A class of retarded nonlinear integral inequalities and its application in nonlinear differential-integral equation

Abstract

In this paper, we discuss a class of retarded nonlinear integral inequalities and give an upper bound estimation of an unknown function by the integral inequality technique. This estimation can be used as a tool in the study of differential-integral equations with the initial conditions.

MSC:26D10, 26D15, 26D20, 34A12, 34A40.

1 Introduction

Gronwall-Bellman inequalities [1, 2] can be used as important tools in the study of existence, uniqueness, boundedness, stability, and other qualitative properties of solutions of differential equations, integral equations, and integral-differential equations. There can be found a lot of generalizations of Gronwall-Bellman inequalities in various cases from literature (e.g., [3–13]).

Lemma 1 (Abdeldaim and Yakout [4])

We assume that$u(t)$and$f(t)$are nonnegative real-valued continuous functions defined on$I=[0,\mathrm{\infty })$and they satisfy the inequality

$$u^{p+1}(t)\le u_0+{({\int }_0^{t}f(s)u^p(s)ds)}^2+2{\int }_0^{t}f(s)u^p(s)[u^p(s)+{\int }_0^{s}f(\lambda )u^p(\lambda )d\lambda ]ds,$$

(1.1)

for all$t\in I$, where$u_0>0$and$p\in (0,1)$are constants. Then

$$u(t)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}{\int }_0^{t}f(s)D_2(s)ds,\mathrm{\forall }t\in I,$$

(1.2)

where

$$D_2(t)=\beta (t){[u_0^{\frac{1-p}{1+p}}+2\frac{1-p}{1+p}{\int }_0^{t}f(s)exp(-2\frac{1-p}{p}{\int }_0^{s}f(\lambda )d\lambda )ds]}^{\frac{p}{1-p}},$$

(1.3)

and$\beta (t)=exp(2{\int }_0^{t}f(s)ds)$, for all$t\in I$.

In this paper, we discuss a class of retarded nonlinear integral inequalities and give an upper bound estimation of an unknown function by the integral inequality technique.

2 Main result

In this section, we discuss some retarded integral inequalities of Gronwall-Bellman type. Throughout this paper, let $I=[0,\mathrm{\infty })$.

Theorem 1 Suppose$\alpha \in C^1(I,I)$is increasing function with$\alpha (t)\le t$, $\alpha (0)=0$, $\mathrm{\forall }t\in I$. We assume that$u(t)$and$f(t)$are nonnegative real-valued continuous functions defined on I, and they satisfy the inequality

$$\begin{array}{rl}{u}^{p+1}(t)\le & u_0+{({\int }_0^{\alpha (t)}f(s)u^p(s)ds)}^2\\ +2{\int }_0^{\alpha (t)}f(s)u^p(s)[u^p(s)+{\int }_0^{s}f(\lambda )u^p(\lambda )d\lambda ]ds,\mathrm{\forall }t\in I,\end{array}$$

(2.1)

where$u_0>0$and$p\in (0,1)$are constants. Then

$$u(t)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}{\int }_0^{\alpha (t)}f(s){\theta }_1({\alpha }^{-1}(s))ds,\mathrm{\forall }t\in I,$$

(2.2)

where

$${\theta }_1(t)={\beta }_1(t){[u_0^{\frac{1-p}{1+p}}+2\frac{1-p}{1+p}{\int }_0^{\alpha (t)}f(s)exp(-2\frac{1-p}{p}{\int }_0^{s}f(\lambda )d\lambda )ds]}^{\frac{p}{1-p}},$$

(2.3)

and${\beta }_1(t)=exp(2{\int }_0^{\alpha (t)}f(s)ds)$, for all$t\in I$.

Remark 1 If$\alpha (t)=t$, then Theorem 1 reduces Lemma 1.

