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On multi-step methods for singular fractional q-integro-differential equations

  • Sayyedeh Narges Hajiseyedazizi , Mohammad Esmael Samei , Jehad Alzabut and Yu-ming Chu EMAIL logo
Published/Copyright: December 31, 2021
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Open Mathematics
From the journal Open Mathematics Volume 19 Issue 1

Abstract

The objective of this paper is to investigate, by applying the standard Caputo fractional q-derivative of order α , the existence of solutions for the singular fractional q-integro-differential equation D q α [ k ] ( t ) = Ω ( t , k 1 , k 2 , k 3 , k 4 ) , under some boundary conditions where Ω is singular at some point 0 ≤ t ≤ 1 , on a time scale T t 0 = { t : t = t 0 q n } ∪ { 0 } , for n ∈ N where t 0 ∈ R and q ∈ ( 0 , 1 ) . We consider the compact map and avail the Lebesgue dominated theorem for finding solutions of the addressed problem. Besides, we prove the main results in context of completely continuous functions. Our attention is concentrated on fractional multi-step methods of both implicit and explicit type, for which sufficient existence conditions are investigated. Finally, we present some examples involving graphs, tables and algorithms to illustrate the validity of our theoretical findings.

Keywords: singularity; multi-step methods; q-integro-differential equation
MSC 2010: 34A08; 34B16; 39A13

1 Introduction

The field of fractional calculus plays a fundamental role in mathematical analysis. It provides efficient techniques to solve fractional differential equations and inclusions [1,2, 3,4,5, 6,7,8, 9,10]. On the other hand, one of the most interesting topics is q-difference equations which were introduced by Jackson in [11]. Later, many researchers studied and presented their significant applications [12,13,14, 15,16,17, 18,19,20, 21,22].

In 2007, Atici and Eloe studied discrete fractional calculus and considered a family of finite fractional linear difference equations. They developed the theory of linear finite fractional difference equations analogously to the theory of finite difference equations. In [23], the fractional problem

D σ [ k ] ( r ) + w ( r , k ( r ) , D ξ [ k ] ( r ) ) = 0 ,

with boundary conditions k ( 0 ) = k ( 1 ) = 0 was investigated, where 0 < r < 1 , 1 < σ < 2 , 0 < ξ ≤ σ − 1 , D σ is the standard Riemann-Liouville fractional derivative, w satisfies the Carathéodory conditions on [ 0 , 1 ] × ( 0 , ∞ ) × R , w is positive and w ( t , k , l ) is singular at t = 0 . In [24,25], the fractional differential equation D σ [ k ] ( r ) + w ( r , k ( r ) ) = 0 with boundary conditions k ( 0 ) = k ″ ( 0 ) = 0 and k ( 1 ) = λ ∫ 0 1 k ( s ) d s was studied, where 0 < r < 1 , 2 < σ < 3 , 0 < λ < 2 , D σ is the Caputo fractional derivative and w : [ 0 , 1 ] × [ 0 , ∞ ) → [ 0 , ∞ ) is a continuous function. In [26], the singular fractional problem

D α c [ k ] ( r ) + w ( r , k ( r ) , D σ c [ k ] ( r ) ) = 0 ,

with boundary conditions k ( 0 ) = k ′ ( 0 ) = 0 and k ′ ( 1 ) = D σ c [ k ] ( 1 ) was considered, where 0 < r < 1 , 2 < α < 3 , 0 < σ < 1 , w : ( 0 , 1 ] × R × R → R is continuous with w ( t , k , l ) may be singular at t = 0 and D α c is the Caputo derivative.

In 2015, Zhang et al. and through the spectral analysis and fixed point index theorem obtained the existence of positive solutions of the singular nonlinear fractional differential equation D α [ u ] ( t ) = w ( t , [ u ] ( t ) , D β [ u ] ( t ) ) for 0 < t < 1 , with integral boundary value conditions D β [ u ] ( 0 ) = 0 and D β [ u ] ( 1 ) = ∫ 0 1 D β [ u ] ( r ) d N ( r ) , where α ∈ ( 1 , 2 ] , β ∈ ( 0 , 1 ] , w ( t , u , v ) may be singular at both t = 0 , 1 and u = v = 0 , ∫ 0 1 u ( r ) d N ( r ) denotes the Riemann-Stieltjes integral with a signed measure, in which N : [ 0 , 1 ] → R is a function of bounded variation [27]. Ahmad et al. investigated the existence of solutions for a q-antiperiodic boundary value problem of fractional q-difference inclusions given by

D q α c [ k ] ( t ) ∈ F ( t , k ( t ) , D q [ k ] ( t ) , D q 2 [ k ] ( t ) ) ,

for t ∈ [ 0 , 1 ] , q ∈ ( 0 , 1 ) , 2 < α ≤ 3 , 0 < β ≤ 3 with conditions k ( 0 ) + k ( 1 ) = 0 ,

D q [ k ] ( 0 ) + D q [ k ] ( 1 ) = 0 , D q 2 k ( 0 ) + D q 2 [ k ] ( 1 ) = 0 ,

where D q α c denotes Caputo fractional q-derivative of order α and F : [ 0 , 1 ] × R 3 → P ( R ) is a multivalued map with P ( R ) a class of all subsets of R [24]. In 2019, Ntouyas et al. in [20], by applying definition of the fractional q-derivative of the Caputo-type and the fractional q-integral of the Riemann-Liouville-type, studied the existence and uniqueness of solutions for a multi-term nonlinear fractional q-integro-differential equations under some boundary conditions

D q σ c [ k ] ( r ) = Ω ( r , k ( r ) , ( φ 1 k ) ( r ) , ( φ 2 k ) ( r ) , D q β 1 c [ k ] ( r ) , D q β 2 c [ k ] ( r ) , … , D q β n c [ k ] ( t ) ) .

In [21], Liang et al. investigated the existence of solutions for a nonlinear problem regular and singular fractional q-differential equation

D q σ c [ k ] ( t ) = Ω ( r , k ( r ) , k ′ ( r ) , D q β c [ k ] ( r ) ) ,

with conditions k ( 0 ) = c 1 k ( 1 ) , k ′ ( 0 ) = c 2 D q β c [ k ] ( 1 ) and k ( m ) ( 0 ) = 0 for 2 ≤ m ≤ n − 1 , here n − 1 < σ < n with n ≥ 3 , β , q , c 1 ∈ ( 0 , 1 ) , c 2 ∈ ( 0 , Γ q ( 2 − β ) ) , function Ω is a L κ -Carathéodory, Ω ( r , k 1 , k 2 , k 3 ) may be singular and D q σ c the fractional Caputo-type q-derivative. Furthermore, they discussed the existence of solutions for the fractional q-derivative inclusions

D q σ c [ k ] ( r ) ∈ ℱ ( r , k ( r ) , k ′ ( r ) , D q β c [ k ] ( r ) ) ,

under conditions

k ( 0 ) + k ′ ( 0 ) + D q β c [ k ] ( 0 ) = ∫ 0 η 1 k ( s ) d s , k ( 1 ) + k ′ ( 1 ) + D q β c [ k ] ( 1 ) = ∫ 0 η 2 k ( s ) d s ,

for any t ∈ I and q , η 1 , η 2 , β ∈ ( 0 , 1 ) , where ℱ maps I × R 3 into 2 R is a compact valued multifunction and D q σ c is the fractional Caputo-type q-derivative operator of order σ ∈ ( 1 , 2 ] , and

Γ q ( 2 − β ) ( η 1 2 η 2 − η 1 η 2 2 − η 1 2 + η 2 2 + 4 η 1 − 2 η 2 − 2 ) + 2 ( 1 − η 1 ) ≠ 0 ,

such that σ − β > 1 [16]. Relevant results have been presented in other studies, for example [27,28,29, 30,31].

In this paper and motivated by the aforementioned achievements, we investigate the singular fractional q-integro-differential equation of the form

(1) D q σ [ k ] ( t ) = Ω t , k ( t ) , k ′ ( t ) , D q ζ [ k ] ( t ) , ∫ 0 t f ( r ) k ( r ) d r ,

for 0 < t < 1 under boundary conditions k ( 0 ) = 0 and k ( 1 ) = k ′ ( 1 ) = ⋯ = k ( n ) ( 1 ) = D q η [ k ] ( τ ) , where k ∈ C 1 ( J ) , n = [ η ] + 1 , σ ≥ 2 , ζ , η , τ ∈ ( 0 , 1 ) , f ∈ L 1 ( J ) is nonnegative with ∥ f ∥ 1 = m , Ω ( t , k 1 , k 2 , k 3 , k 4 ) is singular at some points of t ∈ J ≔ ( 0 , 1 ) and D q σ is the Caputo fractional q-derivative of order σ . Existence of solutions is studied via multi-step methods. We prove the main results in context of completely continuous functions and by the help of the Lebesgue dominated theorem. Examples are presented and MATLAB routines [32] are implemented to demonstrate the validity of the proposed results.

The rest of the paper is organized as follows: Section 2 recalls some preliminary concepts and fundamental results of q-calculus. Sections 3 and 4 are devoted to the main results and examples illustrating the obtained results and some algorithms for the addressed problem, respectively.

2 Essential preliminaries

This section is devoted to starting some notations and essential preliminaries that are acting as necessary prerequisites for the results of the subsequent sections.

2.1 q-Fractional derivative and integral

Throughout this article, we shall apply the time scale calculus notations [17]. In fact, we consider the fractional q-calculus on the specific time scale

T t 0 = { 0 } ∪ { t : t = t 0 q n } ,

for n ∈ N , t 0 ∈ R and q ∈ ( 0 , 1 ) . If there is no confusion concerning t 0 we shall denote T t 0 by T . Let a ∈ R . Define [ a ] q = ( 1 − q a ) / ( 1 − q ) [11]. The q-factorial function ( x − y ) q ( n ) with n ∈ N 0 is defined by

(2) ( x − y ) q ( n ) = ∏ k = 0 n − 1 ( x − y q k ) ,

and ( x − y ) q ( 0 ) = 1 , where x and y are real numbers and N 0 ≔ { 0 } ∪ N [12].

Algorithm 1

MATLAB lines for calculation of q-factorial function ( x − y ) q ( n )

1 function p = qfunction(x, y, q, n)
2 if n==0
3 s=1;
4 else
5 s=1;
6 for k=0:n-1
7 s = s∗(x-y∗q^k);
8 end;
9 p=s;
10 end;
11 end

Also, for σ ∈ R and a ≠ 0 , we have

(3) ( x − y ) q ( σ ) = x σ ∏ k = 0 ∞ x − y q k x − y q σ + k .

Algorithms 1 and 2 simplify q-factorial functions ( x − y ) q ( n ) and ( x − y ) q ( σ ) , respectively. In the previous study [34], the authors proved ( x − y ) q ( σ + ν ) = ( x − y ) q ( σ ) ( x − q σ y ) q ( ν ) and ( a x − a y ) q ( σ ) = a σ ( x − y ) q ( σ ) .

Algorithm 2

MATLAB lines for calculation of q-factorial function ( x − y ) q ( σ )

1 function p = qfunctionreal(x,y,q,sigma,n)
2 if n==0
3 p=1;
4 else
5 s=1;
6 for k=0:n-1
7 s = s∗(x-y∗q^k)/(x-y∗q^(sigma+k));
8 end;
9 p=s∗x^sigma;
10 end;
11 end

If y = 0 , then it is clear that x ( σ ) = x σ . The q-Gamma function is given by

Γ q ( z ) = ( 1 − q ) 1 − z ( 1 − q ) q ( z − 1 ) ,

where z ∈ R \ { ⋯ , − 2 , − 1 , 0 } [11]. In fact, by using (3), we have

(4) Γ q ( z ) = ( 1 − q ) 1 − z ∏ k = 0 ∞ 1 − q k + 1 1 − q z + k − 1 .

