Design and experimentation of sampled-data controller in T-S fuzzy systems with input saturation through the use of linear switching methods

1.   Introduction

This paper considers a fractional coupled system on an infinite interval involving the Erdélyi-Kober derivative:

where $\delta_{1}, \delta_{2}\in (1, 2]$, $\gamma\in (-2, -1)$, and $\beta > 0$. $D^{\gamma, \delta_{1}}_{\beta}$, $D^{\gamma, \delta_{2}}_{\beta}$ are Erdélyi-Kober fractional derivatives (EKFDs for short), and $I^{\delta_{1}+\gamma, 2-\delta_{1}}, I^{\delta_{2}+\gamma, 2-\delta_{2}}$ are the Erdélyi-Kober fractional integrals. $F, G$ are continuous functions. We discuss the existence of positive solutions for (1.1).

During the past several decades, fractional equations have been studied widely; see ^{[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36]} for instance. From the literature, we can see that there are many fractional derivatives used in differential equations. Among these various definitions, the widely used ones are the Riemann-Liouville and Caputo fractional derivatives, in many works. To generalize the Riemann-Liouville fractional derivative, Erdélyi-Kober defined a new fractional derivative, and we call it the Erdélyi-Kober fractional derivative. Moreover, the Erdélyi-Kober operator is very useful; we can refer to ^{[6,9,14,15,16,17]} and the references therein. The Erdélyi-Kober operator is a fractional integration operation which was given by Arthur Erdélyi and Hermann Kober in 1940 ^[23]. Some of these definitions and results were given in Samko et al. ^[3], Kiryakova ^[19], and McBride ^[20].

Nowadays, the theory of fractional operators in the Erdélyi-Kober frame has attracted much interest from researchers. The study of fractional systems is also very important, as these systems appear in various applications, especially in biological sciences. Recently, some problems of Erdélyi-Kober type fractional differential equations on infinite intervals received widespread attention from many scholars; see ^[8,21,22] for example.

Recently, in ^[8], the authors investigated the following equation:

where $\sigma\in(1, 2)$, $\vartheta\in(1, 2)$, $\theta > 0$, and $F$ is a given continuous function, $D_{\theta}^{\vartheta, \sigma}$ denotes the EKFD, and $I^{\sigma+\vartheta, 2-\sigma}$ denotes the Erdélyi-Kober fractional integral. The authors studied the existence and nonexistence of positive solutions for this problem by utilizing a fixed point result which uses the strongly positive-like operators and eigenvalue criteria.

In ^[9], the authors studied a fractional coupled system:

where $^{c}D^{\varrho}, ^{c}D^{\varsigma_{1}}, ^{c}D^{\varsigma}, ^{c}D^{\varrho_{1}}$ are the Liouville-Caputo fractional derivatives of order $2 < \varrho, \varsigma\leq3$, $1 < \varsigma_{1}, \varrho_{1} < 2$. $I^{\xi}, I^{\zeta}$ are the Riemann-Liouville fractional integrals of order $1 < \xi, \zeta < 2$. $J_{\rho}^{\upsilon, \varpi}, J_{v}^{\omega, \theta}$ are the Erdélyi-Kober fractional integrals of order $\varpi, \theta > 0$, with $v, \omega > 0$, $\rho, \ \vartheta\in(-\infty, +\infty)$. $F, G:K\times(-\infty, +\infty)^{4}\rightarrow (-\infty, +\infty)$ and $\phi_{1}, \phi_{2}:C(K, (-\infty, +\infty))\rightarrow (-\infty, +\infty)$ are continuous functions. $\gamma, \delta, \varepsilon_{1}, \varepsilon_{2}$ are positive real constants. The existence result was given by the Leray-Schauder alternative, and the uniqueness result was obtained due to Banach's fixed-point theorem. By the same methods, Arioua and Titraoui ^[18] studied system (1.1). Moreover, In ^[10], Arioua and Titraoui also investigated a new fractional problem involving the Erdélyi-Kober derivative. Inspired by the above articles, we use different methods to consider the fractional coupled system involving Erdélyi-Kober derivative (1.1). We employ the Guo-Krasnosel'skii fixed point theorem to discuss (1.1) in a special Banach space, and we also use the monotone iterative technique to study this system. Some existence results of positive solutions for system (1.1) are obtained, including the existence results of at least two positive solutions.

