R E S E A R C H
Open Access
Existence of solutions to fractional
boundary value problems at resonance in
Hilbert spaces
Phan Dinh Phung
1*and Ha Binh Minh
2*Correspondence: [email protected] 1Institute for Research and Development, Duy Tan University, 03 Quang Trung, Da Nang, Vietnam Full list of author information is available at the end of the article
Abstract
We study the existence of solutions to a nonlinear fractional differential equation in Hilbert spaces associated with three-point boundary conditions at resonance
x(0) =
θ
, Dα–1x(1) =ADα–1x(η)by using Mawhin’s continuation theorem. We propose a new technique to improve the conditions onAwhich have been used previously. In addition, a necessary and sufficient condition for that the fractional differential operator is Fredholm with zero-index is established, especially for the first time when the fractional differential operator takes values in an arbitrary Hilbert space.
MSC: 34A08; 34B10; 34B15
Keywords: coincidence degree; three-point boundary value problem; fractional differential equation; resonance
1 Introduction
Fractional differential equations arise in various areas of physics and applied mathematics and have recently become a powerful tool in modeling of many physical phenomena (for instance, see [–] and the references therein). In this paper, we are concerned with the existence of solutions to the following fractional three-point boundary value problem at resonance:
⎧ ⎨ ⎩
Dαx(t) =f(t,x(t),Dα–x(t)), a.e.t∈(, ), <α≤,
x() =θ, Dα–x() =ADα–x(η), (.)
whereDαis the Riemann-Liouville differential operator of orderα;θis zero element in a
Hilbert spaceH;η∈(, ) is a given constant;A∈L(H) is a bounded linear operator on
Hsuch that
(A) I–Ais a Fredholm operator with zero-index;
andf : [, ]×H×H→His a function satisfying the following assumptions:
(A) f(·,x,y)is Lebesgue measurable on[, ]for every(x,y)∈H×H; (A) f(t,·,·)is continuous onH×Hfor almost everyt∈[, ];
(A) for each bounded set⊂H×H, the functiont→ϕ(t) =sup{f(t,x,y)H: (x,y)∈
}is Lebesgue integrable on[, ], and the set{f(t,x,y) : (x,y)∈}is relatively com-pact inH, here · Hstands for the norm induced by the scalar product·,· inH.
Problem (.) is usually written in the following form:
Lx=Nx, (.)
whereLandNare operators from a Banach spaceXto another Banach spaceY(hereLis linear). IfLis invertible, orkerL={}, (.) is callednon-resonant problem. Otherwise, if kerLis not a trivial space, then it is calledresonant problem. For this problem, we focus on
an important class of resonant problems whenLis a Fredholm operator with zero-index, as a prerequisite for applying coincidence degree theory [].
Different from non-resonant problems being studied for a long time, the solvability of resonant problems has been extensively studied for the last decade. The reason is that res-onant problems are rather complicated due to the non-invertibility ofL. Non-invertibility leads to the difficulty of constructing a suitable continuous projection on a complement ofImL. Due to the fact that constructing that projection becomes more difficult when the dimension ofkerLis large, most of works are investigated mainly fordim kerL= . We refer the reader to [–] and to the references therein.
For the case thatdim kerLcan be arbitrary, recently in [], we use the Mawhin con-tinuation theorem to investigate the existence of solutions for boundary value problems (briefly, BVPs) at resonance. We consider the following three-point BVP:
⎧ ⎨ ⎩
x(t) =f(t,x(t),x(t)) a.e.t∈(, )
x() = , x() =Ax(η), (.)
whereA:Rn→Rnis a continuous linear operator satisfying ⎧
⎨ ⎩
A=A, or
A=I (hereIstands for the identity operator). (.)
It seems that [] is one of premier papers on the case thatdim kerLis arbitrary. Later, problem (.) in [] was extended to fractional BVP (.) in []. Most recently, the result in [] was generalized to an infinite-dimensional space l in []. We emphasize that condition (.) is an essential condition for the techniques in [–], with the note that in [] the condition is slightly different in order to adapt the boundary conditions.
