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Article

A Study on Solutions for a Class of Higher-Order System of Singular Boundary Value Problem

1
Center of Data Sciences, Siksha ‘O’ Anusandhan, ITER College, Bhubaneswar 751030, Odisha, India
2
Department of Mathematics, Indian Institute of Technology Patna, Patna 801106, Bihar, India
3
Department of Mathematics, Texas A & M University-Kingsville, 700 University Blvd., MSC 172, Kingsville, TX 78363-8202, USA
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2023, 15(9), 1729; https://doi.org/10.3390/sym15091729
Submission received: 11 August 2023 / Revised: 1 September 2023 / Accepted: 5 September 2023 / Published: 8 September 2023
(This article belongs to the Special Issue Symmetry in Nonlinear and Convex Analysis)

Abstract

:
In this article, we propose a fourth-order non-self-adjoint system of singular boundary value problems (SBVPs), which arise in the theory of epitaxial growth by considering hte equation 1 r β r β 1 r β ( r β Θ ) = 1 2 r β K 11 μ Θ 2 + 2 μ Θ Θ + K 12 μ φ 2 + 2 μ φ φ + λ 1 G 1 ( r ) , 1 r β r β 1 r β ( r β φ ) = 1 2 r β K 21 μ Θ 2 + 2 μ Θ Θ + K 22 μ φ 2 + 2 μ φ φ + λ 2 G 2 ( r ) , where λ 1 0 and λ 2 0 are two parameters, μ = p r 2 β 2 , p R + , G 1 , G 2 L 1 [ 0 , 1 ] such that M 1 * G 1 ( r ) M 1 > 0 , M 2 * G 2 ( r ) M 2 > 0 and K 12 > 0 , K 11 0 , and K 21 > 0 , K 22 0 are constants that are connected by the relation ( K 12 + K 22 ) ( K 11 + K 21 ) and β > 1 . To study the governing equation, we consider three different types of homogeneous boundary conditions. We use the transformation t = r 1 + β 1 + β to deduce the second-order singular boundary value problem. Also, for β = p = G 1 ( r ) = G 2 ( r ) = 1 , it admits dual solutions. We show the existence of at least one solution in continuous space. We derive a sign of solutions. Furthermore, we compute the approximate bound of the parameters to point out the region of nonexistence. We also conclude bounds are symmetric with respect to two different transformations.

1. Introduction

In epitaxial growth, various methods are used to produce thin films [1], such as solar cells and microwaves from crystalline substrates under high vacuum. This process is very useful in the semiconductor industry. There are many types of epitaxial growth techniques, such as chemical beam epitaxy [2], hybrid vapor phase epitaxy [3], and molecular beam epitaxy [1]. Here, we focus on the model resulting from molecular beam epitaxy. We [4,5,6,7] consider the following equation
F t = 1 + ( F ) 2 1 2 δ I ( F ) δ F + η ( x , t ) ,
where [4,5,6,7] I ( F ) is given as:
I ( F ) = Ω R ( z ) 1 + ( F ) 2 1 2 d x ,
and F is the epitaxial growth function such that F : Ω R N × R + R . We choose N = 2 . Therefore, after similar calculation as in [4,5,6,7], we have
Δ 2 F = det ( D 2 F ) + λ H ( x ) , x Ω R 2 ,
where H ( x ) equivalent to η ( x , t ) is well known [8]. Equations (1) and (3) have a clear geometrical meaning [4,5,6,7]). We set the following two types of limits and analyze Equation (3):
The Dirichlet boundary condition:
F = 0 , F n = 0 on Ω ,
and Navier boundary condition:
F = 0 , Δ F = 0 on Ω .
Let r = | x | and F ( x ) = Θ ( | x | ) . Equation (3) is then simplified to form
1 r r 1 r ( r Θ ) = 1 r Θ Θ + λ H ( r ) ,
where = d d r . The differential Equation (6) is non-self-adjoint, nonlinear, and singular, and its solution is not known. So, it is difficult to study the solution. Verma et al. [9] analyzed the solutions of Equation (6) with a monotone iterative technique and computed the bounds of the growth parameter λ . Recently, they [10] developed the Adomian decomposition method and simulated the solution of (6). All these results and properties make Equation (6) very interesting and attractive for researchers. In [11], Verma et al. proposed a new system of SBVPs.
1 r r 1 r ( r Θ ) = 1 r K 11 Θ Θ + K 12 φ φ + λ 1 , 1 r r 1 r ( r φ ) = 1 r K 21 Θ Θ + K 22 φ φ + λ 2 ,
They [11] studied the existence of solutions as well as derived the bounds λ 1 and λ 2 . Here, we try to generalize Equation (7). In view of [12] and (7), we propose the system of fourth-order singular differential equation given by
1 r β r β 1 r β ( r β Θ ) = 1 2 r β K 11 μ Θ 2 + 2 μ Θ Θ + K 12 μ φ 2 + 2 μ φ φ + λ 1 G 1 , 1 r β r β 1 r β ( r β φ ) = 1 2 r β K 21 μ Θ 2 + 2 μ Θ Θ + K 22 μ φ 2 + 2 μ φ φ + λ 2 G 2 ,
where λ 1 0 and λ 2 0 are two parameters, μ ( r ) = p r 2 β 2 , p R + , G 1 ( r ) , G 2 ( r ) L 1 [ 0 , 1 ] such that M 1 * G 1 ( r ) M 1 > 0 , M 2 * G 2 ( r ) M 2 > 0 and K 12 > 0 , K 11 0 , and K 21 > 0 , K 22 0 are constants that are connected by the relation ( K 12 + K 22 ) ( K 11 + K 21 ) and β > 1 . It is trivial to show that Equations (7) and (8) are the same for β = 1 and G 1 = G 2 = 1 .
To analyze (8), we consider the following three types of homogeneous boundary conditions
lim r 0 r β Θ ( r ) = 0 , Θ ( 0 ) = 0 , Θ ( 1 ) = 0 , Θ ( 1 ) = 0 , lim r 0 r β φ ( r ) = 0 , φ ( 0 ) = 0 , φ ( 1 ) = 0 , φ ( 1 ) = 0 ,
and
lim r 0 r β Θ ( r ) = 0 , Θ ( 0 ) = 0 , Θ ( 1 ) = 0 , Θ ( 1 ) = 0 , lim r 0 r β φ ( r ) = 0 , φ ( 0 ) = 0 , φ ( 1 ) = 0 , φ ( 1 ) = 0 ,
and
lim r 0 r β Θ ( r ) = 0 , Θ ( 0 ) = 0 , Θ ( 1 ) = 0 , Θ ( 1 ) + Θ ( 1 ) = 0 , lim r 0 r β φ ( r ) = 0 , φ ( 0 ) = 0 , φ ( 1 ) = 0 , φ ( 1 ) + φ ( 1 ) = 0 .
The vanishing derivatives Θ ( 0 ) = 0 and φ ( 0 ) = 0 show that the origin is a point of maxima or minima. The fourth boundary condition, i.e., lim r 0 r β Θ ( r ) = 0 and lim r 0 r β φ ( r ) = 0 , assures a higher regularity at the origin. These conditions restrict β to be greater than 1. The conditions lim r 0 r β Θ ( r ) = 0 , lim r 0 r β φ ( r ) = 0 give rise to the following system.
