Existential and Uniqueness Results for Initial Value Problems associated with Non-linear Singular Interface Problems on Time Scales using Fixed Point Theorems

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tial Value Problems associated with

Non-linear Singular Interface Problems on Time

Scales using Fixed Point Theorems

D. K. K. Vamsi, K.N.V.S.D. Dwarakanath, I. Aditya and P. K.

Baruah

The authors dedicate this work to the Founder Chancellor of Sri Sathya Sai Institute of Higher Learning, Bhagwan Sri Sathya Sai Baba.

Abstract. In this paper we present existential and uniqueness results for IVPs associated with general nonlinear singular interface problems on Time Scales. We discuss these results for a fourth order IVP associated with nonlinear singular interface problems using the classical fixed point theorems of Banach and Schauder.

Mathematics Subject Classification (2010).Primary 47H10, 14B05, 34A12; Secondary 65R20.

Keywords.Regular problems, Singular problems, Singular interface prob-lems, Fixed point theorems.

1. Introduction

In literature we find a class of problems wherein two different differential equations are defined on adjacent intervals with a common point of interface. We term these problems as interface problems.

If the interface problem has a well defined boundary, we call the prob-lem to be aregular boundary value problem(RBVP). The interface problem with a boundary that has singularity at the end points is called a singular boundary value problem(SBVP).If there is a singularity at the point of inter-face, we term the problem to be asingular interface problem(SIP). Solving these type of boundary value problems with singularities remains a challenge for mathematicians.

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Whileregular boundary value problems, those over finite intervals with well-behaved coefficients pose no difficulties, the problems wherein the do-main of the problem is not well defined, or the continuity and/or smoothness of the functions, coefficients involved are not guaranteed in some parts of the domain, sometimes in the boundary or parts of the boundary are difficult to tackle. There are quite a number of different approaches that we come across in the literature to tackle thesesingularproblems [1],[4],[6],[8],[9],[10].

In literature, we see that work has been done on initial and boundary value problems associated with a pair of linear differential operators with con-ditions at the interface for both regular and singular cases. Some publications include [15]-[22],[26]-[31].

The singular interface problem requires a special mention. In this case existing theory based on the conventional analysis may not come handy.

We feel that the new framework of dynamic equations on time scales(an arbitrary closed subset of real numbers)[2] with facilities of the two jump op-erators with various definitions of continuity and derivatives makes one’s job simple to study these singular interface problems. These dynamic equations are nothing but the differential equations when T = R and are difference

equations whenT=Z.

Our preliminary investigation about the feasibility of this study for lin-ear second order interface problems has resulted in the work [16, 17, 25].

From the above we observe that substantial amount of work has been done for regular and singular boundary value problems involving linear dif-ferential operators. It is clear that there is a need for these singular interface problems to be discussed for the case where the problem involves nonlinear differential operators.

A systematic study of Initial Value Problems, Boundary Value Problems and eigen Value problems associated with these nonlinear singular interface problems involving nonlinear second order pair of dynamic equations is done in [34]-[41].

In this paper we study the existence and uniqueness of solution for a fourth order Initial Value Problem associated with these nonlinear singular interface problems. Schauder and Banach’s fixed point theorems are used for proving the exsistential and uniqueness results.

2. Mathematical Preliminaries

Definition 2.1. Let T be a time scale(an arbitrary closed subset of real

numbers). Fort∈Twe define theforward jump operator σ:T→Tby σ(t) := inf{s∈T:s > t},

while thebackward jump operatorρ:_T→_Tis defined by

ρ(t) := sup{s∈T:s < t}.

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If σ(t) > t,we say that t is right-scattered, while ρ(t) < t we say that t is left-scattered. Points that are right-scattered and left-scattered at the same time are calledisolated.Also, ift <supTandσ(t) =t,thentis called

right-dense, and ift > infT andρ(t) =t,thent is calledleft-dense. Points that

are right-dense and left-dense at the same time are calleddense. Finally, the graininess functionµ:T→[0,∞) is defined by

µ(t) :=σ(t)−t.

Definition 2.2. _Tκ₌

T− {m} if sup T<∞ T if sup T=∞

wheremis the left

scat-tered maximum of_T.

Definition 2.3. Let f be a function defined on T. We say that f is delta

differentiable at t ∈ Tκ provided there exists an α such that for all > 0

there is a neighborhoodN around t with

|f(σ(t)−f(s)−α(σ(t)−s)| ≤|σ(t)−s| for alls∈ N.

Definition 2.4. For a function f : T → R we shall talk about the second

derivativef∆∆_provided _f∆_{is differentiable on}

Tκ2= (Tκ)κ with derivative f∆∆ _{= (}_f∆₎∆ _:

Tκ2 → R. Similarly we define the higher order derivatives f∆n _:

Tκn→R.

Theorem 2.5. (Banach Contraction Mapping Theorem)

If T :X →X is contractive on a complete metric space X thenT has a unique fixed point inX.

Theorem 2.6. (Schauder’s Fixed Point Thorem)

LetLbe a convex subset of a normed linear spaceE.Then each compact map T :L→L has a fixed point.

LetI= [c, d] with c < ρ(d). We defineIc= [c,∞) in casesupT= +∞.

ByCB

T S(Ia) we mean the linear space of all continuous functions f :Ic →R

such thatsupt∈Ic|f(t)|<∞.

Now we quote the time scales version of the Arzela-Ascoli theorem [1].

Theorem 2.7. (Arzela-Ascoli Theorem) Let X be a subset of CB

T S(Ia) having the following properties.

(i) X is bounded.

