The existence of solutions of Hadamard fractional differential equations with integral and discrete boundary conditions on infinite interval

1.   Introduction

The two most important aspects for studying differential equations are fitting and solving. First, for a practical problem, we want to determine whether we can fit a suitable equation that can realistically portray the practical problem. Second, for a practical problem, we want to determine whether we can find suitable solutions to verify the reasonableness of the equation, thus reflecting the practicability of the practical problem.

Fractional differential equations (FDEs) are popular in the fields of physics, engineering, and biology because they can well characterize complex processes such as heritability and memory properties; see the related literature ^[1,2,3,4,5]. In order to better fit the constructed equations, we can add corresponding boundary-value problems (BVPs) to the equations, such as the integral BVP and the multipoint BVP; we can also regard the problem as a system of equations by means of coupling, or we can add the semilinear Laplace operators. In these ways, different practical problems can be better transformed into equations ^[6,7,8]. After we obtain the equation, we need to verify the practical significance of the solutions. And, there are a number of ways to determine the properties of solutions, such as the monotonic iterative methods^[9,10], and others. More detailed studies are as follows.

Hadamard ^[11] established the concept of fractional derivatives in 1892. Hadamard derivatives differ significantly from Riemann-Liouville and Caputo derivatives in terms of fractional powers; specifically, the kernel of the integral contains the logarithmic function of an arbitrary exponent. Hadamard derivatives have stable characterizations in terms of expansion and well matched the problems on the half-open interval; see ^[4]. At the same time, the Hadamard FDE has an important role in the mechanical behavior of viscoelastic materials and turbulence phenomena in fluid dynamics; see ^[12,13]. Since the problem we discuss in this paper is restricted to half-open intervals, we consider the use of Hadamard fractional derivatives.

Integral boundary conditions are instrumental in computational fluid dynamics studies related to blood flow problems. When dealing with these problems, the usual approach is to assume that the cross-section of the blood vessels is circular, which is not always reasonable. In order to optimize this detailed problem and make the results more detailed and convincing, the integral boundary conditions can be included to develop an efficient and applicable method. More details can be found in ^[14]. In addition, integral boundary conditions have other uses in physics and biology; see ^[15].

The authors of ^[16] found that the study of fractional BVPs for $m$-points on infinite intervals is almost non-existent, so they studied the related problem by referencing ^[17] for the first time. The authors of ^[17] found that the intrinsic equations of viscoelastic fluids in the models of physics and biology are closely connected with the FDEs; see ^[18]. Therefore, multipoint boundary problems are beginning to be studied.

In recent years, Hao et al. ^[19] considered a Hadamard FDE with integral boundary conditions. By applying Schauder's fixed-point theorem and Banach's contraction principle, they obtained the unique solution of the equation. Li et al. ^[20] considered the two integral boundary conditions of the Riemann-Liouville FDE. By employing Krasnoselskii's fixed-point theorem and Banach's contraction principle, they obtained the existence of the solution.

Zhang and Liu ^[21] applied Banach's contraction mapping principle, the monotone iterative method, and the Avery-Peterson fixed-point theorem to show the existence, uniqueness, and multiplicity results of solutions:

where $^{H}D_{1+}^{\alpha}$ is the Hadamard-type fractional derivative of order $\alpha$; $1 < \eta < \xi_{i} < +\infty;$ $b, \alpha_{i}, \sigma_{j}\geq0$ $(i, j = 1, 2, \ldots, n).$

In ^[22], the authors investigated the fractional BVP with

where $n-1 < \alpha\leq n$; $0\leq k \leq n-2$, $\eta\in(1, +\infty)$; $\lambda_{i}$, $\beta_i > 0$ $(i = 1, 2, \ldots, m)$; $g\in C\; ([1, +\infty), (0, +\infty))$. By employing the Bai-Ge fixed-point theorem, they obtained the solutions of the Hadamard-type FDE.

