1 Introduction

Recently, fractional calculus (FC, for short) has been attractive to many researchers due to the nonlocalization properties of the fractional derivatives and integrals, contrary to the integer-order derivatives. For more details on the FC, one can see [17, 23, 24]. In fact, many researchers have devoted themselves to investigate fractional order differential equations and systems with different boundary conditions, for more details, the reader can see the works [1, 3, 4, 6, 9, 10, 14, 20, 22, 29, 37,38,39, 41].

In the present paper, we are interested, in general, to study a differential system of fractional order (SFO, for short) that can induce a limiting differential case of fourth order system (4OS, for short). So, we have, first, to cite some related papers on SFO and 4OS, as well as some other one dimensional differential equations, that are often found in mathematical models of various physical and biological problems, in bridge design and deformation of structures, as well as in settlement of soils. The paper are given in [5, 8, 11, 12, 18, 19, 25, 30,31,32, 34, 35, 40]. After citing the above “SFO and 4OS papers”, we present, in some details, other SFO and 4OS papers that have motivated the present work. We begin by [2] where thanks to the Leray–Schauder alternative and the Banach contraction principle, the authors investigated the existence and uniqueness of solutions for the following nonlinear sequential SFO with integro-differential termes:

$$\begin{aligned} \left\{ \begin{array}{ll} (^{c}D^{\alpha }+\lambda ^{c}D^{\alpha -1})x(t)=f(t,x(t),y(t),I_{0^{+}}^{P_{1}}x(t),I_{0^{+}}^{q_{1}}y(t)), &{}t\in \left( 0,1\right) \\ (^{c}D^{\beta }+\mu ^{c}D^{\beta -1})y(t)=f(t,x(t),y(t),I_{0^{+}}^{p_{2}}x(t),I_{0^{+}}^{q_{2}}y(t)),&{}t\in \left( 0,1\right) \end{array} \right. \end{aligned}$$

under the conditions:

$$\begin{aligned} \left\{ \begin{array}{ll} x(0)=x^{\prime }(0)=x^{\prime \prime }(0)=0,x(1)=\int _{0}^{1}x(s)d\mathfrak {H }_{1}(s)+\int _{0}^{1}y(s)d{\mathfrak {H}}_{2}(s), \\ y(0)=y^{\prime }(0)=y^{\prime \prime }(0)=0,y(1)=\int _{0}^{1}x(s)d\mathfrak {K }_{1}(s)+\int _{0}^{1}y(s)d{\mathfrak {K}}_{2}(s), \end{array} \right. \end{aligned}$$

where \(p_{1},p_{2}\in \left( 0,1\right) \), \(q_{1},q_{2}>0\), \(\alpha ,\beta \in \left( 3,4\right] \), \(\lambda ,\mu >0\) and the Riemann–Stieltjes integrals with given bounded variation functions \({\mathfrak {H}}_{1},\mathfrak {H}_{2},{\mathfrak {K}}_{1},{\mathfrak {K}}_{2}\).

We cite also the work in [36], where the authors studied the existence of positive solutions for the following nonlinear 4OS that is used to describe the deformation of an elastic beam supported at its two end points:

$$\begin{aligned} u^{(4)}(t)+\beta _{1}u^{\prime \prime }(t)-\alpha _{1}u(t)&=f_{1}(t,u(t),v(t)),\quad t\in (0,1) \\ v^{(4)}(t)+\beta _{2}v^{\prime \prime }(t)-\alpha _{2}v(t)&=f_{2}(t,u(t),v(t)),\quad t\in (0,1) \\ u(0)&=u(1)=u^{\prime \prime }(0)=u^{\prime \prime }(1)=0, \\ v(0)&=v(1)=v^{\prime \prime }(0)=v^{\prime \prime }(1)=0, \end{aligned}$$

where \(f_{i}\in C(\left[ 0,1\right] \times \mathbb {R^{+}}\times \mathbb {R^{+} },\mathbb {R^{+}})\), \(\mathbb {R^{+}}=\left[ 0,+\infty \right) \), and the parameters \(\beta _{i},\alpha _{i}\in {\mathbb {R}}\) satisfy the conditions: \( \beta _{i}<2\pi ^{2}\), \(-\beta _{i}^{2}/4\leqslant \alpha _{i}\), \(\alpha _{i}/\pi ^{4}+\beta _{i}/\pi ^{2}<1\), \(i=1,2\). By constructing a cone P in \(C\left( \left[ 0,1\right] \right) \times C\left( \left[ 0,1\right] \right) \) and using some fixed point theorems, the authors proved a first existence of positive solution results. Then, by constructing a product cone along with the fixed point theory on a product cone, they established another positive solution result.

In a recent paper [33], the authors investigated the existence and uniqueness of solutions for the following SFO which contain sequential Caputo derivatives, integral terms, and associated with some Riemann–Stieltjes boundary conditions:

$$\begin{aligned} \left\{ \begin{array}{ll} (^{c}D^{\alpha }+\lambda ^{c}D^{\alpha -1})x(t)=f(t,x(t),y(t),I_{0^{+}}^{P_{1}}x(t),I_{0^{+}}^{P_{2}}y(t)),&{} t\in \left( 0,1\right) \\ \\ (^{c}D^{\beta }+\mu ^{c}D^{\beta -1})y(t)=f(t,x(t),y(t),I_{0^{+}}^{q_{1}}x(t),I_{0^{+}}^{q_{2}}y(t)),&{} t\in \left( 0,1\right) \end{array} \right. \end{aligned}$$

under the conditions:

$$\begin{aligned} \left\{ \begin{array}{ll} x(0)=x^{\prime }(0)=0,x^{\prime }(1)=0,x(1)=\int _{0}^{1}x(s)d{\mathfrak {H}} _{1}(s)+\int _{0}^{1}y(s)d{\mathfrak {H}}_{2}(s), \\ \\ y(0)=y^{\prime }(0)=0,y^{\prime }(1)=0,y(1)=\int _{0}^{1}x(s)d{\mathfrak {K}} _{1}(s)+\int _{0}^{1}y(s)d{\mathfrak {K}}_{2}(s), \end{array} \right. \end{aligned}$$

where \(\alpha ,\beta \in \left( 3,4\right] \), \(\lambda ,\mu >0\), \( p_{1},q_{1},p_{2},q_{2}>0\), \(f,g:\left[ 0,1\right] \times {\mathbb {R}} ^{4}\rightarrow {\mathbb {R}}\) are continuous functions, \(I_{0^{+}}^{J}\) denotes the fractional Riemann–Liouville integral of order \(\nu \), with \(\nu =(p_{1},q_{1},p_{2},q_{2}\)) and the Riemann–Stieltjes integrals with given bounded variation functions \({\mathfrak {H}}_{1},{\mathfrak {H}}_{2},{\mathfrak {K}} _{1},{\mathfrak {K}}_{2}\). The system of this type can be applied in applied nature, mainly in biosciences, see [3] and its reference. The authors obtained existence and uniqueness results for the solutions of their system, and the roof of their results is based on both the Banach fixed point theorem and the Leray–Schauder alternative. Very recently, in [7], by considering a limiting case of the above problem, the authors have been concerned with the existence of solutions and their Ulam-stability for the following non sequential SFO:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\alpha }u(x)=f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,u^{\prime \prime }(x)), \\ \\ D^{\beta }v(x)=f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,v^{\prime \prime }(x)), \end{array} \right. \end{aligned}$$

under the following supported end-conditions:

$$\begin{aligned} \left\{ \begin{array}{l} u(0)=u(1)=u^{\prime \prime }(0)=u^{\prime \prime }(1)=0, \\ \\ v(0)=v(1)=v^{\prime \prime }(0)=v^{\prime \prime }(1)=0, \end{array} \right. \end{aligned}$$

where, \(x\in \left[ 0,1\right] \), \(3<\alpha ,\beta \leqslant 4\), \(D^{\alpha }\) and \(D^{\beta }\) denote the fractional derivatives in the sense of Caputo, and \(f_{i}\in C(\left[ 0,1\right] \times {\mathbb {R}}\times {\mathbb {R}}, {\mathbb {R}})\), \(g_{i},h_{i}\in C(\left[ 0,1\right] \times {\mathbb {R}},{\mathbb {R}})\), \(a_{i},b_{i}\in {\mathbb {R}}\).

In the present paper, we shall be concerned, first, with the following Caputo SFO of sequential derivatives, with the absence of semi group and commutativity properties on the derivatives. So we take the system:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\alpha _{1}}D^{\alpha _{2}}u(x)=f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,D^{\delta }u(x)), \\ D^{\beta _{1}}D^{\beta _{2}}v(x)=f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,D^{\delta }v(x)), \end{array} \right. \end{aligned}$$
(1.1)

under the following flexible/fixed end-conditions:

$$\begin{aligned} \left\{ \begin{array}{l} u(0))=u(1)=a\in {\mathbb {R}}, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=b\in {\mathbb {R}}, \\ v^{\prime }(0)=v^{\prime }(1)=0, \\ \end{array} \right. \end{aligned}$$
(1.2)

where \(D^{\alpha _{i}},D^{\beta _{i}},D^{\delta }\) are Caputo fractional derivatives. The conditions on absence of semi group and commutativity properties are guaranteed by the conditions: \(0<\alpha _{1},\beta _{1}\le 1,:2<\alpha _{2},\beta _{2}\le 3.\) We impose also that: \(1<\delta \le 2\),: For \(i=1,2,f_{i}:[0,1]\times {\mathbb {R}}^{2}\rightarrow {\mathbb {R}}\), \( g_{i}:[0,1]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) and \(h_{i}:[0,1]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) are some given functions that will be specified later.

The reader can observe first that our problem is more general than the above two problems of [7, 36], also they can see that there are differences between our system and the above two others related to the dependence of the nonlinearities \(f_{i}\) and \(g_{i}\) on various fractional derivatives. Another important difference between the three problems is given by the initial conditions, which leads to a significant difference between our system and the associated integral representations presented in [7, 36]. So the novelty of our problem is given by the existence of derivative terms in both sides of the considered system, and by the sequential concept on the derivatives of the left hand side of the same system. The absence of the semi group and the commutativity on the derivatives (of the half hand sides for our system) is also to be noted in this study.