(ii) On every compact subinterval J of[c,∞)we have: For any >0 there existsδ >0 such thatt1, t2∈J, |t1−t2|< δ implies|f(t1)−f(t2)|< for

allf ∈X.

(iii) For every >0there existsb∈Ic such that t1, t2∈[b,∞)implies

|f(t1)−f(t2)|< for allf ∈X.

ThenX is relatively compact.

LetT1andT2be two time scales. LetC(T) denote the space of all continuous

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Definition 2.8. By (t1, t2)∈T1×T2 we mean thatt1∈T1andt2∈T2with

the product topology onT1×T2.

Definition 2.9. Let Z(_T) be a function space on the time scale_T. For x∈

Z(_T) we define

kxk=supt∈T|x(t)|.

Definition 2.10. By (x1, x2)∈X(T1)×Y(T2) whereX andY are function

spaces, we mean thatx1∈X(T1) andx2∈Y(T2) with the product topology

onX(T1)×Y(T2).For (t1, t2)∈T1×T2we define

(x1, x2)(t1, t2) = (x1(t1), x2(t2)).

Definition 2.11. For (t1, t2)∈T1×T2, we define

k(t1, t2)k=kt1k+kt2k=|t1|+|t2|.

Definition 2.12. For (x1, x2)∈C(T1)×C(T2), we define

k(x1, x2)k = kx1k+kx2k

= sup_t

1∈T1|x1(t1)|+supt2∈T2|x2(t2)|.

Definition 2.13. Let (y11, y12)∈ X(T1)×Y(T2). We say that (y11, y12) is

continuous onT1×T2 if for >0 there existsδ >0 such that for arbitrarily

fixed (t01, t02)∈T1×T2 and (t1, t2)∈T1×T2such that

k (t1, t2)−(t01, t02)k< δ

⇒ k(y11, y12)(t1, t2)−(y11, y12)(t01, t02)k< . (2.1)

Definition 2.14. A sequence (yn1, yn2)∈C(T1)×C(T2) is said to be cauchy

sequence if for every > 0 there exists N such that ∀n1, n2, m1, m2 > N

implies

k(xn1, xn2)−(xm1, xm2)k< .

Definition 2.15. A sequence (yn1, yn2)∈X(T1)×Y(T2) is said to be

equicon-tinuous if for every >0 there is aδ >0,depending only on,such that for all (yn1, yn2) and all (t1, t2),(t1

0

, t2

0

)∈T1×T2satisfying

k (t1, t2)−(t1

0

, t2

0 )k< δ

⇒ k(yn1, yn2)(t1, t2)−(yn1, yn2)(t1

0

, t2

0

)k< . (2.2)

Definition 2.16. The spaceX(_T1)×Y(T2) is said to convex if for every

(y11, y12),(y21, y22)∈X(T1)×Y(T2),we have

α(y11, y12) + (1−α)(y21, y22)∈X(T1)×Y(T2) for 0< α <1.

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3. Existence and Uniqueness of Solution for a Fourth Order

Initial Value Problem associated with Nonlinear Singular

Interface Problem

3.1. Definition of the Problem

Let T1 = [0, a]_T(a time scale with end points 0 and a), K1 = [σ(a), l]_T(a

time scale with end points σ(a) and l), T2 =K1κ

4

where a, σ(a), l < +∞.

Also let (f1, f2) be nonlinear function tuple inC(T1×R3)× C(T2×R3). In

this paper we consider the following IVP associated with singular interface problem(IVP-SIP).

y∆∆∆∆₁ (t) = f1(t, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 ), t∈T1 (3.1)

y∆∆∆∆₂ (t) = f2(t, y2, y∆2, y ∆∆ 2 , y

∆∆∆

2 ), t∈T2κ

4

(3.2)

with the initial conditions

y1(0) = 0 (3.3)

y₁∆(0) = 0 (3.4)

y∆∆₁ (0) = 0 (3.5)

y∆∆∆₁ (0) = 0 (3.6)

followed by the matching interface conditions

ρ1y1(a) = ρ2y2(σ(a)), (3.7)

ρ3y1∆(a) = ρ4y2∆(σ(a)), (3.8)

ρ5y∆∆1 (a) = ρ6y2∆∆(σ(a)), (3.9)

ρ7y∆∆∆1 (a) = ρ8y2∆∆∆(σ(a)), ρi>0, i= 1,2,3,4,5,6,7,8. (3.10)

3.2. Existence of Solution using Schauder’s Fixed Point Theorem

In this section we prove the existence of solution for the IVP-SIP using Schauder’s fixed point theorem.

Theorem 3.1. If (f1, f2)is continuous and bounded, then there exists atleast

one solution for the 4th order IVP-SIP.(3.3)-(3.10)

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Then,

y₁∆∆∆∆(t) = f1(t, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )

y∆∆∆₁ (t) =

Z t

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s+c11

y∆∆1 (t) =

Z t

0

Z m

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m+

Z t

0

c11∆s+c12

y∆₁(t) =

Z t

0

Z m

0

Z r

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆r∆m

+

Z t

0

Z m

0

c11∆s∆m+

Z t

0

c12∆s+c13

y1(t) =

Z t

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

+

Z t

0

Z m

0

Z r

0

c11∆s∆r∆m+

Z t

0

Z m

0

c12∆s∆m+

Z t

0

c13∆s+c14

wherec11,c12,c13andc13are constants to be determined. By using the initial

conditions (3.3),(3.4),(3.5)(3.6) we get

y1(0) = 0 ⇒ c14= 0

y∆₁(0) = 0 ⇒ c13= 0

y₁∆∆(0) = 0 ⇒ c12= 0

y1∆∆∆(0) = 0 ⇒ c11= 0

⇒y1(t) =

Z t

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m.

