Henderson and Luca Boundary Value Problems (2016) 2016:61 DOI 10.1186/s13661-016-0569-8
RESEARCH
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Positive solutions for a system of semipositone coupled fractional boundary value problems Johnny Henderson1 and Rodica Luca2* *
Correspondence:
[email protected] 2 Department of Mathematics, Gh. Asachi Technical University, Iasi, 700506, Romania Full list of author information is available at the end of the article
Abstract 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. MSC: 34A08; 45G15 Keywords: Riemann-Liouville fractional differential equations; coupled integral boundary conditions; positive solutions; sign-changing nonlinearities
1 Introduction 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 (S)
Dα+ u(t) + λf (t, u(t), v(t)) = , β D+ v(t) + μg(t, u(t), v(t)) = ,
t ∈ (, ), n – < α ≤ n, 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),
β
where n, m ∈ N, n, m ≥ , Dα+ , and D+ 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 © 2016 Henderson and Luca. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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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),
with f , g nonnegative), in [] (problem ( were investigated in [] (problem (S)-(BC) S) with f , g nonnegative, singular or not), and in [] (problem (S)-(BC) with f , g sign(BC) changing 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) = t α– 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 dt
n
n–α I+ f
n t d f (s) (t) = ds, (n – α) dt (t – s)α–n+
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 for t > , and if α = then D+ f (t) = f (t) for t > .
(m) Dm (t) + f (t) = f
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We consider now the fractional differential system
Dα+ u(t) + x˜ (t) = , t ∈ (, ), n – < α ≤ n, β D+ v(t) + y˜ (t) = , t ∈ (, ), m – < β ≤ m,
()
with the coupled integral boundary conditions
u() = u () = · · · = u(n–) () = , v() = v () = · · · = v(m–) () = ,
u() = v(s) dH(s), v() = u(s) dK(s),
()
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 ⎧ α– ⎪ G (t, s) = g (t, s) + t ( τ β– dH(τ ))( g (τ , s) dK(τ )), ⎪ ⎪ ⎪ ⎨ G (t, s) = tα– g (τ , s) dH(τ ), β– ⎪ G (t, s) = g (t, s) + t ( τ α– dK(τ ))( g (τ , s) dH(τ )), ⎪ ⎪ ⎪ ⎩ G (t, s) = tβ– g (τ , s) dK(τ ), ∀t, s ∈ [, ]
()
and
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ g (t, s) =
(α)
⎪ ⎪ ⎪ ⎪ ⎩ g (t, s) =
(β)
t α– ( – s)α– – (t – s)α– , ≤ s ≤ t ≤ , ≤ t ≤ s ≤ , t α– ( – s)α– , t β– ( – s)β– – (t – s)β– , ≤ s ≤ t ≤ , ≤ t ≤ s ≤ , t β– ( – s)β– ,
()
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) ∈ (, ) × (, ). α– and (b) g (t, s) ≤ h (s), g (t, s) ≤ h (s) for all (t, s) ∈ [, ] × [, ], where h (s) = s(–s) (α–) β–
h (s) = s(–s) for all s ∈ [, ]. (β–) (c) g (t, s) ≥ k (t)h (s), g (t, s) ≥ k (t)h (s) for all (t, s) ∈ [, ] × [, ], where tα– , ≤ t ≤ , ( – t)t α– t α– α– k (t) = min , = (–t)t α– α– α– , ≤ t ≤ , α– tβ– , ≤ t ≤ , ( – t)t β– t β– β– , = (–t)t k (t) = min β– β – β – , ≤ t ≤ . β–
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(d) For any (t, s) ∈ [, ] × [, ], we have g (t, s) ≤
( – t)t α– t α– ≤ , (α – ) (α – )
g (t, s) ≤
( – t)t β– t β– ≤ . (β – ) (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– ( – β– τ ) dK(τ ) > , τ ( – τ ) dH(τ ) > . Then the functions Gi , i = , . . . , satisfy the inequalities: (a ) G (t, s) ≤ σ h (s), ∀(t, s) ∈ [, ] × [, ], where σ = +
K() – K()
τ β– dH(τ ) > .
(a ) G (t, s) ≤ δ t α– , ∀(t, s) ∈ [, ] × [, ], where
β– α– δ = + τ dH(τ ) ( – τ )τ dK(τ ) > . (α – ) (a ) G (t, s) ≥ t α– h (s), (t, s) ∈ [, ] × [, ], where =
τ β– dH(τ )
k (τ ) dK(τ ) > .
(b ) G (t, s) ≤ σ h (s), ∀(t, s) ∈ [, ] × [, ], where σ = (H() – H()) > . β– (b ) G (t, s) ≤ δ t α– , ∀(t, s) ∈ [, ] × [, ], where δ = (β–) dH(τ ) > . ( – τ )τ (b ) G (t, s) ≥ t α– h (s), ∀(t, s) ∈ [, ] × [, ], where = k (τ ) dH(τ ) > . (c ) G (t, s) ≤ σ h (s), ∀(t, s) ∈ [, ] × [, ], where H() – H() σ = +
τ α– dK(τ ) > .