2.   Preliminaries

Definition 2.1. (see ^[2]) Let $\alpha\in(-\infty, +\infty)$. $C_{\alpha}^{n}$, $n\in N$, denotes a set of all functions $f(t), t > 0$, with $f(t) = t^{p}f_{1}(t)$ with $p > \alpha$ and $f_{1}\in C^{n}[0, \infty)$.

Definition 2.2. (see ^[1,2]) For a function $u\in C_{\alpha}$, the $\sigma$-order right-hand Erdélyi-Kober fractional integral is

in which, $\Gamma$ is the Euler gamma function.

Definition 2.3. (see ^[2]) Let $n-1 < \delta\leq n, n\in N$, and for $u\in C_{\alpha}$, the $\sigma$-order right-hand Erdélyi-Kober fractional derivative is

where

Lemma 2.1. (see ^[10]) Let $1 < \sigma\leq 2$, $-2 < \gamma < -1$, $\beta > 0$, and $h\in C_{\alpha}^{2}$, with $\int_0^\infty s^{\beta(\gamma+m)-1}h(\tau)d\tau < \infty$, $m = {1, 2}$. The fractional problem

has a unique solution given by $u(x) = \int_0^\infty G_{\sigma}(x, s)s^{\beta(\gamma+1)-1}h(s)ds$, where

Lemma 2.2. (see ^[10]) For $1 < \sigma\leq2$, $-2 < \gamma < -1$, and $\beta > 0$, the function $G_{\sigma}$, defined in $(2.1)$, has the following properties:

$(i)$ $\frac{G_{\sigma}(x, s)}{1+x^{-\beta(1+\gamma)}} > 0$, for $x, s > 0$;

$(ii)$ $\frac{G_{\sigma}(x, s)}{1+x^{-\beta(1+\gamma)}}\leq \frac{\beta}{\Gamma(\sigma)}$, for $x, s > 0$;

$(iii)$ for $0 < \frac{\tau}{\lambda}\leq x\leq \tau$ and $s > \frac{\tau}{\lambda^{2}}$, where $\lambda > 1, \tau > 0$, we have

where $p(\tau) = \frac{(\sigma-1)\tau^{-\beta(1+\gamma)}}{\lambda^{\beta(1+\gamma)}(1+\tau^{-\beta(1+\gamma)})}$.

Lemma 2.3. (see ^[18]) Let $0 < \sigma_{1}, \sigma_{2}\leq 1$ and $F, G\in C_{\alpha}^{2}$ with

Then, $(1.1)$ has a unique solution given by

where

The following result is our main tool.

Lemma 2.4. (Guo-Krasnosel'skii fixed point theorem; see ^[37]) $P$ is a cone in a Banach space $E$, and $D_{1}$ and $D_{2}$ are bounded open sets in $E$ with $\theta\in D_{1}$, $\overline{D_{1}}\subset D_{2}$. $A:P\cap(\overline{D_{2}}\setminus D_{1})\rightarrow P$ is a completely continuous operator. Consider the following conditions (ⅰ), (ⅱ):

$(i)$ $\|Aw\|\leq \|w\|$ for $w\in P\cap\partial D_{1}$, $\|Aw\|\geq \|w\|$ for $w\in P\cap\partial D_{2}$;

$(ii)$ $\|Aw\|\geq \|w\|$ for $w\in P\cap\partial D_{1}$, $\|Aw\|\leq \|w\|$ for $w\in P\cap\partial D_{2}$.

If one of the preceding conditions (ⅰ), (ⅱ) holds, then $A$ has at least one fixed point in $P\cap(\overline{D_{2}}\setminus D_{1})$.

Next, we present some hypotheses that will play an important role in the subsequent discussion:

$(H_{1})$ $F, G:(0, +\infty)\times(-\infty, +\infty)\times(-\infty, +\infty)\rightarrow (0, +\infty)$ are continuous and nondecreasing with respect to the second, third variables on $(0, +\infty)$.