r β 1 r β ( r β Θ ) = μ 2 K 11 ( Θ ) 2 + K 12 ( φ ) 2 + 0 r λ 1 t β G 1 ( t ) d t , r β 1 r β ( r β φ ) = μ 2 K 21 ( Θ ) 2 + K 22 ( φ ) 2 + 0 r λ 2 t β G 2 ( t ) d t .
We put x = r β Θ and y = r β φ . Therefore, from Equation (12), we have
r 2 x β r x = p K 11 x 2 + K 12 y 2 2 + r 2 0 r λ 1 t β G 1 ( t ) d t , r 2 y β r y = p K 21 x 2 + K 22 y 2 2 + r 2 0 r λ 2 t β G 2 ( t ) d t .
Therefore, the homogeneous boundary condition is (9)
x ( 0 ) = 0 = y ( 0 ) , x ( 1 ) = 0 = y ( 1 ) ,
Equation (11) becomes
x ( 0 ) = 0 = y ( 0 ) β 1 x ( 1 ) = x ( 1 ) , β 1 y ( 1 ) = y ( 1 ) ,
and Equation (10) leads to the following
x ( 0 ) = 0 = y ( 0 ) β x ( 1 ) = x ( 1 ) , β y ( 1 ) = y ( 1 ) .
Now, we set δ ˜ = 1 1 + β , t = r 1 + β δ ˜ , x ( r ) = Ψ ( t ) , y ( r ) = Φ ( t ) , and then (13) gives rise to
Ψ = K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , Φ = K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s ,
where β ˜ = 2 β 1 + β , L ( t ) = ( 1 + β ) 1 1 + β t 1 1 + β , and K i ( t , Ψ , Φ ) = p K i 1 Ψ 2 + K i 2 Φ 2 2 ( 1 + β ) 2 t 2 , i = 1 , 2 for t 0 , δ ˜ . Thus, (17) gives rise to the following boundary value problems (BVPs):
System 1 : Ψ = K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , Φ = K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , Ψ δ ˜ = 0 , lim t 0 + t β δ ˜ Φ ( t ) = 0 , Φ δ ˜ = 0 ,
System 2 : Ψ = K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , Φ = K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , ( β 1 ) Ψ δ ˜ = Ψ δ ˜ , lim t 0 + t β δ ˜ Φ ( t ) = 0 , ( β 1 ) Φ δ ˜ = Φ δ ˜ ,
and
System 3 : Ψ = K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , Φ = K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , β Ψ δ ˜ = Ψ δ ˜ , lim t 0 + t β δ ˜ Φ ( t ) = 0 , β Φ δ ˜ = Φ δ ˜ ,
for t 0 , δ ˜ . Therefore, we can obtain the following integral equation (IE):
  • Corresponding to (18), the IE is
    Ψ ( t ) = δ ˜ t 0 t s ( 1 + β ) K 1 ( s , Ψ , Φ ) d s t t δ ˜ ( 1 + β ) K 1 ( s , Ψ , Φ ) δ ˜ s d s δ ˜ t 0 t ( 1 + β ) γ ˜ s γ ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y d s t t δ ˜ ( 1 + β ) γ ˜ s β ˜ δ ˜ s 0 L ( s ) λ 1 y β G 1 ( y ) d y d s , 0 t δ ˜ , Φ ( t ) = δ ˜ t 0 t s ( 1 + β ) K 2 ( s , Ψ , Φ ) d s t t δ ˜ ( 1 + β ) K 2 ( s , Ψ , Φ ) δ ˜ s d s δ ˜ t 0 t ( 1 + β ) γ ˜ s γ ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y d s t t δ ˜ ( 1 + β ) γ ˜ s β ˜ δ ˜ s 0 L ( s ) λ 2 y β G 2 ( y ) d y d s , 0 t δ ˜ ,
  • Corresponding to (19), the IE is
    Ψ ( t ) = 1 + ( β 1 ) ( β + 1 ) 2 t 0 t s K 1 ( s , Ψ , Φ ) d s t t δ ˜ K 1 ( s , Ψ , Φ ) 1 + ( β 1 ) ( β + 1 ) 2 s d s 1 + ( β 1 ) ( β + 1 ) 2 t 0 t ( 1 + β ) β ˜ s γ ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y d s t t δ ˜ ( 1 + β ) β ˜ s β ˜ 1 + ( β 1 ) ( β + 1 ) 2 s 0 L ( s ) λ 1 y β G 1 ( y ) d y d s , 0 t δ ˜ , Φ ( t ) = 1 + ( β 1 ) ( β + 1 ) 2 t 0 t s K 2 ( s , Ψ , Φ ) d s t t δ ˜ K 2 ( s , Ψ , Φ ) 1 + ( β 1 ) ( β + 1 ) 2 s d s 1 + ( β 1 ) ( β + 1 ) 2 t 0 t ( 1 + β ) β ˜ s γ ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y d s t t δ ˜ ( 1 + β ) β ˜ s β ˜ 1 + ( β 1 ) ( β + 1 ) 2 s 0 L ( s ) λ 2 y β G 2 ( y ) d y d s , 0 t δ ˜ ,
  • Corresponding to (20), the IE is
    Ψ ( t ) = t + 1 ( β + 1 ) β 0 t s β ( 1 + β ) K 1 ( s , Ψ , Φ ) d s t t δ ˜ β ( 1 + β ) K 1 ( s , Ψ , Φ ) × s + 1 ( β + 1 ) β d s t + 1 ( 1 + β ) β 0 t β ( 1 + β ) γ ˜ s γ ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y d s t t δ ˜ β ( 1 + β ) γ ˜ s β ˜ s + 1 ( 1 + β ) β 0 L ( s ) λ 1 y α G 1 ( y ) d y d s , 0 t δ ˜ , Φ ( t ) = t + 1 ( β + 1 ) β 0 t s β ( 1 + β ) K 2 ( s , Ψ , Φ ) d s t t δ ˜ β ( 1 + β ) K 2 ( s , Ψ , Φ ) × s + 1 ( β + 1 ) β d s t + 1 ( 1 + β ) β 0 t β ( 1 + β ) γ ˜ s γ ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y d s t t δ ˜ β ( 1 + β ) γ ˜ s β ˜ s + 1 ( 1 + β ) β 0 L ( s ) λ 2 y β G 2 ( y ) d y d s , 0 t δ ˜ ,
    where γ ˜ = 1 β 1 + β . Now, we define
C l o c 2 0 , δ ˜ ; R = Ψ : 0 , δ ˜ R | Ψ C 2 [ a , b ] , R for   each   interval   [ a , b ] 0 , δ ˜ .
We are looking for a system of solutions ( Ψ , Φ ) E × E that satisfy (17), where E × E = C l o c 2 0 , 1 2 , R × C l o c 2 0 , 1 2 , R . Again, we denote X ˜ = C 0 , 1 2 , R .
These results [11] inspired us to consider (8) for β > 1 subject to boundary conditions (9)–(11) and analyze them. The differential Equation (8) is non-self-adjoint, nonlinear, and singular, and its solution is not known. So, it is difficult to study the solution. In this paper, we prove there is at least one solution in E × E . We derive some results to establish the sign of solutions. We also estimate of the parameters λ 1 and λ 2 to show the nonexistence of solutions.
Many researchers such as Omari et al. [13,14,15,16,17,18], Pandey et al. [19,20], Xinguange et al. [21,22], Shah et al. [23], Li [24], Dunninger et al. [25,26], Cabada et al. [27,28,29,30,31], Zhang [32], Barilla et al. [33], etc., have studied the existence and nonexistence of solutions for a class of singular boundary value problems with the help of upper and lower solutions. Readers can refer to Reference [34] to learn more about the literature regarding the origin of singular BVPs and the various methodologies.
We have organized the rest of the paper as follows. We derive a sign of solutions in Section 2. We show the existence of solutions in Section 3. Section 4 is placed by the derivation of the region of the nonexistence of solutions. Finally, we conclude the work in Section 5.