(c ) G (t, s) ≤ δ t β– , ∀(t, s) ∈ [, ] × [, ], where δ =
+ τ α– dK(τ ) ( – τ )τ β– dH(τ ) > . (β – )
(c ) G (t, s) ≥ t β– h (s), ∀(t, s) ∈ [, ] × [, ], where =
τ
α–
dK(τ )
k (τ ) dH(τ ) > .
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(d ) G (t, s) ≤ σ h (s), ∀(t, s) ∈ [, ] × [, ], where σ = (K() – K()) > . α– dK(τ ) > . (d ) G (t, s) ≤ δ t β– , ∀(t, s) ∈ [, ] × [, ], where δ = (α–) ( – τ )τ β– (d ) G (t, s) ≥ t h (s), ∀(t, s) ∈ [, ] × [, ], where = k (τ ) dK(τ ) > . Proof From the assumptions of this lemma, we obtain
τ
α–
τ α– ( – τ ) dK(τ ) > ,
dK(τ ) ≥
( – τ )τ
α–
( – τ )τ α– dK(τ ) > ,
dK(τ ) ≥
τ α– ( – τ ) dK(τ ) > , α– τ β– dH(τ ) ≥ τ β– ( – τ ) dH(τ ) > ,
k (τ ) dK(τ ) ≥
( – τ )τ
β–
( – τ )τ β– dH(τ ) > ,
dH(τ ) ≥
K() – K() =
β –
k (τ ) dH(τ ) ≥
τ β– ( – τ ) dH(τ ) > ,
dK(τ ) ≥
τ α– ( – τ ) dK(τ ) > ,
H() – H() =
dH(τ ) ≥
τ β– ( – τ ) dH(τ ) > .
By using Lemma ., we deduce, for all (t, s) ∈ [, ] × [, ]: (a ) G (t, s) = g (t, s) + ≤ h (s) +
t α–
g (τ , s) dK(τ )
τ β– dH(τ )
τ β– dH(τ )
h (s) dK(τ )
β– K() – K() τ dH(τ ) = σ h (s). = h (s) + (a )
t α– t α– ( – τ )τ α– β– + dK(τ ) G (t, s) ≤ τ dH(τ ) (α – ) (α – )
α– β– α– τ dH(τ ) ( – τ )τ dK(τ ) = δ t α– . =t + (α – ) (a ) G (t, s) ≥
t α–
= t α– h (s)
k (τ )h (s) dK(τ )
τ β– dH(τ )
τ β– dH(τ )
k (τ ) dK(τ ) = t α– h (s).
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(b ) t α– g (τ , s) dH(τ ) ≤ h (s) dH(τ ) H() – H() h (s) = σ h (s). =
G (t, s) =
(b ) G (t, s) ≤
t α–
( – τ )τ β– dH(τ ) = δ t α– . (β – )
(b ) G (t, s) ≥
t α–
k (τ )h (s) dH(τ ) =
t α– h (s)
k (τ ) dH(τ ) = t α– h (s).
(c )
t β– α– τ dK(τ ) g (τ , s) dH(τ ) G (t, s) = g (t, s) +
≤ h (s) + τ α– dK(τ ) h (s) dH(τ ) α– H() – H() τ dK(τ ) = σ h (s). = h (s) + (c )
( – t)t β– t β– ( – τ )τ β– τ α– dK(τ ) + dH(τ ) (β – ) (β – )
t β– + τ α– dK(τ ) ( – τ )τ β– dH(τ ) = δ t β– . ≤ (β – )
G (t, s) ≤
(c ) t β– G (t, s) ≥
τ
= t β– h (s)
α–
dK(τ )
k (τ )h (s) dH(τ )
τ α– dK(τ )
k (τ ) dH(τ ) = t β– h (s).
(d ) g (τ , s) dK(τ ) ≤ h (s) dK(τ ) = h (s) K() – K() = σ h (s).
t β– G (t, s) =
(d ) G (t, s) ≤
t β–
( – τ )τ α– dK(τ ) = δ t β– . (α – )
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(d ) t β–
G (t, s) ≥
k (τ )h (s) dK(τ ) = t β– h (s)
k (s) dK(τ ) = t β– h (s).
