An Efficient Method for Solving Second-Order Fuzzy Order Fuzzy Initial Value Problems

Abstract

In this paper, we present an accurate numerical approach based on the reproducing kernel method (RKM) for solving second-order fuzzy initial value problems (FIVP) with symmetry coefficients such as symmetric triangles and symmetric trapezoids. Finding the exact solution of FIVP is not an easy task since the definition will produce a complicated optimization problem. To overcome this difficulty, a numerical method is developed to solve this type of problems. We start by introducing the necessary definitions and theorems about the fuzzy logic. Then, we derived the kernels for two Hilbert spaces. The RKM is derived for the second-order IVP in the Boolean sense, and then we generalize it for the fuzzy sense. Numerical and theoretical results will be given to obtain the accuracy of the developed technique. We solved four linear and non-linear fuzzy IVPs numerically using the proposed method, and we compute the error in each case to show the efficiency of the method. The absolute error was very small in the four examples.

1. Introduction

In recent years, fuzzy logic has become attractive to many researchers due to its potential applications in various fields, such as computer science [1], information science [2], mathematics [3], engineering [4], economics [5], and business and finance [6]. Fuzzy logic and fuzzy sets are powerful mathematical tools in modeling entropy systems, for example, in industry, nature, and the humanities. Several researchers have studied the fuzzy boundary value problems. For example, Sanchez et al. [7] discussed the fuzzy solution for nonlinear fuzzy boundary value problem. Gong [8] illustrated discontinuous FIVPs. Zhou et al. [9] illustrated numerous duality outcomes for fuzzy problems with fuzzy coefficients. Tapaswini et al. [10] used the Galerkin method for solving $n\mathrm{th}$-order FBVPs. Gumah et al. [11] used hybrid FODE with a RKM for FDE. Patel and Desai [12] used Laplace transform to solve FIVPs. Diniz et al. [13] investigated the necessary conditions to solve a fuzzy problems using Zadeh’s extension. Wu and Feng [14] used control problems to explain a boundary-disturbed uncertain beam equation. Suhhiem [15] introduced a modified method for solving second-order FDEs. Niu et al. [16] proposed a simplified RKM to solve singular boundary value problems. Shah and Wang [17] developed a numerical technique to solve fuzzy FDEs. Pradip [18] discussed a class of singular BVPs. Wasques et al. [19] illustrated a numerical technique for higher FIVPs. Wasques et al. [20] investigated FIVPs. Jeyaraj and Rajan [21] used RKM#4 to study FIVPs. Al-Refai et al. [22] used the IHBM to solve FIVPs.

Several researchers study the RKM. For example, Kashkari and Syam [23] used the RKM for solving Fredholm integro-differential equation (FIDE). Du et al. [24] developed an RKM for solving FIDEs. Akgül [25] applied the RKM to FDEs. Akgül [26] investigated the boundary layer flow of a Powell–Eyring non-Newtonian fluid over a stretching sheet by RKM. A singular kernel is implemented by Sadamoto et al. [27] Gholami et al. [28] studied the fuzzy inner product space. Mei [29] simplified the RKM to solve integral equations. Geng [30] investigated a class of singularly perturbed delay BVPs by RKM. Li and Wu [31] constructed and applied reproducing kernels with polynomials. Moradi and Javadi [32] used the RKM for studying oscillators under the damping effect. Alvandi and Paripour [33] implemented the RKM for FIDEs. Arqub et al. [34] used RKM for studying FIDEs. Saadeh et al. [35] implemented an iterative RKM to solve BVPs. Qi et al. [36] introduced an RKM for solving FBVPs.

The purpose of this paper is to find an accurate numerical solution of the fuzzy initial value problems of a second order. The proposed methods, which are given in this article, are of high orders of precision and are very close to the exact solutions. Another advantage to these methods is that they can be implemented even when it is impossible to find the exact solution in the closed form. To reach this target, the reproducing kernel method will be applied to second-order initial value problems. Then, the suggested methods will be elaborated to solve the fuzzy type of these problems using some properties of fuzzy operations. Moreover, the convergence of the suggested method will be examined. In addition, various examples to represent the accuracy of the suggested methods are demonstrated. It is worth mentioning that these numerical methods can be used for other fuzzy problems such as fuzzy eigenvalue problems and fuzzy boundary value problems when we combine them with the shooting method.

We divide this paper into five sections. This section is devoted to the literature review, while, in Section 2, we present the necessary preliminaries which we will use. We present the RKM for solving second-order fuzzy IVPs with symmetric coefficients, such as symmetric triangles and symmetric trapezoids, in Section 3. In Section 4, we present some numerical results, and finally, we draw some conclusions in Section 5.

2. Preliminaries

In this section, the definition of fuzzy number and the differentiation of fuzzy functions with related definitions and theorems will be illustrated

Definition 1

([37]). Let ℜ be the set of real numbers. A fuzzy number is a function $\mu :\Re \to [0,1]$ with the following:

• μ is normal, i.e., $\mu \left(c\right)=1$ for some $c\in \Re$;
• μ is a fuzzy convex, i.e., $\mu (\theta c_1+(1-\theta )c_2)\ge min\{\mu \left(c_1\right),\mu \left(c_2\right)\}$ for all $\theta \in [0,1],c_1,c_2\in \Re$;
• μ is upper semi-continuous on $\Re ,$;
• $\overline{\{x\in \Re :\mu (x)>0\}}$ is compact.

Let $F_{\Re }$ be the collection of all fuzzy numbers.

Next, the $\alpha -cut$ of a fuzzy number will be illustrated.

Definition 2

([2]). Let $\mu \in F_{\Re }.$ Then, the $\alpha -$cut set ${\mu }_{\alpha }$ for $\alpha \in (0,1]$ is

$${\mu }_{\alpha }=\{x\in \Re :\mu \left(x\right)\ge \alpha \},$$

and

$${\mu }_0=\overline{\{x\in \Re :\mu (x)>0\}}.$$

Definition 3

([3]). Let A and B be two subsets of ℜ. Then, the Hausdorff metric $d_H$ is given as:

$$d_H(A,B)=max\left\{\underset{x\in A}{sup}\underset{y\in B}{inf}∥x-y∥,\underset{y\in B}{sup}\underset{x\in A}{inf}∥x-y∥\right\}.$$

Then, the metric $d_F$ on $F_{\Re }$ is given as:

$$\begin{array}{ccc}\hfill d_F(u,v)& =& \underset{\alpha \in [0,1]}{sup}\{d_H(u_{\alpha },v_{\alpha }),u_{\alpha },v_{\alpha }\in F_{\Re }\},\hfill \\ & =& \underset{\alpha \in [0,1]}{sup}max\{\left|{\underline{u}}_{\alpha }-{\underline{v}}_{\alpha }\right|,\left|{\overline{u}}_{\alpha }-{\overline{v}}_{\alpha }\right|\}.\hfill \end{array}$$

Theorem 1

([4]). $(F_{\Re },d_F)$ is a complete metric space such that for all $u,v,w,z\in F_{\Re },\lambda \in \Re$:

$$d_F(u\oplus w,v\oplus w)=d_F(u\oplus v),$$

$$d_F(\lambda \odot u,\lambda \odot w)=\left|\lambda \right|d_F(u,w),$$

$$d_F(u\oplus v,w\oplus z)\le d_F(u\oplus w)+d_F(v\oplus z).$$

Definition 4

([37]). The Hukhara difference of two fuzzy numbers, a and b, is $c=a{\ominus }_{H}b$ if

$$c\oplus b=a.$$

Definition 5

([5]). Let X be a subset of $\Re .$ A fuzzy function $F:X\to F_{\Re }$ is called H-differentiable at $x_0\in X$ if and only if the following limit exists and equals $Df\left(x_0\right)$:

$$\begin{array}{ccc}\hfill Df\left(x_0\right)& =& \underset{\Delta \to 0^{+}}{lim}\frac{1}{\Delta }\odot \left(f(x_0+h){\ominus }_{H}f\left(x_0\right)\right),\hfill \\ & =& \underset{\Delta \to 0^{+}}{lim}\frac{1}{\Delta }\odot \left(f\left(x_0\right){\ominus }_{H}f(x_0-\Delta )\right).\hfill \end{array}$$

If f is Hukuhara differentiable for all $x\in X,$ then f is H-differentiable on X.