2. Sign of Solutions

For non-negative values of the parameters λ 1 and λ 2 , here, we derive some properties of the system of solution ( Ψ , Φ ) E × E , which satisfy the inequality
Ψ K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , Φ K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s ,
for t 0 , δ ˜ and denotes d d t .
Lemma 1.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  lim t 0 + t β δ ˜ Φ ( t ) = 0  and  lim t 0 + t β δ ˜ Ψ ( t ) = 0  along with (21), then
lim t 0 + Ψ ( t ) = 0 & lim t 0 + Φ ( t ) = 0 .
Proof. 
The proof is similar as in Lemma 2.0.1 in References [11,35]. □
Lemma 2.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  Ψ δ ˜ = 0 ,  Φ δ ˜ = 0  and Inequality (21) along with (22), then
Ψ ( t ) 0 & Φ ( t ) 0 , t 0 , δ ˜ .
Proof. 
It follows from Lemma 2.0.2 in References [11,35]. □
Lemma 3.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  β Φ δ ˜ = Φ δ ˜ ,  β Ψ δ ˜ = Ψ δ ˜  and Inequality (21) along with (22), then  Ψ ( t ) 0 & Φ ( t ) 0   t 0 , δ ˜ .
Proof. 
It follows from Lemma 2.0.3 in References [11,35]. □
Lemma 4.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  ( β 1 ) Φ δ ˜ = Φ δ ˜ ,  ( β 1 ) Ψ δ ˜ = Ψ δ ˜ , Inequality (21) and Equation (22), then  Ψ ( t ) 0  and  Φ ( t ) 0   t 0 , δ ˜ .
Proof. 
It follows from Lemma 3. □
Corollary 1.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  Ψ ( t ) 0 ,  Φ ( t ) 0  and Inequality (21) along with (22), then  lim t 0 + t β δ ˜ Ψ ( t ) = 0  and  lim t 0 + t β δ ˜ Φ ( t ) = 0  if and only if  lim t 0 + Ψ ( t ) t δ ˜ = 0  and  lim t 0 + Φ ( t ) t δ ˜ = 0 .
Proof. 
It follows from Corollary 2.0.3 in References [11,35]. □
Lemma 5.
Assume the system of solutions  ( Ψ , Φ ) E × E  of (17) satisfies the conditions  lim t 0 + Ψ ( t ) t δ ˜ = 0  and  lim t 0 + Φ ( t ) t δ ˜ = 0 . Then, for every  μ [ 0 , 1 ) , we have
lim t 0 + t 1 μ t δ ˜ Ψ 2 s 2 d s = 0 & lim t 0 + t 1 μ t δ ˜ Φ 2 s 2 d s = 0 .
Proof. 
It follows from Lemma 2.0.5 in References [11,35]. □
Lemma 6.
If the system of solutions of (18) is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies IE (21) along with the conditions
lim t 0 + | Ψ ( t ) | t < + & lim t 0 + | Φ ( t ) | t < + .
Proof. 
Assume, ( Ψ , Φ ) E × E is the solutions of (18). Thus, Green’s function of (18) is ( 1 + β ) s δ ˜ t when 0 s t and ( 1 + β ) t δ ˜ s for all t s δ ˜ . So, from (18), we can have (21). Conversely, let ( Ψ , Φ ) X ˜ × X ˜ satisfy (21). Therefore, we have (18) and ( Ψ , Φ ) E × E . Now, we take F 1 ( t ) = Ψ 2 t δ ˜   and F 2 ( t ) = 1 t β δ ˜ , t 0 , δ ˜ . Since lim t 0 + Ψ ( t ) t δ ˜ = 0 and lim t 0 + Ψ ( t ) = 0 , we have F 1 L 0 , δ ˜ . Hence, F 1 F 2 = Ψ 2 t ( 1 + β ) δ ˜ = Ψ 2 t L 0 , δ ˜ i.e.,
0 δ ˜ Ψ 2 s d s < .
Again, we take F 3 ( t ) = F 1 F 2 × 1 t μ = Ψ 2 t 1 + μ   and F 4 ( t ) = 1 t 1 μ , t 0 , δ ˜ , μ ( 0 , 1 ) . So, we have F 3 L 0 , δ ˜ and F 4 L 0 , δ ˜ . Consequently, F 3 F 4 L 0 , δ ˜ subject to the condition μ ( 0 , 1 ) , i.e.,
0 δ ˜ Ψ 2 s 2 d s < .
By a similar analysis, we have
0 δ ˜ Φ 2 s d s < and 0 δ ˜ Φ 2 s 2 d s < .
Now,
lim t 0 + | Ψ ( t ) | t = 0 δ ˜ ( 1 + β ) K 1 ( s , Ψ , Φ ) δ ˜ s d s + 0 δ ˜ ( 1 + β ) γ ˜ s β ˜ δ ˜ s 0 L ( s ) λ 1 y β G 1 d y d s , lim t 0 + | Φ ( t ) | t = 0 δ ˜ ( 1 + β ) K 2 ( s , Ψ , Φ ) δ ˜ s d s + 0 δ ˜ ( 1 + β ) γ ˜ s β ˜ δ ˜ s 0 δ ˜ λ 2 y β G 2 d y d s .
After substituting the values of K 1 ( s , Ψ , Φ ) and K 2 ( s , Ψ , Φ ) and using Inequalities (26)–(28), we have (25). □
The following proof of two lemmas follows from Lemma 6.
Lemma 7.
The system of solutions of (19) is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies IE (21) along with Condition (25).
Lemma 8.
The system of solutions of (20) is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies IE (21) along with Condition (25).

