Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 281, pp. 1–24.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NONLINEAR INITIAL BOUNDARY-VALUE PROBLEMS WITH
RIESZ FRACTIONAL DERIVATIVE
MARTÍN P. ÁRCIGA-ALEJANDRE
Abstract. We consider an initial boundary-value problem for a nonlinear
partial differential equation with fractional derivative of Riesz type on a halfline. We study local and global existence of solutions in time, as well as the
asymptotic behavior of solutions for large time.
1. Introduction
We study the existence of local and global solutions, and the asymptotic behaviour, for the initial boundary-value problem
ut + N (u) + |∂x |α u = 0, t > 0, x > 0;
u(x, 0) = u0 (x),
u(0, t) = h(t),
x > 0,
(1.1)
t > 0.
σ
where N (u) = |u| u, α < σ < α + 1, 2/5 < α < 1, and |∂x |α is a fractional
derivative of Riesz type defined by
|∂x |α u = R1−α+[α] ∂x[α]+1 u.
Here, [α] denotes the integer part of the number α > 0, α ∈
/ Z, and R
modified Riesz Potential (see [15]),
Z +∞
1
sgn(x − y)
Rα u =
u(y)dy.
π
2Γ(α) sin( 2 α 0
|x − y|1−α
(1.2)
α
is the
The Cauchy problem for nonlinear nonlocal dissipative equations has been extensively studied. In particular, the large time asymptotic behavior for the Cauchy
problem for different nonlinear equations is investigated in [10] and the references
therein.
Boundary value problems arise in many applications and play an important
role in the contemporary mathematical physics. For the study of the effect of the
boundary data on the qualitative properties of the solution the reader is referred
to [2, 3, 4, 5, 6, 7, 16] and references therein.
The general theory of nonlinear nonlocal equations on a half-line was developed
in [9], where the pseudodifferential operator K (|∂x |α in our case) defined on a
2010 Mathematics Subject Classification. 35S11, 47G30, 58J40.
Key words and phrases. Riesz fractional derivative; Riemann-Hilbert problem;
dissipative nonlinear evolution equation; large time asymptotic.
c
2015
Texas State University.
Submitted September 26, 2013. Published November 10, 2015.
1
2
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
half-line was introduced by virtue of the inverse Laplace transformation; it is given
by
Z i∞+σ
[αj ]
X
1 X
px αj
Ku =
u
b(p) −
∂xl−1 u(0)p−l dp.
Cj
e p
2πi j
−i∞+σ
l=1
P
αj
Note that the symbol K(p) =
C
p
is
analytic
in the complex right halfj j
plane. We emphasize that the operator |∂x |α in equation (1.2) has a nonanalytic
nonhomogeneous symbol K(p) = |p|α and the general theory in [9] can not be
applied to problem (1.1) directly.
As far as we know there are few results on the initial-boundary value problems
with pseudodifferential equations having a nonanalytic symbol. The case of rational
symbols K(p) with poles in the complex right half-plane was studied in [11, 12],
where it was proposed a new method for constructing the Green operator based
on the introduction of necessary conditions at the singular points of the symbol
K(p). In [13] the initial-boundary value problem for a pseudodifferential equation
with a nonanalytic homogeneous symbol K(p) = |p|1/2 was studied, a theory of
sectionally analytic functions was implemented for proving that the initial-boundary
value problem is well-posed. Since the symbol K(p) = |p|1/2 does not grow fast at
infinity, no boundary data is needed. Finally, the case of K(p) = |p|α , α ∈ (1/2, 1),
was studied in [1]. In the present work we consider the nonlinear version of the
problem [14], considering α ∈ (2/5, 1). In order to construct a Green operator we
follow the methods used in [1, 14].
To state precisely the results of the present paper we give some notations. We
t
denote hti = 1 + t, {t} = hti
. Here and below pα is the main branch of the complex
analytic function in the complex half-plane Re(p) ≥ 0, so that 1α = 1 (we make a
cut along the negative real axis (−∞, 0)). Note that due to the analyticity of pα
for all Re(p) > 0 the inverse Laplace transform gives us 0 for all x < 0. The direct
Laplace transformation Lx→p is given by
Z +∞
u
b(p) ≡ Lx→p {u} =
e−px u(x)dx
0
and the inverse Laplace transformation
L−1
p→x
is defined by
Z i∞
1
epx u(p)dp.
u(x) ≡ L−1
u} =
p→x {b
2πi −i∞
The norm in weighted Lebesgue space Ls,µ (R+ ) = {ϕ ∈ S 0 ; kϕk
by
Z +∞
1
kϕkLs,µ = (
xµs |ϕ(x)|s dx) s
Ls,µ
< ∞} is given
0
for µ > 0, 1 < s < ∞ and
kϕkL∞ = ess supx∈R+ |ϕ(x)|.
We now introduce the spaces for initial data and solutions on R+ :
Zν = {φ ∈ L1 ∩ L1,ν ∩ L∞ : kφkZν < ∞},
with the norm
kφkZν = kφkL1 + kφkL1,ν + kφkL∞ ,
X = {ϕ ∈ C([0, ∞); L1 ) ∩ C((0, ∞); Ls ∩ Ls,µ ∩ L∞ ) : kϕkX µ < ∞},
µ
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
3
where s ≥ 1, with the norm
1
1
1
1
1
kϕkX µ = sup(t α (1− s ) kϕkLs + t α (1− s −µ) kϕkLs,µ + {t}γ hti α kϕkL∞ )
t>0
where 0 < γ < 1 and |1 −
1
s
− µ| < α. We also define the space
B α = {h ∈ C 1 (0, ∞) : khkB α < ∞}
for the boundary data, where the norm
1
khkB α = suphti α (|h(t)| + h1 + ti|h0 (t)|).
t>0
Now, to state the main results we introduce Λ(s), Υ(s) ∈ L∞ (R+ ) by the formulae
i∞
Z
ξ
Λ(s) = C
i∞
Z
eps
dξe
−i∞
−i∞
pα/2
L(p, ξ)dp
K(p) + ξ
(1.3)
and
i∞
1
2πi
Z
1
+
2πi
Z
Υ(s) = −
dξeξ
−i∞
i∞
Z
i∞
eps
−i∞
e
Y + (p, ξ) +
I (p, ξ)dp
K(p) + ξ
ps−K(p) K(p)
dp,
p
−i∞
(1.4)
where
1
1
)2 + Γ+
,
ξξ +
K(p) + ξ
(K(p) + ξ)2
Z i∞
1
1
K(q)
1
I(z, ξ) =
dq.
2πi −i∞ q − z Y + (q, ξ) q
L(p, ξ) = (Γ+
ξ −
+
Here, Y + (p, ξ) = (−i)α eΓ
Γ+ (p, ξ) = −
(p,ξ)
1
2πi
Z
,
i∞
ln− (q − p)(
−i∞
K10 (q)
K 0 (q)
−
)dq,
K(q) + ξ
K1 (q) + ξ
K(p) = |p|α , K1 (p) = pα and ln− (z) = ln |z| + i arg− (z) for − 23 π < arg− (z) ≤
π
2.
Theorem 1.1. Let the initial data u0 ∈ Zµ and the boundary value h ∈ B α be
such that the correspondent norms are sufficiently small. Also, suppose that h(t) =
btβ + O(tβ−δ ), where β < − α1 − 12 − γ and δ, γ > 0. Then, the initial-boundary
value problem (1.1) has unique global solution u ∈ X µ and the following asymptotic
is valid
1
1
1
1
u(x, t) = t− α − 2 (aΛ(xt−1/α ) + bΥ(xt−1/α )) + O(t− α − 2 −γ ),
for t → ∞ in L∞ (R+ ), where
Z
a=
+∞
y
0
α/2
Z
u0 (y)dy −
+∞
Z
dτ
0
0
+∞
y α/2 N (u(τ ))dy.
4
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
2. Preliminaries
Definition 2.1. A function φ(q) is said to satisfy the Hölder condition on the curve
L, if there exist positive constants C and λ such that
|φ(q1 ) − φ(q2 )| ≤ C|q1 − q2 |λ ,
(2.1)
for any q1 , q2 ∈ L.
The proof of the next theorem can be found in [8].
Theorem 2.2. Let φ(q) be a complex function, which obeys the Hölder condition
(2.1) for all finite q and tends to a finite limit φ(∞) as q → ∞, such that for large
q the following inequality holds
|φ(q) − φ(∞)| ≤ C|q|µ ,
µ > 0.
Then, the Cauchy integral
Z
1
F (z) =
2πi
i∞
−i∞
φ(q)
dq
q−z
defines an analytic function in the left and right semi-planes. Here and below these
functions will be denoted by F + (z) and F − (z), respectively. These functions have
the limiting values F + (p) and F − (p) at all points of the imaginary axis, Re(p) = 0,
on approaching the contour from the left and from the right, respectively. These
limiting values are expressed by the Sokhotski-Plemelj formulae,
Z i∞
1
1
φ(q)
±
−
dq ± φ(p).
