Coincidence points of mappings and relations with applications

Abstract

We obtain coincidence points of mappings and relations under a contractive condition in a metric space. As applications, we achieve an existence and uniqueness theorem of solution for a general class of nonlinear integral equations.

2010 Mathematics Subject Classification: 47H10; 54H25; 54C60.

1 Introduction

The advancement and the rich growth of fixed point theorems in metric spaces has important theoretical and practical applications. This developments in the last three decades were tremendous. For most of them, their reference result is the Banach contraction theorem, which states that if X is a complete metric space and T : X → X a contractions mapping on X (i.e., d(Tx, Ty) ≤ λd(x, y) for all x, y ∈ X, where 0 < λ < 1), then T has a unique fixed point in X (see also [1], Lemma 1]). This theorem looks simple but plays a fundamental role in fixed point theory [2]. Jungck [3] studied coincidence and common fixed points of commuting mappings and improved the Banach contraction principle. The coincidence and common fixed points generalizations were further studied by many authors (e.g., see [4–6]). In addition, see Kirk [7], Murthy [8], Park [9, 10], and Rhoades [11, 12], for a survey of this subject. Currently Aydi et al. [13] established some coincidence and common fixed point results for three self-mappings on a partially ordered cone metric space satisfying a contractive condition and proved an existence theorem of a common solution of integral equations. In the same way, Shatanawi et al. [14] studied some new real generalizations on coincidence points for weakly decreasing mappings satisfying a weakly contractive condition in an ordered metric space.

On the other hand Haghi et al. [15] showed that some coincidence point and common fixed point generalizations for two mappings in fixed point theory are not real generalizations and they obtained some coincidence and common fixed point results for two self mappings from their corresponding fixed point theorems.

In the present article, we prove the existence of a coincidence point of a mapping and a relation under a contractive condition which is an innovative and real generalization of the Banach contraction theorem. Moreover, a result is deduced on existence of a unique coincidence point for two nonself mappings under a contractive condition. As applications, we achieve an existence and uniqueness theorem of solution for a class of nonlinear integral equations.

2 Preliminaries

Let A and B be arbitrary nonempty sets. A relation R from A to B is a subset of A × B and is denoted by R : A ⇝ B. The statement (x, y) ∈ R is read "x is R-related to y", and is denoted by xRy. A relation R : A ⇝ B is called left-total if for all x ∈ A there exists a y ∈ B such that xRy. R is called right-total if for all y ∈ B there exists an x ∈ A such that xRy. R is known as functional, if xRy, xRz implies that y = z, for x ∈ A and y, z ∈ B. A mapping T : A → B is a relation from A to B which is both functional and left-total. For R : A ⇝ B, E ⊂ A we define

$$\begin{array}{c}R\left(E\right)=\left\{y\in B:xRy\mathsf{\text{for}}\mathsf{\text{some}}x\in E\right\}.\\ \mathsf{\text{dom}}\mathsf{\text{(}}R\mathsf{\text{)}}=\left\{x\in A:R\left(\left\{x\right\}\right)\ne \varphi \right\},\\ \mathsf{\text{Range}}\mathsf{\text{(}}R\mathsf{\text{)}}=\left\{y\in B:y\in R\left(\left\{x\right\}\right)\mathsf{\text{for}}\mathsf{\text{some}}x\in \mathsf{\text{dom}}\left(R\right)\right\}.\end{array}$$

For convenience, we denote R ({x}) by R {x}. The class of relations from A to B is denoted by $ℛ\left(A,B\right).$. Thus the collection $ℳ\left(A,B\right)$ of all mappings from A to B is a proper sub collection of $ℛ\left(A,B\right).$. An element w ∈ A is called coincidence point of T : A → B and R : A ⇝ B if Tw ∈ R {w}. In the following, we always suppose that X is nonempty set and (Y, d) is a metric space. For R : X ⇝ Y and u, υ ∈ dom (R), we define

$$D\left(R\left\{u\right\},R\left\{v\right\}\right)=\underset{uRx,vRy}{\text{inf}}d\left(x,y\right)$$

A function Ψ : [0, ∞) → [0,1) is said to have property (p) [16–18] if for t > 0, there exists δ(t) > 0, s(t) < 1 such that

$$0\le r-t<\delta \left(t\right)\Rightarrow \Psi \left(r\right)\le s\left(t\right).$$

3 Coincidence points

Theorem 3.1 Let X be a nonempty set and (Y, d) be a metric space. Let T : X → Y, R : X ⇝ Y be such that R is left-total, Range (T) ⊆ Range (R) and Range (T) or Range (R) is complete. If there exists a non-decreasing function Ψ : [0, ∞) → [0, 1) having property (p) such that for all x, y ∈ X

$$d\left(Tx,Ty\right)\le \Psi \left(D\left(R\left\{x\right\},R\left\{y\right\}\right)\right)D\left(R\left\{x\right\},R\left\{y\right\}\right).$$

(1)

Then there exists w ∈ X such that Tw ∈ R {w}.