In our new research [] condition (.) can be omitted completely thanks to the fruit-ful features of a continuous linear operator onRnwhich can be regarded as matrices. The
breakthrough results in [] allow us to ask the first question:For extending problem(.)
Our next question is:What is the condition for a Hilbert space? In order to find the suitable condition for a Hilbert space, we observe that there is no change in the proofs in [–] if we replace (.) by the following more generalized assumption:
κ(I–A) is idempotent forκ∈R\ {},
or equivalently,
(I–A)=κ–(I–A) forκ∈R\ {}. (.)
Clearly, condition (.) can be derived from condition (.) by takingκ= orκ=. We remark that condition (.) leads to some useful qualitative properties:Im(I–A) is closed andker(I–A) is isomorphic to the complement ofIm(I–A). It suggests us to generalize (.) to the condition for a Hilbert space as follows.
I–A is a Fredholm operator with zero-index onH, (.)
which is actually assumption (A) for operatorA. We discover that in this paper condition (.) is the best generalization on a Hilbert space in the following senses:
() Condition (.) is not only a sufficient condition but also a necessary condition for thatL=Dαis the Fredholm operator with zero-index.
() WhenHis finite-dimensional, (.) automatically holds, therefore the result in [] is a special case of our result in this paper.
() WhenHis infinite-dimensional, it is well known that the rank-nullity theorem no longer holds for an operatorA∈L(H). Condition (.) helps us to overcome this difficulty by providing another characterization of dimension in a Hilbert space, as to be seen in (.).
() Condition (.) gives a unified approach and viewpoint for solving this kind of BVPs, fordimH≤ ∞andL=Dαwith <α≤.
The structure of the paper is organized as follows. In Section , we introduce some nec-essary background of fractional calculus as well as coincidence degree theory. In addition, we establish some essential lemmas needed for our main result later. Section is devoted to presenting the statement and the proof of the main result. Finally, we give a specific example to illustrate.
2 Preliminaries 2.1 Fractional calculus
In this subsection, we recall some definitions and results of the fractional calculus. See [, , ] for more details.
Definition . Givenf: [, ]→Handα> . Then
The fractional integralof orderαof the functionf is given by
Iαf(t) :=
(α) t
The Riemann-Liouville fractional derivativeof orderαoff is given by
Dαf(t) := d
m
dtn
Im–αf(t) =
(m–α)
dm dtm
t
(t–s)m–α–f(s)ds fort> ,
wherem= [α] + ,
provided that the terms on the right-hand side of the above equalities are pointwise de-fined on [, ]. In the above expressions, the sign ‘’ denotes the Bochner integral.
Fork∈N, we denote byACk([, ];H) the space of all functionsf : [, ]→Hwhich have
absolutely continuous derivative up to orderk– . Set
AC[, ];H:=AC[, ];H.
The following lemma concerning basic properties of fractional calculus is needed after-ward. The proof can be found in [], Lemmas ., ., or [], Theorem ..
Lemma . Letϕ∈L([, ];H)andα> ,m= [α] + .
(i) The equality(DαIαϕ)(t) =ϕ(t)holds for almost everyt∈[, ].
(ii) IfDα–mϕ∈ACm([, ];H),then
IαDαϕ(t) =ϕ(t) –
m
j=
Dα–jϕ() (α–j+ )t
α–j
for almost everyt∈[, ].
2.2 Mawhin’s continuation theorem
LetXandYbe two Banach spaces.
Definition . A linear operatorL:dom(L)⊂X→Y is calledFredholm operator with zero-indexifImLis closed inYand
codim ImL=dim kerL<∞.
It is known that whenLis a Fredholm operator with zero-index, there exist a projection
PonXand a projectionQonYsuch that
kerL=ImP, ImL=kerQ.
