Electronic Journal of Differential Equations, Vol. 2006(2006), No. 129, pp.... ISSN: 1072-6691. URL: or

advertisement
Electronic Journal of Differential Equations, Vol. 2006(2006), No. 129, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE OF POSITIVE SOLUTIONS FOR MULTI-TERM
NON-AUTONOMOUS FRACTIONAL DIFFERENTIAL
EQUATIONS WITH POLYNOMIAL COEFFICIENTS
AZIZOLLAH BABAKHANI, VARSHA DAFTARDAR-GEJJI
Abstract. In the present paper we discuss the existence of positive solutions
in the case of multi-term non-autonomous fractional differential equations with
polynomial coefficients; the constant coefficient case has been studied in [2].
We consider the equation
n−1
“
”
X
pj (x)Dαn−j y = f (x, y).
D αn −
j=1
We state various conditions on f and pj ’s under which this equation has:
positive solutions, a unique solution which is positive, and a unique solution
which may not be positive.
1. Introduction
Let E be a real Banach space with a cone K ⊂ E. K introduces a partial order
≤ in E : x ≤ y if and only if y −x ∈ E. A cone K is said to be normal, if there exists
a positive constant τ such that θ ≤ f ≤ g implies kf k ≤ τ kgk, where θ denotes the
zero element of K. For x, y ∈ E the order interval hx, yi is define to be [4]:
hx, yi = {z ∈ E : x ≤ z ≤ y}
Theorem 1.1 ([4]). Let K be a normal cone in a partially ordered Banach space
E. Let F be an increasing operator which transforms hx0 , y0 i into itself; i. e.,
F x0 ≥ x0 and F y0 ≤ y0 . Assume further that F is compact and continuous. Then
F has at least one fixed point x∗ ∈ hx0 , y0 i.
Theorem 1.2 (Banach fixed point theorem [4]). Let K be a closed subspace of a
Banach space E. Let F be a contraction mapping with Lipschitz constant k < 1
from K to itself. Then F has a unique fixed point x∗ in K. Moreover if x0 is an
arbitrary point in K and {xn } is defined by xn+1 = F xn , (n = 0, 1, 2, . . . ) then
limn→∞ xn = x∗ ∈ K and d(xn , x∗ ) ≤ (k n /(1 − k)) d(x1 , x0 ).
2000 Mathematics Subject Classification. 26A33, 34B18.
Key words and phrases. Riemann-Liouville fractional derivatives and integrals; normal cone;
semi-ordered Banach space; completely continuous operator; equicontinuous set.
c
2006
Texas State University - San Marcos.
Submitted July 27, 2005. Published October 16, 2006.
1
2
A. BABAKHANI, V. DAFTARDAR-GEJJI
EJDE-2006/129
Definition 1.3. The left sided Riemann-Liouville fractional integral [5, 6, 7] of
order α of a real function f is defined as
Z x
1
y(t)
Iaα+ y(x) =
dt, α > 0, x > a.
(1.1)
Γ(α) a (x − t)1−α
Definition 1.4. The left sided Riemann-Liouville fractional derivative [5, 6, 7] of
order α of a function f is
Daα+ y(x) =
dn n−α
Ia+ y(x) ,
n
dx
n − 1 ≤ α < n,
n ∈ N.
(1.2)
We denote Daα+ by Daα y(x) and Iaα+ y(x) by Iaα y(x). Also Dα y(x) and I α y(x) refers
to D0α+ y(x) and I0α+ y(x), respectively.
Proposition 1.5. (i) If the fractional derivative Daα y(x) is integrable, then
(x − a)α−1
,
Iaα (Daβ y(x)) = Iaα−β y(x) − Ia1−β y(x) x=a
Γ(α)
0 < β ≤ α < 1.
(1.3)
(ii) If y is continuous on [a, b], then Daα y(x) is integrable, I 1−β y(x)|x=a = 0 and
Iaα Daβ y(x) = Iaα−β y(x), 0 < β ≤ α < 1.
(1.4)
Proof. For (i), we refer the reader to [6]. For (ii), let M = maxa≤x≤b y(x) then,
using (1.2) we get
Z x
Z
x α
M (x − a)1−α
M
,
(x − t)−α dt =
Da y(t) dt ≤
Γ(1 − α) a
Γ(2 − α)
a
so Daα y(t) is integrable. On the other hand
hZ x
i
M
M
|Ia1−β y(x)|x=a ≤
(x − a)1−β x=a = 0,
(x − t)−β dt
=
Γ(1 − β) a
Γ(2 − β)
x=a
and hence (1.3) reduces to Iaα (Daβ y(x)) = Iaα−β y(x).
