*Case III.*

*For* l = 5, *κ* ∈ [ 94 , <sup>5</sup>), *we have:*

$$
\omega\left(\mathcal{I}(5), \mathcal{I}(\kappa)\right) = \omega\left(\{0\}, \frac{9}{4}\right) = \frac{9}{4} > 0.
$$

*Therefore,*

$$
\omega\left(\mathcal{I}(5), \mathcal{I}(\kappa)\right) < \max\left\{\omega\left(5, \kappa\right), \frac{\omega\left(5, \mathcal{I}(5)\right) \cdot \omega\left(\leq, \mathcal{I}(\kappa)\right)}{1 + \omega\left(5, \kappa\right)}\right\} = \mathcal{R}\left(5, \kappa\right).
$$

*Similarly to case I, we get:*

$$\left(\pi + \alpha\left(5,\kappa\right)\beta\left(5,\kappa\right)\,\theta\left(\omega\left(\left.\mathcal{I}\left(5\right),\mathcal{I}\left(\kappa\right)\right)\right) \leq \theta\left(\mathcal{R}\left(5,\kappa\right)\right)\right)$$

*Thus*J *is (<sup>α</sup>*, *β)-type rational ϑ-contraction. Moreover all the assumptions of Theorem 6 are satisfied and* 94 *is a fixed point of* J *.*

### **Corollary 5.** *Let:*

*1. The functions ϑ* ∈ Δ *and α* : Ω × Ω → {−∞} ∪ (0, ∞) *and the mapping* J : Ω → Ω *is α-admissible mapping such that for* l, *κ* ∈ Ω *and ω*(J (l), J (*κ*)) > 0 *the inequality:*

$$
\pi + \alpha(\mathfrak{l}, \mathfrak{k}) \vartheta \left( \omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k})) \right) \le \vartheta \left( \mathcal{R}(\mathfrak{l}, \mathfrak{k}) \right),
$$

*holds where*

$$\mathcal{R}(\mathfrak{l},\mathfrak{k}) = \max \left\{ \omega(\mathfrak{l},\mathfrak{k}), \frac{\omega(\mathfrak{l},\mathcal{J}(\mathfrak{l})) \omega(\mathfrak{k},\mathcal{J}(\mathfrak{k}))}{1 + \omega(\mathfrak{l},\mathfrak{k})} \right\}.$$

*2. The conditions (b) and (c) of Theorem 6 are fulfilled.*

*Then the point* l∗ *from condition (c) is a fixed point of the mapping* J *.*

**Proof.** The claim follows from Theorem 6 with *β*(<sup>l</sup>, *κ*) ≡ 1 for l, *κ* ∈Ω.

**Theorem 7.** *Suppose that the assertions of Theorem 5 are satisfied and the property (P) holds. Then the fixed point of the mapping* J *is unique.*

**Proof.** Let l∗, / l ∈ Ω be such that J (l∗) = l∗ and J ( / l) = / l but l∗ = / l. Then by (P), *<sup>α</sup>*(l<sup>∗</sup>, / l) ≥ 1 and *β*(l<sup>∗</sup>, / l) ≥ 1. Thus,

$$\begin{split} \left(\pi + \theta\left(\omega(\mathbb{I}^\*,\widehat{\mathbb{I}})\right)\right) &= \left.\pi + \theta\left(\omega(\mathcal{I}(\mathbb{I}^\*),\mathcal{J}(\widehat{\mathbb{I}}))\right)\right| \leq \pi + \theta\left(a(\mathbb{I}^\*,\widehat{\mathbb{I}})\beta(\mathbb{I}^\*,\widehat{\mathbb{I}})\omega(\mathcal{J}(\mathbb{I}^\*),\mathcal{J}(\widehat{\mathbb{I}}))\right) \\ &\leq \left.\theta\left(\max\{\omega(\mathbb{I}^\*,\widehat{\mathbb{I}}),\frac{\omega\left(\mathbb{I}^\*,\mathcal{J}(\mathbb{I}^\*)\right)\omega(\widehat{\mathbb{I}},\mathcal{J}(\widehat{\mathbb{I}})\right)}{1+\omega(\mathbb{I}^\*,\widehat{\mathbb{I}})}\right)\right| = \theta(\omega(\mathbb{I}^\*,\widehat{\mathbb{I}})). \end{split}$$

The above inequality is a contradiction because *π* > 0. Hence, l∗ is unique.

