CUBO, A Mathematical Journal

Vol. 24, no. 01, pp. 21–35, April 2022

DOI: 10.4067/S0719-06462022000100021

Infinitely many positive solutions for an iterative
system of singular BVP on time scales

K. Rajendra Prasad
1

Mahammad Khuddush
2

K. V. Vidyasagar
3

1Department of Applied Mathematics,

College of Science and Technology,

Andhra University, Visakhapatnam,

530003, India.

rajendra92@rediffmail.com

2Department of Mathematics,

Dr. Lankapalli Bullayya College,

Resapuvanipalem, Visakhapatnam,

530013, India.

khuddush89@gmail.com

3Department of Mathematics,

S. V. L. N. S. Government Degree

College, Bheemunipatnam, Bheemili,

531163, India.

vidyavijaya08@gmail.com

ABSTRACT

In this paper, we consider an iterative system of singular two-

point boundary value problems on time scales. By applying

Hölder’s inequality and Krasnoselskii’s cone fixed point the-

orem in a Banach space, we derive sufficient conditions for

the existence of infinitely many positive solutions. Finally,

we provide an example to check the validity of our obtained

results.

RESUMEN

En este art́ıculo, consideramos un sistema iterativo de pro-

blemas de valor en la frontera singulares de dos puntos en

escalas de tiempo. Aplicando la desigualdad de Hölder y

el teorema de punto fijo cónico de Krasnoselskii en un es-

pacio de Banach, derivamos condiciones suficientes para la

existencia de una cantidad infinita de soluciones positivas.

Finalmente, entregamos un ejemplo para verificar la validez

de nuestros resultados.

Keywords and Phrases: Iterative system, time scales, singularity, cone, Krasnoselskii’s fixed point theorem,

positive solutions.

2020 AMS Mathematics Subject Classification: 34B18, 34N05.

Accepted: 15 October, 2021

Received: 18 January, 2021

c©2022 K. R. Prasad et al. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
https://dx.doi.org/10.4067/S0719-06462022000100021
https://orcid.org/0000-0001-8162-1391
https://orcid.org/0000-0002-1236-8334
https://orcid.org/0000-0003-4532-8176
mailto:rajendra92@rediffmail.com
mailto:khuddush89@gmail.com
mailto:vidyavijaya08@gmail.com


22 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

1 Introduction

The theory of time scales was created to unify continuous and discrete analysis. Difference and

differential equations can be studied simultaneously by studying dynamic equations on time scales.

A time scale is any closed and nonempty subset of the real numbers. So, by this theory, we can

extend known results from continuous and discrete analysis to a more general setting. As a

matter of fact, this theory allows us to consider time scales which possess hybrid behaviours (both

continuous and discrete). These types of time scales play an important role for applications, since

most of the phenomena in the environment are neither only discrete nor only continuous, but

they possess both behaviours. Moreover, basic results on this issue have been well documented

in the articles [1, 2] and the monographs of Bohner and Peterson [6, 7]. There is a great deal of

research activity devoted to existence of solutions to the dynamic equations on time scales, see for

example [8,9,13,16–19] and references therein.

In [14], Liang and Zhang studied countably many positive solutions for nonlinear singular m–point

boundary value problems on time scales,

(

ϕ(υ∆(t))
)∇

+ a(t)f
(

υ(t)
)

= 0, t ∈ [0, T]T,

υ(0) =

m−2
∑

i=1

aiυ(ξi), υ
∆(T) = 0,

by using the fixed-point index theory and a new fixed-point theorem in cones.

In [12], Khuddush, Prasad and Vidyasagar considered second order n-point boundary value problem

on time scales,

υ
∆∇
i (t) + λ(t)gℓ

(

υi+1(t)
)

= 0, 1 ≤ i ≤ n, t ∈ (0, σ(a)]T,

υn+1(t) = υ1(t), t ∈ (0, σ(a)]T,

υ
∆
i (0) = 0, υi(σ(a)) =

n−2
∑

k=1

ckυi(ζk), 1 ≤ i ≤ n,

and established existence of positive solutions by applying Krasnoselskii’s fixed point theorem.