(2.2)
F (p) =
2πi −i∞ q − p
2
Subtracting the formulae (2.2) we obtain the following equivalent formula
F + (p) − F − (p) = φ(p),
(2.3)
which will be frequently employed hereafter. Now, we consider the following linear
initial-boundary value problem on half-line
ut + |∂x |α u = 0,
u(x, 0) = u0 (x),
u(0, t) = h(t),
x, t > 0;
x > 0,
(2.4)
t > 0.
Setting K(q) = |q|α , K1 (q) = q α , we define
Z +∞
G(t)φ =
G(x, y, t)φ(y)dy,
0
Z t
H(x)h =
H(x, t − τ )h(τ )dτ, H(x, t) = ∂t ∂y−1 G(x, y, t)|y=0 ,
0
where the function G(x, y, t) is given by
Z i∞
Z i∞
1
Y + (p, ξ) −
ξt
G(x, y, t) = −
dξe
epx
Z (p, ξ, y)dp,
2
(2πi) −i∞
K(p) + ξ
−i∞
where
1
Z (p, ξ, y) = z→p
lim
2πi
Rez>0
−
Z
i∞
−i∞
1
1
e−qy dq.
q − z Y + (q, ξ)
(2.5)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
5
+
for x > 0, y > 0, t > 0. Here, Y + (p, ξ) = (−i)α eΓ (p,ξ) , where
Z i∞
1
K 0 (q)
K10 (q)
Γ+ (p, ξ) = −
ln− (q − p)(
−
)dq
2πi −i∞
K(q) + ξ
K1 (q) + ξ
and ln− (z) = ln |z| + i arg− (z), − 23 π < arg− (z) ≤
understood in the sense of the principal values.
π
2,
where the integrals are
Proposition 2.3. If u0 (x) ∈ Zν , there exist a unique solution u(x, t) for the initialboundary value problem (2.4), which has an integral representation
u = G(t)u0 + H(x)h,
x > 0, t > 0.
(2.6)
Proof. To obtain an integral representation for solutions of problem (2.4) we suppose that there exist a solution u(x, t), which is continued by zero outside of x > 0:
u(x, t) = 0,
for all x < 0.
Let φ(q) be a function of the complex variable q, which obeys the Hölder condition
(2.1) for all q, such that Re(q) = 0. We define the operator P by
Z i∞
1
1
φ(q)dq, Re(z) 6= 0.
Pq→z {φ(q)} = −
2πi −i∞ q − z
Applying the Laplace transform with respect to x to |∂x |α u, for Re(q) > 0 we
obtain
u(0, t)
Lx→q {|∂x |α u(x, t)} = Pp→q {K(p)(b
u(p, t) −
)}.
(2.7)
p
Since u
b(q, t) is analytic, for all Re(q) > 0, we have
u
b(q, t) = Pp→q {b
u(p, t)}.
(2.8)
Applying the Laplace transform with respect to x to problem (2.4) and using (2.7)
and (2.8), for Re(p) = 0, we obtain
h(t) Pp→q u
bt (p, t) + K(p) u
b(p, t) −
= 0,
p
(2.9)
u
b(p, 0) = u
b0 (p),
We rewrite (2.9) in the form
K(p)
h(t) = Φ(p, t);
p
u
b(p, 0) = u
b0 (p),
u
bt (p, t) + K(p)b
u(p, t) −
(2.10)
with some function Φ(p, t) such that for Re(p) = 0
Pp→q {Φ(p, t)} = 0
and |Φ(p, t)| ≤ C|p|α−1−γ ,
|p| > 1, γ > 0.
(2.11)
Applying the Laplace transform with respect to time variable to (2.10) we find
1
K(p) b
b
b ξ) ,
u
b(p, ξ) =
u
b0 (p) +
h(ξ) + Φ(p,
(2.12)
K(p) + ξ
p
b
b ξ) are
where Re(p) = 0 and Re(ξ) > 0. Here, the functions u
b(p, ξ), b
h(ξ), and Φ(p,
the Laplace transforms for u
b(p, t), h(t), and Φ(p, t) with respect to time, respectively. In order to obtain an integral formula for solutions to the problem (2.4) it
b ξ) using the
is necessary to know the function Φ(p, t). We find the function Φ(p,
b
analytic properties of the function u
b in the right-half complex planes Re(p) > 0 and
6
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
Re(ξ) > 0. The equation (2.8) and the Sokhotski-Plemelj formulae (2.2) imply, for
Re(p) = 0,
Z i∞
1
1 b
b
u
b(p, ξ) = −
u
b(q, ξ)dq.
(2.13)
−
πi −i∞ q − p
In view of Sokhotski-Plemelj formulae (2.2), via (2.12) the condition (2.13) can be
written as
Θ+ (p, ξ) = −Λ+ (p, ξ),
(2.14)
where the sectionally analytic functions Θ(p, ξ) and Λ(p, ξ) are given by Cauchy
type integrals
Z i∞
1
1
1
b ξ)dq,
Φ(q,
(2.15)
Θ(z, ξ) =
2πi −i∞ q − z K(q) + ξ
Z i∞
1
1
1
K(q) b
Λ(z, ξ) =
(b
u0 (q) +
h(ξ))dq.
(2.16)
2πi −i∞ q − z K(q) + ξ
q
To perform the condition (2.14) in the form of a nonhomogeneous Riemann-Hilbert
problem we introduce the sectionally analytic function
Z i∞
1
1
K(q) b
Ω(z, ξ) =
Φ(q, ξ)dq.
(2.17)
2πi −i∞ q − z K(q) + ξ
Taking into account the assumed condition (2.11), we obtain
1
Θ− (p, ξ) = − Ω− (p, ξ).
ξ
Also observe that from (2.15) and (2.17) by Sokhotski-Plemelj formulae,
K(p)(Θ+ − Θ− ) = Ω+ − Ω− =
K(p) b
Φ.
K(p) + ξ
(2.18)
(2.19)
Substituting (2.14) and (2.18) into this equation we obtain for Re p = 0
Ω+ (p, ξ) = W (p, ξ)Ω− (p, ξ) + g(p, ξ),
(2.20)
K(p)+ξ
ξ
where W =
and g = −K(p)Λ+ . Equation (2.20) is the boundary condition
for a nonhomogeneous Riemann-Hilbert problem. It is required to find two functions for some fixed point ξ, Re ξ > 0: Ω+ (z, ξ), analytic in the left-half complex
plane Re z < 0 and Ω− (z, ξ), analytic in the right-half complex plane Re z > 0,
which satisfy on the contour Re p = 0 the relation (2.20).
Note that bearing in mind formula (2.19) we can find the unknown function
b ξ), which involved in the formula (2.12), by the relation
Φ(p,
b ξ) = K(p) + ξ (Ω+ (p, ξ) − Ω− (p, ξ)).
Φ(p,
K(p)
(2.21)
The method for solving the Riemann problem F + (p) = φ(p)F − (p) + ϕ(p) is
based on the following results. The proofs may be found in [8].
Lemma 2.4. An arbitrary function ϕ(p) given on the contour Re p = 0, satisfying
the Hölder condition, can be uniquely represented in the form ϕ(p) = U + (p) −
U − (p), where U ± (p) are the boundary values of the analytic functions U ± (z) and
the condition U ± (∞) = 0 holds. These functions are determined by formula
Z i∞
1
1
U (z) =
ϕ(q)dq
2πi −i∞ q − z
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
7
Lemma 2.5. An arbitrary function φ(p) given on the contour Re p = 0, satisfying
the Hölder condition, and having zero index,
Z i∞
1
ind φ(p) :=
d ln φ(p) =,
2πi −i∞
is uniquely representable as the ratio of the functions X + (p) and X − (p), constituting the boundary values of functions, X + (z) and X − (z), analytic in the left and
right complex semi-plane and having in these domains no zero. These functions are
determined to within an arbitrary constant factor and given by formula
Z i∞
1
1
±
Γ± (z)
X (z) = e
, Γ(z) =
ln φ(q)dq.
2πi −i∞ q − z
In the formulations of Lemmas 2.4 and 2.5 the coefficient φ(p) and the free term
ϕ(p) of the Riemann problem are required to satisfy the Hölder condition on the
contour Re p = 0. This restriction is essential. On the other hand, it is easy
to observe that both functions W (p, ξ) and g(p, ξ) do not have limiting value as
p → ±i∞. So we can not find the solution using ln W (p, ξ). The principal task
now is to get an expression equivalent to the boundary value problem (2.20), such
that the conditions of Lemmas are satisfied. First, we introduce the function
K(p) + ξ w− (p)
1
φ(p, ξ) =
, w± (p) =
,
(2.22)
K1 (p) + ξ w+ (p)
(p ∓ z0 )α/2
where K(p) = |p|α , K1 (p) = pα and z0 > 0. We make a cut in the plane z:
(−∞, −z0 ] ∪ [z0 , ∞). Owing to the manner of performing the cut the functions
w− (z), K1 (z) are analytic for Re z > 0 and the function w+ (z) is analytic for Re
z < 0.
