#### Positive solutions for a system of semipositone coupled fractional boundary value problems

Henderson and Luca Boundary Value Problems
Positive solutions for a system of semipositone coupled fractional boundary value problems
Johnny Henderson 1
Rodica Luca 0
0 Department of Mathematics, Gh. Asachi Technical University , Iasi, 700506 , Romania
1 Department of Mathematics, Baylor University , Waco, TX 76798-7328 , USA
We study the existence of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations with sign-changing nonlinearities, subject to coupled integral boundary conditions. Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology, economics, control theory, signal and image processing, aerodynamics, viscoelasticity, electromagnetics, and so on (see [-]). Integral boundary conditions arise in thermal conduction problems, semiconductor problems and hydrodynamic problems. We consider the system of nonlinear fractional differential equations
Riemann-Liouville fractional differential equations; coupled integral boundary conditions; positive solutions; sign-changing nonlinearities
1 Introduction
(S)
Dα+u(t) + λf (t, u(t), v(t)) = , t ∈ (, ), n – < α ≤ n,
β
D+v(t) + μg(t, u(t), v(t)) = , t ∈ (, ), m – < β ≤ m,
with the coupled integral boundary conditions
(BC)
u() = u () = · · · = u(n–)() = ,
v() = v () = · · · = v(m–)() = ,
u() = v(s) dH(s),
v() = u(s) dK (s),
where n, m ∈ N, n, m ≥ , Dα+, and Dβ+ denote the Riemann-Liouville derivatives of
orders α and β, respectively, the integrals from (BC) are Riemann-Stieltjes integrals, and f ,
g are sign-changing continuous functions (that is, we have a so-called system of
semipositone boundary value problems). These functions may be nonsingular or singular at t =
and/or t = . The boundary conditions above include multi-point and integral boundary
conditions and sum of these in a single framework.
We present intervals for parameters λ and μ such that the above problem (S)-(BC) has
at least one positive solution. By a positive solution of problem (S)-(BC) we mean a pair
of functions (u, v) ∈ C([, ]) × C([, ]) satisfying (S) and (BC) with u(t) ≥ , v(t) ≥ for
all t ∈ [, ] and u(t) > , v(t) > for all t ∈ (, ). In the case when f and g are
nonnegative, problem (S)-(BC) has been investigated in [] by using the Guo-Krasnosel’skii fixed
point theorem, and in [] where λ = μ = and f (t, u, v) and g(t, u, v) are replaced by f˜(t, v)
and g˜(t, u), respectively (denoted by (S)). In [], the authors study two cases: f and g are
nonsingular and singular functions and they used some theorems from the fixed point
index theory and the Guo-Krasnosel’skii fixed point theorem. The systems (S) and (S) with
uncoupled boundary conditions
(BC)
u() = u () = · · · = u(n–)() = ,
v() = v () = · · · = v(m–)() = ,
u() = u(s) dH(s),
v() = v(s) dK (s),
were investigated in [] (problem (S)-(BC) with f , g nonnegative), in [] (problem
(S)(BC) with f , g nonnegative, singular or not), and in [] (problem (S)-(BC) with f , g
signchanging functions). We also mention paper [], where the authors studied the existence
and multiplicity of positive solutions for system (S) with α = β, λ = μ, and the
boundary conditions u(i)() = v(i)() = , i = , . . . , n – , u() = av(ξ ), v() = bu(η), ξ , η ∈ (, ),
with ξ , η ∈ (, ), < abξ η < , and f , g are sign-changing nonsingular or singular
functions.
The paper is organized as follows. Section contains some preliminaries and lemmas.
The main results are presented in Section , and finally in Section some examples are
given to support the new results.
2 Auxiliary results
We present here the definitions of Riemann-Liouville fractional integral and
RiemannLiouville fractional derivative and then some auxiliary results that will be used to prove
our main results.
Definition . The (left-sided) fractional integral of order α > of a function f : (, ∞) →
R is given by
α
I+f (t) =
(α)
t
(t – s)α–f (s) ds, t > ,
provided the right-hand side is pointwise defined on (, ∞), where (α) is the Euler
gamma function defined by (α) = ∞ tα–e–t dt, α > .
Definition . The Riemann-Liouville fractional derivative of order α ≥ for a function
f : (, ∞) → R is given by
Dα+f (t) =
d n
dt
In+–αf (t) =
d
(n – α) dt
n t f (s)
(t – s)α–n+ ds, t > ,
where n = α + , provided that the right-hand side is pointwise defined on (, ∞).
The notation α stands for the largest integer not greater than α. If α = m ∈ N then
Dm+f (t) = f (m)(t) for t > , and if α = then D+f (t) = f (t) for t > .
We consider now the fractional differential system
Dα+u(t) + x˜(t) = ,
β
D+v(t) + y˜(t) = ,
t ∈ (, ), n – < α ≤ n,
t ∈ (, ), m – < β ≤ m,
with the coupled integral boundary conditions
where n, m ∈ N, n, m ≥ , and H, K : [, ] → R are functions of bounded variation.