Inspired by the aforementioned works, in this paper by applying Hölder’s inequality and Kras-

noselskii’s cone fixed point theorem in a Banach space, we establish the existence of infinitely

many positive solutions for the iterative system of two-point boundary value problems with n–

singularities on time scales,

υ
∆∆
ℓ (t) + λ(t)gℓ

(

υℓ+1(t)
)

= 0, 1 ≤ ℓ ≤ m, t ∈ (0, T)T,

υm+1(t) = υ1(t), t ∈ (0, T)T,







(1.1)

υℓ(0) = υ
∆
ℓ (0), 1 ≤ ℓ ≤ m,

υℓ(T) = −υ∆ℓ (T), 1 ≤ ℓ ≤ m,







(1.2)



CUBO
24, 1 (2022)

Infinitely many positive solutions for an iterative system of... 23

where m ∈ N, λ(t) =
∏k

i=1 λi(t) and each λi(t) ∈ L
pi
∆([0, T]T) (pi ≥ 1) has n–singularities in the

interval (0, T)
T
.

We assume the following conditions are true throughout the paper:

(H1) gℓ : [0, +∞) → [0, +∞) is continuous.

(H2) lim
t→ti

λi(t) = ∞, where 0 < tn < tn−1 < · · · < t1 < T.

2 Preliminaries

In this section, we introduce some basic definitions and lemmas which are useful for our later

discussions.

Definition 2.1 ( [6]). A time scale T is a nonempty closed subset of the real numbers R. T has

the topology that it inherits from the real numbers with the standard topology. It follows that the

jump operators σ, ρ : T → T, and the graininess µ : T → [0, +∞) are defined by

σ(t) = inf{τ ∈ T : τ > t},

ρ(t) = sup{τ ∈ T : τ < t},

and

µ(t) = σ(t) − t,

respectively.

• The point t ∈ T is left-dense, left-scattered, right-dense, right-scattered if ρ(t) = t, ρ(t) < t,
σ(t) = t, σ(t) > t, respectively.

• If T has a right-scattered minimum m, then Tκ = T\{m}; otherwise Tκ = T.

• If T has a left-scattered maximum m, then Tκ = T\{m}; otherwise Tκ = T.

• A function f : T → R is called rd-continuous provided it is continuous at right-dense points in T
and its left-sided limits exist (finite) at left-dense points in T. The set of all rd-continuous functions

f : T → R is denoted by Crd = Crd(T) = Crd(T, R).

• A function f : T → R is called ld-continuous provided it is continuous at left-dense points in
T and its right-sided limits exist (finite) at right-dense points in T. The set of all ld-continuous

functions f : T → R is denoted by Cld = Cld(T) = Cld(T, R).

• By an interval time scale, we mean the intersection of a real interval with a given time scale, i.e.,
[a, b]T = [a, b] ∩ T. Other intervals can be defined similarly.



24 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

Definition 2.2 ([5,11]). Let µ∆ and µ∇ be the Lebesgue ∆−measure and the Lebesgue ∇−measure
on T, respectively. If A ⊂ T satisfies µ∆(A) = µ∇(A), then we call A measurable on T, denoted
µ(A) and this value is called the Lebesgue measure of A. Let P denote a proposition with respect

to t ∈ T.

(i) If there exists Γ1 ⊂ A with µ∆(Γ1) = 0 such that P holds on A\Γ1, then P is said to hold
∆–a.e. on A.

(ii) If there exists Γ2 ⊂ A with µ∇(Γ2) = 0 such that P holds on A\Γ2, then P is said to hold
∇–a.e. on A.

Definition 2.3 ( [4,5]). Let E ⊂ T be a ∆−measurable set and p ∈ R̄ ≡ R ∪ {−∞, +∞} be such
that p ≥ 1 and let f : E → R̄ be a ∆−measurable function. We say that f belongs to Lp∆(E)
provided that either

∫

E

|f|p(s)∆s < ∞ if p ∈ [1, +∞),

or there exists a constant M ∈ R such that

|f| ≤ M, ∆ − a.e. on E if p = +∞.