Proof Let $z_1^{p+1}(t)$ denote the function on the right-hand side of (2.1), which is a positive and nondecreasing function on I with $z_1(0)=u_0^{\frac{1}{p+1}}$. Then (2.1) is equivalent to

$$u(t)\le z_1(t),u(\alpha (t))\le z_1(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.4)

Differentiating $z_1^{p+1}(t)$ with respect to t, using (2.4) we have

$$\begin{array}{rl}(p+1)z_1^p(t)\frac{d{z}_1(t)}{dt}=& 2{\alpha }^{\prime }(t)f(\alpha (t))u^p(\alpha (t)){\int }_0^{\alpha (t)}f(s)u^p(s)ds\\ +2{\alpha }^{\prime }(t)f(\alpha (t))u^p(\alpha (t))[u^p(\alpha (t))+{\int }_0^{\alpha (t)}f(\lambda )u^p(\lambda )d\lambda ]\\ \le & 2{\alpha }^{\prime }(t)f(\alpha (t))z_1^p(t)[z_1^p(t)+2{\int }_0^{\alpha (t)}f(\lambda )z_1^p(\lambda )d\lambda ],\mathrm{\forall }t\in I.\end{array}$$

(2.5)

Since $z_1^p(t)>0$, from (2.5) we have

$$\frac{d{z}_1(t)}{dt}\le \frac{2}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_1(t),\mathrm{\forall }t\in I,$$

(2.6)

where

$$Y_1(t):=z_1^p(t)+2{\int }_0^{\alpha (t)}f(\lambda )z_1^p(\lambda )d\lambda ,\mathrm{\forall }t\in I.$$

(2.7)

Then $Y_1(t)$ is a positive and nondecreasing function on I with $Y_1(0)=u_0^{p/(p+1)}$ and

$$z_1(t)\le Y_1{(t)}^{1/p}.$$

(2.8)

Differentiating $Y_1(t)$ with respect to t, and using (2.6), (2.7) and (2.8), we get

$$\begin{array}{rl}\frac{d{Y}_1(t)}{dt}& \le \frac{2p}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))z_1^{p-1}(t)Y_1(t)+2{\alpha }^{\prime }(t)f(\alpha (t))z_1^p(\alpha (t))\\ \le \frac{2p}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_1^{\frac{2p-1}{p}}(t)+2{\alpha }^{\prime }(t)f(\alpha (t))Y_1(t),\mathrm{\forall }t\in I.\end{array}$$

(2.9)

From (2.9), we have

$$Y_1^{\frac{1-2p}{p}}(t)\frac{d{Y}_1(t)}{dt}-2{\alpha }^{\prime }(t)f(\alpha (t))Y_1^{\frac{1-p}{p}}(t)\le \frac{2p}{p+1}{\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.10)

Let $S_1(t)=Y_1^{\frac{1-p}{p}}(t)$, then $S_1(0)=u_0^{\frac{1-p}{p+1}}$, from (2.10) we obtain

$$\frac{d{S}_1(t)}{dt}-2\frac{1-p}{p}{\alpha }^{\prime }(t)f(\alpha (t))S_1(t)\le 2\frac{1-p}{p+1}{\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.11)

Consider the ordinary differential equation

$$\{\begin{array}{c}\frac{d{S}_2(t)}{dt}-2\frac{1-p}{p}{\alpha }^{\prime }(t)f(\alpha (t))S_2(t)=2\frac{1-p}{p+1}{\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I,\hfill \\ S_2(0)=u_0^{\frac{1-p}{p+1}}.\hfill \end{array}$$

(2.12)

The solution of Equation (2.12) is

$$\begin{array}{rl}{S}_2(t)=& exp({\int }_0^{\alpha (t)}2\frac{1-p}{p}f(s)ds)(u_0^{\frac{1-p}{p+1}}\\ +{\int }_0^{\alpha (t)}2\frac{1-p}{p+1}f(s)exp(-{\int }_0^{s}2\frac{1-p}{p}f(\tau )d\tau )ds),\end{array}$$

(2.13)

for all $t\in I$. By (2.11), (2.12) and (2.13), we obtain

$$Y_1(t)=S_1^{\frac{p}{1-p}}(t)\le S_2^{\frac{p}{1-p}}(t)={\theta }_1(t),\mathrm{\forall }t\in I,$$

(2.14)

where ${\theta }_1(t)$ as defined in (2.3). From (2.6) and (2.14), we have

$$\frac{d{z}_1(t)}{dt}\le \frac{2}{p+1}{\alpha }^{\prime }(t)f(\alpha (t)){\theta }_1(t),\mathrm{\forall }t\in I.$$

