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R E S E A R C H

Open Access

Basic theory of differential equations with

mixed perturbations of the second type on

time scales

Yige Zhao

1*

, Yibing Sun

1

, Zhi Liu

2

and Zhanbing Bai

3

*Correspondence:

[email protected]

1School of Mathematical Sciences,

University of Jinan, Jinan, P.R. China Full list of author information is available at the end of the article

Abstract

In this paper, we develop the theory of differential equations with mixed

perturbations of the second type on time scales. We give an existence theorem for differential equations with mixed perturbations of the second type on time scales under Lipschitz condition. We also present some fundamental differential inequalities on time scales, which are utilized to investigate the existence of extremal solutions. We establish the comparison principle for differential equations with mixed

perturbations of the second type on time scales. Our results in this paper extend and improve some well-known results.

Keywords: Mixed perturbations; Existence; Differential inequalities; Comparison principle; Time scales

1 Introduction

In this paper, we discuss the following differential equations with mixed perturbations of the second type on time scales (DETS):

⎧ ⎨ ⎩

[u(t)–k(t,u(t))f(t,u(t)) ]=g(t,u(t)), tJ,

u(t0) =u0,

(1)

wherefCrd(J×R,R\ {0}) andk,gCrd(J×R,R).

LetTbe a time scale, and letJ= [t0,t0+a]T= [t0,t0+a]∩Tbe a bounded interval inT

for somet0∈Randa> 0. We denote byCrd(J×R,R) the class of rd-continuous functions

g:J×R→R. For basic definitions and useful lemmas from the theory of calculus on time scales, we refer to [1].

By a solution of the DETS (1) we mean a-differentiable functionusuch that

(i) the functiontu–k(t,u)f(t,u) is-differentiable for eachu∈R, and (ii) usatisfies equations (1).

The theory of time scales has been drawn a lot of attention since 1988 (see [1–9]). In recent years the theory of nonlinear differential equations with perturbations has been a hot research topic; see [10–15]. Dhage [13] discussed the following first-order hybrid

(2)

differential equation with mixed perturbations of the second type:

⎧ ⎨ ⎩

d dt[

x(t)–k(t,x(t))

f(t,x(t)) ] =g(t,x(t)), t∈[t0,t0+a],

x(t0) =x0∈R,

where [t0,t0+a] is a bounded interval inRfor somet0∈Randa> 0,fC([t0,t0+a]× R,R\ {0}), andk,gC([t0,t0+a]×R,R). They developed the theory of hybrid

differ-ential equations with mixed perturbations of the second type and gave some original and interesting results.

As far as we know, there are no results for the DETS (1). From the works mentioned we consider the theory of DETS (1). We give an existence theorem for the DETS (1) under Lipschitz conditions. We also present some fundamental differential inequalities on time scales (DITS), which are utilized to investigate the existence of extremal solutions. We establish the comparison principle for the DETS (1). Our results in this paper extend and improve some well-known results.

The paper is organized as follows. In Sect.2, we give an existence theorem for the DETS (1) under Lipschitz conditions by the fixed point theorem in Banach algebra due to Dhage. In Sect.3, we establish some fundamental DITS to strict inequalities for the DETS (1). In Sect.4, we present existence results of maximal and minimal solutions for HDTS. In Sect.5, we prove the comparison principle for the DETS (1), which is followed by the conclusion in Sect.6.

2 Existence result

In this section, we discuss the existence results for the DETS (1). We place the DETS (1) in the spaceCrd(J,R) of rd-continuous functions defined onJwith the supremum norm · defined as

u=sup

tJ

u(t)

and the multiplication “·” inCrd(J,R) defined as

(u·v)(t) = (uv)(t) =u(t)v(t)

foru,vCrd(J,R). Clearly,Crd(J,R) is a Banach algebra with respect to these norm and

multiplication. ByL1(J,R) we denote the space of Lebesgue-integrable functions onJ

equipped with the norm · L1 defined as

uL1= t0+a

t0

u(s)s.

The following fixed point theorem in a Banach algebra due to Dhage [16] is useful in the proofs of our main results.

Lemma 2.1([16]) Let Q be a closed convex bounded subset of a Banach space P,and let

(3)

(a) AandCare Lipschitz with Lipschitz constantsαandβ,respectively, (b) Bis compact and continuous,

(c) u=AuBv+Cufor allvQuQ,and

(d) αM+β< 1,whereM=B(Q)=sup{B(u):uQ}.