Algorithm 3

MATLAB lines for calculation of Γ q ( x )

1 function p = qGamma(q,x,n)
2 s=1;
3 for k=0:n
4 s=s∗(1-q^(k+1))/(1-q^(x+k-1));
5 end;
6 p = s∗(1-q)^(1-x);
7 end

Algorithm 3 shows the MATLAB lines for calculation of Γ q ( x ) which we tend n to infinity in it. Note that, Γ q ( z + 1 ) = [ z ] q Γ q ( z ) [34, Lemma 1]. For a function w : T → R , the q-derivative of w , is

(5) D q [ w ] ( x ) = d d x q w ( x ) = w ( x ) − w ( q x ) ( 1 − q ) x ,

for all t ∈ T ⧹ { 0 } , and D q [ w ] ( 0 ) = lim x → 0 D q [ w ] ( x ) [12]. Also, the higher order q-derivative of the function w is defined by D q n [ w ] ( x ) = D q [ D q n − 1 [ w ] ] ( x ) , for all n ≥ 1 , where D q 0 [ w ] ( x ) = w ( x ) [12]. In fact

(6) D q n [ w ] ( x ) = 1 x n ( 1 − q ) n ∑ k = 0 n ( 1 − q − n ) q ( k ) ( 1 − q ) q ( k ) q k w ( x q k ) ,

for x ∈ T ⧹ { 0 } [33].

Remark 2.1

By using equation (2), we can change equation (6) as follows:

(7) D q n [ w ] ( x ) = 1 x n ( 1 − q ) n ∑ k = 0 n ∏ i = 0 k − 1 ( 1 − q i − n ) ( 1 − q i + 1 ) q k w ( x q k ) .

Algorithms 4 and 5 show the MATLAB codes for calculation of equations (5) and (7), respectively.

Algorithm 4

MATLAB lines for calculation of D q [ w ] ( x )

1 function p = Dq(q,x,fun)
2 if x==0
3 p=limit((subs(fun,x)-subs(fun,q∗x))/((1-q)∗x),x,0);
4 else
5 p=(eval(subs(fun,x))-eval(subs(fun,q∗x)))/((1-q)∗x);
6 end;
7 end

Algorithm 5

MATLAB lines for calculation of D q n [ w ] ( x )

1 function g = Dqnatural(q,x,n,fun)
2 s=0;
3 for k=0:n
4 p=1;
5 for i=0:k-1
6 p=p∗(1-q^(i-n))/(1-q^(i+1));
7 end;
8 p=p∗q^k∗eval(subs(fun,x∗q^k));
9 s=s+p;
10 end;
11 g=s/(x^n∗(1-q)^n);
12 end

The q-integral of the function w is defined by

(8) ℐ q [ w ] ( x ) = ∫ 0 x w ( s ) d q s = x ( 1 − q ) ∑ k = 0 ∞ q k w ( x q k ) ,

for 0 ≤ x ≤ b , provided the series is absolutely converged [12].

Algorithm 6

MATLAB lines for calculation of ℐ q [ w ] ( t )

1 function p = Iq(q,x,n,fun)
2 s=1;
3 for k=0:n
4 s=s+q^k∗eval(subs(fun,x∗q^k));
5 end;
6 p=x∗(1-q)∗s;
7 end

By using Algorithm 6, we can obtain the numerical results of ℐ q [ w ] ( x ) when n → ∞ . If a in [ 0 , b ] , then

(9) ∫ a b w ( s ) d q s = ℐ q [ w ] ( b ) − ℐ q [ w ] ( a ) = ( 1 − q ) ∑ k = 0 ∞ q k [ b w ( b q k ) − a w ( a q k ) ] ,

whenever the series exists. The operator ℐ q n is given by ℐ q 0 [ w ] ( x ) = w ( x ) and

ℐ q n [ w ] ( x ) = ℐ q [ ℐ q n − 1 [ w ] ] ( x ) ,

for n ≥ 1 and g ∈ C ( [ 0 , b ] ) [12]. It has been proved that D q [ ℐ q [ w ] ] ( x ) = w ( x ) , and ℐ q [ D q [ w ] ] ( x ) = w ( x ) − w ( 0 ) , whenever the function w is continuous at x = 0 [12].

The fractional Riemann-Liouville-type q-integral of the function w is defined by

(10) ℐ q σ [ w ] ( t ) = 1 Γ q ( σ ) ∫ 0 t ( t − s ) q ( σ − 1 ) w ( s ) d q s , ℐ q 0 [ w ] ( t ) = w ( t ) ,

for t ∈ [ 0 , 1 ] and σ > 0 [14,33].

Remark 2.2

By using equations (3), (4) and (8), we obtain

1 Γ q ( σ ) ∫ 0 t ( t − s ) q ( σ − 1 ) w ( s ) d q s = 1 Γ q ( σ ) ∫ 0 t t σ − 1 ∏ i = 0 ∞ t − s q i t − s q σ + i − 1 w ( s ) d q s = t σ ( 1 − q ) σ ∏ i = 0 ∞ 1 − q σ + i − 1 1 − q i + 1 ∑ k = 0 ∞ q k ∏ i = 0 ∞ 1 − q k + i 1 − q σ + k + i − 1 w ( t q k ) .

Therefore,

(11) ℐ q σ [ w ] ( t ) = t σ ( 1 − q ) σ lim n → ∞ ∑ k = 0 n q k ∏ i = 0 n ( 1 − q σ + i − 1 ) ( 1 − q k + i ) ( 1 − q i + 1 ) ( 1 − q σ + k + i − 1 ) w ( t q k ) ,

Algorithm 7 shows the MATLAB codes of numerical technique.

Algorithm 7

MATLAB lines for calculation of I q σ [ w ] ( x )

1 function g = Iq_sigma(q,sigma,t,n,fun)
2 p=0;
3 for k=0:n
4 s=1;
5 for i=0:n
6 s=s∗(1-q^(sigma+i-1))∗(1-q^(k+i))/((1-q^(i+1)) ∗(1-q^(sigma+k+i-1)));
7 end
8 p=p+q^k∗s∗eval(subs(fun,t∗q^k));
9 end;
10 g=round(p∗(t^sigma)∗(1-q)^sigma,6);
11 end

The Caputo fractional q-derivative of the function w is defined by

(12) D q σ C [ w ] ( t ) = ℐ q [ σ ] − σ [ D q [ σ ] [ w ] ] ( t ) = 1 Γ q ( [ σ ] − σ ) ∫ 0 t ( t − s ) q ( [ σ ] − σ − 1 ) D q [ σ ] [ w ] ( s ) d q s

for t ∈ [ 0 , 1 ] and σ > 0 [14,36]. It has been proved that ℐ q ν [ ℐ q σ [ w ] ] ( t ) = ℐ q σ + ν [ w ] ( t ) , and D q σ c [ ℐ q σ [ w ] ] ( t ) = w ( t ) , where σ , ν ≥ 0 [14]. Also,

ℐ q σ [ D q n [ w ] ] ( t ) = D q n [ ℐ q σ [ w ] ] ( t ) − ∑ k = 0 n − 1 t σ + k − n Γ q ( σ + k − n + 1 ) D q k [ w ] ( 0 ) ,

where σ > 0 and n ≥ 1 [14].

Remark 2.3

From equation (4), Remark 2.1 and equation (11) in Remark 2.2, we obtain

1 Γ q ( [ σ ] − σ ) ∫ 0 t ( t − s ) q ( [ σ ] − σ − 1 ) D q [ σ ] [ w ] ( s ) d q s = 1 Γ q ( [ σ ] − σ ) ∫ 0 t t [ σ ] − σ − 1 ∏ i = 0 ∞ t − s q i t − s q [ σ ] − σ − 1 + i 1 t [ σ ] ( 1 − q ) [ σ ] ∑ k = 0 [ σ ] ∏ i = 0 k − 1 ( 1 − q i − [ σ ] ) ( 1 − q i + 1 ) q k w ( x q k ) d q s = 1 t σ ( 1 − q ) σ − [ σ ] ∑ k = 0 ∞ ∏ i = 0 ∞ ( 1 − q [ σ ] − σ + i − 1 ) ( 1 − q k + i ) ( 1 − q i + 1 ) ( 1 − q [ σ ] − σ − 1 + k + i ) ∑ m = 0 [ σ ] ∏ i = 0 m − 1 ( 1 − q i − [ σ ] ) ( 1 − q i + 1 ) q m w ( t q k + m ) .

Thus, we have

(13) D q σ c [ w ] ( t ) = 1 t σ ( 1 − q ) σ − [ σ ] lim n → ∞ ∑ k = 0 n ∏ i = 0 n ( 1 − q [ σ ] − σ + i − 1 ) ( 1 − q k + i ) ( 1 − q i + 1 ) ( 1 − q [ σ ] − σ − 1 + k + i ) ∑ m = 0 [ σ ] ∏ i = 0 m − 1 ( 1 − q i − [ σ ] ) ( 1 − q i + 1 ) q m w ( t q k + m ) .

Algorithm 8 shows the MATLAB codes of numerical technique.

Algorithm 8

MATLAB lines for calculation of D q σ C [ w ] ( t )

1 function g = IqCaputo_sigma(q,sigma,t,n,fun)
2 S=0;
3 for k=0:n
4 p1=1;
5 for i=0:n
6 p1=p1∗(1-q^(floor(sigma)-sigma+i-1)) ∗(1-q^(k+i))/((1-q^(i+1)) …
7 ∗(1-q^(floor(sigma)-sigma+k+i-1)));
8 end;
9 s2=0;
10 for m=0:floor(sigma)
11 p2=1;
12 for i=0:m-1
13 p2=p2∗(1-q^(i-floor(sigma)))/(1-q^(i+1));
14 end;
15 p2=p2∗q^m∗eval(subs(fun,t∗q^(k+m)));
16 s2=s2+p2;
17 end;
18 S=S+p1∗s2;
19 end;
20 g=round(S/( t^sigma∗(1-q)^(sigma-floor(sigma))),6);
21 end

Throughout this article, we consider

‖ k ‖ 1 ≔ ∫ 0 1 ∣ k ( t ) ∣ d t , ‖ k ‖ ≔ sup { ∣ k ( t ) ∣ : t ∈ J } , ‖ k ‖ ∗ ≔ max { ‖ k ‖ , ‖ k ′ ‖ } ,

as the norm of ℒ = L 1 ( J ) , A = C ( J ) and ℬ = C 1 ( J ) , respectively.

The following lemmas are used in the subsequent sections.

Lemma 2.1

[37] Suppose that 0 < n − 1 ≤ σ < n and k ∈ A ∩ ℒ . Then

ℐ q σ [ D q σ C [ k ] ] ( t ) = k ( t ) + ∑ i = 0 n − 1 c i t i ,

for some constants c 0 , … , c n − 1 ∈ R .

Lemma 2.2

[38] If C is a closed, bounded and convex subset of a Banach space X and Φ : C → C is completely continuous, then Φ has a fixed point in C .

Lemma 2.3

[39] Let X be a Banach space, C a closed and convex subset of X , O a relatively open subset of C with 0 ∈ O and Ω : O → C a continuous and compact map. Then either Ω has a fixed point in O ¯ or there exist a ∈ ∂ O and λ ∈ ( 0 , 1 ) such that a = λ Ω ( a ) .