Lemma . ([]) If H, K : [, ] → R are functions of bounded variations, = –
( τ α– dK (τ ))( τ β– dH(τ )) = and x˜, y˜ ∈ C(, ) ∩ L(, ), then the pair of functions
(u, v) ∈ C([, ]) × C([, ]) given by
u(t) = G(t, s)x˜(s) ds + G(t, s)y˜(s) ds,
v(t) = G(t, s)y˜(s) ds + G(t, s)x˜(s) ds,
t ∈ [, ],
t ∈ [, ],
where
and
⎪⎧⎪⎪⎨⎪ GG((tt,,ss)) == gtα(–t, s)+gt(ατ–, s() dHτ β(τ–),dH(τ ))( g(τ , s) dK (τ )),
⎪⎪⎩⎪⎪ GG((tt,, ss)) == gtβ–(t, s) +g(tβτ–,s() dKτ(ατ–),dK∀(τt,)s)(∈[g,(τ] , s) dH(τ )),
⎪⎪⎪⎪⎨⎧ g(t, s) = (α)
⎪⎩⎪⎪⎪ g(t, s) = (β)
tα–( – s)α– – (t – s)α–,
tα–( – s)α–,
tβ–( – s)β– – (t – s)β–,
tβ–( – s)β–,
≤ s ≤ t ≤ ,
≤ t ≤ s ≤ ,
≤ s ≤ t ≤ ,
≤ t ≤ s ≤ ,
is solution of problem ()-().
Lemma . The functions g, g given by () have the properties:
(a) g, g : [, ] × [, ] → R+ are continuous functions, and g(t, s) > , g(t, s) > for all
(t, s) ∈ (, ) × (, ).
(b) g(t, s) ≤ h(s), g(t, s) ≤ h(s) for all (t, s) ∈ [, ] × [, ], where h(s) = s(–s)α–
(α–) and
h(s) = s(–s)β–
(β–) for all s ∈ [, ].
(c) g(t, s) ≥ k(t)h(s), g(t, s) ≥ k(t)h(s) for all (t, s) ∈ [, ] × [, ], where
k(t) = min
k(t) = min
( – t)tα– tα–
α – , α –
( – t)tβ– tβ–
β – , β –
=
=
tα–
α– ,
(–t)tα–
α– ,
tβ–
β– ,
(–t)tβ–
β– ,
≤ t ≤ ,
≤ t ≤ ,
≤ t ≤ ,
≤ t ≤ .
()
()
()
()
()
(d) For any (t, s) ∈ [, ] × [, ], we have
g(t, s) ≤
( – t)tα–
(α – ) ≤
tα–
(α – ) ,
g(t, s) ≤
( – t)tβ–
(β – ) ≤
tβ–
(β – ) .
For the proof of Lemma .(a) and (b) see [], for the proof of Lemma .(c) see [],
and the proof of Lemma .(d) is based on the relations g(t, s) = g( – s, – t), g(t, s) =
g( – s, – t), and relations (b) above.
Lemma . ([]) If H, K : [, ] → R are nondecreasing functions, and > , then Gi,
i = , . . . , given by () are continuous functions on [, ] × [, ] and satisfy Gi(t, s) ≥
for all (t, s) ∈ [, ] × [, ], i = , . . . , . Moreover, if x˜, y˜ ∈ C(, ) ∩ L(, ) satisfy x˜(t) ≥ ,
y˜(t) ≥ for all t ∈ (, ), then the solution (u, v) of problem ()-() given by () satisfies
u(t) ≥ , v(t) ≥ for all t ∈ [, ].
Lemma . Assume that H, K : [, ] → R are nondecreasing functions, > , τ α–( –
τ ) dK (τ ) > , τ β–( – τ ) dH(τ ) > . Then the functions Gi, i = , . . . , satisfy the
inequalities:
(a) G(t, s) ≤ σh(s), ∀(t, s) ∈ [, ] × [, ], where
σ = +
K () – K ()
τ β– dH(τ ) > .
(a) G(t, s) ≤ δtα–, ∀(t, s) ∈ [, ] × [, ], where
δ =
(α– ) +
τ β– dH(τ )
(a) G(t, s) ≥ tα–h(s), (t, s) ∈ [, ] × [, ], where
=
τ β– dH(τ )
k(τ ) dK (τ ) > .
(b) G(t, s) ≤ σh(s), ∀(t, s) ∈ [, ] × [, ], where σ = (H() – H()) > .
(b) G(t, s) ≤ δtα–, ∀(t, s) ∈ [, ] × [, ], where δ = (β–) ( – τ )τ β– dH(τ ) > .
(b) G(t, s) ≥ tα–h(s), ∀(t, s) ∈ [, ] × [, ], where = k(τ ) dH(τ ) > .
(c) G(t, s) ≤ σh(s), ∀(t, s) ∈ [, ] × [, ], where
σ = +
H() – H()
τ α– dK (τ ) > .
(c) G(t, s) ≤ δtβ–, ∀(t, s) ∈ [, ] × [, ], where
δ =
(β– ) +
τ α– dK (τ )
( – τ )τ β– dH(τ )
> .
(c) G(t, s) ≥ tβ–h(s), ∀(t, s) ∈ [, ] × [, ], where
=
(d) G(t, s) ≤ σh(s), ∀(t, s) ∈ [, ] × [, ], where σ = (K () – K ()) > .
(d) G(t, s) ≤ δtβ–, ∀(t, s) ∈ [, ] × [, ], where δ = (α–) ( – τ )τ α– dK (τ ) > .
(d) G(t, s) ≥ tβ–h(s), ∀(t, s) ∈ [, ] × [, ], where = k(τ ) dK (τ ) > .
