In view of Lemma 2.5, we introduce a fixed point problem associated with the problem (1)-(2) as follows:
$$u=\mathcal{H}u,$$(13)

where the operator $\mathcal{H}:\mathcal{E}\to \mathcal{E}$ is
$$\begin{array}{c}(\mathcal{H}u)(t)={v}_{1}(t)+\underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}\left(\underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}f(x,\text{\hspace{0.17em}}u(x))dx\right)ds+{v}_{2}(t)\underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}f(s,\text{\hspace{0.17em}}u(s))ds\\ \text{\hspace{0.17em}}+{v}_{3}(t)\underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}\left(\underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}f(x,\text{\hspace{0.17em}}u(x))dx\right)ds.\end{array}$$(14)

Here $\mathcal{E}=C([0,T],\mathbb{R})$ denotes the Banach space of all continuous functions from [0, *T*] → ℝ endowed with the norm defined by $\Vert u\Vert =\mathrm{sup}\left\{\left|u(t)\right|,t\in [0,T]\right\}$.

Observe that that problem (1)-(2) has solutions if the operator equation (13) has fixed points. For computational convenience, we set the notation:
$$Q=\underset{t\in \left[0,T\right]}{\mathrm{sup}}\left\{\frac{1-{e}^{-kt}}{k\text{\Gamma}(\alpha ){t}^{1-\alpha}}+\frac{\left|{v}_{2}(t)\right|}{\text{\Gamma}(\alpha ){T}^{1-\alpha}}+\frac{\left|{v}_{3}(t)\right|(1-{e}^{-kT})}{k\text{\Gamma}(\alpha ){T}^{1-\alpha}}\right\}.$$(15)

Now we are in a position to present our main results for the problem (1)-(2). The first one deals with Schaefer’s fixed point theorem [24].

**Lemma 3.1:** *Let X be a Banach space. Assume that* $\mathcal{T}:X\to X$ *is a completely continuous operator and the set* $Y=\{u\in X|u=\mu \mathcal{T}u,0<\mu <1\}$ *is bounded. Then* $\mathcal{T}$ *has a fixed point in X*.

**Theorem 3.2:** *Assume that there exists a positive constant* *L*_{1} *such that* $\left|f(t,u(t))\right|\le {L}_{1}$ *for* $t\in [0,T],u\in \mathbb{R}$. *Then the boundary value problem (1)-(2) has at least one solution on* [0, *T*].

**Proof:** *In the first step, we show that the operator $\mathcal{H}$ defined by (14) is completely continuous. Observe that continuity of $\mathcal{H}$ follows from the continuity of **f*. For a positive constant *r*, let ${B}_{r}=\{u\in \mathcal{E}:\Vert u\Vert \le r\}$ be a bounded ball in $\mathcal{E}$. Then for *t* ∈ [0, *T*], we have
$$\begin{array}{l}\left|(\mathcal{H}u)(t)\right|\le \left|{v}_{1}(t)\right|+{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,\text{\hspace{0.17em}}u(x))\right|dx}\right)ds+\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(s,\text{\hspace{0.17em}}u(s))\right|ds}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,\text{\hspace{0.17em}}u(x))\right|dx}\right)ds.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\le \left|{v}_{1}(t)\right|+{L}_{1}\{{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}}dx\right)ds+\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}ds}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx}\right)ds\}\le {L}_{1}Q+\Vert {v}_{1}\Vert ,\end{array}$$
which consequently implies that
$$\u2225(\mathcal{H}u)\u2225\le {L}_{1}Q+\u2225{v}_{1}\u2225,$$
where *Q* is defined by (15).Next we show that *the operator* $\mathcal{H}$ *maps bounded sets into equicontinuous sets of* $\mathcal{E}$. Let *τ*_{1}, *τ*_{2} ∈ [0, *T*] with *τ*_{1} < *τ*_{2} and *u* ∈ *B*_{r}. Then we have
$$\begin{array}{c}\left|(\mathcal{H}u)({\tau}_{2})-(\mathcal{H}u)({\tau}_{1})\right|\le \left|{v}_{1}({\tau}_{2})-{v}_{1}({\tau}_{1})\right|\\ \text{\hspace{0.17em}}+{L}_{1}\left[\left|{e}^{-k{\tau}_{2}}-{e}^{-k{\tau}_{1}}\right|\underset{0}{\overset{{\tau}_{1}}{\int}}{e}^{ks}\left(\underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx\right)ds+\underset{{\tau}_{1}}{\overset{{\tau}_{2}}{\int}}{e}^{-k({\tau}_{2}-s)}\left(\underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx\right)ds\right.\\ \text{\hspace{0.17em}}\left.+\left|{v}_{2}({\tau}_{2})-{v}_{2}({\tau}_{1})\right|\underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}ds+\left|{v}_{3}({\tau}_{2})-{v}_{3}({\tau}_{1})\right|\underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}\left(\underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx\right)ds\right].\end{array}$$
As *τ*_{1} — *τ*_{2} → 0, the right-hand side of the above inequality tends to zero independently of *u* ∈ *B*_{r}. Therefore, by the Arzelá-Ascoli theorem, the operator $\mathcal{H}:\mathcal{E}\to \mathcal{E}$ is completely continuous.Finally, we consider the set $V=\{u\in \mathcal{E}:u=\mu \mathcal{H}u,\text{\hspace{0.17em}}0<\mu <1\}$ and show that *V* is bounded. For *u* ∈ *V* and *t* ∈ [0, *T*], we get
$$\Vert u\Vert \le {L}_{1}Q+\Vert {v}_{1}\Vert .$$
Therefore, *V* is bounded. Hence, by Lemma 3.1, the problem (1)-(2) has at least one solution on [0, *T*].