Proof. Let x₀ be an arbitrary, but fixed element of X. We shall construct sequences{x _n } ⊂ X, {y _n } ⊂ Range (R). Let y₁ = Tx₀, using the fact that Range (T) ⊆ Range (R), we may choose x₁ ∈ X such that

$$x_{1}R{y}_1.$$

Let y₂ = Tx₁, if

$$\Psi \left(D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right)\right)D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right)=0,$$

then by assumptions Tx₀ = Tx₁. It implies that

$$x_{1}R{y}_2.$$

Then x₁ is the point of X we are looking for. If

$$\Psi \left(D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right.\right)D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right)\ne 0,$$

then using inequality (1) we have

$$d\left(T{x}_0,T{x}_1\right)\le \Psi \left(D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right)\right)D\left(R\left\{x_0\right\},R\left\{x_1\right\}\right)\ne 0.$$

Choose x₂ ∈ X such that x₂Ry₂. In the case

$$\Psi \left(D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)\right)D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)=0,$$

x₂ is the required point in X. If

$$\Psi \left(D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)\right)D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)\ne 0,$$

then inequality (1) implies that

$$d\left(T{x}_1,T{x}_2\right)\le \Psi \left(D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)\right)D\left(R\left\{x_1\right\},R\left\{x_2\right\}\right)\ne 0,$$

By induction we produce sequences {x _n } ⊂ X and {y _n } ⊂ Range (R) such that y _n = Tx _{n -1}, x _n Ry _n and

$$d\left(y_n,y_{n+1}\right)\le \Psi \left(D\left(R\left\{x_{n-1}\right\},R\left\{x_n\right\}\right)D\left(R\left\{x_{n-1}\right\},R\left\{x_n\right\}\right)\ne 0,n=1,2,3,\dots .\right.$$

Since x _n Ry _n , x _{n +1}Ry _{n +1} therefore, by definition of D, we have

$$D\left(R\left\{x_n\right\},R\left\{x_{n+1}\right\}\right)\le d\left(y_n,y_{n+1}\right).$$

Thus,

$$d\left(y_{n+1},y_{n+2}\right)\le \Psi \left(d\left(y_n,y_{n+1}\right)d\left(y_n,y_{n+1}\right).\right.$$

It follows that

$$d\left(y_{n+1},y_{n+2}\right)<d\left(y_n,y_{n+1}\right),n=1,2,3,\dots .$$

Thus,

$$\underset{n\to \infty }{\text{lim}}d\left(y_n,y_{n+1}\right)=\text{inf}\left\{d\left(y_n,y_{n+1}\right):n\ge 0\right\}.$$

Assume that

$$\underset{n\to \infty }{\text{lim}}d\left(y_n,y_{n+1}\right)=t.$$

We claim t = 0. Otherwise by property (p) of Ψ, there exists δ(t) > 0, s(t) < 1, such that

$$0\le r-t<\delta \left(t\right)\Rightarrow \Psi \left(r\right)\le s\left(t\right)<1.$$

For this δ(t) > 0, there exists a natural number N such that

$$0\le d\left(y_n,y_{n+1}\right)-t<\delta \left(t\right),\mathsf{\text{ whenever }}n\ge N.$$

Hence,

$$\Psi \left(d\left(y_n,y_{n+1}\right)\right)\le s\left(t\right),\mathsf{\text{ whenever }}n\ge N.$$

Then inequality (1) implies that

$$d\left(y_n,y_{n+1}\right)\le \mathsf{\text{ max }}\left\{\Psi \left(d\left(y_{\mathsf{\text{0}}},y_1\right)\right),\Psi \left(d\left(y_1,y_2\right)\right),\dots ,\Psi \left(d\left(y_{N-1},y_N\right)\right),s\left(t\right)\right\}d\left(y_{n-1},y_n\right).$$

