8.2 Publications and conference lectures
8.2.3 International conference presentations related to the Thesis
(C1) E. Awwad,I. Gy˝ori,On the Boundedness of the Solutions in Nonlinear Discrete Volterra Difference Equations, Conference on Differential and Difference Equations and Applications 2010, (CDDEA 2010), Rajeck´e Teplice, Slovak Republic , 2010.
(C2) E. Awwad, I. Gy˝ori, On the Boundedness of the Solutions of Nonlinear Volterra Inte-gral Equation with Applications, 9th Colloquium on the Qualitative Theory of Differential Equations”, (9QTDE), Szeged, Hungary, 2011.
(C3) E. Awwad, I. Gy˝ori, On the Boundedness of the Solutions of Linear and Nonlinear Volterra Difference Equation with Applications, International Conference on Differential &
Difference Equations and Applications, Azores University, Ponta Delgada, Portugal, 2011.
(C4) E. Awwad,I. Gy˝ori,Sharp Boundedness Conditions and their Applications for Difference Equation, 2nd International Conference on Mathematics & Information Science (ICMIS 2011), Egypt, September 10 - 13, 2011
(C5) E. Awwad,BIBO Stability of Nonlinear Control System with Time Delays, 2nd Egyptian Student Workshop Prague, Czech 2012.
(C6) E. Awwad,Bounded Input Bounded Output Stability of Nonlinear Differential Equations with Time Delays, International Conference on Applied Analysis and Algebra (ICAAA2012) Istanbul, Turkey 2012.
(C7) E. Awwad, Application on The Boundedness of Nonlinear Volterra Integral Equations, CSM - The Second Conference of PhD Students in Mathematics, Szeged, Hungary, 2012.
(C8) E. Awwad, Nonlinear Volterra Difference Equations with Time Delays and their Appli-cations, 18th International Conference on Difference Equations and Applications (ICDEA 2012) Barcelona, Spain, 2012.
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Appendix A
In this Appendix, we give some proofs of our results.
A.1 Proofs of some results in Chapter 3
Proof of Theorem 3.3.2. Suppose (3.1.7) and (3.1.8) are satisfied. Clearly, by Theorem3.3.1the solution of (3.3.2)-(3.3.3) is bounded for all ψ∈C [−˜t0,0],(0,∞)
.
Conversely, let the solution x of (3.3.2)-(3.3.3) be bounded on R+. First we prove that the solution x >0. Assume for the sake of contradiction that x(t)≤0 for some t. Then there exists λ > 0 such that x(t) >0, t ∈ [0, λ) and x(λ) = 0. From (3.3.5) it follows that there exists an >0 such that
Z λ−
0
k(λ, s)ds+h(λ)>0.
From (3.3.2) witht=λwe have 0 =x(λ)
= Z λ
0
k(λ, s)xp(s−τ(s))ds+h(λ)
≥ Z λ−
0
k(λ, s)xp(s−τ(s))ds+h(λ)
≥ min
−˜t0≤µ≤λ−xp(µ) Z λ−
0
k(λ, s)ds+h(λ)
≥
Z λ−
0
k(λ, s)ds+h(λ)
min
min
−˜t0≤µ≤λ−
xp(µ),1
>0, which is a contradiction.
Clearly, the positivity of x and (3.3.2) yield
x(t)≥h(t), for all t≥0, hence condition (3.1.8) is satisfied.
94
Appendix A 95
Therefore the boundedness ofx implies sup
which is finite. We show thatm > 0. Assume for the sake of contradiction that m= 0. In this case we can find a strictly increasing sequence (tn)n≥1,such that
x(tn) = inf
Taking limit of the last inequality, we obtain
n→∞lim
which contradicts to (3.3.4). So m >0. Therefore there existsT∗ ≥0 such that x(t)≥ 1
2m >0, t≥T∗.
Appendix A 96 By the boundedness of the solutionx we get
sup
t≥T∗
Z t T∗
k(t, s)ds < ∞.
This and (A.1.1) imply condition (3.1.7).
Proof of Theorem 3.3.3. Let ψ ∈ C [−t˜0,0],Rd
, and x be the solution of (3.1.1)-(3.1.2). We show thatx is bounded if (i) or (ii) holds. Let
µT := max
−˜t0≤t≤T
kx(t)k.
First we prove that µT has property (PT) under condition (i). By (3.1.4) and (3.1.6) we have IT <∞ andHT <∞. Therefore there existsv≥µT such that
Then µT has property (PT), and hence, by Theorem3.2.1, the solutionx of (3.1.1) is bounded.
Next we prove that µT has property (PT) if (ii) is satisfied. For all t≥T
Appendix A 97 (3.3.7) and (3.3.8) imply
sup
Then there exists v≥µT large enough such that
ThenµT has property (PT), and hence, by Theorem3.2.1, the solutionxof (3.1.1) is bounded.
A.2 Proofs of some results in Chapter 4
Proof of Theorem 4.3.1. Let
˜
r:= lim
t→∞r(t) and kg(x)k ≤K, K >0 for all x∈Rd. Consider the fixed point equation
c= Z ∞
0
W(u) [g(c) + ˜r]du, c∈Rd. (A.2.1) Now we prove by using Banach fixed point theorem that equation (A.2.1) has a solution. We define the functionF : Ω→Rd, where Ω :={c∈Rd:kck ≤ L1(K+ ˜r)},and
Appendix A 98 whereγ :=LR∞
0 kW(u)kdu. From condition (4.3.3) we haveγ <1, soF is a contraction. Then, by Banach fixed point theorem, equation (4.3.2) has a unique solution on Ω. In the rest of the proof letc∈Ω be the unique solution of (A.2.1).
