Neural Computing and Applications

, Volume 22, Supplement 1, pp 133–141

Toward the existence and uniqueness of solutions of second-order fuzzy volterra integro-differential equations with fuzzy kernel

Authors

    • Department of Mathematics, Science and Research BranchIslamic Azad University
  • M. Khezerloo
    • Department of Mathematics, Science and Research BranchIslamic Azad University
  • O. Sedaghatfar
    • Department of Mathematics, Science and Research BranchIslamic Azad University
  • S. Salahshour
    • Department of Mathematics, Mobarakeh BranchIslamic Azad University
Original Article

DOI: 10.1007/s00521-012-0849-x

Cite this article as:
Allahviranloo, T., Khezerloo, M., Sedaghatfar, O. et al. Neural Comput & Applic (2013) 22: 133. doi:10.1007/s00521-012-0849-x
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Abstract

In this paper, we study existence and uniqueness of solutions of second-order fuzzy integro-differential equations with fuzzy kernel under strongly generalized differentiability. To this end, four cases are considered to show the existence of the fuzzy solution mentioned equation. Some theorems are proved, and the uniqueness of the fuzzy solution is discussed step by step. Finally, the illustrated examples are solved to investigate the conditions of theorems.

Keywords

Second-order fuzzyVolterra integro-differential equationsFuzzy valued functionsContinuous solution

1 Introduction

The fuzzy differential and integral equations are important part of the fuzzy analysis theory, and they have the important value of theory and application in control theory.

Seikkala in [19] defined the fuzzy derivative and then, some generalizations of that have been investigated in [6, 1618, 20, 22]. Consequently, the fuzzy integral which is the same as that of Dubois and Prade in [8] and, by means of the extension principle of Zadeh, showed that the fuzzy initial value problem
$$ x'(t)=f(t,x(t)),\quad x(0)=x_0 $$
has an unique fuzzy solution when f satisfies the generalized Lipschitz condition, which guarantees an unique solution of the deterministic initial value problem. Kaleva [11] studied the Cauchy problem of fuzzy differential equation, characterized those subsets of fuzzy sets in which the Peano theorem is valid. Bede et al. in [4, 5] have introduced a more general definition of the derivative for fuzzy mappings, enlarging the class of differentiable. Allahviranlo et al. [2] proposed Euler method to obtain the fuzzy solutions of hybrid fuzzy differential equations under generalized Hukuhara differentiability and in [1], a novel operator method is introduced for solving fuzzy linear differential equations under the assumption of strongly generalized differentiability.
Park et al. in [13] have considered the existence of solution of fuzzy integral equation in Banach space, and Subrahmaniam and Sudarsanam in [21] have proved the existence of solution of fuzzy functional equations. Park and Jeong [12, 14] studied existence of solution of fuzzy integral equations of the form
$$ x(t)=f(t)+\int\limits_{0}^{t}k(t,s,x(s))\,{\rm d}s,\quad t\geq 0, $$
where f(t) and x(t) are fuzzy valued functions and k is a crisp function on real numbers. The existence and uniqueness of solutions of fuzzy Volterra integro-differential equations of the second kind using strongly generalized differentiability have been discussed by Hajighasemi et al. in [10]. We try to improve [10] and we use the proposed method in [10] to prove the existence and uniqueness of solutions of second-order fuzzy Volterra integro-differential equations. In both papers, we consider equations with fuzzy kernel but in fuzzy studies, discussion about the existence of fuzzy derivative and high-order differential equations is not easy, so we consider second-order fuzzy Volterra integro-differential equations, and for future researches, we want to prove the existence and uniqueness of solutions of N-order fuzzy Volterra integro-differential equations. This paper is organized as following: in Sect. 2, the basic concepts are brought which are used throughout the paper. In Sect. 3, the main section of the paper, the existence and uniqueness of the solutions of second-order fuzzy Volterra integro-differential equations are discussed using Hausdorff metric properties and strongly generalized differentiability. Some examples are illustrated the conditions of the theorems in Sect. 4. Finally, conclusion is drawn in Sect. 5.