Assume that

$$M=\text{max}\left\{\Psi \left(d\left(y_0,y_1\right)\right),\Psi \left(d\left(y_1,y_2\right),\dots ,\Psi \left(d\left(y_{N-1},y_N\right)\right),s\left(t\right)\right.\right\}.$$

Then M < 1 and

$$d\left(y_n,y_{n+1}\right)\le Md\left(y_{n-1},y_n\right)\mathsf{\text{for}}n\mathsf{\text{ = 1,2,3,}}\dots$$

Hence

$$d\left(y_n,y_{n+1}\right)\le M^{n}d\left(y_0,y_1\right)\to 0\mathsf{\text{as}}n\to \infty ,$$

this contradicts the assumption that t > 0. Consequently

$$\underset{n\to \infty }{\text{lim}}d\left(y_n,y_{n+1}\right)=0.$$

Now we prove that {y _n } is a Cauchy sequence. Assume that {y _n } is not a Cauchy sequence. Then there exists a positive number t* and subsequences {n(i)}, {m(i)} of the natural numbers with n(i) < m(i) such that

$$d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\ge t^{*},d\left(y_{n\left(i\right)},y_{m\left(i\right)-1}\right)<t^{*},$$

for i = 1, 2, 3, .... Then

$$\begin{array}{c}{t}^{*}\le d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\\ \le d\left(y_{n\left(i\right)},y_{m\left(i\right)-1}\right)+d\left(y_{m\left(i\right)-1},y_{m\left(i\right)}\right).\end{array}$$

Letting i → ∞ and using the fact that d(y _{n (i)}, y _{m (i)-1}) < t*, we obtain

$$\underset{n\to \infty }{\text{lim}}d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)=t^{*}.$$

For this t* > 0, by property (p) of Ψ there exists δ (t*) > 0, s(t*) < 1, such that

$$0\le r-t^{*}<\delta \left(t^{*}\right)\Rightarrow \Psi \left(r\right)\le s\left(t^{*}\right)<1.$$

For this δ(t*) > 0, there exists a natural number N₀ such that

$$0\le d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)-t^{*}<\delta \left(t^{*}\right),\mathsf{\text{whenever}}i\ge N_0.$$

Hence

$$i\ge N_o\Rightarrow \Psi \left(d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\right)\le s\left(t^{*}\right).$$

Now, inequality (1) yields

$$\begin{array}{ll}\hfill d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)& \le d\left(y_{n\left(i\right)},y_{n\left(i\right)+1}\right)+d\left(y_{n\left(i\right)+1},y_{m\left(i\right)+1}\right)+d\left(y_{m\left(i\right)+1},y_{m\left(i\right)}\right)\\ \le d\left(y_{n\left(i\right)},y_{n\left(i\right)+1}\right)+\Psi \left(d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\right)d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)+d\left(y_{m\left(i\right)+1},y_{m\left(i\right)}\right)\\ \le d\left(y_{n\left(i\right)},y_{n\left(i\right)+1}\right)+\Psi \left(d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\right)d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)+d\left(y_{m\left(i\right)+1},y_{m\left(i\right)}\right).\end{array}$$

Thus

$$d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)\le d\left(y_{n\left(i\right)},y_{n\left(i\right)+1}\right)+s\left(t^{*}\right)d\left(y_{n\left(i\right)},y_{m\left(i\right)}\right)+d\left(y_{m\left(i\right)+1},y_{m\left(i\right)}\right)$$

Letting i → ∞, we get

$$t^{*}\le s\left(t^{*}\right)t^{*}<t^{*},$$

a contradiction. Hence {y _n } is a Cauchy sequence in Range (R). By completeness of this space there exists an element z ∈ Range (R) such that y _n → z. It further implies that wRz for some w ∈ X. Now,

$$\begin{array}{ll}\hfill d\left(z,Tw\right)& \le d\left(z,y_n\right)+d\left(y_n,Tw\right)\\ \le d\left(z,y_n\right)+d\left(T{x}_{n-1},Tw\right)\\ \le d\left(z,y_n\right)+\Psi \left(D\left(R\left\{x_{n-1}\right\}\right),R\left\{w\right\}\right)D\left(R\left\{x_{n-1}\right\},R\left\{w\right\}\right)\\ <d\left(z,y_n\right)+D\left(R\left\{x_{n-1}\right\},R\left\{w\right\}\right)\\ <d\left(z,y_n\right)+d\left(y_{n-1},z\right).\end{array}$$

Letting n → ∞, we have d(z, Tw) = 0. It follows that z = Tw. Hence Tw ∈ R{w}. In the case when Range (T) is complete. The fact Range (T) ⊆ Range (R) implies that there exists an element z* ∈ Range (R) such that y _n → z*. The remaining part of the proof is same as in previous case.

