Electronic Journal of Qualitative Theory of Differential Equations 2008, No.27, 1-12;http://www.math.u-szeged.hu/ejqtde/
Existence of solutions to neutral differential equations with
deviated argument
M. Muslim
Department of Mathematics
Indian Institute of Science, Bangalore - 560 012, India Email: [email protected]
and
D. Bahuguna
Department of Mathematics and Statistics Indian Institute of Technology Kanpur, India
Email: [email protected]
Abstract: In this paper we shall study a neutral differential equation with deviated argument in an arbitrary Banach space X. With the help of the analytic semigroups theory and fixed point method we establish the existence and uniqueness of solutions of the given problem. Finally, we give examples to illustrate the applications of the abstract results.
Keywords: Neutral differential equation with deviated argument, Banach fixed point theorem, Analytic semigroup.
AMS Subject Classification: 34K30, 34G20, 47H06.
1
Introduction
We consider the following neutral differential equation with deviated argument in a Banach space X:
d
dt[u(t) +g(t, u(a(t)))] +A[u(t) +g(t, u(a(t)))]
=f(t, u(t), u[h(u(t), t)]), 0< t≤T <∞, u(0) =u0,
(1.1)
Initial results related to the differential equations with deviated arguments can be found in some research papers of the last decade but still a complete theory seems to be missing . For the initial works on existence, uniqueness and stability of various types of solutions of different kind of differential equations, we refer to [1]-[10] and the references cited in these papers.
Adimyet al[1] have studies the existence and stability of solutions of the following general class of nonlinear partial neutral functional differential equations:
d
dt(u(t)−g(t, ut)) = A(u(t)−g(t, ut)) +f(t, ut), t≥0,
u0 = ϕ∈C0, (1.2)
where the operator A is the Hille-Yosida operator not necessarily densely defined on the Banach spaceB. The functionsg andf are continuous from [0,∞)×C0 into B.
In this paper, we use the Banach fixed point theorem and analytic semigroup theory to prove existence and uniqueness of different kind of solutions to the given problem (1.1). The plan of the paper is as follows. In Section 3, we prove the existence and uniqueness of local solutions and in Section 4, the existence of global solution for the problem (1.1) is given. In the last section, we have given an example.
The results presented in this paper easily can be apply to the same problem (1.1) with nonlocal condition under some modified assumptions on the functionf and operator A.
2
Preliminaries and Assumptions
We note that if −A is the infinitesimal generator of an analytic semigroup then for c > 0 large enough, −(A+cI) is invertible and generates a bounded analytic semigroup. This allows us to reduce the general case in which−A is the infinitesimal generator of an analytic semigroup to the case in which the semigroup is bounded and the generator is invertible. Hence without loss of generality we suppose that
kS(t)k ≤M for t≥0 and
0∈ρ(−A),
whereρ(−A) is the resolvent set of−A. It follows that for 0≤α≤1,Aα can be defined as a closed linear invertible operator with domainD(Aα) being dense inX. We have Xκ ֒→Xα for 0< α < κ
and the embedding is continuous. For more details on the fractional powers of closed linear operators we refer to Pazy [11]. It can be proved easily that Xα := D(Aα) is a Banach space with norm
kxkα = kAαxk and it is equivalent to the graph norm of Aα. Also, for each α > 0, we define
X−α= (Xα)∗,the dual space of Xα is a Banach space endowed with the norm kxk−α =kA−αxk.
It can be seen easily that Ctα =C([0, t];Xα),for allt∈[0, T],is a Banach space endowed with
the supremum norm,
kψkt,α := sup
0≤η≤tk
We set,
CTα−1 =C([0, T];Xα−1) ={y∈ CTα :ky(t)−y(s)kα−1 ≤L|t−s|,∀ t, s∈[0, T]},
whereLis a suitable positive constant to be specified later and 0≤α <1. We assume the following conditions:
(A1): 0∈ρ(−A) and−Ais the infinitesimal generator of an analytic semigroup{S(t) : t≥0}.
