Introduction

It is well-known that a lot of problems investigated in engineering, mechanics, mathematical physics, vehicular traffic theory [1, 14, 2, pp. 157–167], queuing theory and also several real world problems can be described with help of various functional differential (integral) equations. The theory of functional differential (integral) equations is highly developed and constitutes a significant and important branch of nonlinear analysis. There have been published, up to now, numerous research papers; see [3,4,5,6,7,8, 10, 12, 13, 15,16,17].

In this paper, we are interested with the initial value problem (IVP) for the functional integro-differential equation

$$\begin{aligned} \frac{dx}{dt}=g(t,x(t),\quad \int _{0}^{t}f(s, x(s))ds), \quad a.e\quad t\in (0, T], \end{aligned}$$
(1)

with the nonlocal condition

$$\begin{aligned} x(0) +\sum _{j=1}^{m} p_j x(\tau _j) = x_0, \quad \sum _{j=1}^{m} p_j >0,\quad \tau _j\in (0, T]. \end{aligned}$$
(2)

The existence of at least and unique solution \(x\in C[0,T]\), under certain conditions, will be proved. The continuous dependence of the solution on the nonlocal-data \(p_j\), on \(x_0\) and on the functional f, will be studied.

As applications, the IVP of Eq. (1) with integral condition

$$\begin{aligned} x(0) +\int _{0}^{T}x(s)dh(s)=x_0,\quad h:[0,T]\rightarrow \mathbb {R}~\text {increasing function} \end{aligned}$$
(3)

will be studied. Also, if \(\sum _{j=1}^{\infty } p_j\) is convergent, the IVP of Eq. (1) with infinite-point condition

$$\begin{aligned} x(0) +\sum _{j=1}^{\infty } p_j x(\tau _j)=x_0, \end{aligned}$$
(4)

will be studied.

Integral Representation

Consider the IVP (1)–(2) with the assumptions:

  1. 1.

    \(g: [0,T]\times \mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}\) satisfies Caratheodory-condition. There exist a function \(c_1\in L^1[0,T]\) and a positive constant \(b_1 >0\), such that

    $$\begin{aligned} |g(t,\alpha ,\beta )|\le c_1(t)+b_1|\alpha |+b_1|\beta |. \end{aligned}$$
  2. 2.

    \(f: [0,1]\times \mathbb {R}\rightarrow \mathbb {R}\) satisfies Caratheodory-condition. There exist a function \(c_2\in L^1[0,T]\) and a positive constant \(b_2 >0\), such that

    $$\begin{aligned} |f(t,\beta )|\le c_2(t)+b_2|\beta |. \end{aligned}$$
  3. 3.
    $$\begin{aligned} \sup _{t\in [0,1]}\int _{0}^{t} c_1(s)ds\le M_1,\quad \sup _{t\in [0,1]}\int _{0}^{t}\int _{0}^{s} c_2(\theta )d\theta ds\le M_2. \end{aligned}$$
  4. 4.

    \(\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) <1\).

Definition 2.1

By a solution of the IVP (1)–(2) we mean a function \(x\in C[0,T]\) that satisfies (1)–(2).

Lemma 2.1

The solution of IVP (1)–(2) if it exist, then it can be represented by the integral-equation

$$\begin{aligned} x(t)= & {} E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \nonumber \\&+ \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds, \end{aligned}$$
(5)

where \(E=(1+\sum _{j=1}^{m} p_j)^{-1}.\)

Proof

Let x be a solution of IVP (1)–(2). Integrating both sides of (1) we obtain

$$\begin{aligned} x(t)=x(0)+\int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds. \end{aligned}$$
(6)

Using the nonlocal condition (2), we get

$$\begin{aligned} \sum _{j=1}^{m} p_j x(\tau _j)=x(0)\sum _{j=1}^{m} p_j+\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds, \end{aligned}$$

since, \(\sum _{j=1}^{m} p_jx(\tau _j)=x_0-x(0)\), we have

$$\begin{aligned} x_0-x(0)=x(0)\sum _{j=1}^{m} p_j+\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds, \end{aligned}$$

then

$$\begin{aligned} x(0)=\frac{1}{1+\sum _{j=1}^{m} p_j} \left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] . \end{aligned}$$
(7)

Using (6) and (7), we obtain

$$\begin{aligned} x(t)= & {} \frac{1}{1+\sum _{j=1}^{m} p_j} \left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \\&+ \int _{0}^{t} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds. \end{aligned}$$

\(\square \)

Existence of Solution

Theorem 3.1

Let the assumptions 1–4 be satisfied. Then the IVP (1)–(2) has at least one solution \(x\in C[0,T]\).