Moreover, the operatorLPdefined asLP:=L|dom(L)∩kerPis invertible. LetKP:=L–P and set
KP,Q:=KP(I–Q),
called the generalized inverse ofL. On the other hand, for isomorphismJ:ImQ→kerL, the operatorKP,Q+JQ:Y→dom(L) is isomorphic. Moreover,
(KP,Q+JQ)–x=
L+J–Px
Definition . Suppose thatLis a Fredholm operator with zero-index andis a bounded subset ofXsuch thatdom(L)∩=∅. An operatorN:X→Yis calledL-compactonif the following two conditions hold:
(i) QN:→Yis continuous andQN()is bounded inY; (ii) KP,QN:→Xis completely continuous.
LetLbe a Fredholm operator with zero-index andNbeL-compact on. It follows from Mawhin’s equivalent theorem [] that the equation
Lx=Nx forx∈
is equivalently converted into the fixed point equation
x= x forx∈,
where =P+ (JQ+KP,QN) is a completely continuous operator. This can be solvable
thanks to the following theorem, called Mawhin’s continuation theorem.
Theorem . Assume thatis a bounded and open subset in X,and L is a Fredholm operator with zero-index,and N is an L-compact operator on.Additionally, suppose that the following three assumptions hold:
(i) Lx=λNxforx∈∂∩(dom(L)\kerL)and forλ∈(, ); (ii) QNx= forx∈∂∩kerL;
(iii) for any isomorphismJfromImQtokerL,
degB(JQN|kerL,∩kerL, )= ,
whereQ:Y→Yis a projection given as above.
Then the equation Lx=Nx has a solution in∩dom(L).
For a comprehensive treatment on the subject of the coincidence degree theory and Mawhin’s continuation theorem as well, we refer the reader to [, , ].
Finally, we restate the following compactness criterion on the space of continuous func-tions C(X,Y) which is regarded as a generalization of the Arzela-Ascoli theorem [], Section ...
Lemma . Let X be a compact topological space and Y be a metric space.A subset of C(X,Y)is relatively compact if and only if it is pointwise relatively compact and equicon-tinuous.
2.3 Key lemmas
Let
X=x∈C[, ];H:x(t) =Iα+–ϕ(t),ϕ∈C
[, ];H
be the Banach space with the norm
where · ∞stands for the usual sup-norm onC([, ];H). Also, letY =L([, ];H) be
the Banach space with the Lebesgue norm
y=
y(t)Hdt.
Next we define a linear operatorL:X→Yby
Lx=Dαx, (.)
in which
x∈dom(L) =x=Iα–
+ ϕ∈X:ϕ∈AC
[, ];H,x() =θ,Dα–x() =ADα–x(η). Moreover, we define a (nonlinear) operatorN:X→Yby
Nx(t) =ft,x(t),Dα–x(t) for almost everyt∈[, ]. (.) Clearly, (.) is equivalent to
Lx=Nx,
in which the operatorsLandNare defined by (.) and (.), respectively.
.. Fredholm property of the fractional differential operator L
We first note that ifx∈dom(L), thenDαx∈YandI–αx∈AC([, ];H). From Lemma . andx() =θ, it follows that
x(t) = ctα–+IαDαx(t)
for everyt∈[, ] and some constant element c∈H. Moreover, in view of
Dα–x() =ADα–x(η), we deduce that
T(c) =φDαx, where
• T=I–A,
• φ:Y→His the following bounded linear operator:
φ(y) =
(α)
A
η
y(t)dt–
y(t)dt
fory∈Y. (.)
Thus, for convenience, a functionx∈dom(L) will be represented equivalently as
x(t) = ctα–+Iαy(t) withT(c) =φ(y). (.)
Remark . Note thatφis a surjective map. Indeed, for each c∈H, setting
y(t) =(α)(η– t)
–η c fort∈[, ],
we haveφ(y) = c, by a straightforward computation.
Due to the definition ofLand (.), by some simple calculations, we can indicate the kernel and the image ofLas follows:
kerL=x∈X:x(t) = ctα–for c∈kerT∼=kerT, (.)
and
ImL=y∈Y:φ(y)∈ImT=φ–(ImT). (.)
Now let us recall the definition of theMoore-Penrose inverseof linear operators. Suppose thatT is a linear operator on a Hilbert space. Then we call some linear operatorT†the Moore-Penrose inverse ofTif
(i) T†TT†=T†;
(ii) TT†T=T;
(iii) TT†andT†Tare self-adjoint operators.