2 Basic concepts

Let P(\(\Re \)) denote the family of all non-empty compact convex subsets of \(\Re \) and define the addition and scalar multiplication in P(\(\Re \)) as usual. Let A and B be two non-empty bounded subsets of \(\Re \). The distance between A and B is defined by the Hausdorff metric,
$$ d(A,B)=\max\left\{\sup_{a \in A}\inf_{b\in B}\|a-b\|,\sup_{b \in B}\inf_{a\in A}\|a-b\|\right\}, $$
where \(\|.\|\) denotes the usual Euclidean norm in \(\Re \). Then it is clear that (P(\(\Re \)), d) becomes a metric space. Puri and Ralescu [16] have proved (P(\(\Re \)), d) is complete and separable. Let \(I=[0,a]\subset\Re\) be a closed and bounded interval and denote
$$ E=\{u:\Re^n\rightarrow[0,1]|u\quad satisfies \quad (i)-(iv) \quad below\}, $$
where
  1. (i)

    u is normal, that is, there exists an \(x_0\in\Re\) such that u(x0) = 1,

     
  2. (ii)

    u is fuzzy convex,

     
  3. (iii)

    u upper semicontinuous,

     
  4. (iv)

    \([u]^0=cl\{x\in\Re|u(x)>0\}\) is compact.

     
For 0 < α ≤ 1 denote \([u]^\alpha=\{x\in\Re|u(x)\geq \alpha\}.\) Then from (i)–(iv), it follows that the α-level set \([u]^\alpha\in P(\Re)\) for all 0 < α ≤ 1.

Also, set E is named as the set of all fuzzy real numbers. Obviously \(\Re\subset E\).

Definition 2.1 [7]

An arbitrary fuzzy number u in the parametric form is represented by an ordered pair of functions \((\underline{u},\overline{u})\) which satisfy the following requirements:
  1. (i)

    \(\overline{u}:\alpha \longrightarrow \overline{u}^{\alpha}\in \Re\) is a bounded left-continuous non-decreasing function over [0, 1],

     
  2. (ii)

    \(\underline{u}:\alpha \longrightarrow \underline{u}^{\alpha}\in \Re\) is a bounded left-continuous non-increasing function over [0, 1],

     
  3. (iii)

    \(\underline{u}^{\alpha} \leq \overline{u}^{\alpha}, \quad 0 \leq \alpha \leq 1\).

     
Let \(D:E\times E\rightarrow \Re^+\cup \{0\}\) be defined by
$$ D(u,v)=\sup _{0\leq \alpha\leq 1}d([u]^\alpha,[v]^\alpha), $$
where d is the Hausdorff metric defined in (P(\(\Re \)), d). Then D is a metric on E. Further, (ED) is a complete metric space [11, 16].

Remark 2.1 [3]

$$ D(u, 0)=D(0, u)=\parallel u \parallel_{\infty} $$
Then, it is easy to see that D has the following properties (see [15]):
  1. (i)

    D(u + wv + w) = D(uv) for every \(u,v,w\in E\),

     
  2. (ii)

    \(D(u+v,\tilde{0})\leq D(u,\tilde{0})+D(v,\tilde{0})\) for every \(u,v\in E\),

     
  3. (iii)

    \(D(u\cdot v,\tilde{0})=\|u\cdot v\|_{\infty}\leq \|u\|_{\infty}\cdot \|v\|_{\infty}=D(u,\tilde{0})D(v,\tilde{0})\) for every \(u,v\in E\),

     
  4. (iv)

    D(u + vw + z) ≤ D(uw) + D(vz) for uvw, and \(z\in E\).

     

Definition 2.2

A mapping \(x:I\rightarrow E\) is bounded, if there exists r > 0 such that
$$ D(x(t),\tilde{0})<r\quad \forall t\in I. $$

Definition 2.3 [9]

Let \(f:{\Re} \rightarrow {E}\) be a fuzzy valued function. If for arbitrary fixed \(t_{0} \in \Re\) and \(\epsilon > 0\), a δ > 0 such that
$$ |t-t_{0}|< \delta \Rightarrow D(f(t), f(t_{0})) < \epsilon, $$
f is said to be continuous.

