#### Nonexistence of positive solutions for a system of coupled fractional boundary value problems

Henderson and Luca Boundary Value Problems
Nonexistence of positive solutions for a system of coupled fractional boundary value problems
Johnny Henderson
Rodica Luca 0
0 Department of Mathematics, Gh. Asachi Technical University , Iasi, 700506 , Romania
We investigate the nonexistence of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations with coupled integral boundary conditions. Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology (such as blood flow phenomena), economics, control theory, signal and image processing, aerodynamics, viscoelasticity, electromagnetics, and so on (see [-]). For some recent developments on the topic, see [-] and the references therein. Coupled boundary conditions appear in the study of reaction-diffusion equations and Sturm-Liouville problems, and they have applications in many fields of sciences and engineering such as thermal conduction and mathematical biology (see for example [-]). In this paper, we consider the system of nonlinear fractional differential equations
Riemann-Liouville fractional differential equations; integral boundary conditions; positive solutions; nonexistence
1 Introduction
Dα+u(t) + λf (t, u(t), v(t)) = , t ∈ (, ),
β
D+v(t) + μg(t, u(t), v(t)) = , t ∈ (, ),
with the coupled integral boundary conditions
u() = u () = · · · = u(n–)() = ,
v() = v () = · · · = v(m–)() = ,
where n – < α ≤ n, m – < β ≤ m, n, m ∈ N, n, m ≥ , Dα+, and Dβ+ denote the
RiemannLiouville derivatives of orders α and β, respectively, and the integrals from (BC) are
Riemann-Stieltjes integrals.
We shall give sufficient conditions on λ, μ, f , and g such that (S)-(BC) has no positive
solutions. By a positive solution of problem (S)-(BC) we mean a pair of functions (u, v) ∈
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C([, ]) × C([, ]) satisfying (S) and (BC) with u(t) ≥ , v(t) ≥ for all t ∈ [, ] and
(u, v) = (, ). The existence of positive solutions for (S)-(BC) has been studied in [] by
using the Guo-Krasnosel’skii fixed point theorem. The multiplicity of positive solutions of
the system (S) with λ = μ = , f (t, u, v) = f˜(t, v) and g(t, u, v) = g˜(t, u) (denoted by (S)), with
the boundary conditions (BC) was investigated in [], where the nonlinearities f and g
are nonsingular or singular functions. In [], the authors used some theorems from the
fixed point index theory and the Guo-Krasnosel’skii fixed point theorem. We also mention
[], where we studied the existence of positive solutions for (S)-(BC) (u(t) ≥ , v(t) ≥
for all t ∈ [, ], and u(t) > , v(t) > for all t ∈ (, )), where f and g are sign-changing
functions. The systems (S) and (S) with uncoupled boundary conditions
u() = u () = · · · = u(n–)() = ,
v() = v () = · · · = v(m–)() = ,
were investigated in [–].
In Section , we present the necessary definitions and properties from the fractional
calculus theory and some auxiliary results from [], which investigates a nonlocal boundary
value problem for fractional differential equations. In Section , we prove some
nonexistence results for the positive solutions with respect to a cone for our problem (S)-(BC).
Finally, two examples are given to illustrate our main results.
2 Auxiliary results
We present here the definitions, some lemmas from the theory of fractional calculus, and
some auxiliary results from [] that will be used to prove our main theorems.
Definition . The (left-sided) fractional integral of order α > of a function f : (, ∞) →
R is given by
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
with the coupled integral boundary conditions (BC), where n – < α ≤ n, m – < β ≤ m,
n, m ∈ N, n, m ≥ , and H, K : [, ] → R from (BC) 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 solution of
problem ()-(BC) is 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,
for all t, s ∈ [, ] 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 (τ ))( g(τ , s) dH(τ )),
Lemma . ([]) The functions g and 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) ≤ g(θ(s), s), g(t, s) ≤ g(θ(s), s), for all (t, s) ∈ [, ] × [, ];
(c) for any c ∈ (, /), we have
t∈m[c,in–c] g(t, s) ≥ γg θ(s), s ,
t∈m[c,in–c] g(t, s) ≥ γg θ(s), s ,
for all s ∈ [, ], where γ = cα–, γ = cβ–,
⎧
θ(s) = ⎨
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 unique solution (u, v) of problem ()-(BC) (given by ())
satisfies u(t) ≥ , v(t) ≥ for all t ∈ [, ].