By taking $t=s$ in the above inequality and integrating it from 0 to t, we get

$$u(t)\le z_1(t)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}{\int }_0^{\alpha (t)}f(s){\theta }_1({\alpha }^{-1}(s))ds,\mathrm{\forall }t\in I.$$

The estimation (2.2) of the unknown function in the inequality (2.1) is obtained. □

Theorem 2 Suppose$\alpha \in C^1(I,I)$is increasing function with$\alpha (t)\le t$, $\alpha (0)=0$, $\mathrm{\forall }t\in I$. We assume that$u(t)$and$f(t)$are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$\begin{array}{rl}{u}^{p+1}(t)\le & u_0+{({\int }_0^{\alpha (t)}f(s)u^p(s)ds)}^2\\ +2{\int }_0^{\alpha (t)}f(s)u^p(s)[u(s)+{\int }_0^{s}f(\lambda )u(\lambda )d\lambda ]ds,\mathrm{\forall }t\in I,\end{array}$$

(2.15)

where$u_0>0$and$p\in (0,1)$are constants. Then

$$u(t)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}{\int }_0^{\alpha (t)}f(s){\theta }_2({\alpha }^{-1}(s))ds,\mathrm{\forall }t\in I,$$

(2.16)

where

$${\theta }_2(t)={\beta }_2(t){[u_0^{\frac{1-p}{p+1}}+{\int }_0^{\alpha (t)}(1-p)f(s)exp(-{\int }_0^s\frac{(1-p)(p+3)}{p+1}f(\tau )d\tau )ds]}^{\frac{1}{1-p}},$$

(2.17)

and${\beta }_2(t)=exp({\int }_0^{\alpha (t)}\frac{p+3}{p+1}f(s)ds)$, for all$t\in I$.

Proof Let $z_2^{p+1}(t)$ denote the function on the right-hand side of (2.15), which is a positive and nondecreasing function on I with $z_2(0)=u_0^{\frac{1}{p+1}}$. Then (2.15) is equivalent to

$$u(t)\le z_2(t),u(\alpha (t))\le z_2(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.18)

Differentiating $z_2^{p+1}(t)$ with respect to t, using (2.18) we have

$$\begin{array}{rl}(p+1)z_2^p(t)\frac{d{z}_2(t)}{dt}=& 2{\alpha }^{\prime }(t)f(\alpha (t))u^p(\alpha (t)){\int }_0^{\alpha (t)}f(s)u^p(s)ds\\ +2{\alpha }^{\prime }(t)f(\alpha (t))u^p(\alpha (t))[u(\alpha (t))+{\int }_0^{\alpha (t)}f(\lambda )u(\lambda )d\lambda ]\\ \le & 2{\alpha }^{\prime }(t)f(\alpha (t))z_2^p(t)[z_2(t)+{\int }_0^{\alpha (t)}f(\lambda )z_2(\lambda )d\lambda \\ +{\int }_0^{\alpha (t)}f(\lambda )z_2^p(\lambda )d\lambda ],\mathrm{\forall }t\in I.\end{array}$$

(2.19)

Since $z_2^p(t)>0$, we have

$$\frac{d{z}_2(t)}{dt}\le \frac{2}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_2(t),\mathrm{\forall }t\in I,$$

(2.20)

where

$$Y_2(t):=z_2(t)+{\int }_0^{\alpha (t)}f(\lambda )z_2(\lambda )d\lambda +{\int }_0^{\alpha (t)}f(\lambda )z_2^p(\lambda )d\lambda ,\mathrm{\forall }t\in I.$$

(2.21)

Then $Y_2(t)$ is a positive and nondecreasing function on I with $Y_2(0)=z_2(0)=u_0^{1/(p+1)}$ and

$$z_2(t)\le Y_2(t).$$

(2.22)

Differentiating $Y_2(t)$ with respect to t, and using (2.20), (2.21) and (2.22), we get

$$\begin{array}{rl}\frac{d{Y}_2(t)}{dt}& \le \frac{2}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_2(t)+{\alpha }^{\prime }(t)f(\alpha (t))z_2(\alpha (t))+{\alpha }^{\prime }(t)f(\alpha (t))z_2^p(\alpha (t))\\ \le \frac{p+3}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_2(t)+{\alpha }^{\prime }(t)f(\alpha (t))Y_2^p(t),\mathrm{\forall }t\in I.\end{array}$$