Proof

Let the operator F associated with the integral-equation (5) by

$$\begin{aligned} Fx(t)= & {} E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \\&+ \int _{0}^{t} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds. \end{aligned}$$

Let \(Q_r=\{x\in \mathbb {R}: ||x||\le r\}\), where   \(r=\frac{E|x_0|+(1+E\sum _{j=1}^{m} p_j)(M_1+b_1M_2)}{1-((1+E\sum _{j=1}^{m} p_j)(b_1T+\frac{1}{2}b_1b_2T^2))}\), it clear that \(Q_r\) is nonempty, closed, bounded and convex subset of C[0, T]. Then we have, for \(x\in Q_r\)

$$\begin{aligned} |Fx(t)|\le & {} E\left[ |x_0|+\sum _{j=1}^{m} p_j\int _{0}^{\tau _j}| g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )|ds\right] \\&+ \int _{0}^{t} |g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )|ds\\\le & {} E\left[ |x_0|+\sum _{j=1}^{m} p_j\int _{0}^{\tau _j}(c_1(s)+b_1|x(s)|+b_1\int _{0}^{s}|f(\theta , x(\theta ))|d\theta )ds\right] \\&+ \int _{0}^{t} (c_1(s)+b_1|x(s)|+b_1\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\\\le & {} E\left[ |x_0|+\sum _{j=1}^{m} p_j(M_1+b_1Tr+b_1\int _{0}^{\tau _j}\int _{0}^{s}|c_2(\theta )+b_2|x(\theta )|d\theta ds)\right] \\&+M_1+b_1Tr+b_1 \int _{0}^{t}\int _{0}^{s}(c_2(\theta )+b_2|x(\theta )|)d\theta ds\\\le & {} E|x_0|+E\sum _{j=1}^{m} p_j(M_1+b_1Tr+b_1M_2+\frac{1}{2}b_1b_2T^2r)\\&+M_1+b_1Tr+b_1 M_2+\frac{1}{2}b_1b_2T^2r\\= & {} E|x_0|+\left( 1+E\sum _{j=1}^{m} p_j\right) \left( M_1+b_1Tr+b_1M_2+\frac{1}{2}b_1b_2T^2r\right) =r. \end{aligned}$$

Then \(F: Q_r\rightarrow Q_r\) and the class of functions \(\{F x\}\) is uniformly bounded in \(Q_r.\)

Now, let \(t_1,t_2\in (0,1)\) s. t \(|t_2-t_1|<\delta \), then

$$\begin{aligned} |Fx(t_2)-Fx(t_1)|= & {} \left| \int _{0}^{t_2} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right. \\&\left. - \int _{0}^{t_1} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right| \\\le & {} \int _{t_1}^{t_2}\left| g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )\right| ds\\\le & {} \int _{t_1}^{t_2}(c_1(s)+b_1|x(s)|+b_1\int _{0}^{s}|f(\theta , x(\theta ))d\theta |)ds\\\le & {} \int _{t_1}^{t_2}c_1(s)ds+(t_2-t_1)b_1r+b_1\int _{t_1}^{t_2}\int _{0}^{s} c_2(\theta )d\theta ds\\&+\frac{1}{2}b_1b_2r\left( t_2^2-t_1^2\right) . \end{aligned}$$

Then the class of functions \(\{Fx\}\) is equi-continuous in \(Q_r.\)

Let \(x_n\in Q_r\), \(x_n\rightarrow x (n\rightarrow \infty )\), then from Assumptions 1–2, we obtain \(g(t,x_n(t),y_n(t))\rightarrow g(t,x(t),y(t))\) and \(f(t,x_n(t))\rightarrow f(t,x(t))\) as \(n\rightarrow \infty .\) Also

$$\begin{aligned} \lim _{n\rightarrow \infty }Fx_n(t)= & {} \lim _{n\rightarrow \infty }\biggl [E\biggl [x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x_n(s),\int _{0}^{s}f(s, x_n(\theta ))d\theta )ds\biggr ] \nonumber \\&+ \int _{0}^{t} g\biggl (s,x_n(s),\int _{0}^{s}f(\theta , x_n(\theta ))d\theta \biggr )ds\biggr ]. \end{aligned}$$
(8)

Using assumptions 1–2 and Lebesgue Dominated convergence Theorem [11], from (8) we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }Fx_n(t)= & {} \biggl [E\biggl [x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} \lim _{n\rightarrow \infty }g(s,x_n(s),\int _{0}^{s}f(\theta , x_n(\theta ))d\theta )ds\biggr ] \\&+ \int _{0}^{t} \lim _{n\rightarrow \infty }g\biggl (s,x_n(s),\int _{0}^{s}f(\theta , x_n(\theta ))d\theta \biggr )ds\biggr ]=Fx(t). \end{aligned}$$

Then \(Fx_n\rightarrow Fx\) as \(n\rightarrow \infty \). Therefore F is continuous.

$$\begin{aligned} \lim _{t\rightarrow 0}x(t)=E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \in C[0,T]. \end{aligned}$$

Then by Schauder fixed point Theorem [9] there exist at least one solution \(x\in C[0,T]\) of the integral-equation (5).