Proposition 1.6. Let y be continuous on [0, λ], λ > 0 and n be a non negative
integer, then
n n X
X
−α n!xn−k α+k
−α k n α+k
I
y(x) =
I α (xn y(x)) =
D x
I
y(x),
k (n − k)!
k
k=0
k=0
(1.5)
where
−α
Γ(α + 1)
α+k−1
Γ(1 − α)
= (−1)k
= (−1)k
=
. (1.6)
n!Γ(α)
k
Γ(k + 1)Γ(1 − α − k)
k
The proof of the above proposition can be found in [5, p. 53].
PNj
Corollary 1.7. Let y ∈ C[0, λ], λ > 0 and pj (x) = k=0
ajk xk , Nj ∈ N ∪ {0},
j = 1, 2, . . . , n. Then
Iα
Nj k
n X
X
X
−αn k!xk−r α+r
pj (x)y(x) =
ajk
[I
y(x)].
r
(k − r)!
r=0
j=1
j=1
n
X
k=0
(1.7)
EJDE-2006/129
EXISTENCE OF POSITIVE SOLUTIONS
3
k!xk−r
(k−r)!
k n α+k
Pn
Proof. Using I α (xn y(x)) = k=0 −α
y(x)] and Dr (xk ) =
k [D x ][I
have
n
n
X
X
Iα
pj (x)y(x) =
I α (pj (x)y(x))
j=1
we
j=1
=
Nj
n X
X
ajk I α (xk y(x))
j=1 k=0
=
Nj
n X
X
ajk
k hX
−α
n
r=0
j=1 k=0
=
Nj k
n X
X
X
ajk
j=1 k=0 r=0
r
(1.8)
r k
(D x )I
α+r
i
y(x)
−αn k!xk−r α+r
[I
y(x)].
(k − r)!
r
2. Existence of Positive Solutions
In this section we discuss conditions under which the following fractional initialvalue probelm has a positive solution.
Dαn −
n−1
X
pj (x)Dαn−j y = f (x, y),
y(0) = 0,
0 ≤ x ≤ λ, λ > 0,
(2.1)
j=1
where 0 < α1 < α2 < · · · < αn < 1; pj (x) =
(2m+1)
PNj
k=0
(2m)
ajk xk , pj
(x) ≥ 0,
N
pj
(x) ≤ 0, m = 0, 1, . . . , [ 2j ], j = 1, 2, . . . , n−1, Dαj is the standard RiemannLiouville fractional derivative and f : [0, 1] × [0, +∞) → [0, +∞) a given continuous
function. Let us denote by Y = C[0, λ], the Banach space of all continuous real
functions on [0, λ] endowed with the sup norm and K be the cone:
K = {y ∈ Y : y(x) ≥ 0, 0 ≤ x ≤ λ}.
Definition 2.1. By a solution of (2.1), we mean a continuous function y ∈ C[0, λ],
that satisfies (2.1).
We remark that in [7] the initial-value problems
D0α y(x) = f (x, y),
0 < α < 1,
I01−α y(x)|x=0
0 < α < 1,
= b,
are studied where D0α denotes the Riemann-Liouville derivative and the underlying
space of functions is C(0, λ). However, in the present paper we are dealing with
the space of functions C[0, λ]. For y(x) ∈ C[0, λ], always I01−α y(x)|x=0 = 0, and is
not free data.
Lemma 2.2. The fractional initial-value problem (2.1) is equivalent to the Volterra
integral equation
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
y(x) =
ajk
I
y(x) + I αn f (x, y(x)).