Definition 2.4 [3]

Consider \(u,v\in E\). If there exists \(w\in E\) such that u = v + w, then w is called the H-difference of u and v and is denoted by \(u\ominus v\).

Definition 2.5 [3]

Let \(f:(a, b)\times E\rightarrow E\) and \(x_{0} \in (a, b)\). We Define the nth-order differential of f as follow:

We say that f is strongly generalized differentiable of the nth-order at x0. If there exists an element \(f^{(s)}(x_{0})\in E , \quad \forall s=1, \ldots, n\), such that
  1. (i)
    for all h > 0 sufficiently small,
    $$ \exists f^{(s-1)}(x_{0}+h)\ominus f^{(s-1)}(x_{0}),\quad \quad \exists f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}-h) $$
    and the limits (in the metric \( d_{\infty} \))
    $$ \lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0}+h)\ominus f^{(s-1)}(x_{0})}{h}=\lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}-h)}{h}=f^{(s)}(x_{0}) $$
    or
     
  2. (ii)
    for all h > 0 sufficiently small,
    $$ \exists f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}+h), \quad \quad \exists f^{(s-1)}(x_{0}-h)\ominus f^{(s-1)}(x_{0}) $$
    and the limits (in the metric \(d_{\infty}\))
    $$ \lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}+h)}{-h}=\lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0}-h)\ominus f(x_{0})}{-h}=f^{(s)}(x_{0}) $$
    or
     
  3. (iii)
    for all h > 0 sufficiently small,
    $$\exists f^{(s-1)}(x_{0}+h)\ominus f^{(s-1)}(x_{0}), \quad \quad \exists f^{(s-1)}(x_{0}-h)\ominus f^{(s-1)}(x_{0})$$
    and the limits (in the metric \(d_{\infty}\))
    $$ \lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0}+h)\ominus f^{(s-1)}(x_{0})}{h}=\lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0}-h)\ominus f^{(s-1)}(x_{0})}{-h}=f^{(s)}(x_{0}) $$
    or
     
  4. (iv)
    for all h > 0 sufficiently small,
    $$ \exists f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}+h), \quad \quad \exists f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}-h) $$
    and the limits (in the metric \(d_{\infty}\))
    $$ \lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}+h)}{-h}=\lim _{h \searrow 0}\frac{f^{(s-1)}(x_{0})\ominus f^{(s-1)}(x_{0}-h)} {h}=f^{(s)}(x_{0}) $$
    (h and −h at denominators mean \(\frac{1}{h}\) and \(\frac{-1} {h}\), respectively \(\forall s=1\ldots n\)).
     

It was proved by Puri and Relescu [16] that a strongly measurable and integrably bounded mapping \(F:I\rightarrow E\) is integrable (i.e., \( \int_IF(t)\,{\rm d}t\in E\)).

Theorem 2.1 [21]

If\(F:I\rightarrow E\)is continuous, then it is integrable.

Theorem 2.2 [21]

Let\(F,G:I\rightarrow E\)be integrable and\(\lambda\in \Re.\)Then
  1. (i)

    \(\int_I(F(t)+G(t))\,{\rm d}t=\int_IF(t)\,{\rm d}t+\int_IG(t)\,{\rm d}t,\)

     
  2. (ii)

    \(\int_I\lambda F(t)\,{\rm d}t=\lambda\int_IF(t)\,{\rm d}t,\)

     
  3. (iii)

    D(FG) is integrable,

     
  4. (iv)

    \(D(\int_IF(t)\,{\rm d}t,\int_IG(t)\,{\rm d}t)\leq \int_ID(F(t),G(t))\,{\rm d}t.\)

     

Theorem 2.3 [5]

For\(t_0\in\Re,\)the fuzzy differential equation\(x'=f(t,x),\quad x(t_0)=x_0\in E,\)where\(f:\Re \times E\rightarrow E\)is supposed to be continuous, is equivalent to one of the integral equations
$$ x(t)=x_0+\int\limits_{t_0}^tf(s,x(s))\,{\rm d}s,\quad \forall t\in[t_0,t_1] $$
or
$$ x(t_0)=x(t)+(-1).\int\limits_{t_0}^tf(s,x(s))\,{\rm d}s, \quad \forall t\in[t_0,t_1] $$
on some interval\((x_0,x_1)\subset \Re,\)under the strong differentiability condition, (i) or (ii), respectively. Here the equivalence between two equation means that any solution of an equation is a solution for the other one.

