Existence of solutions to fractional boundary value problems at resonance in Hilbert spaces
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Abstract
Keywords
coincidence degree threepoint boundary value problem fractional differential equation resonanceMSC
34A08 34B10 34B151 Introduction
 (A1)

\(IA\) is a Fredholm operator with zeroindex;
 (A2)

\(f(\cdot, x,y)\) is Lebesgue measurable on \([0,1]\) for every \((x,y)\in \mathcal{H}\times \mathcal{H}\);
 (A3)

\(f(t,\cdot,\cdot)\) is continuous on \(\mathcal{H}\times \mathcal{H}\) for almost every \(t \in [0, 1]\);
 (A4)

for each bounded set \(\Omega \subset \mathcal{H}\times \mathcal{H}\), the function \(t\mapsto \varphi_{\Omega }(t) = \sup \{ \Vert f(t, x, y)\Vert _{\mathcal{H}} : (x, y) \in \Omega \} \) is Lebesgue integrable on \([0,1]\), and the set \(\{ f(t, x, y):(x, y)\in \Omega \} \) is relatively compact in \(\mathcal{H}\), here \(\Vert \cdot \Vert _{\mathcal{H}}\) stands for the norm induced by the scalar product \(\langle \cdot,\cdot \rangle \) in \(\mathcal{H}\).
Different from nonresonant problems being studied for a long time, the solvability of resonant problems has been extensively studied for the last decade. The reason is that resonant problems are rather complicated due to the noninvertibility of L. Noninvertibility leads to the difficulty of constructing a suitable continuous projection on a complement of ImL. Due to the fact that constructing that projection becomes more difficult when the dimension of kerL is large, most of works are investigated mainly for \(\dim \ker L=1\). We refer the reader to [9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23] and to the references therein.
In our new research [27] condition (1.4) can be omitted completely thanks to the fruitful features of a continuous linear operator on \(\mathbb {R}^{n}\) which can be regarded as matrices. The breakthrough results in [27] allow us to ask the first question: For extending problem ( 1.1 ) from a finitedimensional space \(\mathbb {R}^{n}\) to a Hilbert space \(\mathcal{H}\) , is omitting condition ( 1.4 ) still possible or not? The answer is no, for example, let A be the difference of an identity operator and a finiterange operator, problem (1.1) is unsolvable via Mawhin’s continuation theorem.
 (1)
Condition (1.6) is not only a sufficient condition but also a necessary condition for that \(L=D^{\alpha }\) is the Fredholm operator with zeroindex.
 (2)
When \(\mathcal{H}\) is finitedimensional, (1.6) automatically holds, therefore the result in [27] is a special case of our result in this paper.
 (3)
When \(\mathcal{H}\) is infinitedimensional, it is well known that the ranknullity theorem no longer holds for an operator \(A\in \mathcal{L(H)}\). Condition (1.6) helps us to overcome this difficulty by providing another characterization of dimension in a Hilbert space, as to be seen in (2.10).
 (4)
Condition (1.6) gives a unified approach and viewpoint for solving this kind of BVPs, for \(\dim \mathcal{H}\leq \infty \) and \(L=D^{\alpha }\) with \(1<\alpha \leq 2\).
The structure of the paper is organized as follows. In Section 2, we introduce some necessary background of fractional calculus as well as coincidence degree theory. In addition, we establish some essential lemmas needed for our main result later. Section 3 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 [2, 6, 7] for more details.
Definition 2.1
 ⋄
 The fractional integral of order α of the function f is given by$$I^{\alpha }f(t):= \frac{1}{\Gamma (\alpha)} \int_{0}^{t}(ts)^{ \alpha 1}f(s)\,ds\quad \mbox{for } t>0. $$
 ⋄
 The RiemannLiouville fractional derivative of order α of f is given bywhere \(m = [\alpha ] + 1\),$$D^{\alpha }f(t):= \frac{d^{m}}{dt^{n}} \bigl( I^{m\alpha }f \bigr) (t) = \frac{1}{\Gamma (m  \alpha)} \frac{d^{m}}{dt^{m}} \int_{0}^{t} (ts)^{m  \alpha 1} f(s)\,ds\quad \mbox{for } t>0, $$
Lemma 2.2
 (i)
The equality \(( D^{\alpha }I^{\alpha }\varphi) (t)= \varphi (t)\) holds for almost every \(t\in [0,1]\).
 (ii)If \(D^{\alpha m}\varphi \in AC^{m} ( [0,1]; \mathcal{H} ) \), thenfor almost every \(t\in [0,1]\).$$\bigl( I^{\alpha }D^{\alpha }\varphi \bigr) (t)=\varphi (t)\sum _{j=1} ^{m}\frac{D^{\alpha j}\varphi (0)}{\Gamma (\alpha j+1)}t^{\alpha j} $$
2.2 Mawhin’s continuation theorem
Let X and Y be two Banach spaces.
Definition 2.3
Definition 2.4
 (i)
\(QN: \overline{\Omega } \rightarrow Y\) is continuous and \(QN(\overline{\Omega })\) is bounded in Y;
 (ii)
\(K_{P,Q}N : \overline{\Omega } \rightarrow X\) is completely continuous.
Theorem 2.5
 (i)
\(Lx\neq \lambda Nx\) for \(x \in \partial \Omega \cap ( \operatorname {dom}(L) \backslash \ker L ) \) and for \(\lambda \in (0,1)\);
 (ii)
\(QNx \neq 0\) for \(x \in \partial \Omega \cap \ker L\);
 (iii)for any isomorphism J from ImQ to kerL,where \(Q:Y\rightarrow Y\) is a projection given as above.$$\deg_{B} ( \mathrm{JQN}\vert _{\ker L},\Omega \cap \ker L, 0 ) \neq 0, $$
For a comprehensive treatment on the subject of the coincidence degree theory and Mawhin’s continuation theorem as well, we refer the reader to [8, 28, 29].
Finally, we restate the following compactness criterion on the space of continuous functions \(C(X,Y)\) which is regarded as a generalization of the ArzelaAscoli theorem [30], Section 1.1.3.
Lemma 2.6
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 equicontinuous.
2.3 Key lemmas
2.3.1 Fredholm property of the fractional differential operator L