We observe that the function φ(p, ξ) given on the contour Re p = 0, satisfies the
Hölder condition and K1 (p) + ξ does not vanish for any Re ξ > 0. Also we have
R i∞
1
ind φ(p, ξ) = 2πi
d ln φ(p, ξ) = 0. Therefore in accordance with Lemma 2.5,
−i∞
the function φ(p, ξ) can be represented as
φ(p, ξ) =
where Γ0 (z, ξ) =
obtain
1
2πi
R i∞
X + (p, ξ)
,
X − (p, ξ)
1
−i∞ q−z
X ± (p, ξ) =
lim eΓ0 (z,ξ) ,
z→p
±Rez<0
(2.23)
ln φ(q, ξ)dq. From equations (2.22) and (2.23) we
Y+
K(p) + ξ
=
,
Y−
K1 (p) + ξ
(2.24)
±
where Y ± = eΓ w± . Now, we show that Y ± do not depend of z0 . Integrating by
parts we obtain for Γ0 :
Z i∞
iR
1
1
Γ0 = lim
ln(q − z) ln φ(q, ξ)
−
ln(q − z)∂q ln φ(q, ξ)dq. (2.25)
R→∞ 2πi
2πi −i∞
−iR
Using that ln φ(±iR, ξ) → 0, as R → ∞, we obtain
Z i∞
1
Γ0 (z, ξ) = −
ln(q − z)∂q ln φ(q, ξ)dq.
2πi −i∞
(2.26)
8
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
We define ln± (z) = ln |z| + i arg± (z), where − 32 π < arg− (z) ≤
arg+ (z) < π2 . Then, for Re p = 0,
Z i∞
1
Γ±
(p,
ξ)
=
−
ln∓ (q − p)∂q ln φ(q, ξ)dq.
0
2πi −i∞
Now, since
∂q ln φ(q, ξ) = F (q, ξ) +
π
2
and − 23 π ≤
(2.27)
αz0
,
(q − z0 )(q + z0 )
K 0 (q)
0
K (q)
1
where F (q, ξ) = K(q)+ξ
− K1 (q)+ξ
, and via Cauchy’s Residue Theorem,
Z i∞
αz0
1
α
ln± (q − p)
−
dq = ±iπ + ln± (p ± z0 )α/2 ,
2πi −i∞
(q − z0 )(q + z0 )
2
from (2.27) follows that
Γ± (p, ξ) = ∓iπ
1
α
+ ln∓ (p ∓ z0 )α/2 −
2
2πi
Z
i∞
ln∓ (q − p)F (q, ξ)dq.
−i∞
Therefore,
±
Y ± (p, ξ) = (∓i)α eΓ
(p,ξ)
,
(2.28)
where
Z i∞
1
ln∓ (q − p)F (q, ξ)dq.
2πi −i∞
We note that equation (2.24) is equivalent to
Γ± (p, ξ) = −
Y + K1 (p) + ξ
K(p) + ξ
= −(
).
ξ
Y
ξ
Now, we return to the nonhomogeneous Riemann-Hilbert problem defined by the
boundary condition (2.20). We substitute the above equation in (2.20) and add
− Yξ+ Λ+ in both sides to get
K (p) + ξ Ω− K(p) + ξ Ω+ − ξΛ+
1
=
−
Λ+ .
(2.29)
+
Y
ξ
Y−
Y+
On the other hand, by Sokhotski-Plemelj formulae and (2.16),
1
K(p) b h(ξ) .
Λ+ − Λ− =
u
b0 (p) +
K(p) + ξ
p
Now, we substitute Λ+ from this equation in formula (2.29), then by (2.24) we
arrive to
1 K(p) b Ω+ − ξΛ+
K1 (p) + ξ Ω− − ξΛ− − + u
b0 (p) +
h(ξ) .
(2.30)
=
+
−
Y
ξ
Y
Y
p
In subsequent consideration we shall have to use the following property of the
limiting values of a Cauchy type integral, the statement of which we now quote.
The proofs may be found in [8].
Lemma 2.6. If L is a smooth closed contour and φ(q) a function that satisfies the
Hölder condition on L, then the limiting values of the Cauchy type integral
Z
1
1
Φ(z) =
φ(q)dq
2πi L q − z
also satisfy this condition.
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
Since u
b0 (p) +
K(p) b
p h(ξ)
9
satisfies on Re(p) = 0 the Hölder condition, on basis of
+ K(p) b
h(ξ)) also satisfies this condition. There-
1
u0 (p)
Y + (b
Lemma 2.6 the function
p
fore, in accordance with Lemma 2.4 it can be uniquely represented in the form of
the difference of the functions U + (p, ξ) and U − (p, ξ), constituting the boundary
values of the analytic function U (z, ξ), given by formula
Z i∞
1
K(q) b 1
1
u
b
(q)
+
U (z, ξ) =
h(ξ) dq.
(2.31)
0
2πi −i∞ q − z Y + (q, ξ)
q
Therefore, the equation (2.30) takes the form
Ω+ − ξΛ+
K1 (p) + ξ Ω− − ξΛ− +
+
U
=
+ U −.
Y+
ξ
Y−
The last relation indicates that the function
K1 (p)+ξ Ω− −ξΛ−
( Y− )
ξ
Ω+ −ξΛ+
Y+
+ U + , analytic in Re(z) < 0,
and the function
+ U − , analytic in Re(z) > 0, constitute the
analytic continuation of each other through the contour Re(z) = 0. Consequently,
they are branches of a unique analytic function in the entire plane. Moreover,
this function has a zero in infinite. According to Liouville theorem this function
is identically zero. Thus, we obtain the solution of the Riemann-Hilbert problem
defined by the boundary condition (2.20),
Ω+ (p, ξ) = −Y + (p, ξ)U + (p, ξ) + ξΛ+ (p, ξ)
Ω− (p, ξ) = −
ξ
Y − (p, ξ)U − (p, ξ) + ξΛ− (p, ξ).
K1 (p) + ξ
(2.32)
b
b of (2.12) for the solution u
Now, we proceed to find the unknown function Φ
b of
−
problem (2.4). First, we represent Ω as the limiting value of analytic functions on
the left-hand side complex semi-plane. From equation (2.24) and Sokhotski-Plemelj
ξ
formulae we obtain Ω− = − K(p)+ξ
Y + U + + ξΛ+ . Now, making use of (2.32) and
the last equation, we obtain
Ω+ − Ω− = −
K(p)
Y +U +.
K(p) + ξ
b = −Y + U + . We observe that Φ
b is boundary value of a
Thus, by formula (2.21), Φ
function analytic in the left-hand side complex semi-plane and therefore satisfies our
b and bearing in mind
basic assumption (2.11). Having determined the function Φ
1
b
formula (2.12) we determine the required function: u
b = K(p)+ξ
(b
u0 (p) − Y + U + ).
b
Now we prove that, in accordance with last relation, the function u
b constitutes the
limiting value of an analytic function in Re(z) > 0. In fact, making use of Sokhotski1
b
Plemelj formulae and using (2.24), we obtain u
b = − K1 (p)+ξ
Y − U − . Thus, the
b
function u
b is the limiting value of an analytic function in Re(z) > 0. We note the
b
fundamental importance of the proven fact, the solution u
b constitutes an analytic
function in Re(z) > 0 and, as a consequence, its inverse Laplace transform vanish for
all x < 0. We now return to solution u of the problem (2.4). Taking inverse Laplace
transform with respect to time and space variables, we obtain (2.6). Proposition
2.3 has been proved.
In the following lemma we collect some preliminary estimates for the Green
operator G(t)
10
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
Lemma 2.7. If Λ(s) as defined in (1.3), then exists C > 0 such that
1
1
1
1
kG(t)φ − ϑt− α − 2 Λ((·)t−1/α )kL∞ ≤ Ct− α − 2 −γ kφkL1, α2 ,
−γ
kG(t)φkL∞ ≤ C{t}
−1/α
hti
(kφkL1 + kφkL∞ ),
1 1
( r − 1s −µ)
−α
kG(t)φkLs,µ ≤ Ct
R +∞ α/2
where ϑ = 0 y φ(y)dy, | 1r −
1
s
γ > 0,
0 < γ < 1,
1 1
−α
( r − 1s )
kφkLr + Ct
kφkLr,µ ,
(2.33)
(2.34)
(2.35)
− µ| < α and 1 ≤ r ≤ s ≤ ∞.