To complete the proof, differentiation (5) we obtain

$$\begin{aligned} \frac{dx}{dt}= & {} \frac{d}{dt}\biggl \{E\biggl [x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\biggr ]\nonumber \\&+ \int _{0}^{t} g\biggl (s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \biggr )ds\biggr \}\\= & {} \frac{d}{dt}\int _{0}^{t} g\biggr (s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \biggr )ds\\= & {} g\left( t,x(t),\int _{0}^{t}f(\theta , x(\theta ))d\theta \right) . \end{aligned}$$

Also, from the integral-equation (5), we get

$$\begin{aligned} x(\tau _j)= & {} E\Big [x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\Big (s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \Big )ds\Big ]\nonumber \\&+ \int _{0}^{\tau _j} g\Big (s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \Big )ds\nonumber \\ x(0)= & {} E\Big [x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\Big ], \end{aligned}$$
(9)

and

$$\begin{aligned} \sum _{j=1}^{m} p_j x(\tau _j)= & {} E\sum _{j=1}^{m} p_j \left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \nonumber \\&+\sum _{j=1}^{m} p_j \int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds, \end{aligned}$$
(10)

from (9) and (10) we have

$$\begin{aligned}&x(0) +\sum _{j=1}^{m} p_j x(\tau _j)\\&\quad =E(1+\sum _{j=1}^{m} p_j) \left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \\&\qquad +\sum _{j=1}^{m} p_j \int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds.\\ \end{aligned}$$

Then

$$\begin{aligned} x(0)+\sum _{j=1}^{m} p_j x(\tau _j)=x_0. \end{aligned}$$

Therefor there exist at least one solution \(x \in C[0, T]\) of the IVP (1)–(2). \(\square \)

Nonlocal Integral Condition

Let \(x\in C[0, T]\) be the solution of the IVP (1)–(2). Let \(p_j = h(t_j)- h(t_{j-1})\), h is increasing function, \(\tau _j\in (t_{j-1}, t_j)\), \(0 = t_0< t_1< t_2, \ldots < t_m = 1\) then, as \(m\rightarrow \infty \) the nonlocal-condition (2) will be

$$\begin{aligned} x(0)+\sum _{j=1}^{m} h(t_j)-h(t_{j-1})x(\tau _j)=x_0. \end{aligned}$$

And

$$\begin{aligned} x(0)+\lim _{m\rightarrow \infty }\sum _{j=1}^{m} h(t_j)- h(t_{j-1})x(\tau _j)=x(0)+\int _{0}^{T}x(s)dh(s)=x_0. \end{aligned}$$

Theorem 4.1

Let the assumptions 1–4 be satisfied. Then the IVP of (1)–(3) has at least one solution \(x\in C[0,T]\).

Proof

As \(m\rightarrow \infty \), the solution of the IVP (1)–(2) will be

$$\begin{aligned} x(t)= & {} \lim _{m\rightarrow \infty }\frac{1}{1+\sum _{j=1}^{m} p_j}\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \\&+ \int _{0}^{t} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\\= & {} \frac{1}{1+h(T)-h(0)}\left[ x_0-\lim _{m\rightarrow \infty }\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds(h(t_j)\right. \\&- h(t_{j-1}))]+ \int _{0}^{t} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\\= & {} \frac{1}{1+h(T)-h(0)}\left[ x_0-\int _{0}^{T}\int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\cdot dh(t)\right] \\&+ \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds. \end{aligned}$$

\(\square \)

Infinite-Point Boundary Condition

Theorem 5.1

Let the assumptions 1–4 be satisfied. Then the IVP of (1)–(4) has at least one solution \(x\in C[0,T]\).