In the second part of our paper, we will use the tanh method [13, 21] to find new traveling wave solutions for the following coupled beam problem with conformable fractional Khalil derivatives [16]:

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{t}^{2\alpha }u-T_{x}^{4\beta }u+H(u,v,T_{x}^{2\beta }(u,v))=0, \\ T_{t}^{2\alpha }v-T_{x}^{4\beta }v+L(u,v,T_{x}^{2\beta }(u,v))=0, \end{array}\right. } \end{aligned}$$
(1.3)

where \(0<\alpha ,\beta \le 1\), H, L are given functions.

It is important to say that the above two proposed system (1.1), (1.2) and (1.3) may have a connection between them since both of them can induce a coupled system of beam type of fourth order, see our system (7.9) and the above fourth order cited problem of [36].

The paper is arranged as follows: In Sect. 2, we recall some notions and notation that will be used later, Sect. 3 is concerned with the main existence results for the problem (1.1), (1.2), and in Sect. 4, we present two examples illustrating the two main results. The UH stability of the solutions for the problem (1.1), (1.2) is investigated in Sect. 5, some other examples are discussed in Sect. 6. In Sect. 7, new travelling wave solutions are obtained for a similar conformable system of beam type. At the end a conclusion follows.

2 Preliminaries

For the reader’s convenience, we present some necessary definitions and lemmas that will be used later.

Definition 1

For any function \(h\in C(\left[ 0,1\right] ),\) the fractional integral of order \(\alpha >0\) is defined by

$$\begin{aligned} I^{\alpha }h(t)=\int _{0}^{t}\frac{(t-s)^{\alpha -1}}{\Gamma (\alpha )}h(s)ds, \end{aligned}$$
(2.1)

where \(\Gamma \) is the gamma function.

Definition 2

For any function \(h\in C(\left[ 0,1\right] )\), the Riemann–Liouville fractional derivative of a positive order \(\alpha \) is defined by

$$\begin{aligned} D_{RL}^{\alpha }h(t)=\dfrac{1}{\Gamma (n-\alpha )}\left( \dfrac{d}{dt} \right) ^{n}\int _{0}^{t}(t-s)^{n-\alpha -1}h(s)ds, \end{aligned}$$
(2.2)

where \(n=\left[ \alpha \right] +1\) and \(\left[ \alpha \right] \) denotes the integer part of \(\alpha \).

Definition 3

For any function \(h\in C^{n}(\left[ 0,1\right] )\), the Caputo fractional order derivative of h, is defined by

$$\begin{aligned} D^{\alpha }h(t)=\dfrac{1}{\Gamma (n-\alpha )}\int _{0}^{t}(t-s)^{n-\alpha -1}h^{(n)}(s)ds, \end{aligned}$$
(2.3)

with \(n=\left[ \alpha \right] +1\).

Lemma 1

Let \(\alpha >0\). Then, the differential equation

$$\begin{aligned} D^{\alpha }h(t)=0,t\in [0,1], \end{aligned}$$
(2.4)

has, as general solution, the expression: \( h(t)=c_{0}+c_{1}t+c_{2}t^{2}+\cdots \cdot \cdot +c_{n-1}t^{n-1}\), \(c_{i}\in R\), \( i=0,1,2,\ldots ..,n-1\), \(n=\left[ \alpha \right] +1\).

Lemma 2

Let \(\alpha >0\). Then, we have

$$\begin{aligned} I_{a^{+}}^{\alpha }D^{\alpha }h(t)=h(t)+c_{0}+c_{1}t+c_{2}t^{2}+\cdots +c_{n-1}t^{n-1}, \end{aligned}$$
(2.5)

for some \(c_{i}\in R\), \(i=0,1,2,\ldots ,n-1\), \(n=\left[ \alpha \right] +1\).

Remark 1

One can prove that the quantities \(I^{\alpha }\) and \(D^{\alpha }\) have the following properties for any \(t\in [0,1]\):

  1. (1)

    \(I^{\alpha }I^{\beta }h(t)=I^{\alpha +\beta }h(t),\alpha>0,\beta >0,\)

  2. (2)

    \(D^{\alpha }I^{\alpha }h(t)=h(t),\alpha >0,\)

  3. (3)

    \(D^{\alpha }I^{\beta }h(t)=I^{\beta -\alpha }h(t),\beta>\alpha >0.\)

The following lemma will be used in our proofs.

Lemma 3

(Schaefer Fixed Point Theorem) Let H be a Banach space and \( T:H\rightarrow H\) be a completely continuous operator. If the subset \( G=\left\{ X\in H,X=\lambda TX,0<\lambda <1\right\} \), is bounded, then T has at least one fixed point in H.

Now, we present to the reader the following auxiliary lemme related to the integral representation of the above introduced sequential SFO.

Lemma 4

Let us take the functions \(\varphi _{1},\varphi _{2}\in C(\left[ 0,1\right] ,{\mathbb {R}})\). Hence, the problem

$$\begin{aligned} \left\{ \begin{array}{l} D^{\alpha _{1}}D^{\alpha _{2}}u(x)=\varphi _{1}(x) \\ D^{\beta _{1}}D^{\beta _{2}}v(x)=\varphi _{2}(x), \end{array} \right. \end{aligned}$$
(2.6)

with the conditions:

$$\begin{aligned} \left\{ \begin{array}{l} u(0))=u(1)=a\in {\mathbb {R}}, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=b\in {\mathbb {R}}, \\ v^{\prime }(0)=v^{\prime }(1)=0, \\ \end{array} \right. \end{aligned}$$
(2.7)

admits, as integral representation, the following expression:

$$\begin{aligned} \left\{ \begin{array}{l} u(x)=I^{\alpha _{1}+\alpha _{2}}\varphi _{1}(x)+\frac{2k_{1}-k_{2}}{\alpha _{2}-2}x^{\alpha _{2}}+\frac{k_{2}-\alpha _{2}k_{1}}{\alpha _{2}-2}x^{2}+a,\\ v(x)=I^{\beta _{1}+\beta _{2}}\varphi _{2}(x)+\frac{2k_{3}-k_{4}}{\beta _{2}-2}x^{\beta _{2}}+\frac{k_{4}-\beta _{2}k_{3}}{\beta _{2}-2}x^{2}+b, \end{array} \right. \end{aligned}$$
(2.8)

where,

$$\begin{aligned} k_{1}&=I^{\alpha _{1}+\alpha _{2}}\varphi _{1}(x)\mid _{x=1}, \\ k_{2}&=I^{\alpha _{1}+\alpha _{2}-1}\varphi _{1}(x)\mid _{x=1}, \\ k_{3}&=I^{\beta _{1}+\beta _{2}}\varphi _{2}(x)\mid _{x=1}, \\ k_{4}&=I^{\beta _{1}+\beta _{2}-1}\varphi _{2}(x)\mid _{x=1}. \end{aligned}$$

Proof

We know that

$$\begin{aligned} \left\{ \begin{array}{l} u(x)=I^{\alpha _{1}+\alpha _{2}}\varphi _{1}(x)+I^{\alpha _{2}}c_{0}+c_{1}+c_{2}x+c_{3}x^{2}, \\ v(x)=I^{\beta _{1}+\beta _{2}}\varphi _{2}(x)+I^{\beta _{2}}d_{0}+d_{1}+d_{2}x+d_{3}x^{2}. \end{array} \right. \end{aligned}$$
(2.9)

Hence, thanks to (2.7), we write

$$\begin{aligned} \left\{ \begin{array}{ll} c_{1}=a,&{}\quad d_{1}=b, \\ c_{2}=0,&{}\quad d_{2}=0, \\ c_{0}=\frac{\Gamma (\alpha _{2}+1)(2k_{1}-k_{2})}{\alpha _{2}-2},&{}\quad d_{0}= \frac{\Gamma (\beta _{2}+1)(2k_{3}-k_{4})}{\beta _{2}-2}, \\ c_{3}=\frac{k_{2}-\alpha _{2}k_{1}}{\alpha _{2}-2},&{}\quad d_{3}=\frac{ k_{4}-\beta _{2}k_{3}}{\beta _{2}-2}. \end{array} \right. \end{aligned}$$
(2.10)

The solution of (2.6)–(2.7) can be expressed as follows:

$$\begin{aligned} \left\{ \begin{array}{l} u(x)=I^{\alpha _{1}+\alpha _{2}}\varphi _{1}(x)+\frac{2k_{1}-k_{2}}{\alpha _{2}-2}x^{\alpha _{2}}+\frac{k_{2}-\alpha _{2}k_{1}}{\alpha _{2}-2}x^{2}+a,\\ v(x)=I^{\beta _{1}+\beta _{2}}\varphi _{2}(x)+\frac{2k_{3}-k_{4}}{\beta _{2}-2}x^{\beta _{2}}+\frac{k_{4}-\beta _{2}k_{3}}{\beta _{2}-2}x^{2}+b. \end{array} \right. \end{aligned}$$

This ends the proof of the lemma. \(\square \)

Let us now be placed in the fixed point theory by taking the space:

$$\begin{aligned} E:=\left\{ u\in C(\left[ 0,1\right] ,{\mathbb {R}}),\;D^{\delta }u\in C(\left[ 0,1\right] ,{\mathbb {R}})\right\} \end{aligned}$$

and the norm, for \(\ 1<\delta <2\):

$$\begin{aligned} \parallel u\parallel _{E}=\parallel u\parallel _{\infty }+\parallel D^{\delta }u\parallel _{\infty }, \end{aligned}$$
(2.11)
$$\begin{aligned} \parallel u\parallel _{\infty }=\underset{x\in \left[ 0,1\right] }{\sup } \mid u(x)\mid ,\; \parallel D^{\delta }u\parallel _{\infty }=\underset{x\in \left[ 0,1\right] }{\sup }\mid D^{\delta }u(x)\mid . \end{aligned}$$

For \(E\times E\), we can consider the norm

$$\begin{aligned} \parallel (u,v)\parallel _{E\times E}=\parallel u\parallel _{E}+\parallel v\parallel _{E}. \end{aligned}$$

Now, we are ready the present the main results.

3 Existence and uniqueness of solutions

In the following section, we will present and prove our main results on the existence of solutions. We begin first by proposing some conditions using the continuity, the boundedness or the Lypchitzianity of the given functions. It is also possible for the interested reader, which is familiarised with the fractional differential equations, to propose other conditions related to the measurability, or the “Caratheodority” in particular.