Then the operator equation AuBu+Cu=u has a solution in Q. We present the following hypotheses.

(A0) The functionuu–k(t,u)f(t,u) is increasing inRfor alltJ.

(A1) There exist constantsL1> 0andL2> 0such that

f(t,u) –f(t,v)≤L1|uv|

and

k(t,u) –k(t,v)≤L2|uv|

for alltJandu,v∈R. Moreover,LM.

(A2) There exists a functionhL1(J,R)such that

g(t,u)≤h(t), tJ,

for allu∈R.

Lemma 2.2 Suppose that(A0)holds.Then for any vL1(J,R),the-differentiable

func-tion u is a solufunc-tion of the DETS

u(t) –k(t,u(t))

f(t,u(t))

=v(t), tJ, (2)

and

u(t0) =u0∈R, (3)

if and only if u satisfies the integral equation

u(t) =k t,u(t)+f t,u(t)u0–k(t0,u0)

f(t0,u0)

+

t

t0

v(s)s

, tJ. (4)

Proof Letube a solution of problem (2)–(3). Applying the-integral to (2) fromt0tot,

we obtain

u(t) –k(t,u(t))

f(t,u(t))

u0–k(t0,u0)

f(t0,u0)

=

t

t0

v(s)s,

that is,

u(t) =k t,u(t)+f t,u(t)u0–k(t0,u0)

f(t0,u0)

+

t

t0

v(s)s

, tJ.

(4)

Conversely, suppose thatusatisfies equation (4). By direct differentiation, applying the -derivative to both sides of (4), we get that (2) is satisfied. Thus, substitutet=t0in (4)

implies

u(t0) –k(t0,u(t0))

f(t0,u(t0))

=u0–k(t0,u0)

f(t0,u0)

.

Since the mapuu–k(t,u)f(t,u) is increasing inRfortJ, the mapuu–k(t0,u)

f(t0,u) is injective in

R, andu(t0) =u0. Hence (3) also holds.

Now we will give the following existence theorem for the DETS (1).

Theorem 2.1 Suppose that(A0)(A2)hold.If

L1

u0–k(t0,u0)

f(t0,u0)

+hL1

+L2< 1, (5)

then the DETS(1)has a solution defined on J.

Proof SetU=Crd(J,R) and define the subsetSofUby

S=uU|uN,

where

N= F0(|

u0–k(t0,u0)

f(t0,u0) |+hL1) +K0 1 –L1(|u0f–k(t(t0,u00,u)0)|+hL1) –L2

,

F0=suptJ|f(t, 0)|, andK0=suptJ|k(t, 0)|.

Clearly,Sis a closed, convex, and bounded subset of the Banach spaceU. By Lemma2.2 the DETS (1) is equivalent to the nonlinear integral equation

u(t) =k t,u(t)+f t,u(t)u0–k(t0,u0)

f(t0,u0)

+

t

t0

g s,u(s)s

, tJ. (6)

Define three operatorsA,C:UUandB:SUby

Au(t) =f t,u(t), tJ, (7)

Bu(t) = u0–k(t0,u0)

f(t0,u0)

+

t

t0

g s,u(s)s, tJ, (8)

and

Cu(t) =k t,u(t), tJ. (9) Then equation (6) is transformed into the operator equation as

(5)

Next, we prove that the operatorsA,B, andCsatisfy all the conditions of Lemma2.1. First, we prove that Ais a Lipschitz operator on U with Lipschitz constant L1. Let

u,vU. Then by (A1)

Au(t) –Av(t)=f t,u(t)–f t,v(t)≤L1u(t) –v(t)≤L1uv

for alltJ. Taking the supremum overt, then we have

AuAvL1uv

for allu,vU. This shows thatAis a Lipschitz operator onUwith Lipschitz constantL1.

Similarly, we can get thatCis also a Lipschitz operator onUwith Lipschitz constantL2.

Next, we prove thatBis a compact continuous operator fromSintoU. First, we prove thatBis continuous onS. Let{un}be a sequence inSconverging to a pointuS. Then by the Lebesgue dominated convergence theorem adapted to time scale we have

(6)

On the other hand, lett1,t2∈J. Then for anyuS, we get continuous. Hence, forε> 0, there existsδ> 0 such that

|t1–t2|<δBu(t1) –Bu(t2)<ε

for allt1,t2∈JanduS. This shows thatB(S) is an equicontinuous set inU. Now the set

B(S) is uniformly bounded and equicontinuous set inU, so it is compact by Arzelà–Ascoli theorem. ThusBis a compact operator onS.