Proof

Let the assumptions of Theorem 3.1 be satisfied. Let \({S_m},~S_m=\sum _{j=1}^{m} p_j\) be convergent sequence, then

$$\begin{aligned} x_m(t)= & {} \frac{1}{1+\sum _{j=1}^{m} p_j}\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \nonumber \\&+ \int _{0}^{t} g\left( s,x_m(s),\int _{0}^{s}f(\theta , x_m(\theta ))d\theta \right) ds. \end{aligned}$$
(11)

Take the limit to (11), as \(m\rightarrow \infty \), we have

$$\begin{aligned} \lim _{m\rightarrow \infty }x_m(t)= & {} \lim _{m\rightarrow \infty }\left[ \frac{1}{1+\sum _{j=1}^{m} p_j}\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \right. \nonumber \\&\left. + \int _{0}^{t} g\left( s,x_m(s),\int _{0}^{s}f(\theta , x_m(\theta ))d\theta \right) ds\right] \nonumber \\= & {} \lim _{m\rightarrow \infty }\frac{1}{1+\sum _{j=1}^{m} p_j}\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \nonumber \\&+ \lim _{m\rightarrow \infty }\int _{0}^{t} g\left( s,x_m(s),\int _{0}^{s}f(\theta , x_m(\theta ))d\theta \right) ds. \end{aligned}$$
(12)

Now \(|p_j x(\tau _j)|\le |p_j| \Vert x\Vert ,\) therefore by comparison test \(\sum _{j=1}^{\infty } p_j x(\tau _j)\) is convergent. Also

$$\begin{aligned}&\left| \int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right| \le \int _{0}^{\tau _j}(c_1(s)+b_1|x(s)|\\&\qquad +\,b_1\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\\&\quad \le \int _{0}^{\tau _j}(c_1(s)+b_1|x(s)|+b_1\int _{0}^{s}(c_2(s)+b_2|x(s)|)d\theta )ds\\&\quad \le M_1+b_1\Vert x\Vert +b_1M_2+\frac{1}{2} b_1b_2\Vert x\Vert \le M,\\ \end{aligned}$$

then \(| p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds|\le |p_j|\cdot M\) and by the comparison test \(\sum _{j=1}^{\infty } p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\) is convergent.

Now, \(|g|\le |c_1(s)+b_1\Vert x\Vert +b_1M_2+b_1b_2\Vert x\Vert \), using assumptions 1–2 and Lebesgue Dominated convergence Theorem [11], from (12) we obtain

$$\begin{aligned} x(t)= & {} \frac{1}{1+\sum _{j=1}^{\infty } p_j}\left[ x_0-\sum _{j=1}^{\infty } p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \nonumber \\&+ \int _{0}^{t} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds. \end{aligned}$$
(13)

The Theorem proved. \(\square \)

Uniqueness of the Solution

Let g and f satisfy the following assumptions

  1. 5.

    \(g: [0,T]\times \mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}\) is measurable in t for any \(\alpha ,\beta \in \mathbb {R}\) and satisfies the lipschitz condition

    $$\begin{aligned} |g(t,\alpha ,\beta )-g(t,u,v)|\le b_1|\alpha -u|+b_1|\beta -v|, \end{aligned}$$
    (14)
  2. 6.

    \(f: [0,T]\times \mathbb {R}\rightarrow \mathbb {R}\) is measurable in t for any \(\alpha \in \mathbb {R}\) and satisfies the lipschitz condition

    $$\begin{aligned} |f(t,\alpha )-f(t,u)|\le b_2|\alpha -u|, \end{aligned}$$
    (15)
  3. 7.
    $$\begin{aligned} \sup _{t\in [0,T]}\int _{0}^{t}|f(s, 0, 0)|ds\le L_1, \quad \sup _{t\in [0,T]}\int _{0}^{t}\int _{0}^{s}|g(\theta , 0)|d\theta ds\le L_2. \end{aligned}$$

Theorem 6.1

Let the assumptions 5–7 be satisfied. Then the solution of the IVP (1)–(2) is unique.

Proof

From assumption 5 we have g is measurable in t for any \(x, y\in \mathbb {R}\) and satisfies the lipschitz condition, then it is continuous in \(\alpha ,\beta \in \mathbb {R}\) \(\forall t\in [0,T]\), and

$$\begin{aligned} |g(t,\alpha ,\beta )|\le b_1|\alpha |+b_1|\beta |+|f(t,0,0)|. \end{aligned}$$

Then condition 1 is satisfied. Also by the same way we can show that assumption 2 satisfied by Assumption 6. Now, from Theorem 3.1 the solution of the IVP (1)–(2) exists.