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Case II Lett∈_T2

y2∆∆∆∆(t) = f2(t, y2, y2∆, y2∆∆, y2∆∆∆)

y∆∆∆₂ (t) =

Z t

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )∆s+c21

y∆∆₂ (t) =

Z t

σ(a)

Z m

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )∆s∆m+

Z t

σ(a)

c21∆s+c22

y∆₂(t) =

Z t

σ(a)

Z m

σ(a)

Z r

σ(a)

f2(s, y2, y∆2, y ∆∆ 2 , y

∆∆∆

2 )∆s∆r∆m

+

Z t

σ(a)

Z m

σ(a)

c21∆s∆m+

Z t

σ(a)

c22∆s+c23

y2(t) =

Z t

σ(a)

Z m

σ(a)

Z r

σ(a)

Z d

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )∆s∆d∆r∆m

+

Z t

σ(a)

Z m

σ(a)

Z r

σ(a)

c21∆s∆r∆m

+

Z t

σ(a)

Z m

σ(a)

c22∆s∆m+

Z t

σ(a)

c23∆s+c24

wherec21 ,c22, c23 andc24are constants to be determined.

Now, by (3.7), we get

ρ1y1(a) = ρ2y2(σ(a))

⇒c24 =

ρ1

ρ2

Z a

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

.

Also, by (3.8), we get

ρ3y∆1(a) = ρ4y∆2(σ(a))

⇒c23 =

ρ3

ρ4

Z a

0

Z m

0

Z r

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆r∆m

.

Also,by (3.9),we get

ρ5y∆∆1 (a) = ρ6y∆∆2 (σ(a))

⇒c22 =

ρ5

ρ6

Z a

0

Z m

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

.

Also,by (3.10),we get

ρ7y1∆∆(a) = ρ8y∆∆2 (σ(a))

⇒c21 =

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y∆∆1 , y∆∆∆1 )∆s

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Hence,

y2(t) =

Z t

σ(a)

Z m

σ(a)

Z r

σ(a)

Z d

σ(a)

f2(s, y2, y2∆, y2∆∆, y2∆∆∆)∆s∆d∆r∆m

+

Z t

σ(a)

Z m

σ(a)

Z r

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r∆m

+

Z t

σ(a)

Z m

σ(a)

ρ5

ρ6

Z a

0

Z m

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

+

Z t

σ(a)

ρ3

ρ4

Z a

0

Z m

0

Z r

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆r∆m

∆s0

+ ρ1

ρ2

Z a

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y∆1, y∆∆1 , y∆∆∆1 )∆s∆d∆r∆m

.

We now define the integral operatorT :C(T1)×C(T2)→C(T1)×C(T2).

T(y1, y2) =

Z t1

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

,

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y∆2, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0

∆r0∆m0

+

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

+

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

0_∆_s0 ∆m0

+

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0 ∆m0

∆s0

+ ρ1

ρ2

Z a

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆d

0

∆r0∆m0

where t1, m, r, d ∈ T1 and t2, m

0

, r0, d0 ∈ _T2. It is clear that (y1, y2)

is a solution of IVP-SIP iff (y1, y2) solves the operator equation (y1, y2) =

T(y1, y2).In other words a fixed point for the operator (y1, y2) =T(y1, y2) is

a solution for the IVP-SIP.

We use Schauder’s fixed point theorem to show the existence of a solu-tion.

Claim 3.2. The spaceC(T1)×C(T2) is convex.

Let (y11, y12),(y21, y22) ∈ C(T1)×C(T2). For the space to be convex

we need to show that

α(y11, y12) + (1−α)(y21, y22)∈C(T1)×C(T2) forα <1.

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Since (y11, y12)∈C(T1)×C(T2)(Definition(2.13)) for fixed (t01, t02)∈T1×T2

and (t1, t2)∈T1×T2 such that∀ >0∃δ >0 such that whenever

k (t1, t2)−(t01, t02)k< δ(>0)

i.e., |t1−t01|+|t2−t02|< δ

⇒ k(y11(t1), y12(t2))−(y11(t01), y12(t02))k<

2α

i.e., supt1∈T1|y11(t1)−y11(t01)|+supt2∈T2|y12(t2)−y12(t02)|<

2α. (3.11)

Similarly we can show that

sup_t₁_∈_T₁|y21(t1)−y21(t01)|+supt2∈T2|y22(t2)−y22(t02)|<

2α. (3.12)

Now, let us consider

α(y11, y12) + (1−α)(y21, y22)

=

(αy11, αy12) + ((1−α)y21,(1−α)y22)

=

(αy11+ (1−α)y21, αy12+ (1−α)y22)

We see thatk{(αy11+ (1−α)y21)(t1),(αy12+ (1−α)y22)(t2)}

−{(αy11+ (1−α)y21)(t01),(αy12+ (1−α)y22)(t02)}k

≤ sup_t₁_∈_T₁

(αy11+ (1−α)y21)(t1)−(αy11+ (1−α)y21)(t01)

+ supt2∈T2

(αy12+ (1−α)y22)(t2)−(αy12+ (1−α)y22)(t02)

≤ supt1∈T1α|y11(t1)−y11(t01)|+supt2∈T2α|y12(t2)−y12(t02)|

+ sup_t₁_∈_T₁(1−α)|y21(t1)−y21(t01)|+supt2∈T2(1−α)|y22(t2)−y22(t02)|

< α

2α+ (1−α)

2(1−α) =

whenever,|t1−t01|+|t2−t02|< δ. HenceC(T1)×C(T2) is convex.