2.2 Linear multi-step methods

As in the case of ordinary differential equations, linear multi-step methods for fractional differential equations makes use of approximations of values of k 1 ( t ) , k 2 ( t ) , k 3 ( t ) , k 4 ( t ) and Ω ( t , k 1 ( t ) , k 2 ( t ) , k 3 ( t ) , k 4 ( t ) ) on some points of a partition s 0 < s 1 < ⋯ < s n [32,35]. We can therefore write linear multi-step methods for the solution of (1) in the form

(14) ∑ j = 0 n α j 1 ( k 1 n − j , k 2 n − j , k 3 n − j , k 4 n − j ) = h τ ∑ j = 0 n α j 2 Ω ( s n − j , n − j k 1 , k 2 n − j , k 3 n − j , k 4 n − j ) ,

where α j 1 and α j 2 are real parameters and we will indicate with ρ n 1 ( ξ ) and ρ n 2 ( ξ ) the generating polynomials ρ n k ( ξ ) = ∑ j = 0 n α j k ξ n − j . Numerical methods (14) are requested to be consistent with the original problem (1), in the sense that, as h → 0 , the discretized problem is expected to tend asymptotically to the continuous one [32]. In order to formally introduce the consistency concept and study order conditions, it is usually to introduce, associated with (14), the linear difference operator

L h [ ( z 1 ( t ) , z 2 ( t ) , z 3 ( t ) , z 4 ( t ) ) , t , τ ] = ∑ j = 0 n α j 1 ( z 1 n − j ( t − h j ) , z 2 n − j ( t − h j ) , z 3 n − j ( t − h j ) , z 4 n − j ( t − h j ) ) − h τ ∑ j = 0 n α j 2 D q τ C [ z 1 n − j , z 2 n − j , z 3 n − j , z 4 n − j ] ( t − h j ) ,

where ( z 1 ( t ) , z 2 ( t ) , z 3 ( t ) , z 4 ( t ) ) is a sufficiently smooth function [32]. The linear multi-step method (14) is said to be consistent if, for any initial value problem (1), with exact solution ( k 1 ( t ) , k 2 ( t ) , k 3 ( t ) , k 4 ( t ) ) , it holds

lim h → 0 1 h τ L h [ ( k 1 n − j ( t ) , k 2 n − j ( t ) , k 3 n − j ( t ) , k 4 n − j ( t ) ) , t , τ ] = ( 0 , 0 , 0 , 0 ) ,

with h and n related by t = s 0 + h n . Moreover, the method is said to be of order ℓ if

1 h τ L h [ ( k 1 n − j ( t ) , k 2 n − j ( t ) , k 3 n − j ( t ) , k 4 n − j ( t ) ) , t , τ ] = O ( h ℓ ) ,

as h tends to zero. Under the assumption that ( k 1 ( t ) , k 2 ( t ) , k 3 ( t ) , k 4 ( t ) ) is ( m + 1 ) -times differentiable, t = s n , we can expand the true solution

( k 1 ( t − j h ) , k 2 ( t − j h ) , k 3 ( t − j h ) , k 4 ( t − j h ) ) = ( k 1 ( s 0 + ( n − j ) h ) , k 2 ( s 0 + ( n − j ) h ) , k 3 ( s 0 + ( n − j ) h ) , k 4 ( s 0 + ( n − j ) h ) ) ,

of (1) as

( k 1 ( t − j h ) , k 2 ( t − j h ) , k 3 ( t − j h ) , k 4 ( t − j h ) ) = ( k 1 ( s 0 ) , k 2 ( s 0 ) , k 3 ( s 0 ) , k 4 ( s 0 ) ) + ∑ d = 1 m ( n − j ) d h d d ! ( k 1 d ( s 0 ) , k 2 d ( s 0 ) , k 3 d ( s 0 ) , k 4 d ( s 0 ) ) + h m + 1 d ! ∫ 0 n − j ( n − j − ξ ) q ( m ) ( k 1 m + 1 ( s 0 + h ξ ) , k 2 m + 1 ( s 0 + h ξ ) , k 3 m + 1 ( s 0 + h ξ ) , k 4 m + 1 ( s 0 + h ξ ) ) d q ξ ,

and its τ -fractional q-derivative as

D q τ C [ z 1 n − j , n − j z 2 , z 3 n − j , z 4 n − j ] ( t − h j ) = ∑ d = 1 m h d − τ ( n − j ) d − τ Γ q ( d + 1 − τ ) ( k 1 d ( s 0 ) , k 2 d ( s 0 ) , k 3 d ( s 0 ) , k 4 d ( s 0 ) ) + h m + 1 − τ Γ q ( m + 1 − τ ) ∫ 0 n − j ( n − j − ξ ) q ( m − τ ) ( k 1 m + 1 ( s 0 + h ξ ) , k 2 m + 1 ( s 0 + h ξ ) , k 3 m + 1 ( s 0 + h ξ ) , k 4 m + 1 ( s 0 + h ξ ) ) d q ξ .

In this way, we can write the difference operator

L h [ ( k 1 ( t ) , k 2 ( t ) , z 3 ( t ) , z 4 ( t ) ) , t , τ ] ,

as

L h [ ( k 1 ( t ) , k 2 ( t ) , z 3 ( t ) , z 4 ( t ) ) , t , τ ] = C 0 ( n , τ ) + ∑ d = 1 m h d C d ( n , τ ) ( k 1 d ( s 0 ) , k 2 d ( s 0 ) , z 3 d ( s 0 ) , z 4 d ( s 0 ) ) + h m + 1 R m + 1 ,

where the remainder R m + 1 is obtained from Taylor’s expansions and

C 0 ( n , τ ) = ∑ j = 0 n α j 1 , C d ( n , τ ) = 1 d ! ∑ j = 0 n ( n − j ) d α j 1 − 1 Γ q ( d + 1 − τ ) ∑ j = 0 n α j 2 ( n − j ) d − τ ,

for d = 1 , 2 , … , m .

3 Main results

We employ the multi-step methods to prove the main results in this section. First, we adopt the following lemma.

Lemma 3.1

Let z ∈ ℒ . The unique solution of problem

D q σ [ k ] ( t ) + z ( t ) = 0 ,

with boundary conditions k ( 0 ) = 0 and k ( 1 ) = k ′ ( 1 ) = ⋯ = k ( n ) ( t ) = D q η [ k ] ( τ ) , ( n = [ η ] + 1 ) , is k 0 ( t ) = ∫ 0 1 G q ( t , s ) z ( s ) d q s , where

(15) G q ( t , s ) = t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) , t ≤ s , τ ≤ s , t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − ( t − s ) q ( σ − 1 ) Γ q ( σ ) , τ ≤ s ≤ t , t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − t ( τ − s ) q ( σ − η − 1 ) λ Γ q ( σ − η ) , t ≤ s ≤ τ , t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − t ( τ − s ) q ( σ − η − 1 ) λ Γ q ( σ − η ) − λ ( t − s ) q ( σ − 1 ) λ Γ q ( σ ) , s ≤ t , s ≤ τ ,

for t , s ∈ J , σ ≥ 2 , η , τ ∈ ( 0 , 1 ) where

(16) λ ≔ 1 − τ 1 − η Γ q ( 2 − η ) ≠ 0 .

Proof

Assume that k be a solution for the problem. By applying Lemma 2.1, we get

k ( t ) = − ℐ q σ [ z ] ( t ) + d n − 1 t n − 1 + ⋯ + d 1 t + d 0 ,

where n − 1 ≤ σ < n . By utilizing the boundary conditions, we conclude d 0 = 0 . Hence,

D q η [ k ] ( τ ) = − ℐ q σ − η [ z ] ( τ ) + d 1 τ 1 − η Γ q ( 2 − η )

and k ( 1 ) = − ℐ q σ [ z ] ( 1 ) + d 1 . Since k ( 1 ) = D q η [ k ] ( τ ) , we conclude that

d 1 1 − τ 1 − η Γ q ( 2 − η ) = ℐ q σ [ z ] ( 1 ) − ℐ q σ − η [ z ] ( τ )

and so d 1 = 1 λ [ ℐ q σ [ z ] ( 1 ) − ℐ q σ − η ( τ ) ] . Thus, we have

k ( t ) = − ℐ q σ [ k ] ( t ) + t λ [ ℐ q σ [ z ] ( 1 ) − ℐ q σ − η [ z ] ( τ ) ] .

Therefore, we have two cases.

  1. If t ≤ τ , then we can see that

    k ( t ) = − ℐ q σ [ z ] ( t ) + t λ ℐ q σ [ z ] ( t ) + t λ Γ q ( σ ) ∫ t τ ( 1 − s ) q ( σ − 1 ) z ( s ) d q s + ∫ τ 1 ( 1 − s ) q σ − 1 z ( s ) d q s − t λ ℐ q σ − η [ z ] ( t ) + 1 Γ q ( σ − η ) ∫ t τ ( τ − s ) q ( σ − η − 1 ) z ( s ) d q s = ∫ 0 t t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − ( t − s ) q ( σ − 1 ) Γ q ( σ ) − t ( τ − s ) q ( σ − η − 1 ) Γ q ( σ − μ ) z ( s ) d q s + ∫ t τ t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − t ( τ − s ) q ( σ − η − 1 ) λ Γ q ( σ − η ) z ( s ) d q s + ∫ τ 1 t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) z ( s ) d q s .

  2. If t ≥ τ , then we can see that

    k ( t ) = − ℐ q σ [ z ] ( τ ) − ∫ τ t ( t − s ) q ( σ − 1 ) Γ q ( σ ) z ( s ) d q s + t λ ℐ q σ [ z ] ( τ ) + ∫ τ t t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) z ( s ) d q s + ∫ t 1 t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) z ( s ) d q s − t λ ℐ q σ − η [ z ] ( τ ) = ∫ 0 τ t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − ( t − s ) q ( σ − 1 ) Γ q ( σ ) − t ( τ − s ) q ( σ − η − 1 ) λ Γ q ( σ − η ) z ( s ) d q s + ∫ τ t t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) − λ ( t − s ) q ( σ − 1 ) λ Γ q ( σ ) z ( s ) d q s + ∫ t 1 t ( 1 − s ) q ( σ − 1 ) λ Γ q ( σ ) z ( s ) d q s .

This implies that, k ( t ) = ∫ 0 1 G q ( t , s ) z ( s ) d q s = k 0 ( t ) for each t .□

Remark 3.1

If k ∈ ℬ , then

D q β [ k ] ( t ) = 1 Γ q ( 1 − β ) ∫ 0 t ( t − s ) q − β k ′ ( s ) d q s

and so

∣ D q β [ k ] ( t ) ∣ ≤ ∥ k ′ ∥ Γ ( 1 − β ) ∫ 0 t ( t − q s ) − β d q s = ∥ k ′ ∥ Γ ( 2 − β ) t 1 − β .

Thus, D q β [ k ] ∈ A and

∣ D q β [ k ] ∣ ≤ ∥ k ′ ∥ Γ q ( 2 − β ) .

Since ∫ 0 1 f ( r ) d r = m ∈ ( 0 , ∞ ) ,

∫ 0 t f ( r ) k ( r ) d r ≤ ‖ k ‖ ∫ 0 t f ( r ) d r ≤ m ‖ k ‖ .

Now, we give our main result.

Theorem 3.2

The singular problem (1) has a solution whenever the following assumptions hold.

  1. There exist the maps f i : J → R with ∫ 0 1 f i ( r ) d r < ∞ for all i = 1 , 2 , 3 , 4 such that

    ∣ Ω ( t , k 1 , k 2 , k 3 , k 4 ) − Ω ( t , l 1 , l 2 , l 3 , l 4 ) ∣ ≤ ∑ i = 1 4 f i ( t ) ∣ k i ( t ) − l i ( t ) ∣ ,

    for all ( k 1 , k 2 , k 3 , k 4 ) , ( l 1 , l 2 , l 3 , l 4 ) ∈ R 4 and t ∈ J .