So, let us suppose in the following hypotheses:

  1. (H1):

    For \(i=1,2,\) the functions \(f_{i}:[0,1]\times {\mathbb {R}} ^{2}\rightarrow {\mathbb {R}}\), \(g_{i}:[0,1]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) and \(h_{i}:[0,1]\times {\mathbb {R}}\rightarrow {\mathbb {R}}\) are continuous.

  2. (H2):

    There exist positives functions \(\omega _{i}(x),\) \(i=1,2,3,4,\) such that that for all \(x\in \left[ 0,1\right] \) and \((u,v),(w,z)\in {\mathbb {R}} ^{2}\), we have

    $$\begin{aligned}&\mid f_{1}(x,u,v)-f_{1}(x,w,z)\mid \leqslant \omega _{1}(x)\mid u-w\mid +\omega _{2}(x)\mid v-z\mid , \\&\mid f_{2}(x,u,v)-f_{2}(x,w,z)\mid \leqslant \omega _{3}(x)\mid u-w\mid +\omega _{4}(x)\mid v-z\mid , \end{aligned}$$

    with \(n_{i}={\sup }_{x\in \left[ 0,1\right] }\mid \omega _{i}(x)\mid ,\) \(i=1,2,3,4.\)

  3. (H3):

    There exist positives functions \(\theta _{j}(x);\) \(j=1,2\), such that for all \(x\in \left[ 0,1\right] \) and \((u,v)\in {\mathbb {R}}^{2}\), we have

    $$\begin{aligned}&\mid g_{1}(x,u)-g_{1}(x,v)\mid \leqslant \theta _{1}(x)\mid u-w\mid , \\&\mid g_{2}(x,u)-g_{2}(x,v)\mid \leqslant \theta _{2}(x)\mid u-w\mid , \end{aligned}$$

    with, \(m_{j}={\sup }_{x\in \left[ 0,1\right] }\mid \theta _{j}(x)\mid ,\) \(j=1,2.\)

  4. (H4):

    There exist positives functions \(\varepsilon _{j}(x);\) \(j=1,2\), such that for all \(x\in \left[ 0,1\right] \) and \((u,v)\in {\mathbb {R}}^{2}\), we have

    $$\begin{aligned}&\mid h_{1}(x,u)-h_{1}(x,v)\mid \leqslant \varepsilon _{1}(x)\mid u-w\mid ,\\&\mid h_{2}(x,u)-h_{2}(x,v)\mid \leqslant \varepsilon _{2}(x)\mid u-w\mid , \end{aligned}$$

    with, \(r_{j}={\sup }_{x\in \left[ 0,1\right] }\mid \varepsilon _{j}(x)\mid ;j=1,2.\)

  5. (H5):

    There exist positive constants \(\Upsilon _{i}\), \(i=1,2..6\), such that for all \(x\in \left[ 0,1\right] \) and \((u,v)\in {\mathbb {R}}^{2}\), we have

    $$\begin{aligned}&\mid f_{1}(x,u,v)\mid \leqslant \Upsilon _{1},\quad \mid f_{2}(x,u,v)\mid \leqslant \Upsilon _{2}, \\&\mid g_{1}(x,u,v)\mid \leqslant \Upsilon _{3},\quad \mid g_{2}(x,u,v)\mid \leqslant \Upsilon _{4}, \\&\mid h_{1}(x,u,v)\mid \leqslant \Upsilon _{5},\quad \mid h_{2}(x,u,v)\mid \leqslant \Upsilon _{6}. \end{aligned}$$

    Let us now consider the notations:

    $$\begin{aligned} \Theta _{1}&:={\textit{max}}\left\{ (\lambda _{1}+\lambda _{2})(n_{1}+m_{1}\mid a_{1}\mid +r_{1}\mid b_{1}\mid )\, \,;\, \,:(\lambda _{1}+\lambda _{2})n_{2}\right\} , \\ \Theta _{2}&:={\textit{max}}\left\{ (\lambda _{3}+\lambda _{4})(n_{4}+m_{2}\mid a_{2}\mid +r_{2}\mid b_{2}\mid )\, \,;\, \,:(\lambda _{3}+\lambda _{4})n_{3}\right\} , \\ \Theta&:=\Theta _{1}+\Theta _{2}, \\ K_{1}&:=I^{\alpha _{1}+\alpha _{2}}f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,u^{\prime \prime }(x))\mid _{x=1}, \\ K_{2}&:=I^{\alpha _{1}+\alpha _{2}-1}f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,u^{\prime \prime }(x))\mid _{x=1}, \\ K_{3}&:=I^{\beta _{1}+\beta _{2}}f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,v^{\prime \prime }(x))\mid _{x=1}, \\ K_{4}&:=I^{\beta _{1}+\beta _{2}-1}f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,v^{\prime \prime }(x))\mid _{x=1}. \end{aligned}$$

    with, \(\lambda _{1}=\frac{1}{\Gamma (\alpha _{1}+\alpha _{2}+1)},\) \(\lambda _{2}=\frac{1}{\Gamma (\alpha _{1}+\alpha _{2}-\delta +1)},\) \(\lambda _{3}= \frac{1}{\Gamma (\beta _{1}+\beta _{2}+1)},\) \(\lambda _{4}=\frac{1}{\Gamma (\beta _{1}+\beta _{2}-\delta +1)}.\)

Our first result is given by the following theorem.

Theorem 1

Assume that (H2), (H3) and (H4) are satisfied. If \(\Theta <1\), then, (1.1)–(1.2) admits a unique solution \((u(t),v(t));t\in [0,1]\).

Proof

Let us consider the operator \(P:E\times E\rightarrow E\times E\) defined by

$$\begin{aligned} P(u(x),v(x))=(P_{1}(u(x),v(x)),P_{2}(u(x),v(x))), \end{aligned}$$

where,

$$\begin{aligned} P_{1}(u(x),v(x))&= I^{\alpha _{1}+\alpha _{2}}[f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,D^{\delta }u(x))]\nonumber \\&\quad +\frac{2K_{1}-K_{2}}{\alpha _{2}-2}x^{\alpha _{2}}+\frac{K_{2}-\alpha _{2}K_{1}}{\alpha _{2}-2}x^{2}+a, \end{aligned}$$
(3.1)
$$\begin{aligned} P_{2}(u(x),v(x))&= I^{\beta _{1}+\beta _{2}}[f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,D^{\delta }v(x))]\nonumber \\&\quad +\frac{2K_{3}-K_{4}}{\beta _{2}-2}x^{\beta _{2}}+\frac{K_{4}-\beta _{2}K_{3}}{\beta _{2}-2}x^{2}+b. \end{aligned}$$
(3.2)

We prove that P is an operator that satisfies the Banach contraction principle. First of all, it is to note that the stability of the above product space by the operator P is trivial and hence we omit it.

Let us now take two arbitrary elements \((u_{1},v_{1}),(u_{2},v_{2})\in E\times E\). So, we have

$$\begin{aligned}&\mid P_{1}(u_{2}(x),v_{2}(x))-P_{1}(u_{1}(x),v_{1}(x))\mid \\&\quad \leqslant I^{\alpha _{1}+\alpha _{2}}\mid f_{1}(x,u_{2}(x),v_{2}(x))-f_{1}(x,u_{1}(x),v_{1}(x))\mid \\&\qquad +I^{\alpha _{1}+\alpha _{2}}\mid a_{1}[g_{1}(x,u_{2}(x)-g_{1}(x,u_{1}(x)]\mid \\&\qquad +I^{\alpha _{1}+\alpha _{2}}\mid b_{1}[h_{1}(x,D^{\delta }u_{2}(x))-h_{1}(x,D^{\delta }u_{1}(x))]\mid \\&\quad \leqslant \lambda _{1}n_{1}\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{1}n_{2}\parallel v_{2}-v_{1}\parallel _{\infty } \\&\qquad +\lambda _{1}\mid a_{1}\mid m_{1}\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{1}\mid b_{1}\mid r_{1}\parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty }. \end{aligned}$$

Therefore,

$$\begin{aligned}&\parallel P_{1}(u_{2},v_{2})-P_{1}(u_{1},v_{1})\parallel _{\infty } \nonumber \\&\quad \leqslant \lambda _{1}(n_{1}+\mid a_{1}\mid m_{1})\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{1}\mid b_{1}\mid r_{1}\parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty } \nonumber \\&\qquad +\lambda _{1}n_{2}\parallel v_{2}-v_{1}\parallel _{\infty }. \end{aligned}$$
(3.3)

On the other hand, one can show that

$$\begin{aligned}&\mid D^{\delta }P_{1}(u_{2}(x),v_{2}(x))-D^{\delta }P_{1}(u_{1}(x),v_{1}(x))\mid \\&\quad \leqslant I^{\alpha _{1}+\alpha _{2}-\delta }\mid f_{1}(x,u_{2}(x),v_{2}(x))-f_{1}(x,u_{1}(x),v_{1}(x))\mid \\&\qquad +I^{\alpha _{1}+\alpha _{2}-\delta }\mid a_{1}[g_{1}(x,u_{2}(x)-g_{1}(x,u_{1}(x)]\mid \\&\qquad +I^{\alpha _{1}+\alpha _{2}-\delta }\mid b_{1}[h_{1}(x,D^{\delta }u_{2}(x))-h_{1}(x,D^{\delta }u_{1}(x))]\mid \\&\quad \leqslant \lambda _{2}n_{1}\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{2}n_{2}\parallel v_{2}-v_{1}\parallel _{\infty } \\&\qquad +\lambda _{2}\mid a_{1}\mid m_{1}\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{2}\mid b_{1}\mid r_{1}\parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty }. \end{aligned}$$

Therefore,

$$\begin{aligned}&\parallel D^{\delta }P_{1}(u_{2},v_{2})-D^{\delta }P_{1}(u_{1},v_{1})\parallel _{\infty } \nonumber \\&\quad \leqslant \lambda _{2}(n_{1}+\mid a_{1}\mid m_{1})\parallel u_{2}-u_{1}\parallel _{\infty }+\lambda _{2}\mid b_{1}\mid r_{1}\parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty } \nonumber \\&\qquad +\lambda _{2}n_{2}\parallel v_{2}-v_{1}\parallel _{\infty }. \end{aligned}$$
(3.4)