Claim 3.3. T is a completely continuous map.

We first show that T is continuous. We prove it by showing that T

preserves convergence.

Indeed let (yn1, yn2) be a sequence of functions in C(T1)× C(T2) such

that

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The above equation implies that

limn→∞k(yn1−y1, yn2−y2)k → 0

i.e., limn→∞supt1∈T1|(yn1−y1)(t1)| → 0 andlimn→∞supt2∈T2|(yn2−y2)(t2)| → 0.

Let us consider

kT(yn1, yn2)−T(y1, y2)k

= sup_t

1∈T1

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, yn1, y∆n1, y ∆∆

n1 , y ∆∆∆

n1 )∆s∆d∆r∆m

−

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

+ sup_t₂_∈_T₂

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, yn2, yn∆2, y ∆∆

n2 , y ∆∆∆

n2 )∆s∆d

0_∆_r0 ∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0_∆_r0 ∆m0

+

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )∆s

∆s0∆r0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

+

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )∆s∆m

∆s0∆m0

−

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

+

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, yn1, y∆n1, yn∆∆1 , y∆∆∆n1 )∆s∆r0∆m

0 ∆s0

−

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

+ ρ1

ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, yn1, y∆n1, yn∆∆1 , y∆∆∆n1 )∆s∆d

0

∆r0∆m0

− ρ1

ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆d

0

∆r0∆m0

!

.

(11)

Since (f1, f2) is continuous on C(T1)× C(T2) we have

limn→∞|f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )−f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )| → 0,

limn→∞|f2(s, yn2, yn∆2, y ∆∆

n2 , y ∆∆∆

n2 )−f1(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )| → 0.

NowkT(yn1, yn2)−T(y1, y2)k

≤ sup_t

1∈T1

Z t1

0

Z m

0

Z r

0

Z d

0

|f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )

− f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )|∆s∆d∆r∆m

+ sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

|f2(s, yn2, yn∆2, y ∆∆

n2 , y ∆∆∆

n2 )

− f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )|∆s∆d0∆r

0 ∆m0

+

_ρ

7

ρ8

sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z a

0

|f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )

− f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )|∆s∆s

0

∆r0∆m0

+

_ρ

5

ρ6

sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z a

0

Z m

0

|f1(s, yn1, yn∆1, y ∆∆

n1 , y ∆∆∆

n1 )

− f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )|∆s∆m∆s

0 ∆m0

+

_ρ

3

ρ4

sup_t

2∈T2

Z t2

σ(a)

Z a

0

Z m0

0

Z r0

0

|f1(s, yn1, y∆n1, y ∆∆

n1 , y ∆∆∆

n1 )

− f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )|∆s∆r0∆m

0 ∆s0

+

_ρ

1

ρ2

Z a

0

Z m

0

Z r

0

Z d0

0

|(f1(s, yn1, y∆n1, y ∆∆

n1 , y ∆∆∆

n1 )

− f1(s, y1, y1∆, y1∆∆, y1∆∆∆)|∆s∆d0∆r

0 ∆m0.

Hence,limn→∞kT(yn1, yn2)−T(y1, y2)k →0 proving thatT is continuous.

Let

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 ) ≤ M1, for some M1>0,∀s∈T1,

f2(s, y2, y∆2, y ∆∆ 2 , y

∆∆∆

(12)

We now show that T(C(_T1)× C(T2)) is bounded and equicontinuous subset

ofC(T1)× C(T2).Let us assume thatk(y1, y2)k ≤M. Then

kT(y1, y2)k ≤ supt1∈T1

Z t1

0

Z m

0

Z r

0

Z d

0

|f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )|∆s∆d∆r∆m

+ supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

|f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )|∆s∆d

0

∆r0∆m0

+ supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

|f1(s, y1, y∆1, y ∆∆ 1 ,

y∆∆∆₁ )|∆s

∆s0∆r0∆m0

+ supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

|f1(s, y1, y∆1, y ∆∆ 1 ,

y∆∆∆1 )|∆s∆m

∆s0∆m0

+ supt2∈T2

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

|f1(s, y1, y1∆, y ∆∆ 1 ,

y∆∆∆₁ )|∆s∆r0∆m0

∆s0

+ ρ1

ρ2

Z a

0

Z m

0

Z r

0

Z d

0

|f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )|∆s∆d0∆r

0 ∆m0

.

Since (f1, f2) is bounded we can conclude that there exists aK >0

indepen-dent of choice of (y1, y2) such thatkT(y1, y2)k ≤K.Hence,T(C(T1)× C(T2))

is bounded.

We next prove thatT(C(T1)× C(T2)) is equicontinuous subset of

C(T1)× C(T2).We need to show that ∀ >0∃ δ >0 such that whenever

k (t1, t2)−(t1

0

, t2

0 )k< δ

⇒ kT(y1(t1), y2(t2))−T(y1(t1

0 ), y2(t

0

2))k< .

(13)

Let us assume that|t1−t1

0

|+|t2−t2

0

|< δ.We see that T(y1(t1), y2(t2))−

T(y1(t1

0 ), y2(t

0

2))

=

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

− Z t 0 1 0 Z m 0 Z r 0 Z d 0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m,

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0_∆_r0 ∆m0

−

Z t

0 2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆

2 )∆s∆d0∆r

0 ∆m0

+

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

. +

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

0_∆_s0 ∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m0

∆s0∆m0

. +

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0_∆_m0_∆_s0

−

Z t

0 2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆r0∆m0

∆s0

!

.