Lemma . Assume that H, K : [, ] → R are nondecreasing functions, > , τ α– ( – τ ) dK(τ ) > , τ β– ( – τ ) dH(τ ) > , and x˜ , y˜ ∈ C(, ) ∩ L (, ), x˜ (t) ≥ , y˜ (t) ≥ for all t ∈ (, ). Then the solution (u, v) of problem ()-() given by () satisfies the inequalities u(t) ≥ γ t α– u(t ), v(t) ≥ γ t β– v(t ), for all t, t ∈ [, ], where γ = min{ σ , σ } > , γ = min{ σ , σ } > . Proof By using Lemma ., we obtain
G (t, s)˜x(s) ds +
u(t) =
G (t, s)˜y(s) ds
≥
t α– h (s)˜x(s) ds +
=t
t α– h (s)˜y(s) ds
α–
h (s)˜x(s) ds + h (s)˜y(s) ds
G t , s x˜ (s) ds + G t , s y˜ (s) ds ≥t σ
≥ t α– min , G t , s x˜ (s) ds + G t , s y˜ (s) ds σ σ , > . = γ t α– u t , ∀t, t ∈ [, ], where γ = min σ σ
α–
σ
In a similar way, we deduce
v(t) =
G (t, s)˜y(s) ds +
≥ γ t
G (t, s)˜x(s) ds
β–
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 rela¯ → be a completely continuous operator tively open subset of with ∈ U, and let S : U (continuous and compact). Then either ¯ or () S has a fixed point in U, () 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 ¯ ⊂ and let A : C ∩ ( ¯ \ ) → C be a are bounded open subsets of X with ∈ ⊂ 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 ∩ (
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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 )) × β– α– β– ( τ dH(τ )) > , and τ ( – τ ) dK(τ ) > , τ ( – 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 –p (t) ≤ f (t, u, v) ≤ α (t)β (t, u, v), for all t ∈ (, ), u, v ∈ [, ∞), with < (H) There exists c ∈ (, /) such that f∞ = lim
min
u+v→∞ t∈[c,–c]
f (t, u, v) =∞ u+v
(H) βi∞ = limu+v→∞ maxt∈[,]
βi (t,u,v) u+v
–p (t) ≤ g(t, u, v) ≤ α (t)β (t, u, v),
pi (s) ds < ∞, <
or g∞ = lim
αi (s) ds < ∞, i = , .
min
u+v→∞ t∈[c,–c]
g(t, u, v) = ∞. u+v
= , i = , .
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)) = , < t < , β D+ y(t) + μ(g(t, [x(t) – q (t)]∗ , [y(t) – q (t)]∗ ) + p (t)) = , < t < ,
()
with the integral boundary conditions
x() = y(s) dH(s), y() = x(s) dK(s),
x() = x () = · · · = x(n–) () = , y() = y () = · · · = y(m–) () = ,
()
where z(t)∗ = z(t) if z(t) ≥ , and z(t)∗ = if z(t) < . Here (q , q ) with
q (t) = λ
G (t, s)p (s) ds + μ
q (t) = μ
G (t, s)p (s) ds + λ
G (t, s)p (s) ds,
t ∈ [, ],
G (t, s)p (s) ds,
t ∈ [, ],
is solution of the system of fractional differential equations
Dα+ q (t) + λp (t) = , < t < , β D+ q (t) + μp (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).
()
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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 G (t, s)(f (s, [x(s) – q (s)]∗ , [y(s) – q (s)]∗ ) + p (s)) ds ⎪ ⎪ ⎪ ⎨ + μ G (t, s)(g(s, [x(s) – q (s)]∗ , [y(s) – q (s)]∗ ) + p (s)) ds, t ∈ [, ], ⎪ y(t) = μ G (t, s)(g(s, [x(s) – q (s)]∗ , [y(s) – q (s)]∗ ) + p (s)) ds ⎪ ⎪ ⎩ + λ G (t, s)(f (s, [x(s) – q (s)]∗ , [y(s) – q (s)]∗ ) + p (s)) ds, 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) ≥ γ t α– x, ∀t ∈ [, ] , P = y ∈ X, y(t) ≥ γ t β– 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
∗ ∗ G (t, s) f s, x(s) – q (s) , y(s) – q (s) + p (s) ds
+μ
∗ ∗ G (t, s) g s, x(s) – q (s) , y(s) – q (s) + p (s) ds,
t ∈ [, ],
∗ ∗ G (t, s) g s, x(s) – q (s) , y(s) – q (s) + p (s) ds
Q (x, y)(t) = μ
+λ
∗ ∗ G (t, s) f s, x(s) – q (s) , y(s) – q (s) + p (s) ds,
t ∈ [, ].
It is clear that if (x, y) is a fixed point of operator Q, then (x, y) is a solution of problem ()-(). Lemma . If (H) and (H), or (H) and (H) hold, then operator Q : P → P is a completely continuous operator. Proof The operators Q and Q are well defined. To prove this, let (x, y) ∈ P be fixed with (x, y)Y = L. Then we have ≤ x(s) ≤ x ≤ (x, y)Y = L, ∗ y(s) – q (s) ≤ y(s) ≤ y ≤ (x, y)Y = L,
x(s) – q (s)
∗
∀s ∈ [, ], ∀s ∈ [, ].
If (H) and (H) hold, then we deduce easily that Q (x, y)(t) < ∞ and Q (x, y)(t) < ∞ for all t ∈ [, ].