By considering the above two boundary conditions, research questions and the need for practical problem-solving, this article investigates the existence of positive solutions to the following Hadamard-type FDE with an integral and multipoint discrete BVP:

where $^{H}D^{\theta}_{1+}$ is the Hadamard-type fractional derivative of order $\theta$, ${2} < \theta\leq{3}$; $f\in C\; ([1, +\infty)\times[0, +\infty)\times[0, +\infty)$, $[0, +\infty))$; $\omega\in L^1\; [1, +\infty)$, where $\omega\geq0$; $1 < \eta_1 < \eta_2 < \cdots < \eta_{m} < +\infty$; $\beta_i$ denotes positive real constants $(i = 1, 2, \ldots, m)$.

In previous works, the fixed-point theorem is usually used to determine the existence of the solutions of the equations, while the innovation of this paper is the use of the monotone iterative method to solve Hadamard FDEs containing multiple lower-order derivative terms, which results in not only obtaining the existence of the two positive solutions, but also deriving the error estimation formula for the unique positive solution. This paper enriches the use of monotone iterative methods and has potential application to the development of blood flow modeling and properties of viscoelastic fluids for practical applications.

2.   Preliminaries and lemmas

In this section, it is essential to present some important lemmas.

2.1.   Basic concepts and properties

Definition 1. (^[23]) Let $\varphi > 0$; the Hadamard fractional integral of order $\varphi$ for a function $f:[1, +\infty)\rightarrow \mathbb R$ is defined as

Definition 2. (^[23]) Let $\varphi > 0$; the Hadamard fractional derivative of order $\varphi$ for a function $f:[1, +\infty)\rightarrow \mathbb R$ is defined as

where $n = [\varphi]+1$ and $[\varphi]$ is the integer part of $\varphi$.

Lemma 1. (^[23]) If $\varphi$, $\psi > 0$, then

Lemma 2. (^[23]) Let $\varphi > 0$; the solution of $^{H}D_{1+}^{\varphi}x(t) = 0$ with $x\in C[1, \infty)\cap L^{1}[1, \infty)$ is valid if, and only if,

and the following formula holds:

where $c_{i}\in \mathbb R, \ i = 1, 2, \ldots, n, \ n-1 < \varphi < n.$

Lemma 3. We define $h(t)\in L^1(1, +\infty)$, $0 < \int_{1}^{+\infty}h(s)\frac{ds}{s} < +\infty$, and

where

Then, the solution of the Hadamard-type FDE given by

can be expressed as

where

Proof. Because of Lemma 2, (2.3) has a solution:

From $x(1) = x'(1) = 0$, we know that $c_2 = c_{3} = 0$. From Lemma 1, we have

Considering the boundary condition $^{H}D^{\theta-1}_{1+}x(+\infty) = \int_{1}^{+\infty}\omega(t)x(t)\frac{dt}{t}+\sum_{i = 1}^{m}\beta_ix(\eta_i)$, we conclude that

Consequently, substituting $c_1$, $c_2$, and $c_3$ into (2.8), we have

Next, after piecing together and organizing the first and third terms of the above equation, we get

with the help of (2.2) and some arrangement, we obtain

Then,

Given that $\Upsilon_1 = \Upsilon-\int_{1}^{\infty}\omega(t)(\log t)^{\theta-1}\frac{dt}{t}$, we have

Thus,

Lemma 4. The function $K(t, s)$ defined in (2.5) satisfies the following conditions:

1) $K(t, s)$ is continuous for $(t, s)\in [1, +\infty)\times[1, +\infty)$;

2) $K(t, s)$ is nonnegative on $[1, +\infty)\times[1, +\infty)$;

3) $\frac{K(t, s)}{1+(\log t)^{\theta-1}}\leq\frac{K(t, s)}{(\log t)^{\theta-1}} \leq\frac{1}{\Upsilon_1}$ for all $(t, s)\in [1, +\infty)\times[1, +\infty)$.

Proof. We obviously get that conditions 1) and 2) above. Next, we show that condition 3) holds. For all $(t, s)\in [1, +\infty)\times[1, +\infty)$, we deduce the following:

The proof is completed.