Lemma . ([]) Assume that H, K : [, ] → R are nondecreasing functions and
Then the functions Gi, i = , . . . , , satisfy the inequalities
(a) G(t, s) ≤ J(s), ∀(t, s) ∈ [, ] × [, ], where
(a) for every c ∈ (, /), we have
∀t , s ∈ [, ];
(b) G(t, s) ≤ J(s), ∀(t, s) ∈ [, ] × [, ], where J(s) =
g(τ , s) dH(τ );
(b) for every c ∈ (, /), we have
∀t , s ∈ [, ];
(c) G(t, s) ≤ J(s), ∀(t, s) ∈ [, ] × [, ], where
(c) for every c ∈ (, /), we have
∀t , s ∈ [, ];
(d) G(t, s) ≤ J(s), ∀(t, s) ∈ [, ] × [, ], where J(s) = g(τ , s) dK (τ );
(d) for every c ∈ (, /), we have
t∈m[c,in–c] G(t, s) ≥ γJ(s) ≥ γG t , s ,
∀t , s ∈ [, ].
Lemma . ([]) Assume that H, K : [, ] → R are nondecreasing functions, > , c ∈
(, /), and x, y ∈ C(, ) ∩ L(, ), x(t) ≥ , y(t) ≥ for all t ∈ (, ). Then the solution
(u(t), v(t)), t ∈ [, ] of problem ()-(BC) satisfies the inequalities
3 Main results
We present in this section intervals for λ and μ for which there exists no positive solution
of problem (S)-(BC).
We present the assumptions that we shall use in the sequel.
(H) H, K : [, ] → R are nondecreasing functions and
= – ( τ α– dK (τ ))( τ β– dH(τ )) > .
(H) The functions f , g : [, ] × [, ∞) × [, ∞) → [, ∞) are continuous.
For c ∈ (, /), we introduce the following extreme limits:
fs = lim sup max f (t, u, v) ,
u+v→+ t∈[,] u + v
fi = uli+mv→in+f t∈m[c,in–c] f (ut,+u,vv) ,
f ∞s = lim sup max f (t, u, v) ,
u+v→∞ t∈[,] u + v
f ∞i = uli+mv→in∞f t∈m[c,in–c] f (ut,+u,vv) ,
gi = uli+mv→in+f t∈m[c,in–c] g(ut,+u,vv) ,
g∞s = lim sup max g(t, u, v) ,
u+v→∞ t∈[,] u + v
g∞i = uli+mv→in∞f t∈m[c,in–c] g(ut,+u,vv) .
In the definitions of the extreme limits above, the variables u and v are nonnegative.
By using the functions Gi, i = , . . . , from Section (Lemma .), our problem (S)-(BC)
can be written equivalently as the following nonlinear system of integral equations:
u(t) = λ G(t, s)f (s, u(s), v(s)) ds + μ G(t, s)g(s, u(s), v(s)) ds, t ∈ [, ],
v(t) = μ G(t, s)g(s, u(s), v(s)) ds + λ G(t, s)f (s, u(s), v(s)) ds, t ∈ [, ].
We consider the Banach space X = C([, ]) with supremum norm · , and the Banach
space Y = X × X with the norm (u, v) Y = u + v . We define the cone P ⊂ Y by
P = (u, v) ∈ Y ; u(t) ≥ , v(t) ≥ , ∀t ∈ [, ] and
t∈[icn,f–c] u(t) + v(t) ≥ γ (u, v) Y ,
where γ = min{γ, γ} and γ, γ are defined in Section (Lemma .).
For λ, μ > , we introduce the operators T, T : Y → X, and T : Y → Y defined by
G(t, s)g s, u(s), v(s) ds,
≤ t ≤ ,
G(t, s)f s, u(s), v(s) ds,
≤ t ≤ ,
and T (u, v) = (T(u, v), T(u, v)), (u, v) ∈ Y .
Lemma . ([]) If (H) and (H) hold, and c ∈ (, /), then T : P → P is a completely
continuous operator.
The positive solutions of our problem (S)-(BC) coincide with the fixed points of the
operator T .
Theorem . Assume that (H) and (H) hold, and c ∈ (, /). If fs, f ∞s, gs, g∞s < ∞, then
there exist positive constants λ, μ such that, for every λ ∈ (, λ) and μ ∈ (, μ), the
boundary value problem (S)-(BC) has no positive solution.
Proof From the definitions of fs, f ∞s, gs, g∞s, which are finite, we deduce that there exist
M, M > such that
f (t, u, v) ≤ M(u + v),
g(t, u, v) ≤ M(u + v),
∀t ∈ [, ], u, v ≥ .
J(s)g s, u(s), v(s) ds
J(s) u(s) + v(s) ds
∀t ∈ [, ].