It is well known that ifTis bounded and has closed range, thenT†uniquely exists and it is continuous. Moreover,TT†(resp.T†T) is an orthogonal projection onImT(resp.ImT†). For more details, one can see [].
Lemma . Assume that T is a continuous operator with closed range.Then the following two assertions are true.
(i) There exists a continuous projectionQ:Y→Ysuch that
kerQ=ImL and Y=ImL⊕ImQ.
(ii) Lis a Fredholm operator with zero-index if and only if so isT.
Proof (i) Define a mapQ:Y→Yby setting, fory∈Y, that
Qy(t) =qI–TT†φ(y)tα– fort∈[, ], (.)
whereq=η(αα–+)is a real constant. It is clear thatQis a continuous linear operator. More-over,Qis also a projection. Indeed, we observe that
I–TT†A=I–TT†
since
Then, for everyr∈R, we have
I–TT†(rA–I) =rI–TT†A–I–TT†
=rI–TT†–I–TT†
= (r– )I–TT†. (.)
From (.) and (.)-(.), it follows that
Q(Qy)(t) =qI–TT†φ(Qy)tα–
=qI–TT†qI–TT†(η
αA–I)
α(α) φ(y)
tα–
=qI–TT†φ(y)tα–
=Qy(t)
for everyt∈[, ]. HenceQis a projection as asserted. On the other hand, we have
y∈kerQ ⇔ φ(y)∈kerI–TT†
⇔ φ(y)∈ImTT† ⇔ φ(y)∈ImT
⇔ y∈ImL,
due to the fact thatTT†and (I–TT†) are projections andIm(TT†) =ImT. This means that
kerQ=ImL,
and consequently,Y=ImL⊕ImQ.
(ii) To prove this part, we start with the observation that
ImQ=y∈Y:y(t) = ctα–,t∈[, ] for c∈kerT†. (.)
Indeed, sinceTT†is a projection and
kerTT†=kerT†,
the ‘⊆’ inclusion of (.) can be obtained easily. For the converse inclusion, lety(t) = ctα– for c∈kerT†. Then, by the surjectivity ofφ, there exists z∈Y such thatqφ(z) = c. It follows that
Qz(t) =qI–TT†φ(z)tα–=I–TT†ctα–= ctα–=y(t)
Now, by (.), we can deduce thatImQ, a complement ofImLinY, is isomorphic to kerT†. Besides, sinceImTis closed, so isImL=φ–(ImT), due to the continuity ofφ. We are to prove the main claim of this part. Using the fact that
H=ImTT†⊕kerTT†=ImT⊕kerT†,
we can see thatT is a Fredholm operator with zero-index if and only if
dim kerT†=dim kerT<∞. (.)
This is clear to be a necessary and sufficient condition for whichLis a Fredholm operator with zero-index. Then the proof of lemma is complete.
Remark . If (.) holds, we can give another expression ofQquite simpler than (.). Specifically, we set
Qy(t) =q(I–κT)φ(y)tα–.
This also claims that the projectionQas such is in general not unique.
Remark . Lemma . will be no longer true in the general case thatHis a Banach space. IfHis a Banach space rather than a Hilbert space, for this lemma to remain true, we need to add one more assumption to this lemma. Additional assumption is thatThas the Moore-Penrose inverseT†∈L(H).
Now, we define a linear operatorP:X→Xby
Px(t) =
(α)
I–T†TDα–x()tα–, t∈[, ],x∈X. (.)
Note that Pis the continuous operator on X. Moreover, the following lemma provides some properties ofPas well asKP.
Lemma . The following statements are true.
(i) The operatorPin(.)is projection and satisfies that
kerL=ImP, X=kerP⊕kerL.