By Lemma 3 and (2.4), we get

where

From Lemma 3 and (2.4), we have

where

Lemma 5. The functions $K^{\star}(t, s)$ and $K_{\star}(t, s)$, defined in (2.9) and (2.10), ensure that

Proof. According to (2.9) and (2.10), we can easily get that $K^{\star}(t, s)\geq0$ and $\frac{K_{\star}(t, s)}{1+\log t}\geq0$. Furthermore, for all $t$, $s\in[1, +\infty)$,

By the same steps, we have

The proof is completed.

2.2.   Compactness of the operator

Next, let

Similar to ^[19], $(F, \|\cdot\|)$ is a Banach space.

Lemma 6. Let $U\subset F$ be a bounded set. Then, $U$ is relatively compact in $F$ if the following conditions hold:

1) For all $x(t)\in U$, $\frac{x(t)}{1+(\log t)^{\theta-1}}$, $\frac{{^HD^{\theta-2}_{1+}}x(t)}{1+\log t}$, and ${^HD^{\theta-1}_{1+}}x(t)$ are equicontinuous on any compact interval of $[1, +\infty).$

2) For all $\varepsilon > 0$, there is a constant $T = T(\varepsilon) > 1$, which satifies

and

for any $t_{1}$, $t_{2} > T$ and $x\in U$.

Proof. This proof has been proved by Lemma 4 in ^[19], so it is omitted here. Define a cone $P$ :

Next, we give two circumstances:

$(H_{1})$ $f\in C([1, +\infty)\times[0, +\infty)^{3}, [0, +\infty))$ and $f(t, 0, 0, 0)\not\equiv0$ on any subinterval of $[1, +\infty)$; when $|x|\leq M$, $|y|\leq M$ for any $M > 0$, $f(t, (1+(\log t)^{\theta-1})x, (1+\log t)y, z)$ is bounded on $[1, +\infty)$.

$(H_{2})$ $r(t):[1, +\infty)\rightarrow [0, +\infty)$, $r(t)\not\equiv0$ on any subinterval of $[1, +\infty)$, and $0 < \int_{1}^{+\infty}r(s)\frac{ds}{s} < +\infty$.

For the convenience of description, let

According to Lemma 3, let $T:P\rightarrow F$ be an operator with

Thus, we can get the solution of (1.1) to be the fixed point of the operator $T$.

Lemma 7. Assuming that $(H_1)$ and $(H_2)$ are valid, then $T:P\rightarrow F$ is a continuous operator.

Proof. $Tx(t)\geq0$ is obvious for all $t\in[1, +\infty)$ and $x\in P$. Then, we have the evidence that $T$ is continuous:

and

Thus, we know that $T:P\rightarrow F$.

Next, we give the proof that $T:P\rightarrow F$ is a continuous operator. Let $x_{n}\rightarrow x_{0}$ as $n\rightarrow \infty$ in $P$; there is a constant $k_{0} > 0$ such that $\sup_{n\in\mathbb{N}}||x_n||\leq k_0$, $\|x_{0}\|\leq k_{0}$. Let $B_{k_{0}} = \sup\{f(t, (1+(\log t)^{\theta-1})x, (1+\log t)y, z)|(t, x, y, z)\in[1, +\infty)\times[0, k_{0}]^{3}\}$. Next, we aim to prove that $\|Tx_n-Tx_0\|\rightarrow0$ as $n\rightarrow \infty$. By $(H_{1})$, $(H_{2})$, the Lebesgue dominated convergence theorem, and continuity of $f$, we can get

Therefore,

and

Hence, $T$ is continuous.

Lemma 8. Suppose that ${\int_{1}^{+\infty}(\log t)^{\theta-1}\omega(t)\frac{dt}{t}} < \infty$ and $(H_{1})$ and $(H_{2})$ hold; then, $T:P\rightarrow P$ is a compact operator.

Proof. We will show this lemma in the following three procedures. First, let $\Omega$ be a bounded subset of cone $P$ to prove the truth that $T\Omega$ is a bounded set of cone $P$. There is a constant $k_{1} > 0$, which guarantees that $\|x\|\leq k_{1}$ for all $x\in\Omega$. Let $B_{k_{1}} = \sup\{f(t, (1+\log t)^{\theta-1})x, (1+\log t)y, z)|(t, x, y, z)\in[1, +\infty)\times[0, k_{1}]^{3}\}$. By Lemmas 4 and 5, we get

Hence, $T\Omega$ is bounded in $P$.