3. Results on Existence

Here, we go through the following lemmas to prove the main results. Now, for k , m R , Equations (18)–(20) lead to
System 1 a : Ψ + k Ψ = H 1 ( t , Ψ , Φ ) , Φ + m Φ = H 2 ( t , Ψ , Φ ) , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , Ψ δ ˜ = 0 , lim t 0 + t β δ ˜ Φ ( t ) = 0 , Φ δ ˜ = 0 ,
System 2 a : Ψ + k Ψ = H 1 ( t , Ψ , Φ ) , Φ + m Φ = H 2 ( t , Ψ , Φ ) , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , ( β 1 ) Ψ δ ˜ = Ψ δ ˜ , lim t 0 + t β δ ˜ Φ ( t ) = 0 , ( β 1 ) Φ δ ˜ = Φ δ ˜ ,
and
System 3 a : Ψ + k Ψ = H 1 ( t , Ψ , Φ ) , Φ + m Φ = H 2 ( t , Ψ , Φ ) , lim t 0 + t β δ ˜ Ψ ( t ) = 0 , β Ψ δ ˜ = Ψ δ ˜ , lim t 0 + t β δ ˜ Φ ( t ) = 0 , β Φ δ ˜ = Φ δ ˜ ,
for 0 < t δ ˜ such that
H 1 ( t , Ψ , Φ ) = K 1 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s + k Ψ , H 2 ( t , Ψ , Φ ) = K 2 ( t , Ψ , Φ ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s + m Φ ,
Now, we define
  • H 0 = { k < 0 , m < 0 } ;
  • H 1 * = L < 0 , | L | cosh | L | δ ˜ ( β 1 ) sinh | L | δ ˜ > 0 for   L = k , m ;
  • H 2 * = L < 0 , | L | cosh | L | δ ˜ β sinh | L | δ ˜ > 0 for   L = k , m .
By using Lemma 3.0 . 1 in [11], we can prove the following lemmas.
Lemma 9.
Let  H 0  be true. The solution of System 1 a  is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies the following IE
Ψ ( t ) = 0 δ ˜ H 1 ( s , Ψ , Φ ) G D [ k ] ( s , t ) d s & Φ ( t ) = 0 δ ˜ H 2 ( s , Ψ , Φ ) G D [ m ] ( s , t ) d s
for  0 t δ ˜ . Also, Greens functions  G D [ L ] ( s , t )  are nonpositive for   L = m , k  over the domain  0 s δ ˜  and  0 t δ ˜ .
Lemma 10.
Let  H 1 *  be true. The solution of System  2 a  is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies the following IE
Ψ ( t ) = 0 δ ˜ H 1 ( s , Ψ , Φ ) G N 1 [ k ] ( s , t ) d s & Φ ( t ) = 0 δ ˜ H 2 ( s , Ψ , Φ ) G N 1 [ m ] ( s , t ) d s
for  0 t δ ˜ AlsoGreens functions  G N 1 [ L ] ( s , t )  are nonpositive for  L = m , k  over the domain  0 s δ ˜  and  0 t δ ˜ .
Lemma 11.
Let  H 2 *  be true. The solution of System  3 a  is  ( Ψ , Φ ) E × E  if and only if it lies in  X ˜ × X ˜  and satisfies the following IE
Ψ ( t ) = 0 δ ˜ H 1 ( s , Ψ , Φ ) G N 2 [ k ] ( s , t ) d s & Φ ( t ) = 0 δ ˜ H 2 ( s , Ψ , Φ ) G N 2 [ m ] ( s , t ) d s
for  0 t δ ˜ .  Also, Greens functions  G N 2 [ L ] ( s , t )  are nonpositive for  L = m , k  over the domain  0 s δ ˜  and  0 t δ ˜ .
Now, we define ( u 0 , v 0 ) and ( α 0 , β 0 ) such that u 0 α 0 and v 0 β 0 for Equations (18)–(20).
Definition 1
([36]). Pair functions  ( α 0 , β 0 ) E × E  are called the upper solutions of Equation (18)–(20)) if
α 0 K 1 t , α 0 , β 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , β 0 K 2 t , α 0 , β 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s ,
for all  t 0 , δ ˜  with  lim t 0 + β 0 ( t ) t δ ˜ = 0 , lim t 0 + α 0 ( t ) t δ ˜ = 0 , α 0 δ ˜ 0  and  β 0 δ ˜ 0 (respectively, ( β 1 ) α 0 δ ˜ α 0 δ ˜ , ( β 1 ) β 0 δ ˜ β 0 δ ˜  and  β α 0 δ ˜ α 0 δ ˜ ), β β 0 δ ˜ β 0 δ ˜ ).
Definition 2
([36]). Pair functions  ( u 0 , v 0 ) E × E  are called the lower solutions of Equation (18) (respectively, (19) and (20)) if
u 0 K 1 t , u 0 , v 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , v 0 K 2 t , u 0 , v 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s ,
for all  t 0 , δ ˜  such that  lim t 0 + v 0 ( t ) t δ ˜ = 0 , lim t 0 + u 0 ( t ) t δ ˜ = 0 , u 0 δ ˜ 0  and  v 0 δ ˜ 0 (respectively, ( β 1 ) u 0 δ ˜ ( β + 1 ) u 0 δ ˜ , ( β 1 ) v 0 δ ˜ v 0 δ ˜  and  β u 0 δ ˜ ( β + 1 ) u 0 δ ˜ ), β v 0 δ ˜ v 0 δ ˜ ).
We consider the Banach space X ˜ such that
max t 0 , δ ˜ | Ψ ( t ) | = Ψ .
In view of (37), we consider X ˜ × X ˜ such that
Ψ + Φ = ( Ψ , Φ ) 1 .
Now, we define T D : X ˜ × X ˜ X ˜ × X ˜ corresponding to Equation (29), which is given by
T D ( Ψ , Φ ) ( t ) = X D ( Ψ , Φ ) ( t ) , Y D ( Ψ , Φ ) ( t ) ,
where X D ( Ψ , Φ ) ( t ) , Y D ( Ψ , Φ ) ( t ) : X ˜ × X ˜ X ˜ are the operators defined by
X D ( Ψ , Φ ) ( t ) = 0 δ ˜ H 1 ( s , Ψ , Φ ) G D [ k ] ( s , t ) d s & Y D ( Ψ , Φ ) ( t ) = 0 δ ˜ H 2 ( s , Ψ , Φ ) G D [ m ] ( s , t ) d s
for all 0 t δ ˜ . By a similar manner, we can define T N 1 , T N 2 , X N 1 , X N 2 , Y N 1 , and Y N 2 corresponding to Equations (30) and (31), respectively. If we show T D has a fixed point ( Ψ , Φ ) X ˜ × X ˜ , then by using Lemma 9, we obtain the system of solutions ( Ψ , Φ ) for some values of k and m. Hence, we have ( Ψ , Φ ) is the system of solutions of Equation (18)
Lemma 12.
Let  H 0  be true, then the operator  T D : X ˜ × X ˜ X ˜ × X ˜  is completely continuous.
Proof. 
Let us consider ( ζ n , n ) as any sequence in X ˜ × X ˜ that converge to the pair of functions ( ζ , ) in X ˜ × X ˜ . Therefore, we have lim n ( ζ n , n ) = ( ζ , ) . So, there exists m 1 N such that
( ζ n , n ) ( ζ , ) 1 0   for   all   n m 1 .
Now, in view of the definition of the norm, it is easy to show that ζ n ζ   &   n   as   n . Now,
X D ( ζ n , n ( t ) X D ( ζ , ) ( t )
= 0 δ ˜ H 1 ( s , ζ n , n ) G D [ k ] ( s , t ) d s 0 δ ˜ H 1 ( s , ζ , ) G D [ k ] ( s , t ) d s
= 0 δ ˜ G D [ k ] ( s , t ) p K 11 ( ζ n 2 ζ 2 ) 2 ( 1 + β ) 2 s 2 + p K 12 ( n 2 2 ) 2 ( 1 + β ) 2 s 2 + k ( ζ n ζ ) d s .
By using Lemmas 1 and 5, we have
X D ( ζ n , n ) ( t ) X D ( ζ , ) ( t ) 0   as   n .
Similarly,
Y D ( ζ n , n ) ( t ) Y D ( ζ , ) ( t ) 0   as   n .
Therefore,
T D ( ζ n , n ) ( t ) T D ( ζ , ) ( t ) 1 0   as   n .
It follows that the operator T D is continuous. Now, we consider any bounded subset B of X ˜ × X ˜ such that { ( ζ , ) X ˜ × X ˜ : ( ζ , ) 1 m 2 } . Now,
X D ( ζ , ) ( t ) = max t 0 , δ ˜ 0 δ ˜ H 1 ( s , ζ , ) G D [ k ] ( s , t ) d s max ( t , s , ζ , ) 0 , δ ˜ × 0 , δ ˜ × [ 0 , m 2 ] × [ 0 , m 2 ] G D [ k ] ( s , t ) δ ˜ K 1 s , ζ , + ( 1 + β ) β ˜ s β ˜ × 0 L ( s ) λ 1 y β G 1 ( y ) d y + k ζ , < .
Hence, we have X D ( Ψ , Φ ) ( t ) as uniformly bounded. Similarly, we can show Y D ( Ψ , Φ ) ( t ) and T D is uniformly bounded. Now,
X D ( Ψ , Φ ) ( t 1 ) X D ( Ψ , Φ ) ( t 2 )
= 0 δ ˜ H 1 ( s , Ψ , Φ ) G D [ k ] ( s , t 1 ) d s 0 δ ˜ H 1 ( s , Ψ , Φ ) G D [ k ] ( s , t 2 ) d s ,
Since Green’s function is continuous, we have
X D ( Ψ , Φ ) ( t 1 ) X D ( Ψ , Φ ) ( t 2 ) 0   as   t 1 t 2 .
Similarly,
Y D ( Ψ , Φ ) ( t 1 ) Y D ( Ψ , Φ ) ( t 2 ) 0   as   t 1 t 2 .
So, T D is equicontinuous. It follows that T D : X ˜ × X ˜ X ˜ × X ˜ is completely continuous. Similarly, we can show X D ( Ψ , Φ ) ( t ) , Y D ( Ψ , Φ ) ( t ) : X ˜ × X ˜ X ˜ are completely continuous. Hence, the fixed point of T D lies in X ˜ × X ˜ such that
( Ψ , Φ ) ( t ) = T D ( Ψ , Φ ) ( t ) = X D ( Ψ , Φ ) ( t ) , Y D ( Ψ , Φ ) ( t ) , ( Ψ , Φ ) ( t ) X ˜ × X ˜ .
Hence, the proof is complete. □
Similarly, we prove that T N 1 , T N 2 : X ˜ × X ˜ X ˜ × X ˜ and X N 1 , X N 2 , Y N 1 , Y N 2 : X ˜ × X ˜ X ˜ are also completely continuous. We consider
Ω 1 = Ψ E : u 0 ( t ) Ψ ( t ) α 0 ( t ) , t 0 , δ ˜ ,
Ω 2 = Φ E : v 0 ( t ) Φ ( t ) β 0 ( t ) , t 0 , δ ˜ ,
and
Ω = ( Ψ , Φ ) E × E : ( u 0 , v 0 ) ( t ) ( Ψ , Φ ) ( t ) ( α 0 , β 0 ) ( t ) , t 0 , δ ˜ ,
such that u 0 Ψ α 0 and v 0 Φ β 0 . Again, we assume the following conditions:
lim t 0 + α 0 ( t ) = 0 , lim t 0 + | α 0 ( t ) | t < , α 0 ( t ) 0 ,   for   t 0 , δ ˜ , lim t 0 + β 0 ( t ) = 0 , lim t 0 + | β 0 ( t ) | t < , β 0 ( t ) 0 ,   for   t 0 , δ ˜ , lim t 0 + u 0 ( t ) = 0 , lim t 0 + | u 0 ( t ) | t < , lim t 0 + v 0 ( t ) = 0 , lim t 0 + | v 0 ( t ) | t < .
Lemma 13.
Consider an operator  T D Ω : Ω X ˜ × X ˜ X ˜ × X ˜  by
T D Ω ( Ψ , Φ ) ( t ) = X D Ω ( Ψ , Φ ) ( t ) , Y D Ω ( Ψ , Φ ) ( t ) ,
where  X D Ω ( Ψ , Φ ) ( t )  and  Y D Ω ( Ψ , Φ ) ( t )  are defined by (40) such that m and k are negative real numbers. Then,
i. 
X D Ω ( u 0 , v 0 ) ( t ) , Y D Ω ( u 0 , v 0 ) ( t ) = T D Ω ( u 0 , v 0 ) ( t ) ( u 0 , v 0 ) ( t ) ,
X D Ω ( α 0 , β 0 ) ( t ) , Y D Ω ( α 0 , β 0 ) ( t ) = T D Ω ( α 0 , β 0 ) ( t ) ( α 0 , β 0 ) ( t ) ,
where  ( u 0 , v 0 )  and  ( α 0 , β 0 )  are given by (35) and (36).
ii. 
and  T D Ω ( Ω ) Ω .
Proof. 
From (36), we have
u 0 K 1 ( t , u 0 , v 0 ) + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s ,
for t 0 , δ ˜ . Equivalently, there exists a non-negative function f ( t ) L 1 0 , δ ˜ such that
u 0 + k u 0 = K 1 t , u 0 , v 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s + k u 0 + f ( t ) ,
for t 0 , δ ˜ . By using Lemma 9, we obtain