Example 3.2 Let X = Y = R, d(x, y) = |x - y|. Define T : R → R, R : R ⇝ R as follows:

$$\begin{array}{l}Tx=\left\{\begin{array}{c}1\mathsf{\text{if}}x\in \mathbb{Q}\\ 0\mathsf{\text{if}}x\in {\mathbb{Q}}^{\prime },\end{array}\right.\\ \hfill R& =\left(\mathbb{Q}\times \left[0,4\right]\right)\cup \left({\mathbb{Q}}^{\prime }\times \left[7,9\right]\right)\end{array}$$

Then Range (T) = {0, 1} ⊂ Range (R) = [0, 4] ∪ [7, 9]. For$\Psi \left(t\right)=\frac{1}{3}$, all conditions of the above theorem are satisfied.

From Theorem 3.1, we deduce the following result immediately.

Theorem 3.3 Let X be nonempty set and (Y, d) be a metric space. Let T : X → Y, R : X ⇝ Y be such that R is left-total, Range (T) ⊆ Range (R) and Range (T) or Range (R) is complete. If there exists λ ∈ [0, 1) such that for all x, y ∈ X

$$d\left(Tx,Ty\right)\le \lambda D\left(R\left\{x\right\},R\left\{y\right\}\right).$$

Then there exists w ∈ X such that Tw ∈ R{w}.

In the following theorem, we prove the existence of a unique coincidence point of a pair of nonself mappings under a contractive condition.

Theorem 3.4 Let X be a nonempty set and (Y, d) be a metric space. T, S : X → Y be two mappings such that Range (T) ⊆ Range (S) and Range (T) or Range (S) is complete. If there exists a λ ∈ [0, 1) such that for all x, y ∈ X

$$d\left(Tx,Ty\right)\le \lambda d\left(Sx,Sy\right),$$

Then S and T have a coincidence point in X. Moreover, if either T or S is injective, then S and T have a unique coincidence point in X.

Proof. By Theorem 3.1, we obtain that there exists w ∈ X such that Tw = Sw, where,

$$Sw=\underset{x\to \infty }{\text{lim}}S{x}_n=\underset{x\to \infty }{\text{lim}}T{x}_{n-1},x_0\in X.$$

For uniqueness, assume that w₁, w₂ ∈ X, w₁ ≠ w₂, Tw₁ = Sw₁, and Tw₂ = Sw₂. Then d(Tw₁, Tw₂) ≤ λd(Sw₁, Sw₂). If S or T is injective, then

$$d\left(S{w}_1,S{w}_2\right)>0$$

and

$$d\left(S{w}_1,S{w}_2\right)=d\left(T{w}_1,T{w}_2\right)\le \lambda d\left(S{w}_1,S{w}_2\right),$$

a contradiction.

Remark 3.5 If in the above theorem we choose X = Y, and S = I (the identity mapping on X), we obtain the Banach contraction theorem.

4 Integral equations

The purpose of this section is to study the existence and uniqueness of solution of a general class of Fredholm integral equations of 2nd kind under various assumptions on the functions involved. Theorem 3.4 coupled with a function space (C [a, b], ℝ) and a contractive inequality are used to establish the result. Consider the integral equation:

$$fx\left(t\right)-\mu \underset{a}{\overset{b}{\int }}K\left(t,s\right)hx\left(s\right)ds=g\left(t\right)$$

(2)

were, x : [a, b] → ℝ is unknown, g : [a, b] → ℝ and h, f : ℝ → ℝ are given, μ is a parameter. The kernel K of the integral equation is defined on [a, b] × [a, b]. If f = h = I (the identity mapping on ℝ), then (2) is Known as Fredholm integral equation of 2nd kind (see also [19] and the references cited therein).