Let x, y be two the solution of (1)–(2), then

$$\begin{aligned} |x(t)-y(t)|= & {} |E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \\&+ \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\\&-E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,y(s),\int _{0}^{s}f(\theta , y(\theta ))d\theta \right) ds\right] \\&- \int _{0}^{t} g\left( s,y(s),\int _{0}^{s}f(\theta , y(\theta ))d\theta \right) ds|\\\le & {} E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} |g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) \\&\left. -g\left( s,y(s),\int _{0}^{s}f(\theta , y(\theta ))d\theta \right) \right| ds\\&+ \int _{0}^{t}|g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) \\&-g\left( s,y(s),\int _{0}^{s}f(\theta , y(\theta ))d\theta \right) |ds\\\le & {} E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} \left( b_1\Vert x-y\Vert +b_1\int _{0}^{s}|f(\theta , x(\theta )) \right. \\&\left. -f(\theta , y(\theta ))|d\theta \right) ds+ \int _{0}^{t}(b_1\Vert x-y\Vert \\&\left. +\,b_1\int _{0}^{s}|f(\theta , x(\theta ))-f(\theta , y(\theta ))|d\theta \right) ds\\\le & {} b_1T\Vert x-y\Vert E\sum _{j=1}^{m} p_j+\frac{1}{2} b_1b_2T^2\Vert x-y\Vert E\sum _{j=1}^{m} p_j\\&+\,b_1T\Vert x-y\Vert +\frac{1}{2} b_1b_2T^2\Vert x-y\Vert \\= & {} \left( 1+E\sum _{j=1}^{m} p_j\right) (b_1T+\frac{1}{2}b_1b_2T^2)\Vert x-y\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \left( 1-\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) \right) \Vert x-y\Vert \le 0. \end{aligned}$$

Since \(\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) <1\), then \(x(t)=y(t)\) and the solution of the IVP (1)–(2) is unique. \(\square \)

Continuous Dependence

Continuous Dependence on \(x_0\)

Definition 7.1

The solution \(x\in C[0,1]\) of the IVP (1)–(2) continuously depends on \(x_0\), if

$$\begin{aligned} \forall \epsilon >0, \quad \exists \quad \delta (\epsilon )\quad s.t\quad |x_0-x_0^*|<\delta \Rightarrow ||x-x^*||<\epsilon , \end{aligned}$$

where \(x^*\) is the solution of the IVP

$$\begin{aligned} \frac{dx^*}{dt}=g(t,x^*(t),\int _{0}^{t}f(s, x^*(s))ds), \quad a.e\quad t\in (0,T], \end{aligned}$$
(16)

with the nonlocal condition

$$\begin{aligned} x(0)+ \sum _{j=1}^{m} p_j x^*(\tau _j) = x_0^*,\quad \sum _{j=1}^{m} p_j >0,~\tau _j\in (0,T]. \end{aligned}$$
(17)

Theorem 7.1

Let the assumptions of Theorem 6.1 be satisfied. Then the solution of the IVP (1)–(2) continuously depends on \(x_0\).

Proof

Let \(x,x^*\) be two solutions of the IVP (1)–(2) and (16)–(17) respectively. Then

$$\begin{aligned} |x(t)-x^*(t)|= & {} |E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )ds\right] \\&+ \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\\&-E\left[ x^*_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\right] \\&- \int _{0}^{t} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds|\\\le & {} E|x_0-x_0^*|+E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} |g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) \\&-g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) |ds+ \int _{0}^{t}|g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) \\&-g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) |ds,\\\le & {} E|x_0-x_0^*|+E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} (b_1\Vert x-x^*\Vert \\&+\,b_1\int _{0}^{s}|f(\theta , x^*(\theta )) -f(\theta , x(\theta ))|d\theta )ds+ \int _{0}^{t}(b_1\Vert x-x^*\Vert \\&+\,b_1\int _{0}^{s}|f(\theta , x(\theta ))-f(\theta , x^*(\theta ))|d\theta )ds\\\le & {} E|x_0-x_0^*|+b_1T\Vert x-x^*\Vert E\sum _{j=1}^{m} p_j +\frac{1}{2} b_1b_2T^2\Vert x-x^*\Vert E\sum _{j=1}^{m} p_j \\&+\,b_1T\Vert x-x^*\Vert +\frac{1}{2}T^2 b_1b_2\Vert x-x^*\Vert \\\le & {} E\delta +\left( 1+E\sum _{j=1}^{m} p_j\right) (b_1T+\frac{1}{2}b_1b_2T^2)\Vert x-x^*\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \Vert x-x^*\Vert \le \frac{E\delta }{\left[ 1-\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) \right] }=\epsilon . \end{aligned}$$

Then the solution of the IVP (1)–(2) continuously depends on \(x_0\). \(\square \)

Continuous Dependence on the Nonlocal Data \(p_j\)