(2.2)
r
(k
−
r)!
r=0
j=1
k=0
4
A. BABAKHANI, V. DAFTARDAR-GEJJI
EJDE-2006/129
Proof. Suppose y(x) satisfies (2.1), then
n−1
X
I αn Dαn −
pj (x)Dαn−j y = I αn f (x, y).
j=1
Proposition 1.5 (ii) yields I
y(x)x=0 = 0, I 1−αn −αn−j y(x)x=0 = 0, hence
I αn (Dαn y(x)) = y(x) and using (1.7) we obtain the integral equation (2.2). Conversely, let y(x) satisfy the integral equation (2.2). Then
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
ajk
I
y(x) + I αn f (x, y(x))
r
(k
−
r)!
r=0
j=1
1−αn
k=0
=
Nj
n−1
XX
ajk
n
r
r=0
j=1 k=0
=
k hX
−α
Nj
n−1
XX
i
Dr xk I αn −αn−j +r y(x) + I αn f (x, y(x))
ajk I αn (xk Dαn−j y(x)) + I αn f (x, y(x))
j=1 k=0
=
n−1
X
I αn (pj (x)Dαn−j y(x)) + I αn f (x, y(x))
j=1
n−1
X
= I αn (
[pj (x)Dαn−j y(x))] + f (x, y(x))) = y(x)
j=1
But y(x) = I
αn
(Dαn y(x)), hence y(x) satisfies (2.1), and y(0) = 0.
(2m)
pj (x)
(2m+1)
pj
(x)
N
0, 1, . . . , [ 2j ]
where
Lemma 2.3. If
≥ 0 and
≤ 0 for m =
Nj = deg(pj ), j = 1, 2, . . . , n − 1 and pj ’s are as in (2.1), then F defined as
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
F y(x) =
ajk
I
y(x) + I αn f (x, y(x)), (2.3)
r
(k
−
r)!
r=0
j=1
k=0
is from K to itself.
Proof. The right-hand side of (2.3) can be expressed as:
Nj
n−1
X −αn X
F y(x) =
ajk xk I αn −αn−j y(x)
0
j=1
k=0
+
n−1
X
j=1
+
n−1
X
j=1
+
n−1
X
j=1
=
−αn
1
X
Nj
−αn
2
X
Nj
k=1
k(k − 1)ajk xk−2 I αn −αn−j +2 y(x) + . . .
k=2
−αn
(Nj !ajNj )I αn −αn−j +Nj y(x)
Nj
Nj n−1
XX
j=1 k=0
kajk xk−1 I αn −αn−j +1 y(x)
−αn (k)
pj (x)I αn −αn−j +k y(x)
k
EJDE-2006/129
EXISTENCE OF POSITIVE SOLUTIONS
In view of (1.6), we have
−αn
> 0,
2m
−αn
< 0,
2m + 1
5
m ∈ N.
(k)
Then by assumptions on pj (x), k = 0, 1, . . . , Nj , j = 1, 2, . . . , n−1 we get F y(x) ∈
K.
Furthermore, it is easy to show the following result.
Lemma 2.4. The operator F : K → K defined in Lemma 2.3 is completely continuous.
Lemma 2.5. Let M ⊂ K be bounded; i.e. there exists a positive constant l such
that kyk ≤ l, for all y ∈ M . Then F (M ) is compact.
Proof. Let L = max{1 + f (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ l}. For y ∈ M , we have
Nj k n−1
XX
X −αn k!xαn −αn−j +k
Lxαn
|F (y(x))| ≤
+
.
ajk
r
(k − r)!Γ(αn − αn−j + r + 1) Γ(αn + 1)
r=0
j=1
k=0
Hence
Nj k h n−1
i
XX
X −αn L
k!
kF uk ≤
+
ζ,
ajk
(k − r)!Γ(αn − αn−j + r + 1) Γ(αn + 1)
r
r=0
j=1
k=0
where ζ = max{λαn , λαn −αn−1 , λαn −αn−1 +b } and b = max{N1 , N2 , . . . , Nn−1 }.
Hence F (M ) is bounded. Let y ∈ M, x1 , x2 ∈ [0, λ], x1 < x2 then for given > 0,
choose
nh C(j, k, r) i1/(αn −αn−j +r) h Γ(α + 1) i1/αn o
n
,
(2.4)
δ = min
2
4kf k∞
where j = 1, 2, . . . , n − 1, k = 0, 1, . . . , Nj , r = 0, 1, . . . , k,
C(j, k, r) = Pn−1
i=1
and η = max{1, λ
Nj
(k − r)!