Next, we show that (c) of Lemma2.1is satisfied. LetuUandvSbe such thatu=

Taking the supremum overt, we have

(7)

Finally, we obtain

M=B(S)=supB(u):uSu0–k(t0,u0) f(t0,u0)

+hL1,

and so

L1M+L2≤L1

u0–k(t0,u0)

f(t0,u0)

+hL1

+L2< 1.

Thus all the conditions of Lemma 2.1 are satisfied, and hence the operator equa-tionAuBu+Cu=uhas a solution inS. Therefore the DETS (1) has a solution defined

onJ.

3 Differential inequalities on time scales

In this section, we establish DITS for the DETS (1).

Theorem 3.1 Suppose that(A0)holds.Assume that there exist-differentiable functions

v,w such that

v(t) –k(t,v(t))

f(t,v(t))

g t,v(t), tJ, (10)

and

w(t) –k(t,w(t))

f(t,w(t))

g t,w(t), tJ, (11)

one of the inequalities being strict.Then v(t0) <w(t0)implies

v(t) <w(t) (12)

for all tJ.

Proof Assume that inequality (11) is strict. Suppose that the claim is false. Then there existst1∈J,t1>t0, such thatv(t1) =w(t1) andv(t) <w(t) fort0≤t<t1.

Define

V(t) =v(t) –k(t,v(t))

f(t,v(t)) and W(t) =

w(t) –k(t,w(t))

f(t,w(t))

for alltJ. Then we obtainV(t1) =W(t1), and by (A0) we haveV(t) <W(t) for allt<t1.

SinceV(t1) =W(t1), we get

V(t1+h) –V(t1)

h >

W(t1+h) –W(t1)

h

for sufficiently smallh< 0. This inequality implies that

(8)

because of (A0). Then we obtain

g t1,v(t1)

V(t1)≥W(t1) >g t1,w(t1)

.

This is a contradiction withv(t1) =w(t1). Hence inequality (12) is valid.

The next result is concerned with nonstrict DITS, which needs a Lipshitz condition.

Theorem 3.2 Suppose that the conditions of Theorem3.1hold with inequalities(10)and

(11).Suppose that there exists a real number K> 0such that

g(t,u1) –g(t,u2)≤K sup t0≤st

u1(t) –k(t,u1(t))

f(t,u1(t))

u2(t) –k(t,u2(t))

f(t,u2(t))

, tJ (13)

for all u1,u2∈Rwith u1≥u2.Then v(t0)≤w(t0)implies v(t)≤w(t)for all tJ.

Proof Letε> 0 andK> 0 be given. Define

(t) –k(t,(t))

f(t,(t))

=w(t) –k(t,w(t))

f(t,w(t)) +εe

2L(t–t0),

so that we get

(t) –k(t,(t))

f(t,(t))

>w(t) –k(t,w(t))

f(t,w(t)) ⇒ (t) >w(t). Let(t) =(t)–k(t,wf(t,w ε(t))

ε(t)) , so thatW(t) =

w(t)–k(t,w(t))

f(t,w(t)) fortJ. Then by (11) we obtain

(t) =W

(t) + 2Kεe2L(t–t0)≥g t,w(t)+ 2Lεe2L(t–t0).

Then from (13) we have

g t,(t)

g t,w(t)≤K sup

t0≤st

(s) –W(s)

=Kεe2L(t–t0)

for alltJ, and thus

(t)≥g t,(t)

Kεe2L(t–t0)+ 2Kεe2L(t–t0)>g t,w

ε(t)

,

that is,

(t) –f t,(t)

>g t,(t)

for alltJ. Also, we get(t0) >w(t0) >v(t0). Hence Theorem3.1withw=implies that

v(t) <(t) for alltJ. By the arbitrariness ofε> 0, taking the limits asε→0, we have

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4 Existence of maximal and minimal solutions

In this section, we prove the existence of maximal and minimal solutions for the DETS (1) onJ= [t0,t0+a]T.

Definition 4.1 A solutionrof the DETS (1) is said to be maximal if for any other solution

uto the DETS (1), we haveu(t)≤r(t) for alltJ. Similarly, a solutionρof the DETS (1) is said to be minimal ifρ(t)≤u(t) for alltJ, whereuis any solution of the DETS (1) onJ. We discuss the case of maximal solution only. Similarly, the case of minimal solution can be obtained with the same arguments with appropriate modifications. Given an arbitrary small numberε> 0, we discuss the following initial value problem of DETS:

⎧ ⎨ ⎩

[u(t)–k(t,u(t))f(t,u(t)) ]=g(t,u(t)) +ε, tJ,

u(t0) =u0+ε,

(14)

wherefCrd(J×R,R\ {0}) andk,gCrd(J×R,R).