Proof From the assumptions of this lemma, we obtain
τ α– dK (τ ) ≥
τ α–( – τ ) dK (τ ) > ,
( – τ )τ α– dK (τ ) ≥
( – τ )τ α– dK (τ ) > ,
k(τ ) dK (τ ) ≥ α – τ α–( – τ ) dK (τ ) > ,
τ β– dH(τ ) ≥
τ β–( – τ ) dH(τ ) > ,
( – τ )τ β– dH(τ ) ≥
( – τ )τ β– dH(τ ) > ,
k(τ ) dH(τ ) ≥ β – τ β–( – τ ) dH(τ ) > ,
K () – K () =
H() – H() =
dK (τ ) ≥
dH(τ ) ≥
τ α–( – τ ) dK (τ ) > ,
τ β–( – τ ) dH(τ ) > .
By using Lemma ., we deduce, for all (t, s) ∈ [, ] × [, ]:
(a)
tα–
G(t, s) = g(t, s) +
τ β– dH(τ )
g(τ , s) dK (τ )
≤ h(s) +
= h(s) +
τ β– dH(τ )
k(τ ) dK (τ ) = tα–h(s).
G(t, s) =
G(t, s) ≤
tα–
tα–
(b)
(b)
(b)
(c)
(c)
(c)
(d)
(d)
G(t, s) ≥
G(t, s) =
k(τ ) dH(τ ) = tβ–h(s).
G(t, s) = g(t, s) +
tβ–
tβ–
( –(ατ )–τα)– dK (τ ) = δtβ–.
tα– g(τ , s) dH(τ ) ≤
= H() – H() h(s) = σh(s).
h(s) dH(τ )
( –(βτ )–τβ)– dH(τ ) = δtα–.
G(t, s) ≥
k(τ )h(s) dH(τ ) =
tα–
h(s)
k(τ ) dH(τ ) = tα–h(s).
G(t, s) ≤ ( –(βt)–tβ–) +
≤ (tββ–– ) +
tβ–
( –(βτ )–τβ)– dH(τ )
( – τ )τ β– dH(τ )
= δtβ–.
tβ–
= tβ–h(s)
G(t, s) ≥
tβ–
k(τ )h(s) dK (τ ) = tβ–h(s)
k(s) dK (τ ) = tβ–h(s).
G t , s x˜(s) ds + σ
h(s)x˜(s) ds +
h(s)y˜(s) ds
≥ tα– min
,
σ σ
= γtα–u t ,
∀t, t ∈ [, ], where γ = min
,
σ σ
> .
G t , s y˜(s) ds
G t , s x˜(s) ds +
G t , s y˜(s) ds
In a similar way, we deduce
v(t) =
G(t, s)y˜(s) ds +
G(t, s)x˜(s) ds
≥ γtβ–v t ,
∀t, t ∈ [, ], where γ = min σ , σ
> .
In the proof of our main results we shall use the nonlinear alternative of Leray-Schauder
type and the Guo-Krasnosel’skii fixed point theorem presented below (see [, ]).
Theorem . Let X be a Banach space with ⊂ X closed and convex. Assume U is a
relatively open subset of with ∈ U, and let S : U¯ → be a completely continuous operator
(continuous and compact). Then either
() S has a fixed point in U¯ , or
() there exist u ∈ ∂U and ν ∈ (, ) such that u = νSu.
Theorem . Let X be a Banach space and let C ⊂ X be a cone in X. Assume and
are bounded open subsets of X with ∈ ⊂ ¯ ⊂ and let A : C ∩ ( ¯ \ ) → C be a
completely continuous operator such that either
(i) Au ≤ u , u ∈ C ∩ ∂ , and Au ≥ u , u ∈ C ∩ ∂ , or
(ii) Au ≥ u , u ∈ C ∩ ∂ , and Au ≤ u , u ∈ C ∩ ∂ .
Then A has a fixed point in C ∩ ( ¯ \ ).
()
()
()
()
3 Main results
In this section, we investigate the existence and multiplicity of positive solutions for our
problem (S)-(BC). We present now the assumptions that we shall use in the sequel.
(H) H, K : [, ] → R are nondecreasing functions, = – ( τ α– dK (τ )) ×
( τ β– dH(τ )) > , and τ α–( – τ ) dK (τ ) > , τ β–( – τ ) dH(τ ) > .
(H) The functions f , g ∈ C([, ] × [, ∞) × [, ∞), (–∞, +∞)) and there exist functions
p, p ∈ C([, ], [, ∞)) such that f (t, u, v) ≥ –p(t) and g(t, u, v) ≥ –p(t) for any t ∈
[, ] and u, v ∈ [, ∞).
(H) f (t, , ) > , g(t, , ) > for all t ∈ [, ].
(H) The functions f , g ∈ C((, ) × [, ∞) × [, ∞), (–∞, +∞)), f , g may be singular
at t = and/or t = , and there exist functions p, p ∈ C((, ), [, ∞)), α, α ∈
C((, ), [, ∞)), β, β ∈ C([, ] × [, ∞) × [, ∞), [, ∞)) such that
f∞ = u+lvi→m∞ t∈m[c,in–c] f (ut,+u,vv) = ∞
or g∞ = u+lvi→m∞ t∈m[c,in–c] g(ut,+u,vv) = ∞.