Lemma 2.1 [3]

If the H-difference for arbitrary\(u,v\in E\)and\(u,w\in E\)exists, we have
$$ D(u\ominus v,u\ominus w)=D(v,w),\quad \forall u,v,w\in E $$

Lemma 2.2 [10]

If the H-difference for arbitrary\(u,v\in E\)exists, we have
$$ D(u\ominus v,\tilde{0})=D(u,v),\quad \forall u,v\in E $$

Lemma 2.3 [10]

For arbitrary\(u,v\in E\)we have
$$ D(u\, \circleddash\, v,w\, \circleddash\, z)\leq D(u,w)+D(v,z),\quad \forall u,v,w,z\in E $$

3 Second-order fuzzy Volterra integro-differential equation

We consider the following second-order fuzzy Volterra integro-differential equation (FVIDE)
$$ \left\{ \begin{array}{l} x^{\prime\prime}(t)=f(t)+\int\limits_{0}^{t}k(t,s)g(s,x(s))\,{\rm d}s,\quad t\geq 0\\ x(0)=\tilde{c_1},\quad x^{\prime}(0)=\tilde c_2\quad \tilde{c_1},\tilde c_2\in E, \end{array} \right. $$
(3.1)
where \(f:[0,a]\rightarrow E\) and \(k:\Updelta\rightarrow E\), where \(\Updelta=\{(t,s):0\leq s\leq t\leq a\leq 1\}\), and \(g:[0,a]\times E\rightarrow E\) are continuous.

Theorem 3.1

Consider Eq. (3.1). A mapping\(x:[0,1]\rightarrow E\)is a solution to the second-order FVIDE (3.1) if and only ifxandxare continuous and satisfy one of the following conditions:
  1. (a)

    \(x(z)=\tilde c_1+\tilde c_2z+\int_0^z\int_0^uf(t)\,{\rm d}t\,{\rm d}u+\int_0^z \int_0^u\int_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\)

    wherexandxare (i)-differentiable.

     
  2. (b)

    \(x(z)=\tilde c_1\ominus (-1)\left(\tilde c_2z+\int_0^z\int_0^uf(t)\,{\rm d}t\,{\rm d}u+\int_0^z \int_0^u\int_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\)

    wherexis (ii)-differentiable andxis (i)-differentiable.

     
  3. (c)

    \(x(z)=\tilde c_1+\left(\tilde c_2z\ominus (-1)\int_0^z\int_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int_0^z\int_0^u\int_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\)

    wherexis (i)-differentiable andxis (ii)-differentiable.

     
  4. (d)

    \(x(z)=\tilde c_1\ominus (-1)\left(\tilde c_2z\ominus (-1)\int_0^z\int_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int_0^z\int_0^u\int_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\)

    wherexandxare (ii)-differentiable.

     

Proof

Since right side of Eq. (3.1) is continuous, it must be integrable [5]. So Eq. (3.1) can be written in each case as follows:
  1. (a)
    Let x and x′ are (i)-differentiable. So, for
    $$ x^{\prime\prime}(t)=f(t)+\int\limits_{0}^{t}k(t,s)g(x,x(s))\,{\rm d}s,\quad t\geq 0 $$
    we have equivalently
    $$ x^{\prime}(u)=\tilde c_2+\int\limits_0^uf(t)\,{\rm d}t+\int\limits_0^u \int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t,\quad\quad 0\leq u\leq 1 $$
    and thus for 0 ≤ z ≤ 1,
    $$ x(z)=\tilde c_1+\tilde c_2z+\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u+\int\limits_0^z \int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u $$
     
