A new class of fractional boundary value problems

Abstract

In this paper, a fractional boundary value problem with a new boundary condition is studied. This new boundary condition relates the nonlocal value of the unknown function at ξ with its influence due to a sub-strip $(\eta ,1)$, where $0<\xi <\eta <1$. The main results are obtained with the aid of some classical fixed point theorems and Leray-Schauder nonlinear alternative. A demonstration of applications of these results is also given.

MSC:34A12, 34A40.

1 Introduction

We study a boundary value problem of Caputo-type fractional differential equations with new boundary conditions given by

$$\{\begin{array}{l}{}^{c}D^{q}x(t)=f(t,x(t)),1<q\le 2,t\in [0,1],\\ x(0)=0,x(\xi )=a{\int }_{\eta }^{1}x(s)ds,\end{array}$$

(1.1)

where ${}^{c}D^q$ denotes the Caputo fractional derivative of order q, f is a given continuous function, and a is a positive real constant.

In (1.1), the second condition may be interpreted as a more general variant of nonlocal integral boundary conditions, which states that the integral contribution due to a sub-strip $(\eta ,1)$ for the unknown function is proportional to the value of the unknown function at a nonlocal point $\xi \in (0,1)$ with $\xi <\eta <1$. We emphasize that most of the work concerning nonlocal boundary value problems relates the contribution expressed in terms of the integral to the value of the unknown function at a fixed point (left/right end-point of the interval under consideration), for instance, see [1–3] and references therein.

The recent development in the theory, methods and applications of fractional calculus has contributed towards the popularity and importance of the subject. The tools of fractional calculus have been effectively applied in the modeling of many physical and engineering phenomena. Examples include physics, chemistry, biology, biophysics, blood flow phenomena, control theory, wave propagation, signal and image processing, viscoelasticity, percolation, identification, fitting of experimental data, economics, etc. [4–6]. For some recent work on the topic, we refer to [7–22] and the references therein.

2 Preliminaries

Let us recall some basic definitions on fractional calculus.

Definition 2.1 The Riemann-Liouville fractional integral of order q for a continuous function g is defined as

$$I^{q}g(t)=\frac{1}{\mathrm{\Gamma }(q)}{\int }_0^t\frac{g(s)}{{(t-s)}^{1-q}}ds,q>0,$$

provided the integral exists.

Definition 2.2 For at least n th continuously differentiable function $g:[0,\mathrm{\infty })\to \mathbb{R}$, the Caputo derivative of fractional order q is defined as

$${}^{c}D^{q}g(t)=\frac{1}{\mathrm{\Gamma }(n-q)}{\int }_0^t{(t-s)}^{n-q-1}{g}^{(n)}(s)ds,n-1<q<n,n=[q]+1,$$

where $[q]$ denotes the integer part of the real number q.

Lemma 2.1 For any $y\in C[0,1]$, the unique solution of the linear fractional boundary value problem

$$\{\begin{array}{l}{}^{c}D^{q}x(t)=y(t),1<q\le 2,t\in [0,1],\\ x(0)=0,x(\xi )=a{\int }_{\eta }^{1}x(s)ds\end{array}$$

(2.1)

is

$$\begin{array}{rcl}x(t)& =& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}y(s)ds\\ +\frac{t}{A}\{a{\int }_{\eta }^1({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}y(u)du)ds-{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}y(s)ds\},\end{array}$$

(2.2)

where

$$A=\xi -\frac{a}{2}(1-{\eta }^2)\ne 0.$$

(2.3)

Proof It is well known that the general solution of the fractional differential equation in (2.1) can be written as

$$x(t)=c_0+c_{1}t+{\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}y(s)ds,$$

(2.4)

where $c_0,c_1\in \mathbb{R}$ are arbitrary constants.

Applying the given boundary conditions, we find that $c_0=0$, and

$$c_1=\frac{1}{A}\{a{\int }_{\eta }^1({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}y(u)du)ds-{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}y(s)ds\}.$$

Substituting the values of $c_0$, $c_1$ in (2.4), we get (2.2). This completes the proof. □

3 Existence results

Let $\mathcal{P}=C([0,1],\mathbb{R})$ denote the Banach space of all continuous functions from $[0,1]$ to ℝ endowed with the norm $\parallel x\parallel =sup\{|x(t)|,t\in [0,1]\}$.