$(H_{2})$ For $(x, u, v)\in(0, +\infty)\times(-\infty, +\infty)\times(-\infty, +\infty)$,

such that

with $\omega_{i}, \widetilde{\omega_{i}}\in C((0, +\infty), (0, +\infty))$ nondecreasing and $\varphi_{i}, \psi_{i}\in L^{1}(0, +\infty)$, $i = 1, 2$.

$(H_{3})$ There are positive functions $q_{i}, \widetilde{q}_{i}, i = 1, 2,$ with

such that

for any $u, v, \widetilde{u}, \widetilde{v}\in(-\infty, +\infty)$ and $x\in(0, +\infty)$.

$(H_{4})$ $F, G:(0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow (0, +\infty)$ are continuous, such that

where $a_{1}, a_{2}\in L^{1}((0, +\infty), (0, +\infty))$, $F_{1}, G_{1}\in C((0, +\infty)\times(0, +\infty)\times(0, +\infty), (0, +\infty))$, $0 < \int_\frac{\tau}{\lambda}^\tau a_{1}(x)dx < \infty$, $0 < \int_\frac{\tau}{\lambda}^\tau a_{2}(x)dx < \infty$, with $\tau > 0$, $\lambda > 1$. Moreover, $x^{\beta(1+\gamma)-1}F(x, u, v),$ $x^{\beta(1+\gamma)-1}G(x, u, v):[0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow [0, +\infty)$ also are continuous.

Remark 2.1. These conditions ensure the continuity and integrability of nonlinear terms in an infinite interval, which play a very important role in the proof of completely continuity for the relevant integral operators.

3.   Main results

In this section, we use two Banach spaces defined by

with the norm

and

with the norm

So, $(X\times Y, \|(u, v)\|_{X\times Y})$ is a Banach space, with the norm $\|(u, v)\|_{X\times Y} = \|u\|_{X}+\|v\|_{Y}$.

Lemma 3.1. If $F, G$ are continuous, then $(u, v)\in X\times Y$ is a solution of system (1.1)$\Leftrightarrow (u, v)\in X\times Y$ is a solution of the following equations:

For $(u, v)\in X\times Y$, we define an operator $A:X\times Y\rightarrow X\times Y$ as follows:

where

with $G_{\sigma_{i}}(x, s), i = 1, 2$, given by (2.2) and (2.3).

Remark 3.1. Let $\sigma_{1}, \sigma_{2}, \beta, \gamma, \lambda, \tau\in\mathbb{R}$, such that $1 < \sigma_{1}, \sigma_{2}\leq 2, \beta > 0, -2 < \gamma < -1, \lambda > 1, \tau > 0$. If $(H_{2})$ and $(H_{4})$ hold, then for $(u, v)\in X\times Y$ with $u(x), v(x) > 0,$

where $\eta = \max\{\eta_{1}, \eta_{2}\}$ with $\eta_{1} = 1+\frac{\iota}{\varrho_{1}(\lambda^{2}-1)}, \eta_{2} = 1+\frac{\iota^{\ast}}{\varrho_{2}(\lambda^{2}-1)} > 1$, $\varrho_{1}, \varrho_{2}, \iota, \iota^{\ast} > 0$.

Proof. By $(H_{4})$, for $x\in[\frac{\tau}{\lambda^{2}}, \tau]$, we know that there exist two constants $\varrho_{1}, \varrho_{2} > 0$, such that

So, for $(u, v)\in X\times Y$ with $u(x), v(x) > 0$,

and hence,

By $(H_{4})$, we know that there exist two constants $\iota, \iota^{\ast} > 0$, such that

Thus,

Therefore, we can obtain

Similarly,

Take $\eta = \max\{\eta_{1}, \eta_{2}\}$, and thus

hold. □

4.   A positive solution

Define two cones

Obviously, $K_{1}\times K_{2} = \{(u, v)\in X\times Y\mid u(x) > 0, v(x) > 0, \forall x > 0; \ \ \min\limits_{x\in [\frac{\tau}{\lambda}, \tau]}\frac{u(x)}{1+x^{-\beta(1+\gamma)}}\geq\frac{p(\tau)}{\eta}\|u\|_{X}, \min\limits_{x\in [\frac{\tau}{\lambda}, \tau]}\frac{v(x)}{1+x^{-\beta(1+\gamma)}}\geq\frac{p(\tau)}{\eta}\|v\|_{Y}\}$ is also a cone. For convenience, we first list the following definitions:

Lemma 4.1. If assumptions $(H_{1})$ and $(H_{2})$ hold, then $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$ is completely continuous.