Now we define cyclic (*<sup>α</sup>*-*ϑ*) contraction and derive some results from our main theorems.

**Definition 6.** *Let the functions α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, ϑ* ∈ Δ*, the sets S*1, *S*2 ∈ *CL*(Ω)*, and* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *with* J *S*1 ⊆ *S*2 *and* J *S*2 ⊆ *S*1*. The mapping* J *is cyclic (<sup>α</sup>-ϑ) contraction if there exists a number π* > 0 *such that:*

$$
\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k})) > 0 \Longrightarrow \pi + \mathfrak{a}(\mathfrak{l}, \mathfrak{k}) \theta \left( \omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k})) \right) \leq \theta \left( \omega(\mathfrak{l}, \mathfrak{k}) \right),
$$

*holds for all* l ∈ *S*1 *and κ* ∈ *S*2.

**Theorem 8.** *Let the functions α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, ϑ* ∈ Δ*, the mapping* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *is a cyclic (<sup>α</sup>-ϑ) contraction and the following conditions be satisfied:*


*(c) The mapping* J *is continuous.*

> *Then the mapping* J *has a fixed point in S*1 ∩ *S*2*.*

**Proof.** We take Ω = *S*1 ∪ *S*2. Then (Ω, *ω*) is a complete metric space. Define *β* : Ω × Ω → {−∞} ∪ (0, ∞) by:

$$\beta(\mathfrak{l},\mathfrak{x}) = \begin{cases} \ 1, \text{ if } \mathfrak{l} \in \mathcal{S}\_1 \text{ and } \mathfrak{x} \in \mathcal{S}\_2 \\\ \ 0, \text{ otherwise.} \end{cases}$$

Then J is twisted (*<sup>α</sup>*,*β*)-admissible. Let the point l0 ∈ Ω be defined in condition (b). Then *β*(<sup>l</sup>0, J (<sup>l</sup>0)) ≥ 1 holds. Therefore, the assumptions of Theorem 2 are fulfilled and there exists a point l∗ ∈ *S*1 ∪ *S*2 such that J (l∗) = l<sup>∗</sup>. If l∗ ∈ *S*1, then l∗ = J (l∗) ∈ *S*2 because J *S*1 ⊆ *S*2. Thus ∃ l∗ ∈ *S*1 ∩ *S*2 such that J (l∗) = l<sup>∗</sup>. Similarly, if l∗ ∈ *S*2, then l∗ = J (l∗) ∈ *S*1 because J *S*2 ⊆ *S*1. Thus ∃ l∗ ∈ *S*1 ∩ *S*2 such that J (l∗) = l<sup>∗</sup>.

**Theorem 9.** *Let the functions α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, ϑ* ∈ Δ*, the mapping* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *is a cyclic (<sup>α</sup>-ϑ) contraction and the following conditions be satisfied:*