Definition 7.2

The solution \(x\in C[0,1]\) of the IVP (1)–(2) continuously depends on the nonlocal data \(p_j\), if

$$\begin{aligned} \forall \epsilon >0, \quad \exists \quad \delta (\epsilon )\quad s.t\quad |p_j-p_j^*|<\delta \Rightarrow ||x-x^*|| <\epsilon , \end{aligned}$$

where \(x^*\) is the solution of the IVP

$$\begin{aligned} \frac{dx^*}{dt}=g(t,x^*(t),\int _{0}^{t}f(s, x^*(s))ds), \quad a.e\quad t\in (0,1), \end{aligned}$$
(18)

with the nonlocal condition

$$\begin{aligned} x(0)+\sum _{j=1}^{m} p_j^* x^*(\tau _j) = x_0,\quad \sum _{j=1}^{m} p_j^* >0,\quad \tau _j\in (0,1). \end{aligned}$$
(19)

Theorem 7.2

Let the assumptions of Theorem 6.1 be satisfied. Then the solution of the IVP (1)–(2) continuously depends on the nonlocal data \(p_j\).

Proof

Let \(x,~x^*\) be two the solutions of the IVP (1)–(2) and (18)–(19) respectively. Then

$$\begin{aligned} |x(t)-x^*(t)|&=\,|E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \\&\quad + \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds-E^*\left[ x_0-\sum _{j=1}^{m} p_j^*\int _{0}^{\tau _j} g(s,x^*(s),\right. \\&\qquad \left. \int _{0}^{s}f(\theta , x^*(\theta ))d\theta )ds\right] -\int _{0}^{t} g\left( s,x^*(s),\int _{0}^{s}f(s, x^*(\theta ))d\theta \right) ds\Big |\\&\le \, E E^*m\delta |x_0|+|E^*\sum _{j=1}^{m} p_j^*\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\\&\quad -E\sum _{j=1}^{m} p_j^*\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\\&\quad +E\left[ \sum _{j=1}^{m} p_j^*\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\right. \\&\quad \left. -\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \Big |\\&\quad +b_1T\Vert x-x^*\Vert +\frac{1}{2}T^2 b_1b_2\Vert x-x^*\Vert \\&\le \, E E^*m\delta |x_0|+m\delta \left[ \left( b_1T+\frac{1}{2} b_1b_2T^2\right) \Vert x^*\Vert +TL_1+\frac{1}{2} b_1T^2L_2\right] \sum _{j=1}^{m} p_j^*\\&\quad +E|\left[ \sum _{j=1}^{m} p_j^*\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\right. \\&\quad -\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\\&\quad +\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f(\theta , x^*(\theta ))d\theta \right) ds\\&\quad \left. -\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] |\\&\quad +b_1T\Vert x-x^*\Vert +\frac{1}{2}T^2 b_1b_2\Vert x-x^*\Vert \\&\le \, E E^*m\delta |x_0|+m\delta \left[ \left( b_1T+\frac{1}{2} b_1b_2T^2\right) \Vert x^*\Vert +TL_1+\frac{1}{2} b_1T^2L_2\right] \sum _{j=1}^{m} p_j^*\\&\quad +E\left[ m\delta \left[ \left( b_1T+\frac{1}{2} b_1b_2T^2\right) \Vert x^*\Vert +TL_1+\frac{1}{2} b_1T^2L_2\right] \right. \\&\quad \left. +\sum _{j=1}^{m} p_j\left( b_1T\Vert x-x^*\Vert +\frac{1}{2}T^2 b_1b_2\Vert x-x^*\Vert \right) \right] \\&\quad +b_1T\Vert x-x^*\Vert +\frac{1}{2}T^2 b_1b_2\Vert x-x^*\Vert \\&\le E E^*m\delta |x_0|+m\delta \left[ \left( b_1T+\frac{1}{2} b_1b_2T^2\right) \Vert x^*\Vert +TL_1+\frac{1}{2} b_1T^2L_2\right] \\&\qquad \left( E+\sum _{j=1}^{m} p_j^*\right) +\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}T^2 b_1b_2\right) \Vert x-x^*\Vert . \end{aligned}$$

Hence

$$\begin{aligned}&\Vert x-x^*\Vert \\&\quad \le \frac{E E^*m |x_0|+m\left[ \left( b_1T+\frac{1}{2} b_1b_2T^2\right) \Vert x^*\Vert +TL_1+\frac{1}{2} b_1T^2L_2\right] \left( E+\sum _{j=1}^{m} p_j^*\right) }{1-\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}T^2 b_1b_2\right) }\delta =\epsilon , \end{aligned}$$

where \(E^*=(1+\sum _{j=1}^{m} p_j^*)^{-1}.\) Then the solution of the IVP (1)–(2) continuously depends on the nonlocal data \(p_j\). \(\square \)