  2. (b)
    Let x is (ii)-differentiable and x′ is (i)-differentiable. So, for
    $$ x^{\prime\prime}(t)=f(t)+\int\limits_{0}^{t}k(t,s) g(x,x(s))\,{\rm d}s,\quad t\geq 0 $$
    we have equivalently
    $$ x^{\prime}(u)=\tilde c_2+\int\limits_0^uf(t)\,{\rm d}t+\int\limits_0^u\int\limits_0^t k(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t,\quad\quad 0\leq u\leq 1 $$
    and so for 0 ≤ z ≤ 1,
    $$ x(z)=\tilde c_1\ominus (-1)\left(\tilde c_2z+\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u+\int\limits_0^z \int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right) $$
     
  3. (c)
    Let x is (i)-differentiable and x′ is (ii)-differentiable. So, for
    $$ x^{\prime\prime}(t)=f(t)+\int\limits_{0}^{t} k(t,s)g(x,x(s))\,{\rm d}s,\quad t\geq 0 $$
    we have equivalently
    $$ x^{\prime}(u)=\tilde c_2\ominus (-1)\int\limits_0^uf(t)\,{\rm d}t\ominus (-1)\int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t,\quad\quad 0\leq u\leq 1 $$
    and so for 0 ≤ z ≤ 1,
    $$ x(z)=\tilde c_1+\left(\tilde c_2z\ominus (-1) \int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1) \int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right) $$
     
  4. (d)
    Let x and x′ are (ii)-differentiable. So, for
    $$ x^{\prime\prime}(t)=f(t)+\int\limits_{0}^{t}k(t,s)g(x,x(s))\,{\rm d}s,\quad t\geq 0 $$
    we have equivalently
    $$ x^{\prime}(u)=\tilde c_2\ominus (-1)\int\limits_0^uf(t)\,{\rm d}t \ominus (-1)\int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t,\quad\quad 0\leq u\leq 1 $$
    and so for 0 ≤ z ≤ 1,
    $$ x(z)=\tilde c_1\ominus (-1)\left(\tilde c_2z \ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1) \int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right) $$
    (3.2)
    which complete the proof.
     
\(\square\)

Now, we are going to study existence and uniqueness of the solution to Eq. (3.1). Since the proof procedure is similar for all four cases, we consider case (d) without loss of generality.

Theorem 3.2

LetaandLbe positive numbers. Assume that Eq. (3.1) satisfies the following conditions:
  1. (i)
    \(f:[0,a]\rightarrow E\)is continuous and bounded, i.e., there existsR > 0 such that
    $$ D(f(t),\tilde 0)\leq R $$
     
  2. (ii)
    \(k:\Updelta\rightarrow E\)is continuous where\(\Updelta=\{(t,s):0\leq s\leq t\leq a\leq 1\}\)and there existsM > 0 such that
    $$ \int\limits_0^tD(k(s,t),\tilde 0)\,{\rm d}s\leq M $$
     
  3. (iii)
    \(g:[0,a]\times E\rightarrow E\)is continuous and satisfies the Lipschitz condition, i.e.,
    $$ D(g(t,x(t)),g(t,y(t))) \leq D(g(t,x(t)),\tilde 0)+D(g(t,y(t)),\tilde 0)\leq LD(x(t),y(t)) $$
    (3.3)
    whereL < M−1and\(x,y: [0,a]\rightarrow E.\)
     
  4. (iv)

    \(g(t,\tilde 0)\)is bounded on [0, a].

     
Then there exists an unique solution x(z) of Eq. (3.1) on [0, 1] and the successive iterations
$$ \begin{aligned} x_0(z)&=\tilde c_1\ominus (-1)\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\right)\\ x_{n+1}(z)&=\tilde c_1\ominus (-1)\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_n(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\qquad 0\leq t\leq a\leq 1, \quad 0\leq u\leq 1, \quad 0\leq z\leq 1, \quad \tilde{c_1},\tilde c_2\in E \end{aligned} $$
(3.4)
are uniformly convergent to x(z) on [0,1] if x and x′ are (ii)-differentiable.