(ii) The mapKP:ImL→dom(L)∩kerPis defined by
KPy(t) =T†φ(y)tα–+Iαy(t) fort∈[, ], (.)
satisfying
KP=L–P and KPy ≤Cy,
whereC=(α)( +T†L
Proof (i) We first notice that ⎧
⎨ ⎩
Dα–(tα–)(t) =(α),
Dα(tα–)(t) =
for everyt∈[, ]. Since (I–T†T) is a projection, then so isP, by a straightforward com-putation. Furthermore,Pis ontokerL, that is,ImP=kerL. Indeed, because the inclusion ImP⊂kerLcan be proved simply, we need only prove the converse one:kerL⊂ImP. Suppose thatx(t) = ctα–∈kerL, where
c∈kerT=ImI–T†T,
or
c=I–T†Tc for some c∈H.
Then, by again usingDα–(tα–)(t) =(α), we have
Px(t) =I–T†Tctα–=I–T†
Tctα–=
I–T†Tctα–=x(t). This impliesx∈ImP. As a consequence, we obtainX=kerP⊕kerL.
(ii) Assume thaty∈ImL. Thenφ(y) =Tcfor some c∈H. From (.)-(.), we have
PKPy(t) =
I–T†TT†φ(y) =I–T†TT†Tc= for allt∈[, ],
that meansKPy∈kerP. Moreover, we getKPy∈dom(L) due to the fact that
TT†φ(y)=TT†(Tc) =Tc=φ(y)
combined with (.). HenceKPis well defined.
We next show thatKPis the inverse operator ofLP. Clearly, it suffices to show thatKPis
a left-inverse ofLP. Letx∈dom(L)∩kerP, then
x(t) = ctα–+IαLx(t), where
T(c) =φ(Lx) for c∈kerI–T†T.
Thus
KPLPx(t) =T†φ(Lx)tα–+IαLx(t)
The rest of the proof regarding the continuity ofKPis quite simple. Noting from (.)
that
Dα–KPy
(t) =(α)T†φ(y) + t
y(s)ds for allt∈[, ] (.)
and combining (.), (.)-(.) together, we immediately establish the following esti-mates:
• φ(y)H≤
(α)( +AL(H))y,
• KPy∞≤(α)( +T †L
(H)+AL(H)T†L(H))y,
• Dα–(K
Py)∞≤( +T†L(H)+AL(H)T†L(H))y. These inequalities lead toKPy ≤Cy, whereC=(α)( +T
†L
(H)( +AL(H))) and
thus finish the proof of the lemma.
.. L-complete continuity of the nonlinear operator N Lemma . The operator N is(.)is L-compact.
Proof Assume that⊂Xis bounded. By assumption (A) onf, there is a functionϕ∈Y
such that
Nx(t)H=ft,x(t),Dα–x(t)H≤ϕ(t) (.)
for almost everyt∈[, ] and for allx∈. It follows from the equality
QNx(t) =qI–TT†φ(Nx)tα–
and (.) and (.) thatQN() is bounded. In addition, the continuity ofQNis deduced thanks to the Lebesgue dominated convergence theorem.
Next we prove thatKP,QNis completely continuous. Again using the Lebesgue
domi-nated convergence theorem, we can show thatKP,QN is continuous. It suffices to prove
thatKP,QN is compact. For this purpose, we first observe that, for everyx∈and for
every ≤t<t≤,
KP,QNx(t) –KP,QNx(t)H
≤
(α) t
(t–s)α–– (t
–s)α–Nx(s)Hds +
(α) t
t
(t–s)α–Nx(s)Hds +T†φ(Nx)H|t–t|α– +|q|(α)
(α) φ(Nx)Ht α– –t
α–
. (.)
Using the inequality that
we imply from (.) that
KP,QNx(t) –KP,QNx(t)H
≤
(α)|t–t|
α–ϕ
+
(α)
+AL(H)T†L(H)|t–t|α–ϕ
+ |q|
(α)
+AL(H)T†L(H)tα––t
α–
ϕ. (.)
In addition, due to (.) we have
Dα–(KP,QNx)(t) –Dα–(KP,QNx)(t)H≤ t
t
ϕ(s)ds+Cϕtα–t
α
. (.)
These two estimates (.), (.) show that the familiesKP,QN() andDα–KP,QN() are
equicontinuous inC([, ];H).