Lemma 2.4 ( [20]). Let E ⊂ T be a ∆−measurable set. If f : T → R is ∆−integrable on E, then
∫

E

f(s)∆s =

∫

E

f(s)ds +
∑

i∈IE

(

σ(ti) − ti
)

f(ti) + r(f, E),

where

r(f, E) =















µN(E)f(M), if N ∈ T,

0, if N /∈ T,

IE := {i ∈ I : ti ∈ E} and {ti}i∈I, I ⊂ N, is the set of all right-scattered points of T.

Lemma 2.5. For any y(t) ∈ Crd([0, T]T), the boundary value problem,

υ
∆∆
1 (t) + y(t) = 0, t ∈ (0, T)T, (2.1)

υ1(0) = υ
∆
1 (0), υ1(T) = −υ∆1 (T), (2.2)

has a unique solution

υ1(t) =

∫

T

0

ℵ(t, τ)y(τ)∆τ, (2.3)

where

ℵ(t, τ) = 1
2 + T







(T − t + 1)(σ(τ) + 1), if σ(τ) < t,

(T − σ(τ) + 1)(t + 1), if t < τ.
(2.4)



CUBO
24, 1 (2022)

Infinitely many positive solutions for an iterative system of... 25

Proof. Suppose υ1 is a solution of (2.1), then

υ1(t) = −
∫ t

0

∫

τ

0

y(τ1)∆τ1∆τ + A1t + A2

= −
∫ t

0

(t − σ(τ))y(τ)∆τ + A1t + A2,

where A1 = υ
∆
1 (0) and A2 = υ1(0). By the conditions (2.2), we get

A1 = A2 =
1

2 + T

∫

T

0

(T − σ(τ) + 1)y(τ)∆τ.

So, we have

υ1(t) =

∫ t

0

(t − σ(τ))y(τ)∆τ + 1
2 + T

∫

T

0

(T − σ(τ) + 1)(1 + t)y(τ)∆τ

=

∫

T

0

ℵ(t, τ)y(τ)∆τ.

This completes the proof.

Lemma 2.6. Suppose (H1)–(H2) hold. For ε ∈ (0, T2 )T, let G(ε) =
ε + 1

T + 1
< 1. Then ℵ(t, τ) has

the following properties:

(i) 0 ≤ ℵ(t, τ) ≤ ℵ(τ, τ) for all t, τ ∈ [0, 1]T,

(ii) G(ε)ℵ(τ, τ) ≤ ℵ(t, τ) for all t ∈ [ε, T − ε]T and τ ∈ [0, 1]T.

Proof. (i) is evident. To prove (ii), let t ∈ [ε, T − ε]T and t ≤ τ. Then

ℵ(t, τ)
ℵ(τ, τ) =

t + 1

τ + 1
≥

ε + 1

T + 1
= G(ε).

For τ ≤ t,
ℵ(t, τ)
ℵ(τ, τ) =

T − t + 1
T − τ + 1 ≥

ε + 1

T + 1
= G(ε).

This completes the proof.

Notice that an m−tuple (υ1(t), υ2(t), υ3(t), . . . , υm(t)) is a solution of the iterative boundary value
problem (1.1)–(1.2) if and only if

υℓ(t) =

∫ 1

0

ℵ(t, τ)λ(τ)gℓ(υℓ+1(τ))∆τ, t ∈ (0, T)T, 1 ≤ ℓ ≤ m,

υm+1(t) = υ1(t), t ∈ (0, T)T,

i.e.,

υ1(t) =

∫ 1

0

ℵ(t, τ1)λ(τ1)g1
(
∫ 1

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫ 1

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫ 1

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1.



26 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

Let B be the Banach space Crd((0, T)T, R) with the norm ‖υ‖ = max
t∈(0,T)T

|υ(t)|. For ε ∈
(

0, T
2

)

T
, we

define the cone Kε ⊂ B as

Kε =

{

υ ∈ B : υ(t) is nonnegative and min
t∈[ε, T−ε]T

υ(t) ≥ G(ε)‖υ(t)‖
}

.