Proof

It is easy to see that all xn(z) are bounded on [0,1]. Indeed x0(z) = f(z) is bounded by hypothesis. Assume that xn+1(z) is bounded, from Lemma (2.3), we have
$$ \begin{aligned} D(x_n(z),\tilde 0)&=D\left(\tilde c_1\ominus (-1)\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right),\tilde 0\right)\\ &=D\left(\tilde c_1, (-1)\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\right)\\ &\leq D(\tilde c_1,\tilde 0)+D\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\tilde 0\right)\\ &\leq D(\tilde c_1,\tilde 0)+D\left(\tilde c_2z\ominus (-1)\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u, (-1)\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq D(\tilde c_1,\tilde 0)+D(\tilde c_2z,\tilde 0)+D\left(\int\limits_0^z\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u,\tilde 0\right)\\ &\quad+D\left(\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\tilde 0\right)\\ &\leq D(\tilde c_1,\tilde 0)+D(\tilde c_2z,\tilde 0)+\int\limits_0^z\int\limits_0^uD(f(t),\tilde 0)\,{\rm d}t\,{\rm d}u\\ &\quad+\int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t)g(s,x_{n-1}(s)),\tilde 0)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq D(\tilde c_1,\tilde 0)+D(\tilde c_2z,\tilde 0)+Ruz+\int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t),\tilde 0)D(g(s,x_{n-1}(s)),\tilde 0)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq D(\tilde c_1,\tilde 0)+D(\tilde c_2z,\tilde 0)+Ruz \\ &\quad+\int\limits_0^z\int\limits_0^u\left(\sup_{0\leq t\leq a}D(g(t,x_{n-1}(t)),\tilde 0)\cdot\int\limits_0^tD(k(s,t),\tilde 0)\,{\rm d}s\right)\,{\rm d}t\,{\rm d}u \end{aligned} $$
Taking every assumptions into account
$$ \begin{aligned} D(g(t,x_{n-1}(t)),\tilde 0)&\leq D(g(t,x_{n-1}(t)),g(t,\tilde 0))+D(g(t,\tilde 0,\tilde 0)\\ &\leq LD(x_{n-1}(t),\tilde 0)+D(g(t,\tilde 0),\tilde 0) \end{aligned} $$
Since 0 ≤ u ≤ 1 and 0 ≤ z ≤ 1, we obtain that xn(z) is bounded. Thus, xn(z) is a sequence of bounded functions on [0, 1]. Next, we prove that xn(z) are continuous on [0, 1]. For 0 ≤ z1 ≤ z2 ≤ 1, from Lemma (2.2) and (2.3), we have
$$ \begin{aligned} D(x_n(z_1),x_n(z_2))&=D\left((-1)\int\limits_0^{z_1}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^{z_1}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\right.\\ &\qquad\left. (-1)\int\limits_0^{z_2}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\ominus (-1)\int\limits_0^{z_2}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq D\left(\int\limits_0^{z_1}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u,\int\limits_0^{z_2}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\right)\\ &+D\left(\int\limits_0^{z_1}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u, \int\limits_0^{z_2}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq D\left(\int\limits_0^{z_1}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u,\int\limits_0^{z_1} \int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u+\int\limits_{z_1}^{z_2}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u\right)\\ &\quad+D\left(\int\limits_0^{z_1}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\right.\\ &\left.\qquad \int\limits_0^{z_1}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u+\int\limits_{z_1}^{z_2} \int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq D\left(\int\limits_{z_1}^{z_2}\int\limits_0^uf(t)\,{\rm d}t\,{\rm d}u,\tilde 0\right) +D\left(\int\limits_{z_1}^{z_2}\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\tilde 0\right) \end{aligned} $$
Clearly,
$$D(x_n(z_1),x_n(z_2))\rightarrow 0 \quad as \quad z_1\rightarrow z_2$$
Thus, the sequence xn(z) is continuous on [0, 1]. Condition (iii) and its analog corresponding to n + 1 will give for n ≥ 1 and lemma (2.1), (2.2) and (2.3), we get