(H) βi∞ = limu+v→∞ maxt∈[,] βi(t,u,v) = , i = , .
u+v
We consider the system of nonlinear fractional differential equations
Dα+x(t) + λ(f (t, [x(t) – q(t)]∗, [y(t) – q(t)]∗) + p(t)) = ,
β
D+y(t) + μ(g(t, [x(t) – q(t)]∗, [y(t) – q(t)]∗) + p(t)) = ,
< t < ,
< t < ,
with the integral boundary conditions
x() = x () = · · · = x(n–)() = ,
y() = y () = · · · = y(m–)() = ,
x() = y(s) dH(s),
y() = x(s) dK (s),
where z(t)∗ = z(t) if z(t) ≥ , and z(t)∗ = if z(t) < . Here (q, q) with
q(t) = λ
is solution of the system of fractional differential equations
Dα+q(t) + λp(t) = ,
β
D+q(t) + μp(t) = ,
< t < ,
< t < ,
with the integral boundary conditions
q() = q() = · · · = q(n–)() = ,
q() = q() = · · · = q(m–)() = ,
q() = q(s) dH(s),
q() = q(s) dK (s).
Under the assumptions (H) and (H), or (H) and (H), we have q(t) ≥ , q(t) ≥ for
all t ∈ [, ].
We shall prove that there exists a solution (x, y) for the boundary value problem ()-()
with x(t) ≥ q(t) and y(t) ≥ q(t) on [, ], x(t) > q(t), y(t) > q(t) on (, ). In this case
(u, v) with u(t) = x(t) – q(t) and v(t) = y(t) – q(t), t ∈ [, ] represents a positive solution
of boundary value problem (S)-(BC).
By using Lemma . (relations ()), a solution of the system
⎨⎪⎪⎪⎧ x(t) = λ+ μ G(Gt,s()t(,fs()s(,g[(xs(,s[)x–(s)q–(sq)](∗s,)[]y∗(,s[)y(–s)q–(qs)](∗s))]+∗)p+(sp))(ds)s) ds,
⎪⎩⎪⎪ y(t) = μ+ λ G G(t,(st),(sg)((sf,([sx, ([sx)(s–) q–q(s)(]s∗),][∗y,([sy)(s–) q–q(s)(]s∗))]∗+) p+p(s)()s)d)sds,
t ∈ [, ],
t ∈ [, ],
is a solution for problem ()-().
We consider the Banach space X = C([, ]) with the supremum norm · , and the
Banach space Y = X × X with the norm (u, v) Y = u + v . We define the cones
P = x ∈ X, x(t) ≥ γtα– x , ∀t ∈ [, ] ,
P = y ∈ X, y(t) ≥ γtβ– y , ∀t ∈ [, ] ,
where γ, γ are defined in Section (Lemma .), and P = P × P ⊂ Y .
For λ, μ > , we introduce the operators Q, Q : Y → X and Q : Y → Y defined by
Q(x, y) = (Q(x, y), Q(x, y)), (x, y) ∈ Y with
Q(x, y)(t) = λ
If (H) and (H) hold, then we deduce easily that Q(x, y)(t) < ∞ and Q(x, y)(t) < ∞ for
all t ∈ [, ].
If (H) and (H) hold, we deduce, for all t ∈ [, ],
Q(x, y)(t) ≤ λσ
where M = max{maxt∈[,],u,v∈[,L] β(t, u, v), maxt∈[,],u,v∈[,L] β(t, u, v), }.
Besides, by Lemma ., we conclude that
Q(x, y)(t) ≥ γtα– Q(x, y) ,
Q(x, y)(t) ≥ γtβ– Q(x, y) ,
and so Q(x, y), Q(x, y) ∈ P.
By using standard arguments, we deduce that operator Q : P → P is a completely
continuous operator (a compact operator, that is, one that maps bounded sets into relatively
compact sets and is continuous).
Theorem . Assume that (H)-(H) hold. Then there exist constants λ > and μ >
such that, for any λ ∈ (, λ] and μ ∈ (, μ], the boundary value problem (S)-(BC) has at
least one positive solution.
Proof Let δ ∈ (, ) be fixed. From (H) and (H), there exists R ∈ (, ] such that
f (t, u, v) ≥ δf (t, , ),
g(t, u, v) ≥ δg(t, , ),
∀t ∈ [, ], u, v ∈ [, R].
()
max
t∈[,],u,v∈[,R]
f (t, u, v) + p(t) ≥ max δf (t, , ) + p(t) > ,
t∈[,]
g(t, u, v) + p(t) ≥ max δg(t, , ) + p(t) > ,
t∈[,]
f¯(R) =
g¯(R) =
c = σ
c = σ
λ = max
max
t∈[,],u,v∈[,R]
h(s) ds,
h(s) ds,
c = σ
c = σ
h(s) ds,
h(s) ds,
R , R
cf¯(R) cf¯(R)
μ = max
R , R
cg¯(R) cg¯(R)
We will show that, for any λ ∈ (, λ] and μ ∈ (, μ], problem ()-() has at least one
positive solution.
So, let λ ∈ (, λ] and μ ∈ (, μ] be arbitrary, but fixed for the moment. We define the
set U = {(x, y) ∈ P, (u, v) Y < R}. We suppose that there exist (x, y) ∈ ∂U ( (x, y) Y = R
or x + y = R) and ν ∈ (, ) such that (x, y) = νQ(x, y) or x = νQ(x, y), y = νQ(x, y).