Theorem 4.1 Let K, f, g, h be continuous. Let c ∈ R such that, for all t, s ∈ [a, b]

$$\left|K\left(t,s\right)\right|\le c$$

and for each x ∈ (C[a, b], ℝ) there exists y ∈ (C[a, b], ℝ) such that

$$\left(fy\right)\left(t\right)=g\left(t\right)+\mu \underset{a}{\overset{b}{\int }}K\left(t,s\right)hx\left(s\right)ds.$$

If f is injective, there exists L ∈ R such that for all x, y ∈ R

$$\left|hx-hy\right|\le L\left|fx-fy\right|$$

and {fx : x ∈ (C[a, b], ℝ)} is complete. Then, for$\mu \in \left(-\frac{1}{c\left(b-a\right)L},\frac{1}{c\left(b-a\right)L}\right)$, there exists w ∈ (C[a, b], ℝ) such that for x₀ ∈ (C[a, b], ℝ),

$$fw\left(t\right)=\underset{x\to \infty }{\text{lim}}f{x}_n\left(t\right)=\underset{x\to \infty }{\text{lim}}\left[g\left(t\right)+\mu \underset{a}{\overset{b}{\int }}K\left(t,s\right)h{x}_{n-1}\left(s\right)ds\right],$$

(3)

and w is the unique solution of (2).

Proof. Let X = Y = (C[a, b], ℝ) and $d\left(x,y\right)=\underset{t\in \left[a,b\right]}{\text{max}}\left|x\left(t\right)-y\left(t\right)\right|$ for all x, y ∈ X. Let T, S : X → X be defined as follows:

$$\left(Tx\right)\left(t\right)=g\left(t\right)+\mu \underset{a}{\overset{b}{\int }}K\left(t,s\right)\left(hx\right)\left(s\right)ds,Sx=fx.$$

Then by assumptions SX = {Sx : x ∈ X} is complete. Let x* ∈ TX, then x* = Tx for x ∈ X and x* (t) = Tx (t). By assumptions there exists y ∈ X such that Tx (t) = fy (t), hence TX ⊆ SX. Since,

$$\begin{array}{ll}\hfill \left|\left(Tx\right)\left(t\right)-\left(Ty\right)\left(t\right)\right|& =\left|\mu \right|\left|\underset{a}{\overset{b}{\int }}\left[K\left(t,s\right)\left(hx\right)\left(s\right)\right]ds-\underset{a}{\overset{b}{\int }}\left[K\left(t,s\right)\left(hy\right)\left(s\right)\right]ds\right|\\ \le \left|\mu \right|\underset{b}{\overset{b}{\int }}c\left|\left(hx\right)\left(s\right)-\left(hy\right)\left(s\right)\right|ds\\ \le L\left|\mu \right|c\underset{b}{\overset{b}{\int }}\left|\left(fx\right)\left(s\right)-\left(fy\right)\left(s\right)\right|ds\\ \le L\left|\mu \right|c\underset{b}{\overset{b}{\int }}\left|\left(Sx\right)\left(s\right)-\left(Sy\right)\left(s\right)\right|ds\\ \le \left(\underset{t\in \left[a,b\right]}{\text{sup}}\left|\left(Sx\right)\left(t\right)-\left(Sy\right)\left(t\right)\right|\right)L\left|\mu \right|c\left|\underset{a}{\overset{b}{\int }}ds\right|\\ \le L\left|\mu \right|c\left(b-a\right)d\left(Sx,Sy\right).\end{array}$$

Therefore, for any $\mu \in \left(-\frac{1}{c\left(b-a\right)L},\frac{1}{c\left(b-a\right)L}\right)$, all conditions of Theorem 3.4 are satisfied. Hence, there exists a unique w ∈ X such that

$$fw\left(t\right)=\underset{x\to \infty }{\text{lim}}S{x}_n\left(t\right)=\underset{x\to \infty }{\text{lim}}T{x}_{n-1}\left(t\right)=T\left(w\right)\left(t\right),x_0\in X$$

for all t, which is the unique solution of (2).

Example 4.2 Consider the integral equation:

$${\left[3x\left(t\right)\right]}^3=\text{sin}t+\mu \underset{\mathsf{\text{0}}}{\overset{\frac{\pi }{\mathsf{\text{2}}}}{\int }}{\left[\sqrt{t}x\left(s\right)\right]}^{3}ds$$

(4)