Therefore, we conclude
In a similar manner, we obtain
u ≤ (λMA + μMB) (u, v) Y < (λMA + μMB) (u, v) Y ≤ (u, v) Y .
G(t, s)f s, u(s), v(s) ds
G(t, s)g s, u(s), v(s) ds
J(s)f s, u(s), v(s) ds
J(s) u(s) + v(s) ds
∀t ∈ [, ].
Therefore, we deduce
v ≤ (μMC + λMD) (u, v) Y < (μMC + λMD) (u, v) Y ≤ (u, v) Y .
Hence, by () and (), we conclude
which is a contradiction. So the boundary value problem (S)-(BC) has no positive
solution.
Theorem . Assume that (H) and (H) hold, and c ∈ (, /). If fi, f ∞i > and f (t, u, v) >
for all t ∈ [c, – c], u ≥ , v ≥ , u + v > , then there exists a positive constant λ˜ such that,
for every λ > λ˜ and μ > , the boundary value problem (S)-(BC) has no positive solution.
G(c, s)g s, u(s), v(s) ds
G(c, s)f s, u(s), v(s) ds
Then we conclude
u ≥ u(c) ≥ λmγ γA (u, v) Y > λ˜mγ γA (u, v) Y = (u, v) Y ,
and so (u, v) Y = u + v ≥ u > (u, v) Y , which is a contradiction.
If A < D, then λ˜ = γ γmD , and therefore, we deduce
G(c, s)f s, u(s), v(s) ds
G(c, s)f s, u(s), v(s) ds
Then we conclude
v ≥ v(c) ≥ λmγ γD (u, v) Y > λ˜mγ γD (u, v) Y = (u, v) Y ,
and so (u, v) Y = u + v ≥ v > (u, v) Y , which is a contradiction.
Therefore, the boundary value problem (S)-(BC) has no positive solution.
Theorem . Assume that (H) and (H) hold, and c ∈ (, /). If gi, g∞i > and g(t, u, v) >
for all t ∈ [c, – c], u ≥ , v ≥ , u + v > , then there exists a positive constant μ˜ such that,
for every μ > μ˜ and λ > , the boundary value problem (S)-(BC) has no positive solution.
Proof From the assumptions of the theorem, we deduce that there exists m > such that
g(t, u, v) ≥ m(u + v) for all t ∈ [c, – c] and u, v ≥ . We define μ˜ = min{ γ γmB , γ γmC },
where B = c–c J(s) ds and C = c–c J(s) ds. We shall show that, for every μ > μ˜ and λ > ,
problem (S)-(BC) has no positive solution.
Let μ > μ˜ and λ > . We suppose that (S)-(BC) has a positive solution (u(t), v(t)), t ∈
[, ].
If B ≥ C, then μ˜ = γ γmB , and therefore we obtain
G(c, s)g s, u(s), v(s) ds
G(c, s)g s, u(s), v(s) ds
G(c, s)g s, u(s), v(s) ds
Then we conclude
u ≥ u(c) ≥ μmγ γB (u, v) Y > μ˜ mγ γB (u, v) Y = (u, v) Y ,
and so (u, v) Y = u + v ≥ u > (u, v) Y , which is a contradiction.
If B < C, then μ˜ = γ γmC , and therefore, we deduce
G(c, s)f s, u(s), v(s) ds
Then we conclude
v ≥ v(c) ≥ μmγ γC (u, v) Y > μ˜ mγ γC (u, v) Y = (u, v) Y ,
and so (u, v) Y = u + v ≥ v > (u, v) Y , which is a contradiction.
Therefore, the boundary value problem (S)-(BC) has no positive solution.
Theorem . Assume that (H) and (H) hold, and c ∈ (, /). If fi, f ∞i, gi, g∞i > , and
f (t, u, v) > , g(t, u, v) > for all t ∈ [c, – c], u ≥ , v ≥ , u + v > , then there exist positive
constants λˆ and μˆ such that, for every λ > λˆ and μ > μˆ , the boundary value problem
(S)-(BC) has no positive solution.
Proof From the assumptions of the theorem, we deduce that there exist m, m > such
that f (t, u, v) ≥ m(u + v), g(t, u, v) ≥ m(u + v), for all t ∈ [c, – c] and u, v ≥ .
We define λˆ = γ γmA and μˆ = γ γmC , where A = c–c J(s) ds and C = c–c J(s) ds.
Then, for every λ > λˆ and μ > μˆ , problem (S)-(BC) has no positive solution. Indeed, let
λ > λˆ and μ > μˆ . We suppose that (S)-(BC) has a positive solution (u(t), v(t)), t ∈ [, ].