We deduce that
x(t) – q(t) ∗ = x(t) – q(t) ≤ x(t) ≤ R, if x(t) – q(t) ≥ ,
x(t) – q(t) ∗ = , for x(t) – q(t) < , ∀t ∈ [, ],
y(t) – q(t) ∗ = y(t) – q(t) ≤ y(t) ≤ R, if y(t) – q(t) ≥ ,
y(t) – q(t) ∗ = , for y(t) – q(t) < , ∀t ∈ [, ].
Then by Lemma ., for all t ∈ [, ], we obtain
x(t) = νQ(x, y)(t) ≤ Q(x, y)(t)
≤ λσ
h(s)f¯(R) ds + μσ
h(s)g¯(R) ds
≤ λcf¯(R) + μcg¯(R) ≤ R + R = R ,
y(t) = νQ(x, y)(t) ≤ Q(x, y)(t)
≤ μσ
h(s)g¯(R) ds + λσ
h(s)f¯(R) ds
≤ μcg¯(R) + λcf¯(R) ≤ R + R = R .
Hence x ≤ R and y ≤ R . Then R = (x, y) Y = x + y ≤ R + R = R , which is
a contradiction.
Therefore, by Theorem . (with = P), we deduce that Q has a fixed point (x, y) ∈
U¯ ∩ P. That is, (x, y) = Q(x, y) or x = Q(x, y), y = Q(x, y), and x + y ≤ R
with x(t) ≥ γtα– x and y(t) ≥ γtβ– y for all t ∈ [, ].
Moreover, by (), we conclude
x(t) = Q(x, y)(t)
≥ λ
≥ λ
≥ μ
G(t, s) δf (t, , ) + p(s) ds + μ
G(t, s) δg(t, , ) + p(s) ds
G(t, s) δg(t, , ) + p(s) ds + λ
G(t, s) δf (t, , ) + p(s) ds
≥ μ
Therefore x(t) ≥ q(t), y(t) ≥ q(t) for all t ∈ [, ], and x(t) > q(t), y(t) > q(t) for all
t ∈ (, ). Let u(t) = x(t) – q(t) and v(t) = y(t) – q(t) for all t ∈ [, ]. Then u(t) ≥ ,
v(t) ≥ for all t ∈ [, ], u(t) > , v(t) > for all t ∈ (, ). Therefore (u, v) is a positive
solution of (S)-(BC).
Theorem . Assume that (H), (H), and (H) hold. Then there exist λ∗ > and μ∗ >
such that, for any λ ∈ (, λ∗] and μ ∈ (, μ∗], the boundary value problem (S)-(BC) has at
least one positive solution.
Proof We choose a positive number
R > max , γ
γγ
γγ
,
–
,
with
M = max
M = max
max
t∈[,]
u,v≥,u+v≤R
max
t∈[,]
u,v≥,u+v≤R
β(t, u, v), ,
β(t, u, v), .
x(s) – q(s) ∗ ≤ x(s) ≤ x ≤ R,
y(s) – q(s) ∗ ≤ y(s) ≤ y ≤ R.
Let λ ∈ (, λ∗] and μ ∈ (, μ∗]. Then, for any (x, y) ∈ P ∩ ∂ and s ∈ [, ], we have
Then, for any (x, y) ∈ P ∩ ∂ , we obtain
Q(x, y) ≤ λσ
Therefore
On the other hand, we choose a constant L > such that
λL γc(α–)
μL γc(α–)
–c
–c
c
c
h(s) ds ≥ , λL γc(β–)
h(s) ds ≥ , μL γc(β–)
–c
–c
c
c
h(s) ds ≥ ,
h(s) ds ≥ .
From (H), we deduce that there exists a constant M > such that
Q(x, y) Y = Q(x, y) + Q(x, y) ≤ (x, y) Y ,
∀(x, y) ∈ P ∩ ∂ .
()
f (t, u, v) ≥ L(u + v) or g(t, u, v) ≥ L(u + v),
∀t ∈ [c, – c], u, v ≥ , u + v ≥ M. ()
Now we define
R = max R, γMcα– , γMcβ– , γ
and let = {(x, y) ∈ P, (x, y) Y < R}.
We suppose that f∞ = ∞, that is, f (t, u, v) ≥ L(u + v) for all t ∈ [c, – c] and u, v ≥ ,
u + v ≥ M. Then, for any (x, y) ∈ P ∩ ∂ , we have (x, y) Y = R or x + y = R. We
deduce that x ≥ R or y ≥ R .
We suppose that x
R . Then, for any (x, y) ∈ P ∩ ∂ , we obtain
≥
x(t) – q(t) = x(t) – λ
Therefore, we conclude
x(t) – q(t) ∗ = x(t) – q(t) ≥ x(t) ≥ γtα– x
Hence
≥ γtα–R ≥ γcα–R ≥ M,
∀t ∈ [c, – c].
x(t) – q(t) ∗ + y(t) – q(t) ∗ ≥ x(t) – q(t) ∗ = x(t) – q(t) ≥ M,
∀t ∈ [c, – c]. ()
Then, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], by () and (), we deduce
f t, x(t) – q(t) ∗, y(t) – q(t) ∗
≥ L x(t) – q(t) ∗ + y(t) – q(t) ∗
≥ L x(t) – q(t) ∗
L
≥ x(t),
∀t ∈ [c, – c].