+
Proof. First, we estimate the function Y + = (−i)α eΓ (p,ξ) , where
Z i∞
1
Γ+ (p, ξ) = −
ln− (q − p)F (q, ξ)dq
2πi −i∞
K 0 (q)
K(q)+ξ
−
K10 (q)
K1 (q)+ξ .
Using the relation
q
ln− (q − p) = πi + ln+ (p) + ln+ 1 −
+ 2πiθ(Im p)θ(Im q − Im p),
p
and F (q, ξ) =
where θ is the Heaviside step function, we obtain
Z i∞
1
q
α
+
+
ln+ 1 − F (q, ξ)dq + O(p−αγ ξ γ ),
Γ (p, ξ) = (πi + ln (p)) +
2
2πi −i∞
p
where 0 < γ < 1. Then, separating the integral, |q| < |p| and |q| < |p|, and using
ln(1 − z) = O(z γ ), for |z| < 1 and 0 < γ < 1, and the equation
q
p
ln+ 1 −
= −πi + ln+ (q) − ln+ (p) + ln+ (1 − ) + 2πiς(Im p, Im q),
p
q
where ς(x, y) = θ(x − y)(θ(x)θ(y) − θ(−x)θ(−y)) + θ(−x), we obtain
α
Γ+ (p, ξ) = (πi + ln+ (p)) + O pα(1−γ) ξ γ−1 , 0 < γ < 1.
2
Therefore, from (2.28) and (2.36) follows
3
Y + (p, ξ) = pα/2 + O pα( 2 −γ) ξ γ−1 , 0 < γ < 1,
1
= O(p−α/2 ).
+
Y (p, ξ)
We note that in the above formulas pβ = eβ ln
∂ξj Γ+ (p, ξ)
= O(p
−jαγ j(γ−1)
ξ
),
+
(p)
0 < γ < 1, j = 1, 2.
then integrating by parts
Z
(−1)j+1 j i∞ −
K 0 (q)
=
ln (q − p)
dq
2πi
(K(q) + ξ)j+1
−i∞
Z i∞
(−1)j ln− (q − p) i∞
1
dq
=
−
j
2πi (K(q) + ξ)j −i∞
−i∞ q − p (K(q) + ξ)
= O(p−jαγ ξ j(γ−1) ).
(2.37)
(2.38)
. Now, we show that
In fact, from Cauchy’s Theorem we obtain
Z i∞
1
K10 (q)
ln− (q − p)
dq = 0,
2πi −i∞
(K1 (q) + ξ)2
∂ξj Γ+ (p, ξ)
(2.36)
(2.39)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
11
As a direct consequence of (2.38) and (2.39) we obtain
Γ+
α
1
ξ
(2.40)
)
=
−
= O(p−αγ− 2 ξ γ−1 ),
Y+
Y+
1 1 +
2
−2αγ− α
2 ξ 2(γ−1) .
∂ξ2 + = + (Γ+
(2.41)
ξ ) − Γξξ = O p
Y
Y
Then, using e−qy − 1 = O(q µ y µ ), 0 < µ < 1, (2.40) and (2.41), we obtain for
R i∞ e−qy −1 dq
1
Z0+ (p, ξ, y) = 2πi
,
−i∞ q−p Y + (q,ξ)
∂ξ (
1
∂ξj Z0+ (p, ξ, y) = O(y µ pµ−α(jγ+ 2 ) ξ j(γ−1) ),
1
q−p
1
q
0 < µ < 1,
p
q(q−p) ,
(2.42)
Z0+
= +
we express
in the form
where j = 1, 2. Moreover, using
Z i∞ −qy
Z i∞ −qy
1
e
−1
dq
p
e
−1
dq
Z0+ (p, ξ, y) =
+
.
2πi −i∞
q
Y + (q, ξ) 2πi −i∞ q(q − p) Y + (q, ξ)
Thus, from (2.38) we arrive to
α
Z0+ (p, ξ, y) = C0 y α/2 + O(y µ pµ− 2 ),
where C0 =
1
2πi
e−q −1
dq.
−i∞ q 1+ α
2
R i∞
µ>
α
,
2
Now, we consider the function
(2.43)
+
Y+
K(p)+ξ Z0 .
From
(2.37), (2.39) and (2.42), we obtain
Y+
∂ξ2
Z0+
K(p) + ξ
2
1
1
Y + +
Γξ −
Z0+ + 2 Γ+
∂ξ Z0+
=
ξ −
K(p) + ξ
K(p) + ξ
K(p) + ξ
1
+ Γ+
Z0+ + ∂ξ2 Z0+
ξξ +
2
(K(p) + ξ)
1
=
Y + Z0+ L + O(y µ pµ−2αγ ξ 2(γ−1) ) ,
K(p) + ξ
where
L(p, ξ) = Γ+
ξ −
2
1
1
= O p−2αγ ξ 2(γ−1) .
+ Γ+
ξξ +
K(p) + ξ
(K(p) + ξ)2
Then, using (2.37) and (2.43) we obtain the estimate
Y+
C0 y α/2 α/2
∂ξ2
Z0+ =
p L + O(pµ−2αγ ξ 2(γ−1) ) ,
K(p) + ξ
K(p) + ξ
µ>
α
.
2
(2.44)
On the other hand, by Sokhotski-Plemelj formulae and Cauchy’s Residue Theorem,
we represent the Green function G, (2.5), as
Z i∞
Z i∞
1
Y + (p, ξ) +
ξt
G(x, y, t) = −
dξe
epx
Z (p, ξ, y)dp
2
(2πi) −i∞
K(p) + ξ
−i∞
Z i∞
1
+
ep(x−y)−K(p)t dp.
2πi −i∞
Now, by Cauchy’s Theorem we obtain
Z=
1
+ Z0 ,
Y+
where
Z0 (z, ξ, y) =
1
2πi
Z
i∞
−i∞
e−qy − 1
dq
.
q − z Y + (q, ξ)
12
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
Therefore, we can represent G in the form
Z i∞
Z i∞
1
Y + (p, ξ) +
ξt
G=−
dξe
epx
Z (p, ξ, y)dp
2
(2πi) −i∞
K(p) + ξ 0
−i∞
Z i∞
1
epx−K(p)t (e−py − 1)dp = G1 + G2 .
+
2πi −i∞
(2.45)
Using (e−py − 1) = O(pµ y µ ), 0 < µ < 1, we obtain for the second integral in the
above equation
1
G2 (x, y, t) = O(t− α (1+µ) y µ ).
(2.46)
To estimate G1 , we use the analyticity of the integrand in the left-half semiplane
and integrate by parts two times with respect to ξ to obtain
Z i∞
Z i∞
Y + (p, ξ)
−2
ξt
Z0+ (p, ξ, y) dp.
G1 (x, y, t) = Ct
dξe
epx ∂ξ2
K(p) + ξ
−i∞
−i∞
Then, by (2.44) and (2.46) follows
α
1
1
G(x, y, t) = y α/2 t− α (1+ 2 ) Λ(xt−1/α ) + O(y α/2 t− α (1+µ) ),
µ>
α
,
2
(2.47)
where 2/5 < α < 1 and
Z
i∞
Λ(s) = C
ξ
Z
i∞
eps
dξe
−i∞
−i∞
pα/2
L(p, ξ)dp.
K(p) + ξ
Multiplying by a function φ and integrating with respect to y in (2.47) we obtain
Z +∞
1
α
1
(G(x, y, t) − y α/2 t− α (1+ 2 ) Λ(xt−1/α ))φ(y)dy ≤ Ct− α (1+µ) kφkL1, α2 .
0
Therefore,
G(t)φ − t− α1 (1+ α2 ) Λ((·)t−1/α )
Z
0
+∞
1
y α/2 φ(y)dy L∞ ≤ Ct− α (1+µ) kφkL1, α2 ,
α
2.
where µ >
Thus, the first estimate in Lemma 2.7 has been proved. Now, we
are going to prove the second estimate in Lemma 2.7. First, for large t. From
(2.45), integrating by parts the first summand on the right hand side two times
with respect to ξ we obtain
Z i∞
Z i∞
Y + (p, ξ)
G(x, y, t) = Ct−2
dξeξt
epx ∂ξ2
Z + (p, ξ, y) dp
K(p) + ξ
−i∞
−i∞
(2.48)
Z i∞
1
p(x−y)−K(p)t
e
dp = J1 + J2 .
+
2πi −i∞
We have
∂ξ2 (
Y+
1
Z +) =
O(p−2αγ ξ 2(γ−1) ),
K(p) + ξ
K(p) + ξ
0 < γ < 1, then J1 = O(t−1/α ), provided 13 < α < 1, also J2 = O(t−1/α ).
Therefore, G = O(t−1/α ) and
Z
+∞
kG(t)φkL∞ = sup G(x, y, t)φ(y)dy ≤ t−1/α kφkL1 .
