Gen. Math. Notes, Vol. 19, No. 1, November, 2013, pp.6-15
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2013
www.i-csrs.org
Available free online at http://www.geman.in
Fractional Kirchhoff Equation in Scale
of Banach Spaces
Abdelhamid Mohammed Djaouti1 and Mekki Terbeche2
1,2
Department of mathematics, Faculty of sciences
University of Oran, Algeria
1
Laboratory of Mathematics and Applications
Chlef University, Algeria
1
E-mail: djaouti− abdelhamid@yahoo.fr; a.mohammeddjaouti@univ-chlef.dz
2
E-mail: terbeche2000@yahoo.fr
(Received: 2-8-13 / Accepted: 16-9-13)
Abstract
This paper is devoted to investigate the global in time the existence of a
solution of a fractional Kirchhoff equation in scale of Banach spaces using
Ovsjannikov techniques.
Keywords: Caputo fractional derivative, Kirchhoff equation, Ovsjannikov
theorem, scale of Banach spaces, Tonelli approximation method.
1
Introduction
An abstract form of Cauchy-Kowalewski problem in scale of Banach spaces
was introduced by Ovsjannikov [11] and Yamanaka [20]. Other aspects of the
linear Cauchy problem were extended by Terbeche [15], Treves [16] and Nirenberg [9] have studied the existence and uniquness of such problem. In 1962,
Nirenberg [9] obtained the existence and uniqueness result for the abstract
nonlinear Cauchy problem. In 1977, Nishida [5] has simplified the iteration
procedure in Nirenberg theorem replacing strong differentiability by Lipschitz
conditions.
Using Tonelli aproximations, Reissig [13] has generalized the Ovsjannikov theorem in scale of Banach spaces with completely continuous imbedding. In
Fractional Kirchhoff Equation in Scale...
7
2004, [2] Barkova and Zabreiko have extended the previous results to fractional differential equation in Banach scale.
In this paper, we shall give another condition on the equation studied by
[2] to prove the existence results on Tonelli approximation using Ovsjannikov
method. We apply these results to Cauchy problem for fractional generalized
Kirchhoff equations. We assume in this article another condition on the righthand side of the equation studied in [2], and use the Ovsjannikov technique to
prove an existence results based on the approximation of Tonelli. These results
are applied in the second part to prove the existence of solution of fractional
Kirchhoff equations
2
Fractional Differential Equations in Scale of
Banach Spaces
Let (Es , k.ks )0≤s≤1 be a scale of Banach spaces continuously embedded in a
separable topological linear space X. This scale has the following properties
• (i) Es0 ⊂ Es00 , ∀s00 < s0 ,
• (ii) the canonical injection Es0 → Es00 has a norm less than 1.
Let f (t, u) be a function from [0, T ] × Xs0 to Xs00 satisfying the following
estimate
C
kuks0
(1)
kf (t, u)ks00 ≤ 0
(s − s00 )2
where C is a constant which does not depend on s0 and s00 .
Consider the Cauchy problem
Dα u(t) = f (t, u), α > 2, m − 1 < α ≤ m
u(k) (0) = uk , k = 0, 1, 2, ..., m − 1.
(2)
(3)
The initial values uk belong to X1 , and Dα denotes the Caputo fractional
derivative of order α defined by
Dα u(t) =
Z t
1
(t − z)m−α−1 u(m) (z)dz
Γ(m − α) 0
(4)
where m − 1 < α ≤ m, m ∈ N, t > 0.
Theorem 2.1 Let 0 < s < 1, uk ∈ E1 . If the condition (1) is verified, then
the Cauchy problem (2)-(3) has at least one continuous solution in Es defined
on [0, T].
8
Abdelhamid Mohammed Djaouti et al.
Proof. Let s < σ 0 < 1, and let us prove that the problem (2)-(3) is equivalent
to the following Volterra equation in C([0, T ], Xσ0 )
u(t) = t +
u1 t u2 t2
um−1 tm−1
1 Zt
+
+ ... +
+
(t − z)α−1 f (z, u(z))dz
1!
2!
(m − 1)!