On the other hand, in view of (A), the set{Nx(t) :x∈}is relatively compact inH almost allt∈[, ]. Then, due to the Lebesgue dominated convergence theorem, we can prove that{KP,QNx(t) :x∈}and{Dα–KP,QNx(t) :x∈} are relatively compact inH
for everyt∈[, ]. Therefore, Lemma . guarantees thatKP,QN() is relatively compact
inX. This means that the operatorKP,QNis compact. The lemma is then proved.
3 Main result
The main result of this paper is the following theorem.
Theorem . Let(A)-(A)hold,and let the following assumptions(B)-(B)be satisfied.
(B) There exist positive real functionsa,b,c∈L[, ]with(
(α)+C)(aL+bL) <
such that
f(t,x,y)H≤a(t)xH+b(t)yH+c(t) (.)
for almost everyt∈[, ],for every(x,y)∈H×H,with constantCgiven in
Lemma..
(B) There is a constant> such that ifx∈dom(L)andmint∈[,]Dα–x(t)H>,
then
η
ft,x(t),Dα–x(t)dt∈/ImT. (.)
(B) There is a constant> together with an isomorphismJ:ImQ→kerLsatisfying
either
c, JQNctα–≥, (.)
or
c, JQNctα–≤ (.)
for everyc∈kerTsatisfyingcH>,and for somet∈(, ].
The proof of the theorem needs several auxiliary results. We present them first, in the next three lemmas.
Lemma . Let={x∈dom(L)\kerL:Lx=λNx, <λ≤}. Then is a bounded
subset in X.
Proof Assumex∈, soLx=λNxfor some <λ≤. We have
Nx=
λLx∈ImL,
then, by (.), we haveφ(Nx)∈ImT. Therefore
η
Nx(t)dt= –(α)φ(Nx) –T
η
Nx(t)dt
∈ImT.
It follows from (B) thatDα–x(s)H≤
for somes∈[, ]. We thus obtain
Dα–x()
H≤Dα–x(s)H+ s
Dαx(t)dt
H≤
+Lx≤+Nx, (.)
and hence
Px ≤
(α)D
α–x() H≤
(α)
+Nx
. (.)
Furthermore, sincePis the projection onX, (IdX–P)x∈dom(L)∩kerP, we have
(IdX–P)x=KPL(IdX–P)x≤ KPLx ≤CNx, (.)
where constantCin Lemma . and IdXstands for the identity mapping onX. Combining
(.)-(.) yields
x=Px+ (IdX–P)x≤ Px+(IdX–P)x
≤
(α)+
(α)+C
Nx.
Besides, (B) gives us
Nx =
ft,x(t),Dα–x(t)Hdt
≤ aLx∞+bLDα–u∞+cL
≤aL+bL
x+cL.
Thus
x ≤
(α)+ (
(α)+C)cL
– ((α)+C)(aL+bL)
,
Lemma . Let={x∈kerL:Nx∈ImL}.Thenis a bounded subset in X.
Proof Assumex∈. ThenNx∈ImLand there is some c∈kerT for whichx(t) = ctα–. Similarly, arguing as in the proof of Lemma ., we also haveDα–x(t
)H≤for some
s∈[, ]. This leads tocH≤
(α), and hence x=x∞=cH≤(α),
that is,is bounded, or the proof is complete.
Lemma . Setting
+=x∈kerL:λx+ ( –λ)JQNx=θ, ≤λ≤
and
–=x∈kerL: –λx+ ( –λ)JQNx=θ, ≤λ≤,
we have
(i) The set+is bounded inXif (.)holds.
(ii) The set–is bounded inXif (.)holds.
Proof (i) Assumex∈+. Thenx(t) = ctα–for some c∈kerTand
( –λ)JQNx(t) = –λx(t) for everyt∈[, ], for someλ∈[, ].
This deduces for the case λ= that Nx∈ker(JQ) =kerQ=ImL, thenx∈. Due to Lemma .,xis bounded.
For the caseλ> , we argue by contradiction. Suppose thatcH>. Then, by (.), there existst> for which
≤( –λ)c, JQNctα–=c, –λctα–= –λtα–cH< ,
this cannot happen. Hence, we havex=x∞=cH≤, or–is bounded. (ii) Similarly, the boundedness of+is established. The lemma is proved.