(2.49)
x∈R+
0
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
13
Now, for small t, using the inequality |ez | ≤ |z|−γ , for Re(z) < 0 and γ > 0, we
obtain
Y+
1
∂ξ2 (
Z +) =
O(y −γ p−γ−2αγ1 ξ 2(γ1 −1) ),
K(p) + ξ
K(p) + ξ
1
γ > 0, 0 < γ1 < 1. Then, from (2.48) we obtain J1 = O(y −γ t− α (1−γ) ), γ > 0. On
the other hand, we write the second summand as
Z
1
epr−K(p)t dp, ±r > 0,
J2 (r, t) =
2πi C±
where
π
π
C± = {p ∈ (∞e−i( 2 ±ε) , 0) ∪ (0, ∞ei( 2 ±ε) )},
ε > 0.
−1/α
Making the change of variable p = zt
and using the inequality |ez | ≤ |z|−γ ,
1
(1−γ)
−α
|x − y|−γ ), where x, y > 0.
for Re(z) < 0 and γ > 0, we obtain J2 = O(t
1
− α (1−γ)
−γ
−γ
Therefore, G = O t
(|x − y| + y ) . Thus,
Z +∞
kG(t)φkL∞ ≤ sup
|G(x, y, t)||φ(y)|dy
x∈R+
0
1
≤ Ct− α (1−γ)
Z
+∞
(|x − y|−γ + y −γ )|φ(y)|dy
0
1
≤ Ct− α (1−γ) (kφkL1 + kφkL∞ ).
The second estimate in Lemma 2.7 has been proved. Let us introduce the operators
Z +∞
J1 (t)φ = θ(x)
J1 (x, y, t)φ(y)dy,
(2.50)
0
Z +∞
J2 (t)φ = θ(x)
J2 (x − y, t)φ(y)dy,
(2.51)
0
Then, the operator G(t) can be written in the form
G(t) = J1 (t) + J2 (t).
(2.52)
Now, we are going to prove the third estimate in Lemma 2.7. First, we estimate
the operator J2 . Making the change of variable p = qt−1/α , we obtain
|J2 (r, t)| ≤ Ct−1/α .
(2.53)
−1/α
On the other hand, making z = t
r, we obtain
Z
t−1/α i∞ qz−K(q)
J2 =
e
dq.
2πi −i∞
Then, integrating by parts the last equation we obtain
Z i∞
Z i∞
t−1/α 1
t−1/α α
K(q) qz−K(q)
−K(q) qz
J2 (r, t) =
( )
e
de =
( )
e
dq
2πi z −i∞
2πi z −i∞ q
Thus,
Z
α
1
|J2 (r, t)| ≤ Ct
|q|α−1−γ e−C|q| |dq|,
1+γ
|z|
C±
for ±r > 0, where C± are defined as above. Therefore,
1
|J2 (r, t)| ≤ Ct−1/α 1+γ , γ < α.
|z|
−1/α
(2.54)
14
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
Finally, from the inequalities (2.53) and (2.54) we have
|J2 (r, t)| ≤ C
t−1/α
1+
(t−1/α |r|)1+γ
,
γ < α.
(2.55)
Lets write some well known inequalities:
q
• Young’s inequality. Let f ∈ Lp (R) and
R g ∈ L (R), where 1 ≤ p, q ≤r ∞,
1
1
p + q ≥ 1. Then, the convolution h(x) ≡ R f (x − y)g(y)dy belongs to L (R),
where 1r = p1 + 1q − 1 and the Young’s inequality
khkLr ≤ kf kLp kgkLq
(2.56)
holds.
• Minkowski’s Inequality. Let f, g ∈ Lp and 1 ≤ p ≤ ∞, then
kf + gkLp ≤ kf kLp + kgkLp .
(2.57)
Now, we estimate the operator J2 , defined in (2.50). Using the inequality xµ ≤
|x − y|µ + y µ , where 0 ≤ µ ≤ 1, and Minkowski’s inequality (2.57), we obtain
Z +∞ Z +∞
s 1/s
kJ2 (t)φkLs,µ ≤
|x − y|µ |J2 (x − y, t)||φ(y)|dy dx
0
0
Z +∞ Z +∞
s 1/s
+
y µ |J2 (x − y, t)||φ(y)|dy dx
.
0
0
Then, Young’s inequality (2.56) implies
kJ2 (t)φkLs,µ ≤ kJ2 (·, t)kLp,µ kφkLr + kJ2 (·, t)kLp kφkLr,µ ,
(2.58)
where 1s = p1 + 1r − 1, 1 ≤ p, r ≤ ∞, 1 ≤ p1 + 1r ≤ 2 and 0 ≤ µ ≤ 1. Then, by the
inequality (2.55) and the change of variables x = t−1/α |r|, we obtain
1/p
Z +∞
1
1
|x|µ
p
.
)
dx
(
kJ2 (·, t)kLp,µ ≤ Ct− α (1− p −µ)
1+γ
−∞ (1 + |x|)
1
1
Thus, kJ2 (·, t)kLp,µ ≤ Ct− α (1− p −µ) , provided 1 + γ − µ > p1 . Using
it follows
1
1 1
kJ2 (·, t)kLp,µ ≤ Ct− α ( r − s −µ) ,
1
1
r − s −µ+γ
where
> 0. We note that
(2.59) in (2.58), we obtain
1
1
− α1 ( 1r − 1s −µ)
1
1
s
=
1
p
(2.59)
< 1, since γ < α. Substituting
1
1
1
kJ2 (t)φkLs,µ ≤ Ct− α ( r − s −µ) kφkLr + Ct− α ( r − s ) kφkLr,µ ,
where − α1 ( 1r
operator J1 ,
+ 1r − 1,
(2.60)
1
s
− − µ) < 1, 1 ≤ s, r ≤ ∞ and 0 ≤ µ ≤ 1. Now, we estimate the
defined in (2.51). First, we note that using the Cauchy theorem we
obtain
i∞
Y + (p, ξ) +
Z (p, ξ, y)dp
K(p) + ξ
−i∞
Z +∞
1
1
=
e−px
−
Y + (−p, ξ)Z + (−p, ξ, y)dp.
(−ip)α + ξ
(ip)α + ξ
0
Z
epx
Then, by inequality
1
1
|p|α(1−γ)
≤C
−
,
(−ip)α + ξ
(ip)α + ξ
||p|2α + ξ 2 |1−γ |ξ|γ
(2.61)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
0 < γ < 1, we obtain for J1
Z
Z i∞
iξt
|dξ||e |
|J1 | ≤ C
−i∞
+∞
0
α
e−px pα(1−γ)+ 2
dp 2α
|p + ξ 2 |1−γ |ξ|γ
Z
C3
15
e−C|q|y |dq|
.
|q − p| |q|α/2
R +∞
Thus, by |J2 (t)φ| ≤ 0 |J2 (x, y, t)||φ(y)|dy and
Z +∞
1
e−C|q|y |φ(y)|dy ≤ ke−C|q|(·) kLl kφkLr = C|q|− l kφkLr ,
0
where
1
l
+
1
r
= 1, we obtain
Z
i∞
+∞
Z
|dξ||eiξt |
|J2 (t)φ| ≤ CkφkLr
−i∞
0
1
Then, using ke−p(·) kLs,µ = Cp− s −µ , p > 0, and
Z i∞
Z +∞
iξt
kJ2 (t)φkLs,µ ≤ CkφkLr
|dξ||e |
−i∞
0
e−px pα(1−γ)
|p2α
1
l
+
1
r
dp.
1
+ ξ 2 |1−γ |ξ|γ p l
= 1, we obtain
pα(1−γ)
1
1
|p2α + ξ 2 |1−γ |ξ|γ p1− r + s +µ
dp.
Therefore,
1
1
1
kJ2 (t)φkLs,µ ≤ Ct− α ( r − s −µ) kφkLr ,
(2.62)
1
1
where | r − s − µ| < α, 1 ≤ r ≤ s ≤ ∞ and 0 ≤ µ < 1. Finally, from estimates
(2.60) and (2.62) we obtain the third estimate in Lemma 2.7. Then, we have proved
Lemma 2.7.
As a direct consequence of the above lemma, taking r = 1, we obtain the estimate
kG(t)φkX µ ≤ CkφkZµ .
Now, we collect some estimates for operator H defined by
Z t
H(x)h =
H(x, t − τ )h(τ )dτ, H(x, t) = ∂t ∂y−1 G(x, y, t)|y=0 .
(2.63)
0
Lemma 2.8. The following estimates are valid
1
< α,
s
H(x)h = btβ+γ Λ1 (xt−1/α ) + O(tβ ),
kH(·)hkX µ ≤ CkhkB α ,
µ+
for h(t) = btβ + O(tβ−γ ), where β < 0, γ > 0 and
Z i∞
Z i∞
1
Y + (p, ξ) +
dξeξ
epr
I (p, ξ)dp
Υ(r) = −
2πi −i∞
K(p) + ξ
−i∞
Z i∞
1
K(p)
+
epr−K(p)
dp.