u 0 = 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 + f ( s ) d s ,
for t 0 , δ ˜ . Since G D [ k ] ( s , t ) is nonpositive for all 0 t δ ˜ and 0 s δ ˜ , we have
u 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s , = X D Ω ( u 0 , v 0 ) ( t ) ,   for   t 0 , δ ˜ .
By a similar manner, we can obtain v 0 ( t ) Y D Ω ( u 0 , v 0 ) ( t )   for   all   t 0 , δ ˜ . Again, from (35), there exists a non-negative function f ( t ) L 1 0 , δ ˜ such that
α 0 + k α 0 = K 1 t , α 0 , β 0 + ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s + k α 0 f ( t ) ,
for all t 0 , δ ˜ . Therefore, by a similar analysis, we have
α 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , α 0 , β 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k α 0 d s , = X D Ω ( α 0 , β 0 ) ( t ) ,   for   all   t 0 , δ ˜ .
Similarly, we have β 0 ( t ) Y D Ω ( α 0 , β 0 ) ( t ) ,   for   t 0 , δ ˜ . Hence, we have
X D Ω ( u 0 , v 0 ) ( t ) , Y D Ω ( u 0 , v 0 ) ( t ) = T D Ω ( u 0 , v 0 ) ( t ) ( u 0 , v 0 ) ( t ) ,
X D Ω ( α 0 , β 0 ) ( t ) , Y D Ω ( α 0 , β 0 ) ( t ) = T D Ω ( α 0 , β 0 ) ( t ) ( α 0 , β 0 ) ( t ) ,
Here, we choose an element ( Ψ , Φ ) in Ω . Since u 0 Ψ 0 and v 0 Φ 0 , we have
u 0 2 Ψ 2 & v 0 2 Φ 2 .
So, we have
X D Ω ( Ψ , Φ ) ( t ) = 0 δ ˜ G D [ k ] ( s , t ) K 1 s , Ψ , Φ + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k Ψ d s , 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s u 0 ( t ) ,   for   all   0 < t δ ˜ .
Similarly, we can have Y D Ω ( Ψ , Φ ) ( t ) v 0 ( t ) . Again, since Ψ α 0 0 and Φ β 0 0 , we have
Ψ 2 α 0 2 & Φ 2 β 0 2 .
By using (61), we have X D Ω ( Ψ , Φ ) ( t ) α 0 ( t ) and Y D Ω ( Ψ , Φ ) ( t ) β 0 ( t ) . Hence, we have
( u 0 , v 0 ) ( t ) T D Ω ( Ψ , Φ ) ( t ) ( α 0 , β 0 ) ( t ) , ( Ψ , Φ ) Ω .
Hence, T D Ω ( Ω ) Ω . The proof is complete. □
Again, Ω is a subset of X ˜ × X ˜ . So, by using Lemmas 12 and 13 and Assumption (53), we can define T D Ω as
( Ψ , Φ ) ( t ) = T D Ω ( Ψ , Φ ) ( t ) = ( X D Ω ( Ψ , Φ ) ( t ) , Y D Ω ( Ψ , Φ ) ( t ) ) , ( Ψ , Φ ) Ω .
In similar manner, we conclude that
T N 1 Ω , T N 2 Ω : Ω Ω , X D Ω , X N 1 Ω , X N 2 Ω : Ω Ω 1 , Y D Ω , Y N 1 Ω , Y N 2 Ω : Ω Ω 2 ,
are also completely continuous. Now, we define two sequences { ( u n , v n ) } and { ( α n , β n ) } in the region Ω as follows:
T D Ω ( u n , v n ) ( t ) = ( u n + 1 , v n + 1 ) ( t ) = ( X D Ω ( u n , v n ) ( t ) , Y D Ω ( u n , v n ) ( t ) ) ,
T D Ω ( α n , β n ) ( t ) = ( α n + 1 , β n + 1 ) ( t ) = ( X D Ω ( α n , β n ) ( t ) , Y D Ω ( α n , β n ) ( t ) ) ,
where n = 0 , 1 , , ( u 0 , v 0 ) and ( α 0 , β 0 ) are defined by (58) and (59), respectively, i.e.,
u 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s , v 0 0 δ ˜ G D [ m ] ( s , t ) K 2 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y + m v 0 d s ,
and
α 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , α 0 , β 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k α 0 d s , β 0 0 δ ˜ G D [ m ] ( s , t ) K 2 s , α 0 , β 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y + m β 0 d s ,
for all t 0 , δ ˜ .
Theorem 1.
Assume  H 0  holds (respectively,  H 1  and  H 2 ) and  λ 1 , λ 2 R . Let  ( u 0 , v 0 )  and  ( α 0 , β 0 ) E × E  be the lower and upper system of solutions of System 1 (respectively, System 2 and System 3) that satisfy the properties (53) such that  ( u 0 , v 0 ) ( α 0 , β 0 ) . Then, the sequences  { ( u n , v n ) }  and  { ( α n , β n ) }  defined by (64) and (65), in a given order, converge uniformly to  ( Ψ * , Φ * )  and  ( Ψ * , Φ * ) , which are the solutions of (63) in Ω such that
( Ψ * , Φ * ) ( t ) ( Ψ , Φ ) ( t ) ( Ψ * , Φ * ) ( t ) , t 0 , δ ˜ .
Consequently,  ( Ψ * , Φ * ) ,  ( Ψ * , Φ * ) E × E  are also the system of solutions of System 1 (respectively, System 2 and System 3).
Proof. 
We claim that
( u n , v n )   satisfy   ( 66 ) , ( u n + 1 , v n + 1 ) ( u n , v n )   and   ( α 0 , β 0 ) ( u n + 1 , v n + 1 ) n N .
In view of Lemma 13, we obtain ( u 0 , v 0 ) satisfy (66). From (64),
( u 1 , v 1 ) = T D Ω ( u 0 , v 0 ) ( t ) = ( X D Ω ( u 0 , v 0 ) ( t ) , Y D Ω ( u 0 , v 0 ) ( t ) ) = 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s , 0 δ ˜ G D [ m ] ( s , t ) K 2 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y + m v 0 d s .
Now, in view of Account (66), we have
u 0 u 1 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s , = 0 .
Hence, we have u 0 u 1 . Similarly, we can prove v 0 v 1 . Now,
u 1 α 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u 0 , v 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u 0 d s 0 δ ˜ G D [ k ] ( s , t ) K 1 s , α 0 , β 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k α 0 d s .
Now,
u 0 α 0   &   v 0 β 0   implies   u 0 2 α 0 2   &   v 0 2 β 0 2 .
Since G D [ k ] ( s , t ) is nonpositive, by using (70), we can conclude u 1 α 0 0 . Similarly, we have v 1 β 0 0 and u 1 , v 1 ( t ) α 0 , β 0 ( t ) . Hence, Equation (69) is held for n = 0 . Let Equation (69) be held up to n = m ˜ . Again, from (64),
( u m ˜ + 1 , v m ˜ + 1 ) = T D Ω ( u m ˜ , v m ˜ ) ( t ) = ( X D Ω ( u m ˜ , v m ˜ ) ( t ) , Y D Ω ( u m ˜ , v m ˜ ) ( t ) ) = 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ , v m ˜ + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ d s , 0 δ ˜ G D [ m ] ( s , t ) K 2 s , u m ˜ , v m ˜ + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 2 y β G 2 ( y ) d y + m v m ˜ d s .
So, we have,
u m ˜ + 1 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ + 1 , v m ˜ + 1 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ + 1 d s ,
and
v m ˜ + 1 0 δ ˜ G D [ m ] ( s , t ) K 2 s , u m ˜ + 1 , v m ˜ + 1 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + m v m ˜ + 1 d s .
Now, by using (71), we have
u m ˜ + 1 u m ˜ + 2 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ + 1 , v m ˜ + 1 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ + 1 d s 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ + 1 , v m ˜ + 1 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ + 1 d s , = 0 .
Hence, u m ˜ + 1 is less than equal to u m ˜ + 2 . Similaly, we can prove v m ˜ + 1 is less than equal to v m ˜ + 2 . Now,
u m ˜ + 2 α 0 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ + 1 , v m ˜ + 1 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ + 1 d s 0 δ ˜ G D [ k ] ( s , t ) K 1 s , α 0 , β 0 + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k α 0 d s .
Now,
u m ˜ + 1 α 0 & v m ˜ + 1 β 0 implies u m ˜ + 1 2 α 0 2 & v m ˜ + 1 2 β 0 2 .
Since G D [ k ] ( s , t ) is nonpositive, by using (71), we can conclude u m ˜ + 2 α 0 0 . Similarly, v m ˜ + 2 β 0 0 . Thus, u m ˜ + 2 , v m ˜ + 2 ( t ) α 0 , β 0 ( t ) . Hence, Equation (69) is true for n = m ˜ + 1 . Finally, we conclude Equation (69) holds for all n N . By a similar analysis, we claim that
( α n , β n )   satisfy   ( 67 ) , ( α n + 1 , β n + 1 ) ( α n , β n )   and   ( α n + 1 , β n + 1 ) ( u 0 , v 0 ) n N .
To show that
( u n , v n ) ( α n , β n ) , n N .
We assume Equation (73) is true for n = m ˜ . Now,
u m ˜ + 1 α m ˜ + 1 0 δ ˜ G D [ k ] ( s , t ) K 1 s , u m ˜ , v m ˜ + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k u m ˜ d s 0 δ ˜ G D [ k ] ( s , t ) K 1 s , α m , β m + ( 1 + β ) β ˜ s β ˜ 0 L ( s ) λ 1 y β G 1 ( y ) d y + k α m ˜ d s .
Now,
u m ˜ α m ˜   &   v m ˜ β m ˜   implies   u m ˜ 2 α m ˜ 2   &   v m ˜ 2 β m ˜ 2 .
So, u m + 1 α m + 1 and v m + 1 β m + 1 implies that ( u m + 1 , v m + 1 ) ( α m + 1 , β m + 1 ) . Hence,
( α n , β n ) ( u n , v n ) , n N .
Thus, we obtain the following inequalities
( u 0 , v 0 ) ( u n , v n ) ( α n , β n ) ( α 0 , β 0 ) .
Therefore, the sequences { ( u n , v n ) } and { ( α n , β n ) } are monotone and bounded uniformly for all t 0 , δ ˜ . So, the sequences { ( u n , v n ) } and { ( α n , β n ) } converge uniformly to { ( Ψ * , Φ * ) } and { ( Ψ * , Φ * ) } in Ω , i.e.,
lim n ( u n , v n ) = ( Ψ * , Φ * )   &   lim n ( α n , β n ) = ( Ψ * , Φ * ) .
Hence, from (64) and (65), we obtain
( Ψ * , Φ * ) ( t ) = T D Ω ( Ψ * , Φ * ) ( t ) = ( X D Ω ( Ψ * , Φ * ) ( t ) , Y D Ω ( Ψ * , Φ * ) ) ( t ) , ( Ψ * , Φ * ) Ω , ( Ψ * , Φ * ) ( t ) = T D Ω ( Ψ * , Φ * ) ( t ) = ( X D Ω ( Ψ * , Φ * ) ( t ) , Y D Ω ( Ψ * , Φ * ) ( t ) ) , ( Ψ * , Φ * ) Ω .
Since Ω is a subset of E × E , so from Lemma 9, we show that ( Ψ * , Φ * ) , ( Ψ * , Φ * ) E × E are the solutions of System 1. This completes the proof. □