Now we focus on proving Theorem ..
Proof of Theorem. Our purpose is to apply Theorem . by verifying all its conditions (i)-(iii). Firstly, we setis a bounded and open subset inXfor whichi=i⊂, where
= ⎧ ⎨ ⎩
+ if (.) holds,
–
if (.) holds.
this, we use the Brouwer degree property of the invariance with a continuous homotopy. Namely, consider a homotopyHλ:X→Xdefined by
Hλ(x) =±λx+ ( –λ)JQNx forλ∈[, ],
in whichJ:ImQ→kerLis isomorphic assumed in (B). Lemma . gives us that
Hλ(x)= for allx∈kerL∩∂, for allλ∈[, ].
Therefore,
deg(JQN|kerL,∩kerL, ) =deg(H,∩kerL, )
=deg(H,∩kerL, )
=deg(±Id,∩kerL, )
=±= .
Hence, (iii) holds as claimed. The proof of the theorem is complete.
Remark . IfHis a Banach space, assumption (B) needs to be slightly modified. Pre-cisely, two inequalities (.), (.) are respectively generalized as
˜
c, JQNctα–≥ and
˜
c, JQNctα–≤,
in whichc˜∈H(the dual space ofH) with˜c, c =cH, and·,· now stands for the scalar product for the dualityH,H. Then the proof of Lemma . is modified correspondingly; the result of Theorem . still holds.
4 Example and discussion
This section is to provide one illustrative example of Theorem .. Before presenting the example, we should note that giving such a significant example is a challenging task in the case thatHis infinite dimensional. It is caused by two major reasons. The first one is that it is not trivial to give a suitable nonlinear functionf in an infinite dimensional space. The second one, which is more difficult to overcome, is that so far there has been no effective method to find the Moore-Penrose inverse of a general bounded linear operator on Hilbert spaces or of a Fredholm operator with zero-index. For instance, some recent works [, ] for the first time give the examples in an infinite dimensional space. However, to the best of our knowledge, these examples are still controversial.
Indeed, in [], Section , and also in [], Section , for the caseH=l, there is a gap in setting the expression off. Namely, the functionf(t,u,v) in [], Section , is not continuous with respect to the third variable at points whichvl=(yi)∞i=l≥ such
satisfied. Moreover, in [], Section , finding the Moore-Penrose inverse ofMis not much helpful because the given operator Mdoes not need to be Fredholm with zero-index as a prerequisite. This is derived from another gap in [], Lemma ., that the condition onAis not sufficient to conclude thatImQ=kerL.
In what follows, we give an example for the case thatHis finite dimensional. The exam-ple in an infinite dimensional space would be presented in our future research with more careful consideration.
Example . Consider the existence of solutions to the following fractional differential equations:
D/+
x
x
(t) =
t+
(|x(t)|+|x(t)|) +
t+ (|D
/
+x(t)|+|D/+x(t)|) + √
t+
t+
(x(t) +x(t)) +
t+ (D
/
+x(t) +D/+x(t)) +
√ t+ (.) subject to x x () =
and D/+
x x () = – –
D/+
x x . (.)
Setα=,η=,
A= – – ,
and definef: [, ]×R×R→Rby setting
f(t,x,y) =f(t,x,y),f(t,x,y)
, (.)
wheref,f: [, ]×R×R→Rare functions given by
f(t,x,y) =
t+
|x|+|x|
+t +
|y|+|y|
+√t+ (.)
and
f(t,x,y) =
t+
(x+x) +
t+
(y+y) + √
t+
(.)
fort∈[, ] andx= (x,x),y= (y,y)∈R. Then problem (.)-(.) becomes problem (.). Our purpose is to apply Theorem . in order to establish the existent result of (.)-(.).
It is clear that (A)-(A) are satisfied. We need to verify (B)-(B). From (.)-(.) we have
f(t,x,y)R≤a(t)xR+b(t)yR+c(t)
for every ≤t≤ and for everyx,y∈R, in which
a(t) = √
(t+ )
, b(t) =
√(t+ )
, c(t) =
√
Moreover, some direct calculations give us T= – –
and T†= – – .