Thanks to (3.3),(3.4) and (2.11), we get

$$\begin{aligned}&\parallel P_{1}(u_{2},v_{2})-P_{1}(u_{1},v_{1})\parallel _{E} \\&\quad =\parallel P_{1}(u_{2},v_{2})-P_{1}(u_{1},v_{1})\parallel _{\infty }+\parallel D^{\delta }P_{1}(u_{2},v_{2})-D^{\delta }P_{1}(u_{1},v_{1})\parallel _{\infty } \\&\quad \leqslant (\lambda _{1}+\lambda _{2})(n_{1}+\mid a_{1}\mid m_{1})\parallel u_{2}-u_{1}\parallel _{\infty } \\&\qquad +(\lambda _{1}+\lambda _{2})\mid b_{1}\mid r_{1}\parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty }+(\lambda _{1}+\lambda _{2})n_{2}\parallel v_{2}-v_{1}\parallel _{\infty } \\&\quad \leqslant (\lambda _{1}+\lambda _{2})(n_{1}+m_{1}\mid a_{1}\mid +r_{1}\mid b_{1}\mid )\parallel u_{2}-u_{1}\parallel _{E}+(\lambda _{1}+\lambda _{2})n_{2}\parallel v_{2}-v_{1}\parallel _{E}, \end{aligned}$$
$$\begin{aligned} \parallel P_{1}(u_{2},v_{2})-P_{1}(u_{1},v_{1})\parallel _{E}\leqslant \Theta _{1}[\parallel u_{2}-u_{1}\parallel _{E}+\parallel v_{2}-v_{1}\parallel _{E}]. \end{aligned}$$

In the same way, we have the following estimates on \(P_{2}.\)

$$\begin{aligned}&\parallel P_{2}(u_{2},v_{2})-P_{2}(u_{1},v_{1})\parallel _{\infty } \nonumber \\&\quad \leqslant \lambda _{3}(n_{4}+m_{2}\mid a_{2}\mid )\parallel v_{2}-v_{1}\parallel _{\infty }+\lambda _{3}r_{2}\mid b_{2}\mid \parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty } \nonumber \\&\qquad +\lambda _{3}n_{3}\parallel u_{2}-u_{1}\parallel _{\infty }, \end{aligned}$$
(3.5)

and

$$\begin{aligned}&\parallel D^{\delta }P_{2}(u_{2},v_{2})-D^{\delta }P_{2}(u_{1},v_{1})\parallel _{\infty } \nonumber \\&\quad \leqslant \lambda _{4}(n_{4}+m_{2}\mid a_{2}\mid )\parallel v_{2}-v_{1}\parallel _{\infty }+\lambda _{4}r_{1}\mid b_{2}\mid \parallel D^{\delta }u_{2}-D^{\delta }u_{1}\parallel _{\infty } \nonumber \\&\qquad +\lambda _{4}n_{3}\parallel u_{2}-u_{1}\parallel _{\infty }. \end{aligned}$$
(3.6)

Thanks to (3.5), (3.6) and (2.11), we get

$$\begin{aligned}&\parallel P_{2}(u_{2},v_{2})-P_{2}(u_{1},v_{1})\parallel _{E} \nonumber \\&\quad \leqslant (\lambda _{3}+\lambda _{4})(n_{4}+m_{2}\mid a_{2}\mid +r_{2}\mid b_{2}\mid )\parallel u_{2}-u_{1}\parallel _{E} \nonumber \\&\qquad +(\lambda _{3}+\lambda _{4})n_{3}\parallel v_{2}-v_{1}\parallel _{E}, \end{aligned}$$
(3.7)
$$\begin{aligned} \parallel P_{2}(u_{2},v_{2})-P_{2}(u_{1},v_{1})\parallel _{E}\leqslant \Theta _{2}[\parallel u_{2}-u_{1}\parallel _{E}+\parallel v_{2}-v_{1}\parallel _{E}]. \end{aligned}$$

Thus,

$$\begin{aligned} \parallel P(u_{2},v_{2})-P(u_{1},v_{1})\parallel _{E\times E}\leqslant \Theta \parallel (u_{2},v_{2})-(u_{1},v_{1})\parallel _{E\times E}. \end{aligned}$$

So, we have proved that P is a contraction. Hence, Banach fixed point theorem implies that there exists a unique fixed point which is the solution of (1.1)–(1.2). \(\square \)

The second main result is given by the following theorem.

Theorem 2

Assume that the conditions (H1) and (H5) hold. Then, (1.1)–(1.2) has at least one solution.

Proof

We show that the operator P is completely continuous. To do this, we proceed in three main steps.

Step a: Since the functions \(f_{i}\), \(g_{i}\) and \(h_{i}\) are continuous, then the operator P is continuous.

Step b: Let us take \(\varrho >0\) and \(B_{\varrho }=\{(u,v)\in E^{2},\parallel (u,v)\parallel _{E\times E}\leqslant \varrho \}\). For \( (u,v)\in B_{\varrho }\), and for all \(x\in \left[ 0,1\right] \), we have

$$\begin{aligned}&\mid P_{1}(u(x),v(x))\mid \nonumber \\&\quad \leqslant I^{\alpha _{1}+\alpha _{2}}\mid f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,D^{\delta }u(x))\mid \nonumber \\&\qquad +\frac{\mid 2K_{1}-K_{2}\mid }{\alpha _{2}-2}+\frac{\mid K_{2}-\alpha _{2}K_{1}\mid }{\alpha _{2}-2}+\mid a\mid \nonumber \\&\quad \leqslant \lambda _{1}[\Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}]+\frac{\mid 2K_{1}-K_{2}\mid }{\alpha _{2}-2}+\frac{ \mid K_{2}-\alpha _{2}K_{1}\mid }{\alpha _{2}-2}+\mid a\mid . \end{aligned}$$
(3.8)

Hence, it yields that

$$\begin{aligned} \parallel P_{1}(u,v)\parallel _{\infty }&\leqslant \lambda _{1}[\Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}]+\frac{\mid 2K_{1}-K_{2}\mid }{\alpha _{2}-2}\\&\quad +\frac{\mid K_{2}-\alpha _{2}K_{1}\mid }{ \alpha _{2}-2}+\mid a\mid . \end{aligned}$$

In the same way we find

$$\begin{aligned} \parallel D^{\delta }P_{1}(u,v)\parallel _{\infty }&\leqslant \lambda _{2} \left[ \Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}\right] \\&\quad +\dfrac{\Gamma (\alpha _{2}+1)\mid 2K_{1}-K_{2}\mid }{\Gamma (\alpha _{2}+1-\delta )(\alpha _{2}-2)}+\dfrac{2\mid K_{2}-\alpha _{2}K_{1}\mid }{ \Gamma (3-\delta )(\alpha _{2}-2)}+\mid a\mid . \end{aligned}$$

Then, it follows that

$$\begin{aligned}&\parallel P_{1}(u,v)\parallel _{E} \nonumber \\&\quad \leqslant (\lambda _{1}+\lambda _{2})[\Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}]+\left( \frac{\Gamma (\alpha _{2}+1)}{ \Gamma (\alpha _{2}+1-\delta )}+1\right) \frac{\mid 2K_{1}-K_{2}\mid }{(\alpha _{2}-2)} \nonumber \\&\qquad +\left( \frac{2}{\Gamma (3-\delta )}+1\right) \frac{\mid K_{2}-\alpha _{2}K_{1}\mid }{ (\alpha _{2}-2)}+\mid a\mid :=\varrho _{1}. \end{aligned}$$
(3.9)

In the same manner, we have the following three inequalities

$$\begin{aligned} \parallel P_{2}(u,v)\parallel _{\infty }&\leqslant \lambda _{3}[\Upsilon _{2}+\mid a_{2}\mid \Upsilon _{4}+\mid b_{2}\mid \Upsilon _{6}] \\&\quad +\dfrac{\mid 2K_{3}-K_{4}\mid }{\beta _{2}-2}+\dfrac{\mid K_{4}-\beta _{2}K_{3}\mid }{\beta _{2}-2}+\mid b\mid ,\\ \parallel D^{\delta }P_{2}(u,v)\parallel _{\infty }&\leqslant \lambda _{4}[\Upsilon _{2}+\mid a_{2}\mid \Upsilon _{4}+\mid b_{2}\mid \Upsilon _{6}]\\&\quad +\dfrac{\Gamma (\beta _{2}+1)\mid 2K_{3}-K_{4}\mid }{\Gamma (\beta _{2}+1-\delta )(\beta _{2}-2)}+\dfrac{2\mid K_{4}-\alpha _{2}K_{3}\mid }{ \Gamma (3-\delta )(\beta _{2}-2)}+\mid b\mid , \end{aligned}$$

and

$$\begin{aligned}&\parallel P_{2}(u,v)\parallel _{E} \nonumber \\&\quad \leqslant (\lambda _{3}+\lambda _{4})[\Upsilon _{2}+\mid a_{2}\mid \Upsilon _{4}+\mid b_{2}\mid \Upsilon _{6}]+\left( \frac{\Gamma (\beta _{2}+1)}{ \Gamma (\beta _{2}+1-\delta )}+1\right) \frac{\mid 2K_{3}-K_{4}\mid }{(\beta _{2}-2) } \nonumber \\&\quad +\left( \frac{2}{\Gamma (3-\delta )}+1\right) \frac{\mid K_{4}-\beta _{2}K_{3}\mid }{ (\beta _{2}-2)}+\mid b\mid :=\varrho _{2}. \end{aligned}$$
(3.10)

Thanks to (3.9), (3.10), we can observe that

$$\begin{aligned} \parallel P(u,v)\parallel _{E\times E}\leqslant \varrho _{1}+\varrho _{2}<\infty . \end{aligned}$$
(3.11)

Step c: Let \((u,v)\in B_{\varrho }\) \(x_{1},x_{2}\in \left[ 0,1 \right] \), \(x_{1}<x_{2}\), we have

$$\begin{aligned}&\mid P_{1}(u,v)(x_{2})-P_{1}(u,v)(x_{1})\mid \leqslant \frac{\Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}}{\Gamma (\alpha _{1}+\alpha _{2}+1)}(x_{2}^{\alpha _{1}+\alpha _{2}}-x_{1}^{\alpha _{1}+\alpha _{2}}) \nonumber \\&\qquad +\frac{2K_{1}-K_{2}}{\alpha _{2}-2}(x_{2}^{\alpha _{2}}-x_{1}^{\alpha _{2}})+\frac{K_{2}-\alpha _{2}K_{1}}{\alpha _{2}-2}(x_{2}^{2}-x_{1}^{2}). \end{aligned}$$
(3.12)