For any υ1 ∈ Kε, define an operator Ω : Kε → B by

(Ωυ1)(t) =

∫ 1

0

ℵ(t, τ1)λ(τ1)g1
(
∫ 1

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫ 1

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫ 1

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1.

Lemma 2.7. Assume that (H1)–(H2) hold. Then for each ε ∈
(

0, T
2

)

T
, Ω(Kε) ⊂ Kε and Ω : Kε →

Kε are completely continuous.

Proof. From Lemma 2.6, ℵ(t, τ) ≥ 0 for all t, τ ∈ (0, T)T. So, (Ωυ1)(t) ≥ 0. Also, for υ1 ∈ Kε, we
have

‖Ωυ1‖ = max
t∈(0,T)T

∫ 1

0

ℵ(t, τ1)λ(τ1)g1
(
∫ 1

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫ 1

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫ 1

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1

≤
∫ 1

0

ℵ(τ1, τ1)λ(τ1)g1
(
∫ 1

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫ 1

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫ 1

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1.

Again from Lemma 2.6, we get

min
t∈[ε,T−ε]T

{

(Ωυ1)(t)
}

≥ G(ε)
∫ 1

0

ℵ(τ1, τ1)λ(τ1)g1
(
∫ 1

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫ 1

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫ 1

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1.

It follows from the above two inequalities that

min
t∈[ε,T−ε]T

{

(Ωυ1)(t)
}

≥ G(ε)‖Ωυ1‖.

So, Ωυ1 ∈ Kε and thus Ω(Kε) ⊂ Kε. Next, by standard methods and the Arzela-Ascoli theorem, it
can be proved easily that the operator Ω is completely continuous. The proof is complete.

3 Infinitely many positive solutions

For the existence of infinitely many positive solutions for iterative system of boundary value prob-

lem (1.1)–(1.2), we apply following theorems.

Theorem 3.1 ( [10]). Let E be a cone in a Banach space X and let M1, M2 be open sets with
0 ∈ M1, M1 ⊂ M2. Let A : E ∩ (M2\M1) → E be a completely continuous operator such that



CUBO
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Infinitely many positive solutions for an iterative system of... 27

(a) ‖Av‖ ≤ ‖v‖, v ∈ E ∩ ∂M1, and ‖Av‖ ≥ ‖v‖, v ∈ E ∩ ∂M2, or

(b) ‖Av‖ ≥ ‖v‖, v ∈ E ∩ ∂M1, and ‖Av‖ ≤ ‖v‖, v ∈ E ∩ ∂M2.

Then A has a fixed point in E ∩ (M2\M1).

Theorem 3.2 ( [7,15]). Let f ∈ Lp∇(J) with p > 1, g ∈ L
q
∆(J) with q > 1, and

1
p
+ 1

q
= 1. Then

fg ∈ L1∆(J) and ‖fg‖L1∆ ≤ ‖f‖Lp∆‖g‖Lq∆, where

‖f‖Lp
∆
:=











[

∫

J

|f|p(s)∆s
]

1
p

, p ∈ R,

inf
{

M ∈ R / |f| ≤ M ∆ − a.e. on J
}

, p = ∞,

and J = [a, b)T.

Theorem 3.3 (Hölder’s inequality [3,4,15]). Let f ∈ Lpi∆(J) with pi > 1, for i = 1, 2, . . . , n and
∑n

i=1
1
pi

= 1. Then
∏k

i=1 gi ∈ L1∆(J) and
∥

∥

∥

∏k
i=1 gi

∥

∥

∥

1
≤

∏k
i=1 ‖gi‖pi. Further, if f ∈ L1∆(J) and

g ∈ L∞∆ (J), then fg ∈ L1∆(J) and ‖fg‖1 ≤ ‖f‖1‖g‖∞.

We need the following condition in the sequel:

(H3) There exists δi > 0 such that λi(t) > δi (i = 1, 2, . . . , n) for t ∈ [0, T]T.

Consider the following three possible cases for λi ∈ Lpi∆(0, T)T :
n
∑

i=1

1

pi
< 1,

n
∑

i=1

1

pi
= 1,

n
∑

i=1

1

pi
> 1.