In relation to the given problem, we define an operator $\mathcal{T}:\mathcal{P}\to \mathcal{P}$ as

$$\begin{array}{rcl}(\mathcal{T}x)(t)& =& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\\ +\frac{t}{A}\{a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}f(u,x(u))du\right)ds-{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\},\end{array}$$

where A is given by (2.3). Observe that problem (1.1) has solutions if and only if the operator $\mathcal{T}$ has fixed points.

For the sake of computational convenience, we set

$$\omega =\frac{1}{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}.$$

(3.1)

Theorem 3.1 Let $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be a jointly continuous function satisfying the Lipschitz condition

(H₁) $|f(t,x)-f(t,y)|\le L|x-y|$, $\mathrm{\forall }t\in [0,1]$, $x,y\in \mathbb{R}$, $L>0$.

Then problem (1.1) has a unique solution if $L\omega <1$, where ω is given by (3.1).

Proof Let us denote ${sup}_{t\in [0,1]}|f(t,0)|=\gamma$ and show that $\mathcal{T}{B}_r\subset B_r$, where $B_r=\{x\in \mathcal{P}:\parallel x\parallel \le r\}$ with $r>\omega \gamma {(1-\omega L)}^{-1}$. For $x\in B_r$, $t\in [0,1]$, we have

$$\begin{array}{rcl}\parallel (\mathcal{T}x)\parallel & \le & \underset{t\in [0,1]}{sup}\{{\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}|f(s,x(s))-f(s,0)+f(s,0)|ds\\ +\frac{t}{|A|}[a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}|f(u,x(u))-f(u,0)+f(u,0)|du\right)ds\\ +{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}|f(s,x(s))-f(s,0)+f(s,0)|ds\left]\right\}\\ \le & (Lr+M)\underset{t\in [0,1]}{sup}\{\frac{t^q}{\mathrm{\Gamma }(q+1)}+\frac{t}{|A|}[\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}]\}\\ \le & (Lr+M)\frac{1}{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}\\ =& (Lr+M)\omega \le r,\end{array}$$

which implies that $\mathcal{T}{B}_r\subset B_r$.

Now, for $x,y\in \mathcal{P}$ and for each $t\in [0,1]$, we obtain

$$\begin{array}{rcl}\parallel \mathcal{T}x-\mathcal{T}y\parallel & \le & \underset{t\in [0,1]}{sup}\{{\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}|f(s,x(s))-f(s,y(s))|ds\\ +\frac{t}{|A|}(a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}|f(u,x(u))-f(u,y(u))|du\right)ds\\ +{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}|f(s,x(s))-f(s,y(s))|ds\left)\right\}\le L\omega |x-y|.\end{array}$$

Since $L<1/\omega$, by the given assumption, therefore the operator $\mathcal{T}$ is a contraction. Thus, by Banach’s contraction mapping principle, there exists a unique solution for problem (1.1). This completes the proof. □

The next result is based on Krasnoselskii’s fixed point theorem [23].

Theorem 3.2 Let $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be a continuous function satisfying (H₁) and

(H₂) $|f(t,x)|\le \mu (t)$, $\mathrm{\forall }(t,x)\in [0,1]\times \mathbb{R}$, and $\mu \in \mathbb{c}([0,1],{\mathbb{R}}^{+})$.

Then problem (1.1) has at least one solution on $[0,1]$ if

$$\frac{L}{|A|}\{\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}\}<1.$$

Proof Fixing $r\ge \parallel \mu \parallel \omega$, we consider $B_r=\{x\in \mathcal{P}:\parallel x\parallel \le r\}$ and define the operators Φ and Ψ on $B_r$ as

$$\begin{array}{c}(\mathrm{\Phi }x)(t)={\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds,\hfill \\ \begin{array}{rl}(\mathrm{\Psi }x)(t)=& \frac{t}{A}\{a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}f(u,x(u))du\right)ds\\ -{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\}.\end{array}\hfill \end{array}$$

For $x,y\in B_r$, it is easy to show that $\parallel (\mathrm{\Phi }x)+(\mathrm{\Psi }y)\parallel \le \parallel \mu \parallel \omega \le r$, which implies that $\mathrm{\Phi }x+\mathrm{\Psi }y\in B_r$.