Theorem 2

([6]). Let $f\left(x\right):I\to F_{\Re }$ be a fuzzy function defined by

$$f\left(x\right)=u\odot h\left(x\right),$$

where u is a fuzzy number and $I=(u,v)\subset \Re .$ Let $h:I\to {\Re }_{+}$ be differentiable function at x $\in I.$ If $h^{\prime }\left(x\right)>0,$ then:

$$f^{\prime }\left(x\right)=u\odot h^{\prime }\left(x\right).$$

(1)

Theorem 3

([37]). Let $f:M\to F_{\Re }$ be an H-differentiable at $x_0$ with derivative $f^{{}^{\prime }}\left(x_0\right)$, where $M\subset \Re$ and $x_0\in M$. Then, $f_{\alpha }^{\prime }\left(x_0\right)=\left[\underline{f^{{}^{\prime }}}\left(x_0\right),{\overline{f}}^{\prime }\left(x_0\right)\right]$ and $\underline{f}\left(x\right),\overline{f}\left(x\right)$ are differentiable at $x_0$ for all $\alpha \in [0,1]$.

Now, consider the following fuzzy differential equation:

$$y^{″}=f\left(t,y,y^{\prime }\right),a<t<b,$$

(2)

If Equation (2) is a linear problem, then it can be written as:

$$a\left(t\right)y^{″}+b\left(t\right)y^{\prime }+c\left(t\right)y=r\left(t\right),a<t<b,$$

where $a,b,c,andr$ are fuzzy functions. Since the functions are fuzzy, then the linear fuzzy problem can be written in the $\alpha$-cut format as:

$$\begin{array}{c}\left[{\underline{a}}_{\alpha }\left(t\right),{\overline{a}}_{\alpha }\left(t\right)\left]\right[{\underline{y}}^{″}\left(t\right),{\overline{y}}^{″}\left(t\right)\right]+\left[{\underline{b}}_{\alpha }\left(t\right),{\overline{b}}_{\alpha }\left(t\right)\left]\right[{\underline{y}}_{\alpha }^{\prime }\left(t\right),{\overline{y}}_{\alpha }^{\prime }\left(t\right)\right]\hfill \\ +\left[{\underline{c}}_{\alpha }\left(t\right),{\overline{c}}_{\alpha }\left(t\right)\right]\left[{\underline{y}}_{\alpha }\left(t\right),{\overline{y}}_{\alpha }\left(t\right)\right]=\left[{\underline{r}}_{\alpha }\left(t\right),{\overline{r}}_{\alpha }\left(t\right)\right].\hfill \end{array}$$

Thus, we will obtain two complicated optimization problems:

$$\begin{array}{c}min\left\{{\underline{a}}_{\alpha }\underline{y^{″}},{\underline{a}}_{\alpha }{\overline{y}}^{″},{\overline{a}}_{\alpha }{\underline{y}}^{″},{\overline{a}}_{\alpha }{\overline{y}}^{″}\right\}+min\left\{{\underline{b}}_{\alpha }\underline{y^{\prime }},{\underline{b}}_{\alpha }{\overline{y}}_{\alpha }^{\prime },{\overline{b}}_{\alpha }{\underline{y}}_{\alpha }^{\prime },{\overline{b}}_{\alpha }{\overline{y}}_{\alpha }^{\prime }\right\}\\ +min\left\{{\underline{c}}_{\alpha }{\underline{y}}_{\alpha },{\underline{c}}_{\alpha }{\overline{y}}_{\alpha },{\overline{c}}_{\alpha }{\underline{y}}_{\alpha },{\overline{c}}_{\alpha }{\overline{y}}_{\alpha }\right\}={\underline{r}}_{\alpha }\left(t\right),\end{array}$$

and

$$\begin{array}{c}max\left\{{\underline{a}}_{\alpha }\underline{y^{″}},{\underline{a}}_{\alpha }{\overline{y}}^{″},{\overline{a}}_{\alpha }{\underline{y}}^{″},{\overline{a}}_{\alpha }{\overline{y}}^{″}\right\}+max\left\{{\underline{b}}_{\alpha }{\underline{y}}^{\prime },{\underline{b}}_{\alpha }{\overline{y}}_{\alpha }^{\prime },{\overline{b}}_{\alpha }{\underline{y}}_{\alpha }^{\prime },{\overline{b}}_{\alpha }{\overline{y}}_{\alpha }^{\prime }\right\}\\ +max\left\{{\underline{c}}_{\alpha }{\underline{y}}_{\alpha },{\underline{c}}_{\alpha }{\overline{y}}_{\alpha },{\overline{c}}_{\alpha }{\underline{y}}_{\alpha },{\overline{c}}_{\alpha }{\overline{y}}_{\alpha }\right\}={\overline{r}}_{\alpha }\left(t\right).\end{array}$$

The above min–max problems are difficult to solve and sometimes not possible. For this reason, a numerical method to solve Problem (2) will be given.

3. Second-Order Fuzzy Initial Value Problems

In this section, linear and nonlinear second-order fuzzy initial value problems will be discussed. First, let us define the operation ⊙ for any $a\in \Re$ and the $\alpha$-cut of b as follows:

$$a\odot [b_{1,\alpha },b_{2,\alpha }]=\left\{\begin{array}{cc}[a{b}_{1,\alpha },a{b}_{2,\alpha }],\hfill & a\ge 0\hfill \\ [a{b}_{2,\alpha },a{b}_{1,\alpha }],\hfill & a<0\hfill \end{array}\right..$$

Consider the following linear second order FIVP:

$$y^{″}+a\left(x\right)y^{\prime }+b\left(x\right)y=c\left(x\right),0⩽x⩽1,$$

(3)

$$y\left(0\right)=\widehat{\beta },$$

(4)

$$y^{\prime }\left(0\right)=\widehat{\gamma },$$

(5)

where $\widehat{\beta }$ and $\widehat{\gamma }$ are symmetric fuzzy numbers $a\left(x\right),$$b\left(x\right)$ is a continuous functions on $[0,1]$, and $c\left(x\right)$ is fuzzy function. Let the $\alpha$-cuts of $y^{\prime }\left(x\right),y\left(x\right)$, $c\left(x\right)$, $\widehat{\beta }$, and $\widehat{\gamma }$ be given by:

$$\begin{array}{c}\hfill \begin{array}{c}{y}_{\alpha }^{\prime }\left(x\right)=\left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right],\hfill \\ y_{\alpha }\left(x\right)=\left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right],\hfill \\ \widehat{\beta }=\left[{\beta }_1,{\beta }_2\right],\hfill \\ \widehat{\gamma }=\left[{\gamma }_1,{\gamma }_2\right],\hfill \end{array}\end{array}$$

and

$$c_{\alpha }\left(x\right)=\left[c_{1\alpha }\left(x\right),c_{2\alpha }\left(x\right)\right].$$

To solve Problem (3)–(5), four cases should be implemented.