Firstly, we seek infinitely many positive solutions for the case

n
∑

i=1

1

pi
< 1.

Theorem 3.4. Suppose (H1)–(H3) hold, let {εr}∞r=1 be such that 0 < ε1 < T/2, ε ↓ t∗ and
0 < t∗ < tn. Let {Γr}∞r=1 and {Λr}∞r=1 be such that

Γr+1 < G(εr)Λr < Λr < θΛr < Γr, r ∈ N,

where

θ = max

{

[

G(ε1)
k
∏

i=1

δi

∫

T−ε1

ε1

ℵ(τ, τ)∆τ
]−1

, 1

}

.

Assume that gℓ satisfies

(C1) gℓ(υ) ≤ N1Γr ∀ t ∈ (0, T)T, 0 ≤ υ ≤ Γr, where

N1 <

[

‖ℵ‖Lq
∆

k
∏

i=1

‖λi‖Lpi
∆

]−1

,

(C2) gℓ(υ) ≥ θΛr ∀ t ∈ [εr, T − εr]T, G(εr)Λr ≤ υ ≤ Λr.



28 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

Then the iterative boundary value problem (1.1)–(1.2) has infinitely many solutions

{(υ[r]1 , υ
[r]
2 , . . . , υ

[r]
m )}∞r=1 such that υ

[r]
ℓ (t) ≥ 0 on (0, T)T, ℓ = 1, 2, . . . , m and r ∈ N.

Proof. Let

M1,r = {υ ∈ B : ‖υ‖ < Γr}, M2,r = {υ ∈ B : ‖υ‖ < Λr},

be open subsets of B. Let {εr}∞r=1 be given in the hypothesis and we note that

t∗ < tr+1 < εr < tr <
T

2
,

for all r ∈ N. For each r ∈ N, we define the cone Kεr by

Kεr =
{

υ ∈ B : υ(t) ≥ 0, min
t∈[εr, T−εr]T

υ(t) ≥ G(εr)‖υ(t)‖
}

.

Let υ1 ∈ Kεr ∩ ∂M1,r. Then, υ1(τ) ≤ Γr = ‖υ1‖ for all τ ∈ (0, T)T. By (C1) and for τm−1 ∈
(0, T)T, we have

∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm ≤
∫

T

0

ℵ(τm, τm)λ(τm)gm(υ1(τm))∆τm

≤ N1Γr
∫

T

0

ℵ(τm, τm)
k
∏

i=1

λi(τm)∆τm.

There exists a q > 1 such that
1

q
+

n
∑

i=1

1

pi
= 1. So,

∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm ≤ N1Γr
∥

∥ℵ
∥

∥

L
q

∆

∥

∥

∥

∥

∥

k
∏

i=1

λi

∥

∥

∥

∥

∥

L
pi
∆

≤ N1Γr‖ℵ‖Lq
∆

k
∏

i=1

‖λi‖Lpi
∆

≤ Γr.

It follows in similar manner (for τm−2 ∈ (0, T)T), that
∫

T

0

ℵ(τm−2, τm−1)λ(τm−1)gm−1
(
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

∆τm−1

≤
∫

T

0

ℵ(τm−2, τm−1)λ(τm−1)gm−1(Γr)∆τm−1

≤
∫

T

0

ℵ(τm−1, τm−1)λ(τm−1)gm−1(Γr)∆τm−1

≤ N1Γr
∫

T

0

ℵ(τm−1, τm−1)
k
∏

i=1

λi(τm−1)∆τm−1

≤ N1Γr‖ℵ‖Lq
∆

k
∏

i=1

‖λi‖Lpi
∆

≤ Γr.



CUBO
24, 1 (2022)

Infinitely many positive solutions for an iterative system of... 29

Continuing with this bootstrapping argument, we get

(Ωυ1)(t) =

∫

T

0

ℵ(t, τ1)λ(τ1)g1
(
∫

T

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫

T

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1

≤ Γr.