In view of the assumption $L\{\frac{1}{|A|}(\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)})\}<1$, the operator Ψ is a contraction. The continuity of f implies that the operator Φ is continuous. Also, Φ is uniformly bounded on $B_r$ as $\parallel \mathrm{\Phi }x\parallel \le \parallel \mu \parallel /\mathrm{\Gamma }(q+1)$. Moreover, Φ is relatively compact on $B_r$ as

$$\parallel (\mathrm{\Phi }x)(t_2)-(\mathrm{\Phi }x)(t_1)\parallel \le \frac{f_m}{\mathrm{\Gamma }(q+1)}(|{t_2}^q-{t_1}^q|+2{|t_2-t_1|}^q),$$

where ${sup}_{(t,x)\in [0,1]\times B_r}|f(t,x)|=f_m<\mathrm{\infty }$. Hence, by the Arzelá-Ascoli theorem, Φ is compact on $B_r$. Thus all the assumptions of Krasnoselskii’s fixed point theorem are satisfied. So problem (1.1) has at least one solution on $[0,1]$. This completes the proof. □

Our next result is based on the following fixed point theorem [23].

Theorem 3.3 Let X be a Banach space. Assume that $T:X\to X$ is a completely continuous operator and the set $V=\{u\in X|u=ϵTu,0<ϵ<1\}$ is bounded. Then T has a fixed point in X.

Theorem 3.4 Assume that there exists a positive constant $L_1$ such that $|f(t,x)|\le L_1$ for all $t\in [0,1]$, $x\in \mathcal{P}$. Then there exists at least one solution for problem (1.1).

Proof As a first step, we show that the operator $\mathcal{T}$ is completely continuous. Clearly, the continuity of $\mathcal{T}$ follows from the continuity of f. Let $D\subset \mathcal{P}$ be bounded. Then, $\mathrm{\forall }x\in D$, it is easy to establish that $|(\mathcal{T}x)(t)|\le L_1\omega =L_2$. Furthermore, we find that

$$\begin{array}{rcl}|{(\mathcal{T}x)}^{\prime }(t)|& \le & |{\int }_0^t\frac{{(t-s)}^{q-2}}{\mathrm{\Gamma }(q-1)}f(s,x(s))ds\\ +\frac{1}{A}\{a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}f(u,x(u))du\right)ds-{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\}|\\ \le & L_1[\frac{t^{q-1}}{\mathrm{\Gamma }(q)}+\frac{1}{|A|}\{\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}\}]\\ \le & L_1[\frac{1}{\mathrm{\Gamma }(q)}+\frac{1}{|A|}\{\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}\}]\\ =& L_3.\end{array}$$

Hence, for $t_1,t_2\in [0,1]$, it follows that

$$|(\mathcal{T}x)(t_1)-(\mathcal{T}x)(t_2)|\le {\int }_{t_1}^{t_2}|{(\mathcal{T}x)}^{\prime }(s)|ds\le L_3(t_2-t_1).$$

Therefore, $\mathcal{T}$ is equicontinuous on $[0,1]$. Thus, by the Arzelá-Ascoli theorem, the operator $\mathcal{T}$ is completely continuous.

Next, we consider the set $V=\{x\in \mathcal{P}:x=ϵ\mathcal{T}x,0<ϵ<1\}$. To show that V is bounded, let $x\in V$, $t\in [0,1]$. Then

$$\begin{array}{rcl}x(t)& =& {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\\ +\frac{t}{A}\{a{\int }_{\eta }^1\left({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}f(u,x(u))du\right)ds-{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}f(s,x(s))ds\}\end{array}$$

and $|x(t)|=ϵ|(\mathcal{T}x)(t)|\le L_1\omega =L_2$. Hence, $\parallel x\parallel \le L_2$, $\mathrm{\forall }t\in [0,1]$. So V is bounded. Thus, Theorem 3.3 applies and, in consequence, problem (1.1) has at least one solution. This completes the proof. □

Our final result is based on Leray-Schauder nonlinear alternative.