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If (H) and (H) hold, we deduce, for all t ∈ [, ],
Q (x, y)(t) ≤ λσ
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
+ μσ
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
h (s) α (s) + p (s) ds + μσ h (s) α (s) + p (s) ds < ∞, ≤ M λσ
Q (x, y)(t) ≤ μσ
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
+ λσ
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
≤ M μσ
h (s) α (s) + p (s) ds + λσ
h (s) α (s) + p (s) ds < ∞,
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) ≥ γ t α– Q (x, y),
Q (x, y)(t) ≥ γ t β– Q (x, y),
∀t ∈ [, ],
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 ].
We define f¯ (R ) = g¯ (R ) =
max
f (t, u, v) + p (t) ≥ max δf (t, , ) + p (t) > ,
max
g(t, u, v) + p (t) ≥ max δg(t, , ) + p (t) > ,
t∈[,],u,v∈[,R ]
t∈[,],u,v∈[,R ]
t∈[,]
t∈[,]
c = σ
h (s) ds,
c = σ
h (s) ds,
c = σ
h (s) ds,
λ = max
c = σ
R R , , c f¯ (R ) c f¯ (R )
h (s) ds,
μ = max
R R , . c g¯ (R ) c g¯ (R )
()
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Page 11 of 23
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) = ν Q(x, y) or x = νQ (x, y), y = νQ (x, y). We deduce that
x(t) – q (t) x(t) – q (t)
∗ ∗
y(t) – q (t) y(t) – q (t)
∗ ∗
= x(t) – q (t) ≤ x(t) ≤ R , = ,
for x(t) – q (t) < , ∀t ∈ [, ],
= y(t) – q (t) ≤ y(t) ≤ R , = ,
if x(t) – q (t) ≥ ,
if 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) h (s)f¯ (R ) ds + μσ h (s)¯g (R ) ds ≤ λσ
R R R + = , ≤ λ c f¯ (R ) + μ c g¯ (R ) ≤ y(t) = νQ (x, y)(t) ≤ Q (x, y)(t) h (s)¯g (R ) ds + λσ h (s)f¯ (R ) ds ≤ μσ
R R R ≤ μ c g¯ (R ) + λ c f¯ (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) ≥ γ t α– x and y (t) ≥ γ t β– y for all t ∈ [, ]. Moreover, by (), we conclude x (t) = Q (x , y )(t) ≥λ G (t, s) δf (t, , ) + p (s) ds + μ G (t, s) δg(t, , ) + p (s) ds
≥λ
G (t, s)p (s) ds + μ
x (t) > λ
G (t, s)p (s) ds + μ
G (t, s)p (s) ds = q (t),
∀t ∈ [, ],
G (t, s)p (s) ds = q (t),
∀t ∈ (, ),
y (t) = Q (x , y )(t) G (t, s) δg(t, , ) + p (s) ds + λ G (t, s) δf (t, , ) + p (s) ds ≥μ
Henderson and Luca Boundary Value Problems (2016) 2016:61
≥μ
Page 12 of 23
G (t, s)p (s) ds = q (t),
∀t ∈ [, ],
G (t, s)p (s) ds = q (t),
∀t ∈ (, ).
G (t, s)p (s) ds + λ
y (t) > μ
G (t, s)p (s) ds + λ
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 δ p (s) + δ p (s) ds, δ p (s) + δ p (s) ds, R > max , γ γ
– sβ– dH(s) δ p (s) + δ p (s) ds, γ γ
– δ p (s) + δ p (s) ds , sα– dK(s) γ γ and we define the set = {(x, y) ∈ P, (x, y)Y < R }. We introduce λ∗ = min , R σ M μ∗ = min ,
– R h (s) α (s) + p (s) ds , σ M
– , h (s) α (s) + p (s) ds
R σ M
– R h (s) α (s) + p (s) ds , σ M
– h (s) α (s) + p (s) ds ,
with M = max
max
β (t, u, v), ,
M = max
max
β (t, u, v), .
t∈[,] u,v≥,u+v≤R
t∈[,] u,v≥,u+v≤R
Let λ ∈ (, λ∗ ] and μ ∈ (, μ∗ ]. Then, for any (x, y) ∈ P ∩ ∂ and s ∈ [, ], we have
x(s) – q (s)
∗
y(s) – q (s)
∗
≤ x(s) ≤ x ≤ R , ≤ y(s) ≤ y ≤ R .
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 13 of 23
Then, for any (x, y) ∈ P ∩ ∂ , we obtain Q (x, y) ≤ λσ
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
+ μσ
∗
≤ λ σ M
h (s) α (s) + p (s) ds + μ∗ σ M
h (s) α (s) + p (s) ds
R R R (x, y)Y + = = , ≤ ∗ ∗ Q (x, y) ≤ μσ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
∗ ∗ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
+ λσ
∗
≤ μ σ M
h (s) α (s) + p (s) ds + λ∗ σ M
h (s) α (s) + p (s) ds
R R R (x, y)Y + = = . ≤ Therefore Q(x, y) = Q (x, y) + Q (x, y) ≤ (x, y) , Y Y
∀(x, y) ∈ P ∩ ∂ .