Now

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

− Z t 0 1 0 Z m 0 Z r 0 Z d 0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

(14)

≤

Z t1

0

Z m

0

Z r

0

Z d

0

M1∆s∆d∆r∆m−

Z t

0 1

0

Z m

0

Z r

0

Z d

0

M1∆s∆d∆r∆m

=

Z t1

0

Z m

0

Z r

0

M1d∆d∆r∆m−

Z t

0 1

0

Z m

0

Z r

0

M1d∆d∆r∆m

=

Z t1

0

Z m

0

M1(

r2

2)∆r∆m−

Z t

0 1

0

Z m

0

M1(

r2

2)∆r∆m

=

Z t1

0

M1

m3

6 ∆m−

Z t

0 1

0

M1

m3

6 ∆m

= M1 24

t4₁−t0₁4

= M1 24

(t2₁−t0₁2)(t2₁+t0₁2)

. = M1

24(t1−t 0

1)

1.(t1+t

0

1).(t 2 1+t0

2 1)

Hence, whenever|t1−t

0

1|< δ we have

Z t1

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

−

Z t

0 1

0

Z m

0

Z r

0

Z d

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆d∆r∆m

<

5.

Also,

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0

∆r0∆m0

−

Z t

0 2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0

∆r0∆m0

(15)

≤

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

M2∆s∆d

0

∆r0∆m0

−

Z t

0 2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

M2∆s∆d

0

∆r0∆m0

=

Z t2

σ(a)

Z m0

σ(a)

Z r

0

σ(a)

M2(d

0

−σ(a))∆d0∆r0∆m0

−

Z t0₂

σ(a)

Z m0

σ(a)

Z r0

σ(a)

M2(d

0

−σ(a))∆d0∆r0∆m0

=

Z t2

σ(a)

Z m

0

σ(a)

M2(

r02

2 −σ(a)r 0

+σ(a)

2

2 )∆r 0

∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

M2(

r02

2 −σ(a)r 0

+σ(a)

2

2 )∆r 0

∆m0

=

Z t2

σ(a)

M2(

m03

6 +

σ(a)2m0

2 −

σ(a)m02

2 −

σ(a)3 6 )∆m

0

−

Z t

0 2

σ(a)

M2(

m03

6 +

σ(a)2_m0

2 −

σ(a)m02

2 −

σ(a)3

6 )∆m 0

0

2|< δ we have

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0

∆r0∆m0

−

Z t

0 2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, y2, y2∆, y ∆∆ 2 , y

∆∆∆ 2 )∆s∆d

0

∆r0∆m0

< 5. Now

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

−

Z t

0 2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆r0∆m

0 ∆s0

= ρ3

ρ4

Z t2

σ(a)

Z a

0

M1

_m02

2 ∆m

0_∆_s0 −ρ3

ρ4

Z t02

σ(a)

Z a

0

M1

_m02

2 ∆m

0_∆_s0

= ρ3

ρ4

M1(

a3

6)(t2−t 0

(16)

0

2|< δ we have

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

−

Z t

0 2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

< 5. Now

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

= ρ5

ρ6

M1

a2

2

Z t2

σ(a)

Z m0

σ(a)

∆m0∆s0−

Z t

0 2

σ(a)

Z m0

σ(a)

∆m0∆s0

= ρ5

ρ6

M1

a2

2 (t2−t2

0₎t2+t20

2 −σ(a)(1)

0

2|< δ we have

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆

1 )∆s∆m

∆s0∆m0

< 5. Now

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y1∆, y1∆∆, y1∆∆∆)∆s

∆s0∆r0∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y1∆, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

= ρ7

ρ8

M1a

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

∆s0∆r0∆m0−

Z t

0 2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

∆s0∆r0∆m0

= ρ7

ρ8

M1a(t2−t

0

2)

₁

6(t

2 2+t2t

0

2+t022) +

σ(a)

2 (−1)(t2+t 0

2) +

σ(a)2

2 (1)

(17)

0

2|< δ we have

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

−

Z t

0 2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, y1, y∆1, y ∆∆ 1 , y

∆∆∆ 1 )∆s

∆s0∆r0∆m0

<

5.

So, we see that

kT(y1(t1), y2(t2))−T(y1(t1

0 ), y2(t

0

2))k<

5+

5 +

5+

5 = whenever

k (t1, t2)−(t1

0

, t2

0 )k< δ

i.e., |t1−t1

0

|+|t2−t2

0 |< δ.

So,T(C(_T1)× C(T2)) is equicontinuous subset ofC(T1)× C(T2).

Condition(iii) of Arzela-Acoli theorem can be seen from that fact that anygi(s, zi)∈ C(Ti× C(Ti)), i= 1,2,is uniformly continuous onTi asTi is

compact(since closed and bounded).

ThusT is compact by Arzela-Ascoli theorem. So from Schauder’s fixed point theorem(2.6) a fixed point exists for the operator equation (y1, y2) =

T y.Hence a solution exists for the IVP-SIP.

3.3. Existence and Uniqueness of Solution using Banach’s Fixed Point The-orem

In this section, we prove the existence of a unique solution of IVP-SIP with certain restricted conditions on the interface constants. We use the Banach contraction principle.