Since Γr = ‖υ1‖ for υ1 ∈ Kεr ∩ ∂M1,r, we get

‖Ωυ1‖ ≤ ‖υ1‖. (3.1)

Let t ∈ [εr, T − εr]T. Then,

Λr = ‖υ1‖ ≥ υ1(t) ≥ min
t∈[εr,T−εr]T

υ1(t) ≥ G(εr) ‖υ1‖ ≥ G(εr)Λr.

By (C2) and for τm−1 ∈ [εr, T − εr]T, we have
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm ≥
∫

T−εr

εr

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm

≥ G(εr)θΛr
∫

T−εr

εr

ℵ(τm, τm)λ(τm)∆τm

≥ G(εr)θΛr
∫

T−εr

εr

ℵ(τm, τm)
k
∏

i=1

λi(τm)∆τm

≥ G(ε1)θΛr
k
∏

i=1

δi

∫

T−ε1

ε1

ℵ(τm, τm)∆τm

≥ Λr.

Continuing with the bootstrapping argument, we get

(Ωυ1)(t) =

∫

T

0

ℵ(t, τ1)λ(τ1)g1
(
∫

T

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫

T

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1

≥ Λr.

Thus, if υ1 ∈ Kεr ∩ ∂K2,r, then
‖Ωυ1‖ ≥ ‖υ1‖. (3.2)

It is evident that 0 ∈ M2,k ⊂ M2,k ⊂ M1,k. From (3.1)–(3.2), it follows from Theorem 3.1 that the
operator Ω has a fixed point υ

[r]
1 ∈ Kεr ∩

(

M1,r\M2,r
)

such that υ
[r]
1 (t) ≥ 0 on (0, T)T, and r ∈ N.

Next setting υm+1 = υ1, we obtain infinitely many positive solutions {(υ[r]1 , υ
[r]
2 , . . . , υ

[r]
m )}∞r=1 of

(1.1)–(1.2) given iteratively by

υℓ(t) =

∫

T

0

ℵ(t, τ)λ(τ)gℓ(υℓ+1(τ))∆τ, t ∈ (0, T)T, ℓ = m, m − 1, . . . , 1.

The proof is completed.



30 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

For

n
∑

i=1

1

pi
= 1, we have the following theorem.

Theorem 3.5. Suppose (H1)–(H3) hold, let {εr}∞r=1 be such that 0 < ε1 < T/2, ε ↓ t∗ and
0 < t∗ < tn. Let {Γr}∞r=1 and {Λr}∞r=1 be such that

Γr+1 < G(εr)Λr < Λr < θΛr < Γr, r ∈ N,

where

θ = max

{

[

G(ε1)
k
∏

i=1

δi

∫

T−ε1

ε1

ℵ(τ, τ)∆τ
]−1

, 1

}

.

Assume that gℓ satisfies (C2) and

(C3) gj(υ) ≤ N2Γr ∀ t ∈ (0, T)T, 0 ≤ υ ≤ Γr, where

N2 < min







[

‖ℵ‖L∞
∆

k
∏

i=1

‖λi‖Lpi
∆

]−1

, θ







.

Then the iterative boundary value problem (1.1)–(1.2) has infinitely many solutions

{(υ[r]1 , υ
[r]
2 , . . . , υ

[r]
m )}∞r=1 such that υ

[r]
ℓ (t) ≥ 0 on (0, T)T, ℓ = 1, 2, . . . , m and r ∈ N.

Proof. For a fixed r, let M1,r be as in the proof of Theorem 3.4 and let υ1 ∈ Kεr ∩ ∂M2,r. Again

υ1(τ) ≤ Γr = ‖υ1‖,

for all τ ∈ (0, T)T. By (C3) and for τℓ−1 ∈ (0, T)T, we have

∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm ≤
∫

T

0

ℵ(τm, τm)λ(τm)gm(υ1(τm))∆τm

≤ N1Γr
∫

T

0

ℵ(τm, τm)
k
∏

i=1

λi(τm)∆τm

≤ N1Γr
∥

∥ℵ
∥

∥

L∞
∆

∥

∥

∥

∥

∥

k
∏

i=1

λi

∥

∥

∥

∥

∥

L
pi
∆

≤ N1Γr‖ℵ‖L∞
∆

k
∏

i=1

‖λi‖Lpi
∆

≤ Γr.