*
*Our next existence result is based on Leray-Schauder’s nonlinear alternative.

**Lemma 3.3:** *(Nonlinear alternative for single valued maps [2–5]). **Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and* 0 ∈ *U*. *Suppose that* $\mathcal{A}:\overline{U}\to C$ *is a continuous, compact (that is*, $\mathcal{A}(\overline{U})$ *is a relatively compact subset of C ) map. Then either*

*(i)*

* $\mathcal{A}$ **has a fixed point in $\overline{U}$, or*

*(ii)*

*there is a x* ∈ ∂*U* *(the boundary of U in C) and* *λ* ∈ (0, 1) *with* $x=\text{\lambda}\mathcal{A}(x)$.

*
***Theorem 3.4:** *Assume that*

*(E1)*

*there exists a continuous nondecreasing function* $\chi :[0,\infty )\to (0,\infty )$ *and a function* *p* ∈ *C*([0, *T*], ℝ^{+}) *such that*
$$\left|f(t,\text{\hspace{0.17em}}u)\right|\le p(t)\chi (\Vert u\Vert )\text{\hspace{0.17em}}for\text{\hspace{0.17em}}each\text{\hspace{0.17em}}(t,\text{\hspace{0.17em}}u)\in \left[0,T\right]\times \mathbb{R};$$

*(E2)*

*there exists a constant **N* > 0 such that
$$\frac{N}{\chi (N)\Vert p\Vert Q+\Vert {v}_{1}\Vert}>1,$$(16)

*where Q is given by (15)*.

*
**Then the boundary value problem (1)-(2) has at least one solution on* [0, *T*].The next existence result is based on Leray-Schauder’s degree theory [25].

**Theorem 3.5:** *Let* *f* : [0, *T*] × ℝ → ℝ *be a continuous function. Suppose that* (*E*_{3}) *there exist constants* 0 ≤ ω < *Q*^{-1}, *and* *M*_{1} > 0 *such that*
$$\left|f(t,\text{\hspace{0.17em}}u)\right|\le \omega \left|u\right|+{M}_{1}\text{\hspace{0.17em}}for\text{\hspace{0.17em}}all\text{\hspace{0.17em}(}t,\text{\hspace{0.17em}}u\text{)}\in \left[0,T\right]\times \mathbb{R},$$
*where Q is given by (15)*.*Then the boundary value problem (1)-(2) has at least one solution on* [0, *T*].

Next we show the existence of a unique solution of the problem (1)-(2) by applying Banach’s contraction mapping principle.

**Theorem 3.6:** *Assume that* *f* : [0, *T*] × ℝ → ℝ *is a continuous function satisfying the Lipschitz condition*: (*E*_{4}) *there exists a positive number ℓ such that* $\left|f(t,u)-f(t,\upsilon )\right|\le \ell \left|u-\upsilon \right|,\forall t\in [0,\text{\hspace{0.17em}}T],\text{\hspace{0.17em}}u,\text{\hspace{0.17em}}\upsilon \in \mathbb{R}$.*Then the boundary value problem (1)-(2) has a unique solution on* [0, *T*] *if* *Q* < 1/*ℓ*, *where* *Q* *is given by (15)*.