$$ \begin{aligned} D(x_{n+1}(z),x_n(z))&=D\left(\int\limits_0^z\int\limits_0^u\int\limits_0^t k(s,t)g(s,x_{n}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\int\limits_0^z\int\limits_0^u\int\limits_0^t k(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t)g(s,x_{n}(s)),k(s,t)g(s,x_{n-1}(s)))\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^t\left(D(k(s,t)g(s,x_{n}(s)),\tilde 0)+D(k(s,t)g(s,x_{n-1}(s)),\tilde 0)\right)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t),\tilde 0)\left(D(g(s,x_{n}(s)),\tilde 0)+D(g(s,x_{n-1}(s)),\tilde 0)\right)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\left(\sup_{0\leq t\leq a}\left(D(g(s,x_{n}(t)),\tilde 0)+D(g(s,x_{n-1}(t)),\tilde 0)\right)\int\limits_0^tD(k(s,t),\tilde 0)\,{\rm d}s\right)\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^uML\sup_{0\leq t\leq a}D(x_{n}(t),x_{n-1}(t))\,{\rm d}t\,{\rm d}u \end{aligned} $$
Thus, we get
$$ \sup_{0\leq z\leq 1}D(x_{n+1}(z),x_n(z))\leq \int\limits_0^z\int\limits_0^uML\sup_{0\leq t\leq a}D(x_{n}(t),x_{n-1}(t))\,{\rm d}t\,{\rm d}u $$
(3.5)
For n = 0, we have
$$ \begin{aligned} D(x_{1}(z),x_0(z))&= D\left(\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_0(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u,\tilde 0\right)\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t)g(s,x_0(s)),\tilde 0)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t),\tilde 0)D(g(s,x_0(s)),\tilde 0)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u \left(\sup_{0\leq t\leq a}D(g(t,x_0(t)),\tilde 0)\int\limits_0^tD(k(s,t),\tilde 0)\,{\rm d}s\right)\,{\rm d}t\,{\rm d}u \end{aligned} $$
(3.6)
so,
$$ \sup_{0\leq z\leq 1}D(x_{1}(z),x_0(z))\leq NM\frac{z^2}{2}\leq NMz $$
where \(N=\sup_{0\leq t\leq a}D(g(t,x_0(t)),\tilde 0)\) and \(z\in [0,1]\). Moreover, we derive
$$ \sup_{0\leq z\leq 1}D(x_{n+1}(z),x_n(z))\leq z^{n+1}L^nM^{n+1}N $$
(3.7)
which shows that the series \(\sum_{n=1}^{\infty}D(x_n(z),x_{n-1}(z))\) is dominated, uniformly on [0, 1], by the series \(zMN\sum_{n=1}^{\infty}(zLM)^n\). But (3.3) and z ≤ 1 guarantees the convergence of the last series, implying the uniform the convergence of the sequence xn(z). If we denote \(x(z)=\lim_{n\rightarrow \infty}x_n(z)\), then x(z) satisfies (3.2). It is obviously continuous on [0, 1] and bounded. To prove the uniqueness, let y(z) be a continuous solution of (3.1) on [0, 1]. Then
$$ y^{\prime}(t)=f(t)+\int\limits_0^tk(s,t)g(s,y(s))\,{\rm d}s $$
(3.8)
From (3.3), we obtain for n ≥ 1,
$$ \begin{aligned} D(y(z),x_n(z))&=D\left(\int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,y(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u, \int\limits_0^z\int\limits_0^u\int\limits_0^tk(s,t)g(s,x_{n-1}(s))\,{\rm d}s\,{\rm d}t\,{\rm d}u\right)\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t)g(s,y(s)), k(s,t)g(s,x_{n-1}(s)))\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^t\left(D(k(s,t)g(s,y(s)),\tilde 0)+D(k(s,t)g(s,x_{n-1}(s)),\tilde 0)\right)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\int\limits_0^tD(k(s,t),\tilde 0)\left(D(g(s,y(s)),\tilde 0)+D(g(s,x_{n-1}(s)),\tilde 0)\right)\,{\rm d}s\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^u\left(\sup_{0\leq t\leq a}\left(D(g(s,y(t)),\tilde 0)+D(g(s,x_{n-1}(t)),\tilde 0)\right)\int\limits_0^tD(k(s,t),\tilde 0)\,{\rm d}s\right)\,{\rm d}t\,{\rm d}u\\ &\leq \int\limits_0^z\int\limits_0^uML\sup_{0\leq t\leq a}D(y(t),x_{n-1}(t))\,{\rm d}t\,{\rm d}u\\ &\quad \vdots\\ &\leq \int\limits_0^z\int\limits_0^u(ML)^n\sup_{0\leq t\leq a}D(y(t),x_0(t))\,{\rm d}t\,{\rm d}u \end{aligned} $$
Since ML < 1,
$$ \lim_{n\rightarrow \infty}x_n(z)=y(z)=x(z), \quad \quad 0\leq z\leq 1 $$
which ends the proof of the theorem. \(\square\)