In a similar manner to that used in the proofs of Theorems . and ., we obtain
u ≥ u(c) ≥ λmγ γA (u, v) Y ,
v ≥ v(c) ≥ μmγ γC (u, v) Y ,
(u, v) Y = u + v ≥ λmγ γA (u, v) Y + μmγ γC (u, v) Y
which is a contradiction. Therefore, the boundary value problem (S)-(BC) has no positive
solution.
We can also define λˆ = γ γmD and μˆ = γ γmB , where B = c–c J(s) ds and D =
c–c J(s) ds. Then, for every λ > λˆ and μ > μˆ , problem (S)-(BC) has no positive solution.
Indeed, let λ > λˆ and μ > μˆ . We suppose that (S)-(BC) has a positive solution (u(t), v(t)),
t ∈ [, ]. In a similar manner to that used in the proofs of Theorems . and ., we obtain
v ≥ v(c) ≥ λmγ γD (u, v) Y ,
u ≥ u(c) ≥ μmγ γB (u, v) Y ,
(u, v) Y = u + v ≥ μmγ γB (u, v) Y + λmγ γD (u, v) Y
which is a contradiction. Therefore, the boundary value problem (S)-(BC) has no positive
solution.
Remark . Under the assumptions of Theorem ., we have the following observations.
(a) In the case A ≥ D and B ≤ C, Theorem . gives some supplementary information
for the domain of λ and μ for which there is no positive solution of (S)-(BC), in
comparison to Theorems . and ., because λˆ = ˜ and μˆ = μ˜ .
λ
(b) In the case A ≤ D and B ≥ C, Theorem . gives some supplementary information
for the domain of λ and μ for which there is no positive solution of (S)-(BC), in
comparison to Theorems . and ., because λˆ = ˜ and μˆ = μ˜ .
λ
4 Examples
We consider the system of fractional differential equations
with the boundary conditions
D/+u(t) + λf (t, u(t), v(t)) = ,
D/+v(t) + μg(t, u(t), v(t)) = ,
u() = u () = ,
v() = v () = ,
Then we deduce ≈ . > , θ(s) = –s+s–s , θ(s) = –s+s for all s ∈ [, ]
(see also []). For the functions Ji, i = , . . . , , we obtain
(/) { (–ss(+–ss)–/s)/ + √ [( – s)/ – ( – s)/
+ ( – s)/ – ( – s)/]},
(/) { (–ss(+–ss)–/s)/ + √ [( – s)/ + ( – s)/
– ( – s)/]},
(/) { (–ss(+–ss)–/s)/ + √ [( – s)/ + ( – s)/]},
≤ s < /,
( – s)/ – ( – s)/ – s( – s)/ ,
s ∈ [, ],
s ∈ [, ],
J(s) = ⎨⎪⎪⎧ √√ ((//)) [[(( –– ss))// +– ((––s)s)//–+((––s)s)//],– ( – s)/],
⎪⎪⎩ √ (/) [( – s)/ + ( – s)/],
For c = /, we deduce γ = –/ ≈ ., γ = , γ = γ. After some
computatiJo(nss),dwse≈co.ncludeA=,B =J(s/)/dJs(≈s) ds.≈., A=, C//=J(s)Jds(s≈)ds.≈., B=,
C = // J(s) ds ≈ ., D = J(s) ds ≈ ., D = // J(s) ds ≈
..
Example We consider the functions
for t ∈ [, ], u, v ≥ , where p, p > and q, q > .
We obtain fs = q, gs = q + , f ∞s = p(q + ), g∞s = p(q + ), and then we can apply
Theorem .. So we conclude that there exist λ, μ > such that, for every λ ∈ (, λ)
and μ ∈ (, μ), the boundary value problem (S)-(BC) has no positive solution. By
Theorem ., the positive constants λ and μ are given by λ = min{ MA , MD } = MA and
tive solution. From the proof of Theorem ., the positive constant λ˜ is given by λ˜ =
min{ γ γmA , γ γmD }. For example, if p = and q = , then we deduce m = √ and
Example We consider the functions
f (t, u, v) = pta˜ u + v ,
g(t, u, v) = p( – t)b˜ eu+v – , t ∈ [, ], u, v ≥ ,
where a˜ , b˜, p, p > .
if p = b˜ = , then we deduce m = and μ˜ ≈ ,..
5 Conclusions
linear Riemann-Liouville fractional differential equations (S) with the coupled integral
boundary conditions (BC) has no positive solutions. Some examples which illustrate the
obtained results are also presented.
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 a
grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, project number
PN-II-ID-PCE-2011-3-0557.
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