Γ(α) 0
Consider the sequence of equations
u1 t u2 t2
um−1 tm−1
1 Zt
T
un (t) = t+
+
+...+
+
(t−z)α−1 f (z, un (z− ))dz, n = 1, 2, ...
1!
2!
(m − 1)! Γ(α) 0
n
(5)
The functions un (t), (n = 1, 2, ...) are called the Tonelli approximations for the
Cauchy problem (2)-(3) and defined by
(
un (t) =
A (t)
A0 (t) +
0
1 Rt
(t − z)α−1 f (z, u
Γ(α) 0
if − Tn ≤ t ≤ 0
T
0 ≤ t ≤ Tn
n (z − n ))dz if
m−1
2
t
2t
with A0 (t) = t + u1!1 t + u2!
+ ... + um−1
, and u1 , u2 , ..., um−1 ∈ X1 .
(m−1)!
Now, we shall prove that the sequence un (t) is uniformly bounded in the space
C([0, T ], Xσ0 ). Consider σ such that σ, σ 0 ≤ σ ≤ 1, and set
h(σ) = sup−T ≤t≤T kA0 (t)kσ .
(6)
If 0 ≤ t ≤ Tn , then
t
1
α−1
kun (t)kσ = kA0 (t) + Γ(α)
f (z, un (z − Tn ))dzkσ
0 (t − z)
R
t
1
α−1 C
kun (z − Tn )k1 dz
≤ h(σ) + Γ(α)
0 (t − z)
(1−σ)2
R
t
1
α−1 C
≤ h(σ) + Γ(α)
kA0 (z − Tn )k1 dz
0 (t − z)
(1−σ)2
α
C
1
. tα (1−σ)
≤
h(σ) + Γ(α)
2 h(1).
R
Hence
kun (t)kσ ≤ h(σ) + h(1).
C
tα
.
.
(1 − σ)2 Γ(α + 1)
This inequality holds for all σ such that σ 0 ≤ σ ≤ 1, therefore we can assume
that σ = σ 0 .
Since h(σ 0 ) ≤ h(1) we obtain
C
tα
kun (t)kσ0 ≤ h(1) 1 +
.
.
(1 − σ 0 )2 Γ(α + 1)
"
Similarly, for
T
n
kun (t)kσ ≤
≤
≤
#
≤ t ≤ 2 Tn and σ 0 ≤ σ ≤ s1 < 1, we have
1 Rt
C
T
(t − z)α−1 (s1 −σ)
2 kun (z − n )ks1 dz
Γ(α) 0
h
i
R
t
1
C
C
tα
h(σ) + Γ(α)
− z)α−1 (s1 −σ)
2 h(s1 ) + (1−s )2 . Γ(α+1) h(1) dz
0 (t
1
2
C
tα
t2α
h(σ) + h(s1 ) (s1 −σ)
+ h(1) (1−s1 )C2 (s1 −σ)2 Γ(1+2α)
.
2 . Γ(1+α)
h(σ) +
9
Fractional Kirchhoff Equation in Scale...
Setting σ = σ 0 and using (6), we get
h(σ 0 ) ≤ h(s1 ) ≤ h(1)
and
C2
C
tα
t2α
+
.
kun (t)kσ0 ≤ h(1) 1 +
.
(s1 − σ 0 )2 Γ(1 + α) (1 − s1 )2 (s1 − σ 0 )2 Γ(1 + 2α)
)
(
Proceeding in the same way, we can estimate the Tonelli approximations un (t)
on every interval (j − 1) Tn ≤ t ≤ j Tn for all σ 0 ≤ sj−1 ≤ ... ≤ s1 < 1 to have
n
α
C
. t
+ ...+
(sj−1 −σ 0 )2 Γ(1+α)
o
Cj
tjα
.
.
(sj−1 −σ 0 )2 (sj−2 −sj−1 )2 ...(1−s1 )2 Γ(1+jα)
kun (t)kσ0 ≤
h(1) 1 +
For 0 ≤ t ≤ T, σ 0 ≤ ... ≤ sn−1 ≤ ... ≤ s1 < 1, we obtain the estimate
n
kun (t)kσ0 ≤
h(1) 1 +
α
C
. t
(sn−1 −σ 0 )2 Γ(1+α)
Cn
(sn−1
−σ 0 )2 (s
+ ...+
o
nα
n−2 −sn−1
)2 ...(1−s
1
)2
t
. Γ(1+nα)
+ ...
this series converges for all t ≤ T , provided that
T nα
Cn
.