  2. There exist g ∈ ℒ and Θ ∈ A 4 such that

    ∣ Ω ( t , k 1 , k 2 , k 3 , k 4 ) ∣ ≤ g ( t ) Θ ( k 1 , k 2 , k 3 , k 4 ) ,

    for each ( k 1 , k 2 , k 3 , k 4 ) ∈ R 4 , almost all t ∈ J . Also

    (17) ‖ Θ ‖ A = sup { ∣ Θ ( k 1 , k 2 , k 3 , k 4 ) ∣ : ( k 1 , k 2 , k 3 , k 4 ) ∈ R 4 } < ∞ .

Proof

We first define a map T : ℬ → ℬ by

(18) T k ( t ) = ∫ 0 1 G q ( t , s ) Ω ˜ ( k , s ) d q s = − ℐ q σ [ Ω ˜ ] ( k , t ) + t λ [ ℐ q σ [ Ω ˜ ] ( k , 1 ) − ℐ q σ − η [ Ω ˜ ] ( k , τ ) ] ,

for each k ∈ ℬ and t ∈ J where

Ω ˜ ( z , t ¯ ) = Ω t ¯ , z ( t ¯ ) , z ′ ( t ¯ ) , D q ζ [ z ] ( t ¯ ) , ∫ 0 t ¯ f ( r ) z ( r ) d r .

Suppose that k 1 , k 2 ∈ ℬ . Then we have

∣ T k 1 ( t ) − T k 2 ( t ) ∣ ≤ ℐ q σ [ ∣ Ω ˜ ( k 1 , s ) − Ω ˜ ( k 2 , s ) ∣ ] + t λ ℐ q σ [ ∣ Ω ˜ ( k 1 , 1 ) − Ω ˜ ( k 2 , 1 ) ∣ ] + t λ ℐ q σ − η [ ∣ Ω ˜ ( k 1 , τ ) − Ω ˜ ( k 2 , τ ) ∣ ] ≤ ℐ q σ f 1 ( t ) ∣ k 1 ( t ) − k 2 ( t ) ∣ + f 2 ( t ) ∣ k 1 ′ ( t ) − k 2 ′ ( t ) ∣ + f 3 ( t ) ∣ D q ζ [ k 1 ] ( t ) − D q ζ [ k 2 ] ( t ) ∣ + f 4 ( t ) ∫ 0 t f ( r ) ( k 1 ( r ) − k 2 ( r ) ) d r + t λ ℐ q σ f 1 ( 1 ) ∣ k 1 ( 1 ) − k 2 ( 1 ) ∣ + f 2 ( 1 ) ∣ k 1 ′ ( 1 ) − k 2 ′ ( 1 ) ∣ + f 3 ( 1 ) ∣ D q ζ [ k 1 ] ( 1 ) − D q ζ [ k 2 ] ( 1 ) ∣ + f 4 ( 1 ) ∫ 0 1 f ( r ) ( k 1 ( r ) − k 2 ( r ) ) d r + ℐ q σ − η f 1 ( τ ) ∣ k 1 ( τ ) − k 2 ( τ ) ∣ + f 2 ( τ ) ∣ k 1 ′ ( τ ) − k 2 ′ ( τ ) ∣ + f 3 ( τ ) ∣ D q ζ [ k 1 ] ( τ ) − D q β k 2 ( τ ) ∣ + f 4 ( τ ) ∫ 0 1 f ( r ) ( k 1 ( r ) − k 2 ( r ) ) d r ≤ ‖ k 1 − k 2 ‖ ℐ q α ( f 1 ( t ) + m f 4 ( t ) ) + ‖ k 1 ′ − k 2 ′ ‖ ℐ q σ f 2 ( t ) + f 3 ( t ) Γ q ( 2 − ζ ) + ‖ k 1 − k 2 ‖ t λ ℐ q σ [ f 1 ( 1 ) + m f 4 ( 1 ) ] + ‖ k 1 ′ − k 2 ′ ‖ t λ ℐ q σ f 2 ( 1 ) + f 3 ( 1 ) Γ q ( 2 − ζ ) + ‖ k 1 − k 2 ‖ t λ ℐ q σ − η [ f 1 + m f 4 ] ( τ ) + ‖ k 1 ′ − k 2 ′ ‖ ℐ q σ − η f 2 ( τ ) + f 3 ( τ ) Γ q ( 2 − ζ ) ≤ ‖ k 1 − k 2 ‖ ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) 2 f 1 ( s ) + 2 m f 4 ( s ) Γ q ( σ ) + f 1 ( s ) + m f 4 ( s ) λ Γ q ( σ − η ) d q s + ‖ k 1 ′ − k 2 ′ ‖ ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) 2 f 2 ( s ) Γ q ( σ ) + f 2 ( s ) λ Γ q ( σ − η ) + 2 f 3 ( s ) Γ q ( σ ) Γ q ( 2 − ζ ) + f 3 ( s ) λ Γ q ( σ − η ) Γ q ( 2 − ζ ) d q s ≤ Λ 1 ( ‖ k 1 − k 2 ‖ + ‖ k 1 ′ − k 2 ′ ‖ ) = Λ 1 ‖ k 1 − k 2 ‖ ∗ ,

where

Λ 1 = max ∫ 0 1 ( 1 − s ) q σ − η − 1 2 f 1 ( s ) + 2 m f 4 ( s ) Γ q ( σ ) + f 1 ( s ) + m f 4 ( s ) λ Γ q ( σ − η ) d q s , ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) 2 f 2 ( s ) Γ q ( σ ) + f 2 ( s ) λ Γ q ( σ − η ) + 2 f 3 ( s ) Γ q ( σ ) Γ q ( 2 − ζ ) + f 3 ( s ) λ Γ q ( σ − η ) Γ q ( 2 − ζ ) d q s < ∞ .

On the other hand, we get

∣ T k 1 ′ ( t ) − T k 2 ′ ( t ) ∣ ≤ ∫ 0 1 ∂ G q ( t , s ) ∂ t ∣ Ω ˜ ( k 1 , s ) − Ω ˜ ( k 2 , s ) ∣ d q s ≤ ‖ k 1 − k 2 ‖ ∫ 0 1 ( 1 − s ) q ( σ − 2 ) f 1 ( s ) Γ q ( σ − 1 ) + f 1 ( s ) Γ q ( σ ) + f 1 ( s ) λ Γ ( σ − η ) + m f 4 ( s ) Γ q ( σ − 1 ) + m f 4 ( s ) Γ q ( σ ) + m f 4 ( s ) λ Γ q ( σ − η ) d q s + ‖ k 1 ′ − k 2 ′ ‖ ∫ 0 1 ( 1 − s ) q ( σ − 2 ) f 2 ( s ) Γ q ( σ − 1 ) + f 2 ( s ) Γ q ( σ ) + f 2 ( s ) λ Γ q ( σ − η ) + f 3 ( s ) Γ q ( σ − 1 ) Γ q ( 2 − ζ ) + f 3 ( s ) Γ q ( σ ) Γ q ( 2 − ζ ) + f 3 ( s ) λ Γ q ( σ − η ) Γ q ( 2 − ζ ) d q s ≤ Λ 2 ( ‖ k 1 − k 2 ‖ + ‖ k 1 ′ − k 2 ′ ‖ ) = Λ 2 ‖ k 1 − k 2 ‖ ∗ ,

where

Λ 2 = max ∫ 0 1 ( 1 − s ) q ( σ − 2 ) f 1 ( s ) Γ q ( σ − 1 ) + f 1 ( s ) Γ q ( σ ) + f 1 ( s ) λ Γ q ( σ − η ) + m f 4 ( s ) Γ q ( σ − 1 ) + m f 4 ( s ) Γ q ( σ ) + m f 4 ( s ) λ Γ q ( σ − η ) d q s , ∫ 0 1 ( 1 − s ) q ( σ − 2 ) f 2 ( s ) Γ q ( σ − 1 ) + f 2 ( s ) Γ q ( σ ) + f 2 ( s ) λ Γ q ( σ − η ) + f 3 ( s ) Γ q ( σ − 1 ) Γ q ( 2 − ζ ) + f 3 ( s ) Γ q ( σ ) Γ q ( 2 − β ) + f 3 ( s ) λ Γ q ( σ − η ) Γ q ( 2 − ζ ) d q s < ∞ .

Put

(19) M 1 = 1 Γ q ( σ ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) , M 2 = 1 Γ q ( σ − 1 ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) ,

(20) m 0 = ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s = ( 1 − q ) ∑ k = 0 ∞ q k ( 1 − q k ) q ( σ − η − 1 ) g ( q k ) = ( 1 − q ) ∑ k = 0 ∞ q k g ( q k ) ∏ i = 0 ∞ 1 − q k + i 1 − q k + σ − η − 1 + i ,

and r 0 = m 0 ‖ Θ ‖ A max { M 1 , M 2 } , Λ 0 = max { Λ 1 , Λ 2 } . Since g ∈ ℒ , m 0 < ∞ . Then we have

‖ T k 1 ( t ) − T k 2 ( t ) ‖ ∗ ≤ Λ 0 ‖ k 1 − k 2 ‖ ∗

and so ‖ T k 1 ( t ) − T k 2 ( t ) ‖ ∗ → 0 as ‖ k 1 − k 2 ‖ ∗ → 0 . Consider k ∈ ℬ and

B r 0 = { k ∈ ℬ : ‖ k ‖ ∗ ≤ r 0 } .

Then, we have

∣ T k ( t ) ∣ ≤ ℐ q σ g ( t ) Θ t , k ( t ) , k ′ ( t ) , D q β [ k ] ( t ) , ∫ 0 t f ( r ) k ( r ) d r + 1 λ ℐ q σ g ( 1 ) Θ 1 , k ( 1 ) , k ′ ( 1 ) , D q ζ [ k ] ( 1 ) , ∫ 0 1 f ( r ) k ( r ) d r + ℐ q σ − η g ( τ ) Θ τ , k ( τ ) , k ′ ( τ ) , D q ζ [ u ] ( τ ) , ∫ 0 τ f ( r ) k ( r ) d r ≤ ‖ Θ ‖ A 1 Γ q ( σ ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s = m 0 ‖ Θ ‖ A M 1 ,

for each t ∈ J . Note that, ∫ 0 1 ( 1 − q s ) ( σ − 1 ) g ( s ) d q s ≤ m 0 . Also, we can conclude that

T k ′ ( t ) = ∫ 0 1 ∂ G q ( t , s ) ∂ t Ω s , k ( s ) , k ′ ( s ) , D q ζ [ k ] ( s ) , ∫ 0 s f ( r ) k ( r ) d r d q s = − ℐ q σ − 1 Ω t , k ( t ) , k ′ ( t ) , D q ζ k ( t ) , ∫ 0 t f ( r ) k ( r ) d r + 1 λ ℐ q σ Ω 1 , k ( 1 ) , k ′ ( 1 ) , D q ζ k ( 1 ) , ∫ 0 1 f ( r ) k ( r ) d r + ℐ q σ − η Ω τ , k ( τ ) , k ′ ( τ ) , D q ζ k ( τ ) , ∫ 0 s f ( r ) k ( τ ) d r ,

and so

∣ T k ′ ( t ) ∣ ≤ ‖ Θ ‖ A 1 Γ q ( σ − 1 ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s = m 0 ‖ Θ ‖ A M 2 .