Continuous Dependence on the Functional f

Definition 7.3

The solution \(x\in C[0,T]\) of the IVP (1)–(2) continuously depends on the functional f, if

$$\begin{aligned} \forall \epsilon >0, \quad \exists \quad \delta (\epsilon )\quad s.t\quad |f-f^*|<\delta \Rightarrow ||x-x^*|| <\epsilon , \end{aligned}$$

where \(x^*\) is the solution of the IVP

$$\begin{aligned} \frac{dx^*}{dt}=g\left( t,x^*(t),\int _{0}^{t}f^*(s, x^*(s))ds\right) , \quad a.e\quad t\in (0,T], \end{aligned}$$
(20)

with the nonlocal condition

$$\begin{aligned} x(0)+\sum _{j=1}^{m} p_j x^*(\tau _j) = x_0,\quad \sum _{j=1}^{m} p_j >0,\quad \tau _j\in (0,T]. \end{aligned}$$
(21)

Theorem 7.3

Let the assumptions of Theorem 6.1 be satisfied. Then the solution of the IVP (1)–(2) continuously depends on the functional f.

Proof

Let \(x,x^*\) be two solutions of the IVP (1)–(2) and (20)–(21) respectively. Then

$$\begin{aligned} |x(t)-x^*(t)|&\le |,|E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\right] \\&\quad + \int _{0}^{t} g\left( s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta \right) ds\\&\quad -E\left[ x_0-\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} g\left( s,x^*(s),\int _{0}^{s}f^*(\theta , x^*(\theta ))d\theta \right) ds\right] \\&\left. \quad - \int _{0}^{t} g\left( s,x^*(s),\int _{0}^{s}f^*(\theta , x^*(\theta ))d\theta \right) ds\right| \\&\le E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} \left| g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )\right. \\&\left. \quad -g\left( s,x^*(s),\int _{0}^{s}f^*(\theta , x^*(\theta ))d\theta \right) \right| ds\\&\quad + \int _{0}^{t}|g(s,x(s),\int _{0}^{s}f(\theta , x(\theta ))d\theta )-g\left( s,x^*(s),\int _{0}^{s}f^*(\theta , x^*(\theta ))d\theta \right) \Big |ds,\\&\le E\sum _{j=1}^{m} p_j\int _{0}^{\tau _j} \left( b_1\Vert x-x^*\Vert +b_1\int _{0}^{s}|f^*(\theta , x^*(\theta ))-f(\theta , x(\theta ))|d\theta \right) ds\\&\quad + \int _{0}^{t}\left( b_1\Vert x-x^*\Vert +b_1\int _{0}^{s}|f(\theta , x(\theta ))-f^*(\theta , x^*(\theta ))|d\theta \right) ds\\&\le b_1T\Vert x-x^*\left\| E\sum _{j=1}^{m} p_j +\frac{1}{2} b_1T^2\delta E\sum _{j=1}^{m} p_j +\frac{1}{2} b_1b_2T^2\right\| x-x^*\Vert E\sum _{j=1}^{m} p_j \\&\quad +\frac{1}{2} b_1T^2\delta +b_1T^2\Vert x-x^*\Vert +\frac{1}{2} b_1b_2T^2\Vert x-x^*\Vert \\&\le \left( 1+ E\sum _{j=1}^{m} p_j\right) \frac{1}{2} b_1T^2\delta +\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) \Vert x-x^*\Vert . \end{aligned}$$

Hence

$$\begin{aligned} \Vert x-x^*\Vert \le \frac{\left( 1+ E\sum _{j=1}^{m} p_j\right) \frac{1}{2} b_1T^2\delta }{1-\left( 1+E\sum _{j=1}^{m} p_j\right) \left( b_1T+\frac{1}{2}b_1b_2T^2\right) }=\epsilon . \end{aligned}$$

Then the solution of the IVP (1)–(2) continuously depends on the functional f. \(\square \)

Examples

Example 8.1

Consider the nonlinear integro-differential equation

$$\begin{aligned} \frac{dx}{dt}= & {} t^4 e^{-t}+\frac{\ln (1+x(t))}{4+t^3}\nonumber \\&+\int _{0}^{t}\frac{1}{9}\left( \sin (3s+3)+\frac{ s^4\cos x(s)}{e^{|x(s)|}}\right) dt,\quad a.e \quad t\in (0, 1], \end{aligned}$$
(22)

with infinite point boundary condition

$$\begin{aligned} x(0)+\sum _{j=1}^{\infty }\frac{1}{j^2}x\left( \frac{j-1}{j}\right) =x_0. \end{aligned}$$
(23)