Theorem 3.4. Let R

Tifi∆t <∞ andfi, i= 1,2 satisfy

|f1(t, y1, y∆1, y ∆∆ 1 , y

∆∆∆

1 )−f1(t, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )| ≤ |y1−z1|for allt∈T1

y1(t), z1(t)∈R (3.13)

|f2(t, y2, y∆2, y ∆∆ 2 , y

∆∆∆

2 )−f2(t, z2, z∆2, z ∆∆ 2 , z

∆∆∆

2 )| ≤ |y2−z2|for allt∈T2

y2(t), z2(t)∈R (3.14)

and

ρ1

ρ2

(a

4

24) +

ρ3

ρ4

(a

2_l

6 −

a3σ(a) 6 ) +

ρ5

ρ6

(a

2_l2

4 −

a2σ(a)2

2 +

a2σ(a)2

4 )

+ρ7

ρ8

(al

3

6 −

aσ(a)3

2 +

alσ(a)2

2 −

aσ(a)3

6 ) +

a4

24 < 1 (3.15)

l4

24− 5σ(a)4

24 +

l2σ(a)2

4 < 1 (3.16) Then the IVP-SIP has a unique solution.

(18)

Claim 3.5. The spaceC(_T1)×C(T2) is complete.

Let (xn1, xn2) be a cauchy sequence in C(T1)×C(T2) and let > 0

be given. Let (t01, to2)∈T1×T2 be fixed. From the definition of continuity

(2.13) for (t1, t2)∈T1×T2 we have

k(t1, t2)−(t01, to2)k< δ(>0)

implies that

k(xn1(t1), xn2(t2))−(xn1(t01), xn2(t02))k<

3.

i.e.,

k (t1−t01, t2−t02)k

= |t1−t01|+|t2−t02|< δ

⇒ k(xn1(t1)−xn1(t01), xn2(t2)−xn2(t02))k

= sup_t₁_∈_T₁|xn1(t1)−xn1(t01)|

+ sup_t

2∈T2|xn2(t2)−xn2(t02)|<

3. (3.17)

Now we letlimn→∞(xn1, xn2)→(x1, x2).Then∃ N >0 such that∀n1, n2>

N we have

k (xn1, xn2)−(x1, x2)k

= k(xn1−x1, xn2−xn2)k

= sup_t₁_∈_T₁|(xn1−x1)(t1)|

+ supt2∈T2|(xn2−x2)(t2)|<

3. (3.18)

For the space C(_T1)×C(T2) to be complete we need to show that

(x1, x2) belongs to the space. That is

k (t1, t2)−(t01, t02)k< δ

⇒ k(x1(t1), x2(t2))−(x1(t01), x2(t02))k< .

Now let us considerk(x1(t1), x2(t2))−(x1(t01), x2(t02))k

= k(x1(t1)−x1(t01), x2(t2)−x2(t02))k

= sup_t₁_∈_T₁|x1(t1)−x1(t01)|+supt2∈T2|x2(t2)−x2(t02)| = sup_t

1∈T1|x1(t1)−xn1(t1) +xn1(t1)−xn1(t01) +xn1(t01)−x1(t01)| + sup_t₂_∈_T₂|x2(t2)−xn2(t2) +xn2(t2)−xn2(t02) +xn2(t02)−x2(t02)|

≤ sup_t

<

3 +

3 (since (3.17),(3.18))

= .

Hence,C(T1)×C(T2) is complete.

(19)

Claim 3.6. The mapT is a contraction. We see that

kT(x1, x2)−T(z1, z2)k

= supt1∈T1

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆d∆r∆m

−

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )∆s∆d∆r∆m

+ sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, x2, x∆2, x ∆∆ 2 , x

∆∆∆

2 )∆s∆d0∆r0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, z2, z∆2, z ∆∆ 2 , z

∆∆∆

2 )∆s∆d0∆r0∆m

0

+

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s

∆s0∆r0∆m0

−

Z t2

σ(a)

Z m0

σ(a)

Z r0

σ(a)

ρ7

ρ8

Z a

0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s

∆s0∆r0∆m0

+

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆m

∆s0∆m0

−

Z t2

σ(a)

Z m0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆

1 )∆s∆m

∆s0∆m0

+

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

−

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, z1, z∆1, z1∆∆, z1∆∆∆)∆s∆r0∆m

0 ∆s0

+ ρ1

ρ2

Z a

0

Z m0

0

Z r0

0

Z d0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s∆d

0_∆_r0 ∆m0

− ρ1

ρ2

Z a

0

Z m0

0

Z r0

0

Z d0

0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )∆s∆d0∆r

0 ∆m0

!

.

Now let us consider each of the term in the above equation separately.

sup_t

1∈T1

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆d∆r∆m

−

Z t1

0 Z m 0 Z r 0 Z d 0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )∆s∆d∆r∆m

(20)

≤ sup_t₁_∈_T₁

Z t1

0

Z m

0

Z r

0

Z d

0

|f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )

− f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆

1 )|∆s∆d∆r∆m

≤ supt1∈T1

Z t1

0

Z m

0

Z r

0

Z d

0

|x1−z1|∆s∆d∆r∆m(since (3.13))

≤ sup_t₁_∈_T₁

sup_s_∈_T₁|(x1−z1)(s)|

Z t1

0

Z m

0

Z r

0

Z d

0

∆s∆d∆r∆m

= sup_s_∈T

1|(x1−z1)(s)|supt1∈T1

Z t1

0

Z m

0

r2

2∆r∆m

= sups∈T1|(x1−z1)(s)|supt1∈T1

Z t1

0

m3

6 ∆m

= sup_s_∈_T₁|(x1−z1)(s)|supt1∈T1

_t4 1

24

.