()
On the other hand, we choose a constant L > such that
–c
(α–)
–c
h (s) ds ≥ , λL γ c
(β–)
λL γ c
c
h (s) ds ≥ ,
c
–c
μL γ c(α–)
–c
h (s) ds ≥ , μL γ c(β–)
c
h (s) ds ≥ .
c
From (H), we deduce that there exists a constant M > such that 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 M M R = max R , δ p (s) + δ p (s) ds, , , α– β– γ c γ c γ δ p (s) + δ p (s) ds > , 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 .
Henderson and Luca Boundary Value Problems (2016) 2016:61
We suppose that x ≥
R .
x(t) – q (t) = x(t) – λ
Page 14 of 23
Then, for any (x, y) ∈ P ∩ ∂ , we obtain
G (t, s)p (s) ds – μ
G (t, s)p (s) ds
≥ x(t) – t α– δ p (s) ds + δ p (s) ds ≥ x(t) –
x(t) γ x
δ p (s) + δ p (s) ds
δ p (s) + δ p (s) ds γ x δ p (s) + δ p (s) ds ≥ x(t) ≥ , ≥ x(t) – γ R = x(t) –
∀t ∈ [, ].
Therefore, we conclude
x(t) – q (t)
∗
= x(t) – q (t) ≥ x(t) ≥ γ t α– x α– α– ≥ γ t R ≥ γ c R ≥ M , ∀t ∈ [c, – c].
Hence
∗ ∗ ∗ 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 ≥ L x(t) – q (t) ≥ x(t),
∀t ∈ [c, – c].
It follows that, for any (x, y) ∈ P ∩ ∂ , t ∈ [c, – c], we obtain
Q (x, y)(t) ≥ λ
≥λ
–c
∗ ∗ G (t, s) f s, x(s) – q (s) , y(s) – q (s) + p (s) ds
–c
∗ G (t, s)L x(s) – q (s) ds ≥ λ
c
≥λ
∗ ∗ G (t, s) f s, x(s) – q (s) , y(s) – q (s) + p (s) ds
c
c
–c
G (t, s) Lγ cα– R ds
–c
t α– h (s) Lγ cα– R ds c –c h (s) ds ≥ R . ≥ λc(α–) Lγ R c ≥λ
Then Q (x, y) ≥ (x, y)Y and Q(x, y) ≥ (x, y) , Y Y
∀(x, y) ∈ P ∩ ∂ .
()
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 15 of 23
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
≥μ
∗ ∗ G (t, s) g s, x(s) – q (s) , y(s) – q (s) + p (s) ds
–c
∗ ∗ G (t, s) g s, x(s) – q (s) , y(s) – q (s) + p (s) ds
–c
∗ G (t, s)L x(s) – q (s) ds ≥ μ
c
–c
G (t, s) Lγ cα– R ds c c –c –c α– α– (α–) Lγ R t h (s) Lγ c R ds ≥ μc h (s) ds ≥μ c c
≥μ
∀t ∈ [c, – c].
≥ R ,
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 ¯ \ ), that is, R ≤ (x , y )Y ≤ R . Since (x , y )Y ≥ R , then x ≥ R (x , y ) ∈ P ∩ ( R or y ≥ . We suppose first that x ≥ R . Then we deduce
x (t) – q (t) = x (t) – λ
≥ x (t) – t
G (t, s)p (s) ds – μ
α–
G (t, s)p (s) ds
p (s) ds + δ p (s) ds δ
x (t) δ p (s) + δ p (s) ds γ x δ p (s) + δ p (s) ds x (t) ≥ – γ R δ p (s) + δ p (s) ds γ t α– x ≥ – γ R R ≥ δ p (s) + δ p (s) ds γ t α– – γ R ≥ x (t) –
= t α– ,
∀t ∈ [, ],
and so x (t) ≥ q (t) + t α– for all t ∈ [, ], where = Then y () = x (s) dK(s) ≥ sα– dK(s) > and
y ≥ y () = ≥
γ R
x (s) dK(s) ≥
sα– dK(s) > .
γ sα– x dK(s)
γ R
–
(δ p (s) + δ p (s)) ds > .