(A2): LetU1⊂Dom(f) be an open subset ofR+×Xα×Xα−1and for each (t, u, v)∈U1
there is a neighborhoodV1 ⊂U1of (t, u, v).The nonlinear mapf :R+×Xα×Xα−1 →X
satisfies the following condition,
kf(t, x, ψ)−f(s, y,ψ˜)k ≤Lf[|t−s|θ1 +kx−ykα+kψ−ψ˜kα−1],
where 0 < θ1 ≤ 1, 0 ≤ α < 1, Lf > 0 is a positive constant, (t, x, ψ) ∈ V1, and
(s, y,ψ˜)∈V1.
(A3): Let U2 ⊂Dom(h) be an open subset ofXα×R+ and for each (x, t)∈U2 there
is a neighborhood V2 ⊂U2 of (x, t). The maph:Xα×R+→R+ satisfies the following
condition,
|h(x, t)−h(y, s)| ≤Lh[kx−ykα+|t−s|θ2],
where 0 < θ2 ≤ 1, 0 ≤ α < 1, Lh > 0 is a positive constant, (x, t), (y, s) ∈ V2 and
h(.,0) = 0.
(A4): Let U3 ⊂ Dom(g) be an open subset of [0, T]×Xα−1 and for each (t, x) ∈ U3
there is a neighborhood V3 ⊂ U3 of (x, t). The function g : [0, T]×Xα−1 → Xα is
continuous for (t, u)∈[0, T]×Xα−1 such that
kAαg(t, x)−Aαg(s, y)k ≤Lg{|t−s|+kx−ykα−1},
where 0≤α <1, Lg>0 is a positive constant and (x, t), (y, s)∈V3.
(A5): The functiona: [0, T]→[0, T] satisfies the following two conditions: (i) asatisfies the delay propertya(t)≤t for all t∈[0, T];
(ii) The function a is Lipschitz continuous; that is, there exist a positive constant La
such that
Definition 2.1 A continuous functionu∈CTα−1∩CTαis said to be a mild solution of equation (1.1) ifu is the solution of the following integral equation
u(t) = S(t)[u(0) +g(0, u0)]−g(t, u(a(t)))
+
Z t
0
S(t−s)f(s, u(s), u[h(u(s), s)])ds, t∈[0, T] (2.1)
and satisfies the initial condition u(0) =u0.
Definition 2.2 By a solution of the problem (1.1), we mean a functionu: [0, T]→Xα satisfying
the following four conditions:
(i) u(.) +g(., u(a(.)))∈CTα−1∩C1((0, T), X)∩C([0, T], X),
(ii) u(t)∈D(A), and (t, u(t), u[h(u(t), t)])∈U1,
(iii) dtd[u(t) +g(t, u(a(t)))] +A[u(t) +g(t, u(a(t)))] =f(t, u(t), u[h(u(t), t)]) for all t∈(0, T], (iv) u(0) =u0.
3
Existence of Local Solutions
We can prove that assumptions(A2)–(A3), for 0 ≤α <1, 0< T0 ≤T, and u ∈ CTα0 imply that
f(s, u(s), u[h(u(s), s)]) is continuous on [0, T0].Therefore, we can show that there exists a positive
constant N such that
kf(s, u(s), u[h(u(s), s)])k ≤N=Lf[T0θ1+δ(1 +LLh) +LLhT0θ2] +N0,
where N0 =kf(0, u0, u0)k. Similarly with the help of the assumptions (A4)–(A5), we can show
easily that kAαg(t, u(a(t)))k ≤ L
g[T0+δ] +kg(0, u0)kα = N1. Also, we denote kA−1k = M2 and kA−αk=M3.
Theorem 3.1 Let us assume that the assumptions (A1)-(A5) are hold and u0 ∈ D(Aα) for 0 ≤
α <1. Then, the differential equation (1.1) has a unique local mild solution if
Lg+CαLf[2 +LLh]
T01−α 1−α
<1. (3.1)
Proof. Now for a fixed δ >0,we choose 0< T0 ≤T such that
k(S(t)−I)Aα[u0+g(0, u0)]k+Lg[T0+δ]≤ δ
2, for all t∈[0, T0] (3.2) and
CαN
T01−α 1−α ≤
δ
We set
W ={u∈CTα0 ∩C
α−1
T0 :u(0) =u0, ku−u0kT0,α≤δ}.