(2.23)

From (2.23), we have

$$Y_2^{-p}(t)\frac{d{Y}_2(t)}{dt}-\frac{p+3}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))Y_2^{1-p}(t)\le {\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.24)

Let $S_3(t)=Y_2^{1-p}(t)$, then $S_3(0)=u_0^{\frac{1-p}{p+1}}$, from (2.24) we obtain

$$\frac{d{S}_3(t)}{dt}-\frac{(1-p)(p+3)}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))S_3(t)\le (1-p){\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I.$$

(2.25)

Consider the ordinary differential equation

$$\{\begin{array}{c}\frac{d{S}_4(t)}{dt}-\frac{(1-p)(p+3)}{p+1}{\alpha }^{\prime }(t)f(\alpha (t))S_4(t)=(1-p){\alpha }^{\prime }(t)f(\alpha (t)),\mathrm{\forall }t\in I,\hfill \\ S_4(0)=u_0^{\frac{1-p}{p+1}}.\hfill \end{array}$$

(2.26)

The solution of Equation (2.26) is

$$\begin{array}{rl}{S}_4(t)=& exp({\int }_0^{\alpha (t)}\frac{(1-p)(p+3)}{p+1}f(s)ds)(u_0^{\frac{1-p}{p+1}}\\ +{\int }_0^{\alpha (t)}(1-p)f(s)exp(-{\int }_0^s\frac{(1-p)(p+3)}{p+1}f(\tau )d\tau )ds),\end{array}$$

(2.27)

for all $t\in I$. By (2.25), (2.26) and (2.27), we obtain

$$Y_2(t)=S_3^{\frac{1}{1-p}}(t)\le S_4^{\frac{1}{1-p}}(t)={\theta }_2(t),\mathrm{\forall }t\in I,$$

(2.28)

where ${\theta }_2(t)$ as defined in (2.17). From (2.20) and (2.28), we have

$$\frac{d{z}_2(t)}{dt}\le \frac{2}{p+1}{\alpha }^{\prime }(t)f(\alpha (t)){\theta }_2(t),\mathrm{\forall }t\in I.$$

By taking $t=s$ in the above inequality and integrating it from 0 to t, we get

$$u(t)\le z_2(t)\le u_0^{\frac{1}{p+1}}+\frac{2}{p+1}{\int }_0^{\alpha (t)}f(s){\theta }_2({\alpha }^{-1}(s))ds,\mathrm{\forall }t\in I.$$

The estimation (2.16) of the unknown function in the inequality (2.15) is obtained. □

Theorem 3 Suppose${\varphi }_1,{\varphi }_2,{\varphi }_2/{\varphi }_1,\alpha \in C^1(I,I)$are increasing functions with$\alpha (t)\le t$, ${\varphi }_i(t)>0$, $\mathrm{\forall }t>0$, $i=1,2$, $\alpha (0)=0$. We assume that$u(t)$and$f(t)$are nonnegative real-valued continuous functions defined on I and satisfy the inequality

$$\begin{array}{rl}u(t)\le & u_0+{({\int }_0^{\alpha (t)}f(s){\varphi }_1(u(s))ds)}^2\\ +2{\int }_0^{\alpha (t)}f(s){\varphi }_1(u(s))[u(s)+{\int }_0^{s}g(\lambda ){\varphi }_2(u(\lambda ))d\lambda ]ds,\mathrm{\forall }t\in I,\end{array}$$

(2.29)

where$u_0>0$is a constant. Then

$$u(t)\le {\mathrm{\Phi }}_1^{-1}\left[{\mathrm{\Phi }}_2^{-1}({\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (t)}g(s)ds)+{\int }_0^{\alpha (t)}2f(s)ds)\right],\mathrm{\forall }t<T_1,$$

(2.30)

where

$${\mathrm{\Phi }}_1(r):={\int }_1^r\frac{dt}{{\varphi }_2(t)},r>0,$$

(2.31)

$${\mathrm{\Phi }}_2(r):={\int }_1^r\frac{{\varphi }_2({\mathrm{\Phi }}_1^{-1}(s))ds}{{\varphi }_1({\mathrm{\Phi }}_1^{-1}(s))({\mathrm{\Phi }}_1^{-1}(s)+1)},r>0,$$