Proof. First, we show $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$. By $(H_{1})$ and $(H_{2})$, for $(u, v)\in K_{1}\times K_{2}$,

Similarly,

By $(H_{1})$ and Lemma 2.2, for $(u, v)\in K_{1}\times K_{2}$, we have $A_{1}(u, v)(x) > 0, A_{2}(u, v)(x) > 0, x > 0$. From Lemma 2.2 and Remark 3.1, for $x\in[\frac{\tau}{\lambda}, \tau], \tau > 0$, and $\lambda > 1$,

So, $\frac{A_{1}(u, v)(x)}{1+x^{-\beta(1+\gamma)}}\geq\frac{p(\tau)}{\eta}\|A_{1}(u, v)\|_{X}$. Similarly, $\frac{A_{2}(u, v)(x)}{1+x^{-\beta(1+\gamma)}}\geq\frac{p(\tau)}{\eta}\|A_{2}(u, v)\|_{Y}$. Therefore,

That is, $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$ is true.

Second, it will give a simply prove that $A$ is continuous. Let $D = \{(u, v)|(u, v)\in K_{1}\times K_{2}, \|(u, v)\|_{X\times Y}\leq K, K > 0\}$, a bounded subset in $K_{1}\times K_{2}$. Let $(u_{n}, v_{n})\in D$ be a sequence that converges to $(u, v)$ in $\in K_{1}\times K_{2}$. Then $\|(u_{n}, v_{n})\|_{X\times Y}\leq K$. From Lemma 2.2,

By $(H_{2})$,

By the continuity of $s^{\beta(\gamma+1)-1}F(s, u(s), v(s))$ and the Lebesgue dominated convergence theorem,

Therefore, $\|A_{1}(u_{n}, v_{n})-A_{1}(u, v)\|_{X}\rightarrow 0, n\rightarrow \infty$. Similarly, $\|A_{2}(u_{n}, v_{n})-A_{2}(u, v)\|_{Y}\rightarrow 0, n\rightarrow \infty$.

So, $\|A(u_{n}, v_{n})-A(u, v)\|_{X\times Y}\rightarrow 0, n\rightarrow \infty$. That is, $A$ is continuous in $D$. In the end, we know that $A(D)$ is relatively compact on $(0, \infty)$ and is equi-convergent at $\infty$ by ^[18]. Therefore, $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$ is completely continuous. □

Theorem 4.1. Assume that $(H_{2})$ and $(H_{4})$ hold. If $F_{0} = 0, G_{0}^{\ast} = 0, f_{\infty} = \infty, g_{\infty}^{\ast} = \infty,$ then the system (1.1) has at least one positive solution.

Proof. We divide the proof into several steps.

Step 1. $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$ is completely continuous. This result easily follows from Lemma 4.1.

Step 2. We show that there exist $R_1 > 0$ and $D_{1} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < R_{1}\}$ such that $\|A(u, v)\|_{X\times Y}\leq\|(u, v)\|_{X\times Y}, \ (u, v)\in (K_{1}\times K_{2})\cap\partial D_{1}.$

Because $F_{0} = 0, G_{0}^{\ast} = 0,$ we choose $R_{1} > 0$, such that

for $0 < u+v\leq R_{1}, x > 0$, where $\epsilon_{1}, \epsilon_{2} > 0$ satisfy

So, for $(u, v)\in K_{1}\times K_{2}$ and $\|(u, v)\|_{X\times Y} = R_{1}$, by Lemma 2.2,

By $(H_{4})$,

Similarly,

Therefore,

Let $D_{1} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < R_{1}\}.$ Then,

Step 3. We show that there exist $R_2 > 0$ and $D_{2} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < R_{2}\}$ such that

Because $f_{\infty} = \infty, g_{\infty}^{\ast} = \infty$, there exists $R > 0$, such that

for $u+v\geq R, x > 0$, where $m_{1}, m_{2} > 0$ satisfy

Let $R_{2}\geq\max\{R_{1}, \frac{\eta R}{p(\tau)}\}$, and $D_{2} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < R_{2}\}$. Then, $D_{1}\subset D_{2}$.