So we have

sup_t

1∈T1

Z t1

0

Z m

0

Z r

0

Z d

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆d∆r∆m

−

Z t1

0

Z m

0

Z r

0

Z d

0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )∆s∆d∆r∆m

≤sup_s_∈_T₁|(x1−z1)(s)|supt1∈T1

t14

24. (3.19)

Also,

sup_t₂_∈_T₂

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, x2, x∆2, x ∆∆ 2 , x

∆∆∆ 2 )∆s∆r

0 ∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, z2, z2∆, z ∆∆ 2 , z

∆∆∆ 2 )∆s∆d

0_∆_r0 ∆m0

!

(21)

≤ sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

|f2(s, x2, x∆2, x ∆∆ 2 , x

∆∆∆ 2 )

− f2(s, z2, z2∆, z ∆∆ 2 , z

∆∆∆

2 )|∆s∆d0∆r

0 ∆m0

≤ supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

|x2−z2|∆s∆d0∆r

0

∆m0 (since (3.14))

≤ sup_t₂_∈_T₂

sup_s_∈_T₂|(x2−z2)(s)|

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

∆s∆d0∆r0∆m0

Z t2

σ(a)

Z m0

σ(a)

Z m0

σ(a)

Z r0

σ(a)

(d0−σ(a))∆d0∆r0∆m0.

= sup_s_∈_T₂|(x2−z2)(s)|supt2∈T2

Z t2

σ(a)

Z m0

σ(a)

_r02

2 −σ(a)r

0₊σ(a)2

2

∆r0∆m0.

Z t2

σ(a)

m03

6 −

σ(a)m02

2 +

σ(a)2m0

2 −

σ(a)3 6

∆m0.

= sup_s_∈_T

2|(x2−z2)(s)|supt2∈T2

_t4 2

24−

σ(a)t3 2

6 +

σ(a)2_t2 2

4 −

σ(a)3_t 2

6 +

σ(a)4

24

Hence we have

sup_t₂_∈_T₂

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, x2, x∆2, x ∆∆ 2 , x

∆∆∆ 2 )∆s∆r

0 ∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z d

0

σ(a)

f2(s, z2, z2∆, z ∆∆ 2 , z

∆∆∆ 2 )∆s∆d

0_∆_r0 ∆m0

!

≤sups∈T2|(x2−z2)(s)| supt2∈T2

t42

24−

σ(a)t32

6 +

σ(a)2t22

4 −

σ(a)3t2

6 +

σ(a)4 24

. (3.20)

Now

sup_t₂_∈_T₂

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s

∆s0∆r0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s

∆s0∆r0∆m0

(22)

≤ _ρ 7 ρ8 sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z a

0

|f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )

− f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )|∆s∆s

0

∆r0∆m0

≤

_ρ

7

ρ8

supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

Z a

0

|x1−z1|∆s∆s

0

∆r0∆m0

≤ sup_s_∈_T₁|(x1−z1)(s)|

_ρ

7

ρ8

sup_t₂_∈_T₂(a)

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

∆s0∆r0∆m0

= sups∈T1|(x1−z1)(s)|

_ρ

7

ρ8

supt2∈T2

_at3 2

6 −

aσ(a)t2 2

2 +

aσ(a)2_t 2

2 −

aσ(a)3

6

which implies that

sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s

∆s0∆r0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

Z r

0

σ(a)

ρ7

ρ8

Z a

0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s

∆s0∆r0∆m0

≤sup_s_∈_T₁|(x1−z1)(s)|

_ρ

7

ρ8

sup_t₂_∈_T₂

_at3 2

6 −

aσ(a)t2 2

2 +

aσ(a)2_t 2

2 −

aσ(a)3

6

.(3.21)

Now

sup_t₂_∈_T₂

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆m

∆s0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s

∆s0∆m0

≤ _ρ 5 ρ6 sup_t

2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z a

0

Z m0

0

|f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )

− f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆

1 )|∆s∆m∆s

0 ∆m0

≤

_ρ

5

ρ6

supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

Z a

0

Z m0

0

|x1−z1|∆s∆m∆s

0 ∆m0

≤ sup_s_∈_T₁|(x1−z1)(s)|

_ρ

5

ρ6

sup_t₂_∈_T₂a

2

Z t2

σ(a)

Z m

0

σ(a)

∆m0∆s0

= sups∈T1|(x1−z1)(s)|

_ρ

5

ρ6

supt2∈T2

_a2_t2 2

4 −

a2_σ₍_a₎_t 2

2 +

a2_σ₍_a₎2

4

(23)

which implies that

supt2∈T2

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆

1 )∆s∆m

∆s0∆m0

−

Z t2

σ(a)

Z m

0

σ(a)

ρ5

ρ6

Z a

0

Z m0

0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆ 1 )∆s

∆s0∆m0

≤sups∈T1|(x1−z1)(s)|

ρ5

ρ6

supt2∈T2

a2t22

4 −

a2σ(a)t2

2 +

a2σ(a)2 4

. (3.22)

Now

sup_t₂_∈_T₂

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s∆r

0 ∆m0

∆s0

−

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆ 1 )∆s∆r

0_∆_m0 ∆s0

≤ _ρ 3 ρ4

sup_t₂_∈_T₂

Z t2

σ(a)

Z a

0

Z m0

0

Z r0

0

|f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )

− f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆

1 )|∆s∆r0∆m

0 ∆s0

≤ _ρ 3 ρ4 sup_t

2∈T2

Z t2

σ(a)

Z a

0

Z m0

0

Z r0

0

|x1−z1|∆s∆r0∆m

0 ∆s0

≤ sups∈T1|(x1−z1)(s)|

ρ3

ρ4

supt2∈T2

Z t2

σ(a)