Clearly, W is a closed and bounded subset ofCTα−1. We define a mapF :W → W given by
(Fu)(t) = S(t)[u0+g(0, u0)]−g(t, u(a(t)))
+
Z t
0
S(t−s)f(s, u(s), u[h(u(s), s)])ds, t∈[0, T]. (3.4)
In order to proved this theorem first we need to show that Fu ∈ CTα0−1 for any u∈ CTα0−1. Clearly,
F :CTα → CTα.
Ifu∈ CTα0−1, T > t2 > t1>0, and 0≤α <1,then we get k(Fu)(t2)−(Fu)(t1)kα−1
≤ k(S(t2)−S(t1))(u0+g(0, u0))kα−1
+kA−1kkAαg(t2, u(a(t2)))−Aαg(t1, u(a(t1)))k
+
Z t1
0 k
(S(t2−s)−S(t1−s))Aα−1kkf(s, u(s), u[h(u(s), s)])kds
+
Z t2
t1
kS(t2−s)Aα−1kkf(s, u(s), u[h(u(s), s)])kds. (3.5)
We have,
k(S(t2)−S(t1))(u0+g(0, u0))kα−1 ≤
Z t2
t1
kAα−1S′(s)(u0+g(0, u0))kds
=
Z t2
t1
kAαS(s)(u0+g(0, u0))kds
≤
Z t2
t1
kS(s)k[ku0kα+kg(0, u0)kα]ds
≤ C1(t2−t1), (3.6)
whereC1= [ku0kα+kg(0, u0)kα]M.
Also, we can see that
kAα−1g(t2, u(a(t2)))−Aα−1g(t1, u(a(t1)))k ≤ kA−1kkAαg(t2, u(a(t2)))−Aαg(t1, u(a(t1)))k ≤ kA−1kLg[(t2−t1) +ku(a(t2))−u(a(t1))kα−1]
≤ kA−1k[Lg+LLa](t2−t1). (3.7)
We observe that,
k(S(t2−s)−S(t1−s))kα−1 ≤
Z t2−t1
0 k
≤
Z t2−t1
0 k
S(l)AαS(t1−s)kdl
≤ M Cα(t2−t1)(t1−s)−α. (3.8)
Now we use the inequality (3.8) to get the inequality given below,
Z t1
0 k
(S(t2−s)−S(t1−s))Aα−1kkf(s, u(s), u[h(u(s), s)])kds≤C2(t2−t1), (3.9)
whereC2=N M CαT
1−α 0 1−α.
Similarly,
Z t2
t1
kS(t2−s)Aα−1kkf(s, u(s), u[h(u(s), s)])kds ≤C3(t2−t1), (3.10)
whereC3=kAα−1kM N.
We use the inequalities (3.6) (3.7) (3.9) and (3.10) in inequality (3.5) and get the following inequality,
k(Fu)(t2)−(Fu)(t1)kα−1 ≤L|t2−t1|, (3.11)
where,L= C1+C2+C3+kA−1kLg 1−kA−1kL
a .Hence,F :C
α−1
T0 → C
α−1
T0 .
Our next task is to show thatF :W → W.Now, for t∈(0, T0] and u∈ W, we have k(Fu)(t)−u0kα
≤ k(S(t)−I)Aα[u0+g(0, u0)]k
+kAαg(s, u(a(s)))−Aαg(0, u(a(0)))k +
Z t
0 k
S(t−s)Aαkkf(s, u(s), u[h(u(s), s)])kds
≤ k(S(t)−I)Aα[u0+g(0, u0)]k+Lg[T0+δ] +CαN
T01−α 1−α. Hence, from inequalities (3.2) and (3.3), we get
kFu−u0kT0,α≤δ.
Therefore,F :W → W.