(Ni + 1)(Ni + 2)
×
Γ(αn − αn−j + r + 1)
ajk −αn lηk!
r
, j = 1, 2, . . . , n − 1}. If |x1 − x2 | < δ,
|F y(x1 ) − F y(x2 )|
Nj k
n−1
XXX
≤
j=1 k=0 r=0
h Z x1 ×
Z
0
x2
−
n
|ajk −α
|k!xk−r
1
r
(k − r)!Γ(αn − αn−j + r)
(x1 −
y(t)
t)αn−j −αn −r+1
dt
−
y(t)
(x2 −
t)αn−j −αn −r+1
dt
i
(x2 − t)αn−j −αn −r+1
Z x1
1
+
(x1 − t)αn −1 − (x2 − t)αn −1 f (t, y(t))dt
Γ(αn ) 0
Z x2
1
−
(x2 − t)αn −1 f (t, y(t))dt
Γ(αn ) x1
Nj k
n−1
n
XX
X
|ajk −α
|lk!η
2L(x2 − x1 )αn
r
(x2 − x1 )αn −αn−j +r +
≤
(k − r)!Γ(αn − αn−j + r)
Γ(αn + 1)
r=0
j=1
x1
k=0
6
A. BABAKHANI, V. DAFTARDAR-GEJJI
=
Nj k
n−1
XX
X
j=1 k=0 r=0
≤
EJDE-2006/129
n
|ajk −α
|lk!η
2Lδ αn
r
δ αn −αn−j +r +
(k − r)!Γ(αn − αn−j + r)
Γ(αn + 1)
+ = .
2 2
Hence F (M ) is equicontinuous and Arzela-Ascoli theorem implies that F (M ) is
compact.
Theorem 2.6. Consider the fractional differential equation
n−1
X
Dαn −
pj (x)Dαn−j y = g(y), y(0) = 0, 0 ≤ x ≤ λ, λ > 0,
(2.5)
j=1
PNj
where 0 < α1 < α2 < · · · < αn < 1, pj (x) = k=0
ajk xk , Nj ∈ N ∪ {0}, j =
1, 2, . . . , n − 1, g : [0, 1] × [0, +∞) → [0, +∞) satisfies the Lipschitz condition with
constant L and g(0) < ∞.
(2m)
(2m+1)
N
If pj (x) ≥ 0, pj
(x) ≤ 0, m = 0, 1, . . . , [ 2j ], then (2.5) has a positive
solution.
Proof. In view Lemma 2.2, (2.5) is equivalent to the integral equation
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
I
y(x) + I αn g(y).
y(x) =
ajk
(k
−
r)!
r
r=0
j=1
k=0
In view of Lemma 2.4, y(x) ∈ K. Let T : K → K be defined as
Nj k
n−1
X
XX
−αn k!xk−r αn −αn−j +r
ajk
I
y(x) + I αn g(y).
Ay(x) =
(k
−
r)!
r
r=0
j=1
k=0
A is completely continuous by Lemma 2.3.
Case (i) g(0) 6= 0. Let
n
o
g(0)xαn
B(r) = y(x) ∈ C[0, δ] : y(x) ≥ 0 ky −
k≤r ,
Γ(1 + αn )
be a convex bounded and closed subset of the Banach space C[0, δ] where
n rΓ(α + 1) 1/αn
1 1/αn o
n
δ < min λ,
,
,
2Eg(0)
2E
where
Nj k
n−1
n
XX
X
|ajk −α
|k!
L
r
E=ξ
+
,
(k − r)!Γ(αn − αn−j + r + 1)
Γ(αn + 1)
r=0
j=1
k=0
and
ξ = max xαn , xαn −αn−1 +ρ : 0 ≤ x ≤ δ ,
ρ = max{N1 , . . . , Nn−1 }.
Note that, for all y ∈ B(r),
αn Ay(x) − g(0)x
Γ(1 + αn )
Nj k
h n−1
XX
X |ajk −αn |k!xαn −αn−j +k
Lxαn i
r
≤ kuk
+
(k − r)!Γ(αn − αn−j + r + 1) Γ(αn + 1)
r=0
j=1
k=0
EJDE-2006/129
EXISTENCE OF POSITIVE SOLUTIONS
Nj k
h n−1
XX
X
≤ kuk
j=1 k=0 r=0
7
i
n
|ajk −α
|k!
L
r
ξ.
+
(k − r)!Γ(αn − αn−j) + r + 1 Γ(αn + 1)
Since
kuk ≤
g(0)
g(0)
xαn + r ≤
δ αn + r,
Γ(1 + αn )
Γ(1 + αn )
we have
Ay(x) −
g(0)
r
g(0)
r
xαn ≤ E
δ αn +r ≤ + = r.
Γ(1 + αn )
Γ(αn + 1)
2 2
So we have A(B(r)) ⊆ B(r). It can be seen that A(B(r)) is equicontinuous (the
proof is similar to the proof of Lemma 2.5). Let {yn } be a bounded sequence in B(r).