4. Bounds of λ 1   &   λ 2

In this section, we derive the bounds of λ 1 and λ 2 . From (17), we have
t Ψ Ψ = t K 1 t , Ψ , Φ + ( 1 + β ) β ˜ t γ ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , t Φ Φ = t K 2 t , Ψ , Φ + ( 1 + β ) β ˜ t γ ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s ,
for t 0 , δ ˜ . By using z ( t ) = Φ ( t ) t and w ( t ) = Ψ ( t ) t , we have
z ( t ) 0   &   w ( t ) 0 , t 0 , δ ˜ .
Integrating (78) from 0 to t, we have
t 2 w = 0 t s 3 K 1 s , w , z d s + 0 t ( 1 + β ) β ˜ y γ ˜ 0 L ( y ) λ 1 s β G 1 ( s ) d s d y ,
t 2 z = 0 t s 3 K 2 s , w , z d s + 0 t ( 1 + β ) β ˜ y γ ˜ 0 L ( y ) λ 2 s β G 2 ( s ) d s d y .
So, from (18), we deduce
w δ ˜ = 0   &   z δ ˜ = 0 ,
from (19), we derive
w δ ˜ = 1 2 w δ ˜   &   z δ ˜ = 1 2 z δ ˜ ,
and from (20), we obtain
w δ ˜ = w δ ˜   &   z δ ˜ = z δ ˜ .
Lemma 14.
Assume there exists a solution  u , v  of (17) satisfying  lim t 0 + u ( t ) t δ ˜ = 0 ,  lim t 0 + v ( t ) t δ ˜ = 0 ,  v ( t ) 0  and  u ( t ) 0 . Then,  λ 1 0  and  λ 2 0  satisfies the inequality
λ 1 M 1 + λ 2 M 2 2 ( 1 + β ) 4 ( 3 + β ) π 2 p ( K 11 + K 21 ) .
Proof. 
From Equations (80) and (81), we have
w + z = 1 t 2 0 t p K 11 + K 21 w 2 + K 12 + K 22 z 2 s 2 ( 1 + β ) 2 d s 1 t 2 0 t ( 1 + β ) β ˜ y γ ˜ 0 L ( y ) λ 1 G 1 + λ 2 G 2 s β d s d y .
Now, on account of Equation (79), G 1 ( y ) M 1 and G 2 ( y ) M 2 , from Equation (86), we have
( w + z ) p K 11 + K 21 w 2 + K 12 + K 22 z 2 4 ( 1 + β ) 2
1 t 2 0 t ( 1 + β ) β ˜ s γ ˜ 0 L ( s ) λ 1 M 1 + λ 2 M 2 y β d y d s ,
= p K 11 + K 21 w 2 + K 12 + K 22 z 2 4 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( 1 + β ) γ ˜ t γ ˜ ( 3 + β ) .
Therefore,
w + z p K 11 + K 21 w 2 + K 12 + K 22 z 2 4 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( 3 + β ) , p K 11 + K 21 ( w + z ) 2 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( 3 + β ) , t 0 , δ ˜ .
Now, we can have
w + z ( w + z ) 2 + 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( K 11 + K 21 ) ( 3 + β ) p p ( K 11 + K 21 ) 8 ( 1 + β ) 2 , t 0 , δ ˜ .
From Inequality (91), we obtain
t δ ˜ w + z ( w + z ) 2 + 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( K 11 + K 21 ) ( 3 + β ) p d s p ( K 11 + K 21 ) 8 ( 1 + β ) 2 δ ˜ t
for all t 0 , δ ˜ . Equivalently, we have
w δ ˜ + z δ ˜ w ( t ) + z ( t ) d x x 2 + 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( K 11 + K 21 ) ( 3 + β ) p p ( K 11 + K 21 ) 8 ( 1 + β ) 2 δ ˜ t
for all t 0 , δ ˜ . Thus
0 d x x 2 + 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( K 11 + K 21 ) ( 3 + β ) p lim t 0 + v δ ˜ v ( t ) d x x 2 + 8 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( K 11 + K 21 ) ( 3 + β ) p p ( K 11 + K 21 ) 8 ( 1 + β ) 3 .
Hence, from (94), we have
λ 1 M 1 + λ 2 M 2 2 ( 1 + β ) 4 ( 3 + β ) π 2 p ( K 11 + K 21 ) .
Hence, we obtain the results. □
Lemma 15.
If BVPs (18)–(20) are solvable for some pair  ( λ 1 c , λ 2 c ) ( 0 , 0 ) , where the constants  λ 2 c  and  λ 1 c  are the critical values of the parameters  λ 2  and  λ 1 , in a given order, then the solution of these BVPs also exists for every order pair  ( λ 1 , λ 2 )  such that  ( 0 , 0 ) ( λ 1 , λ 2 ) ( λ 1 c , λ 2 c ) .
Proof. 
It follows from Lemma 4.0.2 in [11]. □
We now upgrade bounds of the parameters λ 1 and λ 2 corresponding to the problems (18)–(20).
Lemma 16.
Assume there exists  ( w , z ) E × E  satisfying (80) and (81) such that  w ( t ) 0 ,  z ( t ) 0 ,  w δ ˜ = 0  and  z δ ˜ = 0 . Then
λ 1 M 1 + λ 2 M 2 12 ( 1 + β ) 4 ( β + 3 ) p ( K 11 + K 21 )
and
w ( t ) + z ( t ) c δ ˜ t , 0 < t δ ˜ .
Proof. 
We can easily see that
w ( t ) 0 & z ( t ) 0 , 0 < t δ ˜ .
Now,
w + z 1 t 2 0 t ( 1 + β ) β ˜ s γ ˜ 0 L ( s ) λ 1 M 1 + λ 2 M 2 y β d y d s ,
( λ 1 M 1 + λ 2 M 2 ) ( β + 3 ) , since   t δ ˜ .
So, we have
w + z λ 1 M 1 + λ 2 M 2 ( β + 3 ) , t 0 , δ ˜ .
Integrate (101) from t to δ ˜ , we have
w δ ˜ + z δ ˜ w ( t ) z ( t ) t δ ˜ ( λ 1 M 1 + λ 2 M 2 ) ( β + 3 ) d t .
Therefore, we have
w ( t ) + z ( t ) ( λ 1 M 1 + λ 2 M 2 ) ( β + 3 ) δ ˜ t , t 0 , δ ˜ .
Let
c 1 = ( λ 1 M 1 + λ 2 M 2 ) ( β + 3 )   and   w ( t ) + z ( t ) c n δ ˜ t , t 0 , δ ˜ .
So, by using (86) and (104), we have
w + z p ( K 11 + K 21 ) 4 ( 1 + β ) 2 t 2 0 t c n 2 δ ˜ s 2 s d s λ 1 M 1 + λ 2 M 2 ( β + 3 ) , = p c n 2 ( K 11 + K 21 ) 16 ( 1 + β ) 2 t 4 3 ( 1 + β ) 2 + 2 9 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( β + 3 ) , p c n 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 λ 1 M 1 + λ 2 M 2 ( β + 3 ) ,
since t 4 3 ( 1 + β ) 2 1 9 ( 1 + β ) 2 for all  0 < t δ ˜ . Thus
w ( t ) + z ( t ) c n + 1 δ ˜ t , t 0 , δ ˜ ,
where
c n + 1 = p c n 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
We claim that the sequence { c n } is convergent. Now, from (107), we obtain
c 2 = p c 1 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) c 1 .
We assume there exists n N such that c n c n 1 0 . From (107), we have
c n + 1 = p c n 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 )
p c n 1 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) = c n .
Therefore, by using (104), we can have the sequence { c n } converge to c. Thus, from (106), we have (97) such that
c = p c 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Hence, we obtain Result (96). □
Lemma 17.
Assume there exists  ( w , z ) E × E  satisfying (79), (80), and (84). Then
λ 1 M 1 + λ 2 M 2 12 ( 1 + β ) 4 ( β + 3 ) p ( K 11 + K 21 ) ( 6 β 2 + 16 β + 11 ) ,
and
w ( t ) + z ( t ) c ( β + 2 ) δ ˜ t , 0 < t δ ˜ .
Proof. 
From (84) and (102), we have
w δ ˜ z δ ˜ w ( t ) z ( t ) t δ ˜ λ 1 M 1 + λ 2 M 2 ( β + 3 ) d t
for all t 0 , δ ˜ . So, we have
w ( t ) z ( t ) λ 1 M 1 + λ 2 M 2 ( β + 3 ) δ ˜ t + w δ ˜ + z δ ˜
for all 0 < t δ ˜ . By using (101), from (115), we obtain
c 1 ( β + 2 ) δ ˜ t w ( t ) + z ( t ) ,   for   all   0 < t δ ˜   where   c 1 = λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Let
c n ( β + 2 ) δ ˜ t w ( t ) + z ( t ) ,   for   all   0 < t δ ˜ .
From (80) and (117), we have
w + z
p ( K 11 + K 21 ) 4 ( 1 + β ) 2 t 2 0 t c n 2 ( β + 2 ) δ ˜ s 2 s d s λ 1 M 1 + λ 2 M 2 ( β + 3 ) ,
= p c n 2 ( K 11 + K 21 ) 16 ( 1 + β ) 2 t 4 ( β + 2 ) 3 ( β + 1 ) 2 + 2 ( β + 2 ) 2 9 ( 1 + β ) 2 λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Since t 4 ( β + 2 ) 3 ( β + 1 ) 4 β 5 3 ( β + 1 ) , we have
w + z p ( 6 β 2 + 16 β + 11 ) c n 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
So, we have
w + z c n + 1 , t 0 , δ ˜ ,
where { c n } is given by
c n + 1 = p ( 6 β 2 + 16 β + 11 ) c n 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
After simplifying (84) and (122),
w ( t ) + z ( t ) c n + 1 ( β + 2 ) δ ˜ t ,   for   all   0 < t δ ˜ ,
Similarly, we obtain the sequence { c n } converge to c in R . Hence, from (123), we have
c = p ( 6 β 2 + 16 β + 11 ) c 2 ( K 11 + K 21 ) 48 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Finally, from (125), we conclude
λ 1 M 1 + λ 2 M 2 12 ( 1 + β ) 4 ( β + 3 ) p ( 6 β 2 + 16 β + 11 ) ( K 11 + K 21 ) .
Hence, the proof is complete. □
Lemma 18.
Assume there exists  ( w , z ) E × E  satisfying (79), (80), and (83). Then,
λ 1 M 1 + λ 2 M 2 24 ( 1 + β ) 4 ( β + 3 ) p ( 3 β 2 + 10 β + 9 ) ( K 11 + K 21 )
and
w ( t ) + z ( t ) c ( β + 3 ) δ ˜ 2 t ,   f o r   a l l   0 < t δ ˜ .
Proof. 
From (83) and (102), we obtain
1 2 w δ ˜ 1 2 z δ ˜ w ( t ) z ( t ) t δ ˜ λ 1 M 1 + λ 2 M 2 ( β + 3 ) d s
for all t 0 , δ ˜ . After calculating as in Lemma 17, we have
w ( t ) + z ( t ) c 1 ( β + 3 ) δ ˜ 2 t ,   for   all   0 < t δ ˜   subject   to   c 1 = λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Again, we assume
w ( t ) + z ( t ) c n ( β + 3 ) δ ˜ 2 t ,   for   all   0 < t δ ˜ .