Case l: Let $a\left(x\right)⩾0,b\left(x\right)⩾0$ for all $x\in [0,1]$. Then,

$$\left[y_{1\alpha }^{″}\left(x\right),y_{2\alpha }^{″}\left(x\right)\right]+a\left(x\right)\odot \left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right]+b\left(x\right)\odot \left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right]=\left[c_{1\alpha }\left(x\right),c_{2\alpha }\left(x\right)\right],$$

which produces the following system of non-homogeneous IVPs:

$$y_{1\alpha }^{″}\left(x\right)+a\left(x\right)y_{1\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{1\alpha }\left(x\right)=c_{1\alpha }\left(x\right),y_1\left(0\right)={\beta }_1,y_1^{\prime }\left(0\right)={\gamma }_1,$$

(6)

and

$$y_{2\alpha }^{″}\left(x\right)+a\left(x\right)y_{2\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{2\alpha }\left(x\right)=c_{2\alpha }\left(x\right),y_2\left(0\right)={\beta }_2,y_2^{\prime }\left(0\right)={\gamma }_2.$$

(7)

Case 2: Let $a\left(x\right)⩾0,b\left(x\right)<0$ for all $x\in [0,1]$. Then,

$$\left[y_{1\alpha }^{″}\left(x\right),y_{2\alpha }^{″}\left(x\right)\right]+a\left(x\right)\odot \left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right]+b\left(x\right)\odot \left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right]=\left[c_{1\alpha }\left(x\right),c_{2\alpha }\left(x\right)\right],$$

which implies that

$$y_{1\alpha }^{″}\left(x\right)+a\left(x\right)y_{1\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{2\alpha }\left(x\right)=c_{1\alpha }\left(x\right),y_{1\alpha }\left(0\right)={\beta }_1,y_{1\alpha }^{\prime }={\gamma }_1,$$

(8)

and

$$y_{2\alpha }^{″}\left(x\right)+a\left(x\right)y_{2\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{1\alpha }\left(x\right)=c_{2\alpha }\left(x\right),y_{2\alpha }\left(0\right)={\beta }_2,y_{2\alpha }^{\prime }={\gamma }_2.$$

(9)

Let

$$\begin{array}{c}\hfill Y_{\alpha }\left(x\right)=\left(\begin{array}{c}{y}_{1\alpha }\left(x\right)\\ y_{2\alpha }\left(x\right)\end{array}\right),C_{\alpha }\left(x\right)=\left(\begin{array}{c}{c}_{1\alpha }\left(x\right)\hfill \\ c_{2\alpha }\left(x\right)\hfill \end{array}\right),\end{array}$$

$$\begin{array}{c}\hfill \lambda =\left(\begin{array}{c}{\beta }_1\hfill \\ {\beta }_2\hfill \end{array}\right),{\lambda }^{\prime }=\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right),B\left(x\right)=\left(\begin{array}{cc}0\hfill & b\left(x\right)\hfill \\ b\left(x\right)\hfill & 0\hfill \end{array}\right).\end{array}$$

Then, Equations (8) and (9) can be written in the matrix form as:

$$Y_{\alpha }^{″}\left(x\right)+a\left(x\right)Y_{\alpha }^{\prime }\left(x\right)+B\left(x\right)Y_{\alpha }\left(x\right)=C_{\alpha }\left(x\right),Y_{\alpha }\left(0\right)=\lambda ,Y_{\alpha }^{\prime }\left(0\right)={\lambda }^{\prime },$$

(10)

where B is symmetric matrix.

Case 3: Let $a\left(x\right)<0,b\left(x\right)⩾0$ for all $x\in [0,1]$. Then,

$$\left[y_{1\alpha }^{″}\left(x\right),y_{2\alpha }^{″}\left(x\right)\right]+a\left(x\right)\odot \left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right]+b\left(x\right)\odot \left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right]=\left[c_{1\alpha }\left(x\right),c_{2\alpha }\left(x\right)\right],$$

which implies that

$$y_{1\alpha }^{″}\left(x\right)+a\left(x\right)y_{2\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{1\alpha }\left(x\right)=c_{1\alpha }\left(x\right),y_{1\alpha }\left(0\right)={\beta }_1,y_{1\alpha }^{\prime }\left(0\right)={\gamma }_1,$$

(11)

$$y_{2\alpha }^{″}\left(x\right)+a\left(x\right)y_{1\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{2\alpha }\left(x\right)=c_{2\alpha }\left(x\right),y_{2\alpha }\left(0\right)={\beta }_2,y_{2\alpha }^{\prime }\left(0\right)={\gamma }_2.$$

(12)

Let

$$\begin{array}{c}\hfill Y_{\alpha }\left(x\right)=\left(\begin{array}{c}{y}_1\alpha \left(x\right)\\ y_{2\alpha }\left(x\right)\end{array}\right),C_{\alpha }\left(x\right)=\left(\begin{array}{c}{c}_{1\alpha }\left(x\right)\hfill \\ c_{2\alpha }\left(x\right)\hfill \end{array}\right),\end{array}$$

$$\begin{array}{c}\hfill \lambda =\left(\begin{array}{c}{\beta }_1\hfill \\ {\beta }_2\hfill \end{array}\right),{\lambda }^{\prime }=\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right),A\left(x\right)=\left(\begin{array}{cc}0\hfill & a\left(x\right)\hfill \\ a\left(x\right)\hfill & 0\hfill \end{array}\right).\end{array}$$

Then, Equations (11) and (12) can be written in the matrix form as:

$$Y_{\alpha }^{″}\left(x\right)+A\left(x\right)Y_{\alpha }^{\prime }\left(x\right)+b\left(x\right)Y_{\alpha }\left(x\right)=C_{\alpha }\left(x\right),Y_{\alpha }\left(0\right)=\lambda ,Y_{\alpha }^{\prime }\left(0\right)={\lambda }^{\prime },$$

(13)

where A is a symmetric matrix.

Case 4: Let $a\left(x\right)<0,b\left(x\right)<0$ for all $x\in [0,1]$.

$$\left[y_{1\alpha }^{″}\left(x\right),y_{2\alpha }^{″}\left(x\right)\right]+a\left(x\right)\odot \left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right]+b\left(x\right)\odot \left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right]=\left[c_{1\alpha }\left(x\right),c_{2\alpha }\left(x\right)\right],$$

which implies that

$$y_{1\alpha }^{″}\left(x\right)+a\left(x\right)y_{2\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{2\alpha }\left(x\right)=c_{1\alpha }\left(x\right),y_{1\alpha }\left(0\right)={\beta }_1,y_{1\alpha }^{\prime }\left(0\right)={\gamma }_1,$$

(14)

$$y_{2\alpha }^{″}\left(x\right)+a\left(x\right)y_{1\alpha }^{\prime }\left(x\right)+b\left(x\right)y_{1\alpha }\left(x\right)=c_{2\alpha }\left(x\right),y_{2\alpha }\left(0\right)={\beta }_2,y_{2\alpha }^{\prime }\left(0\right)={\gamma }_2.$$

(15)

Let

$$\begin{array}{c}\hfill Y_{\alpha }\left(x\right)=\left(\begin{array}{c}{y}_1\alpha \left(x\right)\\ y_{2\alpha }\left(x\right)\end{array}\right),C_{\alpha }\left(x\right)=\left(\begin{array}{c}{c}_{1\alpha }\left(x\right)\hfill \\ c_{2\alpha }\left(x\right)\hfill \end{array}\right),\end{array}$$

$$\begin{array}{c}\hfill \lambda =\left(\begin{array}{c}{\beta }_1\hfill \\ {\beta }_2\hfill \end{array}\right),{\lambda }^{\prime }=\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right),A\left(x\right)=\left(\begin{array}{cc}0\hfill & a\left(x\right)\hfill \\ a\left(x\right)\hfill & 0\hfill \end{array}\right)B\left(x\right)=\left(\begin{array}{cc}0\hfill & b\left(x\right)\hfill \\ b\left(x\right)\hfill & 0\hfill \end{array}\right).\end{array}$$

Then, Equations (14) and (15) can be written in the matrix form as:

$$Y_{\alpha }^{″}\left(x\right)+A\left(x\right)Y_{\alpha }^{\prime }\left(x\right)+B\left(x\right)Y_{\alpha }\left(x\right)=C_{\alpha }\left(x\right),Y_{\alpha }\left(0\right)=\lambda ,Y_{\alpha }^{\prime }\left(0\right)={\lambda }^{\prime }.$$

(16)

Thus, we can summarize the four cases as follows. The linear system of non-homogeneous IVPs is:

$$Y_{\alpha }^{″}\left(x\right)+A\left(x\right)Y_{\alpha }^{\prime }\left(x\right)+B\left(x\right)Y_{\alpha }\left(x\right)=C_{\alpha }\left(x\right),Y_{\alpha }\left(0\right)=\lambda ,Y_{\alpha }^{\prime }\left(0\right)={\lambda }^{\prime },$$