(2.32)

and $T_1$ is the largest number such that

$$\begin{array}{r}{\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (t)}g(s)ds)+{\int }_0^{\alpha (t)}2f(s)ds\le {\int }_1^{\mathrm{\infty }}\frac{{\varphi }_2({\mathrm{\Phi }}_1^{-1}(s))ds}{{\varphi }_1({\mathrm{\Phi }}_1^{-1}(s))({\mathrm{\Phi }}_1^{-1}(s)+1)},\\ {\mathrm{\Phi }}_2^{-1}({\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (t)}g(s)ds)+{\int }_0^{\alpha (t)}2f(s)ds)\le {\int }_1^{\mathrm{\infty }}\frac{dt}{{\varphi }_2(t)}\end{array}$$

for all$t\le T_1$.

Proof Let $z_3(t)$ denote the function on the right-hand side of (2.29), which is a positive and nondecreasing function on I with $z_3(0)=u_0$. Then (2.29) is equivalent to

$$u(t)\le z_3(t),u(\alpha (t))\le z_3(t),\mathrm{\forall }t\in I.$$

(2.33)

Differentiating $z_3(t)$ with respect to t, using (2.33) we have

$$\begin{array}{rl}\frac{d{z}_3(t)}{dt}\le & 2{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(z_3(t)){\int }_0^{\alpha (t)}f(s){\varphi }_1(z_3(s))ds\\ +2{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(z_3(t))[z_3(t)+{\int }_0^{\alpha (t)}g(\lambda ){\varphi }_2(z_3(\lambda ))d\lambda ]ds\\ \le & 2{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(z_3(t))[z_3(t)+{\int }_0^{\alpha (t)}f(s){\varphi }_1(z_3(s))ds\\ +{\int }_0^{\alpha (t)}g(\lambda ){\varphi }_2(z_3(\lambda ))d\lambda ],\mathrm{\forall }t\in I.\end{array}$$

(2.34)

Let

$$Y_3(t):=z_3(t)+{\int }_0^{\alpha (t)}f(s){\varphi }_1(z_3(s))ds+{\int }_0^{\alpha (t)}g(\lambda ){\varphi }_2(z_3(\lambda ))d\lambda ,\mathrm{\forall }t\in I.$$

(2.35)

Then $Y_3(t)$ is a positive and nondecreasing function on I with $Y_3(0)=z_3(0)=u_0$ and

$$z_3(t)\le Y_3(t).$$

(2.36)

Differentiating $Y_3(t)$ with respect to t, and using (2.34), (2.35) and (2.36), we get

$$\begin{array}{rl}\frac{d{Y}_3(t)}{dt}\le & 2{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(z_3(t))Y_3(t)+{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(z_3(t))\\ +{\alpha }^{\prime }(t)g(\alpha (t)){\varphi }_2(z_3(t))\\ \le & 2{\alpha }^{\prime }(t)f(\alpha (t)){\varphi }_1(Y_3(t))(Y_3(t)+1)+{\alpha }^{\prime }(t)g(\alpha (t)){\varphi }_2(Y_3(t)),\end{array}$$

(2.37)

for all $t\in I$. Since ${\varphi }_2(Y_3(t))>0$, $\mathrm{\forall }t>0$, from (2.37) we have

$$\frac{d{Y}_3(t)}{{\varphi }_2(Y_3(t))}\le 2{\alpha }^{\prime }(t)f(\alpha (t))\frac{{\varphi }_1(Y_3(t))(Y_3(t)+1)}{{\varphi }_2(Y_3(t))}+{\alpha }^{\prime }(t)g(\alpha (t)),\mathrm{\forall }t\in I.$$

By taking $t=s$ in the above inequality and integrating it from 0 to t, we get

$$\begin{array}{rl}{\mathrm{\Phi }}_1(Y_3(t))\le & {\mathrm{\Phi }}_1(Y_3(0))+{\int }_0^{t}2{\alpha }^{\prime }(s)f(\alpha (s))\frac{{\varphi }_1(Y_3(s))(Y_3(s)+1)}{{\varphi }_2(Y_3(s))}ds\\ +{\int }_0^t{\alpha }^{\prime }(s)g(\alpha (s))ds,\end{array}$$