Then {A(yn )} ⊂ T (B(r)). Hence {A(yn )} is equicontinuous. Since yn ∈ C[a, b],
Arzela-Ascoli theorem [1, 3] implies that {A(yn )} has a convergent subsequence.
Therefore A : B(r) → B(r) is compact. Hence by Schauder fixed point theorem [4]
it has a fixed point, which is a positive solution of(2.5).
A similar proof can be given for the case g(0) = 0.
Example 2.7. Consider the equation
Dα3 y(x) − (x2 − 3x + 2)Dα2 y(x) − (1 − x)Dα1 y(x) =
1+y
,
1 + y2
y(0) = 0, 0 ≤ x ≤ 1, 0 < α1 < α2 < α3 < 1. Note that p1 (x) = x2 − 3x + 2,
1+y
p2 (x) = 1 − x and g(y) = 1+y
2 satisfy the conditions required in Theorem 2.6,
hence this equation has a positive solution.
Theorem 2.8. Let f : [0, λ] × [0, ∞) → [0, ∞) be continuous and f (x, .) be increasing for each x ∈ [0, λ]. Assume there exist v0 , w0 satisfying L(D)v0 ≤ f (x, v0 ),
L(D)w ≥ f (x, w0 ) and 0 ≤ v0 (x) ≤ w0 (x), 0 ≤ x ≤ 1, where L(D) = Dαn −
Pn−1 0
αn−j
. Then (2.1) has a positive solution.
j=1 pj (x)D
Proof. We need to consider the fixed point of the operator F . Let y1 , y2 ∈ K,
y1 ≤ y2 , then
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
F y1 =
ajk
I
y1 + I αn f (x, y1 (x)) ≤ F y2 ,
(k
−
r)!
r
r=0
j=1
k=0
as f is nondecreasing. Hence F is an increasing operator. Assuming F v0 ≥
v0 , F w0 ≤ w0 , implies that F : hv0 , w0 i → hv0 , w0 i is compact operator in view
of Lemma 2.4 and completely continuous in view of Lemma 2.3. Since K is a
normal cone and F is compact continuous, by Theorem 1.1 F has a fixed point
u∗ ∈ hv0 , w0 i, which is the required positive solution.
Example 2.9. Consider the equation
3
7
3
D1/2 y(x) − (Γ( ))−1 Γ( )xD1/4 y(x) = (Γ( ))−1 f (x, y),
2
4
2
7
where 0 ≤ x ≤ 1, 0 ≤ y ≤ +∞, f (x, y) = µ(x1/2 − x 4 )e2y−x and 0 < µ ≤ 1 which
is equivalent to equation
3
7
Γ( )D1/2 y(x) − Γ( )xD1/4 y(x) = f (x, y).
2
4
8
A. BABAKHANI, V. DAFTARDAR-GEJJI
EJDE-2006/129
7
If we let v0 = 0, w0 = 21 x, then 0 ≤ v0 ≤ w0 , L(D)v0 = 0, L(D)w0 = x1/2 − x 4 ,
L(D)v0 ≤ f (x, 0) and L(D)w0 ≥ f (x, 12 x). Then this equation has a positive
solution .
Theorem 2.10. Let f : [0, λ]×[0, ∞) → [0, ∞) be continuous and f (x, .) increasing
for each x ∈ [0, λ]. If 0 < limy→+∞ f (x, y) < +∞ for each x ∈ [0, λ] then (2.1) has
a positive solution.
Proof. There exist positive constants N, R such that f (x, y) ≤ N , for all x ∈ [0, λ],
and all y ≥ R. Let C = max{f (x, y)|0 ≤ y ≤ λ, 0 ≤ y ≤ R}. Then we have
f ≤ N + C, for all y ≥ 0. Now we consider the equation,
n−1
X
Dαn −
pj (x)Dαn−j w(y) = N + C, w(0) = 0, 0 < x < λ.
j=1
Using Lemma 2.2, the above equation is equivalent to the integral equation
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
w(x) =
ajk
I
y(x) + I αn (N + C).
r
(k
−
r)!
r=0
j=1
k=0
This integral equation has a positive solution w(x) in view of Theorem 2.8. Also
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
w(x) ≥
ajk
I
y(x) + I αn f (x, w(x)) = F w(x).