(17)

where

$$\begin{array}{c}\hfill Y_{\alpha }\left(x\right)=\left(\begin{array}{c}{y}_1\alpha \left(x\right)\\ y_{2\alpha }\left(x\right)\end{array}\right),C_{\alpha }\left(x\right)=\left(\begin{array}{c}{c}_{1\alpha }\left(x\right)\hfill \\ c_{2\alpha }\left(x\right)\hfill \end{array}\right),\end{array}$$

$$\begin{array}{c}\hfill \lambda =\left(\begin{array}{c}{\beta }_1\hfill \\ {\beta }_2\hfill \end{array}\right),{\lambda }^{\prime }=\left(\begin{array}{c}{\gamma }_1\hfill \\ {\gamma }_2\hfill \end{array}\right),A\left(x\right)=\left(\begin{array}{cc}{\mu }_{1}a\left(x\right)\hfill & {\mu }_{2}a\left(x\right)\hfill \\ {\mu }_{2}a\left(x\right)\hfill & {\mu }_{1}a\left(x\right)\hfill \end{array}\right)B\left(x\right)=\left(\begin{array}{cc}{\gamma }_{1}b\left(x\right)\hfill & {\gamma }_{2}b\left(x\right)\hfill \\ {\gamma }_{2}b\left(x\right)\hfill & {\gamma }_{1}b\left(x\right)\hfill \end{array}\right),\end{array}$$

such that ${\mu }_1=1$ and ${\mu }_2=0$ if $a\left(x\right)\ge 0$ while ${\mu }_1=0$ and ${\mu }_2=1$ if $a\left(x\right)<0$. In addition, ${\gamma }_1=1$ and ${\gamma }_2=0$ if $b\left(x\right)\ge 0$, while ${\gamma }_1=0$ and ${\gamma }_2=1$ if $b\left(x\right)<0$, where A and B are symmetric matrices.

Now, consider the following general form of the FIVP:

$$y^{″}=f\left(x,y,y^{\prime }\right),0⩽x\le 1,$$

subject to

$$y\left(0\right)=\widehat{\beta },y^{\prime }\left(0\right)=\widehat{\gamma }.$$

Let the $\alpha$-cut of $y\left(x\right),y^{\prime }\left(x\right),y^{″}\left(x\right),\widehat{\beta },\widehat{\gamma }$, and $f\left(x,y,y^{\prime }\right)$ be given by

$$y_{\alpha }\left(x\right)=\left[y_{1\alpha }\left(x\right),y_{2\alpha }\left(x\right)\right],y_{\alpha }^{\prime }\left(x\right)=\left[y_{1\alpha }^{\prime }\left(x\right),y_{2\alpha }^{\prime }\left(x\right)\right],$$

$$y_{\alpha }^{{}^{″}}\left(x\right)=\left[y_{1\alpha }^{″}\left(x\right),y_{2\alpha }^{″}\left(x\right)\right],\widehat{\beta }=\left[{\beta }_{1\alpha },{\beta }_{2\alpha }\right],\widehat{\gamma }=\left[{\gamma }_{1\alpha },{\gamma }_{2\alpha }\right],$$

$$f_{\alpha }\left(x,y,y^{\prime }\right)=\left[f_{1\alpha }\left(x,y,y^{\prime }\right),f_{2\alpha }\left(x,y,y^{\prime }\right)\right],$$

where

$$f_{1\alpha }\left(x,y,y^{\prime }\right)=min\left\{f(x,u,v):u\in \left[y_{1\alpha },y_{2\alpha }\right],v\in \left[y_{1\alpha }^{{}^{\prime }},y_{2\alpha }^{{}^{\prime }}\right]\right\},$$

and

$$f_{2\alpha }\left(x,y,y^{\prime }\right)=max\left\{f(x,u,v):u\in \left[y_{1\alpha },y_{2\alpha }\right],v\in \left[y_{1\alpha }^{{}^{\prime }},y_{2\alpha }^{{}^{\prime }}\right]\right\}.$$

Then,

$$y_{1\alpha }^{″}=f_{1\alpha }\left(x,y,y^{\prime }\right),y_{1\alpha }\left(0\right)={\beta }_{1\alpha },y_{1\alpha }^{{}^{\prime }}\left(0\right)={\gamma }_{1\alpha },$$

and

$$y_{2\alpha }^{″}=f_{2\alpha }\left(x,y,y^{\prime }\right),y_{2\alpha }\left(0\right)={\beta }_{2\alpha },y_{2\alpha }^{{}^{\prime }}\left(0\right)={\gamma }_{2\alpha }.$$

As we see from the previous discussion, the FIVP will produce a system of two IVPs. To solve this system, we will use the RKM.

Definition 6

([38,39]). Let $F\ne \varnothing$. A function $Q:F\times F\to \mathbb{C}$ is called a reproducing kernel function of the Hilbert space H if and only if:

(a)

$Q(·,t)\in H$, $t\in F$;

(b)

$〈\phi ,Q(·,t)〉=\phi \left(t\right)$, $t\in F$, $\phi \in H$.

A Hilbert space which possesses a reproducing kernel is called a reproducing kernel Hilbert space.

Definition 7.

Let $W_2^1[0,1]=\{w:w$ be an absolutely continuous real-valued function on $[0,1],w^{{}^{\prime }},$

$w^{{}^{″}}\in$$L^2[0,1]\}.$

The inner product in $W_2^1[0,1]$ is symmetric and is defined as:

$${(w\left(x\right),v\left(x\right))}_{W_2^1[0,1]}=w\left(0\right)v\left(0\right)+{\int }_0^1{w}^{\prime }\left(x\right)v^{\prime }\left(x\right)dx,$$

(18)

and its norm is defined as:

$${∥w∥}_{W_2^1[0,1]}=\sqrt{{(w\left(x\right),w\left(x\right))}_{W_2^1[0,1]}},$$

(19)

where $w,v\in W_2^1[0,1]$.

Theorem 4.

The Hilbert space $W_2^1[0,1]$ is a reproducing kernel, and its reproducing kernel function $R_y\left(x\right)$ can be defined by:

$$R_y\left(x\right)=\left\{\begin{array}{cc}1+x,\hfill & x⩽y\hfill \\ 1+y,\hfill & x>y\hfill \end{array}\right..$$

Proof. 

Let

$$w\left(y\right)=\left(w\left(x\right),R_y\left(x\right)\right)=w\left(0\right)R_y\left(0\right)+{\int }_0^1{w}^{\prime }\left(x\right)R_y^{\prime }\left(x\right)dx.$$

(20)

Using integration by parts, we obtain:

$$w\left(y\right)=w\left(0\right)R_y\left(0\right)+R_y^{\prime }\left(1\right)w\left(1\right)-R_y^{\prime }\left(0\right)w\left(0\right)-{\int }_0^{1}w\left(x\right)R_y^{\left(2\right)}\left(x\right)dx.$$

Therefore:

$$R_y\left(0\right)-R_y^{\prime }\left(0\right)=0,$$

(21)

$$R_y^{\prime }\left(1\right)=0.$$

(22)

Thus:

$$w\left(y\right)=〈w\left(x\right),R_y\left(x\right)〉=-{\int }_0^{1}w\left(x\right)R_y^{\left(2\right)}\left(x\right)dx.$$

Hence:

$$-R_y^{\left(2\right)}\left(x\right)=\delta (x-y)=\left\{\begin{array}{c}1,\mathrm{if}x=y\\ 0,\mathrm{if}x\ne y\end{array}\right..$$

Hence:

$$R_y\left(x\right)=\left\{\begin{array}{cc}{C}_1\left(y\right)+C_2\left(y\right)x,\hfill & y\ge x\hfill \\ d_1\left(y\right)+d_2\left(y\right)x,\hfill & y<x\hfill \end{array}\right..$$

Since $-R_y^{\left(2\right)}\left(x\right)=\delta (x-y),$ then

$$R_y\left(y^{+}\right)=R_y\left(y^{-}\right),$$

(23)

and

$$\frac{\partial R_y(y^{+)}}{\partial y}-\frac{\partial R_y\left(y^{-}\right)}{\partial y}=-1.$$

(24)

For simplicity, let $C_i\left(y\right)=C_i$ and $d_i\left(y\right)=d_i$ for $i=1,2$. Solve system (20)–(23) to obtain:

$$\begin{array}{ccc}\hfill C_1-C_2& =& 0,\hfill \\ \hfill C_1+C_{2}y& =& d_1+d_{2}y,\hfill \\ \hfill d_2-C_2& =& -1,\hfill \\ \hfill d_2& =& 0.\hfill \end{array}$$

Then,

$$C_1\left(y\right)=1,C_2\left(y\right)=1,d_1\left(y\right)=1+y,d_2\left(y\right)=0.$$

Then, the kernel is symmetric and given by:

$$R_y\left(x\right)=\left\{\begin{array}{cc}1+x,& y⩾x\\ 1+y,& x>y\end{array}\right..$$

□

Definition 8.