Thus, for $\ (u, v)\in K_{1}\times K_{2}$, $\|(u, v)\|_{X\times Y} = R_{2}$, we have

So,

By $(H_{4})$, for $x\in [\frac{\tau}{\lambda}, \tau]$, we can obtain

Similarly, $\frac{A_{2}(u, v)(x)}{1+x^{-\beta(1+\gamma)}}\geq \frac{1}{2}\|(u, v)\|_{X\times Y}$. Therefore,

Finally, by Lemma 2.4, $A$ has a fixed point in $(K_{1}\times K_{1})\cap\partial(\overline{D_{2}}\backslash D_{1})$. So, (1.1) has at least one positive solution. □

Theorem 4.2. Assume that $(H_{2})$ and $(H_{4})$ hold. If $f_{0} = \infty, g_{0}^{\ast} = \infty, F_{\infty} = 0, G_{\infty}^{\ast} = 0,$ then (1.1) has at least one positive solution.

Proof. We divide the proof into several steps.

Step 1. $A:K_{1}\times K_{2}\rightarrow K_{1}\times K_{2}$ is completely continuous. This result easily follows from Lemma 4.1.

Step 2. We show that there exist $r_1 > 0$ and $D_{1} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < r_{1}\}$ such that

Because $f_{0} = \infty, g_{0}^{\ast} = \infty$, there exists $r_{1} > 0$ such that

for $0 < u+v\leq r_{1}, x > 0$, where $M_{1}, M_{2} > 0$, satisfy

Let $D_{1} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < r_{1}\}$. So, for $(u, v)\in K_{1}\times K_{2}$ with $\|(u, v)\|_{X\times Y} = r_{1}$, and $x\in [\frac{\tau}{\lambda}, \tau]$, then by $(H_{4})$,

Similarly, $\frac{A_{2}(u, v)(x)}{1+x^{-\beta(1+\gamma)}}\geq \frac{1}{2}\|(u, v)\|_{X\times Y}$. Thus,

Step 3. We show that there exist $r_2 > 0$ and $D_{2} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < r_{2}\}$ such that

Because $F_{\infty} = 0, G_{\infty}^{\ast} = 0$, there exists $r > 0$, such that

for $u+v > r, x > 0$, where $\epsilon_{1}, \epsilon_{2} > 0$ satisfy

Let $D_{2} = \{(u, v)\in X\times Y, \|(u, v)\|_{X\times Y} < r_{2}\},$ where $r_{2} > \max\{r_{1}, r\}$. Then $D_{1}\subset D_{1}$. We define two functions $U_{1}, U_{2}$ as follows:

For $(u, v)\in K_{1}\times K_{2}$ and $\|(u, v)\|_{X\times Y} = r_{2}$,

By Lemma 2.2 and $(H_{4})$,

Similarly, $\frac{A_{2}(u, v)(x)}{1+x^{-\beta(1+\gamma)}}\leq\frac{1}{2}\|(u, v)\|_{X\times Y}$. Therefore, $\|A(u, v)\|_{X\times Y}\leq\|(u, v)\|_{X\times Y}$, for $(u, v)\in (K_{1}\times K_{2})\cap\partial D_{2}$. Finally, by Lemma 2.4, $A$ has a fixed point in $(K_{1}\times K_{1})\cap\partial(\overline{D_{2}}\backslash D_{1})$. So, the system (1.1) has at least one positive solution. □

5.   Multiple positive solutions

In the section, we obtain the multiplicity of positive solution of (1.1) by using the monotone iterative technique.