(k
−
r)!
r
r=0
j=1
k=0
Now for v(x) ≡ 0, F (v(x)) = I αn f (x, v(x)) ≥ v(x). Hence in view of Theorem 2.8,
the result follows.
It is easy to prove the following existence theorem using Theorems 2.8 and 2.10.
Theorem 2.11. Let f : [0, λ]×[0, ∞) → [0, ∞) be continuous and f (x, .) increasing
for each x ∈ [0, λ]. If
f (x, y)
0 ≤ lim max
< +∞.
y→∞ 0≤x≤λ
y
Then(2.1) has a positive solution.
Example 2.12. (1) f (x, y) = x(1 + e−y )−1 , satisfies the condition required in
Theorem 2.10.
(2) f (x, y) = x ln(1 + y) satisfies the conditions required in Theorem 2.11.
3. Uniqueness and existence of solutions
In this section we give conditions on f and pj ’s, which render unique positive
solution to (2.1).
Theorem 3.1. Let f : [0, λ] × [0, ∞) → [0, ∞) be continuous and Lipschitz with
(2m)
(2m+1)
respect to the second variable with constant L. If (i) pj (x) ≥ 0 and pj
(x) ≤
Nj
2 ],
Nj = deg(pj ) j = 1, 2, . . . , n − 1; and (ii)
Nj k
n−1
XX
X |ajk −αn |k!λαn −αn−j +k
Lλαn
r
+
< 1,
0<
Γ(αn + 1) j=1
(k − r)!Γ(αn − αn−j + r + 1)
r=0
0, m = 0, 1, . . . , [
k=0
then (2.1) has unique solution which is positive.
EJDE-2006/129
EXISTENCE OF POSITIVE SOLUTIONS
9
Proof. As pointed out in the preceding section, (2.1) is equivalent to (2.2). For
y1 , y2 ∈ K we have
F (y1 (x)) − F (y2 (x))
Nj k
n−1
XX
X |ajk −αn k!xk−r
r
≤
I αn −αn−j +r |y1 (x) − y2 (x)| + LI αn |y1 (x) − y2 (x)|
(k
−
r)!
r=0
j=1
k=0
≤ ky1 (x) − y2 (x)k
h
Nj k
n−1
XX
X |ajk −αn |k!xαn −αn−j +k i
Lxαn
r
,
+
Γ(αn + 1) j=1
(k − r)!Γ(αn − αn−j + r + 1)
r=0
k=0
where F is given in (2.3). Hence
kF y1 − F y2 k
Nj k
n−1
X |ajk −αn |k!λαn −αn−j +k i
XX
Lλαn
r
≤
+
ky1 (x) − y2 (x)k.
Γ(αn + 1) j=1
(k − r)!Γ(αn − αn−j + r + 1)
r=0
h
k=0
In view of Theorem 1.2, F has unique fixed point in K, which is the unique positive
solution of (2.1).
In the following, we omit the condition on pj (x)0 s and study the equation
Dnα −
n−1
X
pj (x)Dαn−j y = f (x, y),
y(0) = 0,
0 ≤ x ≤ λ, λ > 0,
(3.1)
j=1
PNj
k
where 0 < α1 < α2 < · · · < αn < 1, pj (x) =
k=0 ajk x , Nj ∈ N ∪ {0}, j =
1, 2, . . . , n − 1. Using Banach fixed point theorem for F : C[0, λ] → C[0, λ] we
obtain the following result.
Example 3.2. Consider the equation
1
1
D1/2 − (A − x)(B − x)D1/4 − (C − x)D1/6 − M D1/8 y = Ly + ex , (3.2)
60
40
where y(0) = 0, 0 ≤ x ≤ 1, A ≥ 1 and B ≥ 1. (3.2) is equivalent to the integral
equation
Nj k
3 X
X
X
1
−1/2
k!
y(x) =
ajk
xk−r I 2 −αn−j +r y(x) + I 1/2 (Ly + ex ).
r
(k
−
r)!