**Proof:** *Consider a set ${B}_{r}=\{u\in \mathcal{E}:\Vert u\Vert \le r\}$ with $r\ge \frac{QM+\Vert {v}_{1}\Vert}{1-\ell Q}$, where $M={\mathrm{sup}}_{t\in [0,T]}\left|f(t,0)\right|$ and **Q* is given by (15). In the first step, we show that $\mathcal{H}{B}_{r}\subset {B}_{r}$, where the operator $\mathcal{H}$ is defined by (14). For any *u* ∈ *B*_{r}, *t* ∈ [0, *T*], observe that
$$\left|f(t,\text{\hspace{0.17em}}u(t))\right|=\left|f(t,\text{\hspace{0.17em}}u(t))-f(t,\text{\hspace{0.17em}}0)+f(t,\text{\hspace{0.17em}}0)\right|\le \left|f(t,\text{\hspace{0.17em}}u(t))-f(t,\text{\hspace{0.17em}}0)\right|+\left|f(t,\text{\hspace{0.17em}}0)\right|\le \ell \Vert u\Vert +M\le \ell r+M,$$
where we have used the assumption (*E*_{4}). Then, for *u* ∈ *B*_{r}, we obtain
$$\begin{array}{l}\Vert (\mathcal{H}u)\Vert \le \underset{t\in \left[0,T\right]}{\mathrm{sup}}\{\left|{v}_{1}(t)+\right|{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,\text{\hspace{0.17em}}u(x))\right|dx}\right)ds\\ +\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(s,\text{\hspace{0.17em}}u(s))\right|}ds\\ +\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,\text{\hspace{0.17em}}u(x))\right|dx}\right)ds.\}\\ \le \left(\ell r+M\right)\underset{t\in \left[0,T\right]}{\mathrm{sup}}\{{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx}\right)ds\\ +\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}ds+}\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx}\right)ds\}+\Vert {v}_{1}\Vert \\ \le \left(\ell r+M\right)Q+\Vert {v}_{1}\Vert \le r,\end{array}$$
which implies that $\mathcal{H}u\in {B}_{r}$. Thus $\mathcal{H}{B}_{r}\subset {B}_{r}$. Next we show that the operator $\mathcal{H}$ is a contraction. Using the assumption (*E*_{4}) and (15), we get
$$\begin{array}{l}\Vert \mathcal{H}u-\mathcal{H}v\Vert \le \underset{t\in [0,T]}{\mathrm{sup}}\{{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,u(x))-f(x,v(x))\right|dx}\right)}ds\\ +\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}}\left|f(s,u(s))-f(s,v(s))\right|ds\\ +\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}\left|f(x,u(x))-f(x,v(x))\right|dx}\right)ds}\}\\ \le \ell \Vert u-v\Vert \underset{t\in [0,T]}{\mathrm{sup}}\{{\displaystyle \underset{0}{\overset{t}{\int}}{e}^{-k(t-s)}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx}\right)ds}\\ +\left|{v}_{2}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}\frac{{(T-s)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}ds}+\left|{v}_{3}(t)\right|{\displaystyle \underset{0}{\overset{T}{\int}}{e}^{-k(T-s)}\left({\displaystyle \underset{0}{\overset{s}{\int}}\frac{{(s-x)}^{\alpha -2}}{\text{\Gamma}(\alpha -1)}dx}\right)ds}\}\\ \le \ell Q\Vert u-v\Vert .\end{array}$$
In view of the assumption: *Q* < 1/ℓ, it follows that the operator $\mathcal{H}$ is a contraction. Thus, by Banach’s contraction mapping principle, we deduce that the operator $\mathcal{H}$ has a fixed point, which in turn implies that there exists a unique solution for the problem (1)-(2) on [0, *T*].

*
*In the following theorem, we show the existence of solutions for the problem (1)-(2) by applying Krasnoselskii’s fixed point theorem.

**Lemma 3.7:** *(Krasnoselskii’s fixed point theorem [24]). **Let Y be a closed bounded, convex and nonempty subset of a Banach space X. Let* *B*_{1}, *B*_{2} *be the operators such that (i)* *B*_{1}*y*_{1} + *B*_{2}*y*_{2} *∈* *Y* *whenever* *y*_{1},*y*_{2} *∈* *Y;* *(ii)* *B*_{1} *is compact and continuous and (iii)* *B*_{2} *is a contraction mapping. Then there exists* *z* *∈* *Y* *such that* *z* = *B*_{1}*z* + *B*_{2}*z*.