Let$X=Y=\left(C\left[0,\frac{\pi }{2}\right],ℝ\right),d\left(x,y\right)=\underset{t\in C\left[0,\frac{\pi }{\mathsf{\text{2}}}\right]}{\text{max}}\left|x\left(t\right)-y\left(t\right)\right|$for all x, y ∈ X. Since,

$$\left|K\left(t,s\right)\right|=\left|t^{\frac{3}{2}}{s}^0\right|\le {\left(\frac{\pi }{2}\right)}^{\frac{3}{2}}$$

and

$$\left|x^3\left(t\right)-y^3\left(t\right)\right|\le \frac{1}{27}\left|{\left[3x\left(t\right)\right]}^3-{\left[3y\left(t\right)\right]}^3\right|$$

for all x, y ∈ R, therefore all conditions of Theorem 4.2 are satisfied for$c={\left(\frac{\pi }{\mathsf{\text{2}}}\right)}^{\frac{3}{2}},h\left(x\right)=x^3,f\left(x\right)=27x^3,g\left(x\right)=\text{sin}x,L=\frac{1}{27}$Hence for$\mu \in \left(-\frac{27}{{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}},\frac{27}{{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}}\right)$, there exists a unique solution of (4). We approximate the solution, by constructing the iterative sequences:

$$S{x}_n=T{x}_{n-1},x_0\in X,n=1,2,3\dots$$

in connection with the mappings S, T : X → X defined as follows:

$$\left(Tx\right)\left(t\right)=\text{sin}t+\mu \underset{\mathsf{\text{0}}}{\overset{\frac{\pi }{\mathsf{\text{2}}}}{\int }}{\left[\sqrt{t}x\left(s\right)\right]}^{3}ds,\left(Sx\right)\left(t\right)=27x^3\left(t\right).$$

Let$x_0:\left[0,\frac{\pi }{\mathsf{\text{2}}}\right]\to ℝ$be defined as x₀ (t) = 0. Then

$$\left(T{x}_0\right)\left(t\right)=\text{sin}t=\left(S{x}_1\right)\left(t\right).$$

It follows that

$$x_1\left(t\right)=\frac{1}{3}{\left(\text{sin}t\right)}^{\frac{1}{3}}.$$

Now

$$\begin{array}{ll}\hfill \left(T{x}_1\right)\left(t\right)& =\text{sin}t+\mu \underset{\mathsf{\text{0}}}{\overset{\frac{\pi }{\mathsf{\text{2}}}}{\int }}{\left[\sqrt{t}\frac{1}{3}{\left(\text{sin}s\right)}^{\frac{1}{3}}\right]}^{3}ds\\ =\text{sin}t+\mu \frac{1}{27}\underset{\mathsf{\text{0}}}{\overset{\frac{\pi }{\mathsf{\text{2}}}}{\int }}{t}^{\frac{3}{2}}\text{sin}sds\\ =\text{sin}t+\frac{1}{27}\mu t^{\frac{3}{2}}=\left(S{x}_2\right)\left(t\right).\end{array}$$

It implies that

$$x_2\left(t\right)=\frac{1}{3}{\left[\text{sin}t+\frac{1}{27}\mu t^{\frac{3}{2}}\right]}^{\frac{1}{3}}.$$

Similarly,

$$\left(T{x}_n\right)\left(t\right)=\text{sin}t+\sum _{j=1}^n{t}^{\frac{3}{2}}\frac{1}{{27}^j}{\mu }^j{\left(\frac{2}{5}\right)}^{\left(j-1\right)}{\left(\frac{\pi }{2}\right)}^{\left(j-1\right)\frac{5}{2}}=\left(S{x}_{n+1}\right)\left(t\right).$$

As$\mu \in \left(-\frac{27}{{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}},\frac{27}{{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}}\right)$, the series$\sum _{j=1}^{\infty }{t}^{\frac{3}{2}}\frac{1}{{27}^j}{\mu }^j{\left(\frac{\mathsf{\text{2}}}{\mathsf{\text{5}}}\right)}^{\left(j-1\right)}{\left(\frac{\pi }{2}\right)}^{\left(j-1\right)\frac{5}{2}}$is convergent and

$$\underset{x\to \infty }{\text{lim}}S{x}_n\left(t\right)=\text{sin}t+\left(\frac{\frac{1}{27}\mu t^{\frac{3}{2}}}{1-\frac{1}{27}\mu \frac{2}{5}{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}}\right)=Sw\left(t\right).$$

Hence,

$$\frac{1}{3}{\left[\text{sin}t+\left(\frac{\frac{1}{27}\mu t^{\frac{3}{2}}}{1-\frac{1.2}{27.5}{\left(\frac{\pi }{2}\right)}^{\frac{5}{2}}}\right)\right]}^{\frac{1}{3}}=w\left(t\right)$$

is the required solution.