Second, we prove that condition 1) in Lemma 6 is established. For all $t_{1}$, $t_{2}\in [L_{1}, L_{2}]$ with $t_{1} < t_{2}$ and $x\in\Omega$ mentioned above, we get

Because $\frac{K(t, s)}{1+(\log t)^{\theta-1}}, \frac{\ (\log t)^{\theta-1}}{1+(\log t)^{\theta-1}}$, and $\frac{\log t}{1+\log t}$ are uniformly continuous on any compact set $[L_{1}, L_{2}]\; \times$ $[L_{1}, L_{2}]$, $[L_{1}, L_{2}]$, and $[L_{1}, L_{2}]$, respectively, we have

Furthermore, in view of

the function $K^{\star}(t, s)\in C([1, +\infty)\times[1, +\infty))$ does not depend on $t$. Hence, we know that condition 1) in Lemma 6 is established.

Finally, we prove that condition 2) in Lemma 6 is established. For any $\varepsilon > 0$, since $0 < \int_{1}^{+\infty}r(s)\frac{ds}{s} < +\infty$, we can get that there exists $M_{1} > 1$ such that

In view of

for all $\varepsilon > 0$, there exist two constants $M_{2} > 0$, $M_{3} > M_{1}$ such that, for all $t_{1}$, $t_{2} > M_{2}$,

for all $t_{1}$, $t_{2} > M_{3}$ and $1\leq s\leq M_{1}$,

Since

for all $\varepsilon > 0$, there exist constants $M_{4} > 0$ and $M_{5} > M_{1}$ such that, for all $t_{1}$, $t_{2} > M_{4}$,

for all $t_{1}$, $t_{2} > M_{5}$ and $1\leq s\leq M_{1}$,

Let $M > \max\{M_{2}, M_{3}, M_{4}, M_{5}\}$; for all $t_{1}$, $t_{2} > M$, we deduce the following result by substituting (2.5)–(2.7) and applying Lemma 4:

Next, we substitute the specific form of $K_\star(t, s)$ and apply the properties of $K_\star(t, s)$. We have

In the same way, considering the the properties of $K^\star(t, s)$, we obtain

Hence, we know that condition 2) in Lemma 6 is established. By Lemma 6, $T$ is a compact operator. The proof is completed.

In consequence, according to Lemmas 7 and 8, the operator $T$ is completely continuous.

3.   Existence of solutions

$(H_3)$ Let $\zeta > 0$ be a constant, which satisfies

for all $(t, x, y, z)\in[1, +\infty)\times[0, \zeta]^{3}$, where $\Upsilon_1$ is shown by Lemma 3.

$(H_4)$ Regardless of the values of $t\in[1, +\infty)$ and $x, y, z\in[0, +\infty)$, $f(t, x, y, z)$ is nondecreasing and continuous, which satisfies

where $x_1 > x_2$, $y_1 > y_2$, $z_1 > z_2$.

Theorem 1. (^[24]) Assume that $(H_1)$–$(H_4)$ hold; there are two positive solutions $x^{\ast}$, $y^{\ast}$ of (1.1), where $\|x^\ast\|$, $\|y^\ast\| \in (0, \zeta]$. Actually, the solutions can be established by applying the sequences $\{x_n\}_{n = 1}^\infty$ and $\{y_n\}_{n = 1}^\infty$, which satisfy

where

with the initial values $x_{0}(t) = \frac{\zeta(\log t)^{\theta-1}}{\Gamma(\theta)}$ and $y_{0}(t) = 0$. At the same time, for all $t\in(1, +\infty)$, $\{x_n\}_{n = 1}^\infty$ and $\{y_n\}_{n = 1}^\infty$ converge to $x^*$, $y^*$, separately.