2πi −i∞
p
Proof. First, we note that
H(x, t) = H1 (x, t) + H2 (x, t),
where
H1 (x, t) = −
1
(2πi)2
Z
H2 (x, t) =
i∞
dξeξt
Z
−i∞
1
2πi
i∞
−i∞
Z
epx
Y + (p, ξ) +
I (p, ξ)dp,
K(p) + ξ
i∞
−i∞
epx−K(p)t
K(p)
dp.
p
(2.64)
16
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
Here,
I(z, ξ) =
1
2πi
Z
i∞
−i∞
1
1
K(q)
dq.
+
q − z Y (q, ξ) q
For H2 , making the change of variable q = pt−1/α , we obtain |H2 | ≤ Ct−1 . On the
other hand, for H1 , we consider two domains: |p| < 1 and |p| > 1. For |p| < 1,
we use Y + I + = O(p−1+α ); and for |p| > 1, we integrate by parts two times with
respect to ξ and use
∂ξ2 (
p−1+α
Y+
I +) =
O(p−2αγ ξ 2(γ−1) )
K(p) + ξ
K(p) + ξ
to obtain |H1 | ≤ Ct−1 . Therefore,
|H(x, t)| ≤ Ct−1 .
By the Cauchy’s Residue Theorem, for Re(z) < 0, we have
Z i∞
Z i∞
1 K(q) + ξ dq
1
1
1
dq
1
−
ξ
I=
+
+
2πi −i∞ q − z Y (q, ξ) q
2πi −i∞ q − z Y (q, ξ) q
1
ξ
1
1 1
1
1 1
−
ξ
+
=−
2 (−z) Y + (0, ξ)
z Y + (z, ξ) 2 (−z) Y + (0, ξ)
ξ
ξ
=
−
.
zY + (0, ξ) zY + (z, ξ)
Also, integrating H with respect to t and substituting (2.66) we obtain
Z i∞
Z i∞
+
eξt
1
−1
px Y (p, ξ) dp
dξ
.
∂t H(x, t) = −
e
(2πi)2 −i∞ Y + (0, ξ) −i∞
K(p) + ξ p
(2.65)
(2.66)
(2.67)
Then,
|∂t−1 H(x, t)| ≤ C
(2.68)
Now, we divide the integration domain and integrate by parts (2.63) to obtain
Z t/2
t
H(x)h =
H(x, t − τ )h(τ )dτ − h(τ )∂τ−1 H(x, t − τ )t/2
0
Z t
+
h0 (τ )∂τ−1 H(x, t − τ )dτ.
t/2
Therefore, from (2.65), (2.68) and last equation we have
kH(·)hkL∞ ≤ Chti−1/α khkB α .
(2.69)
s,µ
We estimate the operator H in norm L . First, we estimate H1 from (2.64),
Z
Z
|dp|
1
−λ|ξ|t
s,µ
.
kH1 (·, t)kL ≤ C
|dξ|e
1
1−α+
|K(p)
+
ξ|
s +µ
|p|
C1
C2
Making the change of variables p = p1 t−1/α , ξ = ξ1 t−1 , we obtain
1 1
1
kH1 (·, t)kLs,µ ≤ Ct α ( s +µ)−1 ,
+ µ < α.
s
On the other hand for H2 we have
Z i∞
|dp|
kH2 (·, t)kLs,µ ≤ C
e−K(p)t 1−α+ 1 +µ .
s
|p|
−i∞
(2.70)
(2.71)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
17
Thus,
1
1
kH2 (·, t)kLs,µ ≤ Ct α ( s +µ)−1 ,
1
+ µ < α.
s
(2.72)
Therefore, from (2.70) and (2.72) we conclude
Z t
1
1
1 1
(t − τ ) α ( s +µ)−1 hti−1/α dτ ≤ Chti− α (1− s −µ) khkB α ,
kH(·)hkLs,µ ≤ CkhkB α
0
where 1s + µ < α. Finally, we estimate the operator H in norm L1 . By Cauchy’s
Residue Theorem we have for Re(z) > 0,
Z i∞
Z i∞
1
1 K(q) + ξ dq
1
1
1
dq
I(z, ξ) =
−ξ
+
+
2πi −i∞ q − z Y (q, ξ) q
2πi −i∞ q − z Y (q, ξ) q
K(z) + ξ
ξ
=− +
+
.
zY (z, ξ) zY + (0, ξ)
Then, we obtain the formula
Z i∞
Z i∞
dp
Y + (p, ξ)ξ
1
ξt
px
dξe
e
−
1
.
(2.73)
H(x, t) =
(2πi)2 −i∞
(K(p) + ξ)Y + (0, ξ)
p
−i∞
Now, we substitute the equation
Y + (p, ξ) − Y + (0, ξ) Y + (p, ξ)ξ
K(p)
−
1
=
ξ−
+
+
(K(p) + ξ)Y (0, ξ)
Y (0, ξ)(K(p) + ξ)
K(p) + ξ
in (2.73), and change ξ and p by ξt−1 and pt−1/α , respectively, to obtain
H(x, t) = t−1 J1 (s) + t−1 J2 (s),
s = xt−1/α ,
(2.74)
where
1
J1 (s) =
(2πi)2
Z
i∞
−i∞
dξ
eξ
Y + (0, ξ)
Z
i∞
−i∞
eps
Y + (p, ξ) − Y + (0, ξ) dp
ξ
K(p) + ξ
p
(2.75)
and
Z i∞
Z i∞
1
K(p) dp
ξ
dξe
eps
.
(2πi)2 −i∞
K(p)
+ξ p
−i∞
We estimate J2 . First, for x < 1, we obtain
Z 1
1
|J2 (xt−1/α )|dx ≤ Ct α .
J2 (s) = −
(2.76)
(2.77)
0
On the other hand, for x > 1, applying Cauchy’s Theorem, we write
Z
Z i∞
1
K(p) dp
ξ
,
J2 (s) = −
dξe
eps
(2πi)2 C1
K(p)
+ξ p
−i∞
where
iθ
C1 = {ξ ∈ (−i∞, −i) ∪ S ∪ (i, i∞)},
K(p)+ξ
π
[− 2 , − 3π
− K(p)
2 ]}. Substituting 1 =
ξ
ξ
S = {ξ = e : θ ∈
in last equation,
we obtain
Z
Z
Z
Z
1
dξ ξ i∞ ps
dp
1
dξ ξ i∞ ps K(p)2 dp
J2 = −
e
e
K(p)
+
e
e
.
(2πi)2 C1 ξ
p
(2πi)2 C1 ξ
K(p) + ξ p
−i∞
−i∞
By Cauchy’s Theorem, using the analyticity with respect to ξ, the first summand
in last equation is zero. Thus,
J2 (s) = J21 (s) + J22 (s),
(2.78)
18
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
where
Z
Z
1
dξ ξ
K(p)2 dp
e
eps
,
2
(2πi) C1 ξ
K(p) + ξ p
C2
2
Z
Z
1
dξ ξ i/s ps K(p)2 dp
e
e
.
J22 (s) =
(2πi)2 C1 ξ
K(p) + ξ p
−i/s2
J21 (s) =
For > 0 small enough,
1
π
π
1
C2 = p ∈ 2 − i + ∞ei( 2 +) , −i ∪ 2 i, i + ∞ei( 2 +) .
s
s
Taking the L1 norm with respect to x, we obtain
Z
Z
K(p)2 |dp|
|dξ|
≤ Ct−1/α .
kJ21 (·)kL1 ≤ Ct−1/α
2
C2 |K(p) + ξ| |p|
C1 |ξ|
(2.79)
(2.80)
(2.81)
Now, for J22 we have
Z
|J22 (s)| ≤ C
C1
|dξ|
|ξ|
Z
i/s2
−i/s2
K(p)2 |dp|
1
≤ C 2.
|K(p) + ξ| |p|
s
From (2.78), (2.81) and (2.82), we obtain for t > 0 and x > 1,
Z ∞
1
|J2 (xt−1/α )|dx ≤ Ct α .
(2.82)
(2.83)
1
Thus, from (2.77) and (2.83) we obtain
1
kJ2 (·)kL1 ≤ Ct α .