**Theorem 3.8:** *Let* *f* : [0,*T*] × ℝ → ℝ *be a continuous function satisfying the condition* (*E*_{4}) *and that* $\left|f\left(t,x\right)\right|\le g\left(t\right)$ *∀*(*t*,*x*) ∈ [0, *T*] × ℝ *with* *g* ∈ *C*([0, *T*], ℝ^{+}) *and* ${\mathrm{sup}}_{t\in \left[0,T\right]}\left|g\left(t\right)\right|=\Vert g\Vert $ . *In addition, it is assumed that* ${Q}_{1}<1/\ell $, *where*$${Q}_{1}=\underset{t\in \left[0,T\right]}{\mathrm{sup}}\left\{\frac{|{v}_{2}(t)}{\text{\Gamma}(\alpha ){T}^{1-\alpha}}+\frac{\left|{v}_{3}(t)\right|(1-{e}^{-kT})}{k\text{\Gamma}(\alpha ){T}^{1-\alpha}}\right\}.$$(19)
*Then problem (1)-(2) has at least one solution on [0, **T*].

**Example 3.9:** *Consider the following anti-periodic fractional boundary value problem*:$$\{\begin{array}{l}\left({}^{c}{D}^{3/2}+{1.5}^{c}{D}^{1/2}\right)u(t)=f(t,u(t)),\text{\hspace{0.17em}}t\in \left[0,2\right],\\ 1.25u(0)+4u(2)=-1,\text{\hspace{0.17em}}0.5{u}^{\prime}(0)-2{u}^{\prime}(2)=\mathrm{2.5.}\end{array}$$(20)
Here *T* = 2, *k* = 1.5, *α*_{1} = 1.25, *ρ*_{1} = 4, *β*_{1} = —1, *α*_{2} = 0.5, *ρ*_{2} = —2, *β*_{2} = 2.5. With the given values, we find that *Q* ≈ 7.742915 (*Q* is given by (15)).**(a)** Let$$f(t,u)={e}^{-{u}^{2}}\left(\frac{{\mathrm{cos}}^{2}(3u+2)}{\sqrt{{t}^{2}+9}}+\frac{t\mathrm{sin}t}{1+{u}^{2}}+\frac{2}{t+3}\right).$$(21)
Clearly $\left|f\left(t,u\left(t\right)\right)\right|\le 3={L}_{1}$ for all *t ∈* [0, 2], *u ∈* ℝ. Thus, by Theorem 3.2, the problem (20) with *f*(*t*,*u*) given by (21) has at least one solution on [0, 2].**(b)** Letting$$f(t,u)=\frac{{e}^{-t}}{27}\left(\frac{{\left|u\right|}^{3}}{1+{\left|u\right|}^{3}}+\frac{\left|u\right|}{1+\left|u\right|}+\frac{1}{t+1}\right),$$(22)
we have $\left|f\left(t,u\right)\right|\le {e}^{-t}/9=p\left(t\right)\chi \left(\Vert u\Vert \right)$ . Selecting $\chi \left(\Vert u\Vert \right)=1$ and $p\left(t\right)={e}^{-t}/9\left(\Vert p\Vert =1/9\right)$, we find that the assumption (*E*_{2}) holds true for *N* > 4.064141. As all the conditions of Theorem 3.4 are satisfied, there exists at least one solution of the problem (20) with *f*(*t*, *u*) given by (22) on [0, 2].**(c)** Let us take$$f(t,u)=\frac{1}{\sqrt{{t}^{2}+100}}\mathrm{sin}u+\frac{1}{t+2}.$$(23)
Then $\left|f\left(t,u\right)\right|\le \left(1/10\right)u+1/2$ implies that *ω* = 1/10, *M*_{1} = 1/2. Clearly *ω* < 1/*Q*(*Q* ≈ 7.742915). In consequence, the conclusion of Theorem 3.5 applies and the problem (20) with *f*(*t*, *u*) given by (23) has a solution on [0, *T*].**(d)** Let us choose$$f(t,u)=\frac{1}{10}{\mathrm{tan}}^{-1}u(t)+\mathrm{cos}t.$$(24)
Clearly $\ell =1/10$ as $\left|f\left(t,u\right)-f\left(t,\upsilon \right)\right|\le \frac{1}{10}\left|u-\upsilon \right|$ and $\ell Q\approx 0.774292<1$ . Thus all the conditions of Theorem 3.6 are satisfied. Hence we deduce by the conclusion of Theorem 3.6 that there exists a unique solution for the problem
(20) with *f*(*t*, *u*) given by (24).For the applicability of Theorem 3.8, we find that $\left|f\left(t,u\right)\right|\le g\left(t\right)=\pi /20+\mathrm{cos}t$ with $\Vert g\Vert =\left(20+\pi \right)/20$ and *Q*_{1} ≈ 6.732035 (*Q*_{1} is given by (19)). Obviously $\ell {Q}_{1}\approx 0.673203<1$ . Thus all the conditions of Theorem 3.8 are satisfied. Hence the conclusion of Theorem 3.8 implies that the problem (20) with *f*(*t*, *u*) given by (24) has at least one solution on [0, 2].

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