On the other hand, we have

$$\begin{aligned}&\mid D^{\delta }P_{1}(u,v)(x_{2})-D^{\delta }P_{1}(u,v)(x_{1})\mid \nonumber \\&\quad \leqslant \frac{\Upsilon _{1}+\mid a_{1}\mid \Upsilon _{3}+\mid b_{1}\mid \Upsilon _{5}}{\Gamma (\alpha _{1}+\alpha _{2}+1-\delta )}(x_{2}^{\alpha _{1}+\alpha _{2}-\delta }-x_{1}^{\alpha _{1}+\alpha _{2}-\delta }) \nonumber \\&\qquad +\frac{\Gamma (\alpha _{2}+1)(2K_{1}-K_{2})}{\Gamma (\alpha _{2}+1-\delta )(\alpha _{2}-2)}(x_{2}^{\alpha _{2}-\delta }-x_{1}^{\alpha _{2}-\delta })\nonumber \\&\qquad +\frac{2(K_{2}-\alpha _{2}K_{1})}{\Gamma (3-\delta )(\alpha _{2}-2)} (x_{2}^{2-\delta }-x_{1}^{2-\delta }). \end{aligned}$$
(3.13)

As before, we can establish the inequalities:

$$\begin{aligned}&\mid P_{2}(u,v)(x_{2})-P_{2}(u,v)(x_{1})\mid \nonumber \\&\quad \leqslant \frac{\Upsilon _{2}+\mid a_{2}\mid \Upsilon _{4}+\mid b_{2}\mid \Upsilon _{6}}{\Gamma (\beta _{1}+\beta _{2}+1)}(x_{2}^{\beta _{1}+\beta _{2}}-x_{1}^{\beta _{1}+\beta _{2}}) \nonumber \\&\qquad +\frac{2K_{3}-K_{4}}{\beta _{2}-2}(x_{2}^{\beta _{2}}-x_{1}^{\beta _{2}})+ \frac{K_{4}-\beta _{2}K_{3}}{\beta _{2}-2}(x_{2}^{2}-x_{1}^{2}) \end{aligned}$$
(3.14)

and

$$\begin{aligned}&\mid D^{\delta }P_{2}(u,v)(x_{2})-D^{\delta }P_{2}(u,v)(x_{1})\mid \nonumber \\&\quad \leqslant \frac{\Upsilon _{1}+\mid a_{2}\mid \Upsilon _{4}+\mid b_{2}\mid \Upsilon _{6}}{\Gamma (\beta _{1}+\beta _{2}+1-\delta )}(x_{2}^{\beta _{1}+\beta _{2}-\delta }-x_{1}^{\beta _{1}+\beta _{2}-\delta }) \nonumber \\&\qquad +\frac{\Gamma (\beta _{2}+1)(2K_{3}-K_{4})}{\Gamma (\beta _{2}+1-\delta )(\beta _{2}-2)}(x_{2}^{\beta _{2}-\delta }-x_{1}^{\beta _{2}-\delta })\nonumber \\&\qquad +\frac{2(K_{4}-\beta _{2}K_{3})}{\Gamma (3-\delta )(\beta _{2}-2)} (x_{2}^{2-\delta }-x_{1}^{2-\delta }). \end{aligned}$$
(3.15)

The right-hand sides of (3.12), (3.13), (3.14) and (3.15) are independent of (u, v) and tend to zero as \(x_{2}\rightarrow x_{1}\). Consequently, P is an equicontinuous operator. By Arzella–Ascolli theorem and thanks to the above three steps, we confirm that P is completely continuous. Next, we consider the subset:

$$\begin{aligned} L:=\left\{ (u,v)\in E,(u,v)=\tau P(u,v),0<\tau <1\right\} . \end{aligned}$$

We need to show that L is bounded. For \((u,v)\in L\) and \(x\in \left[ 0,1 \right] \), we have

$$\begin{aligned} (u,v)(x)=\tau P(u,v)(x). \end{aligned}$$

The inequality (3.11) allows us to write

$$\begin{aligned} \parallel (u,v)\parallel \leqslant \tau (\varrho _{1}+\varrho _{2})<\infty . \end{aligned}$$

Hence, L is bounded. Using Lemma 3, we conclude that system (1.1)–(1.2) has at least one solution on \(\left[ 0,1\right] \). Theorem 2 is thus proved. \(\square \)

4 Applications on existence of solutions

In this section, we present some examples to illustrate the applicability of the above two main results.

Example 1

Let us consider the following system:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\frac{4}{5}}D^{\frac{29}{10}}u(x)=\frac{u(x)}{(3+x^{2})e^{x}}+\frac{v(x)}{ 2x^{2}+1}+\frac{1}{2}\frac{u(x)}{3x+7}+\frac{1}{3}\frac{D^{\frac{9}{5}}u(x)}{ 5x^{2}+2}, \\ D^{\frac{17}{20}}D^{\frac{23}{8}}v(x)=\frac{e^{-2x}}{x^{2}+1}u(x)+\frac{ e^{-x}}{(x+1)e^{3x}+1}v(x)+\frac{1}{5}\frac{v(x)}{x+9}+\frac{1}{4}\frac{D^{ \frac{9}{5}}v(x)}{x^{2}+1}, \\ u(0))=u(1)=a\in {\mathbb {R}}, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=b\in {\mathbb {R}}, \\ v^{\prime }(0)=v^{\prime }(1)=0, \end{array} \right. \end{aligned}$$
(4.1)

it is clear that

$$\begin{aligned} \alpha _{1}&=\frac{4}{5},\alpha _{2}=\frac{29}{10},\beta _{1}=\frac{17}{20} ,\beta _{2}=\frac{23}{8},\delta =\frac{9}{5} \\ f_{1}(x,u(x),v(x))&=\frac{1}{(3+x^{2})e^{x}}u(x)+\frac{1}{2x^{2}+1}v(x), \\ f_{2}(x,u(x),v(x))&=\frac{e^{-2x}}{x^{2}+1}u(x)+\frac{e^{-x}}{(x+1)e^{3x}+1} v(x), \\ g_{1}(x,u(x))&=\frac{1}{3x+7}u(x),\quad g_{2}(x,v(x))=\frac{1}{x+9}v(x), \\ h_{1}(x,D^{\delta }u(x))&=\frac{1}{5x^{2}+2}D^{\delta }u(x),\quad h_{2}(x,D^{\frac{9}{5}}v(x))=\frac{1}{x^{2}+1}D^{\frac{9}{5}}v(x). \end{aligned}$$

Also, for all \(x\in \left[ 0,1\right] \) and \((u,v),(w,z)\in {\mathbb {R}}^{2}\), we have

$$\begin{aligned} \mid f_{1}(x,u,v)-f_{1}(x,w,z)\mid&\leqslant \frac{1}{(3+x^{2})e^{x}}\mid u-w\mid +\frac{1}{2x^{2}+1}\mid v-z\mid \\&\leqslant \frac{1}{4}\mid u-w\mid + \frac{1}{3}\mid v-z\mid , \\ \mid f_{2}(x,u,v)-f_{2}(x,w,z)\mid&\leqslant \frac{e^{-2x}}{x^{2}+1}\mid u-w\mid +\frac{e^{-x}}{(x+1)e^{3x}+1}\mid v-z\mid \\&\leqslant \frac{1}{2}\mid u-w\mid +\frac{1}{3}\mid v-z\mid , \\ \mid g_{1}(x,u)-g_{1}(x,v)\mid&\leqslant \frac{1}{3x+7}\mid u-v\mid \leqslant \frac{1}{10}\mid u-v\mid , \\ \mid g_{2}(x,u)-g_{2}(x,v)\mid&\leqslant \frac{1}{x+9}\mid u-v\mid \leqslant \frac{1}{10}\mid u-v\mid , \\ \mid h_{1}(x,u)-h_{1}(x,v)\mid&\leqslant \frac{1}{5x^{2}+2}\mid u-v\mid \leqslant \frac{1}{7}\mid u-v\mid , \\ \mid h_{2}(x,u)-h_{2}(x,v)\mid&\leqslant \frac{1}{x^{2}+1}\mid u-v\mid \leqslant \frac{1}{2}\mid u-v\mid . \end{aligned}$$

We also have

$$\begin{aligned} \theta _{1}= & {} {\textit{max}}\left\{ 0.212743\quad ;\quad 0.204\right\} ,\\ \theta _{2}= & {} {\textit{max}}\left\{ 0.3974705\quad ;\quad 0.19923\right\} ,\\ \theta= & {} \theta _{1}+\theta _{2}=0.6102135<1. \end{aligned}$$

The conditions of Theorem 1 hold. Therefore, problem (4.1) has a unique solution on \(\left[ 0,1\right] \).

Example 2

For \(x\in \left[ 0,1\right] \), we consider the following system:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\frac{6}{7}}D^{\frac{20}{7}}u(x)=\frac{{\textit{sin}}(u)+{\textit{cos}}(v)}{1+x^{2}}+\frac{ {\textit{sin}}(u)}{(x+1)e^{x}}+\frac{3e^{-x}}{2+{\textit{sin}}^{2}(D^{\frac{17}{9}}(u))}, \\ D^{\frac{8}{9}}D^{\frac{17}{6}}=\frac{e^{-x}{\textit{sin}}(u+v)}{x+2}+\frac{e^{-x}{\textit{cos}}(v) }{e^{x}+1}+\frac{2}{2+cos(D^{\frac{17}{9}}(v))}, \\ u(0))=u(1)=a\in {\mathbb {R}}, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=b\in {\mathbb {R}}, \\ v^{\prime }(0)=v^{\prime }(1)=0. \end{array} \right. \end{aligned}$$
(4.2)

We have

$$\begin{aligned} \alpha _{1}&=\frac{6}{7},\alpha _{2}=\frac{20}{7},\beta _{1}=\frac{8}{9} ,\beta _{2}=\frac{17}{6},\delta =\frac{17}{9}, \\ f_{1}(x,u,v)&=\frac{{\textit{sin}}(u)+{\textit{cos}}(v)}{1+x^{2}},\quad f_{2}(x,u,v)=\frac{ e^{-x}{\textit{sin}}(u+v)}{x+2} \\ g_{1}(x,u)&=\frac{{\textit{sin}}(u)}{(x+1)e^{x}},\quad g_{2}(x,v)=\frac{ e^{-x}{\textit{cos}}(v)}{e^{x}+1}, \\ h_{1}(x,D^{\frac{17}{9}}(u))&=\frac{3e^{-x}}{{\textit{sin}}^{2}(D^{\frac{17}{9}}u)+2} ,\quad h_{2}(x,D^{\delta }v)=\frac{2}{2+{\textit{cos}}(D^{\frac{17}{9}}v)}. \end{aligned}$$

For \(x\in \left[ 0,1\right] \), it is easy to state that \(f_{1}\), \(f_{2}\), \( g_{1}\), \(g_{2}\), \(h_{1}\) and \(h_{2}\) are continuous and bounded functions. Using Theorem 2, we conclude that system (4.2) has at least one solution on \(\left[ 0,1\right] \).