(sn−1 − σ 0 )2 (sn−2 − sn−1 )2 ...(1 − s1 )2 Γ(1 + nα)
limn→∞
!1
n
< 1.
let us define the sequence sn by
γ(k)
sn−1 − sn =
where γ(k) =
1
k
12
1
+
k
+ ... +
22
1
k
k
(1 − σ 0 )
n2
−1
+ ...
.
n2
Then
limn→∞
nα
Cn
. T
(sn−1 −σ 0 )2 (sn−2 −sn−1 )2 ...(1−s1 )2 Γ(1+nα)
C n (n!)k
T nα
2n (1−σ 0 )2n . Γ(1+nα)
(γ(k))
k
α
(n!) n
.
limn→∞ (γ(k))CT
1
2 (1−σ 0 )2
Γ(1+nα) n
= limn→∞
=
= limn→∞
= limn→∞
n
1
n
k
2n k α
CT α
. (2πn)1 n eα k
(γ(k))2 (1−σ 0 )2 (παn) 2n
(nα) e
k
1
k
2n α
CT α
. (2πn)1 eα k . nnα
(γ(k))2 (1−σ 0 )2 (παn) 2n
α e
= 0.
hence the Tonelli approximations un (t) are uniformly bounded in C([0, T ], Xσ0 ).
10
Abdelhamid Mohammed Djaouti et al.
Now, we shall prove that the approximations are equicontinuous in the
space C([0, T ], Xσ0 ).
let t ∈]0, T [, ∆t ∈ [0, T − ∆t], s < σ 0 < 1, then
R
1
kun (t + ∆t) − un (t)ks ≤
k 0t+∆t (t + ∆t − z)α−1 f (z, un (z − Tn ))dz
Γ(α)
R
− 0t (t − z)α−1 f (z, un (z − Tn ))dzks
R
1
≤ Γ(α)
k 0t [(t + ∆t − z)α−1 − (t − z)α−1 ] f (z, un (z − Tn ))dzks
R
T
α−1
+k tt+∆t
(t + ∆t − αz) α f (z, un (z − n ))dzks
(t+∆t) −t
C
≤
kun (z − Tn )kσ0
(σ 0 −s)2
Γ(α+1)
α−1
C
T
∆t
≤
kun (z − Tn )kσ0
(σ 0 −s)2 Γ(α+1)
but the Tonelli sequence un (t) is uniformly bounded in the space C([0, T ], Xσ0 ),
hence obtain the equicontinuity in C([0, T ], Xs ). Consequently the sequence
un (t) is precompact in C([0, T ], Xs ) for t ∈ [0, T ], and s < σ 0 .
According to Ascoli-Arzela theorem, there is a subsequence unk (t) of un (t)
which converges to u∗ (t) in C([0, T ], Xs ) and the sequence unk (t − Tn ) converges
also to u∗ (t) in C([0, T ], Xs ).
Finally to complete the proof of existence we need to prove the continuity
of the following integral operator A
1 Zt
(t − z)α−1 f (z, u(z))dz
A(t) = A0 (t) +
Γ(α) 0
(7)
where
A : C([0, T ], Xs ) → C([0, T ], Xσ00 )
for σ 00 < s.
Indeed, if un (t) ∈ C([0, T ], Xs ) converges to a function u∗ (t) ∈ C([0, T ], Xs )
then f (t, un (t)) converges to f (t, u∗ (t)) because f is continuous with respect
to t for all t ∈ [0, T ], otherwise we have σ 00 < s
kf (t, un (t))kσ00 ≤
C
kun (t)ks
(s − σ 00 )2
and since un (t) is bounded in C([0, T ], Xs ) we get
kf (t, un (t))kσ00 ≤
C
M
(s − σ 00 )2
where M is a constant.