So, from (80) and (131), we have
w ( t ) + z ( t )
p ( K 11 + K 21 ) 4 ( 1 + β ) 2 t 2 0 t c n 2 ( β + 3 ) δ ˜ 2 s 2 s d s λ 1 M 1 + λ 2 M 2 ( β + 3 ) ,
= p c n 2 ( K 11 + K 21 ) 16 ( 1 + β ) 2 t 2 ( β + 3 ) 3 ( β + 1 ) 2 + ( β + 3 ) 2 18 ( β + 1 ) 2 λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Since t 2 ( β + 3 ) 3 ( β + 1 ) 2 β 3 3 ( β + 1 ) , we obtain
w + z p ( 3 β 2 + 10 β + 9 ) c n 2 ( K 11 + K 21 ) 96 ( 1 + β ) 4 λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
So, we have
w + z c n + 1 , t 0 , δ ˜ ,
where
c n + 1 = p ( 3 β 2 + 10 β + 9 ) c n 2 ( K 11 + K 21 ) 96 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Therefore, from (83) and (136),
w ( t ) + z ( t ) c n + 1 ( β + 3 ) δ ˜ 2 t ,   for   all   0 < t δ ˜ .
Thus, { c n } converges to c. Hence, from (137), we obtain
c = p ( 3 β 2 + 10 β + 9 ) c 2 ( K 11 + K 21 ) 96 ( 1 + β ) 4 + λ 1 M 1 + λ 2 M 2 ( β + 3 ) .
Finally, from Equation (139), we have
λ 1 M 1 + λ 2 M 2 24 ( 1 + β ) 4 ( β + 3 ) p ( 3 β 2 + 10 β + 9 ) ( K 11 + K 21 ) .
This completes the proof. □
Lemma 19.
Assume  λ 1  and  λ 2  satisfy conditions
0 λ 1 C 2 δ ˜ ( 2 + β ) M 1 * ( 1 + β ) δ ˜ , 0 λ 2 C 2 δ ˜ ( 2 + β ) M 2 * ( 1 + β ) δ ˜
a n d 0 C 8 ( 2 + β ) ( 1 + β ) 2 p 2 δ ˜ ( 1 + β ) β δ ˜ min 1 K 11 + K 12 , 1 K 21 + K 22 .
Then, there is at least one function  Ψ , Φ E × E  satisfying (36), (53),  Ψ δ ˜ = 0 ,  Φ δ ˜ = 0 ,  Ψ ( t ) 0 , and  Φ ( t ) 0 .
Proof. 
Set
Ψ ( t ) = C t A ( 2 t ) δ ˜ , Φ ( t ) = C t B ( 2 t ) δ ˜ , and C 0
for all t 0 , δ ˜ . Clearly, Ψ ( t ) and Φ ( t ) satisfies (53). Now, Ψ δ ˜ = 0 and Φ δ ˜ = 0 implies A = 2 δ ˜ δ ˜ δ ˜ and B = 2 δ ˜ δ ˜ δ ˜ . Therefore, Ψ ( t ) 0 and Φ ( t ) 0 are also fulfilled. Now, we have
Ψ K 1 t , Ψ , Φ ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , 2 δ ˜ C t β δ ˜ ( 2 + β ) ( 1 + β ) 2 p ( K 11 + K 12 ) C 2 2 δ ˜ δ ˜ δ ˜ ( 2 t ) δ ˜ 2 2 ( 1 + β ) 2 λ M 1 * ( 1 + β ) β ˜ t γ ˜ C t β δ ˜ 2 δ ˜ ( 2 + β ) ( 1 + β ) ( 1 + 2 β ) δ ˜ 1 ( 1 + β ) δ ˜ λ M 1 * ( 1 + β ) δ ˜ t δ ˜ C 2 δ ˜ ( 2 + β ) p ( K 11 + K 12 ) C 2 2 2 δ ˜ 2 ( 1 + β ) 2 δ ˜ δ ˜ ( t ) δ ˜ 2 C t β δ ˜ 2 δ ˜ ( 2 + β ) ( 1 + β ) ( 1 + 2 β ) δ ˜ 1 ( 1 + β ) δ ˜ t δ ˜ p ( K 11 + K 12 ) C 2 2 2 δ ˜ 2 ( 1 + β ) 2 δ ˜ δ ˜ ( t ) δ ˜ 2 , Since   λ C 2 δ ˜ ( 2 + β ) M 1 * ( 1 + β ) δ ˜ = A 1 2 ( 2 + β ) ( 1 + β ) 2 ( K 11 + K 12 ) P C 2 δ ˜ ( 1 + β ) ( 2 + β ) δ ˜ 1 ( 1 + β ) δ ˜ t δ ˜ + t δ ˜ 2 ,
where
A 1 = ( K 11 + K 12 ) δ ˜ δ ˜ t δ ˜ p C 2 2 γ ˜ ( 1 + β ) ( 1 + 3 β ) δ ˜ t δ ˜ .
So, from (144), we obtain
Ψ K 1 t , Ψ , Φ ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 1 s β G 1 ( s ) d s , A 1 t δ ˜ 1 2 δ ˜ δ ˜ 2 + 8 ( 2 + β ) ( 1 + β ) 2 ( K 11 + K 12 ) p C 2 δ ˜ ( 1 + β ) β δ ˜ ( K 11 + K 12 ) 4 p C 2 δ ˜ ( 1 + β ) ( β + 2 ) δ ˜ 0 ,   since   C 8 ( 2 + β ) ( 1 + β ) 2 p 2 δ ˜ ( β + 1 ) β δ ˜ min 1 K 11 + K 12 , 1 K 21 + K 22 .
Again, we have
Φ K 2 t , Ψ , Φ ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s , C t β δ ˜ 2 δ ˜ ( 2 + β ) ( 1 + β ) ( 1 + 2 β ) δ ˜ 1 ( 1 + β ) δ ˜ λ M 2 * ( 1 + β ) δ ˜ t δ ˜ C 2 δ ˜ ( 2 + β ) p ( K 21 + K 22 ) C 2 2 2 δ ˜ 2 ( 1 + β ) 2 δ ˜ δ ˜ ( t ) δ ˜ 2 C t β δ ˜ 2 δ ˜ ( 2 + β ) ( 1 + β ) ( 1 + 2 β ) δ ˜ 1 ( 1 + β ) δ ˜ t δ ˜ p ( K 21 + K 22 ) C 2 2 2 δ ˜ 2 ( 1 + β ) 2 δ ˜ δ ˜ ( t ) δ ˜ 2 , Since   λ C 2 δ ˜ ( 2 + β ) M 2 * ( 1 + β ) δ ˜
= A 1 * 2 ( 2 + β ) ( 1 + β ) 2 ( K 21 + K 22 ) P C 2 δ ˜ ( 1 + β ) ( 2 + β ) δ ˜ 1 ( 1 + β ) δ ˜ t δ ˜ + t δ ˜ 2 ,
where
A 1 * = ( K 21 + K 22 ) 1 ( 1 + β ) δ ˜ t δ ˜ p C 2 2 γ ˜ ( 1 + β ) ( 1 + 3 β ) δ ˜ t β δ ˜ .
Similarly, from (147), we derive
Φ K 2 t , Ψ , Φ ( 1 + β ) β ˜ t β ˜ 0 L ( t ) λ 2 s β G 2 ( s ) d s , A 1 * t δ ˜ 1 2 δ ˜ δ ˜ 2 + 8 ( 2 + β ) ( 1 + β ) 2 ( K 21 + K 22 ) p C 2 δ ˜ ( 1 + β ) β δ ˜ ( K 21 + K 22 ) 4 p C 2 δ ˜ ( 1 + β ) ( β + 2 ) δ ˜ 0 ,   since   C 8 ( 2 + β ) ( 1 + β ) 2 p 2 δ ˜ ( β + 1 ) β δ ˜ min 1 K 11 + K 12 , 1 K 21 + K 22 .
Hence, the proof is complete. □
Lemma 20.
Assume  λ 1  and  λ 2  satisfy conditions
0 λ 1 ( 2 + β ) C M 1 * ( β + 1 ) δ ˜ 2 β δ ˜ , 0 λ 2 ( 2 + β ) C M 2 * ( β + 1 ) δ ˜ 2 β δ ˜
and
C ( β + 2 ) ( 1 + β ) ( β + 2 ) δ ˜ p 2 δ ˜ min 1 K 11 + K 12 , 1 K 21 + K 22 .
Then, there is at least one function  ( Ψ , Φ ) E × E  satisfying (36), (53),  β Ψ δ ˜ = Ψ δ ˜ ,  β Φ δ ˜ = Φ δ ˜ ,  Ψ ( t ) 0 , and  Φ ( t ) 0 .
Proof. 
We put
Ψ ( t ) = C t A ( 2 t ) δ ˜ , Φ ( t ) = C t B ( 2 t ) δ ˜ ,   and   C 0 .
for all t 0 , δ ˜ . Thus, Condition (53) is held. Now, β Ψ δ ˜ = Φ δ ˜ and β Φ δ ˜ = Φ δ ˜ implies
A = 2 ( β + 2 ) δ ˜ δ ˜ δ ˜   &   B = 2 ( 2 + β ) δ ˜ δ ˜ δ ˜ .
Therefore, Ψ ( t ) 0 and Φ ( t ) 0 are also fulfilled. Hence, we obtain the results. □
Lemma 21.
Assume  λ 1  and  λ 2  satisfy conditions
0 λ 1 C 2 ( β + 2 ) δ ˜ ( 2 + β ) 3 M 1 * ( β + 1 ) δ ˜ , 0 λ 2 C 2 ( β + 2 ) δ ˜ ( 2 + β ) 3 M 2 * ( β + 1 ) δ ˜ ,
and
C 2 ( 6 β + 5 ) δ ˜ ( β + 2 ) ( 1 + β ) ( β + 2 ) δ ˜ 27 p min 1 K 11 + K 12 , 1 K 21 + K 22 .
Then, there is a solution  ( Ψ , Φ ) E × E  satisfying (36), (53),  ( β 1 ) Ψ δ ˜ = ϕ δ ˜ ,  ( β 1 ) Φ δ ˜ = Φ δ ˜ ,  Ψ ( t ) 0 , and  Φ ( t ) 0 .
Proof. 
We put
Ψ ( t ) = C t A ( 2 t ) δ ˜ , Φ ( t ) = C t B ( 2 t ) δ ˜ ,   and   C 0 .
for t 0 , δ ˜ . Now, the boundary conditions ( β 1 ) Φ δ ˜ = Φ δ ˜ and ( β 1 ) Ψ δ ˜ = Ψ δ ˜ provide the following values as
A = 3 × 2 β δ ˜ δ ˜ δ ˜   &   B = 3 × 2 β δ ˜ δ ˜ δ ˜ .
Therefore, Ψ ( t ) 0 and Φ ( t ) 0 are also fulfilled. Hence, after calculating as in Lemma 19, we obtain the results. □
Theorem 2.
Assume the conditions  λ 1 c > λ 1 0  and  λ 2 c > λ 1 0  hold. Then, the differential Equation (13) with respect to (14)–(16) is solvable. Also, the solutions of these problems do not exist if  λ 1 > λ 1 c  and  λ 2 > λ 2 c . Furthermore, every solution of (13) corresponding to (14)–(16) is nonpositive and gives the value zero at the origin.
Proof. 
The proof can be achived by using Lemmas 1–4, 14 and 15. □
Proposition 1.
Bounds of  λ 1 c  and  λ 2 c  for the system (14) are defined by the following estimation
λ 1 c 8 ( 2 + β ) 2 ( 1 + β ) L ˜ 1 p M 1 * , λ 2 c 8 ( 2 + β ) 2 ( 1 + β ) L ˜ 1 p M 2 *
&   λ 1 c M 1 + λ 2 c M 2 12 ( 1 + β ) 4 ( β + 3 ) p ( K 11 + K 21 ) ,
where
L ˜ 1 = min 1 K 11 + K 12 , 1 K 21 + K 22 .
Proof. 
The results can be deduced from Lemmas 16 and 19. □
Proposition 2.
Bounds of  λ 1 c  and  λ 2 c  for the system (15) are defined by the following estimation
λ 1 c 128 ( 2 + β ) 2 ( β + 1 ) L ˜ 1 81 p M 1 * , λ 2 c 128 ( 2 + β ) 2 ( β + 1 ) L ˜ 1 81 p M 2 *
a n d   λ 1 c M 1 + λ 2 c M 2 24 ( 1 + β ) 4 ( β + 3 ) p ( 3 β 2 + 10 β + 9 ) ( K 11 + K 21 ) .
Proof. 
The results can be deduced from Lemmas 18 and 21. □
Proposition 3.
Bounds of  λ 1 c  and  λ 2 c  for the system (16) are defined by the following estimation
λ 1 c ( 2 + β ) 2 ( β + 1 ) L ˜ 1 2 p M 1 * , λ 2 c ( 2 + β ) 2 ( β + 1 ) L ˜ 1 2 p M 2 *
a n d   λ 1 c M 1 + λ 2 c M 2 12 ( 1 + β ) 4 ( β + 3 ) p ( K 11 + K 21 ) ( 6 β 2 + 16 β + 11 ) .
Proof. 
The results can be deduce from Lemmas 17 and 20. □
Remark 1.
We can also reduce Equation (13) into a second-order differential equation similar to (17) by using the transformation  t = r 1 + β 2 . Hence, for every  β > 1  the domain is transformed to  0 , 1 2  instead of  0 , δ ˜ , and similar results can be obtained. An important point to note is that the bounds of the parameters remain the same under this transformation.
Remark 2.
For  β = 1 , it is easy to see that the bounds of the parameters  λ 1  and  λ 2  are well matched with respective bounds of the same parameters in the reference [11]. In [11], the authors placed the numerical examples to verify the theoretical results.