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 16 of 23
Therefore, we obtain
y (t) – q (t) = y (t) – μ
G (t, s)p (s) ds – λ
δ p (s) + δ p (s) ds
β–
≥ y (t) –
G (t, s)p (s) ds
≥ y (t) – t
y (t) γ y
δ p (s) + δ p (s) ds
– δ p (s) + δ p (s) ds sα– dK(s) γ γ R
– δ p (s) + δ p (s) ds sα– dK(s) ≥ γ t β– y – γ γ R γ γ R β– ≥ t sα– dK(s)
– δ p (s) + δ p (s) ds × – sα– dK(s) γ γ R ≥ y (t) –
= t β– ,
∀t ∈ [, ],
where = γ γ R sα– dK(s) – (δ p (s) + δ p (s)) ds > . Hence y (t) ≥ q (t) + t β– for all t ∈ [, ]. If y ≥ R , then by a similar approach, we deduce that y (t) ≥ q (t) + t β– and x (t) ≥ q (t) + t α– for all t ∈ [, ], where = γR – (δ p (s) + δ p (s)) ds > and = γ γ R sβ– dH(s) – (δ p (s) + δ p (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) ≥ t α– and v (t) ≥ t β– 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 –p (t) ≤ f (t, u, v) ≤ α (t)β (t, u, v), for all t ∈ [, ], u, v ∈ [, ∞), with
–p (t) ≤ g(t, u, v) ≤ α (t)β (t, u, v),
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
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 17 of 23
(H) there exists c ∈ (, /) such that i = lim inf min f (t, u, v) > L f∞ u+v→∞ t∈[c,–c] u,v≥
i or g∞ = lim inf min g(t, u, v) > L , u+v→∞ t∈[c,–c] u,v≥
where δ p (s) + δ p (s) ds, δ p (s) + δ p (s) ds, L = max γ γ
– δ p (s) + δ p (s) ds, sα– dK(s) γ γ
– δ p (s) + δ p (s) ds sβ– dH(s) γ γ – –c –c × min cα– h (s) ds, cα– h (s) ds , c
c
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
–
– M M . δ p (s) + δ p (s) ds , β– δ p (s) + δ p (s) ds λ∗ = max α– c c We assume now λ ≥ λ∗ . Let λ λ δ p (s) + δ p (s) ds, δ p (s) + δ p (s) ds, R = max γ γ
– λ α– δ p (s) + δ p (s) ds, s dK(s) γ γ
– λ β– δ p (s) + δ p (s) ds , s dH(s) γ γ and = {(x, y) ∈ P, (x, y)Y < R }. i We suppose first that f∞ > L , that is, f (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 /. We assume that x ≥ R /. Then for all t ∈ [, ] we deduce x(t) – q (t) ≥ γ t
α–
x – λt
α–
p (s) ds – λt
δ
α–
p (s) ds
δ
γ R δ p (s) + δ p (s) ds –λ δ p (s) + δ p (s) ds – λ δ p (s) + δ p (s) ds ≥ t α– λ ≥ t α–
Henderson and Luca Boundary Value Problems (2016) 2016:61
= t α– λ
Page 18 of 23
δ p (s) + δ p (s) ds
≥ t α– λ∗
M δ p (s) + δ p (s) ds ≥ α– t α– ≥ . c
Therefore, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we have
x(t) – q (t)
∗
∗ ∗ M + y(t) – q (t) ≥ x(t) – q (t) = x(t) – q (t) ≥ α– t α– ≥ M . c
()
Hence, for any (x, y) ∈ P ∩ ∂ and t ∈ [c, – c], we conclude
Q (x, y)(t) ≥ λ
∗ ∗ G (t, s) f s, x(s) – q (s) , y(s) – q (s) + p (s) ds
∗ ∗ h (s)f s, x(s) – q (s) , y(s) – q (s) ds
–c
≥ λ t α– c
–c
≥ λL t α–
–c
h (s) ds ≥ λL cα–
c
c
h (s) ds ≥ R = (x, y)Y .
Therefore we obtain Q (x, y) ≥ R for all (x, y) ∈ P ∩ ∂ , and so Q(x, y) ≥ R = (x, y) , Y Y
∀(x, y) ∈ P ∩ ∂ .
()
If y ≥ R /, then by a similar approach we deduce again relation (). i > L , that is, g(t, u, v) ≥ L for all t ∈ [c, – c] and u, v ≥ , We suppose now that g∞ u + v ≥ M . Let (x, y) ∈ P ∩ ∂ . Then (x, y)Y = R , so x ≥ R / or y ≥ R /. If x ≥ i R /, then we obtain in a similar manner as in the first case above (f∞ > L ) that x(t) – M α– q (t) ≥ cα– t ≥ 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
Q (x, y)(t) ≥ λ
∗ ∗ G (t, s) g s, x(s) – q (s) , y(s) – q (s) + p (s) ds
–c
≥ λ t α– c
≥ λL t α– c
∗ ∗ h (s)g s, x(s) – q (s) , y(s) – q (s) ds
–c
h (s) ds ≥ λL cα– c
–c
h (s) ds ≥ R = (x, y)Y .