An existence theorem for the DETS (14) can be stated as follows.

Theorem 4.1 Suppose that(A0)(A2)and inequality(5)hold.Then for every small

num-berε> 0,the DETS(14)has a solution defined on J.

Proof By hypothesis, since

L1

u0–k(t0,u0)

f(t0,u0)

+hL1

+L2< 1,

there existsε0> 0 such that

L1

u0+εk(t0,u0+ε)

f(t0,u0+ε)

+hL1+εa

+L2< 1

for all 0 <εε0. The rest of the proof is similar to that of Theorem2.1, and we omit

it.

Our main existence theorem for maximal solution for the DETS (1) is the following:

Theorem 4.2 Suppose that(A0)(A2)and inequality(5)hold.Then the DETS(1)has a

maximal solution defined on J.

Proof Let{εn}0 be a decreasing sequence of positive numbers such thatlimn→∞εn= 0,

whereε0is a positive number satisfying the inequality

L1

u0+ε0–k(t0,u0+ε0)

f(t0,u0+ε0)

+hL1+ε0a

+L2< 1.

Such a numberε0exists in view of inequality (5). Then for any solutionxof the DETS (1),

by Theorem4.1we get

(10)

for alltJandn∈N∪ {0}, wherer(t,εn) defined onJis a solution of the DETS

(11)

·u0+εnk(t0,u0+εn)

f(t0,u0+εn)

+hL1+εna

+Fp(t1) –p(t2)+|t1–t2|εn

,

whereF=sup(t,u)J×[–N,N]|f(t,u)|andp(t) =tt 0h(s)s.

Sincef andkare continuous on the compact setJ×[–N,N], they are uniformly con-tinuous. Hence

f t1,r(t1,εn)

f t2,r(t2,εn)→0 ast1→t2

and

k t1,r(t1,εn)

k t2,r(t2,εn)→0 ast1→t2

uniformly for alln∈N. Similarly, since the functionpis continuous on the compact setJ, it is uniformly continuous, and hence

p(t1) –p(t2)→0 ast1→t2.

Therefore we obtain

r(t1,εn) –r(t2,εn)→0 ast1→t2

uniformly for alln∈N. Therefore

r(t,εn)→r(t) asn→ ∞

for alltJ.

Next, we prove that the functionr(t) is a solution of the DETS (1) defined onJ. Since

r(t,εn) is a solution of the DETS (15), we get

r(t,εn) =

f t,r(t,εn)

u0+εnk(t0,u0+εn)

f(t0,u0+εn)

+

t

t0

g s,r(s,εn)

s

+

t

t0 εns

+k t,r(t,εn)

(17)

for alltJ. Taking the limit asn→ ∞in (17) implies

r(t) =f(t,r(t)u0–k(t0,u0)

f(t0,u0)

+

t

t0

g s,r(s)s

+k t,r(t)

for alltJ. Thus the functionris a solution of the DETS (1) onJ. Finally, from inequality (15) it follows thatx(t)≤r(t) for alltJ. Hence the DETS (1) has a maximal solution

(12)

5 Comparison theorems on time scales

The main problem of the DITS is to estimate a bound for the solution set for the DITS related to the DETS (1). In this section, we present the maximal and minimal solutions serving as bounds for the solutions of the related DITS to the DETS (1) onJ= [t0,t0+a]T.

Theorem 5.1 Suppose that(A0)(A2)and inequality(5)hold.Assume that there exists a

-differentiable function u such that

⎧ ⎨ ⎩

[x(t)–k(t,x(t))f(t,x(t)) ]g(t,x(t)), tJ,

x(t0)≤u0.

(18)

Then

x(t)≤r(t) (19)

for all tJ,where r is a maximal solution of the the DETS(1)on J.

Proof Letε> 0 be arbitrarily small. By Theorem4.2,r(t,ε) is a maximal solution of the DETS (14), the limit

r(t) =lim

ε→0r(t,ε) (20)

is uniform onJ, and the functionris a maximal solution of the DETS (1) onJ. Hence we have

⎧ ⎨ ⎩

[r(t,ε)–k(t,r(t,ε))f(t,r(t,ε)) ]=g(t,r(t,ε)) +ε, tJ,

r(t0,ε) =u0+ε.