*Then the mapping* J *has a fixed point in S*1 ∩ *S*2*.*

**Proof.** We take Ω = *S*1 ∪ *S*2. As in the proof of Theorem 8 we define the function *β* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>). Then J is twisted (*<sup>α</sup>*,*β*)-admissible. Let the point l0 ∈ Ω be defined in condition (b). Then *β*(<sup>l</sup>0, J (<sup>l</sup>0)) ≥ 1 holds. Let {<sup>l</sup>*j*} ⊆ Ω such that *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 and *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 for all *j* ∈ N ∪ {0} and l*j* → l∗ as *j* → +<sup>∞</sup>. Then l*j* ∈ *S*1 and <sup>l</sup>*j*+<sup>1</sup> ∈ *S*2. Now as *S*2 is closed, so l∗ ∈ *S*2 and hence *<sup>α</sup>*(<sup>l</sup>*j*, l∗) ≥ 1 and *β*(<sup>l</sup>*j*, l∗) ≥ 1. Therefore, the assumptions of Theorem 3 are fulfilled and ∃ l∗ ∈ *S*1 ∪ *S*2 such that J (l∗) = l<sup>∗</sup>. If l∗ ∈ *S*1, then l∗ = J (l∗) ∈ *S*2 because J *S*1 ⊆ *S*2. Thus ∃ l∗ ∈ *S*1 ∩ *S*2 such that J (l∗) = l<sup>∗</sup>. Similarly, if l∗ ∈ *S*2, then l∗ = J (l∗) ∈ *S*1 because J *S*2 ⊆ *S*1. Thus ∃ l∗ ∈ *S*1 ∩ *S*2 such that J (l∗) = l<sup>∗</sup>.

**Corollary 6.** *Let the function ϑ* ∈ Δ*, the sets S*1, *S*2 ∈ *CL*(Ω)*, and* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *with* J *S*1 ⊆ *S*2 *and* J *S*2 ⊆ *S*1 *is continuous and the inequality:*

> *ω*(J (l), J (*κ*)) > 0 =⇒ *π* + *ϑ ω*(J (l), J (*κ*)) ≤ *ϑ <sup>ω</sup>*(<sup>l</sup>, *κ*)

*holds for all* l ∈ *S*1 *and κ* ∈ *S*2. *Then the mapping* J *has a fixed point in S*1 ∩ *S*2*.*

**Proof.** The claim follows from Theorem 8 with *<sup>α</sup>*(<sup>l</sup>, *κ*) = 1 for all l <sup>∈</sup>*S*1 and *<sup>κ</sup>*∈*S*2.

Now we define cyclic ordered (*<sup>α</sup>*-*ϑ*) contraction and derive some results from our main theorems.

**Definition 7.** *Let (*Ω*, ω,* #*) be an ordered metric space and S*1, *S*2 ∈ *CL*(Ω)*, and* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *with* J *S*1 ⊆ *S*2 *and* J *S*2 ⊆ *S*1*. The mapping* J *is a cyclic ordered (<sup>α</sup>-ϑ) contraction if there exists a number π* > 0 *and α* : Ω × Ω → {−∞} ∪ (0, ∞) *such that:*

$$\omega\left(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k})\right) > 0 \Longrightarrow \pi + \mathfrak{a}\left(\mathfrak{l}, \mathfrak{k}\right) \theta\left(\omega\left(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{k})\right)\right) \leq \theta\left(\omega(\mathfrak{l}, \mathfrak{k})\right)$$

*holds for all* l ∈ *S*1 *and κ* ∈ *S*2 *with* l # , *where ϑ* ∈ Δ. **Theorem 10.** *Let the functions α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, ϑ* ∈ Δ*, the mapping* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *is decreasing continuous cyclic ordered (<sup>α</sup>-ϑ) contraction and the following conditions be satisfied:*


*Then* ∃l∗ ∈ *S*1 ∩ *S*2 *such that* l∗ = J (l<sup>∗</sup>)*.*

**Proof.** We take Ω = *S*1 ∪ *S*2. Then (Ω, *ω*) is a complete metric space. Define *β* : Ω × Ω → {−∞} ∪ (0, ∞) by:

$$\beta(\mathfrak{l},\kappa) = \begin{cases} \ 1, \text{ if } \mathfrak{l} \in \mathcal{S}\_1 \text{ and } \kappa \in \mathcal{S}\_{2\prime} \text{ with } \mathfrak{l} \preceq \kappa^\* \\\ \qquad 0, \text{ otherwise.} \end{cases}$$