Let $W_2^3[0,1]=\{w:w,w^{\prime },w^{{}^{\left(2\right)}}$ be absolutely continuous real-valued functions on $[0,1],w^{{}^{\left(i\right)}}\in$$L^2[0,1]$ for $i=3,4,5,6$, $w\left(0\right)=w^{\prime }\left(0\right)=0\}$ with the following inner product:

$$〈w\left(x\right),v\left(x\right)〉=w\left(0\right)v\left(0\right)+w^{\prime }\left(0\right)v^{\prime }\left(0\right)+w\left(1\right)v\left(1\right)+{\int }_0^1{w}^{\left(3\right)}\left(x\right)v^{\left(3\right)}\left(x\right)dx,w,v\in W_2^3[0,1],$$

and the norm

$${\parallel w\parallel }_{W_2^3}=\sqrt{{\langle w,w\rangle }_{W_2^3}}.$$

Theorem 5.

The Hilbert space $W_2^3[0,1]$ is a reproducing kernel space and its reproducing kernel function $R_y\left(x\right)$ can be defined by

$$\begin{array}{c}{R}_y\left(x\right)=\left\{\begin{array}{cc}-\frac{x^2}{y^2},\hfill & x\le y\hfill \\ \frac{1}{120}{y}^5-\frac{1}{24}x{y}^4+\frac{y^5-12}{12y^2}{x}^2-\frac{1}{12}{x}^3{y}^2\hfill & \\ +\frac{1}{24}{x}^4{y}^2-\frac{1}{120}{x}^5{y}^2,\hfill & x>y\hfill \end{array}\right..\hfill \end{array}$$

Proof. 

Let

$$w\left(y\right)=〈w\left(x\right),R_y\left(x\right)〉=w\left(0\right)R_y\left(0\right)+w^{\prime }\left(0\right)R_y^{\prime }\left(0\right)+w\left(1\right)R_y\left(1\right)+{\int }_0^1{w}^{\left(3\right)}\left(x\right)R_y^{\left(3\right)}\left(x\right)dx.$$

Integrate by parts three times to obtain:

$$\begin{array}{ccc}\hfill 〈w,R_y〉& =& w\left(0\right)R_y\left(0\right)+w^{\prime }\left(0\right)R_y^{\prime }\left(0\right)+w\left(1\right)R_y\left(1\right)+w^{\left(2\right)}\left(1\right)R_y^{{}^{\left(3\right)}}\left(1\right)-w^{{}^{\left(2\right)}}\left(0\right)R_y^{{}^{\left(3\right)}}\left(0\right)\hfill \\ & & -w^{{}^{\prime }}\left(1\right)R_y^{\left(4\right)}\left(1\right)+w^{\prime }\left(0\right)R_y^{\left(4\right)}\left(0\right)+w\left(1\right)R_y^{\left(5\right)}\left(1\right)-w\left(0\right)R_y^{\left(5\right)}\left(0\right)-{\int }_0^{1}w\left(x\right)R_y^{\left(6\right)}\left(x\right)dx.\hfill \end{array}$$

Substitute the conditions $w\left(0\right)=w^{\prime }\left(0\right)=0$ to obtain:

$$\begin{array}{ccc}\hfill w\left(y\right)& =& w\left(1\right)R_y\left(1\right)+w^{{}^{\left(2\right)}}\left(1\right)R_y^{{}^{\left(3\right)}}\left(1\right)-w^{{}^{\left(2\right)}}\left(0\right)R_y^{{}^{\left(3\right)}}\left(0\right)-w^{{}^{\prime }}\left(1\right)R_y^{\left(4\right)}\left(1\right)\hfill \\ & & +w\left(1\right)R_y^{\left(5\right)}\left(1\right)-{\int }_0^{1}w\left(x\right)R_y^{\left(6\right)}\left(x\right)\hfill \\ \hfill dx.\end{array}$$

Let

$$R_y\left(1\right)-R_y^{\left(5\right)}\left(1\right)=0,$$

(25)

$$R_y^{\left(3\right)}\left(1\right)=0,$$

(26)

$$R_y^{\left(3\right)}\left(0\right)=0,$$

(27)

$$R_y^{{}^{\left(4\right)}}\left(1\right)=0.$$

(28)

Thus, under these conditions, we obtain:

$$w\left(y\right)=-{\int }_0^{1}w\left(x\right)R_y^{\left(6\right)}\left(x\right)dx.$$

(29)

Then,

$$-R_y^{\left(6\right)}\left(x\right)=\delta (x-y),$$

where

$$\delta (x-y)=\left\{\begin{array}{c}1\mathrm{if}x=y\\ 0\mathrm{if}x\ne y\end{array}\right..$$

Thus,

$$R_y\left(x\right)=\left\{\begin{array}{cc}\sum _{i=1}^6{C}_1\left(y\right)x^{i-1}\hfill & x⩽y\hfill \\ \sum _{i=1}^6{d}_i\left(y\right)x^{i-1}\hfill & x>y\hfill \end{array}\right..$$

Since

$$-R_y^{\left(6\right)}\left(x\right)=\delta (x-y),$$

then

$$\frac{{\partial }^k{R}_y\left(y^{+}\right)}{\partial y^k}=\frac{{\partial }^K{R}_y\left(y^{-}\right)}{\partial y^k},\mathrm{for}k=0,1,2,3,4,$$

(30)

and

$$\frac{{\partial }^5{R}_y\left(y^{+}\right)}{\partial y^5}-\frac{{\partial }^5{R}_y\left(y^{-}\right)}{\partial y^5}=-1.$$

(31)

Since $R_y\in W_2^3[0,1]$, then

$$R_y\left(0\right)=0,$$

(32)

and

$$R_y^{\prime }\left(0\right)=0.$$

(33)

Then, we obtain the following system:

$$\begin{array}{ccc}\hfill 6C_4& =& 0,\hfill \\ \hfill 4!d_5+5!d_6& =& 0,\hfill \\ \hfill 6d_4+24d_{5}y+60d_6{y}^2& =& 0,\hfill \\ \hfill d_1+d_{2}y+d_3{y}^2+d_4{y}^3+d_5{y}^4+d_6{y}^5-5!d_6& =& 0,\hfill \\ \hfill C_1+C_{2}y+C_3{y}^2+C_4{y}^3+C_5{y}^4+C_6{y}^5& =& d_1+d_{2}y+d_3{y}^2+d_4{y}^3+d_5{y}^4+d_6{y}^5,\hfill \\ \hfill C_2+2C_{3}y+3C_4{y}^2+4C_5{y}^3+5C_6{y}^4& =& d_2+2d_{3}y+3d_4{y}^2+4d_5{y}^3+5d_6{y}^4,\hfill \\ \hfill 2C_3+6C_4{y}_{+}12C_5{y}^2+20C_6{y}^3& =& 2d_3+6d_{4}y+12d_5{y}^2+20d_6{y}^3,\hfill \\ \hfill 6C_4+24C_{5}y+60C_6{y}^2& =& 6d_4+24d_{5}y+60d_6{y}^2,\hfill \\ \hfill 24C_5+120C_{6}y& =& 24d_5+120d_{6}y,\hfill \\ \hfill 120d_6-120C_6& =& -1,\hfill \\ \hfill C_1& =& 0,\hfill \\ \hfill C_2& =& 0.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill \begin{array}{c}{C}_1\left(y\right)=0,C_2\left(y\right)=0,C_3\left(y\right)=\frac{-1}{y^2},C_4\left(y\right)=0,\hfill \\ C_5\left(y\right)=0,C_6\left(y\right)=0,d_1\left(y\right)=\frac{1}{120}{y}^5,\hfill \\ d_2\left(y\right)=-\frac{1}{24}{y}^4,d_3\left(y\right)=\frac{y^5-12}{12y^2},d_4\left(y\right)=-\frac{1}{12}{y}^2,\hfill \\ d_5\left(y\right)=\frac{1}{24}{y}^2,d_6\left(y\right)=-\frac{1}{120}{y}^2.\hfill \end{array}\end{array}$$