Proof. Set $P_{\zeta} = \{x\in P, \ \|x\|\leq\zeta\}$ if $x\in P_{\zeta}$, where $\|x\|\leq\zeta$. According to $(H_{3})$ and Lemmas 4 and 5, we have

From the above inequalities, we prove that $\|Tx\|\leq\zeta$. Thus, $T(P_{\zeta})\subset P_{\zeta}$, $T:P_\zeta \rightarrow P_\zeta$.

First, let $x_{0}(t) = \frac{\zeta(\log t)^{\theta-1}}{\Gamma(\theta)}$, $x_1 = Tx_0$, and $x _2 = T^2x_0 = Tx_1$. It is clear that $\|x_0\|\leq\zeta$, so $x_{0}(t)\in P_{\zeta}$. Furthermore, we know that $T:P_{\zeta}\rightarrow P_{\zeta}$, which means that $x_1\in T(P_\zeta)\subset P_\zeta$ and $x_2\in T(P_\zeta)\subset P_\zeta$. By $(H_{3})$ and Lemmas 4 and 5, we have

Hence,

Suppose that the following holds:

Then, we show that

By $(H_4)$, we can get

In the same way, according to $(H_4)$, (2.9) and (2.10), we get

By induction, we obtain that $\{x_n\}_{n = 1}^{\infty}\subset T(P_{\zeta})\subset P_{\zeta}$ and $x_{n+1} = Tx_{n}$. For all $t\in[1, \infty)$, we derive the following:

Due to the complete continuity of $T$ and the existence of $x^{*}$ in $P_{\zeta}$, we obtain that $x_{n_k}\to x^*$ as $n\rightarrow \infty$. So, we can get that there is a convergent subsequence $\{x_ {n_{k}}\}_{k = 1}^{\infty}$ of $\{x_ {n}\}_{n = 1}^{\infty}$. This demonstrates that $\lim_{n\to\infty}x_{n} = x^{*}$. Given that $x_{n+1} = Tx_{n}$ and $T$ is continuous, we prove that $Tx^{*} = x^{*}$, which demonstrates that $x^{*}$ is a fixed point of $T$.

Since $y_0(t) = 0$ and $\|y_0\|\leq\zeta$, we can easily get that $y_{0}(t)\in P_{\zeta}$, which implies that $T:P_{\zeta}\rightarrow P_{\zeta}$. Similarly, we can get that $y_{n+1} = Ty_{n}$ and $\{y_ {n}\}_{n = 1}^{\infty}\subset T(P_{\zeta})\subset P_{\zeta}$. By induction, for any $t\in[1, \infty)$, it can be seen that

Because of the properties of $T$, there exists a $y^{*}\in P_{\zeta}$, which satifies that $y_{n_k}\to y^*$ as $n\rightarrow \infty$. So, we can get that there is a convergent subsequence $\{y_ {n_{k}}\}_{k = 1}^{\infty}$ of $\{y_ {n}\}_{n = 1}^{\infty}$. This demonstrates that $\lim_{n\to\infty}y_{n} = y^{*}$. Given that $y_{n+1} = Ty_{n}$ and $T$ is continuous, we have that $Ty^{*} = y^{*}$, which implies that $y^{*}$ is a fixed point of $T$. By induction, we obtain

According to (3.7), (3.10), and (3.13), we have

This, together with $x^{*} = \lim_{n\to\infty}x_{n}$ and $y^{*} = \lim_{n\to\infty}y_{n}$, yields that

Because $f(t, 0, 0, 0)\not\equiv0$, it implies that $x = 0$ is not a solution of (1.1). Therefore, $x^{\ast}$ and $y^{\ast}$ are the solutions of (1.1).

4.   Error estimate

$(H_5)$ There exist constants $L_i > 0$, $i = 1, 2, 3$, $\forall t\in[1, +\infty)$, as well as $x, y, z\in[0, +\infty)$. Thus, $f(t, x, y, z)$ satisfies the conditions of the following inequality:

where $x_1 > x_2$, $y_1 > y_2$, $z_1 > z_2$.