Let *β*(<sup>l</sup>, *κ*) ≥ 1, for all l, *κ* ∈ Ω, then l <sup>∈</sup>*S*1 and *<sup>κ</sup>*∈*S*2 with l # *κ*. It follows that J (l)∈*S*2 and J (*κ*)∈*S*1 with J (*κ*) # J (l), since J is decreasing. Therefore *β*(J (*κ*), J (l)) ≥ 1, that is, J is twisted (*<sup>α</sup>*,*β*)-admissible. Now, let *<sup>α</sup>*(<sup>l</sup>0, J (<sup>l</sup>0)) ≥ 1, with l0 ∈ *S*1 and l0 # J (<sup>l</sup>0). From l0 ∈ *S*1, we have J (<sup>l</sup>0) ∈ *S*2 with l0 # J (<sup>l</sup>0), that is *β*(<sup>l</sup>0, J (<sup>l</sup>0)) ≥ 1. Then all assumptions of Theorem 2 are satisfied and J has a fixed point l∗ in *S*1 ∪ *S*2. The remaining proof is identical to the proof of Theorem 9.

**Theorem 11.** *Let the functions α* : Ω × Ω → {−∞} ∪ (0, <sup>∞</sup>)*, ϑ* ∈ Δ*, the mapping* J : *S*1 ∪ *S*2 → *S*1 ∪ *S*2 *is a cyclic ordered (<sup>α</sup>-ϑ) contraction and the following conditions be satisfied:*


*Then* ∃l∗ ∈ *S*1 ∩ *S*2 *such that* l∗ = J (l<sup>∗</sup>)*.*

**Proof.** We take Ω = *S*1 ∪*S*2. As in the proof of Theorem 10 we define the function *β* : Ω × Ω → [0, <sup>+</sup>∞). Then J is twisted (*<sup>α</sup>*,*β*)-admissible. Let {<sup>l</sup>*j*} ⊆ Ω such that *<sup>α</sup>*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 and *β*(<sup>l</sup>*j*, <sup>l</sup>*j*+<sup>1</sup>) ≥ 1 for all *j* ∈ N ∪ {0} and l*j* → l∗ as *j* → +<sup>∞</sup>. Then l*j* ∈ *S*1 and <sup>l</sup>*j*+<sup>1</sup> ∈ *S*2. Now as *S*2 is closed, so l∗ ∈ *S*2 and hence l*j* # l∗ and *β*(<sup>l</sup>*j*, l∗) ≥ 1. Therefore, the assumptions of Theorem 3 are fulfilled and ∃ l∗ ∈ *S*1 ∪ *S*2 such that J (l∗) = l<sup>∗</sup>. The remaining proof is identical to the proof of Theorem 6.

#### **4. Applications to Caputo Fractional Differential Equations**

Recently, many researchers have studied the existence of solutions of varies types of fractional differential equations. In this paper we will emphasize our study of Caputo fractional differential equations of the fractional order in (1, 2) and the integral boundary condition. Note that similar problems are studied in [25–27] but the main condition is connected with enough small Lipschitz constant of the right hand side part of the equation. Based on the obtained fixed points theorems we can use weaker conditions for the right hand side part of the equation (see Example 5).

We will apply some of the proved above Theorems to investigate the existence of the solutions of the nonlinear Caputo fractional differential equation:

$$^G\_a D\_t^q(\mathbf{x}(t)) = f(t, \mathbf{x}(t)) \quad \text{ for } t \in (a, b) \tag{26}$$

with the integral boundary condition:

$$\mathbf{x}(a) = 0, \ x(b) = \int\_{a}^{\lambda} \mathbf{x}(s) ds \qquad (a < \lambda < b) \tag{27}$$

where *x* ∈ R, *q* ∈ (1, <sup>2</sup>), *Ca Dqt x*(*t*) = 1 <sup>Γ</sup>(<sup>2</sup>−*q*) *t a* (*t* − *s*)<sup>1</sup>−*<sup>q</sup> x* (*s*)*ds* represents the Caputo fractional derivative, and *a*, *b* : 0 ≤ *a* < *b* are given real numbers.