Now, if t∈(0, T0] and u, v ∈ W,then k(Fu)(t)−(Fv)(t)kα
≤ kAαg(t, u(a(t)))−Aαg(t, v(a(t)))k +
Z t
0 k
We have the following inequalities,
kAαg(t, u(a(t)))−Aαg(t, v(a(t)))k ≤ LgkA−1kku−vkT0,α, (3.13)
kf(s, u(s), u[h(u(s), s)])−f(s, v(s), v[h(v(s), s)])k ≤ Lf[2 +LLh]ku−vkT0,α. (3.14)
We use the inequalities (3.13) and (3.14) in the inequality (3.12) and get
k(Fu)(t)−(Fv)(t)kα ≤
LgkA−1k+CαLf[2 +LLh]
T01−α 1−α
ku−vkT0,α. (3.15)
Hence from inequality (3.1), we get the following inequality given below
kFu− FvkT0,α<ku−vkT0,α.
Therefore, the mapF has a unique fixed pointu∈ W which is given by, u(t) = S(t)[u0+g(0, u0)]−g(t, u(a(t)))
+
Z t
0
S(t−s)f(s, u(s), u[h(u(s), s)])ds, t∈[0, T0]. (3.16)
Hence, the mild solutionuof equation (1.1) is given by the equation (3.16) and belong toCα T0 ∩C
α−1
T0 . Theorem 3.2 Let us assume that the assumptions (A1)-(A5) are hold andu0 ∈D(Aα)for0≤α <
1.Then, the differential equation (1.1) has a unique local solution in the sense of the Definition 2.2.
Proof. In order to prove this theorem, we first need to prove that the mild solution u is H¨older continuous on (0, T0]. From Theorem 2.6.13 in Pazy [11], it follows that for every 0 < β <1−α,
t > s >0 and every 0< h <1, we have
k(S(h)−I)AαS(t−s)k ≤ CβhβkAα+βS(t−s)k
≤ Chβ(t−s)−(α+β), (3.17) whereC=CβCα+β.
For 0< t < t+h≤T0, we have ku(t+h)−u(t)kα
≤ k((S(h)−I)S(t)(u0+g(0, u0))kα
+kAαg(t2, u(a(t2)))−Aαg(t1, u(a(t1)))k
+
Z t
0 k
(S(h)−I)AαS(t−s)kkf(s, u(s), u[h(u(s), s)])kds
+
Z t+h
t k
We calculate the first term of the above inequality (3.18) as follows;
k(S(h)−I)S(t)Aα(u0+g(0, u0))k ≤ Ct−(α+β){ku0k+kg(0, u0)k}hβ
≤ M1hβ, (3.19)
whereM1=Ct−(α+β){ku0k+kg(0, u0)k} depends ontand blows up as tdecreases to zero.
Second term of the above inequality (3.18) we calculate as follows,
kAαg(t+h, u(a(t+h)))−Aαg(t, u(a(t)))k
≤ kAαg(t+h, u(a(t+h)))−Aαg(t, u(a(t)))k
≤Lg[h+kAkku(a(t+h))−u(a(t))kα−1] ≤Lg[h+kAkLLah]
≤M2h, (3.20)
whereM2=Lg[1 +kAkLLa] is a constant independent of t.
Third and the fourth term of the inequality (3.18) can be calculated as follows:
Z t
0 k
(S(h)−I)AαS(t−s)kkf(s, u(s), u[h(u(s), s)])kds
≤ChβN
Z t
0
(t−s)−(α+β)ds
≤M3hβ, (3.21)
Z t+h
t k
AαS(t+h−s)kkf(s, u(s), u[h(u(s), s)])kds, ≤ CαN
Z t+h
t
(t+h−s)−αds
≤ M4hβ, (3.22)
whereM3 and M4 can be chosen to be independent oft.
Therefore,
ku(t+h)−u(t)kα≤C
′
hβ,
whereC′ is a positive constant. Thus,u is locally H¨older continuous on (0, T0].