Z a

0

Z m0

0

Z r0

0

∆s∆r0∆m0∆s0

= sup_s_∈_T

1|(x1−z1)(s)|

_ρ

3

ρ4

sup_t

2∈T2

_t

2a3

6 −

σ(a)a3

6

which implies that

supt2∈T2

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, x1, x∆1, x∆∆1 , x∆∆∆1 )∆s∆r

0 ∆m0

∆s0

−

Z t2

σ(a)

ρ3

ρ4

Z a

0

Z m0

0

Z r0

0

f1(s, z1, z∆1, z ∆∆ 1 , z

∆∆∆

1 )∆s∆r0∆m

0 ∆s0

≤sup_s_∈T

1|(x1−z1)(s)|

_ρ

3

ρ4

sup_t

2∈T2

_t

2a3

6 −

σ(a)a3

6

. (3.23)

Finally we see that

ρ1 ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s∆d

0_∆_r0 ∆m0

−ρ1

ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s∆d

0_∆_r0 ∆m0

(24)

≤

_ρ

1

ρ2

sup_s_∈T

1|(x1−z1)(s)|

Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

∆s∆d0∆r0∆m0.

So we have

ρ1 ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, x1, x∆1, x ∆∆ 1 , x

∆∆∆ 1 )∆s∆d

0_∆_r0_∆_m0

−ρ1

ρ2 Z a 0 Z m 0 0 Z r 0 0 Z d 0 0

f1(s, z1, z1∆, z ∆∆ 1 , z

∆∆∆ 1 )∆s∆d

0_∆_r0 ∆m0

≤sup_s_∈_T₁|(x1−z1)(s)|

_ρ

1

ρ2

_a4

24. (3.24)

From the equations (3.19)-(3.24) we can conclude that kT(x1, x2)−T(z1, z2)k

≤ sup_s_∈T

1|(x1−z1)(s)|supt1∈T1

t14

24

+ sup_s_∈_T₂|(x2−z2)(s)|supt2∈T2

_t4 2

24−

σ(a)t3 2

6 +

σ(a)2_t2 2

4 −

σ(a)3_t 2

6 +

σ(a)4

24

.

+ sup_s_∈T

1|(x1−z1)(s)|

_ρ

7

ρ8

sup_t

2∈T2

_at3 2

6 −

aσ(a)t2 2

2 +

aσ(a)2_t 2

2 −

aσ(a)3

6

+ sup_s_∈_T₁|(x1−z1)(s)|

_ρ

5

ρ6

sup_t₂_∈_T₂

_a2_t2 2

4 −

a2σ(a)t2

2 +

a2σ(a)2 4

.

+ sup_s_∈T

1|(x1−z1)(s)|

_ρ

3

ρ4

sup_t

2∈T2

_t

2a3

6 −

σ(a)a3

6

+ sup_s_∈_T₁|(x1−z1)(s)|

ρ1 ρ2 a4 24.

⇒ kT(x1, x2)−T(z1, z2)k ≤K1

sups∈T1|(x1−z1)(s)|

+K2

sups∈T2|(x2−z2)(s)|

where

K1 = supt1∈T1

t14

24 + _ρ 3 ρ4 sup_t

2∈T2

_t

2a3

6 −

σ(a)a3

6 . + ρ5 ρ6

supt2∈T2

a2t22

4 −

a2σ(a)t2

2 +

a2σ(a)2 4 . + _ρ 7 ρ8

sup_t₂_∈_T₂

_at3 2

6 −

aσ(a)t2 2

2 +

aσ(a)2_t 2

2 −

aσ(a)3

6

.

+ supt2∈T2

_ρ

1

ρ2

_a4

24,

K2 = supt2∈T2

_t4 2

24−

σ(a)t3 2

6 +

σ(a)2_t2 2

4 −

σ(a)3_t 2

6 +

σ(a)4

24

.

LetK=max{K1, K2}.ThenK <1(since (3.15)-(3.16)). We now have

(25)

kT(x1, x2)−T(z1, z2)k

≤ K

sup_s_∈T

1|(x1−z1)(s)|+sups∈T2|(x2−z2)(s)|

= Kk(x1−z1, x2−z2)k

= Kk(x1, x2)−(z1, z2)k.

Since K < 1, by Banach Contraction mapping theorem (2.5) we have a unique fixed point for (y1, y2) =T(y1, y2).Hence a unique solution exists for

the IVP-SIP.

Remark 3.7. The above theorems can be proved for IVPs for the Interface II and Interface III with suitable changes in the notations.

References

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with applications, Birkhuser Boston, Inc., Boston, MA, 2001.

[3] C. Allan Boyles,Acoustic Waveguides, Applications to Oceanic Sciences, John-Wiley and Sons, 1978.

[4] C. T. Fulton,Parametrizations of Titchmarshsm(λ)-functions in the limit circle case, Trans. Amer. Math. Soc.229(1977), 51-63.

[5] P. K. Ghosh,The Mathematics of Waves and Vibrations, MacMillan, 1975. [6] D. B. Hinton and J. K. Shaw, Titchmarsh-Weyl theory for Hamiltonian

sys-tems. Spectral theory of differential operators (Birmingham, Ala., 1981), North-Holland Math. Stud., 219-231, 55, North-North-Holland, Amsterdam- New York, 1981. [7] John Von Neumann,Allgemeine Eigenwerttheorie Hermitescher

Funktionalop-eratoren, Math. Ann.102(1929), 49–131.

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D. K. K. Vamsi

Department of Mathematics and Computer Science Sri Sathya Sai Institute of Higher Learning Prasanthinilayam, Puttaparthi, Andhra-Pradesh India

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K.N.V.S.D. Dwarakanath

e-mail:[email protected]

I. Aditya

P. K. Baruah