Let Ω = *<sup>C</sup>*([*<sup>a</sup>*, *b*], R) with a norm *x*[*<sup>a</sup>*,*b*] = sup*s*<sup>∈</sup>[*<sup>a</sup>*,*b*] |*x*(*s*)| . For any *x*, *y* ∈ Ω we define *<sup>ω</sup>*(*<sup>x</sup>*, *y*) = *x* − *<sup>y</sup>*[*<sup>a</sup>*,*b*].

Consider the linear fractional differential equation:

$$\, \_a^C D\_t^q(\mathbf{x}(t)) = \mathbf{g}(t) \quad \text{ for } t \in (a, b) \tag{28}$$

with the integral boundary condition (27) where *g* ∈ Ω.

**Lemma 1.** *Let g* ∈ Ω*. Then the boundary value problem (28), (27) has a solution:*

$$\begin{split} x(t) &= \frac{1}{\Gamma(q)} \int\_{a}^{t} (t-s)^{q-1} \mathbf{g}(s) ds \\ &+ \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{b} (b-s)^{q-1} \mathbf{g}(s) ds \\ &- \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{\lambda} \int\_{a}^{s} (s-\xi)^{q-1} \mathbf{g}(\xi) d\xi ds. \end{split} \tag{29}$$

The proof of Lemma 1 is based on the presentation of the solution *x*(*t*) = 1 <sup>Γ</sup>(*q*) *ta* (*t* − *s*)*<sup>q</sup>*−<sup>1</sup>*g*(*s*)*ds* − *d*1 − *d*2(*<sup>t</sup>* − *a*) given in [28].

Based on the presentation (29) we will define a mild solution of (26) and (27).

**Definition 8.** *The function x* ∈ Ω *is a mild solution of the boundary value problem (26) and (27) if it satisfies:*

$$\begin{split} \mathbf{x}(t) &= \frac{1}{\Gamma(q)} \int\_{a}^{t} (t-s)^{q-1} \, f \left(s, \mathbf{x}(s)\right) ds \\ &+ \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{b} (b-s)^{q-1} \, f \left(s, \mathbf{x}(s)\right) ds \\ &- \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{\lambda} \int\_{a}^{s} (s-\xi)^{q-1} \, f \left(\xi, \mathbf{x}(\xi)\right) d\xi ds, \quad t \in [a,b]. \end{split}$$

For any function *u* ∈ Ω, we define the mapping J : Ω → Ω by:

$$\begin{split} \mathcal{I}(u)(t) &= \frac{1}{\Gamma(q)} \int\_{a}^{t} (t-s)^{q-1} f\left(s, u(s)\right) ds \\ &+ \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{b} (b-s)^{q-1} f\left(s, u(s)\right) ds \\ &- \frac{2(t-a)}{((\lambda-a)^{2} - 2(b-a))\Gamma(q)} \int\_{a}^{\lambda} \int\_{a}^{s} (s-\overline{\xi})^{q-1} f\left(\overline{\xi}, u(\overline{\xi})\right) d\overline{\xi} ds, \\ &\quad \text{for} \quad t \in [a, b]. \end{split} \tag{30}$$

Now, we establish the existence result as follows.

#### **Theorem 12.** *Suppose that:*

(i) The function *f* ∈ *<sup>C</sup>*([*<sup>a</sup>*, *b*] × R, R) and there exists a constant *K* such that:

$$\frac{K(b-a)^q}{\Gamma(1+q)}\left(1+\frac{2K(b-a)}{(2(b-a)-(\lambda-a)^2)}\left(1+\frac{\lambda-a}{1+q}\right)\right)\in(0,1)\tag{31}$$

and a number *p* ∈ (0, 1] such that:

$$|f(t, \mathbf{x}) - f(t, y)| \le K|\mathbf{x} - y|^p, \quad \mathbf{x}, y \in \mathbb{R}, \ t \in [a, b]\varphi$$


Then the boundary value problem (26),(27) has a mild solution.