(2.38)

for all $t\in I$, where ${\mathrm{\Phi }}_1$ is defined by (2.31). From (2.38), we have

$$\begin{array}{rl}{\mathrm{\Phi }}_1(Y_3(t))\le & {\mathrm{\Phi }}_1(Y_3(0))+{\int }_0^T{\alpha }^{\prime }(s)g(\alpha (s))ds\\ +{\int }_0^{t}2{\alpha }^{\prime }(s)f(\alpha (s))\frac{{\varphi }_1(Y_3(s))(Y_3(s)+1)}{{\varphi }_2(Y_3(s))}ds,\end{array}$$

(2.39)

for all $t<T$, where $0<T<T_1$ is chosen arbitrarily. Let $Y_4(t)$ denote the function on the right-hand side of (2.39), which is a positive and nondecreasing function on I with $Y_4(0)={\mathrm{\Phi }}_1(u_0)+{\int }_0^T{\alpha }^{\prime }(s)g(\alpha (s))ds$ and

$$Y_3(t)\le {\mathrm{\Phi }}_1^{-1}(Y_4(t)),\mathrm{\forall }t<T.$$

(2.40)

Differentiating $Y_4(t)$ with respect to t, using the hypothesis on ${\varphi }_2/{\varphi }_1$, from (2.40) we have

$$\begin{array}{rl}\frac{d{Y}_4(t)}{dt}& \le 2{\alpha }^{\prime }(t)f(\alpha (t))\frac{{\varphi }_1(Y_3(t))(Y_3(t)+1)}{{\varphi }_2(Y_3(t))}\\ \le 2{\alpha }^{\prime }(t)f(\alpha (t))\frac{{\varphi }_1({\mathrm{\Phi }}_1^{-1}(Y_4(t)))({\mathrm{\Phi }}_1^{-1}(Y_4(t))+1)}{{\varphi }_2({\mathrm{\Phi }}_1^{-1}(Y_4(t)))},\mathrm{\forall }t<T.\end{array}$$

(2.41)

By the definition of ${\mathrm{\Phi }}_2$ in (2.32), from (2.41) we obtain

$$\begin{array}{rl}{\mathrm{\Phi }}_2(Y_4(t))& \le {\mathrm{\Phi }}_2(Y_4(0))+{\int }_0^{t}2{\alpha }^{\prime }(s)f(\alpha (s))ds\\ \le {\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (T)}g(s)ds)+{\int }_0^{\alpha (t)}2f(s)ds,\mathrm{\forall }t<T.\end{array}$$

(2.42)

Let $t=T$, from (2.42) we have

$${\mathrm{\Phi }}_2(Y_4(T))\le {\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (T)}g(s)ds)+{\int }_0^{\alpha (T)}2f(s)ds.$$

(2.43)

Since $0<T<T_1$ is chosen arbitrarily, from (2.33), (2.36), (2.40) and (2.43), we have

$$u(t)\le {\mathrm{\Phi }}_1^{-1}\left[{\mathrm{\Phi }}_2^{-1}({\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(u_0)+{\int }_0^{\alpha (t)}g(s)ds)+{\int }_0^{\alpha (t)}2f(s)ds)\right],\mathrm{\forall }t<T_1.$$

This proves (2.30). □

3 Application

In this section, we apply our Theorem 3 to the following differential-integral equation

$$\{\begin{array}{c}\frac{dx(t)}{dt}=F(t,x(\alpha (t)))+H(t,x(\alpha (t))),\mathrm{\forall }t\in I,\hfill \\ x(0)=x_0,\hfill \end{array}$$

(3.1)

where $F\in C(I\times I,\mathbf{R})$, $H\in C(I^3,\mathbf{R})$, $|x_0|>0$ is a constant satisfying the following conditions

$$\left|F(t,x(t))\right|\le f^2(t){\varphi }_1^2\left(|x(t)|\right),$$

(3.2)

$$\left|H(t,x(t))\right|\le 2f(t){\varphi }_1\left(|x(t)|\right)(|x(t)|+{\int }_0^{t}g(s){\varphi }_2\left(|x(s)|\right)ds),$$

(3.3)

where f, g is nonnegative real-valued continuous function defined on I.