Set

$$\begin{aligned} g(t,x(t),\int _{0}^{t}f(s, x(s))ds)= & {} t^4 e^{-t}+\frac{\ln (1+x(t))}{4+t^3}\nonumber \\&+\int _{0}^{t}\frac{1}{9}\left( \sin (3s+3)+\frac{ s^4\cos x(s)}{e^{|x(s)|}}\right) dt.\\ \end{aligned}$$

Then

$$\begin{aligned} \left| g(t,x(t),\int _{0}^{t}f(s, x(s))ds)\right|\le & {} t^4 e^{-t}\\&+\frac{1}{4}\left( |x|+\frac{1}{4}\int _{0}^{t}\frac{4}{9}|\left( \cos (3s+3)+\frac{ s^4\cos x(s)}{e^{|x(s)|}}\right) dt|\right) , \end{aligned}$$

and also

$$\begin{aligned} |f(s, x(s))|\le \frac{4}{9}|\cos (3s+3)|+\frac{4}{9} |x(s)|. \end{aligned}$$

The assumptions 1–4 of Theorem 3.1 are satisfied with \(c_1(t)=t^3 e^{-t}\in L^1[0,1]\), \(c_2(t)= \frac{1}{2}|\cos (3t+3)|\in L^1[0,1]\), \(b_1=\frac{1}{3}\), \(b_2=\frac{4}{9}\), \(\left( 1+\frac{\sum _{j=1}^{\infty }\frac{1}{j^2}}{1+\sum _{j=1}^{\infty }\frac{1}{j^2}}\right) \left( b_1+\frac{1}{2} b_1b_2\right) =\left( 1+\frac{\frac{\pi ^2}{6}}{1+\frac{\pi ^2}{6}}\right) \left( \frac{1}{3}+\frac{2}{27}\right) <1,\) and the series: \(\sum _{j=1}^{\infty }\frac{1}{j^2}\) is convergent. Therefore, by applying to Theorem 3.1, the given IVP (22)–(23) has a solution \(x\in [0, 1]\).

Example 8.2

Consider the nonlinear integro-differential equation

$$\begin{aligned} \frac{dx}{dt}= & {} t^5 +t^2+1+\frac{x(t)}{\sqrt{t+9}}\nonumber \\&+\int _{0}^{t}\frac{1}{4}\left( \sin ^2(3s+3)+\frac{s x(s)}{2(1+x(s))}\right) dt,\quad a.e \quad t\in (0, 1], \end{aligned}$$
(24)

with infinite point boundary condition

$$\begin{aligned} x(0)+\sum _{j=1}^{\infty }\frac{1}{j^4}x\left( \frac{j^2+j-1}{j^2+j}\right) =x_0. \end{aligned}$$
(25)

Set

$$\begin{aligned} g(t,x(t),\int _{0}^{t}f(s, x(s))ds)= & {} t^5 +t^2+1+\frac{x(t)}{\sqrt{t+9}}\nonumber \\&+\frac{1}{4}\int _{0}^{t}\left( \sin ^2(3s+3)+\frac{s x(s)}{2(1+x(s))}\right) dt.\\ \end{aligned}$$

Then

$$\begin{aligned} \left| g(t,x(t),\int _{0}^{t}f(s, x(s))ds)\right|\le & {} t^5 +t^2+1+\frac{1}{3}|x|\\&+\frac{1}{3}\int _{0}^{t}\frac{3}{4}\left| \left( \sin ^2(3s+3)+\frac{s x(s)}{2(1+x(s))}\right) \right| dt, \end{aligned}$$

and also

$$\begin{aligned} |f(s, x(s))|\le \frac{3}{4}\left| \left( \sin ^2(3s+3)\right) \right| +\frac{3}{8} |x(s)|. \end{aligned}$$

The assumptions 1–4 of Theorem 3.1 are satisfied with \(c_1(t)= t^5 +t^2+1\in L^1[0,1]\), \(c_2(t)= \frac{3}{4}|(\sin ^2(3s+3)|\in L^1[0,1]\), \(b_1=\frac{1}{3}\), \(b_2=\frac{3}{8}\), \(\left( 1+\frac{\sum _{j=1}^{\infty }\frac{1}{j^4}}{1+\sum _{j=1}^{\infty }\frac{1}{j^4}}\right) \left( b_1+\frac{1}{2} b_1b_2\right) =\left( 1+\frac{\frac{\pi ^4}{90}}{1+\frac{\pi ^4}{90}}\right) \left( \frac{1}{3}+\frac{1}{16}\right) <1,\) and the series: \(\sum _{j=1}^{\infty }\frac{1}{j^4}\) is convergent Therefore, by applying to Theorem 3.1, the given IVP (24)–(25) has a solution \(x\in [0, 1]\).