Therefore we obtain Q (x, y) ≥ R , and so Q(x, y)Y ≥ R = (x, y)Y for all (x, y) ∈ 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 h (s)α (s) ds , h (s)α (s) ds , λσ λσ
–
– . h (s)α (s) ds , h (s)α (s) ds λσ λσ
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 19 of 23
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 h (s) α (s) + p (s) ds, R = max R , λσ max{M , }
h (s) α (s) + p (s) ds,
h (s) α (s) + p (s) ds,
λσ max{M , }
λσ max{M , }
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
∗ ∗ σ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
Q (x, y)(t) ≤ λ
∗ ∗ σ h (s) α (s)β s, x(s) – q (s) , y(s) – q (s) + p (s) ds
∗ ∗ + p (s) ds h (s) α (s) M + ε x(s) – q (s) + y(s) – q (s)
+λ
≤ λσ
+ λσ
∗ ∗ + p (s) ds h (s) α (s) M + ε x(s) – q (s) + y(s) – q (s)
≤ λσ max{M , }
h (s) α (s) + p (s) ds + λσ εR
h (s)α (s) ds
+ λσ max{M , }
h (s) α (s) + p (s) ds + λσ εR
≤
R R R R R (x, y)Y + + + = = ,
h (s)α (s) ds
∀t ∈ [, ],
Y and so Q (x, y) ≤ (x,y) for all (x, y) ∈ P ∩ ∂ . Y for all t ∈ [, ], and so Q (x, y) ≤ In a similar way we obtain Q (x, y)(t) ≤ (x,y) for all (x, y) ∈ P ∩ ∂ . Therefore, we deduce
Q(x, y) ≤ (x, y) , Y Y
∀(x, y) ∈ P ∩ ∂ .
(x,y)Y
()
¯\ By Theorem ., (), and (), we conclude that Q has a fixed point (x , y ) ∈ P ∩ ( ). Since (x , y ) ≥ R then x ≥ R / or y ≥ R /.
Henderson and Luca Boundary Value Problems (2016) 2016:61
Page 20 of 23
We suppose that x ≥ R /. Then x (t) – q (t) ≥ y () =
x (s) dK(s) ≥ γ x
sα– dK(s) ≥
M α– t cα–
γ R
for all t ∈ [, ]. Besides
sα– dK(s) > ,
and then y ≥ y () =
x (s) dK(s) ≥
γ R
sα– dK(s) > .
Therefore, we deduce that, for all t ∈ [, ], y (t) – q (t) ≥ y (t) – λδ
t
β–
δ p (s) + δ p (s) ds
γ γ R β– t sα– dK(s) – λt β– β– δ p (s) + δ p (s) ds ≥ λt ≥
t β– p (s) ds
≥ γ t β– y – λt β–
p (s) ds – λδ
δ p (s) + δ p (s) ds
≥ λ∗ t β–
M δ p (s) + δ p (s) ds ≥ β– t β– . c
α– If y ≥ R /, then by a similar approach we conclude again that x (t) – q (t) ≥ cM α– t M β– and y (t) – q (t) ≥ cβ– t for all t ∈ [, ]. t α– and Let u (t) = x (t) – q (t) and v (t) = y (t) – q (t) for all t ∈ [, ]. Then u (t) ≥ M M β– t for all t ∈ [, ], where = α– , = β– . Hence we deduce that (u , v ) is v (t) ≥ c c 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 i = lim inf min f (t, u, v) > L f∞ u+v→∞ t∈[c,–c] u,v≥
or
i g∞ = lim inf min g(t, u, v) > L , u+v→∞ t∈[c,–c] u,v≥
where L = max δ p (s) + δ p (s) ds, δ p (s) + δ p (s) ds, γ γ
– δ p (s) + δ p (s) ds, sα– dK(s) γ γ
– δ p (s) + δ p (s) ds sβ– dH(s) γ γ – –c –c β– β– × min c h (s) ds, c h (s) ds , c
c
Henderson and Luca Boundary Value Problems (2016) 2016:61
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then there exists λ∗ > such that for any λ ≥ λ∗ problem (S)-(BC) (with λ = μ) has at least one positive solution. Theorem . Assume that λ = μ, and (H), (H), and (H) hold. In addition if (H) there exists c ∈ (, /) such that fˆ∞ = lim
min f (t, u, v) = ∞
u+v→∞ t∈[c,–c] u,v≥
or gˆ∞ = lim
min g(t, u, v) = ∞,
u+v→∞ t∈[c,–c] 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) = s u(s) ds and v(s) dH(s) = sv(s) ds. We consider the system of fractional differential equations (S )
t ∈ (, ), t ∈ (, ),
D/ + u(t) + λf (t, u(t), v(t)) = , D/ + v(t) + μg(t, u(t), v(t)) = ,
with the boundary conditions (BC )
u() = u () = , v() = v () = ,
u() = sv(s) ds, v() = s u(s) ds.
> Then we obtain = – ( sα– dK(s))( sβ– dH(s)) = > , τ α– ( – τ ) dK(τ ) = β– , τ ( – τ ) dH(τ ) = > . The functions H and K are nondecreasing, and so assumption (H) is satisfied. Besides, we deduce
t / ( – s)/ – (t – s)/ , ≤ s ≤ t ≤ , ≤ t ≤ s ≤ , t / ( – s)/ , t / ( – s)/ – (t – s)/ , ≤ s ≤ t ≤ , g (t, s) = (/) t / ( – s)/ , ≤ t ≤ s ≤ , / / τ g (τ , s) ds, G (t, s) = t τ g (τ , s) dτ , G (t, s) = g (t, s) + t / / t τ g (τ , s)dτ , G (t, s) = t τ g (τ , s) dτ . G (t, s) = g (t, s) + g (t, s) = √ π
We also obtain h (s) = k (t) =
√ s( – s)/ , π
/ t , ( – t)t / ,
≤ t ≤ /, / ≤ t ≤ ,
In addition, we have σ = , δ = ,
δ = √
, (/)
( –) √
=
√ – √ ,
≈ ..
h (s) =
σ =
s( – s)/ , (/)
k (t) =
/ t , ( – t)t / ,
≤ t ≤ /, / ≤ t ≤ .