Lemma 3.1 (Nonlinear alternative for single-valued maps [24])

Let E be a Banach space, $E_1$ be a closed, convex subset of E, V be an open subset of $E_1$ and $0\in V$. Suppose that $\mathcal{U}:\overline{V}\to E_1$ is a continuous, compact (that is, $\mathcal{U}(\overline{V})$ is a relatively compact subset of $E_1$) map. Then either

• (i) $\mathcal{U}$ has a fixed point in $\overline{V}$, or

• (ii) there are $x\in \partial V$ (the boundary of V in $E_1$) and $\kappa \in (0,1)$ with $x=\kappa \mathcal{U}(x)$.

Theorem 3.5 Let $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be a continuous function. Assume that

(H₃) there exist a function $p\in \mathcal{C}([0,1],{\mathbb{R}}^{+})$ and a nondecreasing function $\psi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ such that $|f(t,x)|\le p(t)\psi (\parallel x\parallel )$, $\mathrm{\forall }(t,x)\in [0,1]\times \mathbb{R}$;

(H₄) there exists a constant $M>0$ such that

$$M{\left[\frac{\psi (M)\parallel p\parallel }{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}\right]}^{-1}>1.$$

Then problem (1.1) has at least one solution on $[0,1]$.

Proof Let us consider the operator $\mathcal{T}:\mathcal{P}\to \mathcal{P}$ defined by (3.1) and show that $\mathcal{T}$ maps bounded sets into bounded sets in $\mathcal{P}$. For a positive number r, let $B_r=\{x\in \mathcal{P}:\parallel x\parallel \le r\}$ be a bounded set in $\mathcal{T}$. Then, for $x\in B_r$ together with (H₃), we obtain

$$\begin{array}{rcl}|(\mathcal{T}x)(t)|& \le & {\int }_0^t\frac{{(t-s)}^{q-1}}{\mathrm{\Gamma }(q)}p(s)\psi (\parallel x\parallel )ds+\frac{t}{|A|}\{a{\int }_{\eta }^1({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}p(u)\psi (\parallel x\parallel )du)ds\\ +{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}p(s)\psi (\parallel x\parallel )ds\}\\ \le & \parallel p\parallel \psi (\parallel x\parallel )\left[\frac{1}{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}\}\right]\\ \le & \parallel p\parallel \psi (r)\omega .\end{array}$$

Next, it will be shown that $\mathcal{T}$ maps bounded sets into equicontinuous sets of $\mathcal{P}$. Let $t_1,t_2\in [0,1]$ with $t_1<t_2$ and $x\in B_r$. Then

$$\begin{array}{rcl}|(\mathcal{T}x)(t_2)-(\mathcal{T}x)(t_1)|& \le & |{\int }_0^{t_2}\frac{{(t_2-s)}^{q-1}}{\mathrm{\Gamma }(q)}p(s)\psi (r)ds-{\int }_0^{t_1}\frac{{(t_1-s)}^{q-1}}{\mathrm{\Gamma }(q)}p(s)\psi (r)ds\\ +\frac{(t_2-t_1)}{|A|}\{a{\int }_{\eta }^1({\int }_0^s\frac{{(s-u)}^{q-1}}{\mathrm{\Gamma }(q)}p(u)\psi (r)du)ds\\ -{\int }_0^{\xi }\frac{{(\xi -s)}^{q-1}}{\mathrm{\Gamma }(q)}p(s)\psi (r)ds\left\}\right|\\ \le & \psi (r)\parallel p\parallel [\frac{|t_2^q-t_1^q|}{\mathrm{\Gamma }(q+1)}+\frac{|t_2-t_1|}{|A|}\{\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}\}].\end{array}$$

Clearly, the right-hand side tends to zero independently of $x\in B_r$ as $t_2\to t_1$. Thus, by the Arzelá-Ascoli theorem, the operator $\mathcal{T}$ is completely continuous.

Let x be a solution for the given problem. Then, for $\lambda \in (0,1)$, following the method of computation used in proving that $\mathcal{T}$ is bounded, we have

$$\parallel x(t)\parallel =\parallel \lambda (\mathcal{T}x)(t)\parallel \le \left[\frac{\psi (\parallel x\parallel )\parallel p\parallel }{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}\right],$$

which implies that

$$\parallel x\parallel {\left[\frac{\psi (\parallel x\parallel )\parallel p\parallel }{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}\right]}^{-1}\le 1.$$

In view of (H₄), there exists M such that $\parallel x\parallel \ne M$. Let us choose $\mathcal{N}=\{x\in \mathcal{P}:\parallel x\parallel <M+1\}$.