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Infinitely many positive solutions for an iterative system of... 31

It follows in similar manner (for τm−2 ∈ [0, 1]T), that
∫

T

0

ℵ(τm−2, τm−1)λ(τm−1)gm−1
(
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

∆τm−1

≤
∫

T

0

ℵ(τm−2, τm−1)λ(τm−1)gm−1(Γr)∆τm−1

≤
∫

T

0

ℵ(τm−1, τm−1)λ(τm−1)gm−1(Γr)∆τm−1

≤ N1Γr
∫

T

0

ℵ(τm−1, τm−1)
k
∏

i=1

λi(τm−1)∆τm−1

≤ N1Γr‖ℵ‖L∞
∆

k
∏

i=1

‖λi‖Lpi
∆

≤ Γr.

Continuing with this bootstrapping argument, we get

(Ωυ1)(t) =

∫

T

0

ℵ(t, τ1)λ(τ1)g1
(
∫

T

0

ℵ(τ1, τ2)λ(τ2)g2
(
∫

T

0

ℵ(τ2, τ3) · · ·

× gm−1
(
∫

T

0

ℵ(τm−1, τm)λ(τm)gm(υ1(τm))∆τm
)

· · · ∆τ3
)

∆τ2

)

∆τ1

≤ Γr.

Since Γr = ‖υ1‖ for υ1 ∈ Kεr ∩ ∂M1,r, we get

‖Ωυ1‖ ≤ ‖υ1‖. (3.3)

Now define M2,r = {υ1 ∈ B : ‖υ1‖ < Λr}. Let υ1 ∈ Kεr ∩ ∂M2,r and let τ ∈ [εr, T − εr]T. Then, the
argument leading to (3.2) can be done to the present case. Hence, the theorem.

Lastly, the case
n
∑

i=1

1

pi
> 1.

Theorem 3.6. Suppose (H1)–(H3) hold, let {εr}∞r=1 be such that 0 < ε1 < T/2, ε ↓ t∗ and
0 < t∗ < tn. Let {Γr}∞r=1 and {Λr}∞r=1 be such that

Γr+1 < G(εr)Λr < Λr < θΛr < Γr, r ∈ N,

where

θ = max

{

[

G(ε1)
k
∏

i=1

δi

∫

T−ε1

ε1

ℵ(τ, τ)∆τ
]−1

, 1

}

.

Assume that gℓ satisfies (C2) and

(C4) gj(υ) ≤ N3Γr ∀ t ∈ (0, T)T, 0 ≤ υ ≤ Γr, where N3 < min
{

[

‖ℵ‖L∞
∆

∏k
i=1 ‖λi‖L1

∆

]−1

, θ

}

.

Then the iterative boundary value problem (1.1)–(1.2) has infinitely many solutions

{(υ[r]1 , υ
[r]
2 , . . . , υ

[r]
m )}∞r=1 such that υ

[r]
ℓ (t) ≥ 0 on (0, T)T, ℓ = 1, 2, . . . , m and r ∈ N.

Proof. The proof is similar to the proof of Theorem 3.4. So, we omit the details here.



32 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

4 Example

In this section, we present an example to check validity of our main results.

Example 4.1. Consider the following boundary value problem on T = R.

υ
′′
ℓ (t) + λ(t)gℓ(υℓ+1(t)) = 0, ℓ = 1, 2,

υ3(t) = υ1(t),







(4.1)

υℓ(0) = υ
′
ℓ(0),

υℓ(1) = −υ′ℓ(1),







(4.2)

where

λ(t) = λ1(t)λ2(t)

in which

λ1(t) =
1

|t − 1
4
| 12

and λ2(t) =
1

|t − 3
4
| 12

,

g1(υ) = g2(υ) =











































































1
5
× 10−4, υ ∈ (10−4, +∞),

25×10−(4r+3)− 1
5
×10−4r

10−(4r+3)−10−4r
(υ − 10−4r)+

1
5
× 10−8r, υ ∈

[

10−(4r+3), 10−4r
]

,

25 × 10−(4r+3), υ ∈
(

1
5
× 10−(4r+3), 10−(4r+3)

)

,

25×10−(4r+3)− 1
5
×10−8r

1
5
×10−(4r+3)−10−(4r+4)

(υ − 10−(4r+4))+
1
5
× 10−8r, υ ∈

(

10−(4r+4), 1
5
× 10−(4r+3)

]

,

0, υ = 0.