Now we show thatx(t)→cast→ ∞. First we rewrite the equation (A.2.1) as
Hence by Lipschitz condition (4.3.2), we get kX(t)k ≤ kz(t)k+
Proof of Corollary 4.3.2. Under condition (4.2.19) Gy˝ori and Ladas [54] proved that for all i= 1, . . . , dthe positivity of the fundament solution wi of equation
˙
zi(t) =−diz(t−τ), t≥0
Appendix A 99
Clearly, from the proof of Theorem 4.3.1 with condition L≤ kbk∞ the solution x defined by (4.3.1) is convergent.
A.3 Proofs of some results in Chapter 5
Proof of Lemma 5.4.2. Let the nonnegative constantµ have property (P0) (N = 0 in Definition 5.2.1). Then the conditions (5.2.1) and (5.2.2) are satisfied for some positivev and for alln≥1,
Appendix A 100 so
α0 : = max
1≤i≤dsup
n≥1
ai(n,0)<∞, β0 := max
1≤i≤dsup
n≥1 n
X
j=1
ai(n, j)<∞, (A.3.1) and γ0 : = max
1≤i≤dsup
n≥1
|hi(n)|<∞, (A.3.2)
therefore conditions (5.2.8) and (5.2.9) are satisfied.
Conversely, we assume (5.2.8), (5.2.9), and we prove thatµ:=kψkτ has property (P0). Clearly, (5.2.9) is equivalent toα0 <∞, β0<∞ and (5.2.8) impliesγ0 <∞.
Since p∈(0,1), it is clear that (5.2.4) and α0µp
v +β0vp−1+ γ0
v ≤1 (A.3.3)
are satisfied for anyv large enough. From (A.3.3) we get α0µp+β0vp+γ0≤v.
Hence for alln≥1, we have fori= 1, . . . , d ai(n,0)φ(||ψ||τ) +
n
X
j=1
ai(n, j)φ(v) +|hi(n)| ≤α0µp+β0vp+γ0 ≤v
that (5.2.5) is satisfied with
µ:=kψkτ := max
−τ≤n≤0kx(n)k.
Then by Definition 2.1,µ has property (P0).
Proof of Theorem 5.4.3. Assume (5.2.8) and (5.2.9) are satisfied. Clearly, by Theorem5.4.1 the solution of (5.4.1)-(5.4.2) is bounded.
Conversely, let the solutionx(n) =x(n;ψ) of (5.4.1) be bounded onR+, withψ∈S([−τ,0],(0,∞)).
Under condition (5.4.4), by mathematical induction we show that x(n) >0, n≥ 0. For n= 0 this is clear. Suppose that required inequality is not satisfied for all n ≥ 0. Then there exists index`≥0 such thatx(0)>0, . . . , x(`)>0 andx(`+ 1)≤0. But, by condition (5.4.4), we get
x(`+ 1) =
`
X
j=0
a(`, j)xp(j) +h(`)
≥a(`, j`)xp(j`) +h(`)>0, 0≤j`≤`,
Appendix A 101 which is finite. First we show thatm >0.
Assume for the sake of contradiction thatm= 0. In this case we can find a strictly increasing sequence of positive integers (Nk)k≥1,such that
x(Nk) = min
Appendix A 102
But the solutionx(n) is a bounded sequence, and hence sup
This and (A.3.4) imply condition (5.2.9).
Proof of Lemma 5.4.6. Necessity. We show that (PN) implies (5.2.9) and either (i) or (ii). Sup-pose a positive constant µ has property (PN) with an integer N ≥0, hence (5.2.1) and (5.2.2) are satisfied for somev≥µ.
The last inequality implies two cases with respect to the value of βN. The first case is βN <1.
In this case the condition (i) is satisfied. Consider now the second case whereβN = 1.Clearly, if
n
X
j=N+1
ai(n, j) = 1, i= 1, . . . , d,
then from (A.3.5) we get
N
X
j=0
ai(n, j)µ+hi(n) = 0, n∈Γ(1)N , i= 1, . . . , d,
or equivalently (5.4.6). But if
n
X
j=N+1
ai(n, j)<1, i= 1, . . . , d,
Appendix A 103 and hence (5.4.7) and (5.4.8) are satisfied, i.e., condition (ii) holds.
Sufficiency. We show that if (5.2.9) and one of the conditions (i) and (ii) are satisfied with some µ≥ 0 andN ≥ 0, then µ has property (PN). It is easy to observe that (5.2.9) yields for Let condition (i) of Theorem5.4.5 be satisfied, that isβN <1, so
1−
Hence (5.2.2) is satisfied and (5.2.1) also is satisfied for all vlarge enough, hence µhas property PN.
Now suppose βN = 1, and (5.4.6) holds. Then, clearly (5.2.1) and (5.2.2) are satisfied for any v≥0 and n∈Γ(1)N .
Appendix A 104 Then there exists v >0 large enough such that
1−
n
X
j=N+1
ai(n, j)
−1
N
X
j=0
ai(n, j)µ+|hi(n)|
≤v, n∈Γ(2)N ,
since 1−Pn
j=N+1ai(n, j)>0, for alln∈Γ(2)N . Therefore
N
X
j=0
ai(n, j)µ+|hi(n)| ≤
1−
n
X
j=N+1
ai(n, j)
v, i= 1, . . . , d
i.e. for all v large enough the conditions (5.2.1) and (5.2.2) are satisfied. Hence µhas property (PN).