We estimate J1 . Substituting
ξ
p
=
K(p)+ξ
p
−
K(p)
p
(2.84)
in (2.75), we obtain
J1 (s) = J11 (s) + J12 (s),
(2.85)
where
Z i∞
Z i∞
1
dξ
dp
ξ
J11 (s) =
e
eps (Y + (p, ξ) − Y + (0, ξ)) ,
(2πi)2 −i∞ Y + (0, ξ)
p
−i∞
Z i∞
Z i∞
+
+
1
dξ
dp
Y (p, ξ) − Y (0, ξ)
J12 (s) = −
eξ
K(p) .
eps
2
+
(2πi) −i∞ Y (0, ξ)
K(p) + ξ
p
−i∞
The integrand in J11 is an analytic function in the left half-plane, with respect to
p, and the residue at p = 0 is zero, then by Cauchy’s Residue Theorem we conclude
that J11 (s) = 0. Now, we estimate J12 . Using
Z p
+
α( 23 −γ) γ−1
ξ
),
Y + (p, ξ) − Y + (0, ξ) =
Γ+
q (q, ξ)Y (q, ξ)dq = O(p
0
1
we obtain kJ12 (·)kL1 ≤ Ct α . Therefore,
kJ1 (·)kL1 ≤ Ct1/α .
(2.86)
Then, from (2.84) and (2.86) we obtain
1
kH(·, t)kL1 ≤ Ct−1+ α .
Finally, integrating with respect to time we obtain
Z t
1
kH(·)hkL1 ≤ CkhkB α
(t − τ ) α −1 hti−1/α dτ ≤ CkhkB α .
0
(2.87)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
19
Then, first estimate of Lemma 2.8 has been proved. Now we are going to obtain
an asymptotic representation for H. First, we split the integral,
Z t−1
Z t
H(x)h =
H(x, t − τ )h(τ )dτ +
H(x, t − τ )h(τ )dτ
(2.88)
0
t−1
For the first integral in (2.88) we have
Z t−1
H(x, t − τ )h(τ )dτ
0
Z
t−1
Z
H(x, t − τ )dτ.
H(x, t − τ )(h(t) − h(τ ))dτ + h(t)
=−
(2.89)
t−1
0
0
Making the change t − τ = s, we obtain
Z t−1
Z t
Z t
ds
H(x, t − τ )dτ =
H(x, s)ds =
Υ(xs−1/α )
s
0
1
1
Z t
ds
=
(Υ(xs−1/α ) − Υ(xt−1/α )) + ln(t)Υ(xt−1/α )
s
1
Z t
t
=
(H(x, s) − H(x, t))ds + ln(t)Υ(xt−1/α ),
s
1
where Υ is given by (1.4). From (2.89) and (2.90) we obtain
Z t−1
H(x, t − τ )h(τ )dτ
0
Z t
t
−1/α
= h(t) ln(t)Υ(xt
) + h(t)
H(x, s) − H(x, t) ds
s
1
Z t−1
−
H(x, t − τ )(h(t) − h(τ ))dτ.
(2.90)
(2.91)
0
Now, since |H(x, t)| ≤ Ct−1 and h(t) = btβ + O(tβ−γ ), we obtain
Z
t
t
(H(x, s) − H(x, t))ds ≤ ln(t),
s
1
Z t−1
H(x, t − τ )(h(t) − h(τ ))dτ = btβ + O(tβ−δ ).
0
Then, from (2.91) we obtain
Z t−1
H(x, t − τ )h(τ )dτ = btβ+γ Υ(xt−1/α ) + O(tβ ).
(2.92)
0
On the other hand, for the second integral in (2.88), integrating by parts and using
|∂t−1 H(x, t)| ≤ C, we have
Z t
H(x, t − τ )h(τ )dτ = O(tβ ).
(2.93)
t−1
Thus, by (2.88), (2.92) and (2.93) we obtain
H(x)h = btβ+γ Υ(xt−1/α ) + O(tβ ).
Therefore, Lemma 2.8 has been proved.
(2.94)
20
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
The following theorem is a local version of Theorem 1.1, its proof is similar and
will be omitted.
Theorem 2.9. Let the initial data u0 ∈ Zµ and the boundary value h ∈ B α such
that the correspondent norms are sufficiently small. Then, for some T > 0, there
exist a unique solution
u ∈ C([0, T ]; L1 ) ∩ C((0, T ]; Ls ∩ Ls,µ ∩ L∞ ),
s ≥ 1,
to the initial boundary-value problem (1.1).
3. Proof of Theorem 1.1
By the Local Existence Theorem 2.9, it follows that the global solution (if it
exist) is unique. Indeed, on the contrary, we suppose that there exist two global
solutions with the same initial data. And these solutions are different at some time
t > 0. By virtue of the continuity of solutions with respect to time, we can find
a maximal time segment [0, T ], where the solutions are equal, but for t > T they
are different. Now, we apply the local existence theorem taking the initial time
T and obtain that these solutions coincide on some interval [T, T1 ], which give us
a contradiction with the fact that T is the maximal time of coincidence. So our
main purpose in the proof of Theorem 1.1 is to show the global in time existence
of solutions.
First, we note that Lemma 2.7 imply for the Green operator G : Zµ → X µ the
inequality kG(t)u0 kX µ ≤ Cku0 kZµ . Now, we are going to show the estimate
Z t
k
G(t − τ )(N (u(τ )) − N (v(τ )))dτ kX µ
(3.1)
0
σ
≤ Cku − vkX µ (kukX µ + kvkX µ ) ,
for all u, v ∈ X, where N (u) = |u|σ u and α < σ < 1 + α. In fact, using the
inequality
||u|σ u − |v|σ v| ≤ C|u − v|(|u|σ + |v|σ ),
we obtain
kN (u) − N (v)kL1,ν
≤ Cku − vkL∞ kukσLσ, σν + kvkσLσ, σν
≤ C{τ }
−γ
1
−1/α − α
(σ−1−ν)
hτ i
τ
(3.2)
σ
ku − vkX µ (kukX µ + kvkX µ ) ,
where µ = ν/σ ≥ 0, and
kN (u) − N (v)kL∞
≤ Cku − vkL∞ (kukσL∞ + kvkσL∞ )
≤ C{τ }
−γ(σ+1)
hτ i
1
−α
(σ+1)
(3.3)
σ
ku − vkX µ (kukX µ + kvkX µ ) .
Then, estimates (3.2), (3.3), and Lemma 2.7, imply
kG(t − τ )(N (u(τ )) − N (v(τ )))kLs,µ
µ
µ
1
1
1
≤ C(t − τ )− α (1− s ) τ − α (σ−1) {τ }−γ hτ i−1/α (t − τ ) α + τ α
× ku − vkX µ (kukX µ + kvkX µ )σ ,
(3.4)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
21
where 0 ≤ µ < 1, and
kG(t − τ )(N (u)(τ ) − N (v)(τ ))kL∞
1
≤ C{t − τ }−γ ht − τ i−1/α {τ }−γ hτ i−1/α τ − α (σ−1) + {τ }−γσ hτ i−σ/α
(3.5)
σ
× ku − vkX µ (kukX µ + kvkX µ ) .
Now, we integrate with respect to τ inequalities (3.4) and (3.5),
Z t
kG(t − τ )(N (u)(τ ) − N (v)(τ ))kLs,µ
(3.6)
0
1
(σ− 1s −µ)−γ
1− α
≤ C{t}
1
1− α
(1+σ− 1s −µ)
hti
ku − vk
Xµ
(kuk
Xµ
+ kvk
Xµ
σ
) ,
1
1−α ,
provided σ < 1 + α, s <
and
Z t
kG(t − τ )(N (u)(τ ) − N (v)(τ ))kL∞
(3.7)
0
≤ C{t}
1
1− α
(σ−1)−2γ
hti
1
1− α
(σ+1)
ku − vk
Xµ
(kuk
Xµ
+ kvk
Xµ
σ
) ,
whenever σ < 1 + α. Then, for α < σ < 1 + α, the definition of the norm
in the space X µ and the estimates (3.6) and (3.7), imply (3.1). Now, we apply
the Contraction Mapping Principle on a ball with ratio ρ > 0 in the space X µ ,
Xρµ = {φ ∈ X µ : kφkX µ ≤ ρ}, where ρ = 3C max{ku0 kZµ , khkB α }. First, we show
that
kM(u)kX µ ≤ ρ,
(3.8)
where u ∈ Xρµ . Indeed, from the integral formula
Z t
M(u) = G(t)u0 −
G(t − τ )N (u(τ ))dτ + H(x)h
(3.9)
0
and the estimate (3.1) (with v ≡ 0) we obtain
Z t
kM(u)kX µ ≤ kG(t)u0 kX µ + k
G(t − τ )N (u)(τ )dτ kX µ + kH(·)hkX µ
0
α
≤ C ku0 kZµ + kukσ+1
+
khk
µ
B
X
ρ
ρ
σ+1
+ ≤ ρ,
≤ + Cρ
3
3
provided ρ > 0 is sufficient small. Therefore, the operator M transforms a ball of
ratio ρ > 0 into itself, in the space X µ . In the same way we estimate the difference
of two functions u, v ∈ Xρµ ,
Z t
kM(u) − M(v)kX µ ≤ k
G(t − τ )(N (u)(τ ) − N (v)(τ ))dτ kX µ
0
≤ Cku − vkX µ (kukX µ + kvkX µ )σ ≤ C(2ρ)σ ku − vkX µ
1
≤ ku − vkX µ
2
whenever ρ > 0 is sufficient small. Thus, M is a contraction mapping in Xρµ .