**Proof.** Note that any fixed point of the mapping J is a mild solution of the boundary value problem (26) and (27).

Now, let *x*, *y* ∈ Ω be such that *<sup>ω</sup>*(*<sup>x</sup>*, *y*) > 0. By condition (i) of Theorem 12 we obtain:



or

with

$$
\omega(\mathcal{J}(\mathbf{x}), \mathcal{J}(y)) \le \Lambda(\omega(\mathbf{x}, y))^p. \tag{32}
$$

From (32) applying condition (ii) we get:

$$\ln \left( \omega(\mathcal{J}(\mathbf{x}), \mathcal{J}(y)) \right) \le \ln(\Lambda) + p \ln \left( \omega(\mathbf{x}, y) \right) \dots$$

Thus,

$$
\ln \left( \frac{1}{\Lambda} \right)^{\frac{1}{p}} + \frac{1}{p} \ln \left( \omega(\mathcal{J}(\mathbf{x}), \mathcal{J}(y)) \right) \le \ln(\omega(\mathbf{x}, y)).
$$

Therefore, the operator J is (*<sup>α</sup>*, *β*)-type *ϑ*-contraction with *<sup>ϑ</sup>*(*u*) = ln *u* ∈ Δ (see Example 1), *π* = ln 1Λ 1 *p* > 0, and the mappings *α*, *β* : Ω × Ω → {−∞} ∪ (0, <sup>+</sup>∞) are defined by:

$$\alpha(\mathbf{x}, y) = \begin{cases} \frac{1}{p} & \text{if } \omega(\mathbf{x}, y) > 0, \\ 0.1 & \text{otherwise} \end{cases} \quad \beta(\mathbf{x}, y) = \begin{cases} 1 & \text{if } \omega(\mathbf{x}, y) > 0, \\ -\infty & \text{otherwise}. \end{cases}$$

Therefore, the assumption (i) of Theorem 3 is satisfied.

The operator J is twisted (*<sup>α</sup>*,*β*)-admissible because for any *x*, *y* ∈ Ω if *<sup>α</sup>*(*<sup>x</sup>*, *y*) ≥ 1 and *β*(*<sup>x</sup>*, *y*) ≥ 1 then from definitions of *α*, *β* it follows that *<sup>ω</sup>*(*<sup>x</sup>*, *y*) > 0 and from condition (iii) of Theorem 12 the inequality *ω*(J (*x*)(*t*), J (*y*)(*t*)) > 0 holds. Thus, *α*(J (*x*), J (*y*)) ≥ 1 and *β*(J (*x*), J (*y*)) ≥ 1. Therefore, the condition (a) of Theorem 2 is satisfied.

From condition (ii) of Theorem 12, there exists a point *x*0 ∈ Ω such that *<sup>ω</sup>*(*<sup>x</sup>*0, J (*<sup>x</sup>*0)) > 0 and therefore, *<sup>α</sup>*(*<sup>x</sup>*0, J (*<sup>x</sup>*0)) = 1*p* ≥ 1 and *β*(*<sup>x</sup>*0, J (*<sup>x</sup>*0)) = 1 ≥ 1. Thus condition (b) of Theorem 2 is satisfied.

According to Theorem 3 the operator J has a fixed point in Ω, i.e., there exists a function *x*<sup>∗</sup> ∈ *<sup>C</sup>*([*<sup>a</sup>*, *b*], R) such that *x*<sup>∗</sup> = J (*x*<sup>∗</sup>). This function *x*<sup>∗</sup> is a mild solution of the boundary value problem for (26) and (27).

**Remark 2.** *Note that the condition (i) of Theorem 12 for the function f*(*<sup>t</sup>*, *x*) *is less restrictive than the Lipschitz condition used in many existence results (see, for example [25]).*

Now we will provide an example to demonstrate the existence result.