5 Ulam–Hyers stability results

In this subsection, we will deal with Ulam–Hyers stability. This stability provides a framework to quantify the sensitivity to initial condition parameters and assess the stability of solutions. It helps us to determine whether small change in initial conditions or parameters results in bounded or unbounded deviations from the original solution. So, let us introduce the following definition.

Definition 4

The solution of the system (1.1)–(1.2) is Ulam–Hyers stable if there exists \(\Lambda >0;\) such that for any \(\epsilon _{1}>0,\epsilon _{2}>0 \) and for each solution \((u,v)\in E^{2}\) of the differential inequality

$$\begin{aligned} \left\{ \begin{array}{l} \mid D^{\alpha _{1}}D^{\alpha _{2}}u(x)-f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,D^{\delta }u(x))\mid \leqslant \epsilon _{1}, \\ \mid D^{\beta _{1}}D^{\beta _{2}}v(x)-f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,D^{\delta }v(x))\mid \leqslant \epsilon _{2}, \end{array} \right. \end{aligned}$$
(5.1)

with

$$\begin{aligned} \left\{ \begin{array}{l} u(0))=u(1)=a\in {\mathbb {R}}, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=b\in {\mathbb {R}}, \\ v^{\prime }(0)=v^{\prime }(1)=0, \\ \end{array} \right. \end{aligned}$$
(5.2)

there exists a solution \((u^{*},v^{*})\in E\times E\) of system (1.1)–(1.2), such that one has the following estimate:

$$\begin{aligned} \parallel (u-u^{*},v-v^{*})\parallel _{E\times E}\leqslant (\epsilon _{1}+\epsilon _{2})\Lambda . \end{aligned}$$

We present to the reader the following result.

Theorem 3

Suppose that the conditions of Theorem 1 are satisfied. Then, problem (1.1)–(1.2) is Ulam–Hyers stable.

Proof

Let \((u,v)\in E\times E\) be a solution of (5.1)–(5.2), and let, by Theorem 1, \((u^{*},v^{*})\in E\times E\) be the unique solution of (1.1)–(1.2). We integrate the differential inequality (5.1), we can write

$$\begin{aligned}&\bigg \vert u(x)-I^{\alpha _{1}+\alpha _{2}}[f_{1}(x,u(x),v(x))+a_{1}g_{1}(x,u(x))+b_{1}h_{1}(x,D^{\delta }u(x))] \nonumber \\&\left. \qquad - \frac{2K_{1}-K_{2}}{\alpha _{2}-2}x^{\alpha _{2}}+\frac{ K_{2}-\alpha _{2}K_{1}}{\alpha _{2}-2}x^{2}+a\right| \nonumber \\&\quad \leqslant \frac{1}{\Gamma (\alpha _{1}+\alpha _{2})}\epsilon _{1}, \end{aligned}$$
(5.3)

and

$$\begin{aligned}&\bigg \vert v(x)-I^{\beta _{1}+\beta _{2}}[f_{2}(x,u(x),v(x))+a_{2}g_{2}(x,v(x))+b_{2}h_{2}(x,D^{\delta }v(x))]\nonumber \\&\left. \qquad -\frac{2K_{3}-K_{4}}{\beta _{2}-2}x^{\beta _{2}}+\frac{K_{4}-\beta _{2}K_{3}}{\beta _{2}-2}x^{2}+b\right| \nonumber \\&\quad \leqslant \frac{1}{\Gamma (\beta _{1}+\beta _{2})}\epsilon _{2}. \end{aligned}$$
(5.4)

Using (5.1), (5.3) and (5.4), we have

$$\begin{aligned} \parallel u-u^{*}\parallel _{\infty }&\leqslant \lambda _{1}\epsilon _{1}+\lambda _{1}(n_{1}+\mid a_{1}\mid m_{1})\parallel u-u^{*}\parallel _{\infty } \\&\quad +\lambda _{1}\mid b_{1}\mid r_{1}\parallel D^{\delta }u-D^{\delta }u^{*}\parallel _{\infty }+\lambda _{1}n_{2}\parallel v-v^{*}\parallel _{\infty } \end{aligned}$$

and

$$\begin{aligned} \parallel v-v^{*}\parallel _{\infty }&\leqslant \lambda _{3}\epsilon _{2}+\lambda _{3}(n_{4}+\mid a_{2}\mid m_{2})\parallel v-v^{*}\parallel _{\infty } \\&\quad +\lambda _{3}\mid b_{2}\mid r_{2}\parallel D^{\delta }v-D^{\delta }v^{*}\parallel _{\infty }+\lambda _{3}n_{3}\parallel u-u^{*}\parallel _{\infty }. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \parallel D^{\delta }u-D^{\delta }u^{*}\parallel _{\infty }&\leqslant \lambda _{2}\epsilon _{1}+\lambda _{2}(n_{1}+\mid a_{1}\mid m_{1})\parallel u-u^{*}\parallel _{\infty } \\&\quad +\lambda _{2}\mid b_{1}\mid r_{1}\parallel D^{\delta }u-D^{\delta }u^{*}\parallel _{\infty }+\lambda _{2}n_{2}\parallel v-v^{*}\parallel _{\infty } \end{aligned}$$

and

$$\begin{aligned} \parallel D^{\delta }v-D^{\delta }v^{*}\parallel _{\infty }&\leqslant \lambda _{4}\epsilon _{2}+\lambda _{4}(n_{4}+\mid a_{2}\mid m_{2})\parallel v-v^{*}\parallel _{\infty } \\&\quad +\lambda _{4}\mid b_{2}\mid r_{2}\parallel D^{\delta }v-D^{\delta }v^{*}\parallel _{\infty }+\lambda _{4}n_{3}\parallel u-u^{*}\parallel _{\infty }. \end{aligned}$$

So, we can write

$$\begin{aligned} \parallel u-u^{*}\parallel _{E}&\leqslant (\lambda _{1}+\lambda _{2})\epsilon _{1}+(\lambda _{1}+\lambda _{2})(n_{1}+\mid a_{1}\mid m_{1}+\mid b_{1}\mid r_{1})\parallel u-u^{*}\parallel _{E} \\&\quad +(\lambda _{1}+\lambda _{2})n_{2}\parallel v-v^{*}\parallel _{E}, \end{aligned}$$

and

$$\begin{aligned} \parallel v-v^{*}\parallel _{E}&\leqslant (\lambda _{3}+\lambda _{4})\epsilon _{2}+(\lambda _{3}+\lambda _{4})(n_{4}+\mid a_{2}\mid m_{2}+\mid b_{2}\mid r_{2})\parallel v-v^{*}\parallel _{E} \\&\quad +(\lambda _{3}+\lambda _{4})n_{3}\parallel u-u^{*}\parallel _{E}. \end{aligned}$$

Then, we get

$$\begin{aligned} \parallel (u-u^{*},v-v^{*})\parallel _{E\times E}\leqslant \epsilon \Xi +\Theta \parallel (u-u^{*},v-v^{*})\parallel _{E\times E}, \end{aligned}$$

where,

$$\begin{aligned} \epsilon =\epsilon _{1}+\epsilon _{2},\quad \Xi ={\textit{max}}\left\{ \lambda _{1}+\lambda _{2};\lambda _{3}+\lambda _{4}\right\} . \end{aligned}$$

Consequently, we have

$$\begin{aligned} \parallel (u-u^{*},v-v^{*})\parallel _{E\times E}\leqslant \frac{ \epsilon \Xi }{1-\Theta }:=\epsilon \Lambda ,\quad \Lambda =\frac{\Xi }{ 1-\Theta }. \end{aligned}$$

(Remark that \(1>\Theta \) since the condition of Theorem 1 are supposed satisfied.) Therefore, the solution of (1.1)–(1.2) is Ulam Hyers stable. \(\square \)

6 Two examples on stability

Example 3

The first example is given in Example 1 of Sect. 4. Since it satisfies all conditions of Theorem 1, then the associated system is Ulam Hyers stable.