Theorem 5.1. If $(H_{1})$ and $(H_{2})$ hold, then (1.1) has two positive solutions $(u^{\ast}, v^{\ast})$ and $(w^{\ast}, z^{\ast})$ satisfying $0\leq\|(u^{\ast}, v^{\ast})\|_{X\times Y}\leq \Upsilon$ and $0\leq\|(w^{\ast}, z^{\ast})\|_{X\times Y}\leq \Upsilon$, where $\Upsilon$ is a positive preset constant. Moreover, $\lim\limits_{n\rightarrow \infty}(u_{n}, v_{n}) = (u^{\ast}, v^{\ast})$ and $\lim\limits_{n\rightarrow \infty}(w_{n}, z_{n}) = (w^{\ast}, z^{\ast})$, where $(u_{n}, v_{n})$ and $(w_{n}, z_{n})$ are given by

with

and

with $(w_{0}(x), z_{0}(x)) = (0, 0)$. In addition,

Proof. First, from Lemma 4.1, $A(K_{1}\times K_{2})\subset K_{1}\times K_{2}$ for $(u, v)\in K_{1}\times K_{2}$. Let

and $\Upsilon\geq\Upsilon_{1}+\Upsilon_{2}$ with $D_{\Upsilon} = \{(u, v)\in K_{1}\times K_{2}:\|(u, v)\|_{X\times Y}\leq\Upsilon\}$. For any $(u, v)\in D_{\Upsilon}$, from $(H_{2})$ and Lemma 2.2,

Similarly, $\|A_{2}(u, v)\|_{Y}\leq\Upsilon_{2}$ for $(u, v)\in D_{\Upsilon}$. Thus,

That is, $A(D_{\Upsilon})\subset D_{\Upsilon}$. We construct two sequences as follows:

Obviously, $(u_{0}(x), v_{0}(x)), (w_{0}(x), z_{0}(x))\in D_{\Upsilon}$. Because $A(D_{\Upsilon})\subset D_{\Upsilon}$, $(u_{n}, v_{n}), (w_{n}, z_{n})\in D_{\Upsilon}, n = 1, 2, \ldots$. We need to show that there exist $(u^{\ast}, v^{\ast})$ and $(w^{\ast}, z^{\ast})$ satisfying $\lim\limits_{n\rightarrow \infty}(u_{n}, v_{n}) = (u^{\ast}, v^{\ast})$ and $\lim\limits_{n\rightarrow \infty}(w_{n}, z_{n}) = (w^{\ast}, z^{\ast})$ which are two monotone sequences for approximating positive solutions of the system (1.1).

For $x\in(0, +\infty), (u_{n}, v_{n})\in D_{\Upsilon}$, from Lemma 2.2 and (5.1),

and

that is,

So, by the condition $(H_{1})$,

For $x\in(0, +\infty)$, the sequences $\{(u_{n}, v_{n})\}_{n = 0}^{\infty}$ satisfy $(u_{n+1}(x), v_{n+1}(x))\leq(u_{n}(x), v_{n}(x))$. By the iterative sequences $(u_{n+1}, v_{n+1}) = A(u_{n}, v_{n})$ and the complete continuity of the operator $A$, $(u_{n}, v_{n})\rightarrow (u^{\ast}, v^{\ast})$, and $A(u^{\ast}, v^{\ast}) = (u^{\ast}, v^{\ast})$.

Similarly, for the sequences $\{(w_{n}, z_{n})\}_{n = 0}^{\infty}$, we have

Then, by the condition $(H_{1})$,

Analogously, for $x\in(0, +\infty)$, we have $(w_{n+1}(x), z_{n+1}(x))\geq(w_{n}(x), z_{n}(x))$. By the iterative sequences $(w_{n+1}, z_{n+1}) = A(w_{n}, z_{n})$ and the complete continuity of the operator $A$, $(w_{n}, z_{n})\rightarrow (w^{\ast}, z^{\ast})$, and $A(w^{\ast}, z^{\ast}) = (w^{\ast}, z^{\ast})$.