It follows that, for any (x, y) ∈ P ∩ ∂ , t ∈ [c, – c], we obtain
If y ≥ R , then by a similar approach, we obtain again relation ().
We suppose now that g∞ = ∞, that is, g(t, u, v) ≥ L(u + v), for all t ∈ [c, – c] and u, v ≥ ,
u + v ≥ M. Then, for any (x, y) ∈ P ∩ ∂ , we have (x, y) Y = R. Hence x ≥ R or
y ≥ R .
If x ≥ R , then for any (x, y) ∈ P ∩ ∂ we deduce in a similar manner as above that
x(t) – q(t) ≥ x(t) for all t ∈ [, ] and
Q(x, y)(t) ≥ μ
G(t, s)L x(s) – q(s) ∗ ds ≥ μ
c
–c
G(t, s) Lγcα–R ds
tα–h(s) Lγcα–R ds ≥ μc(α–)
LγR
∀t ∈ [c, – c].
c
–c
Hence we obtain relation ().
If y ≥ R , then in a similar way as above, we deduce again relation ().
Therefore, by Theorem ., relations () and (), we conclude that Q has a fixed point
(x, y) ∈ P ∩ ( ¯ \ ), that is, R ≤ (x, y) Y ≤ R. Since (x, y) Y ≥ R, then x ≥ R
or y ≥ R .
We suppose first that x ≥ R . Then we deduce
x(t) – q(t) = x(t) – λ
G(t, s)p(s) ds – μ
G(t, s)p(s) ds
≥ x(t) – tα– δ
≥ x(t) – γx(xt)
≥ –
γR
≥ –
γR
R –
≥
γR
=
tα–,
and so x(t) ≥ q(t) + tα– for all t ∈ [, ], where = γR – (δp(s) + δp(s)) ds > .
Then y() = x(s) dK (s) ≥ sα– dK (s) > and
γR
≥
Therefore, we obtain
y(t) – q(t) = y(t) – μ
G(t, s)p(s) ds – λ
G(t, s)p(s) ds
≥ y(t) – tβ–
≥ y(t) –
≥ y(t) –
y(t)
γ y
–
≥
=
≥ γtβ– y –
γγR tβ–
γγR
–
–
where = γγR sα– dK (s) – (δp(s) + δp(s)) ds > .
Hence y(t) ≥ q(t) + tβ– for all t ∈ [, ].
If y ≥ R , then by a similar approach, we deduce that y(t) ≥ q(t) + tβ– and
x(t) ≥ q(t) + tα– for all t ∈ [, ], where = γR – (δp(s) + δp(s)) ds > and
= γγR sβ– dH(s) – (δp(s) + δp(s)) ds > .
Let u(t) = x(t) – q(t) and v(t) = y(t) – q(t) for all t ∈ [, ]. Then (u, v) is a positive
solution of (S)-(BC) with u(t) ≥ tα– and v(t) ≥ tβ– for all t ∈ [, ], where =
min{ , } and = min{ , }. This completes the proof of Theorem ..
Theorem . Assume that (H), (H), (H), and
(H ) The functions f , g ∈ C([, ] × [, ∞) × [, ∞), (–∞, +∞)) and there exist functions
p, p, α, α ∈ C([, ], [, ∞)), β, β ∈ C([, ] × [, ∞) × [, ∞), [, ∞)) such that
for all t ∈ [, ], u, v ∈ [, ∞), with pi(s) ds > , i = , ,
hold. Then the boundary value problem (S)-(BC) has at least two positive solutions for λ >
and μ > sufficiently small.
Proof Because assumption (H ) implies assumptions (H) and (H), we can apply
Theorems . and .. Therefore, we deduce that, for < λ ≤ min{λ, λ∗} and < μ ≤
min{μ, μ∗}, problem (S)-(BC) has at least two positive solutions (u, v) and (u, v) with
(u + q, v + q) Y ≤ and (u + q, v + q) Y > .
Theorem . Assume that λ = μ, and (H), (H), and (H) hold. In addition if
(H) there exists c ∈ (, /) such that
δp(s) + δp(s) ds, γ
–
,
then there exists λ∗ > such that for any λ ≥ λ∗ problem (S)-(BC) (with λ = μ) has at least
one positive solution.
Proof By (H) we conclude that there exists M > such that
f (t, u, v) ≥ L
or g(t, u, v) ≥ L,
∀t ∈ [c, – c], u, v ≥ , u + v ≥ M.
We define
λ∗ = max
We assume now λ ≥ λ∗. Let
R = max λ
γ
M
cα–
.
λ
δp(s) + δp(s) ds, γ
–
x(t) – q(t) ≥ γtα– x – λtα–δ
p(s) ds – λtα–δ
δp(s) + δp(s) ds – λ
δp(s) + δp(s) ds
= tα–λ
≥ tα–λ∗
δp(s) + δp(s) ds ≥ cMα– tα– ≥ .
Therefore, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we have
Hence, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we conclude
c
–c
c
–c
Therefore we obtain Q(x, y) ≥ R for all (x, y) ∈ P ∩ ∂ , and so
Q(x, y) Y ≥ R = (x, y) Y ,
∀(x, y) ∈ P ∩ ∂ .