Example 4

Consider the following system:

$$\begin{aligned} \left\{ \begin{array}{l} D^{\frac{7}{8}}D^{\frac{34}{12}}u(x)=\dfrac{e^{-x}\sin x}{x^{2}+1}u(x)+ \dfrac{\cos x}{3x^{2}+1}v(x)+\dfrac{1}{7}\dfrac{\sin x\cos x}{x^{2}+9}u(x)\\ \quad + \dfrac{1}{8}\dfrac{\cos x}{e^{2x}+3}D^{\frac{9}{5}}u(x) \\ D^{\frac{4}{5}}D^{\frac{29}{10}}v(x)=\dfrac{\sin x+\cos x}{e^{x}+3}u(x)+ \dfrac{2\sin x-\cos x}{4e^{x}+5}v(x)+\dfrac{1}{3}\dfrac{\sin ^{2}x}{2x^{2}+7} v(x)\\ \quad +\dfrac{1}{5}\dfrac{\sin x}{x^{2}+5}D^{\frac{9}{5}}v(x), \\ u(0))=u(1)=0, \\ u^{\prime }(0)=u^{\prime }(1)=0, \\ v(0)=v(1)=0, \\ v^{\prime }(0)=v^{\prime }(1)=0. \end{array} \right. \nonumber \\ \end{aligned}$$
(6.1)

For this example, we have

$$\begin{aligned} \alpha _{1}&=\frac{7}{8},\alpha _{2}=\frac{34}{12},\beta _{1}=\frac{4}{5} ,\beta _{2}=\frac{29}{10},\delta =\frac{9}{5} \\ f_{1}(x,u(x),v(x))&=\dfrac{e^{-x}\sin x}{x^{2}+1}u(x)+\dfrac{\cos x}{ 3x^{2}+1}v(x), \\ f_{2}(x,u(x),v(x))&=\dfrac{\sin x+\cos x}{e^{x}+3}u(x)+\dfrac{2\sin x-\cos x }{4e^{x}+5}v(x), \\ g_{1}(x,u(x))&=\dfrac{\sin x\cos x}{x^{2}+9}u(x),\quad g_{2}(x,v(x))=\dfrac{ \sin ^{2}x}{2x^{2}+7}v(x), \\ h_{1}(x,D^{\delta }u(x))&=\dfrac{\cos x}{e^{2x}+3}D^{\frac{9}{5}}u(x),\quad h_{2}(x,D^{\frac{9}{5}}v(x))=\dfrac{\sin x}{x^{2}+5}D^{\frac{9}{5}}v(x). \end{aligned}$$

It is clear that, for all \(x\in \left[ 0,1\right] \) and \((u,v),(w,z)\in {\mathbb {R}}^{2}\), the following inequalities hold:

$$\begin{aligned} \mid f_{1}(x,u,v)-f_{1}(x,w,z)\mid&\leqslant \dfrac{e^{-x}\sin x}{x^{2}+1} \mid u-w\mid +\dfrac{\cos x}{3x^{2}+1}\mid v-z\mid \\&\leqslant \frac{1}{2}\mid u-w\mid +\frac{1}{3}\mid v-z\mid , \\ \mid f_{2}(x,u,v)-f_{2}(x,w,z)\mid&\leqslant \dfrac{\sin x+\cos x}{e^{x}+3} \mid u-w\mid +\dfrac{2\sin x-\cos x}{4e^{x}+5}\mid v-z\mid \\&\leqslant \frac{1 }{3}\mid u-w\mid +\frac{1}{5}\mid v-z\mid , \\ \mid g_{1}(x,u)-g_{1}(x,v)\mid&\leqslant \dfrac{\sin x\cos x}{x^{2}+9} u(x)\mid u-v\mid \leqslant \frac{1}{9}\mid u-v\mid , \\ \mid g_{2}(x,u)-g_{2}(x,v)\mid&\leqslant \dfrac{\sin ^{2}x}{2x^{2}+7}\mid u-v\mid \leqslant \frac{1}{7}\mid u-v\mid , \\ \mid h_{1}(x,u)-h_{1}(x,v)\mid&\leqslant \dfrac{\cos x}{e^{2x}+3}\mid u-v\mid \leqslant \frac{1}{3}\mid u-v\mid , \\ \mid h_{2}(x,u)-h_{2}(x,v)\mid&\leqslant \dfrac{\sin x}{x^{2}+5}\mid u-v\mid \leqslant \frac{1}{5}\mid u-v\mid . \end{aligned}$$

We also have

$$\begin{aligned} \Theta _{1}= & {} {\textit{max}}\left\{ 0.3456\quad ;\quad 0.203047\right\} ,\\ \Theta _{2}= & {} {\textit{max}}\left\{ 0.1113\quad ;\quad 0.2254\right\} ,\\ \Theta= & {} \Theta _{1}+\Theta _{2}=0.5710<1. \end{aligned}$$

The reader remarks that the conditions of Theorem 1 hold. Therefore, the unique solution of the problem (6.1) is Ulam–Hyers stable.

7 New travelling wave solutions for a beam deflection system

In this section, we will use the Tanh method [13, 21] for finding traveling wave solutions to a coupled system, that involves conformable fractional derivatives, of type:

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{t}^{2\alpha }u(t,x)+T_{x}^{4\beta }u(t,x)+H(u,v,T_{x}^{2\beta }(u,v))(t,x)=0, \\ T_{t}^{2\alpha }v(t,x)+T_{x}^{4\beta }v(t,x)+L(u,v,T_{x}^{2\beta }(u,v))(t,x)=0, \end{array}\right. } \end{aligned}$$
(7.1)

where \(0<\alpha ,\beta \le 1,\) and H and L are two given functions, while \(T_{t}^{\alpha }u(t,x)\) is the conformable fractional derivatives in the sense of Khalil of the unknown function u with respect to t, (see [16, 28]). The Khalil derivative is given by the following expression:

$$\begin{aligned} \begin{array}{l} T_{t}^{\alpha }u(t,x)=\frac{\partial ^{\alpha }u\left( t,x\right) }{\partial t^{\alpha }}=\underset{\varepsilon \rightarrow 0}{\lim }\left( \frac{u\left( t+\varepsilon t^{1-\alpha },x\right) -u\left( t,x\right) }{\varepsilon } \right) ,\quad 0<\alpha \le 1. \end{array} \end{aligned}$$

(In the same manner, it can be defined \(T_{x}^{\beta }u(t,x).\)) The reader can observe that in the case where \(\alpha =\beta =1,\) system (7.1) can be transformed into the following classical coupled beam equations [15]:

$$\begin{aligned} {\left\{ \begin{array}{ll} u_{tt}+u_{xxxx}+H(u,v,(u,v)_{xx})=0, \\ \\ v_{tt}+v_{xxxx}+L(u,v,(u,v)_{xx})=0. \end{array}\right. } \end{aligned}$$

Let us now recall the main steps of Tanh method for the case of Khalil derivatives [13, 26,27,28].

  1. 1.

    We begin by considering the following general form of coupled equations:

    $$\begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {F}}_{1}\left( u,v,T_{t}^{\alpha }u,T_{x}^{\beta }u,T_{t}^{\alpha }v,T_{x}^{\beta }v,T_{t}^{2\alpha }u,T_{t}^{\alpha }(T_{x}^{\beta }u),T_{x}^{2\beta }u,T_{t}^{2\alpha }v,T_{t}^{\alpha }(T_{x}^{\beta }v),T_{x}^{2\beta }v,\ldots \right) =0, \\ {\mathcal {F}}_{2}\left( u,v,T_{t}^{\alpha }u,T_{x}^{\beta }u,T_{t}^{\alpha }v,T_{x}^{\beta }v,T_{t}^{2\alpha }u,T_{t}^{\alpha }(T_{x}^{\beta }u),T_{x}^{2\beta }u,T_{t}^{2\alpha }v,T_{t}^{\alpha }(T_{x}^{\beta }v),T_{x}^{2\beta }v,\ldots \right) =0. \end{array}\right. } \nonumber \\ \end{aligned}$$
    (7.2)
  2. 2.

    Then, tanks to the change of variable (with its two constants k and \( \omega \))

    $$\begin{aligned} \begin{array}{l} \xi =\frac{k}{\alpha }t^{\alpha }+\frac{\omega }{\beta }x^{\beta }, \end{array} \end{aligned}$$
    (7.3)

    our general form can be transformed into

    $$\begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {G}}_{1}\left( U,V,U^{^{\prime }},V^{^{\prime }},U^{^{\prime \prime }},V^{^{\prime \prime }},U^{^{\prime \prime \prime }},V^{^{\prime \prime \prime }},\ldots \right) =0, \\ {\mathcal {G}}_{2}\left( U,V,U^{^{\prime }},V^{^{\prime }},U^{^{\prime \prime }},V^{^{\prime \prime }},U^{^{\prime \prime \prime }},V^{^{\prime \prime \prime }},\ldots \right) =0. \end{array}\right. } \end{aligned}$$
    (7.4)
  3. 3.

    Now, we shall use the following second change of variable

    $$\begin{aligned} \psi =tanh(\xi ), \end{aligned}$$
    (7.5)

    which allows us to obtain the following derivatives

    $$\begin{aligned} \frac{d}{d\xi }&= \left( 1-\psi ^{2}\right) \frac{d}{d\psi }, \nonumber \\ \frac{d^{2}}{d\xi ^{2}}&= -2\psi \left( 1-\psi ^{2}\right) \frac{d}{d\psi } +\left( 1-\psi ^{2}\right) ^{2}\frac{d^{2}}{d\psi ^{2}}, \nonumber \\ \frac{d^{3}}{d\xi ^{3}}&= 2\left( 1-\psi ^{2}\right) \left( 3\psi ^{2}-1\right) \frac{d}{d\psi }-6\psi \left( 1-\psi ^{2}\right) ^{2}\frac{ d^{2}}{d\psi ^{2}}+\left( 1-\psi ^{2}\right) ^{3}\frac{d^{3}}{d\psi ^{3}},\nonumber \\ \frac{d^{4}}{d\xi ^{4}}&= -8\psi \left( 1-\psi ^{2}\right) \left( 3\psi ^{2}-2\right) \frac{d}{d\psi }+4\left( 1-\psi ^{2}\right) ^{2}\left( 9\psi ^{2}-2\right) \frac{d^{2}}{d\psi ^{2}},\nonumber \\&\quad -12\psi \left( 1-\psi ^{2}\right) ^{3}\frac{d^{3}}{d\psi ^{3}}+\left( 1-\psi ^{2}\right) ^{4}\frac{d^{4}}{d\psi ^{4}}. \end{aligned}$$
    (7.6)
  4. 4.

    We shall also suppose that our travelling wave solution (u, v), if there exists, it can be expressed in the form

    $$\begin{aligned} {\left\{ \begin{array}{ll} u(x,t)=U(\xi )=F(\psi )=\sum _{i=0}^{b}a_{i}\psi ^{i}, \\ v(x,t)=V(\xi )=S(\psi )=\sum _{i=0}^{d}b_{i}\psi ^{i}. \end{array}\right. } \end{aligned}$$
    (7.7)
  5. 5.

    Determining \(a_{i}\) and \(b_{i}\) by “term-balancing” and by solving some associated algebraic system, see [13, 21], we obtain the desired solutions.

7.1 An application

As application, we propose to find travelling wave solutions for the following coupled problem of beam type:

$$\begin{aligned} {\left\{ \begin{array}{ll} T_{t}^{2\alpha }u+T_{x}^{4\beta }u+mT_{x}^{2\beta }(u^{2})-T_{x}^{2\beta }(v^{2})-T_{x}^{2\beta }u=0, \\ T_{t}^{2\alpha }v+T_{x}^{4\beta }v+nT_{x}^{2\beta }(uv)-T_{x}^{2\beta }v=0, \end{array}\right. } \end{aligned}$$
(7.8)

where m, n are real constants.