So we can pass to the limit in the formula
T
1 Zt
un (t) = A0 (t) +
(t − z)α−1 f (z, un (z − ))dz
Γ(α) 0
n
11
Fractional Kirchhoff Equation in Scale...
we find
1 Zt
(t − z)α−1 f (z, u∗ (z))dz
Γ(α) 0
then u∗ (t) is the solution of the problem (2)-(3) and the proof is achieved.
u∗ (t) = A0 (t) +
3
3.1
Fractional Kirchhoff Equation
Preliminary
Let Ω be an open subset of Rn , we denote by G(Ω) the class of real functions
of C ∞ (Ω) satisfying
β!
L|β|
where β = (β1 , β2 , ..., βn ) ∈ N n , kuk = sup {ku(x)k, x ∈ Ω} , β! = β1 !β2 !...βn !,
and |β| = β1 + β2 + ... + βn .
For any 0 < s < 1 we denote by Es the space of all functions u ∈ C ∞ (Ω) such
that
X
s|β|
< ∞.
(8)
kuks =
kDβ uk
β!
β∈N n
∃K > 0, ∃L > 0, kDβ uk ≤ K
(Es , k.ks ) form a scale of Banach spaces, furthermore G(Ω) =
S
s>0
Es .
Lemma 3.1 The scale (Es , k.ks )0<a≤s<1 has the following properties:
(a) if u, v ∈ Es then u.v ∈ Es and ku.vks ≤ kuks kvks ,
(b) there exists a constant C > 0 independent of s and s0 , such that
k∆uks ≤
C
kuks0
(s0 − s)2
∆ is the Laplacian operator and 0 < a ≤ s ≤ s0 < 1.
Proof. (a) Let u ∈ Es and v ∈ Es
P
|β|
β∈N n
kDβ (uv)k sβ!
|β|
β
β−γ
=
vkCγβ sβ!
β∈N n
γ≤α kD ukkD
P
P
β!
s|β|
β
β−γ
=
vk γ!(β−γ)!
β∈N n
γ≤α kD ukkD
β!
P
≤
≤
<
P
P
β∈N n
|β|
P
γ≤α
then uv ∈ Es and kuvks ≤ kuks kvks .
(b) Set β + 2 = (β1 + 2, β2 , ..., βn ) then
β ∂ 2 u s|β|
D
β!
2
∂x
=
1
|β|
kDβ+2 uk sβ!
0|β|+2
≤
|β|+2
(β1 +2)(β1 +1)
s2
s0|β+2|
4s02
β+2
kD uk (β+2)! e2 s2 (s0 −s)2 .
s
= kDβ+2 uk (β+2)!
|β−γ|
s
kDβ uk sγ! kDβ−γ vk (β−γ)!
kuks kvks
∞
s
s0
12
Abdelhamid Mohammed Djaouti et al.
Since
|β|+2
(β1 +2)(β1 +1)
s
s0
|β|+2
(β1 +2)2
s
≤
≤
s2
s0
s2
4s02
e2 s2 (s0 −s)2
then
P
s|β|
β
β∈N n kD ∆uk β!
P
β ∂2u
∂2u
∂ 2 u s|β|
D
β!
+
+
...
+
n
2
2
β∈N
∂x2n
∂x1
∂x2
P
β+2
β+2
β+2 s|β|
β!
u
+
D
u
+
...
+
D
u
n Dx
β∈N
x2
xn
1
n
|β|
|β|
k∆uks =
=
=
|β|
o
β∈N n
kDxβ+2
u sβ! k + kDxβ+2
u sβ! k + ... + kDxβ+2
u sβ! k
n
1
2
β∈N n
s
4s
s
4s
kDxβ+2
uk (β+2)!
+ kDxβ+2
uk (β+2)!
1
2
e2 s2 (s0 −s)2
e2 s2 (s0 −s)2
≤
P
≤
P
0|β+2|
02
0|β+2|
0|β+2|
02
02
s
4s
uk (β+2)!
+... + kDxβ+2
n
e2 s2 (s0 −s)2
2
nkuks0
s0
0 2
es2 (s−s )
nkuks0
1
4 ea (s−s0 )2 .