()
If y ≥ R/, then by a similar approach we deduce again relation ().
We suppose now that g∞i > L, that is, g(t, u, v) ≥ L for all t ∈ [c, – c] and u, v ≥ ,
u + v ≥ M. Let (x, y) ∈ P ∩ ∂ . Then (x, y) Y = R, so x ≥ R/ or y ≥ R/. If x ≥
R/, then we obtain in a similar manner as in the first case above (f ∞i > L) that x(t) –
q(t) ≥ cMα– tα– ≥ for all t ∈ [, ].
Therefore, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we deduce inequalities ().
Hence, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we conclude
Therefore we obtain Q(x, y) ≥ R, and so
P ∩ ∂ , that is, we have relation ().
By a similar approach we obtain relation () if y ≥ R/.
On the other hand, we consider the positive number
ε = min
λσ
λσ
–
–
h(s)α(s) ds
–
, λσ
h(s)α(s) ds
–
Q(x, y) Y ≥ R = (x, y) Y for all (x, y) ∈
Then by (H) we deduce that there exists M > such that
βi(t, u, v) ≤ ε(u + v),
∀t ∈ [, ], u, v ≥ , u + v ≥ M, i = , .
Therefore we obtain
βi(t, u, v) ≤ M + ε(u + v),
∀t ∈ [, ], u, v ≥ , i = , ,
where M = maxi=,{maxt∈[,],u,v≥,u+v≤M βi(t, u, v)}.
We define now
R = max R, λσ max{M, }
h(s) α(s) + p(s) ds,
λσ max{M, }
h(s) α(s) + p(s) ds,
λσ max{M, }
h(s) α(s) + p(s) ds,
λσ max{M, }
h(s) α(s) + p(s) ds ,
and let = {(x, y) ∈ P, (x, y) Y < R}.
For any (x, y) ∈ P ∩ ∂ , we have
Q(x, y)(t) ≤ λ
σh(s) α(s)β s, x(s) – q(s) ∗, y(s) – q(s) ∗ + p(s) ds
+ λ
σh(s) α(s)β s, x(s) – q(s) ∗, y(s) – q(s) ∗ + p(s) ds
≤ λσ
h(s) α(s) M + ε x(s) – q(s) ∗ + y(s) – q(s) ∗
+ p(s) ds
+ λσ
h(s) α(s) M + ε x(s) – q(s) ∗ + y(s) – q(s) ∗
+ p(s) ds
≤ λσ max{M, }
h(s) α(s) + p(s) ds + λσεR
≤ R + R + R + R = R = (x,y) Y ,
and so Q(x, y) ≤ (x,y) Y for all (x, y) ∈ P ∩ ∂ .
In a similar way we obtain Q(x, y)(t) ≤ (x,y) Y for all t ∈ [, ], and so Q(x, y) ≤ (x,y) Y
for all (x, y) ∈ P ∩ ∂ .
Therefore, we deduce
Q(x, y) Y ≤ (x, y) Y ,
∀(x, y) ∈ P ∩ ∂ .
()
By Theorem ., (), and (), we conclude that Q has a fixed point (x, y) ∈ P ∩ ( ¯ \
). Since (x, y) ≥ R then x ≥ R/ or y ≥ R/.
≥ λtβ–
≥ λ∗tβ–
sα– dK (s) – λtβ–
Therefore, we deduce that, for all t ∈ [, ],
y(t) – q(t) ≥ y(t) – λδ
tβ–p(s) ds – λδ
tβ–p(s) ds
≥ γtβ– y – λtβ–
If y ≥ R/, then by a similar approach we conclude again that x(t) – q(t) ≥ cMα– tα–
and y(t) – q(t) ≥ cMβ– tβ– for all t ∈ [, ].
Let u(t) = x(t) – q(t) and v(t) = y(t) – q(t) for all t ∈ [, ]. Then u(t) ≥ tα– and
v(t) ≥ tβ– for all t ∈ [, ], where = cMα– , = cMβ– . Hence we deduce that (u, v) is
a positive solution of (S)-(BC), which completes the proof of Theorem ..
In a similar manner as we proved Theorem ., we obtain the following theorems.
Theorem . Assume that λ = μ, and (H), (H), and (H) hold. In addition if
(H ) there exists c ∈ (, /) such that
f ∞i = lim inf min f (t, u, v) > L or g∞i = lim inf min g(t, u, v) > L,
u+u,vv→≥∞ t∈[c,–c] u+u,vv→≥∞ t∈[c,–c]
where
Theorem . Assume that λ = μ, and (H), (H), and (H) hold. In addition if
(H) there exists c ∈ (, /) such that
fˆ∞ = u+ul,viv→m≥∞ t∈m[c,in–c] f (t, u, v) = ∞
or
gˆ∞ = u+ul,viv→m≥∞ t∈m[c,in–c] g(t, u, v) = ∞,
then there exists λ˜∗ > such that for any λ ≥ λ˜∗ problem (S)-(BC) (with λ = μ) has at least
one positive solution.
4 Examples
Let α = / (n = ), β = / (m = ), H(t) = t, K (t) = t. Then u(s) dK (s) = su(s) ds
and v(s) dH(s) = sv(s) ds.