Hence, ‖ T k ‖ ∗ = max { ‖ T k ‖ , ‖ T k ′ ‖ } ≤ r 0 . Therefore, T maps B r 0 into B r 0 . Similarly, one can check that T maps bounded sets into bounded sets. Let t 1 , t 2 ∈ J with t 1 ≤ t 2 . Then, we have

‖ T k ( t 1 ) − T k ( t 2 ) ‖ ≤ 1 Γ q ( σ ) ∫ 0 t 1 [ ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) ] ∣ Ω ˜ ( k , s ) ∣ d q s + 1 Γ q ( σ ) ∫ t 1 t 2 ( t 1 − s ) q ( σ − 1 ) ∣ Ω ˜ ( k , s ) ∣ d q s + ∣ t 2 − t 1 ∣ Δ Γ q ( σ ) ∫ 0 1 ( 1 − s ) q ( σ − 1 ) ∣ Ω ˜ ( k , s ) ∣ d q s + ∣ t 2 − t 1 ∣ λ Γ q ( σ − η ) ∫ 0 μ ( μ − s ) q ( σ − η − 1 ) ∣ Ω ˜ ( k , s ) ∣ d q s ≤ ‖ Θ ‖ A 1 Γ q ( σ ) ∫ 0 t 1 [ ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) ] g ( s ) d q s + 1 Γ q ( σ ) ∫ t 1 t 2 ( t 1 − s ) q σ − 1 g ( s ) d q s + ∣ t 2 − t 1 ∣ 1 λ q Γ q ( σ ) + 1 λ Γ q ( σ − η ) ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s .

Since g ∈ ℒ , ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s < ∞ . Also, we have

sup t 1 , t 2 ∈ J ∫ 0 t 1 [ ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) ] g ( s ) d q s ≤ ∫ 0 1 ( 1 − s ) q ( σ − 1 ) g ( s ) d q s < ∞ .

Since ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) → 0 , as t 2 → t 1 , for each ε > 0 there exists δ > 0 such that ∣ t 2 − t 1 ∣ < δ implies

( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) < ε .

If 0 < δ < ε and ∣ t 2 − t 1 ∣ < δ , then

∫ 0 t 1 [ ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) ] g ( s ) d q s ≤ ε ∫ 0 1 g ( s ) d q s ,

and so

∫ 0 t 1 [ ( t 2 − s ) q ( σ − 1 ) − ( t 1 − s ) q ( σ − 1 ) ] g ( s ) d q s → 0 ,

as t 2 → t 1 . Similarly, we conclude that

∫ t 1 t 2 ( t 1 − s ) q ( σ − 1 ) g ( s ) d q s

and

∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s

tend to 0 as t 2 → t 1 . Thus, ∣ T k ( t 2 ) − T k ( t 1 ) ∣ → 0 as t 2 → t 1 . Note that

∣ T k ′ ( t 2 ) − T k ′ ( t 1 ) ∣ ≤ ‖ Θ ‖ A Γ q ( σ − 1 ) ∫ 0 t 1 [ ( t 2 − s ) q ( σ − 2 ) − ( t 1 − s ) q ( σ − 2 ) ] g ( s ) d q s + ∫ t 1 t 2 ( t 1 − s ) q ( σ − 2 ) g ( s ) d q s .

By using a similar way, we conclude that ∣ T k ′ ( t 2 ) − T k ′ ( t 1 ) ∣ → 0 as t 2 → t 1 . Hence,

‖ T k ( t 2 ) − T k ( t 1 ) ‖ ∗ → 0 ,

as t 2 → t 1 and so T is equi-continuous on B r 0 . Hence, T : B r 0 → B r 0 is completely continuous. At present, Lemma 2.2 implies that T has a fixed point on B r 0 which is the solution of the problem (1). The proof is completed.□

Note that in Theorem 3.2, the map Ω ( t , . , . , . , . ) could be discontinuous at points of a subset of J of measure zero. One can obtain solutions of the problem (1) under some different conditions. For example in next result, the map Ω ( t , . , . , . , . ) could be discontinuous at t = 0 .

Theorem 3.3

Let Ω : J × ℬ 4 → R be a map. Then the problem (1) has a solution, whenever the following assumptions hold for all ( k 1 , k 2 , k 3 , k 4 ) ∈ ℬ 4 and almost all t ∈ J .

  1. Ω ( t , . , . , . , . ) : J → R is continuous and Ω ( t , k 1 , k 2 , k 3 , k 4 ) ≥ 0 .

  2. There exist g ∈ ℒ and Θ 1 , Θ 2 : R 4 → [ 0 , ∞ ) such that Θ 1 and Θ 2 are nondecreasing in all components,

    lim k → ∞ Θ 1 ( k , k , k , k ) / k = 0 , lim l → ∞ Θ 2 ( l , l , l , l ) = ℓ < ∞

    and

    Ω ( t , k 1 , k 2 , k 3 , k 4 ) ≤ g ( t ) Θ 1 ( k 1 , k 2 , k 3 , k 4 ) + Θ 2 ( k 1 , k 2 , k 3 , k 4 ) .

Proof

For each k ∈ ℬ and i ≥ 1 define

( k ) i ( t ) = min − 1 i , k ( t ) ,

whenever k ( t ) < 0 and ( k ) i ( t ) = max 1 i , k ( t ) whenever k ( t ) ≥ 0 . Put

Ω i ( t , k 1 , k 2 , k 3 , k 4 ) = Ω ( t , ( k 1 ) i , ( k 2 ) i , ( k 3 ) i , ( k 4 ) i ) ,

for all i , t and k 1 , k 2 , k 3 , k 4 . By simple method, we conclude that ( k ) i ( t ) → k ( t ) and each Ω i is a regular function on J . A regular function at a point a is a function that is regular in some neighborhood of a . For each i , consider the regular fractional q-integro-differential equation

(21) D q σ [ k ] ( t ) + Ω i t , k ( t ) , k ′ ( t ) , D q ζ [ k ] ( t ) , ∫ 0 t f ( r ) k ( r ) d r = 0 ,

under the boundary condition of the problem (1). Suppose that ‖ g ‖ 1 = m > 0 and ε 0 > 0 be given. Choose r 1 > 0 and r 2 > 0 such that ℓ k < 1 2 ε 0 for each ∣ k ∣ > r 1 and

Θ 1 ( k , k , k , k ) k < 1 2 ‖ g ‖ 1 ε 0 ,

for each ∣ k ∣ > r 2 , respectively. Take r 0 ≔ max { r 1 , r 2 } , then for all ∣ k ∣ > r 0 , we obtain

ℓ + ‖ g ‖ 1 Θ 1 ( k , k , k , k ) k < ε 0 .

Put Λ 0 ≔ max { M 1 , M 2 } , here M 1 and M 2 are defined in equation (19), and ε 0 = 1 Λ 0 . If

r > r 0 max 1 , 1 Γ q ( 2 − ζ ) , m ,

then

(22) 1 r ℓ + ‖ g ‖ 1 Θ r , r , r Γ q ( 2 − ζ ) , m r < 1 Λ 0 .

At present, consider the set

B r = { k ∈ ℬ : ‖ k ‖ ∗ < r } .

For each i ≥ 1 , define T i : B ¯ r → ℬ as (18) in which we replaced Ω by Ω i . If { k i } is a convergent sequence in B ¯ r , then k i → k and k i ′ → k ′ uniformly on J . Since

‖ D q ζ [ k i ] ( t ) − D q ζ [ k ] ( t ) ‖ ≤ ‖ k i − k ′ ‖ Γ q ( 2 − ζ )

and D q ζ [ k i ] ( t ) → D q ζ [ k ] ( t ) . Also, we have

∫ 0 t f ( r ) k i ( r ) d r − ∫ 0 t f ( r ) k ( r ) d r ≤ ∫ 0 t f ( r ) ∣ k i ( r ) − k ( r ) ∣ d r ≤ m ‖ k i − k ‖

and so

lim i → ∞ ∫ 0 t f ( r ) k i ( r ) d r = ∫ 0 t f ( r ) k ( r ) d r .

Thus, lim i → ∞ Ω ˜ i ( k i , t ) = Ω ˜ i ( k , t ) . Note that

∣ T n [ k i ] ( t ) − T n [ k ] ( t ) ∣ ≤ ∫ 0 1 ( t − s ) q ( σ − 1 ) Γ q ( σ ) + t ( t − s ) q ( σ − 1 ) λ Γ q ( σ ) + t ( τ − s ) q ( σ − η − 1 ) λ Γ q ( σ − η ) ∣ Ω ˜ n ( k i , s ) − Ω ˜ n ( k , s ) ∣ d q s ≤ M 1 ∫ 0 1 ∣ Ω ˜ n ( k i , s ) − Ω ˜ n ( k , s ) ∣ d q s .

By using a similar method, we have

∣ T n ′ [ k i ] ( t ) − T n ′ [ k ] ( t ) ∣ ≤ M 2 ∫ 0 1 ∣ Ω ˜ n ( k i , s ) − Ω ˜ n ( k , s ) ∣ d q s .

Thus, ‖ T n [ k i ] ( t ) − T n [ k ] ( t ) ‖ ∗ → 0 as k i → k . Hence, { T n [ k i ] } i = 1 ∞ is relatively compact in B ¯ r and so T i is a completely continuous operator on B ¯ r for all i . Suppose that i ≥ 1 be given and there exist z ∈ ∂ B r and 0 < c < 1 such that z = c T i [ z ] . Since ‖ z ‖ ∗ = r , ‖ z ‖ ≤ r , ‖ z ′ ‖ ≤ r ,

‖ D q ζ [ z ] ‖ ≤ ‖ z ′ ‖ Γ q ( 2 − ζ ) ≤ r Γ q ( 2 − ζ )

and ∫ f ( r ) z ( r ) d r ≤ m r . By using the assumption, we have

∣ z ( t ) ∣ = ∣ c T n [ z ] ( t ) ∣ = c ∫ 0 1 G q ( t , q s ) Ω ˜ i ( z , s ) d q s < M 1 ∫ 0 1 Θ 2 r , r , r Γ q ( 2 − ζ ) , m r d q s + ∫ 0 1 f ( s ) Θ 1 z ( s ) , z ′ ( s ) , D q σ [ z ] ( s ) , ∫ 0 s f ( r ) z ( r ) d r d q s ≤ M 1 ℓ + ‖ g ‖ 1 Θ 1 r , r , r Γ q ( 2 − ζ ) , m r

and

∣ z ′ ( t ) ∣ = ∣ c T n ′ [ z ] ( t ) ∣ = c ∫ 0 1 ∂ G q ( t , s ) ∂ t Ω ˜ i ( z , s ) d q s < M 2 ℓ + ‖ g ‖ 1 Θ 1 r , r , r Γ q ( 2 − ζ ) , m r .

Hence,

‖ z ‖ ∗ < max { M 1 , M 2 } ℓ + ‖ g ‖ 1 Θ 1 r , r , r Γ q ( 2 − β ) , m r ,

and so

r < Λ 0 ℓ + ‖ g ‖ 1 Θ 1 r , r , r Γ q ( 2 − ζ ) , m r .

Thus,

ℓ + ‖ g ‖ 1 Θ 1 r , r , r Γ q ( 2 − ζ ) , m r > r Λ 0 ,

which is a contradiction to (22). This implies that z ∉ ∂ B r . By employing Lemma 2.3, T i has a fixed point k i ∈ B ¯ r for each i , that is the problem (21) has a solution. Let ( k ) i be the solution of the problem (21). As we proved, { ( k ) i } is relatively compact and ( k ) i → k for some k ∈ ℬ . Thus, k ∈ B ¯ r . Similar to last result, we can show that lim i → ∞ D q β [ k i ] ( t ) = D q ζ [ k ] ( t ) , lim i → ∞ k i ′ ( t ) = k ( t ) and

lim i → ∞ ∫ 0 t f ( r ) k i ( r ) d r = ∫ 0 t f ( r ) k ( r ) d r ,

for each t ∈ J . Consequently, we get lim i → ∞ Ω ˜ i ( k , t ) = Ω ˜ ( k , t ) and

∣ G q ( t , s ) Ω ˜ i ( k , t ) − Ω ˜ ( k , t ) ∣ ≤ M 1 g ( s ) Θ 2 r , r , r Γ q ( 2 − ζ ) , m r < ∞ .