Suppose that $(H_{1})$–$(H_5)$ hold and (1.1) has a unique solution $z^{*}\in(0, \frac{\zeta}{\Gamma(\theta)}(\log t)^{\theta-1}]$, which relies on the following sequence:

The initial value of (4.1) is $z_0(t) = 0$ or $z_0(t) = \frac{\zeta}{\Gamma(\theta)}(\log t)^{\theta-1}$. Furthermore, the error estimate can be defined as

where $Z = \frac{\Gamma(\theta)}{\Upsilon_1}(L_1+L_2+L_3)\int_{1}^{+\infty}r(s)(1+\log s)^{\theta-1}\frac{ds}{s} < 1$.

Proof. According to Theorem 1, we prove that (1.1) has two positive solutions, which can be established by $x_{n+1} = Tx_{n}$ and $y_{n+1} = Ty_n$, with the initial values $x_ {0}(t) = \frac{\zeta(\log t)^{\theta-1}}{\Gamma(\theta)}$ and $y_{0}(t) = 0$, $t\in[1, +\infty).$

If $\|x_n-y_n\| = \sup_{t\in[1, +\infty)}\frac{|x_n(t)-y_n(t)|}{1+(\log t)^{\theta-1}}$, by $(H_5)$ and (3.1), we derive the following:

Thus,

By induction, we can get

If $\|x_n-y_n\| = \sup_{t\in[1, +\infty)}\frac{|{^HD_{1+}^{^{\theta-2}}}x_n(t)-{^HD_{1+}^{^{\theta-2}}}y_n(t)|}{1+\log t}$, by $(H_5)$ and (3.1), we have

Thus, by the same arrangement and induction, we obtain

If $\|x_n-y_n\| = \sup_{t\in[1, +\infty)}{|^HD_{1+}^{^{\theta-1}}x_n(t)-{^HD_{1+}^{^{\theta-1}}}y_n(t)|}$, by $(H_5)$ and (3.1), we have

Thus, by using the same method, we get

From the above steps, we have

Consequently, we show that

We suppose that (1.1) has a solution $z\in[y_{0}, x_{0}]$. It is apparent that $Tz = z$ and $y_0(t)\leq z(t)\leq x_0(t)$. Since $T$ is continuous and nondecreasing, together with (3.1), we can get

Furthermore, by (4.4), we know that

This shows that $x^* = y^*\triangleq z^*$. Hence, (1.1) has a unique solution $z^*\in[y_0, x_0]$. In consequence, in view of (4.5), it is obvious that (4.2) holds. The proof is completed.

5.   Example

where $\theta = \frac{5}{2}$, $r(t) = \frac{1}{e^{\log t}}$, $\omega(t) = \frac{1}{3e^{\log t^{2}}}$, $\theta_1 = \frac{1}{5}$, $\theta_2 = \frac{2}{5}$, $\theta_3 = \frac{3}{5}$, $\theta_4 = \frac{4}{5}$, $\eta_1 = \frac{8}{7}$, $\eta_2 = \frac{7}{6}$, $\eta_3 = \frac{6}{5}$, $\eta_4 = \frac{5}{4}$,

Thus,

From the above steps, the conditions $(H_1)$–$(H_5)$ hold; according to Theorem 1, the BVP (1.1) has twin positive solutions $x^{\ast}$ and $y^{\ast}$ such that $\|x^*\|$, $\|y^*\| \in (0, 8]$.

6.   Conclusions

In this article, we first transformed the solutions of the equation into fixed points of the operator by means of a nonlinear alternative, while proving that the operator is completely continuous. Then, we applied two iterative sequences to find two positive solutions of the equation via the monotone iterative method. Finally, we derived the unique solution of the Hadamard fractional BVP. The significance of this article lies in the fact that we can use the monotone iterative method to discuss the existence and uniqueness of the positive solution to the equation containing multiple fractional-order derivative terms.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported by the National Natural Science Foundation of China under grant (62073186), the Taishan Scholar Project of Shandong Province of China, and the NSF of Shandong Province (ZR2021MA097).

Conflict of interest

The authors declare there are no conflicts of interest.