**Example 5.** *Consider the nonlinear Caputo fractional differential equation:*

$$\, \_{2}^{C}D\_{t}^{1.75}(\mathbf{x}(t)) = \frac{1}{\sqrt{t+14}} \arctan(\sqrt{|\mathbf{x}(t)|} + e^{t}\cos t) + \sin t \quad \text{for } t \in (2,3) \tag{33}$$

*with the integral boundary condition:*

$$\mathbf{x}(2) = 0 \ x(3) = \int\_0^{2.5} \mathbf{x}(s) ds. \tag{34}$$

*In this case f*(*<sup>t</sup>*, *u*) = √ 1*t*+14 arctan(.|*u*| + *et* cos *t*) + sin *t and* | *f*(*<sup>t</sup>*, *x*) − *f*(*<sup>t</sup>*, *u*)| ≤ 0.25.|*x* − *y*|*. The condition (31) is reduced to:*

$$\frac{K(b-a)^q}{\Gamma(1+q)} \left( 1 + \frac{2K(b-a)}{(2(b-a) - (\lambda-a)^2)} \left( 1 + \frac{\lambda-a}{1+q} \right) \right)$$

$$= \frac{K}{\Gamma(2.\mathcal{T}5)} \left( 1 + \frac{2K}{1.\mathcal{T}5} \frac{3.25}{2.\mathcal{T}5} \right) = 0.215998 \in (0,1)$$

*with K* = 0.25*.*

> *According to Theorem 12 the boundary value problem (33) and (34) has a solution.*

**Remark 3.** *Note that the boundary value problem (33) and (34) is studied in [25], but the absolute value is missing under the square root. Also, the function f*(*<sup>t</sup>*, *x*) *is assumed as Lipschitz, but it is not (see Figure 1 for the particular value t* = 2.2 ∈ (2, 3)*). At the same time the function f*(*<sup>t</sup>*, *x*) *satisfies the condition 1 with k* = 0.25 *(see Figure 2 for the particular value t* = 2.2 ∈ (2, 3)*), and by one of the fixed point theorems proved in this paper the existence of the solution follows.*

## **5. Discussion**

In fixed point theory, the contractive inequality and underlying space play a significant role. A pioneer result in this theory is a Banach contraction principle that consists of compete metric space (Ω, *ω*) as underlying space and the following contractive inequality:

$$
\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{x})) \le \pi \omega(\mathfrak{l}, \mathfrak{x}) \tag{35}
$$

in which J is a self mapping and *π* ∈ [0, <sup>1</sup>). Over the years, many mathematicians have generalized and extended above contractive inequality in different ways.

In 2012, Wardowski ([10]) initiated the application of a mapping J : (<sup>Ω</sup>, *ω*) → (<sup>Ω</sup>, *ω*) and *π* > 0 such that:

$$
\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{x})) > 0 \implies \pi + \theta\left(\omega(\mathcal{J}(\mathfrak{l}), \mathcal{J}(\mathfrak{x}))\right) \le \theta\left(\omega(\mathfrak{l}, \mathfrak{x})\right) \tag{36}
$$

for l, *κ* ∈ Ω, where *ϑ* : (0, <sup>+</sup>∞) → R satisfies the following conditions:


As it is pointed out in [10] the introduced mapping and inequality (36) are a generalization of Banach contraction (35) with *ϑ*(l) = ln(l), for l > 0.

In this paper, we generalized the mapping used in [10] by introducing two new notions (*<sup>α</sup>*, *β*)-type *ϑ*-contraction and (*<sup>α</sup>*, *β*)-type rational *ϑ*-contraction.

As a partial case of some of our results, we obtained known results in the literature. For example, if *<sup>α</sup>*(<sup>l</sup>, *κ*) = *β*(<sup>l</sup>, *κ*) = 1 in Theorem 2 then we obtain Theorem 1 ([10]) by which one can derive the result of [1].