It follows that
(α)+C
aL+bL
≈. < .
Hence, (B) holds. On the other hand, observe that
f
t,x(t),D/+x(t)
>f
t,x(t),D/+x(t)
, ∀t∈[, ],
for allx∈dom(L). This implies that
η
ft,x(t),D/+x(t)
dt∈/ImT
due to the fact thatImT=(, ) ={(c,c) :c∈R}; that is, (B) holds. Finally, to check (B)
we see that
kerL=x: [, ]→R:x(t) =c√t(, ),t∈[, ],c∈R
and
ImQ=y: [, ]→R:y(t) =c√t(, –),t∈[, ],c∈R,
andJ:ImQ→kerLisomorphism defined as
(Jz)(t) =
z(t), z∈ImQ.
Thus, we get
JQ(z)(t) = √
π√t
√ –
– –
φ(z) (.)
for allz∈L([, ],R). BykerT=(, ) , suppose thatγ = (a, a)∈kerTfor somea∈R. Then
Nγt/(t) =
f(t,γt/,
√
π
γ)
f(t,γt/,
√
π
γ) =
t+ √t+ √
π
t+
|a| a
+
√
and hence ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ /
N(γt/)(t)dt= (
√
+
√π
) ⎡ ⎣|a|
a ⎤ ⎦+√– ⎡ ⎣ ⎤ ⎦,
N(γt/)(t)dt= ( +
√π
) ⎡ ⎣|a|
a ⎤ ⎦+ √ – ⎡ ⎣ ⎤ ⎦. (.)
It follows from (.) and (.) that
φNγt/
=
(α)
A
/
Nγt/(t)dt–
Nγt/(t)dt
=√ π ( √ – +
√π
)|a|– ( √
+ √π
)a– √ + √+ ( √ +
√π
)|a|– ( √
+ √π
+ )a+
√ – √ + . (.)
Combining (.) and (.), we obtain
JQNγt/= √
t
(√ – )
( – √ + √π)a–|a|+ √
+ √ – . Therefore,
γ, JQNγtα–=θ(t)( – √ + √π)a–|a|a
+ (√ + √ – )a,
whereθ(t) =
√ t
(√–) ≤ for every ≤t≤. This deduces thatγ, JQN(γt
α–) ≤ for allt∈[, ], for|a|large enough. Hence, (B) holds. Problem (.)-(.) thus has at least one solution.
Remark . We should comment on the computation of the Moore-Penrose inverse ma-trix T† in the above example since this step is crucial. Basically, the common method for computing the Moore-Penrose inverse of matrix is the singular value decomposition (SVD) method (see []). This method is accurate but time-intensive since it requires a large amount of computational resources, especially in the case of large scale matrix. An-other approach for computing the Moore-Penrose inverse of matrix has been proposed recently, i.e., the method based on Tensor-product matrix (see []). The new method can be applied for rectangular matrix with full-rank and square matrix with rank-deficient only. However, the method works well compared to SVD method in some aspects: larger dimension matrix and less time-consuming.
5 Conclusions
sufficient and necessary condition for which some Riemann-Liouville fractional differen-tial operators associated with three-point boundary condition are Fredholm with zero-index. This means that we have handled the problem with a wide range of resonant con-ditions in which Mawhin’s continuation theorem can be applied. Our result is a natural generalization of some recent ones [–].
Acknowledgements
We are thankful to the editor and the anonymous reviewers for many valuable suggestions to improve this paper.
Funding
This research is partially supported by Duy Tan University and Banking University of Ho Chi Minh City.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed to each part of this study equally and read and approved the final version of the manuscript.
Author details
1Institute for Research and Development, Duy Tan University, 03 Quang Trung, Da Nang, Vietnam.2Faculty of Management Information Systems, Banking University of Ho Chi Minh City, 39 Ham Nghi Street, District 1, Ho Chi Minh City, Vietnam.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 3 December 2016 Accepted: 27 June 2017
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