√ √ – – √ , = √ √ , , σ = , δ = , = (/) π √ √ – – , δ = √π , = √ , γ = √ ≈ .,
σ = γ =
Henderson and Luca Boundary Value Problems (2016) 2016:61
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Example We consider the functions (u + v) f (t, u, v) = √ + ln t, t( – t)
g(t, u, v) =
+ sin(u + v) + ln( – t), √ t( – t)
t ∈ (, ), u, v ≥ .
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 (δ p (s) + δ p (s)) ds ≈ ., (δ p (s) + δ p (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∗ ].
We have p (t) = – ln t, p (t) = – ln( – t), α (t) = α (t) =
√ t(–t)
Example We consider the functions f (t, u, v) = (u + v) + cos u,
t ∈ [, ], u, v ≥ .
g(t, u, v) = (u + v)/ + cos 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¯ () =
max
f (t, u, v) + p (t) ≈ .,
max
g(t, u, v) + p (t) ≈ ..
t∈[,],u,v∈[,]
t∈[,],u,v∈[,]
We also obtain c ≈ ., c ≈ ., c ≈ ., c ≈ ., and then λ = max{ c Rf¯ (R , R } ≈ . and μ = max{ cRg¯ (R ) , cRg¯ (R ) } ≈ ) c f¯ (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) where a ∈ (, ).
ln( + u + v) , g(t, u, v) = –√ –t t( – t)
t ∈ (, ), u, v ≥ ,
Henderson and Luca Boundary Value Problems (2016) 2016:61
Here we have p (t) =
√ , t
p (t) =
√ , –t
Page 23 of 23
α (t) = √
t (–t)
, α (t) = √
t(–t)
for all t ∈ (, ),
β (t, u, v) = (u + v) , β (t, u, v) = ln( + u + v) for all t ∈ [, ], u, v ≥ . For c ∈ (, /) fixed, 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. a
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. Author details 1 Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA. 2 Department of Mathematics, Gh. Asachi Technical University, Iasi, 700506, Romania. 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. Received: 8 October 2015 Accepted: 2 March 2016 References 1. Das, S: Functional Fractional Calculus for System Identification and Controls. Springer, New York (2008) 2. Graef, JR, Kong, L, Kong, Q, Wang, M: Uniqueness of positive solutions of fractional boundary value problems with non-homogeneous integral boundary conditions. Fract. Calc. Appl. Anal. 15(3), 509-528 (2012) 3. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies, vol. 204. Elsevier, Amsterdam (2006) 4. Podlubny, I: Fractional Differential Equations. Academic Press, San Diego (1999) 5. Sabatier, J, Agrawal, OP, Machado, JAT (eds.): Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Dordrecht (2007) 6. Samko, SG, Kilbas, AA, Marichev, OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Yverdon (1993) 7. Henderson, J, Luca, R: Positive solutions for a system of fractional differential equations with coupled integral boundary conditions. Appl. Math. Comput. 249, 182-197 (2014) 8. Henderson, J, Luca, R, Tudorache, A: On a system of fractional differential equations with coupled integral boundary conditions. Fract. Calc. Appl. Anal. 18(2), 361-386 (2015) 9. Henderson, J, Luca, R: Positive solutions for a system of nonlocal fractional boundary value problems. Fract. Calc. Appl. Anal. 16(4), 985-1008 (2013) 10. Henderson, J, Luca, R: Existence and multiplicity of positive solutions for a system of fractional boundary value problems. Bound. Value Probl. 2014, 60 (2014) 11. Luca, R, Tudorache, A: Positive solutions to a system of semipositone fractional boundary value problems. Adv. Differ. Equ. 2014, 179 (2014) 12. Yuan, C, Jiang, D, O’Regan, D, Agarwal, RP: Multiple positive solutions to systems of nonlinear semipositone fractional differential equations with coupled boundary conditions. Electron. J. Qual. Theory Differ. Equ. 2012, 13 (2012) 13. Yuan, C: Multiple positive solutions for (n – 1, 1)-type semipositone conjugate boundary value problems of nonlinear fractional differential equations. Electron. J. Qual. Theory Differ. Equ. 2010, 36 (2010) 14. Agarwal, RP, Meehan, M, O’Regan, D: Fixed Point Theory and Applications. Cambridge University Press, Cambridge (2001) 15. Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones. Academic Press, New York (1988)