Then,

$$\begin{array}{c}{R}_y\left(x\right)=\left\{\begin{array}{c}-\frac{x^2}{y^2},x\le y\hfill \\ \frac{1}{120}{y}^5-\frac{1}{24}x{y}^4+\frac{y^5-12}{12y^2}{x}^2-\frac{1}{12}{x}^3{y}^2\hfill \\ +\frac{1}{24}{x}^4{y}^2-\frac{1}{120}{x}^5{y}^2,x>y\hfill \end{array}\right..\hfill \end{array}$$

□

Consider the following system of a second-order initial value problem:

$$u^{{}^{″}}+A\left(x\right)u^{\prime }+B\left(x\right)u=f(x,u,u^{\prime }),0\le x\le 1,$$

(34)

with

$$u\left(0\right)=0,u^{{}^{\prime }}\left(0\right)=0,$$

(35)

when A and B are symmetric matrices, $p,q$∈$C^2(0,1),$ and $f\in L^2[0,1].$ We can obtain homogeneous initial conditions by using a change in variables. Thus,

$$\left(Lu\right)\left(x\right)=f(x,u,u^{\prime }),0\le x\le 1$$

(36)

with

$$u\left(0\right)=u^{{}^{\prime }}\left(0\right)=0,$$

(37)

where

$$Lu=\frac{d^{2}u}{d{x}^2}+A\left(x\right)\frac{du}{dx}+B\left(x\right)u^{\prime },$$

(38)

and

$$L:W_2^3[0,1]\to W_2^1[0,1].$$

(39)

Then,

$$\begin{array}{ccc}\hfill L(\mu u+v)\left(x\right)& =& \frac{d^2}{d{x}^2}(\mu u+v)\left(x\right)+A\left(x\right)\frac{d}{dx}(\mu u+v)\left(x\right)+B\left(x\right)(\mu u+v)\left(x\right),\hfill \\ & =& \mu \frac{d^2}{d{x}^2}u\left(x\right)+\mu A\left(x\right)\frac{d}{dx}u\left(x\right)+\mu B\left(x\right)u\left(x\right)+\frac{d^2}{d{x}^2}v\left(x\right),\hfill \\ & & +A\left(x\right)\frac{d}{dx}v\left(x\right)+B\left(x\right)v\left(x\right),\hfill \\ & =& \mu L\left(u\right)\left(x\right)+L\left(v\right)\left(x\right),\hfill \end{array}$$

where $\mu$ is constant. Then, L is the linear operator.

Theorem 6.

The linear operator L is a bounded linear operator.

Proof. 

By Equation (19), we have:

$${\parallel Lw\parallel }_{W_2^1}^2={\langle Lw,Lw\rangle }_{W_2^1}={\int }_0^1\parallel L{w}^{\prime }{\left(x\right)\parallel }^2\mathrm{d}x+{\parallel Lw\left(0\right)\parallel }^2.$$

By Theorem (5), we have:

$$w\left(x\right)={〈w(·),R_x(·)〉}_{W_2^3},$$

and

$$Lw\left(x\right)={〈w(·),L{R}_x(·)〉}_{W_2^1}.$$

By the Cauchy–Schwartz inequality,

$$\parallel Lw\left(x\right)\parallel \le {\parallel w\parallel }_{W_2^3}{∥L{R}_x∥}_{W_2^1}=M_1{\parallel w\parallel }_{W_2^3},$$

where $M_1>0$ is a positive constant. Thus,

$${\parallel \left(Lw\right)\left(0\right)\parallel }^2\le M_1^2{\parallel w\parallel }_{W_2^3}^2.$$

Since

$${\left(Lw\right)}^{\prime }\left(x\right)={〈w(·),{\left(L{R}_x\right)}^{\prime }(·)〉}_{W_2^1},$$

then

$${\parallel \left(Lw\right)}^{\prime }{\left(x\right)\parallel \le \parallel w\parallel }_{W_2^3}{∥{\left(L{R}_x\right)}^{\prime }∥}_{W_2^3}=M_2{\parallel w\parallel }_{W_2^3},$$

where $M_2>0$ is a positive constant. We have

$${\parallel \left(Lw\right)}^{\prime }{\left(t\right)\parallel }^2\le M_2^2{\parallel w\parallel }_{W_2^3}^2,$$

and

$${\int }_0^1{\parallel \left(Lw\right)}^{\prime }{\left(x\right)\parallel }^2\mathrm{d}x\le M_2^2{\parallel w\parallel }_{W_2^3}^2,$$

which implies that

$${\parallel Lw\parallel }_{W_2^1}^2\le M{\parallel w\parallel }_{W_2^3}^2,$$

where $M=M_1^2+M_2^2$ □

Let ${\left\{x_i\right\}}_{i=1}^{\infty }$ be a countable dense subset of $[0,1]$. Let ${\phi }_i\left(x\right)=R_{x_i}\left(x\right)$ and ${\psi }_i\left(x\right)=L^{\ast }{\phi }_i\left(x\right)$, where $L^{\ast }$ is conjugate operator of L. Let ${\left\{{\widehat{\Psi }}_i\left(x\right)\right\}}_{i=1}^{\infty }$⊂ $W_2^3[0,1]$ be an orthonormal set of functions produced from ${\left\{{\psi }_i\left(x\right)\right\}}_{i=1}^{\infty }$. Then,

$${\widehat{\psi }}_i\left(x\right)=\sum _{k=1}^i{\beta }_{ik}{\psi }_k\left(x\right),{\beta }_{ii}>0,i=1,2,\dots .$$

Theorem 7.

The exact solution of

$$u^{″}=G(x,u,u^{\prime }),0<x<1,$$

(40)

with

$$u\left(0\right)=u_0,u^{\prime }\left(0\right)=u_1,$$

(41)

then

$$u\left(x\right)=\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}G\left(x_k,u_k,u_k^{\prime }\right){\widehat{\Psi }}_i\left(x\right),$$

(42)

where ${\left\{x_i\right\}}_{i=1}^{\infty }$ is dense in $[0,1].$

Proof. 

Simple calculations imply that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill u\left(x\right)& =\sum _{i=1}^{\infty }{〈u\left(x\right),{\widehat{\Psi }}_i\left(x\right)〉}_{W_2^3}{\widehat{\Psi }}_i\left(x\right),\hfill \\ & =\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}{〈u\left(x\right),{\Psi }_k\left(x\right)〉}_{W_2^3}{\widehat{\Psi }}_i\left(x\right),\hfill \\ & =\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}{〈u\left(x\right),L^{\ast }{\phi }_k\left(x\right)〉}_{W_2^3}{\widehat{\Psi }}_i\left(x\right),\hfill \\ & =\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}{〈Lu\left(x\right),{\phi }_k\left(x\right)〉}_{W_2^1}{\widehat{\Psi }}_i\left(x\right),\hfill \\ & =\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}{〈G(x,u,u^{\prime }),T_{x_k}〉}_{W_2^1}{\widehat{\Psi }}_i\left(x\right),\hfill \\ & =\sum _{i=1}^{\infty }\sum _{k=1}^i{\beta }_{ik}G\left(x_k,u_k,u_k^{\prime }\right){\widehat{\Psi }}_i\left(x\right).\hfill \end{array}\end{array}$$

□

4. Results

In this section, four examples will be discussed.

Example 1.