r=0
j=1
k=0
P2
1
1
Here p1 (x) = k=0 a1k xk = 60
[x2 − (A + B)x + AB], hence N1 = 2; a10 = 60
AB,
P1
−1
1
1
k
a11 = 60 (A + B), a12 = 60 , p2 (x) =
a
x
=
(C
−
x),
so
N
=
1;
2k
2
k=0
40
P0
1
−1
k
a20 = 40 C, a21 = 40 , p3 (x) = k=0 a3k x = M , so N3 = 0, a30 = M . Hence
h−1/2 1 1
−1/2 1 − 1
−1/2 1 − 1 +1 i
y(x) = a10
I 2 4 y + a11
xI 2 − 4 y +
I2 4 y
0
0
1
h−1/2
1
1
1
1
−1/2
−1/2 1 − 1 +2 i
+ a12
x2 I 2 − 4 y + 2
xI 2 − 4 +1 y + 2
I2 4 y
0
1
2
−1 h−1/2 1 1
1
2 2
−1/2 1 − 1 +1 i
− 16
−6
2
2
+ a20
I
y + a21
xI
y+
I2 6 y
0
0
1
1
−1/2 1 − 1
+ a30
I 2 8 y + LI 2 y + I 1/2 ex
0
10
A. BABAKHANI, V. DAFTARDAR-GEJJI
EJDE-2006/129
i
AB 1 − 1
A + B h 1−1
1 1 1
xI 2 4 y − I 2 − 4 +1 y
I2 4y −
60
60
2
i
1
1 h 2 1−1
3 1 − 1 +2
− 14 +1
2
4
2
y − xI
y + I2 4 y +
x I
+
60
4
1
1
1 h 1−1
1 1 − 1 +1 i
C 1−1
xI 2 6 y − I 2 6 y + M I 2 − 8 y + LI 1/2 y + I 1/2 ex .
+ I2 6y −
40
40
2
1
If 1 ≤ A ≤ 3, 1 ≤ B ≤ 3, 0 < M ≤ 40
and 0 < L ≤ 41 in the above equation satisfy
the conditions required in Theorem 3.1. The iterated sequence is
=
y1 (x) = I 1/2 ex = x1/2 E1, 34 (x)
y2 (x) =
h AB
1
3
1
3
A+B
1 2 1/4
xI 4 − I 5/4 +
x I
− xI 4 + I 9/4
60
60
2
60
4
i
C 1/3
1
1 4/3 1/3
3/8
1/2
y1 + y 1 ,
+ I
−
xI
− I
+ MI
+ LI
40
40
2
I 1/4 −
and
yn+1 (x) =
n h
X
AB
A+B
1
1 2 1/4
3
xI 1/4 − I 5/4 +
x I
− xI 5/4 + I 9/4
60
60
2
60
4
k=0
i
n−k
1
C
1
1
+ I 1/3 −
xI 3 − I 4/3 + M I 3/8 + LI 1/2
y1 ,
40
40
2
I 1/4 −
1
n = 1, 2, 3, . . . , where I α y1 = xα+ 2 E1,α+ 34 (x), α > 0. y(x) = limn→∞ yn (x) is the
unique positive solution.
Theorem 3.3. Let f : [0, λ] × [0, ∞) → [0, ∞) be continuous and Lipschitz with
respect to the second variable with constant L. Let ajk ’s satisfy
Nj k
n−1
XX
X |ajk −αn |k!λαn −αn−j +k
Lλαn
r
+
< 1.
0<
Γ(αn + 1) j=1
(k − r)!Γ(αn − αn−j + r + 1)
r=0
k=0
Then (3.1) has unique solution, which may not necessarily be positive.
Proof. Using Lemma 2.2, (3.1) is equivalent to the integral equation
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
y(x) =
ajk
I
y(x) + I αn f (x, y(x)).
r
(k
−
r)!
r=0
j=1
k=0
We define an operator F : C[0, λ] → C[0, λ] as
Nj k
n−1
XX
X
−αn k!xk−r αn −αn−j +r
F y(x) =
ajk
I
y(x) + I αn f (x, y(x)),
r
(k
−
r)!
r=0
j=1
k=0
For y1 , y2 ∈ C[0, λ],
kF y1 − F y2 k
≤
h
Nj k
n−1
XX
X |ajk −αn |k!λαn −αn−j +k i
Lλαn
r
+
ky1 (x) − y2 (x)k.
Γ(αn + 1) j=1
(k − r)!Γ(αn − αn−j + r + 1)
r=0
k=0
Hence in view of Theorem 1.2, F will have unique fixed point in C[0, λ], which is
the unique solution of (3.1). This solution is not necessarily positive one.
EJDE-2006/129
EXISTENCE OF POSITIVE SOLUTIONS
11
Example 3.4. Consider the euation
1
D1/2 − ax2 D1/4 − bxD1/6 − cD 8 y = Ly + ex ,
y(0) = 0, 0 ≤ x ≤ 1.