≤
4
≤
If we take C = 4n
3.2
1
ea
2
then the result follows.
Cauchy Problem for Fractional Generalized Kirchhoff Equations
Consider the following Cauchy problem
Dtα u(t, x) = f (t, x,
Z
P
|∇x u|2 dx)∆x u(t, x), α > 2, m − 1 < α ≤ m
uk (0, x) = uk (x), k = 0, 1, 2, ..., m − 1
(9)
(10)
where (t, x) ∈ ΩT = [0, T ] × Ω and P, Ω are open subsets in Rn with P ⊂ Ω
bounded.
Lemma 3.2 Assume the following hypotheses
(A) Dxβ f (t, x, u) ∈ C ∞ (ΩT × R+ , R),
(B) There are two constants L > 0 and K > 0 such that
β
Dx f (t, x, u)
≤K
β!
.
L|β|
If (A) and (B) are satisfied, then the operator Bu(t) = f (t, x, P |∇x u|2 dx)
is continuous from C 1 ([0, T ], Es ) into C 1 ([0, T ], E1 ) with 0 < s < 1, and
kBu(t)k1 is finite.
R
Proof. The operator B can be written as
B =G◦F
13
Fractional Kirchhoff Equation in Scale...
where
and
F : C 1 ([0, T ], Es ) →
7
C([0,R T ], R)
u(t)
7
→
F u(t) = P |∇x u|2 dx
G : C([0, T ], R) →
7
C([0, T ], E1 )
u(t)
7
→
Gu(t) = f (t, x, u(t)).
Consider Va = {u ∈ C 1 ([0, T ], Es ), kuks ≤ a} .
We have
s|β|
kDβ uk
≤ kuks
β!
for all β ∈ N n . If βi = 1 and βj = 0 for all j = 0, 1, ..., n with j 6= i; then
k
therefore
k
∂u
ks ≤ kuks
∂xi
1
a
∂u
k ≤ kuks ≤ .
∂xi
s
s
For all u, v ∈ Va we have
∂u 2
∂xi
−
2 ∂v
∂xi
∂u
= ∂xi
≤
≤
∂u
∂v ∂x − ∂x
i
i ∂u
2a
∂v
− ∂x
∂x
s
+
∂v
∂xi
i
2a
ku
s2
i
− vks
therefore
P
n
∂u 2
i=1 ( ∂x
)
i
hence
−
Pn
∂v 2 i=1 ( ∂xi ) = k
≤
=
2
2
∂u
∂v
−
i=1 ∂xi
∂xi
Pn 2a
ku
−
vk
s
i=1 s2
2na
ku − vks
s2
Pn
k
kF u − F vk = k PR|∇x u|2 dx − P |∇x v|2 dxk
2
2
=
k P (|∇
x u| − |∇x v| )dxk
R 2na
≤
k P s2 ku − vks dxk
2namesP
≤
ku − vks
s2
R
R
which proves that F is lipschitzian and its continuity is deduced.
To end the proof we need to show the boundedness of the operator G. In fact,
we have
1
β
kGu(t)k1 =
β∈N n kD f (t, x, u(t))k β!
P
β! 1
≤
(by hypothesis)
β∈N n K L|β| . β!
P
≤
for all t ∈ [0, T ] and u ∈ C([0, T ], R).
Consequently kBu(t)k1 is finite.
K
P
β∈N n
|β|
1
L
14
Abdelhamid Mohammed Djaouti et al.
Theorem 3.3 Let 0 < s < 1, uk ∈ E1 , if the hypotheses (A) and (B)
be satisfied. Then the Cauchy problem (9)-(10) has at least one continuous
solution in C 1 ([0, T ], Es ).
Proof. Let us consider the scale of Banach spaces defined in lemma 3.1. The
Cauchy problem (9)-(10) has the same form as (2)-(3) and we have
kf (t, x,
|∇x u|2 dx)∆x u(t, x)ks ≤ kf (t, x, P |∇x u|2 dx)ks k∆x u(t, x)ks
C
≤
M. (s0 −s)
2 ku(t, x)ks0 .
R
R
P
hence the assumptions of the Theorem 2.1 are verified and the proof is completed.
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