Consider the following problem:

$$y^{{}^{″}}+y^{{}^{\prime }}=x^2,0\le x\le 1,$$

subject to

$$\begin{array}{ccc}\hfill y\left(0\right)& =& \widehat{\beta },\hfill \\ \hfill y^{{}^{\prime }}\left(0\right)& =& \widehat{\gamma },\hfill \end{array}$$

where $\widehat{\beta }=(0,1,2)$ and $\widehat{\gamma }=(1,2,3)$. Then, the $\alpha -cut$ of $y,y^{\prime },y^{″},\widehat{\beta }$, and $\widehat{\gamma }$ are

$$\begin{array}{ccc}\hfill y_{\alpha }& =& [y_{1\alpha },y_{2\alpha }],y_{\alpha }^{{}^{\prime }}=[y_{1\alpha }^{{}^{\prime }},y_{2\alpha }^{{}^{\prime }}],y_{\alpha }^{{}^{″}}=[y_{1\alpha }^{{}^{″}},y_{2\alpha }^{{}^{″}}],\hfill \\ \hfill \widehat{\beta }& =& [\alpha ,2-\alpha ],\widehat{\gamma }=[\alpha +1,3-\alpha ].\hfill \end{array}$$

Then,

$$\begin{array}{c}{y}_{1\alpha }^{{}^{{}^{″}}}+y_{1\alpha }^{{}^{\prime }}=x^2,y_{\alpha }\left(0\right)=\alpha ,y_{\alpha }^{\prime }\left(0\right)=\alpha +1,\hfill \\ y_{2\alpha }^{{}^{″}}+y_{2\alpha }^{\prime }=x^2,y_2\alpha \left(0\right)=2-\alpha ,y_{2\alpha }^{\prime }\left(0\right)=3-\alpha .\hfill \end{array}$$

Then, the exact solution is

$$y_{\alpha }\left(x\right)=\left[2\alpha +1-(\alpha +1)e^{-x}-\frac{1}{3}{x}^3-x^2,5-2\alpha -(3-\alpha )e^{-x}-\frac{1}{3}{x}^3-x^2\right].$$

Using $n=8$, the absolute error in $y_{1\alpha }$ and $y_{2\alpha }$ are given in

Table 1. The absolute errors in Example 1.

${\mathit{x}}_{\mathit{k}}$	$\mathit{Abs}.\mathit{Error}\mathit{of}{\mathit{y}}_{1\mathit{\alpha }}$	$\mathit{Abs}.\mathit{Error}\mathit{of}{\mathit{y}}_{2\mathit{\alpha }}$
0	0	0
$0.1$	$2.3$×${10}^{-14}$	$2.4\times {10}^{-14}$
$0.2$	$2.4\times {10}^{-14}$	$2.5\times {10}^{-14}$
$0.3$	$2.5\times {10}^{-14}$	$2.6\times {10}^{-14}$
$0.4$	$2.6\times {10}^{-14}$	$2.7\times {10}^{-14}$
$0.5$	$2.7\times {10}^{-14}$	$2.8\times {10}^{-14}$
$0.6$	$2.8\times {10}^{-14}$	$3.1\times {10}^{-14}$
$0.7$	$3.0\times {10}^{-14}$	$3.2\times {10}^{-14}$
$0.8$	$3.1\times {10}^{-14}$	$3.4\times {10}^{-14}$
$0.9$	$3.3\times {10}^{-14}$	$3.6\times {10}^{-14}$
1	$3.5\times {10}^{-14}$	$3.8\times {10}^{-14}$

.

Example 2.

Consider the following problem:

$$y^{″}+(-1)\odot y^{\prime }+y=1,$$

subject to

$$\begin{array}{c}y\left(0\right)=\widehat{\beta },\hfill \\ y^{\prime }\left(0\right)=\widehat{\gamma },\hfill \end{array}$$

where $\widehat{\beta }=(-1,0,1)$ and $\widehat{\gamma }=(0,1,2)$. Then, the α-cut of $y,g^{\prime },y^{″},\widehat{\beta }$, and $\widehat{\gamma }$ are:

$$\begin{array}{c}{y}_{\alpha }=\left[y_{1\alpha },y_{2\alpha }\right],y_{\alpha }^{\prime }=\left[y_{1\alpha }^{\prime },y_{2\alpha }^{{}^{\prime }}\right],y_{\alpha }^{″}\left[y_{1\alpha }^{″},y_{2\alpha }^{{}^{″}}\right],\hfill \\ \widehat{\beta }=[\alpha -1,1+\alpha ],\widehat{\gamma }=[\alpha ,2-\alpha ].\hfill \end{array}$$

Then,

$$\begin{array}{c}{y}_{1\alpha }^{{}^{″}}-y_{2\alpha }^{{}^{\prime }}+y_{1\alpha }=1,y_{1\alpha }\left(0\right)=\alpha -1,y_{1\alpha }^{\prime }\left(0\right)=\alpha ,\hfill \\ y_{2\alpha }^{{}^{\prime }}-y_{1\alpha }^{\prime }+y_{2\alpha }=1,y_{2\alpha }\left(0\right)=1+\alpha ,y_2^{\prime }\left(0\right)=2-\alpha .\hfill \end{array}$$

Then, using Mathematica, the exact solution is

$$\begin{array}{ccc}\hfill y_{\alpha }\left(x\right)& =& \left[-\frac{1}{3}{e}^{-x/2}\left(3cos\left[\frac{\sqrt{3}x}{2}\right]+3e^{x}cos\left[\frac{\sqrt{3}x}{2}\right]-3\alpha e^{x}cos\left[\frac{\sqrt{3}x}{2}\right]-3e^{x/2}cos{\left[\frac{\sqrt{3}x}{2}\right]}^2\right.\right.\hfill \\ & & +3\sqrt{3}sin\left[\frac{\sqrt{3}x}{2}\right]-2\sqrt{3}\alpha sin\left[\frac{\sqrt{3}x}{2}\right]-3\sqrt{3}{e}^{x}sin\left[\frac{\sqrt{3}x}{2}\right]+\sqrt{3}\alpha e^{t}sin\left[\frac{\sqrt{3}t}{2}\right]\hfill \\ & & \left.-3e^{x/2}sin{\left[\frac{\sqrt{3}x}{2}\right]}^2\right),-\frac{1}{3}{e}^{-x/2}\left(-3cos\left[\frac{\sqrt{3}x}{2}\right]+3e^{x}cos\left[\frac{\sqrt{3}x}{2}\right]-3\alpha e^{t}cos\left[\frac{\sqrt{3}x}{2}\right]\right.\hfill \\ & & -3e^{x/2}cos{\left[\frac{\sqrt{3}x}{2}\right]}^2-3\sqrt{3}sin\left[\frac{\sqrt{3}x}{2}\right]+2\sqrt{3}\alpha sin\left[\frac{\sqrt{3}x}{2}\right]-3\sqrt{3}{e}^{x}sin\left[\frac{\sqrt{3}x}{2}\right]\hfill \\ & & \left.\left.+\sqrt{3}\alpha e^{x}sin\left[\frac{\sqrt{3}x}{2}\right]-3e^{x/2}sin{\left[\frac{\sqrt{3}x}{2}\right]}^2\right)\right].\hfill \end{array}$$

Using $n=8$, the absolute error in $y_{1\alpha }$ and $y_{2\alpha }$ are given in

Table 2. The absolute errors in Example 2.

${\mathit{x}}_{\mathit{k}}$	$\mathit{Abs}.\mathit{Error}\mathit{of}{\mathit{y}}_{1\mathit{\alpha }}$	$\mathit{Abs}.\mathit{Error}\mathit{of}{\mathit{y}}_{2\mathit{\alpha }}$
0	0	0
$0.1$	$3.4\times {10}^{-13}$	$3.6\times {10}^{-13}$
$0.2$	$3.6\times {10}^{-13}$	$3.8\times {10}^{-13}$
$0.3$	$3.9\times {10}^{-13}$	$4.1\times {10}^{-13}$
$0.4$	$4.2\times {10}^{-13}$	$4.4\times {10}^{-13}$
$0.5$	$4.5\times {10}^{-13}$	$4.6\times {10}^{-13}$
$0.6$	$4.8\times {10}^{-13}$	$4.9\times {10}^{-13}$
$0.7$	$5.2\times {10}^{-13}$	$5.3\times {10}^{-13}$
$0.8$	$5.6\times {10}^{-13}$	$5.7\times {10}^{-13}$
$0.9$	$5.9\times {10}^{-13}$	$6.0\times {10}^{-13}$
1	$6.2\times {10}^{-13}$	$6.3\times {10}^{-13}$

.

Example 3.