Let

tr =
31

64
−

r
∑

k=1

1

4(k + 1)4
, εr =

1

2
(tr + tr+1), r = 1, 2, 3, . . .,

then

ε1 =
15

32
− 1

648
<

15

32
,

and

tr+1 < εr < tr, εr >
1

5
.

Therefore,

G(εr) =
εr + 1

T + 1
=

εr + 1

2
>

1

5
, r = 1, 2, 3, . . .

It is clear that

t1 =
15

32
<

1

2
, tr − tr+1 =

1

4(r + 2)4
, r = 1, 2, 3, . . .

Since
∞
∑

x=1

1

x4
=

π4

90
and

∞
∑

x=1

1

x2
=

π2

6
, it follows that

t∗ = lim
r→∞

tr =
31

64
−

∞
∑

k=1

1

4(r + 1)4
=

47

64
− π

4

360
= 0.4637941914,



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Infinitely many positive solutions for an iterative system of... 33

λ1, λ2 ∈ Lp[0, 1] for all 0 < p < 2, and δ1 = δ2 = (4/3)1/4 ,

G(ε1) = 0.7336033951.
∫

T−ε1

ε1

ℵ(τ, τ)∆τ =
∫ 1− 15

32
+ 1

648

15
32

− 1
648

(2 − τ)(1 + τ)
3

dτ = 0.04918197801.

Thus, we get

θ = max

{

[

G(ε1)
k
∏

i=1

δi

∫

T−ε1

ε1

ℵ(τ, τ)∇τ
]−1

, 1

}

= max

{

1

0.04166167167
, 1

}

= 24.00287746.

Next, let 0 < a < 1 be fixed. Then λ1, λ2 ∈ L1+a[0, 1] and 21+a > 1 for 0 < a < 1. It follows that

k
∏

i=1

‖λi‖Lpi
∆

≈ π − ln(7 − 4
√
3),

and also ‖ℵ‖∞ = 23. So, for 0 < a < 1, we have

N1 <

[

‖ℵ‖∞
k
∏

i=1

‖λi‖Lpi
∆

]−1

≈ 0.2597173925.

Taking N1 =
1
4
. In addition if we take

Γr = 10
−4r, Λr = 10

−(4r+3),

then

Γr+1 = 10
−(4r+4) <

1

5
× 10−(4r+3) < G(εr)Λr < Λr = 10−(4r+3) < Γr = 10−4r,

θΛr = 24.00287746×10−(4r+3) < 14 ×10
−4r = N1Γr, r = 1, 2, 3, . . . , and g1, g2 satisfy the following

growth conditions:

g1(υ) = g2(υ) ≤ N1Γr =
1

4
× 10−4r, υ ∈

[

0, 10−4r
]

,

g1(υ) = g2(υ) ≥ θΛr = 24.00287746 × 10−(4r+3), υ ∈
[

1

5
× 10−(4r+3), 10−(4r+3)

]

.

Then all the conditions of Theorem 3.4 are satisfied. Therefore, by Theorem 3.4, the iterative

boundary value problem (1.1) has infinitely many solutions {(υ[r]1 , υ
[r]
2 )}∞r=1 such that υ

[r]
ℓ (t) ≥ 0

on [0, 1], ℓ = 1, 2 and r ∈ N.

Acknowledgements

The authors would like to thank the referees for their valuable suggestions and comments for the

improvement of the paper.



34 K. R. Prasad, M. Khuddush & K. V. Vidyasagar CUBO
24, 1 (2022)

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	Introduction
	Preliminaries
	 Infinitely many positive solutions
	Example