Using (7.3) to change (7.8) into the following nonlinear problem

$$\begin{aligned} {\left\{ \begin{array}{ll} k^{2}U_{\xi \xi }+\omega ^{4}U_{\xi \xi \xi \xi }+m\omega ^{2}(U^{2})_{\xi \xi }-\omega ^{2}(V^{2})_{\xi \xi }-\omega ^{2}U_{\xi \xi }=0, \\ k^{2}V_{\xi \xi }+\omega ^{4}V_{\xi \xi \xi \xi }+n\omega ^{2}(UV)_{\xi \xi }-\omega ^{2}V_{\xi \xi }=0. \end{array}\right. } \end{aligned}$$
(7.9)

Integrating (7.9) twice and neglecting constants of integration, we find

$$\begin{aligned} {\left\{ \begin{array}{ll} k^{2}U+\omega ^{4}U_{\xi \xi }+m\omega ^{2}(U^{2})-\omega ^{2}(V^{2})-\omega ^{2}U=0, \\ k^{2}V+\omega ^{4}V_{\xi \xi }+n\omega ^{2}(UV)-\omega ^{2}V=0. \end{array}\right. } \end{aligned}$$
(7.10)

Substituting (7.6) and (7.7) into (7.10), so the first equation of (7.10) is transformed into the following equation:

$$\begin{aligned} \omega ^{4}\left[ -2 \psi (1-\psi ^{2}) \frac{dF}{d\psi }+(1-\psi ^{2})^{2}\frac{d^{2 }F}{d\psi ^{2}}\right] +m\omega ^{2}F^{2}-\omega ^{2}S^{2}+(k^{2}-\omega ^{2})F=0. \end{aligned}$$
(7.11)

The second equation of (7.10) can be transformed into the equation:

$$\begin{aligned} \omega ^{4}\left[ -2 \psi (1-\psi ^{2}) \frac{dS}{d\psi }+(1-\psi ^{2})^{2}\frac{d^{2 }S}{d\psi ^{2}}\right] +n\omega ^{2}(FS)+(k^{2}-\omega ^{2})S=0. \end{aligned}$$
(7.12)

Now, in (7.11), we balance \(\psi ^{4}\frac{d^{2}F}{d\psi ^{2}}\) with \( S^{2}\) to get

$$\begin{aligned} 4+b-2=2d. \end{aligned}$$

Using the same technique with (7.12); we balance \(\psi ^{4}\frac{ d^{2}F}{d\psi ^{2}}\) with SF. We can write

$$\begin{aligned} 4+d-2=d+b. \end{aligned}$$

Consequently, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} F(\psi )=a_{0}+a_{1}\psi +a_{2}\psi ^{2}, \\ S(\psi )=b_{0}+b_{1}\psi +b_{2}\psi ^{2}. \end{array}\right. } \end{aligned}$$
(7.13)

Substituting (7.13) into (7.11), we observe that

$$\begin{aligned}&\omega ^{4}\Big [-2 \psi (1-\psi ^{2})(a_{1}+2a_{2}\psi )+2a_{2}(1-\psi ^{2})^{2}\Big ]+m \omega ^{2}(a_{0}+a_{1}\psi +a_{2}\psi ^{2})^{2}\nonumber \\&\quad -\omega ^{2}(b_{0}+b_{1}\psi +b_{2}\psi ^{2})^{2}+(k^{2}- \omega ^{2})(a_{0}+a_{1}\psi +a_{2}\psi ^{2})=0. \end{aligned}$$
(7.14)

Also, replacing (7.13) into (7.12), we get

$$\begin{aligned}&\omega ^{4}\Big [-2 \psi (1-\psi ^{2})(b_{1}+2b_{2}\psi )+2a_{2}(1-\psi ^{2})^{2}\Big ]\nonumber \\&\quad +n \omega ^{2}(a_{0}+a_{1}\psi +a_{2}\psi ^{2})(b_{0}+b_{1}\psi +b_{2}\psi ^{2})\nonumber \\&\quad +(k^{2}-\omega ^{2})(b_{0}+b_{1}\psi +b_{2}\psi ^{2})=0. \end{aligned}$$
(7.15)

Therefore, we obtain the following two sets:

Set 1:

$$\begin{aligned} \left\{ \begin{array}{l} \psi ^{0}:m\omega ^{2}a_{0}^{2}+2\omega ^{4}a_{2}-\omega ^{2}b_{0}^{2}+k^{2}a_{0}-\omega ^{2}a_{0}=0, \\ \psi ^{1}:2m\omega ^{2}a_{0}a_{1}-2\omega ^{4}a_{1}-2\omega ^{2}b_{0}b_{1}+k^{2}a_{1}-\omega ^{2}a_{1}=0, \\ \psi ^{2}:2m\omega ^{2}a_{0}a_{2}+m\omega ^{2}a_{1}^{2}-8\omega ^{4}a_{2}-2\omega ^{2}b_{0}b_{2}-\omega ^{2}b_{1}^{2}+k^{2}a_{2}-\omega ^{2}a_{2}=0, \\ \psi ^{3}:2m\omega ^{2}a_{1}+2\omega ^{4}a_{1}-2\omega ^{2}b_{1}b_{2}=0, \\ \psi ^{4}:m\omega ^{2}a_{2}^{2}+6\omega ^{4}a_{2}-\omega ^{2}b_{2}^{2}=0. \end{array} \right. \end{aligned}$$
(7.16)

Set 2:

$$\begin{aligned} \begin{array}{l} \left\{ \begin{array}{l} \psi ^{0}:n\omega ^{2}a_{0}b_{0}+2\omega ^{4}b_{2}+k^{2}b_{0}-\omega ^{2}b_{0}=0, \\ \psi ^{1}:n\omega ^{2}a_{0}b_{1}+n\omega ^{2}a_{1}b_{0}-2\omega ^{4}b_{1}+k^{2}b_{1}-\omega ^{2}b_{1}=0, \\ \psi ^{2}:n\omega ^{2}a_{0}b_{2}+n\omega ^{2}a_{1}b_{1}+n\omega ^{2}a_{2}b_{0}-8\omega ^{4}b_{2}+k^{2}b_{2}-\omega ^{2}b_{2}=0, \\ \psi ^{3}:n\omega ^{2}a_{1}b_{2}+n\omega ^{2}a_{2}b_{1}+2\omega ^{4}b_{1}=0, \\ \psi ^{4}:n\omega ^{2}a_{2}b_{2}+6\omega ^{4}b_{2}=0. \end{array} \right. \end{array} \end{aligned}$$
(7.17)

Solving (7.16) and (7.17) with the aid of Maple, we obtain:

Case 1:

$$\begin{aligned}&\omega =\omega , \quad k=\omega \sqrt{4\omega ^{2}+1}, \quad a_{0}= \frac{2\omega ^{2}}{n}, \quad a_{1}=0,\quad a_{2}=-\frac{6\omega ^{2}}{n},\nonumber \\&b_{0}=\pm \frac{\omega ^{2}\sqrt{4m-4n}}{n},\quad b_{1}=0,\quad b_{2}=\pm \frac{3\omega ^{2}\sqrt{4m-4n}}{n}. \end{aligned}$$
(7.18)

Substituting (7.18) into (7.13), the following travelling wave solution of (7.8) is obtained:

$$\begin{aligned}&u(x,t)=\frac{2\omega ^{2}}{n}-\frac{6\omega ^{2}}{n}\tanh ^{2}(\xi ), \end{aligned}$$
(7.19)
$$\begin{aligned}&v(x,t)=\pm \frac{2\omega ^{2}\sqrt{m-n}}{n}\pm \frac{6\omega ^{2}\sqrt{m-n}}{n} \tanh ^{2}(\xi ). \end{aligned}$$
(7.20)
Fig. 1
figure 1

Plots of (7.19, 7.20), with: \( 0\le x\le 10\), \(0\le t\le 10\) and \(m=5, n=-3, \omega =\frac{1}{9}, \) \(\alpha =\frac{3}{5}, \beta =\frac{1}{2}\)

Full size image
Fig. 2
figure 2

Plots of (7.22, 7.23) with: \( 0\le x\le 10\), \(0\le t\le 10\) and \(m=2, n=-1, \omega =\frac{1}{5}, \) \(\alpha =\frac{9}{10}, \beta =\frac{9}{10}\)

Full size image

Case 2:

$$\begin{aligned}&\omega =\omega , \quad k=\omega \sqrt{1-4\omega ^{2}}, \quad a_{0}= \frac{6\omega ^{2}}{n}, \quad a_{1}=0,\quad a_{2}=-\frac{6\omega ^{2}}{n},\nonumber \\&b_{0}=\pm \frac{\omega ^{2}\sqrt{36m-36n}}{n},\quad b_{1}=0,\quad b_{2}=\pm \frac{\omega ^{2}\sqrt{36m-36n}}{n}. \end{aligned}$$
(7.21)

Substituting (7.21) into (7.13), the following travelling wave solution of (7.8) is obtained (Figs. 1 and 2):

$$\begin{aligned}&u(x,t)=\frac{6\omega ^{2}}{n}(1-\tanh ^{2}(\xi )), \end{aligned}$$
(7.22)
$$\begin{aligned}&v(x,t)=\pm \frac{6\omega ^{2}\sqrt{m-n}}{n}\pm \frac{6\omega ^{2}\sqrt{m-n}}{n} \tanh ^{2}(\xi ). \end{aligned}$$
(7.23)

8 Conclusion

We have analysed a more general case of a coupled system of sequential differential equations of beam type. For our studied problem, the classical mechanical model “system” that involves the deflection of a long beam, is seen as a limiting case. The approach of Caputo has been chosen to investigate our system. We have first proved the existence of a unique solution for the sequential system. Subsequently, we have extended our study to explore the existence of at least one solution for the same system. Our detailed examples, presented in this paper, support the existence of solutions to various hypotheses that have been imposed in the paper. The Ulam–Hyers stability, illustrated with some examples, has also been considered in the paper. New travelling wave solutions, using Khalil conformable approach and the Tanh method, have been obtained and some graphs of the “waves” have been plotted.

The obtained results have implications for applications in engineering, and mathematical modeling, where sequential beam systems play a crucial role.

Further research can build upon this work by considering additional properties, other stability analysis, or exploring specific applications in long beam deflection problems.