5. Conclusions

We proposed a system of fourth-order ordinary differential equations that arise in real life. We successfully converted the fourth-order SBVP into a system of integral equations. We classified the nature of solutions with respect to the space C l o c 2 0 , 1 2 . We showed the existence of at least one solution. We observed that the parameters λ 1 and λ 2 are dependent with each other. We calculated the relationship between these two parameters, which helps us find areas where there is no solution. We noticed that the range of the parameters is symmetric with respect to the transformations t = r 1 + β 1 + β and t = r 1 + β 2 .

Author Contributions

Conceptualization, A.K.V. and B.P.; validation, R.P.A. and B.P.; writing—original draft preparation, A.K.V. and B.P.; writing—review and editing, A.K.V., B.P. and R.P.A.; visualization, A.K.V., B.P. and R.P.A.; supervision, R.P.A.; project administration, A.K.V. All authors have read and agreed to the published version of the manuscript.

Funding

This work is supported by the project SB/S4/MS/805/12 and INSPIRE Program Division, Department of Science & Technology, New Delhi, India-110016.

Institutional Review Board Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

This work is supported by the project SB/S4/MS/805/12 and INSPIRE Program Division, Department of Science & Technology, New Delhi, India-110016.

Conflicts of Interest

The authors declare no conflict of interest.

Abbreviations

The following abbreviations are used in this manuscript:
SBVPSingular boundary value problem
BVPBoundary value problem
IEIntegral equation

References

  1. Barabasi, A.L.; Stanley, H.E. Fractal Concepts in Surface Growth; Cambridge University Press: Cambridge, UK, 1995. [Google Scholar]
  2. Foord, J.S.; Davies, G.J.; Tsang, W.T. Chemical Beam Epitaxy and Related Techniques; John Wiley and Sons Ltd.: Chichester, UK, 1997. [Google Scholar]
  3. Lourdudoss, S.; Kjebon, O. Hybrid vapor phase epitaxy revisited. IEEE J. Sel. Top. Quantum Electron. 1997, 3, 749–767. [Google Scholar] [CrossRef]
  4. Escudero, C. Geometric principles of surface growth. Phys. Rev. Lett. 2008, 101, 196102. [Google Scholar] [CrossRef]
  5. Escudero, C.; Hakl, R.; Peral, I.; Torres, P.J. On radial stationary solutions to a model of nonequilibrium growth. Eur. J. Appl. Math. 2013, 24, 437–453. [Google Scholar] [CrossRef]
  6. Escudero, C.; Hakl, R.; Peral, I.; Torres, P.J. Existence and nonexistence result for a singular boundary value problem arising in the theory of epitaxial growth. Math. Methods Appl. Sci. 2014, 37, 793–807. [Google Scholar] [CrossRef]
  7. Escudero, C.; Korutcheva, E. Origins of scaling relations in nonequilibrium growth. J. Phys. Math. Theor. 2012, 45, 125005. [Google Scholar] [CrossRef]
  8. Marsili, M.; Maritan, A.; Toigo, F.; Banavar, J.R. Stochastic growth equations and reparametrization invariance. Rev. Mod. Phys. 1996, 68, 963–983. [Google Scholar] [CrossRef]
  9. Verma, A.K.; Pandit, B. Existence and nonexistence results of radial solutions to singular bvps arising in epitaxial growth theory. Mathematics 2020, 9, 774. [Google Scholar] [CrossRef]
  10. Verma, A.K.; Pandit, B.; Agarwal, R.P. On multiple solutions for a fourth order nonlinear singular boundary value problems arising in epitaxial growth theory. Math. Methods Appl. Sci. 2021, 44, 5418–5435. [Google Scholar] [CrossRef]
  11. Verma, A.K.; Pandit, B.; Agarwal, R.P. Existence and nonexistence results for a class of fourth-order coupled singular boundary value problems arising in the theory of epitaxial growth. Math. Methods Appl. Sci. 2020. [Google Scholar] [CrossRef]
  12. Keller, J.B. Electrohydrodynamics i. the equilibrium of a charged gas in a container. J. Ration. Mech. Anal. 1956, 5, 715–724. [Google Scholar] [CrossRef]
  13. Gomez, J.L.; Omari, P.; Rivetti, S. Bifurcation of positive solutions for a one-dimensional indefinite quasilinear Neumann problem. Nonlinear Anal. 2017, 155, 1–51. [Google Scholar] [CrossRef]
  14. Habets, P.; Omari, P. Multiple positive solution of a one-dimensional prescribed mean curvature problem. Commun. Contemp. Math. 2006, 9, 701–730. [Google Scholar] [CrossRef]
  15. Obersnel, F.; Omari, P. Positive solutions of the Dirichlet problem for the prescribed mean curvature equation. J. Differ. Equ. 2010, 249, 1674–1725. [Google Scholar] [CrossRef]
  16. Obersnel, F.; Omari, P. Existence, regularity and boundary behaviour of bounded variation solutions of a one-dimensional capillarity equation. Am. Inst. Math. Sci. 2013, 33, 305–320. [Google Scholar] [CrossRef]
  17. Obersnel, F.; Omari, P.; Rivetti, S. Existence, regularity and stability properties of periodic solutions of capillarity equation in the presence of lower and upper solutions. Nonlinear Anal. Real World Appl. 2012, 13, 2830–2852. [Google Scholar] [CrossRef]
  18. Omari, P.; Trombetta, M. Remarks on the lower and upper solutions method for second and third order periodic boundary value problems. Appl. Math. Comput. 1992, 50, 1–21. [Google Scholar] [CrossRef]
  19. Pandey, R.K. On a class of weakly regular singular two point boundary value problems, II. J. Differ. Equ. 1996, 127, 110–123. [Google Scholar] [CrossRef]
  20. Pandey, R.K. On a class of regular singular two point boundary value problems. J. Math. Anal. Appl. 1997, 208, 388–403. [Google Scholar] [CrossRef]
  21. Zhang, X.; Tian, H.; Wu, Y.; Wiwatanapataphee, B. Existence of positive solutions for third-order semipositone boundary value problems on time scales. Nonlinear Anal. Model. Control 2023, 28, 133–151. [Google Scholar] [CrossRef]
  22. Zhang, X.; Tian, H.; Wu, Y.; Wiwatanapataphee, B. Existence of Positive Solutions for a Singular Second-Order Changing-Sign Differential Equation on Time Scales. Fractral Fract. 2022, 6, 315. [Google Scholar]
  23. Shah, K.; Abdeljawad, T.; Abdalla, B. On a coupled system under coupled integral boundary conditions involving non-singular differential operator. AIMS Math. 2023, 8, 9890–9910. [Google Scholar] [CrossRef]
  24. Li, P. Existence of Analytic Solutions for Some Classes of Singular Integral Equations of Non-normal Type with Convolution Kernel. Acta Appl. Math. 2022, 181, 5. [Google Scholar] [CrossRef]
  25. Dunninger, D.R.; Kurtz, J.C. Existence of solutions for some nonlinear singular boundary value problems. J. Math. Anal. Appl. 1986, 115, 396–405. [Google Scholar] [CrossRef]
  26. Dunninger, D.R.; Kurtz, J.C. A priori bounds and existence of positive solutions for singular nonlinear boundary value problems. SIAM J. Math. Anal. 1986, 17, 595–609. [Google Scholar] [CrossRef]
  27. Cabada, A.; Cid, J.A.; Sanchez, L. Positivity and lower and upper solutions for fourth order boundary value problems. Nonlinear Anal. 2007, 67, 1599–1612. [Google Scholar] [CrossRef]
  28. Cabada, A.; Saavedra, L. Existence of solutions for nth order nonlinear differential boundary value problems by means of fixed point theorems. Nonlinear Anal. Real World Appl. 2018, 42, 180–206. [Google Scholar] [CrossRef]
  29. Cabada, A.; Somoza, L.L.; Tojo, F.A.F. Existence of solutions of integral equations with asymptotic conditions. Nonlinear Anal. Real World Appl. 2018, 42, 140–159. [Google Scholar] [CrossRef]
  30. Cabada, A.; Tojo, F.A.F. Existence results for a linear equation with reflection, non-constant coefficient and periodic boundary conditions. J. Math. Anal. Appl. 2014, 412, 529–546. [Google Scholar] [CrossRef]
  31. Wang, G.; Agarwal, R.P.; Cabada, A. Existence results and the monotone iterative technique for systems of nonlinear fractional differential equations. Appl. Math. Lett. 2012, 25, 1019–1024. [Google Scholar] [CrossRef]
  32. Zhang, Y. Existence of solutions of a kind of singular boundary value problem. Nonlinear Anal. Theory Methods Appl. 1993, 35, 153–199. [Google Scholar] [CrossRef]
  33. Barilla, D.; Bohner, M.; Heidarkhani, S.; Moradi, S. Existence results for dynamic Sturm–Liouville boundary value problems via variational methods. Appl. Math. Comput. 2021, 409, 125614. [Google Scholar] [CrossRef]
  34. Verma, A.K.; Pandit, B.; Verma, L.; Agarwal, R.P. A review on a class of second order nonlinear singular bvps. Mathematics 2020, 8, 1045. [Google Scholar] [CrossRef]
  35. Verma, A.K.; Pandit, B.; Agarwal, R.P. Existence and nonexistence results for a class of non-self-adjoint fourth-order singular boundary value problems arising in real life. Math. Methods Appl. Sci. 2023, 46, 6077–6110. [Google Scholar]
  36. Cabada, A. An overview on the lower and upper solutions method with nonlinear boundary value problem. Bound. Value Probl. 2011, 2011, 893753. [Google Scholar]
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Pandit, B.; Verma, A.K.; Agarwal, R.P. A Study on Solutions for a Class of Higher-Order System of Singular Boundary Value Problem. Symmetry 2023, 15, 1729. https://doi.org/10.3390/sym15091729

AMA Style

Pandit B, Verma AK, Agarwal RP. A Study on Solutions for a Class of Higher-Order System of Singular Boundary Value Problem. Symmetry. 2023; 15(9):1729. https://doi.org/10.3390/sym15091729

Chicago/Turabian Style

Pandit, Biswajit, Amit K. Verma, and Ravi P. Agarwal. 2023. "A Study on Solutions for a Class of Higher-Order System of Singular Boundary Value Problem" Symmetry 15, no. 9: 1729. https://doi.org/10.3390/sym15091729

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