Consider the following problem:

$$y^{″}=-{\left(y^{\prime }\left(x\right)\right)}^2,$$

subject to

$$y\left(0\right)=\widehat{\beta },y^{\prime }\left(0\right)=\widehat{\gamma },$$

where

$$\widehat{\beta }=[\alpha ,2-\alpha ],\widehat{\gamma }=[1+\alpha ,3-\alpha ].$$

Then,

$$y_{1\alpha }^{″}\left(x\right)=f_{1\alpha }\left(x,y,y^{\prime }\right),y_{1\alpha }\left(0\right)=\alpha ,y_{1\alpha }^{{}^{\prime }}\left(0\right)=1+\alpha ,$$

and

$$y_{2a}^{″}\left(x\right)=f_{2\alpha }\left(x_{1}y,y^{\prime }\right),y_{2\alpha }\left(0\right)=2-\alpha ,y_{2\alpha }^{\prime }\left(0\right)=3-\alpha ,$$

where

$$f\left(x,y,y^{\prime }\right)=-{\left(y^{\prime }\left(x\right)\right)}^2.$$

Then, the exact solution is:

$$y_{\alpha }\left(x\right)=[ln\left(\left(\alpha e^{\alpha }+e^{\alpha }\right)x+e^x\right),ln\left((3e^{2-\alpha }-\alpha e^{2-\alpha }x)+e^{2\alpha }\right)].$$

Then, using the method proposed in the previous section, one obtains:

$$y_{1\alpha n}\left(x\right)=\sum _{i=1}^n\sum _{k=1}^i{\beta }_{ik}{\widehat{\psi }}_i\left(x\right)f_{1\alpha }\left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right),$$

$$y_{2\alpha n}\left(x\right)=\sum _{i=1}^n\sum _{k=1}^i{\beta }_{ik}{\widehat{\psi }}_i\left(x\right)f_{2\alpha }\left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right),$$

where

$$f_{1\alpha }\left(x_k,y_{{\alpha }_k},y_{{\alpha }_k}^{\prime }\right)=min\left\{-{\nu }^2:\nu \in [y_{1\alpha }^{\prime }\left(x_k),y_{2\alpha }^{\prime }\left(x_k\right)\right]\right\},$$

and

$$f_{2\alpha }\left(x_k,y_{{\alpha }_k},y_{{\alpha }_k}^{\prime }\right)=max\left\{-{\nu }^2:\nu \in [y_{1\alpha }^{\prime }\left(x_k),y_{2\alpha }^{\prime }\left(x_k\right)\right]\right\}.$$

Let $n=8$. Let $E_1\left(x_k\right)$ and $E_2\left(x_k\right)$ be the absolute error in $y_{1\alpha }$ and $y_{2\alpha }$, respectively. The results are reported in

Table 3. The absolute error of Example 3.

${\mathit{x}}_{\mathit{k}}$	${\mathit{E}}_1\left({\mathit{x}}_{\mathit{k}}\right)$	${\mathit{E}}_2\left({\mathit{x}}_{\mathit{k}}\right)$
0	0	0
$0.1$	$3.1\times {10}^{-12}$	$2.9\times {10}^{-12}$
$0.2$	$3.3\times {10}^{-12}$	$3.2\times {10}^{-12}$
$0.3$	$3.7\times {10}^{-12}$	$3.5\times {10}^{-12}$
$0.4$	$4.1\times {10}^{-12}$	$3.9\times {10}^{-12}$
$0.5$	$4.5\times {10}^{-12}$	$4.3\times {10}^{-12}$
$0.6$	$4.8\times {10}^{-12}$	$4.7\times {10}^{-12}$
$0.7$	$5.2\times {10}^{-12}$	$5.1\times {10}^{-12}$
$0.8$	$5.5\times {10}^{-12}$	$5.4\times {10}^{-12}$
$0.9$	$5.8\times {10}^{-12}$	$5.7\times {10}^{-12}$
1	$6.2\times {10}^{-12}$	$6.0\times {10}^{-12}$

.

Example 4.

Consider the following problem:

$$y^{″}=x^2\odot y^{\prime }\left(x\right)\oplus 2x\odot y\left(x\right)\oplus x\odot \widehat{\beta },$$

subject to

$$y\left(0\right)=\widehat{\beta },y^{\prime }\left(0\right)=\widehat{\gamma },$$

where

$$\widehat{\beta }=[1+\alpha ,3-\alpha ],\widehat{\gamma }=[0,0].$$

Then,

$$y_1^{{}^{″}}\left(x\right)=f_{1\alpha }\left(x,y,y^{\prime }\right),y_{1\alpha }\left(0\right)=1+\alpha ,y_{1\alpha }^{\prime }\left(0\right)=0,$$

and

$$y_{2\alpha }^{{}^{″}}\left(x\right)=f_{2\alpha }\left(x,y,y^{\prime }\right),y_{2x}\left(0\right)=3-\alpha ,y_{2\alpha }\left(0\right)=0,$$

where

$$f\left(x,y,y^{\prime }\right)=x^2\odot y^{\prime }\left(x\right)\oplus 2x\odot y\left(x\right)\oplus x\odot \widehat{\beta }.$$

Then, the exact solution is given by

$$\left[\left(e^{\left(x^3/3\right)}-1\right)(1-\alpha ),\left(2e^{\left(x^3/3\right)}-1\right)(3-\alpha )\right].$$

Then using the method which proposed in the previous section, one obtains:

$$y_{1\alpha n}\left(x\right)=\sum _{i=1}^n\sum _{k=1}^i{\beta }_{ik}{\widehat{\psi }}_i\left(x\right)f_1\alpha \left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right),$$

$$y_{2an}\left(x\right)=\sum _{i=1}^n\sum _{k=1}^i{\beta }_{ik}{\widehat{\psi }}_i\left(x\right)f_{2\alpha }\left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right),$$

where

$$f_{1\alpha }\left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right)=x^2{y}_{2\alpha }^{\prime }+2x{y}_{1\alpha }+x{\beta }_1,$$

$$f_{2\alpha }\left(x_k,y_{\alpha k},y_{\alpha k}^{\prime }\right)=x^2{y}_{2\alpha }^{\prime }+2x{y}_{1\alpha }+x{\beta }_2.$$

Let $n=8$. Let $E_1\left(x_k\right)$ and $E_2\left(x_k\right)$ be the absolute error in $y_{1\alpha }$ and $y_{2\alpha }$, respectively. The results are reported in

Table 4. The absolute error of Example 4.

${\mathit{x}}_{\mathit{k}}$	${\mathit{E}}_1\left({\mathit{x}}_{\mathit{k}}\right)$	${\mathit{E}}_2\left({\mathit{x}}_{\mathit{k}}\right)$
0	0	0
$0.1$	$2.7\times {10}^{-14}$	$2.5\times {10}^{-14}$
$0.2$	$2.9\times {10}^{-14}$	$2.7\times {10}^{-14}$
$0.3$	$3.1\times {10}^{-14}$	$2.8\times {10}^{-14}$
$0.4$	$3.4\times {10}^{-14}$	$3.0\times {10}^{-14}$
$0.5$	$3.5\times {10}^{-14}$	$3.2\times {10}^{-14}$
$0.6$	$3.7\times {10}^{-14}$	$3.4\times {10}^{-14}$
$0.7$	$3.9\times {10}^{-14}$	$3.7\times {10}^{-14}$
$0.8$	$4.2\times {10}^{-14}$	$3.9\times {10}^{-14}$
$0.9$	$4.4\times {10}^{-14}$	$4.1\times {10}^{-14}$
1	$4.7\times {10}^{-14}$	$4.4\times {10}^{-14}$

.

5. Conclusions

In this article, the reproducing kernel method has been presented for second-order fuzzy initial value problems. It started with preliminaries about fuzzy numbers and differentiation and then highlighted the direct method for solving fuzzy problems. When the direct method was used to solve the problem, complicated optimization problems that are difficult to solve appeared. The proposed method was based on an RKM and the Gram Schmidt process. The structure of the RKM was explained and supported by several examples. The numerical results showed the efficiency of the proposed method. The absolute errors were computed using Mathematica. For future work, the fuzzy boundary value problems should be investigated using RKM by implementing the shooting method. Moreover, several applications for this method should be investigated, such as Fuzzy Strum–Liouville problems and the delay fuzzy initial value problems. In addition, this approach can be used to solve several fuzzy integral and differential equations such as fuzzy integro-differential equations, fractional IVPs, fuzzy Cahn–Allen equations, and fuzzy duffing modesl.