By applying the Lebesgue dominated theorem, we obtain

k ( t ) = ∫ 0 1 G q ( t , s ) Ω ˜ ( k , s ) d q s ,

for all t ∈ J . This completes the proof.□

4 Illustrative examples via computational results

In this section, we present two illustrative examples. For problems for which the analytical solution is not known, we will use, as reference solution, the numerical approximation obtained with a tiny step h by the implicit trapezoidal PI rule, which, as we will see, usually shows an excellent accuracy [32]. For this purpose, we need to present a simplified analysis that is able to execute the values of the q-Gamma function. We provided a pseudo-code description of the method for calculation of the q-Gamma function of order n in Algorithms 3, 4, 6 and 7; for more details see https://www.dm.uniba.it/members/garrappa/software.

All the experiments are carried out in MATLAB Ver. 8.5.0.197613 (R2015a) on a computer equipped with a CPU AMD Athlon(tm) II X2 245 at 2.90 GHz running under the operating system Windows 7.

Example 4.1

Consider the fractional q-integro-differential problem

(23) D q 25 9 [ k ] ( t ) + g ( t ) ∣ k ( t ) ∣ 3 + ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ 3 + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ 3 + ∣ D q 9 14 [ k ] ( t ) ∣ + ∣ z k ( t ) ∣ 3 + ∣ z k ( t ) ∣ = 0 ,

for t ∈ J , k ∈ C 1 ( J ) and for each q ∈ ( 0 , 1 ) , under conditions k ( 0 ) = 0 and

k ( 1 ) = D q 5 7 [ k ] 8 9 ,

where

z k ( t ) = ∫ 0 t f ( r ) k ( r ) d r ,

f ( t ) = 3 t + 1 ∈ ℒ and m = ‖ f ‖ 1 = 3 . Clearly in the problem σ = 25 9 ≥ 2 , ζ = 9 14 ∈ ( 0 , 1 ) , τ = 8 9 ∈ ( 0 , 1 ) , η = 5 7 ∈ ( 0 , 1 ) . We define g ( t ) by

g ( t ) = 1 t p 1 , t ∈ ( 0 , γ 1 ] , 1 ( t − γ 1 ) p 2 , t ∈ ( γ 1 , γ 2 ] , ⋮ ⋮ 1 ( t − γ k ) p N 0 + 1 , t ∈ ( γ k , 1 ) ,

where p 1 , … , p N 0 + 1 ∈ ( 0 , 1 ) ( k ≥ 1 ) , and γ 1 , γ 2 , … , γ N 0 be real numbers such that

0 < γ 1 < γ 2 < ⋯ < γ N 0 < 1 .

For N 0 = 4 , we take

(24) g ( t ) = 1 t 1 8 , t ∈ 0 , 1 4 , 1 t − 1 4 3 7 , t ∈ 1 4 , 1 2 , 1 t − 1 2 5 7 , t ∈ 1 2 , 3 4 , 1 t − 3 4 9 10 , t ∈ 3 4 , 1 .

Now, define

Θ ( k 1 , k 2 , k 3 , k 4 ) = ∑ i = 1 4 ∣ k i ∣ 3 + ∣ k i ∣ ,

for ( k 1 , k 2 , k 3 , k 4 ) ∈ R 4 . One can see that Θ satisfies in equation (17). Then we have

∣ Ω ( t , k 1 , k 2 , k 3 , k 4 ) − Ω ( t , l 1 , l 2 , l 3 , l 4 ) ∣ = g ( t ) ∣ k 1 ( t ) ∣ 3 + ∣ k 1 ( t ) ∣ + ∣ k ′ ( t ) ∣ 3 + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k 3 ] ( t ) ∣ 3 + ∣ D q 9 14 [ k 3 ] ( t ) ∣ + ∣ z k 4 ( t ) ∣ 3 + ∣ z k 4 ( t ) ∣ − g ( t ) ∣ l 1 ( t ) ∣ 3 + ∣ l 1 ( t ) ∣ + ∣ l ′ ( t ) ∣ 3 + ∣ l ′ ( t ) ∣ + ∣ D q 9 14 [ l 3 ] ( t ) ∣ 3 + ∣ D q 9 14 [ l 3 ] ( t ) ∣ + ∣ z l 4 ( t ) ∣ 3 + ∣ z l 4 ( t ) ∣ ≤ ∣ g ( t ) ∣ 1 3 ∣ k 1 ( t ) − l 1 ( t ) ∣ + 1 3 ∣ k 2 ( t ) − l 2 ( t ) ∣ + 1 3 ∣ k 3 ( t ) − l 3 ( t ) ∣ + 1 3 ∣ k 4 ( t ) − l 4 ( t ) ∣ ≤ ∣ g ( t ) ∣ ∑ i = 1 4 ∣ k i ( t ) − l i ( t ) ∣ .

Thus, by applying the definition of g ( t ) , we get

∣ Ω ( t , k 1 , k 2 , k 3 , k 4 ) − Ω ( t , l 1 , l 2 , l 3 , l 4 ) ∣ ≤ 1 t 1 8 ∑ i = 1 4 ∣ k i ( t ) − l i ( t ) ∣ , t ∈ 0 , 1 4 , 1 ( t − 1 4 ) 3 7 ∑ i = 1 4 ∣ k i ( t ) − l i ( t ) ∣ , t ∈ 1 4 , 1 2 , 1 t − 1 2 5 7 ∑ i = 1 4 ∣ k i ( t ) − l i ( t ) ∣ , t ∈ 1 2 , 3 4 , 1 t − 3 4 9 10 ∑ i = 1 4 ∣ k i ( t ) − l i ( t ) ∣ , t ∈ 3 4 , 1 .

Therefore,

f i ( t ) = 1 t 8 , 1 ( t − 0.25 ) 3 7 , 1 ( t − 0.5 ) 5 7 , 1 ( t − 0.75 ) 9 10 ,

for t ∈ 0 , 1 4 , 1 4 , 1 2 , 1 2 , 3 4 , 3 4 , 1 , respectively, for i = 1 , 2 , 3 , 4 . In addition by using equations (16) and (19), we obtain

λ = 1 − τ 1 − η Γ q ( 2 − η ) = 1 − 8 9 2 7 Γ q 9 7 ≅ 0.0062 , q = 1 10 , − 0.0412 , q = 1 2 , − 0.0664 , q = 6 7 .

On the other hand,

M 1 = 1 Γ q ( σ ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) = 1 Γ q 25 9 + 1 λ Γ q 25 9 + 1 λ Γ q 130 63 , M 2 = 1 Γ q ( σ − 1 ) + 1 λ Γ q ( σ ) + 1 λ Γ q ( σ − η ) = 1 Γ q 16 9 + 1 λ Γ q 25 9 + 1 λ Γ q 130 63

and from equation (20),

m 0 = ∫ 0 1 ( 1 − s ) q ( σ − η − 1 ) g ( s ) d q s = ∫ 0 1 ( 1 − s ) q 25 9 − 5 7 − 1 g ( s ) d q s = ∫ 0 1 ( 1 − s ) q 67 63 g ( s ) d q s = ( 1 − q ) ∑ k = 0 ∞ q k g ( q k ) ∏ i = 0 ∞ 1 − q k + i 1 − q k + σ − η − 1 + i = ( 1 − q ) ∑ k = 0 ∞ q k g ( q k ) ∏ i = 0 ∞ 1 − q k + i 1 − q k + i + 67 63 .

Thus, we have M 1 ≅ 313.0401 , − 41.1026 , − 23.6644 , M 2 ≅ 313.1262 , − 40.7920 , − 23.2307 , m 0 ≅ 0.1372 , 0.6360, 1.5717, r 0 = ‖ Θ ‖ A × ( 42.9762 ) , ‖ Θ ‖ A × ( − 25.9439 ) , ‖ Θ ‖ A × ( − 36.5123 ) for q = 1 10 , 1 2 , 6 7 , respectively. These results are obtained by Algorithms 9, 10 and 11. Now, for showing the numerical results, we consider the problem (23) as follows:

(25) D q 25 9 [ k ] ( t ) + g ( t ) ∣ k ( t ) ∣ 3 + ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ 3 + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ 3 + ∣ D q 9 14 [ k ] ( t ) ∣ + ∣ z k ( t ) ∣ 3 + ∣ z k ( t ) ∣ ≤ D q 25 9 [ k ] ( t ) + g ( t ) ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ + ∣ z k ( t ) ∣ = 0 .

Let t 1 = 1 8 , t 2 = 4 11 , t 3 = 5 9 and t 4 = 16 19 . Then from definition of g ( t ) in equation (24), we have g ( t 1 ) = 1.2668 , g ( t 2 ) = 2.5396 , g ( t 3 ) = 7.8817 and g ( t 4 ) = 8.5535 , which, upon substitution in equation (25), leads to

(26) D q 25 9 [ k ] ( t ) + 1 8 8 ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ = − 1 8 8 ∫ 0 t f 1 ( r ) k ( r ) d r , D q 25 9 [ k ] ( t ) + 1 5 44 3 7 ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ = − 1 5 44 3 7 ∫ 0 t f 2 ( r ) k ( r ) d r , D q 25 9 [ k ] ( t ) + 1 1 18 5 7 ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ = − 1 1 18 5 7 ∫ 0 t f 3 ( r ) k ( r ) d r , D q 25 9 [ k ] ( t ) + 1 7 76 9 10 ∣ k ( t ) ∣ + ∣ k ′ ( t ) ∣ + ∣ D q 9 14 [ k ] ( t ) ∣ = − 1 7 76 9 10 ∫ 0 t f 4 ( r ) k ( r ) d r .

Table 4 shows numerical values of k ( t ) for each equations in (26). Also, one can see that the curve of k ( t ) with respect to t in Figure 1 for t ∈ 0 , 1 4 , 1 4 , 1 2 , 1 2 , 3 4 , 3 4 , 1 , respectively (Algorithm 12). By using Theorem 3.3, one can see that the singular q-integro-differential problem (23) has a solution.