We consider the system of fractional differential equations
(S)
(BC)
with the boundary conditions
D/+u(t) + λf (t, u(t), v(t)) = ,
D/+v(t) + μg(t, u(t), v(t)) = ,
t ∈ (, ),
t ∈ (, ),
u() = u () = ,
v() = v () = ,
u() = sv(s) ds,
v() = su(s) ds.
Then we obtain = – ( sα– dK (s))( sβ– dH(s)) = > , τ α–( – τ ) dK (τ ) = >
, τ β–( – τ ) dH(τ ) = > . The functions H and K are nondecreasing, and so
assumption (H) is satisfied. Besides, we deduce
g(t, s) = √π
t/( – s)/ – (t – s)/,
t/( – s)/,
≤ s ≤ t ≤ ,
≤ t ≤ s ≤ ,
g(t, s) =
(/)
t/( – s)/ – (t – s)/,
t/( – s)/,
G(t, s) = g(t, s) + t/
τ g(τ , s) ds,
G(t, s) = g(t, s) + t/
≤ s ≤ t ≤ ,
≤ t ≤ s ≤ ,
G(t, s) = t/
We also obtain h(s) = √π s( – s)/, h(s) = (/) s( – s)/,
k(t) =
t/, ≤ t ≤ /,
( – t)t/, / ≤ t ≤ ,
k(t) =
t/, ≤ t ≤ /,
( – t)t/, / ≤ t ≤ .
In addition, we have σ = , δ = √π , = √√– , σ = , δ = √(/) , = √– , σ =
, δ = (/) , = √√– , σ = , δ = √π , = √ √– , γ = √– ≈ .√, γ =
(√ √–) ≈ ..
τ g(τ , s)dτ ,
G(t, s) = t/
τ g(τ , s) dτ .
τ g(τ , s) dτ ,
Example We consider the functions
g(t, u, v) =
+ ln( – t), t ∈ (, ), u, v ≥ .
We have p(t) = – ln t, p(t) = – ln( – t), α(t) = α(t) = √t(–t) for all t ∈ (, ), β(t, u, v) =
(u + v), β(t, u, v) = + sin(u + v) for all t ∈ [, ], u, v ≥ , p(t) dt = , p(t) dt = ,
αi(t) dt = π , i = , . Therefore, assumption (H) is satisfied. In addition, for c ∈ (, /)
fixed, assumption (H) is also satisfied (f∞ = ∞).
After some computations, we deduce (δp(s) + δp(s)) ds ≈ ., (δp(s) +
δp(s)) ds ≈ ., h(s)(α(s) + p(s)) ds ≈ ., h(s)(α(s) + p(s)) ds ≈
.. We choose R = , which satisfies the condition from the beginning of
the proof of Theorem .. Then M = R, M = , λ∗ ≈ . · –, and μ∗ = . By
Theorem ., we conclude that (S)-(BC) has at least one positive solution for any λ ∈ (, λ∗]
and μ ∈ (, μ∗].
Example We consider the functions
f (t, u, v) = (u + v) + cos u,
g(t, u, v) = (u + v)/ + cos v, t ∈ [, ], u, v ≥ .
We have p(t) = p(t) = for all t ∈ [, ], and then assumption (H) is satisfied. Besides,
assumption (H) is also satisfied, because f (t, , ) = and g(t, , ) = for all t ∈ [, ].
Let δ = < and R = . Then
f (t, u, v) ≥ δf (t, , ) = ,
g(t, u, v) ≥ δg(t, , ) = ,
∀t ∈ [, ], u, v ∈ [, ].
In addition,
f¯(R) = f¯() =
g¯(R) = g¯() =
We also obtain c ≈ ., c ≈ ., c ≈ ., c ≈ .,
R R R R
and then λ = max{ cf¯(R) , cf¯(R) } ≈ . and μ = max{ cg¯(R) , cg¯(R) } ≈
..
By Theorem ., for any λ ∈ (, λ] and μ ∈ (, μ], we deduce that problem (S)-(BC)
has at least one positive solution.
Because assumption (H ) is satisfied (α(t) = α(t) = , β(t, u, v) = (u+v) +, β(t, u, v) =
(u + v)/ + for all t ∈ [, ], u, v ≥ ) and assumption (H) is also satisfied (f∞ = ∞), by
Theorem . we conclude that problem (S)-(BC) has at least two positive solutions for
λ and μ sufficiently small.
Example We consider λ = μ and the functions
(u + v)a
f (t, u, v) = t( – t) – √t ,
g(t, u, v) =
ln( + u + v)
t( – t)
– √
– t
, t ∈ (, ), u, v ≥ ,
where a ∈ (, ).
Here we have p(t) = √t , p(t) = √–t , α(t) = √t(–t) , α(t) = √t(–t) for all t ∈ (, ),
the assumptions (H), (H), and (H) are satisfied (βi∞ = for i = , and fˆ∞ = ∞).
Then by Theorem ., we deduce that there exists λ˜∗ > such that for any λ ≥ λ˜∗ our
problem (S)-(BC) (with λ = μ) has at least one positive solution.
Competing interests
The authors declare that no competing interests exist.
Authors’ contributions
The authors contributed equally to this paper. Both authors read and approved the final manuscript.
Acknowledgements
The authors thank the referees for their valuable comments and suggestions. The work of R Luca was supported by the
CNCS grant PN-II-ID-PCE-2011-3-0557, Romania.
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