Hence,
kf(t, u(t), u[h(u(t), t)])−f(s, u(s), u[h(u(s), s)])k
≤Lf{|t−s|θ1 +ku(t)−u(s)kα+L|h(u(t), t)−h(u(s), s)|}
≤Lf{|t−s|θ1 +ku(t)−u(s)kα+LLh[|t−s|θ2+ku(t)−u(s)kα]}
≤Lf{|t−s|θ1 +C
′
|t−s|β +LLh[|t−s|θ2 +C
′
|t−s|β]}. (3.23) Hence, the mapt7→f(t, u(t), u[h(u(t), t)]) is locally H¨older continuous. Therefore,
f(t, u(t), u[h(u(t), t)]) ∈C([0, T], X)∩Cβ′((0, T], X),
where 0 < β′ < min{θ1, β, θ2}. Similarly, we can prove that u(.) +g(., u(a(.))) is also H¨older
continuous on (0, T0].Therefore from Theorem 3.1 pp. 110 and Corollary 3.3, pp. 113, Pazy [11], the
functionu(.) +g(., u(a(.))) ∈CTα0−1∩C1((0, T0), X)∩C([0, T0], X) and u(.) is the unique solution
4
Existence of Global Solutions
then the initial value problem (1.1) has a unique solution which exists for allt∈[0, T].
Proof: By theorem (3.1) we can continue the solution of equation (1.1) as long as ku(t)kα stays
bounded. It is therefore sufficient to show that ifuexists on [0, T[ thenku(t)kαis bounded ast↑T.
1−k3(T) . Hence by applying the Gronwall’s lemma
5
Examples
LetX=L2(0,1).We consider the following partial differential equations with deviated argument,
∂t[w(t, x) +∂xf1(t, w(a(t), x))]−∂x2[w(t, x) +f1(t, w(a(t), x))]
=f2(x, w(t, x)) +f3(t, x, w(t, x)), x∈(0,1), t >0,
w(t,0) =w(t,1) = 0, t∈[0, T], 0< T <∞, w(0, x) =u0, x∈(0,1),
(5.1)
where
f2(x, w(t, x)) =
Z x
0
K(x, s)w(s, h(t)(a1|w(s, t)|+b1|ws(s, t)|))ds.
The function f3 :R+×[0,1]×R→ R is measurable in x, locally H¨older continuous in t, locally
Lipschitz continuous inw and uniformly in x.Further we assume that a1, b1 ≥0,(a1, b1) 6= (0,0),
h : R+ → R+ is locally H¨older continuous in t with h(0) = 0 and K : [0,1]×[0,1] → R. The functionf1:R+×R→Ris locally H¨older continuous int, locally Lipschitz continuous inw.
We define an operatorA,as follows,
Au=−u′′ with u∈D(A) ={u∈H01(0,1)∩H2(0,1) : u′′∈X}. (5.2) Here clearly the operatorAis self-adjoint with compact resolvent and is the infinitesimal generator of an analytic semigroup S(t). Now we take α = 1/2, D(A1/2) = H01(0,1) is the Banach space endowed with the norm,
kxk1/2 :=kA1/2xk, x∈D(A1/2)
and we denote this space byX1/2.Also, fort∈[0, T], we denote
Ct1/2=C([0, t];D(A1/2)), endowed with the sup norm
kψkt,1/2 := sup
0≤η≤tk
ψ(η)kα, ψ∈ Ct1/2.
We observe some properties of the operators A and A1/2 defined by (5.2). For u ∈D(A) and λ∈R, withAu=−u′′=λu, we have hAu, ui =hλu, ui; that is,
−u′′, u
=|u′|2L2 =λ|u|2
L2
soλ >0. A solution u ofAu=λuis of the form
u(x) =Ccos(√λx) +Dsin(√λx)
and the conditionsu(0) =u(1) = 0 imply that C = 0 and λ=λn =n2π2,n∈N. Thus, for each
n∈N, the corresponding solution is given by
un(x) =Dsin(
p
We havehun, umi= 0 for n6=m and hun, uni= 1 and hence D=
The equation (5.1) can be reformulated as the following abstract equation inX =L2(0,1): d
withQ(., t)∈X andQis continuous in its second argument. We can easily verify that the function f satisfies the assumptions (H1)-(H4). For more details see [6].
For the function awe can take
(i)a(t) =kt, where t∈[0, T] and 0< k≤1.
(ii) a(t) =ktnfort∈I = [0,1] k∈(0,1] andn∈N;
(iii) a(t) =ksint fort∈I = [0,π2],and k∈(0,1].
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