Table 1

Some numerical results of λ , M 1 , M 2 , m 0 , r 0 in Example 4.1 for t ∈ J ¯ and q = 1 10

q = 1 10 n λ M 1 M 2 m 0 r 0
1 0.0067 287.9505 288.0367 0.1189 34.2533
2 0.0062 310.3480 310.4341 0.1348 41.8468
3 0.0062 312.7840 312.8701 0.1369 42.8388
4 0.0062 313.0398 313.1259 0.1372 42.9627
5 0.0062 313.0401 313.1262 0.1372 42.9746
6 0.0062 313.0401 313.1262 0.1372 42.9762
7 0.0062 313.0401 313.1262 0.1372 42.9762
Table 2

Some numerical results of λ , M 1 , M 2 , m 0 , r 0 in Example 4.1 for t ∈ J ¯ and q = 1 2

q = 1 2 n λ M 1 M 2 m 0 r 0
1 0.0083 178.5039 178.8292 0.3858 68.9912
2 − 0.0172 − 92.2014 − 91.8824 0.4974 − 45.7011
3 − 0.0294 − 55.8206 − 55.5059 0.559 − 31.0280
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
9 − 0.0410 − 41.2640 − 40.9543 0.6338 − 25.9558
10 − 0.0411 − 41.1831 − 40.8735 0.6348 − 25.9454
11 − 0.0412 − 41.1423 − 40.8327 0.6353 − 25.9421
12 − 0.0412 − 41.1219 − 40.8123 0.6356 − 25.9416
13 − 0.0412 − 41.1122 − 40.8027 0.6358 − 25.9423
14 − 0.0412 − 41.1069 − 40.7973 0.6359 − 25.9427
15 − 0.0412 − 41.1042 − 40.7947 0.6359 − 25.9430
16 − 0.0412 − 41.1034 − 40.7939 0.6360 − 25.9437
17 − 0.0412 − 41.1025 − 40.7929 0.6360 − 25.9437
18 − 0.0412 − 41.1025 − 40.7930 0.6360 − 25.9441
19 − 0.0412 − 41.1026 − 40.7930 0.6360 − 25.9443
20 − 0.0412 − 41.1016 − 40.7920 0.6360 − 25.9437
21 − 0.0412 − 41.1016 − 40.7920 0.6360 − 25.9438
22 − 0.0412 − 41.1016 − 40.7920 0.6360 − 25.9438
23 − 0.0412 − 41.1016 − 40.7920 0.6360 − 25.9439
24 − 0.0412 − 41.1016 − 40.7920 0.6360 − 25.9439
Table 3

Some numerical results of λ , M 1 , M 2 , m 0 , r 0 in Example 4.1 for t ∈ J ¯ and q = 6 7

q = 6 7 n λ M 1 M 2 m 0 r 0
1 0.2188 2.2717 2.6290 0.6443 1.6939
2 0.1616 3.9488 4.3474 0.7476 3.2501
3 0.1192 6.4012 6.8259 0.9636 6.5775
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
52 − 0.0663 − 23.6855 − 23.2518 1.5712 − 36.5341
53 − 0.0664 − 23.6825 − 23.2489 1.5713 − 36.5307
54 − 0.0664 − 23.6798 − 23.2461 1.5713 − 36.5275
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
67 − 0.0664 − 23.6666 − 23.2329 1.5716 − 36.5139
68 − 0.0664 − 23.6663 − 23.2327 1.5717 − 36.5137
69 − 0.0664 − 23.666 − 23.2324 1.5717 − 36.5135
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
84 − 0.0664 − 23.6647 − 23.231 1.5717 − 36.5126
85 − 0.0664 − 23.6643 − 23.2307 1.5717 − 36.5120
86 − 0.0664 − 23.6643 − 23.2307 1.5717 − 36.5121
87 − 0.0664 − 23.6643 − 23.2307 1.5717 − 36.5121
88 − 0.0664 − 23.6644 − 23.2307 1.5717 − 36.5121
89 − 0.0664 − 23.6644 − 23.2307 1.5717 − 36.5121
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
98 − 0.0664 − 23.6644 − 23.2307 1.5717 − 36.5122
99 − 0.0664 − 23.6644 − 23.2307 1.5717 − 36.5123
100 − 0.0664 − 23.6644 − 23.2307 1.5717 − 36.5123
Table 4

Some numerical results of T k ( t ) = ∫ 0 1 G q ( t , s ) Ω ˜ ( k , s ) d q s in Example 4.1 for t ∈ J ¯

0 , 1 4 1 4 , 1 2 1 2 , 3 4 3 4 , 1
t k ( t ) t k ( t ) t k ( t ) t k ( t )
0 0 0.2500 0 0.500 0 0.7500 0
0.0156 0 0.2656 0 0.5156 0 0.7656 0
0.0313 0 0.2813 0 0.5313 0 0.7813 0
0.0469 0.0001 0.2969 0 0.5469 0 0.7969 0
0.0625 0.0001 0.3125 0.0001 0.5625 0.0001 0.8125 0
0.0781 0.0002 0.3281 0.0001 0.5781 0.0001 0.8281 0
0.0938 0.0004 0.3438 0.0002 0.5938 0.0002 0.8438 0.0001
0.1094 0.0006 0.3594 0.0004 0.6094 0.0003 0.8594 0.0001
0.1250 0.0008 0.375 0.0005 0.6250 0.0005 0.8750 0.0001
0.1406 0.0011 0.3906 0.0007 0.6406 0.0007 0.8906 0.0002
0.1563 0.0015 0.4063 0.0010 0.6563 0.0011 0.9063 0.0002
0.1719 0.0019 0.4219 0.0013 0.6719 0.0015 0.9219 0.0003
0.1875 0.0024 0.4375 0.0017 0.6875 0.0021 0.9375 0.0003
0.2031 0.0029 0.4531 0.0021 0.7031 0.0027 0.9531 0.0004
0.2188 0.0036 0.4688 0.0027 0.7188 0.0035 0.9688 0.0005
0.2344 0.0043 0.4844 0.0033 0.7344 0.0042 0.9844 0.0006
Figure 1 k ( t ) k\left(t) with respect to t t for Equations in (26) in Example 4.1 for t ∈ 0 , 1 4 t\in \left(0,\frac{1}{4}\right] , 1 4 , 1 2 \left(\frac{1}{4},\frac{1}{2}\right] , 1 2 , 3 4 \left(\frac{1}{2},\frac{3}{4}\right] , 3 4 , 1 \left(\frac{3}{4},1\right) , respectively, according to Table 4.
Figure 1

k ( t ) with respect to t for Equations in (26) in Example 4.1 for t ∈ 0 , 1 4 , 1 4 , 1 2 , 1 2 , 3 4 , 3 4 , 1 , respectively, according to Table 4.

In the next example we consider the discontinuous map Ω ( t , . , . , . , . ) at points of a subset of J of measure zero. Then, we obtain solutions of the problem (1) under some different conditions in Theorem 3.3 when the map Ω ( t , . , . , . , . ) is discontinuous at t = 0 .

Example 4.2

Consider the singular fractional q-integro-differential problem

(27) D q 10 3 [ k ] ( t ) + 1 t 5 1 2 ∣ k ∣ 1 3 + 8 5 ∣ k ′ ∣ 2 5 + 1 10 ∣ D q 4 15 [ k ] ( t ) ∣ 3 4 + 15 6 ∣ z k ( t ) ∣ 7 9 + 3 2 t 2 + Γ q 4 3 1 1 + k 2 ( t ) + 1 2 + [ k ′ ( t ) ] 2 + 1 1 + ( D q 4 15 [ k ] ( t ) ) 2 + 1 2 + [ z k ( t ) ] 2 = 0 ,

for t ∈ J and q ∈ ( 0 , 1 ) , with boundary conditions k ( 0 ) = 0 and

k ( 1 ) = D q 6 11 [ k ] 5 8 .

It is clear that σ = 10 3 ≥ 2 , ζ = 4 15 ∈ ( 0 , 1 ) , η = 6 11 ∈ ( 0 , 1 ) , τ = 5 8 ∈ ( 0 , 1 ) and

z k ( t ) = ∫ 0 t f ( r ) k ( r ) d r .

Put g ( t ) = 1 t 5 and

Θ 1 ( k 1 , k 2 , k 3 , k 4 ) = ∑ i = 1 4 β i ∣ k i ∣ p i , Θ 2 ( k 1 , k 2 , k 3 , k 4 ) = ∑ i = 1 4 3 ( t 2 + 5 ) 2 + k i 2 ,

for t ∈ J . Hence, we get m = ‖ g ( t ) ‖ 1 = 5 4 ,

lim k → ∞ Θ 1 ( k , k , k , k ) k = 0 ,

ℓ = lim k → ∞ Θ 2 ( k , k , k , k ) = 6 1 + Γ q 4 3 < ∞

and

Ω ( t , k 1 , k 2 , k 3 , k 4 ) ≤ 1 t 5 Θ 1 ( k 1 , k 2 , k 3 , k 4 ) + 6 1 + Γ q 4 3 .

One can see that in Problem (27) γ = 1 5 ∈ ( 0 , 1 ) , β 1 = 1 2 , β 2 = 8 5 , β 3 = 1 10 , β 4 = 15 6 ∈ [ 0 , ∞ ) , p 1 = 1 3 , p 2 = 2 5 , p 3 = 3 4 , p 4 = 7 9 ∈ [ 0 , 1 ) . At first by using Eqs (16) and (19), we obtain

λ = 1 − τ 1 − η Γ q ( 2 − η ) = 1 − 5 8 5 11 Γ q 5 11 ≅ 0.1644 , q = 1 8 , − 0.1222 , q = 1 2 , − 0.1074 , q = 9 13 ,

M 1 ≅ 11.5855 , 11.1921, 11.1201; M 2 ≅ 11.6995 , 11.5296, 11.5264; ε 0 ≅ 0.0855 , 0.0867, 0.0868; 1 Γ q ( 5 11 ) = 1.0224 , 1.0605, 1.0743 and

Λ 0 = max { M 1 , M 2 } = 11.6995 , 11.5206 , 11.5264 ,

for q = 1 8 , 1 2 , 9 13 , respectively, which are shown in Tables 5, 6 and 7. Note that the value of r must be more than

r 0 max 1 , 1 Γ q ( 2 − ζ ) , m = 1.2500 ,

for q ∈ ( 0 , 1 ) according to Tables 5, 6 and 7. These results are obtained by Algorithm 13. Now, for showing the numerical results, we consider the problem (27) as follows (Figure 2):

D q 10 3 [ k ] ( t ) + 1 t 5 1 2 ∣ k ∣ 1 3 + 8 5 ∣ k ′ ∣ 2 5 + 1 10 ∣ D q 4 15 [ k ] ( t ) ∣ 3 4 + 15 6 ∣ z k ( t ) ∣ 7 9 + 3 2 t 2 + Γ q 4 3 1 1 + k 2 ( t ) + 1 2 + [ k ′ ( t ) ] 2 + 1 1 + ( D q 4 15 [ k ] ( t ) ) 2 + 1 2 + [ z k ( t ) ] 2 ≤ D q 10 3 [ k ] ( t ) + 1 2 ∣ k ∣ 1 3 + 8 5 ∣ k ′ ∣ 2 5 + 1 10 ∣ D q 4 15 [ k ] ( t ) ∣ 3 4 + 15 6 ∣ z k ( t ) ∣ 7 9 + 6 t 2 + Γ q 4 3 = 0 .

Thus,

(28) D q 10 3 [ k ] ( t ) + 8 5 ∣ k ′ ∣ 2 5 + 1 10 ∣ D q 4 15 [ k ] ( t ) ∣ 3 4 + 1 2 ∣ k ∣ 1 3 = − 15 6 ∣ z k ( t ) ∣ 7 9 − 6 t 2 + Γ q 4 3 .

Table 8 shows numerical values of k ( t ) in equation (27). Furthermore, one can see that the curve of k ( t ) with respect to t in Table 8 (Algorithm 14)). We can see that Θ 1 , Θ 2 and g satisfy the conditions of Theorem 3.3. Thus, the problem (27) has a solution.