By the above inequality it implies that

⎧ ⎨ ⎩

[r(t,ε)–k(t,r(t,ε))f(t,r(t,ε)) ]>g(t,r(t,ε)), tJ,

r(t0,ε) >u0.

(21)

Now, applying Theorem3.2to inequalities (18) and (21), we conclude thatx(t) <r(t,ε) for

alltJ. Thus (20) implies that inequality (19) holds onJ.

Theorem 5.2 Suppose that(A0)(A2)and inequality(5)hold.Assume that there exists a

-differentiable function u such that

⎧ ⎨ ⎩

[y(t)–k(t,y(t))f(t,y(t)) ]g(t,y(t)), tJ,

y(t0)≥u0.

Then

ρ(t)≤y(t)

(13)

Note that Theorem5.1is useful to prove the boundedness and uniqueness of the solu-tions for the DETS (1) onJ. We have a following result.

Theorem 5.3 Suppose that(A0)(A2)and inequality(5)hold.Assume that there exists a

function G:J×R+R+such that

g(t,u1) –g(t,u2)≤G

t,max

s∈[t0,t]

u1(s) –k(s,u1(s))

f(s,u1(s))

u2(s) –k(s,u2(s))

f(s,u2(s))

, tJ,

for all u1,u2∈Rwith u1≥u2.If the identically zero function is the only solution of the

differential equation

m(t) =G t,m(t), tJ, m(t0) = 0, (22)

then the DETS(1)has a unique solution on J.

Proof By Theorem2.1the DETS (1) has a solution defined onJ. Suppose that there are two solutionsx1andx2of the DETS (1) existing onJwithx1>x2. Definem:J→R+by

m(t) =u1(t) –k(t,u1(t))

f(t,u1(t))

u2(t) –k(t,u2(t))

f(t,u2(t))

.

Since (|x(t)|)≤ |x(t)|, we obtain that

m(t)≤

u1(t) –k(t,u1(t))

f(t,u1(t))

u2(t) –k(t,u2(t))

f(t,u2(t))

=g(t,x1) –g(t,x2) ≤G

t,u1(t) –k(t,u1(t))

f(t,u1(t))

u2(t) –k(t,u2(t))

f(t,u2(t))

=G t,m(t) fortJandm(t0) = 0.

Now we apply Theorem5.1withk(t,u)≡0 andf(t,u)≡1 to get thatm(t)≤0 for all

tJ, where the identically zero function is the only solution of the DETS (22), which is a contradiction withm(t) > 0. This implies

u1(t) –k(t,u1(t))

f(t,u1(t))

=u2(t) –k(t,u2(t))

f(t,u2(t))

for alltJ. Then we havex1=x2.

Remark5.1 Whenk≡0,f ≡1, andT=Rin our results, we obtain the differential in-equalities and other related results of Lakshmikantam and Leela [17] for the IVP of ordi-nary nonlinear differential equation

u(t) =g t,u(t), t∈[t0,t0+a], u(t0) =u0.

(14)

6 Conclusion

In this paper, we have developed the theory of the DETS (1). By the fixed point theorem in Banach algebra due to Dhage we have presented an existence theorem for the DETS (1) under Lipschitz conditions. We have also established some DITS for the DETS (1), which are used to investigate the existence of extremal solutions, and the comparison principle for the DETS (1). Our results in this paper extend and improve some well-known results.

Acknowledgements

The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original manuscript.

Funding

This research is supported by the National Natural Science Foundation of China (61703180, 61803176, 61877028, 61807015, 61773010), the Natural Science Foundation of Shandong Province (ZR2019MF032, ZR2017BA010, ZR2017LF012), the Project of Shandong Province Higher Educational Science and Technology Program (J18KA230, J17KA157), and the Scientific Research Foundation of University of Jinan (1008399, 160100101).

Availability of data and materials Not applicable.

Competing interests

The authors declare that they have no competing interests. Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.

Author details

1School of Mathematical Sciences, University of Jinan, Jinan, P.R. China.2School of Automation and Electrical Engineering, Key Laboratory of Complex systems and Intelligent Computing in Universities of Shandong, Linyi University, Linyi, P.R. China.3College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao, P.R. China.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 8 May 2019 Accepted: 20 June 2019

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13. Dhage, B.C.: Basic results in the theory of hybrid differential equations with mixed perturbations of second type. Funct. Differ. Equ.19, 87–106 (2012)

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