Therefore, there exist a unique solution u ∈ X µ to the Cauchy problem (1.1). Now
we prove the asymptotic formula
1
1
1
1
(3.10)
u(x, t) = t− α − 2 aΛ(xt−1/α ) + bΥ(xt−1/α ) + O(t− α − 2 −γ ),
22
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
where γ > 0 and
+∞
Z
y
a=
α/2
Z
+∞
u0 (y)dy −
Z
1
1
Z
+∞
y α/2 N (u(τ ))dy.
0
0
0
+∞
dτ
We denote G0 (t) = t− α − 2 Λ(xt−1/α ) and G1 (t) = btβ+γ Υ(xt−1/α ). From Lemmas
2.7 and 2.8, we have for all t > 1,
1
1
y α/2 φ(y)dykL∞ ≤ Chti− α − 2 −γ kφkZ1, α2 ,
kG(t)φ − G0 (t)
0
(3.11)
kH(·)h − G1 (t)kL∞ ≤ Chtiβ ,
(3.12)
for h(t) = btβ + O(tβ−γ ), β < 0, γ > 0. Also, in view of the definition of the norm
X µ we have
Z
+∞ α/2
y N (u(τ ))dy ≤ kN (u(τ ))k 1, α ≤ ku(τ )kσL∞ ku(τ )k 1, α
L
0
L
2
σ
−α
≤ C{τ }−γσ hτ i
2
1
2
τ kukσ+1
.
X α/2
By a direct calculation, for t > 1, we have
Z
t
2
Z
k
+∞
dτ (G0 (t − τ ) − G0 (t))
0
y α/2 N (u(τ ))dykL∞
0
≤ Ckukσ+1
X α/2
≤ Chti
1
−α
− 21
t
2
Z
σ
1
kG0 (t − τ ) + G0 (t)kL∞ {τ }−γσ hτ i− α τ 2 dτ
(3.13)
0
kukσ+1
X α/2
Z
t
2
0
σ
1
1
1
σ
3
{τ }−γσ hτ i− α τ 2 dτ ≤ Chti− α − 2 −( α − 2 ) kukσ+1
,
X α/2
where σ > 3α/2, provided γσ < 32 , and in the same way
∞
Z
kG0 (t)
Z
dτ
t/2
0
+∞
1
1
σ
3
,
y α/2 N (u(τ ))dykL∞ ≤ Chti− α − 2 −( α − 2 ) kukσ+1
X α/2
(3.14)
provided σ > 3α/2. Also we have for all t > 1,
Z
k
t
2
Z
(G(t − τ )N (u(τ )) − G0 (t − τ )
0
Z t
+k
G(t − τ )N (u(τ ))dτ kL∞
t
2
t
2
Z
≤C
0
Z
+C
1
y α/2 N (u(τ ))dy)dτ kL∞
0
1
(t − τ )− α − 2 −γ kN (u(τ ))kL1, α2 dτ
t
{t − τ }−γ ht − τ i−1/α (kN (u(τ ))kL1 + kN (u(τ ))kL∞ )dτ
t
2
1
σ
−α
− 12 −( α
− 32 )
≤ Chti
+∞
kukσ+1
,
X α/2
(3.15)
EJDE-2015/281
NONLINEAR BVP WITH FRACTIONAL DERIVATIVE
provided γσ <
3
2
23
and σ > 23 α. By virtue of the integral equation (3.9) we obtain
ku(t) − aG0 (t) − G1 (t)kL∞
Z +∞
≤ kG(t)u0 − G0 (t)
y α/2 u0 (y)dykL∞
0
t
2
Z
+k
Z
+∞
y α/2 N (u(τ ))dy)dτ kL∞
Z ∞ Z +∞
dτ
y α/2 N (u(τ ))dykL∞
+ kG0 (t)
(G(t − τ )N (u(τ )) − G0 (t − τ )
0
0
t
Z
G(t − τ )N (u(τ ))dτ kL∞
+k
t
2
Z
t
2
t
2
Z
dτ (G0 (t − τ ) − G0 (t))
+k
0
0
+∞
y α/2 N (u(τ ))dykL∞
0
+ kH(·)h − G1 (t)kL∞ .
(3.16)
All summands in the right-hand side of (3.16) are estimated, via (3.13)-(3.15), by
1
1
Chti− α − 2 −γ (ku0 kZα/2 + kukσ+1
),
X α/2
σ
where 0 < γ < α
− 32 , provided β < − α1 − 12 − γ. Thus by last estimate the
asymptotic (3.10) is valid. Theorem 1.1 has been proved.
References
[1] Árciga-Alejandre, M. P.; Asymptotics for Nonlinear Evolution Equation with ModuleFractional Derivative on a Half-Line, Boundary Value Problems, Volume 2011, Article ID
946143, 29 pages, doi:10.1155/2011/946143.
[2] Bashir, A.; Nieto, J. J.; Existence Results for Nonlinear Boundary Value Problems of Fractional Integrodifferential Equations with Integral Boundary Conditions, Boundary Value
Problems, vol. 2009, Article ID 708576, 11 pages, (2009). doi:10.1155/2009/708576.
[3] Bashir, A.; Otero-Espinar, V.; Existence of Solutions for Fractional Differential Inclusions
with Antiperiodic Boundary Conditions, Boundary Value Problems, vol. 2009, Article ID
625347, 11 pages, (2009). doi:10.1155/2009/625347.
[4] Belmekki, M.; Nieto, J. J.; Rodrguez-López, R.; Existence of Periodic Solution for a Nonlinear Fractional Differential Equation, Boundary Value Problems, vol. 2009, Article ID
324561, 18 pages, (2009). doi:10.1155/2009/324561.
[5] Benchohra, M.; Cabada, A.; Seba, D.; An Existence Result for Nonlinear Fractional Differential Equations on Banach Spaces, Boundary Value Problems, vol. 2009, Article ID 628916,
11 pages, (2009). doi:10.1155/2009/628916.
[6] Benchohra, M., Hamani, S.; Nonlinear boundary value problems for differential inclusions
with Caputo fractional derivative. Topol. Methods Nonlinear Anal. 32 (2008), no. 1, 115–130.
[7] Caballero-Mena, J.; Harjani, J.; Sadarangani, K.; Existence and Uniqueness of Positive and Nondecreasing Solutions for a Class of Singular Fractional Boundary Value
Problems, Boundary Value Problems, vol. 2009, Article ID 421310, 10 pages, (2009).
doi:10.1155/2009/421310.
[8] Gakhov, F. D.; Boundary value problems, Pergamon Press, ISBN 0-486-66275-6, (1966), pp.
562.
[9] Hayashi, N.; Kaikina, E. I.; Nonlinear theory of pseudodifferential equations on a half-line.
North-Holland Mathematics Studies 194. Elsevier Science B.V., Amsterdam, (2004), 319 pp.
[10] Hayashi, N.; Kaikina, E. I.; Naumkin, P. I.; Shishmarev, I. A.; Asymptotics for dissipative
nonlinear equations. Lecture Notes in Mathematics 1884. Springer-Verlag, Berlin, (2006),
557 pp.
[11] Kaikina, E. I.; Subcritical pseudodifferential equation on a half-line with nonanalytic symbol.
Differential Integral Equations 18, (2005), no. 12, 1341–1370.
[12] Kaikina, E. I.; Nonlinear pseudoparabolic type equations on a half-line with large initial data.
Nonlinear Anal. 67, (2007), no. 10, 2839–2858.
24
M. P. ÁRCIGA-ALEJANDRE
EJDE-2015/281
[13] Kaikina, E. I.; Pseudodifferential operator with a nonanalytic symbol on a half-line, J. of
Mathematical Physics 48, (2007), no. 11, 1341–1370.
[14] Kaikina, E. I.; Fractional derivative of Abel type on a half-line. Trans. Amer. Math. Soc.,
364 (2012), no. 10, 5149-5172.
[15] Samko, S. G.; Kilbas, A. A.; Marichev, O. I.; Fractional Integrals and Derivatives. Theory
and Applications, Gordon and Breach, Yverdon, 1993. MR1347689 (96d:26012)
[16] Spence, E. A.; Fokas, A. S.; A new transform method I: domain-dependent fundamental
solutions and integral representations. Proceedings of the Royal Society A, (2010) 466, 22592281 doi:10.1098/rspa.2009.0512.
Martı́n P. Árciga-Alejandre
Instituto de Matemáticas UNAM, Campus Morelia, Apartado Postal 61-3 (Xangari),
C.P. 58089, Morelia, Michoacán, México
E-mail address: mparciga@gmail.com