Corollary 1 Consider the nonlinear system (3.1) and suppose that F, H satisfy the conditions (3.2) and (3.3), and${\varphi }_1,{\varphi }_2,{\varphi }_2/{\varphi }_1,\alpha \in C^1(I,I)$are increasing functions with$\alpha (t)\le t$, ${\varphi }_i(t)>0$, $\mathrm{\forall }t>0$, $i=1,2$, $\alpha (0)=0$. Then all solutions of Equation (3.1) exist on I and satisfy the following estimation

$$|x(t)|\le {\mathrm{\Phi }}_1^{-1}\left[{\mathrm{\Phi }}_2^{-1}({\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(|x_0|)+{\int }_0^{\alpha (t)}\frac{g({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds)+{\int }_0^{\alpha (t)}2\frac{f({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds)\right],$$

(3.4)

for all$t<T_2$, where

$$\begin{array}{r}{\mathrm{\Phi }}_1(r):={\int }_1^r\frac{dt}{{\varphi }_2(t)},r>0,\\ {\mathrm{\Phi }}_2(r):={\int }_1^r\frac{{\varphi }_2({\mathrm{\Phi }}_1^{-1}(s))ds}{{\varphi }_1({\mathrm{\Phi }}_1^{-1}(s))({\mathrm{\Phi }}_1^{-1}(s)+1)},r>0,\end{array}$$

and $T_2$ is the largest number such that

$$\begin{array}{r}{\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(|x_0|)+{\int }_0^{\alpha (t)}\frac{g({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds)+{\int }_0^{\alpha (t)}2\frac{f({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds\\ \le {\int }_1^{\mathrm{\infty }}\frac{{\varphi }_2({\mathrm{\Phi }}_1^{-1}(s))ds}{{\varphi }_1({\mathrm{\Phi }}_1^{-1}(s))({\mathrm{\Phi }}_1^{-1}(s)+1)},\\ {\mathrm{\Phi }}_2^{-1}({\mathrm{\Phi }}_2({\mathrm{\Phi }}_1(|x_0|)+{\int }_0^{\alpha (t)}\frac{g({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds)+{\int }_0^{\alpha (t)}2\frac{f({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}ds)\le {\int }_1^{\mathrm{\infty }}\frac{dt}{{\varphi }_2(t)}\end{array}$$

for all$t\le T_2$.

Proof Integrating both sides of Equation (3.1) from 0 to t, we get

$$x(t)=x_0+{\int }_0^{t}F(s,x(\alpha (s)))ds+{\int }_0^{t}H(s,x(\alpha (s)))ds,\mathrm{\forall }t\in I.$$

(3.5)

Using the conditions (3.2) and (3.3), from (3.5) we obtain

$$\begin{array}{rl}|x(t)|\le & |x_0|+{\int }_0^t{f}^2(s){\varphi }_1^2\left(\left|x(\alpha (s))\right|\right)ds\\ +2{\int }_0^{t}f(s){\varphi }_1\left(\left|x(\alpha (s))\right|\right)(\left|x(\alpha (s))\right|+{\int }_0^{s}g(\tau ){\varphi }_2\left(\left|x(\alpha (\tau ))\right|\right)d\tau )ds\\ \le & |x_0|+{\left({\int }_0^{\alpha (t)}\frac{f({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}{\varphi }_1\left(|x(s)|\right)ds\right)}^2\\ +2{\int }_0^{\alpha (t)}\frac{f({\alpha }^{-1}(s))}{{\alpha }^{\prime }({\alpha }^{-1}(s))}{\varphi }_1\left(|x(s)|\right)(|x(s)|+{\int }_0^s\frac{g({\alpha }^{-1}(\tau ))}{{\alpha }^{\prime }({\alpha }^{-1}(\tau ))}{\varphi }_2\left(|x(\tau )|\right)d\tau )ds,\end{array}$$

(3.6)

for all $t\in I$. Applying Theorem 3 to (3.6), we get the estimation (3.4). This completes the proof of the Corollary 1. □