Finally, we prove that $(u^{\ast}, v^{\ast})$ and $(w^{\ast}, z^{\ast})$ are the minimal and maximal positive solutions of (1.1). Assume that $(\varsigma(x), \mu(x))$ is any positive solution of (1.1). Then, $A(\varsigma(x), \mu(x)) = (\varsigma(x), \mu(x))$, and

Therefore,

That is, $(w_{1}(x), z_{1}(x))\leq(\varsigma(x), \mu(x))\leq(u_{n}(x), v_{n}(x))$. So, (5.3) holds. By $(H_{1})$, $(0, 0)$ is not a solution of (1.1). From (5.1), $(w^{\ast}, z^{\ast})$ and $(u^{\ast}, v^{\ast})$ are two extreme positive solutions of (1.1), which can be constructed via limitS of two monotone iterative sequences in (5.1) and (5.2). □

6.   Examples

Example 6.1. We consider the following system:

where $\sigma_{1} = \frac{5}{3}, \sigma_{2} = \frac{3}{2}, \gamma = -\frac{3}{2}, \beta = 1$,

First, for $F_{1}(x, u, v) = x^{\beta(1+\gamma)-1}F(x, (1+x^{-\beta(1+\gamma)})u, (1+x^{-\beta(1+\gamma)})v) = e^{-x}(u^{2}, v^{2})$, we choose $\omega_{1}(u) = u^{2}\in C((0, +\infty), (0, +\infty))$, $\omega_{2}(v) = v^{2}\in C((0, +\infty), (0, +\infty))$, and $\varphi_{1}(x) = \psi_{1}(x) = e^{-x}\in L^{1}(0, +\infty)$. Then,

Similarly, for $F_{2}(x, u, v) = x^{\beta(1+\gamma)-1}G(x, (1+x^{-\beta(1+\gamma)})u, (1+x^{-\beta(1+\gamma)})v) = xe^{-2x^{2}}[u^{2}\ln(u^{2}+1)+ v^{2}\ln(v^{2}+1)]$, we choose $\widetilde{\omega_{1}}(u) = u^{2}\ln(u^{2}+1)\in C((0, +\infty), (0, +\infty))$, $\widetilde{\omega_{2}}(v) = v^{2}\ln(v^{2}+1)\in C((0, +\infty), (0, +\infty))$, and $\varphi_{2}(x) = \psi_{2}(x) = xe^{-2x^{2}}\in L^{1}(0, +\infty)$. Then,

So, the condition $(H_{2})$ holds. Obviously, $F, G:(0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow (0, +\infty)$ are continuous.

where $a_{1}(x) = e^{-x}, a_{2}(x) = xe^{-2x^{2}}$, $F_{1}(x, u, v) = (\frac{u}{1+x^{\frac{1}{2}}})^{2}+(\frac{v}{1+x^{\frac{1}{2}}})^{2}$, $G_{1}(t, u, v) = (\frac{u}{1+x^{\frac{1}{2}}})^{2}\ln(1+(\frac{u}{1+x^{\frac{1}{2}}})^{2}) +(\frac{u}{1+x^{\frac{1}{2}}})^{2}\ln(1+(\frac{u}{1+x^{\frac{1}{2}}})^{2})$. So, $x^{-\frac{3}{2}}f(x, u, v), x^{-\frac{3}{2}}G(x, u, v):[0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow [0, +\infty)$ are continuous. Hence, the condition $(H_{4})$ holds. Finally,

Therefore, from Theorem 4.1, (6.1) has at least one positive solution $(u(x), v(x))$. Further,

Example 6.2. We consider the following system:

where $\sigma_{1} = \frac{3}{2}, \sigma_{2} = \frac{7}{6}, \gamma = -\frac{3}{2}, \beta = 1$,

First, for

we choose $\omega_{1}(u) = \arctan u^{2}+\frac{1}{\pi}\in C((0, +\infty), (0, +\infty)), \omega_{2}(v) = \arctan v^{2}+\pi\in C((0, +\infty), (0, +\infty))$, and $\varphi_{1}(x) = \psi_{1}(x) = xe^{-2x^{2}+1}\in L^{1}(0, +\infty)$. Then,

Similarly, for

we choose $\widetilde{\omega_{1}}(u) = \arctan(\ln(u^{2}+1))+\frac{3}{2}\pi\in C((0, +\infty), (0, +\infty))$, $\widetilde{\omega_{2}}(v) = \arctan(\ln(v^{2}+1))+1\in C((0, +\infty), (0, +\infty))$, and $\varphi_{2}(x) = \psi_{2}(x) = e^{-x}\in L^{1}(0, +\infty)$. Then,