Observe that the operator $\mathcal{T}:\overline{\mathcal{N}}\to \mathcal{P}$ is continuous and completely continuous. From the choice of $\mathcal{N}$, there is no $x\in \partial \mathcal{N}$ such that $x=\lambda \mathcal{T}(x)$ for some $\lambda \in (0,1)$. Consequently, by Lemma 3.1, we deduce that the operator $\mathcal{T}$ has a fixed point $x\in \overline{\mathcal{N}}$ which is a solution of problem (1.1). This completes the proof. □

4 Examples

Example 4.1 Consider a fractional boundary value problem given by

$$\{\begin{array}{l}{}^{c}D^{\frac{3}{2}}x(t)=\frac{1}{{(t^2+2)}^2}(\frac{|x|}{1+|x|})-{sin}^{2}t,t\in [0,1],\\ x(0)=0,x(1/4)=2{\int }_{1/2}^{1}x(s)ds.\end{array}$$

(4.1)

Here, $q=3/2$, $a=2$, $\eta =1/2$, $\xi =1/4$ and $f(t,x)=\frac{1}{{(t^2+2)}^2}(\frac{|x|}{1+|x|})-{sin}^{2}t$. With the given values, $\omega \simeq 1.931151$ and $L=1/4$ as $|f(t,x)-f(t,y)|\le 1/4|x-y|$. Clearly, $L\omega \simeq 0.482788<1$. Therefore, by Theorem 3.1, there exists a unique solution for problem (4.1).

Example 4.2 Consider a fractional boundary value problem given by

$$\{\begin{array}{l}{}^{c}D^{\frac{3}{2}}x(t)=\frac{1}{\pi \sqrt{1+t}}({tan}^{-1}x+\frac{\pi }{2}),t\in [0,1],\\ x(0)=0,x(1/2)={\int }_{3/4}^{1}x(s)ds.\end{array}$$

(4.2)

Observe that $|f(t,x)-f(t,y)|\le \frac{1}{\pi }|x-y|$ implies that $L=1/\pi$ and $|f(t,x)|\le \frac{1}{\sqrt{1+t}}=\mu (t)$. With the given data, it is found that $A=9/32$ and

$$\frac{L}{|A|}\{\frac{a(1-{\eta }^{q+1})}{\mathrm{\Gamma }(q+2)}+\frac{{\xi }^q}{\mathrm{\Gamma }(q+1)}\}=\frac{128}{27\pi \sqrt{\pi }}[\frac{2}{5}(1-{\left(\frac{3}{4}\right)}^{5/2})+\frac{1}{2\sqrt{2}}]\simeq 0.475662<1.$$

Clearly, all the conditions of Theorem 3.2 are satisfied. Hence there exists a solution for problem (4.2).

Example 4.3 Consider the problem

$$\{\begin{array}{l}{}^{c}D^{3/2}x(t)=\frac{1}{{(t+3)}^2}(|x|+1),t\in [0,1],\\ x(0)=0,x(1/2)={\int }_{3/4}^{1}x(s)ds.\end{array}$$

(4.3)

Here, $q=3/2$, $\xi =1/2$, $a=1$, $\eta =3/4$, $|f(t,x)|\le \frac{1}{9}(|x|+1)$ and $\omega \simeq 2.246588$. Let us fix $p(t)=1$, $\psi (|x|)=\frac{1}{9}(|x|+1)$. Further, by the condition

$$M{\{\psi (M)\parallel p\parallel \left(\frac{1}{|A|}\{\frac{1}{\mathrm{\Gamma }(q+1)}(|A|+{\xi }^q)+\frac{a}{\mathrm{\Gamma }(q+2)}(1-{\eta }^{q+1})\}\right)\}}^{-1}>1,$$

it is found that $M>M_1$ with $M_1\simeq 0.33266$. Thus, Theorem 3.5 applies and there exists a solution for problem (4.3) on $[0,1]$.