1}{2}}(x,y)\leq\frac{1}{2}u(x,y)+\frac{1}{2}. \end{aligned}$$

(61)

Substituting (61) to (60), we get

$$\begin{aligned} u(x,y)\leq{}& v_{0}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta }-t^{\zeta} \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}f(t,s) \biggl(\frac{1}{2}u(t,s)+ \frac{1}{2} \biggr)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{} +\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta_{i}u(x_{i}-0,y_{i}-0) \\ \leq{} &v_{0}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta}-t^{\zeta } \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}\frac {f(t,s)}{2}u(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{}+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta}-t^{\zeta} \bigr)^{\gamma -1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}\frac{f(t,s)}{2}\,{\mathrm{d}}s\,{\mathrm{d}}t+\sum _{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta _{i}u(x_{i}-0,y_{i}-0) \\ \leq{}& a(x,y)+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta}-t^{\zeta } \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}\frac {f(t,s)}{2}u(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ & {}+\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta_{i}u(x_{i}-0,y_{i}-0), \end{aligned}$$

that is,

$$\begin{aligned} u(x,y)\leq{}& a(x,y)+ \int_{x_{0}}^{x} \int_{y_{0}}^{y}\bigl(x^{\zeta}-t^{\zeta } \bigr)^{\gamma-1}\bigl(y^{\zeta}-s^{\zeta} \bigr)^{\gamma-1}\frac {f(t,s)}{2}u(t,s)\,{\mathrm{d}}s\,{\mathrm{d}}t \\ &{} +\sum_{x_{0}< x_{i}< x,y_{0}< y_{i}< y}\zeta_{i}u(x_{i}-0,y_{i}-0), \end{aligned}$$

(62)

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. □