That is, $(H_{2})$ holds. Second, $F, G:(0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow (0, +\infty)$ are continuous. And

where $a_{1}(x) = xe^{-2x^{2}+1}, a_{2}(x) = e^{-x}$, $F_{1}(x, u, v) = \arctan(\frac{u}{1+x^{\frac{1}{2}}})^{2}+\frac{1}{\pi}+\arctan(\frac{v}{1+x^{\frac{1}{2}}})^{2}+\pi$, $G_{1}(x, u, v) = \arctan(\ln((\frac{u}{1+x^{\frac{1}{2}}})^{2}+1))+\frac{3}{2}\pi+\arctan(\ln((\frac{v}{1+x^{\frac{1}{2}}})^{2}+1))+1$. So, $x^{-\frac{3}{2}}F(x, u, v), x^{-\frac{3}{2}}G(x, u, v):[0, +\infty)\times(0, +\infty)\times(0, +\infty)\rightarrow [0, +\infty)$ are continuous. That is, $(H_{4})$ holds. In addition,

Therefore, from Theorem 4.2, (6.2) has at least one positive solution $(u(x), v(x))$. Further,

Example 6.3. We consider the following system:

where $\sigma_{1} = \frac{5}{3}, \sigma_{2} = \frac{3}{2}, \gamma = -\frac{3}{2}, \beta = 1$,

Obviously, $F, G:(0, +\infty)\times(-\infty, +\infty)\times(-\infty, +\infty)\rightarrow (0, +\infty)$ are continuous and nondecreasing with respect to the second and the third variables on $(0, +\infty)$. That is, $(H_{1})$ holds. Next,

We choose $\omega_{1}(u) = \mid u\mid\in C((0, +\infty), (0, +\infty))$, $\omega_{2}(v) = \ln(\mid v\mid+1)\in C((0, +\infty), (0, +\infty))$, and $\varphi_{1}(x) = \frac{e^{-x}}{3}, \psi_{1}(x) = \frac{xe^{-2x^{2}+1}}{10}\in L^{1}(0, +\infty)$. Then,

Similarly, for

we choose $\widetilde{\omega_{1}}(u) = \arctan(\mid u\mid+\frac{1}{\sqrt{\pi}})\in C((0, +\infty), (0, +\infty))$, $\widetilde{\omega_{2}}(v) = \mid v\mid\in C((0, +\infty), (0, +\infty))$, and $\varphi_{2}(x) = xe^{-2x^{2}+1}, \psi_{2}(x) = x\frac{e^{-2x^{2}+1}}{5}\in L^{1}(0, +\infty)$. Then,

That is, $(H_{2})$ holds. Therefore, from Theorem 5.1, (6.3) has two positive solutions $(u^{\ast}, v^{\ast})$ and $(w^{\ast}, z^{\ast})$ with $(0, 0)\leq (u^{\ast}(x), v^{\ast}(x)), (w^{\ast}(x), z^{\ast}(x))\leq ((1+x^{\frac 12})\Upsilon_{1}, (1+x^{\frac 12})\Upsilon_{2})$, where $\Upsilon_{1}+\Upsilon_{2}\leq \Upsilon$, and $\Upsilon$ satisfies

7.   Conclusions

This paper studies the Erdélyi-Kober fractional coupled system (1.1), where the variable is in an infinite interval. We give some proper conditions and set a special Banach space. We obtain the existence of at least one positive solution for (1.1) by using the Guo-Krasnosel'skii fixed point theorem, and we get the existence of at least two positive solutions for (1.1) by using the monotone iterative technique. Our methods and results are different from ones in ^[18]. Moreover, we give three examples to show the plausibility of our main results. For future work, we intend to use other fixed point theorems to solve some Erdélyi-Kober fractional differential equations.

Use of AI tools declaration

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

Acknowledgments

This paper is supported by the Fundamental Research Program of Shanxi Province (202303021221068).

Conflict of interest

The authors declare that they have no competing interests.