Table 5

Some numerical results of λ , M 1 , M 2 , ε 0 , 1 Γ q ( 2 − ζ ) in Example 4.2 for t ∈ J ¯ and q = 1 8

q = 1 8 n λ M 1 M 2 ε 0 1 Γ q ( 2 − ζ )
1 0.1655 11.4862 11.6001 0.0862 1.0206
2 0.1645 11.573 11.6869 0.0856 1.0221
3 0.1644 11.5839 11.6979 0.0855 1.0223
4 0.1644 11.5853 11.6993 0.0855 1.0224
5 0.1644 11.5855 11.6995 0.0855 1.0224
6 0.1644 11.5855 11.6995 0.0855 1.0224
Table 6

Some numerical results of λ , M 1 , M 2 , ε 0 , 1 Γ q ( 2 − ζ ) in Example 4.2 for t ∈ J ¯ and q = 9 13

q = 9 13 n λ M 1 M 2 ε 0 1 Γ q ( 2 − ζ )
1 0.2735 2.3829 2.6859 0.3723 0.7858
2 0.2180 3.6932 4.0335 0.2479 0.8780
3 0.1821 5.0443 5.4078 0.1849 0.9399
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
21 0.1075 11.1067 11.5129 0.0869 1.0741
22 0.1074 11.1108 11.5170 0.0868 1.0741
23 0.1074 11.1136 11.5199 0.0868 1.0742
24 0.1074 11.1156 11.5218 0.0868 1.0742
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
31 0.1074 11.1197 11.526 0.0868 1.0742
32 0.1074 11.1198 11.5261 0.0868 1.0743
33 0.1074 11.1200 11.5262 0.0868 1.0743
34 0.1074 11.1200 11.5262 0.0868 1.0743
35 0.1074 11.1200 11.5262 0.0868 1.0743
36 0.1074 11.1200 11.5263 0.0868 1.0743
37 0.1074 11.1200 11.5263 0.0868 1.0743
38 0.1074 11.1200 11.5263 0.0868 1.0743
39 0.1074 11.1201 11.5264 0.0868 1.0743
40 0.1074 11.1201 11.5264 0.0868 1.0743
Table 7

Some numerical results of λ , M 1 , M 2 , ε 0 , 1 Γ q ( 2 − ζ ) in Example 4.2 for t ∈ J ¯ and q = 1 2

q = 1 2 n λ M 1 M 2 ε 0 1 Γ q ( 2 − ζ )
1 0.1839 6.2473 6.5585 0.1525 0.9529
2 0.1524 8.2455 8.5706 0.1167 1.0072
3 0.1371 9.5643 9.8958 0.1011 1.034
⋮ ⋮ ⋮ ⋮ ⋮ ⋮
9 0.1224 11.1638 11.5011 0.0869 1.0601
10 0.1223 11.1779 11.5154 0.0868 1.0603
11 0.1222 11.185 11.5225 0.0868 1.0604
12 0.1222 11.1886 11.5260 0.0868 1.0604
13 0.1222 11.1903 11.5278 0.0867 1.0604
14 0.1222 11.1912 11.5286 0.0867 1.0605
15 0.1222 11.1916 11.5291 0.0867 1.0605
16 0.1222 11.1919 11.5294 0.0867 1.0605
17 0.1222 11.1920 11.5295 0.0867 1.0605
18 0.1222 11.1920 11.5295 0.0867 1.0605
19 0.1222 11.1920 11.5295 0.0867 1.0605
20 0.1222 11.1921 11.5296 0.0867 1.0605
21 0.1222 11.1921 11.5296 0.0867 1.0605
Figure 2 k ( t ) k\left(t) with respect to t t for Equations in (27) in Example 4.2 for q ∈ 1 8 , 1 2 , 9 13 q\in \left\{\frac{1}{8},\frac{1}{2},\frac{9}{13}\right\} , respectively, according to Table 8.
Figure 2

k ( t ) with respect to t for Equations in (27) in Example 4.2 for q ∈ 1 8 , 1 2 , 9 13 , respectively, according to Table 8.

Table 8

Some numerical results of k ( t ) in Example 4.2 for t ∈ J ¯ , q ∈ 1 8 , 1 2 , 9 13 and n = 1 , 2 , … , 10

q = 1 8 q = 1 2 q = 9 13
t k ( t ) t k ( t ) t k ( t )
( n = 1 )
 1 0 0 0 0 0 0
 1 0.0156 0 0.0156 0 0.0156 0
 1 0.0313 0 0.0313 0 0.0313 0
 1 0.0469 0 0.0469 0 0.0469 0
 1 0.0625 0.0001 0.0625 0.0001 0.0625 0.0001
 1 0.0781 0.0002 0.0781 0.0002 0.0781 0.0002
  ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
 1 0.9531 0.7127 0.9531 0.7175 0.9531 0.7632
 1 0.9688 0.7528 0.9688 0.7579 0.9688 0.806
 1 0.9844 0.7945 0.9844 0.7999 0.9844 0.8506
( n = 2 )
 2 0 0 0 0 0 0
 2 0.0156 0 0.0156 0 0.0156 0
 2 0.0313 0 0.0313 0 0.0313 0
 2 0.0469 0 0.0469 0 0.0469 0
 2 0.0625 0.0001 0.0625 0.0001 0.0625 0.0001
 2 0.0781 0.0002 0.0781 0.0002 0.0781 0.0002
  ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
 2 0.9688 0.7523 0.9688 0.7422 0.9688 0.7738
 2 0.9844 0.794 0.9844 0.7834 0.9844 0.8166
 2 1 0.8373 1 0.8261 1 0.8611
  ⋮
( n = 10 )
 10 0 0 0 0 0 0
 10 0.0156 0 0.0156 0 0.0156 0
 10 0.0313 0 0.0313 0 0.0313 0
 10 0.0469 0 0.0469 0 0.0469 0
 10 0.0625 0.0001 0.0625 0.0001 0.0625 0.0001
 10 0.0781 0.0002 0.0781 0.0002 0.0781 0.0002
  ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
 10 0.9531 0.7121 0.9531 0.6894 0.9531 0.6842
 10 0.9688 0.7522 0.9688 0.7282 0.9688 0.7228
 10 0.9844 0.7939 0.9844 0.7686 0.9844 0.7629
 10 1 0.8372 1 0.8106 1 0.8045

5 Conclusion

The q-integro-differential boundary equations and their applications represent a matter of high interest in the area of fractional q-calculus due to their various applications in areas of science and technology. Indeed, the q-integro-differential boundary value problems often occur in mathematical modeling of a variety of physical operations. In this context, we prove the existence of a solution for a new class of singular q-integro-differential equations (18) and (27) on a time scale. The results are verified by constructing two examples along with their numerical simulations that demonstrated perfect consistency with the theoretical findings. To this end, the authors investigated a complicated case by utilizing an appropriate basic theory which facilitates a particular interest in this paper.

Acknowledgements

The authors would like to thank the editors and the anonymous reviewers for their constructive comments and suggestions that have helped to improve the present paper. Sayyedeh Narges Hajiseyedazizi and Mohammad Esmael Samei were supported by Bu-Ali Sina University. Alzabut would like to thank Prince Sultan University for supporting this work.

  1. Author contributions: Authors declare that the study was realized in collaboration with equal responsibility. All authors read and approved the final manuscript.

  2. Conflict of interest: Authors declare that they have no competing interests.

  3. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2020-09-23
Revised: 2021-08-18
Accepted: 2021-08-19
Published Online: 2021-12-31

© 2021 Sayyedeh Narges Hajiseyedazizi et al., published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2021-0093
Keywords for this article
singularity; multi-step methods; q-integro-differential equation
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  1. Regular Articles
  2. Sharp conditions for the convergence of greedy expansions with prescribed coefficients
  3. Range-kernel weak orthogonality of some elementary operators
  4. Stability analysis for Selkov-Schnakenberg reaction-diffusion system
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  81. Symmetric pairs and pseudosymmetry of Θ-Yetter-Drinfeld categories for Hom-Hopf algebras
  82. A new characterization of the automorphism groups of Mathieu groups
  83. The role of w-tilting modules in relative Gorenstein (co)homology
  84. Primitive and decomposable elements in homology of ΩΣℂP∞
  85. The G-sequence shadowing property and G-equicontinuity of the inverse limit spaces under group action
  86. Classification of f-biharmonic submanifolds in Lorentz space forms
  87. Some new results on the weaving of K-g-frames in Hilbert spaces
  88. Matrix representation of a cross product and related curl-based differential operators in all space dimensions
  89. Global optimization and applications to a variational inequality problem
  90. Functional equations related to higher derivations in semiprime rings
  91. A partial order on transformation semigroups with restricted range that preserve double direction equivalence
  92. On multi-step methods for singular fractional q-integro-differential equations
  93. Compact perturbations of operators with property (t)
  94. Entire solutions for several complex partial differential-difference equations of Fermat type in ℂ2
  95. Random attractors for stochastic plate equations with memory in unbounded domains
  96. On the convergence of two-step modulus-based matrix splitting iteration method
  97. On the separation method in stochastic reconstruction problem
  98. Robust estimation for partial functional linear regression models based on FPCA and weighted composite quantile regression
  99. Structure of coincidence isometry groups
  100. Sharp function estimates and boundedness for Toeplitz-type operators associated with general fractional integral operators
  101. Oscillatory hyper-Hilbert transform on Wiener amalgam spaces
  102. Euler-type sums involving multiple harmonic sums and binomial coefficients
  103. Poly-falling factorial sequences and poly-rising factorial sequences
  104. Geometric approximations to transition densities of Jump-type Markov processes
  105. Multiple solutions for a quasilinear Choquard equation with critical nonlinearity
  106. Bifurcations and exact traveling wave solutions for the regularized Schamel equation
  107. Almost factorizable weakly type B semigroups
  108. The finite spectrum of Sturm-Liouville problems with n transmission conditions and quadratic eigenparameter-dependent boundary conditions
  109. Ground state sign-changing solutions for a class of quasilinear Schrödinger equations
  110. Epi-quasi normality
  111. Derivative and higher-order Cauchy integral formula of matrix functions
  112. Commutators of multilinear strongly singular integrals on nonhomogeneous metric measure spaces
  113. Solutions to a multi-phase model of sea ice growth
  114. Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms
  115. Bernstein-Walsh type inequalities for derivatives of algebraic polynomials in quasidisks
  116. Review Article
  117. Semiprimeness of semigroup algebras
  118. Special Issue on Problems, Methods and Applications of Nonlinear Analysis (Part II)
  119. Third-order differential equations with three-point boundary conditions
  120. Fractional calculus, zeta functions and Shannon entropy
  121. Uniqueness of positive solutions for boundary value problems associated with indefinite ϕ-Laplacian-type equations
  122. Synchronization of Caputo fractional neural networks with bounded time variable delays
  123. On quasilinear elliptic problems with finite or infinite potential wells
  124. Deterministic and random approximation by the combination of algebraic polynomials and trigonometric polynomials
  125. On a fractional Schrödinger-Poisson system with strong singularity
  126. Parabolic inequalities in Orlicz spaces with data in L1
  127. Special Issue on Evolution Equations, Theory and Applications (Part II)
  128. Impulsive Caputo-Fabrizio fractional differential equations in b-metric spaces
  129. Existence of a solution of Hilfer fractional hybrid problems via new Krasnoselskii-type fixed point theorems
  130. On a nonlinear system of Riemann-Liouville fractional differential equations with semi-coupled integro-multipoint boundary conditions
  131. Blow-up results of the positive solution for a class of degenerate parabolic equations
  132. Long time decay for 3D Navier-Stokes equations in Fourier-Lei-Lin spaces
  133. On the extinction problem for a p-Laplacian equation with a nonlinear gradient source
  134. General decay rate for a viscoelastic wave equation with distributed delay and Balakrishnan-Taylor damping
  135. On hyponormality on a weighted annulus
  136. Exponential stability of Timoshenko system in thermoelasticity of second sound with a memory and distributed delay term
  137. Convergence results on Picard-Krasnoselskii hybrid iterative process in CAT(0) spaces
  138. Special Issue on Boundary Value Problems and their Applications on Biosciences and Engineering (Part I)
  139. Marangoni convection in layers of water-based nanofluids under the effect of rotation
  140. A transient analysis to the M(τ)/M(τ)/k queue with time-dependent parameters
  141. Existence of random attractors and the upper semicontinuity for small random perturbations of 2D Navier-Stokes equations with linear damping
  142. Degenerate binomial and Poisson random variables associated with degenerate Lah-Bell polynomials
  143. Special Issue on Fractional Problems with Variable-Order or Variable Exponents (Part I)
  144. On the mixed fractional quantum and Hadamard derivatives for impulsive boundary value problems
  145. The Lp dual Minkowski problem about 0 < p < 1 and q > 0
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