4 Numerical examples

Example 4.1

Consider the following second-order fuzzy Volterra integro-differential equation
$$ \left\{\begin{array}{l} x^{\prime\prime}(t)=\left((3r-3)\sin t+\frac{3} {20}(r^2-6r+5)t\hbox{cos}^2 t, (9-9r)\sin t-\frac{9} {20}(r^2-6r+5)t\hbox{cos}^2 t\right)\\ \qquad +\int\limits_{0}^{t}(0.1+0.3r,0.5-0.1r)\, t\, \hbox{cos} s \,x(s)\,{\rm d}s,\quad 0\leq t\leq1\\ x(0)=(0,0),\quad x^{\prime}(0)=(-9+9r,3-3r) \end{array} \right. $$
where holds in condition of Theorem (3.2). Using case (c) in Theorem (3.1), we get:
$$ x(t)=(-9+9r,3-3r)\sin t $$
It is clear that x is (i)-differentiable and x′ is (ii)-differentiable.

Example 4.2

Consider the following second-order fuzzy Volterra integro-differential equation
$$ \left\{\begin{array}{l} x^{\prime\prime}(t)=\left((-10+5r)e^{1-t}-(1+0.5r)t, (-3-2r)e^{1-t}-(0.6+0.1r-0.2r^2)t\right)\\ \qquad +\int\limits_{0}^{t}(-0.2+0.1r,-0.1)e^{s-1}x(s)\,{\rm d}s,\quad 0\leq t\leq 1\\ x(0)=(-10+5r,-3-2r)e,\quad x^{\prime}(0)=(3+2r,10-5r)e \end{array} \right. $$
where holds in condition of Theorem (3.2). Using case (d) in Theorem (3.1), we get:
$$ x(t)=(-10+5r,-3-2r)e^{1-t} $$
It is clear that x and x′ are (ii)-differentiable.

Example 4.3

Consider the following second-order fuzzy Volterra integro-differential equation
$$ \left\{ \begin{array}{l} x^{\prime\prime}(t)=\left((-11+2r)\hbox{cos} t-(0.1r+0.8r^2)t, (-1-8r)\hbox{cos} t-(2.75-2.15r+0.3r^2)t\right)\\ \qquad +\int\limits_{0}^{t}(0.1r,0.25-0.15r)\,\frac{1}{\hbox{cos} s}\,x(s)\,{\rm d}s,\quad 0 \leq t\leq 1\\ x(0)=(1+8r,11-2r),\quad x^{\prime}(0)=(0,0) \end{array} \right. $$
where holds in condition of Theorem (3.2). Using case (b) in Theorem (3.1), we get:
$$ x(t)=(1+8r,11-2r)\hbox{cos} t $$
It is clear that x is (ii)-differentiable and x′ is (i)-differentiable.

5 Conclusion

In this work, we proved the existence and uniqueness of solution of second-order fuzzy Volterra integro-differential equations with fuzzy kernel under strongly generalized differentiability, which is the first attempt in the fuzzy literatures. To do this, we considered four cases and discussed about kind of differentiability. Also, we used three illustrated examples to investigate the conditions of theorems.

Acknowledgments

The authors would like to present their sincere thanks to the Editor in Chief for his valuable suggestions.

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© Springer-Verlag London Limited 2012