(3.3)
This equation is equivalent to the integral equation
1
Nj k
3 X
X
X
− 2 k!xk−r 1 −αn−j +r
y(x) =
I2
y(x) + I 1/2 (Ly + ex ).
ajk
r
(k
−
r)!
r=0
j=1
k=0
P2
Here p1 (x) = k=0 a1k xk = ax2 , then N1 = 2, a10 = a11 = 0, a12 = a, p2 (x) =
P1
P0
k
k
k=0 a2k x = bx, then N2 = 1, a20 = a21 = b, and p3 (x) =
k=0 a3k x = c, then
N3 = 0, a30 = c. Hence
h−1/2 1 1
−1/2 1 − 1 +1 i
−1/2 1 − 1
−4
2
4
2
y + a11
xI
y+
I2 4 y
y(x) = a10
I
0
1
0
i
h−1/2
1
−1/2
−1/2 1 − 1 +2
− 41 +1
2 12 − 41
2
y+2
xI
y+2
I2 4 y +
+ a12
x I
1
2
0
−1 h
1
1
1
1
−1/2 1 − 1 +1 i
2 2
−1/2
I 2 − 6 y + a21
+ a20
I2 6 y
xI 2 − 6 y +
0
1
0
1
−1/2 1 − 1
+ a30
I 2 8 y + LI 2 y + I 1/2 ex .
0
√
√
√
4 π
−3
In view of (1.6) and that Γ( 21 ) = π, Γ( −1
2 ) = −2 π and Γ( 2 ) = 3 we obtain
h
i
h
i
3
1
y(x) = a x2 I 1/4 y(x) − I 5/4 y(x) + I 9/4 y(x) + b xI 1/3 y(x) − I 4/3 y(x)
4
2
3/8
1/2
1/2 x
+ cI y(x) + LI y(x) + I e .
If |a| ≤ 53 , |b| ≤ 25 , |c| ≤ 51 , 0 < L ≤ 45 in the above equation satisfy the conditions
required in Theorem 3.3. The iterated sequence is
y1 (x) = I 1/2 ex = x1/2 E1, 43 (x),
i
1
3 9
y2 (x) = a(x2 I 1/4 − I 5/4 + I 4 ) + b(xI 1/3 − I 4/3 ) + cI 3/8 + LI 1/2 y1 + y1 ,
4
2
n h
in−k
X
1
3
y1 ,
yn+1 =
a x2 I 1/4 − I 5/4 + I 9/4 + b xI 1/3 − I 4/3 + cI 3/8 + LI 1/2
4
2
h
k=0
1
for n = 1, 2, 3, . . . , where I α y1 = xα+ 2 E1,α+ 34 (x), α > 0. y(x) = limn→∞ yn (x) is
the unique solution, which may not be positive.
V. Daftardar-Gejji acknowledges University Grants Commission, N. Delhi, India
for the support through Major Research Project.
References
[1] C. D. Aliprantis, O. Burkinshaw; Principles of Real Analysis, (Second Edition), Academic
Press, New York, 1990.
[2] A. Babakhani, V. Daftardar-Gejji; Existence of Positive Solutions of Nonlinear Fractional
Differential Equations, 278 J. Math. Anal. Appl. (2003) 434-442.
[3] R. R. Goldberg; Methods of Real Analysis, Oxford and IBH Publishing Company, New Delhi,
1970.
[4] M. C. Joshi, R. K. Bose; Some Topics in Nonlinear Functional Analysis, Wiley Eastern
Limited, New Delhi, 1985.
12
A. BABAKHANI, V. DAFTARDAR-GEJJI
EJDE-2006/129
[5] K. S. Miller, B. Ross; An Introduction to the Fractional Calculus and Fractional Differential
Equations, Wiley Interscience, New York, 1993.
[6] I. Podlubny; Fractional Differential Equations, Academic Press, San Diego, 1999.
[7] S. G. Samko, A. A. Kilbas, O. I. Marichev; Fractional Integrals and Derivatives: Theory and
Applications, Gordon and Breach, Yverdon, 1993.
Azizollah Babakhani
Department of Mathematics, University of Mazanderan, Babol, Iran
E-mail address: [email protected]
Varsha Daftardar-Gejji
Department of Mathematics, University of Pune, Ganeshkhind, Pune - 411007, India
E-mail address: [email protected]
Download