\(T = I  A\),
 \(\phi: Y \rightarrow H\) is the following bounded linear operator:$$ \phi ( y) = \frac{1}{\Gamma (\alpha)} \biggl( A \int_{0}^{ \eta }y(t)\,dt  \int_{0}^{1} y(t)\,dt \biggr) \quad \mbox{for } y \in Y. $$(2.3)
Remark 2.7
 (i)
\(T^{\dagger }TT^{\dagger }= T^{\dagger }\);
 (ii)
\(TT^{\dagger }T =T\);
 (iii)
\(TT^{\dagger }\) and \(T^{\dagger }T\) are selfadjoint operators.
Lemma 2.8
 (i)There exists a continuous projection \(Q:Y\rightarrow Y\) such that$$\ker Q=\operatorname {Im}L \quad \textit{and}\quad Y=\operatorname {Im}L \oplus \operatorname {Im}Q. $$
 (ii)
L is a Fredholm operator with zeroindex if and only if so is T.
Proof
Remark 2.9
Remark 2.10
Lemma 2.8 will be no longer true in the general case that \(\mathcal{H}\) is a Banach space. If \(\mathcal{H}\) is 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 that T has the MoorePenrose inverse \(T^{\dagger }\in \mathcal{L(H)}\).
Lemma 2.11
 (i)The operator P in (2.11) is projection and satisfies that$$\ker L=\operatorname {Im}P,\qquad X = \ker P \oplus \ker L. $$
 (ii)The map \(K_{P} : \operatorname {Im}L \rightarrow \operatorname {dom}(L) \cap \ker P\) is defined bysatisfying$$ K_{P}y(t) = T^{\dagger } \phi (y)t^{\alpha 1} + I^{\alpha }y(t)\quad \textit{for } t \in [0, 1], $$(2.12)where \(C =\frac{1}{\Gamma (\alpha)} ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ( 1 + \Vert A\Vert _{\mathcal{L(H)}} )) \).$$K_{P} = L_{P}^{1}\quad \textit{and} \quad \Vert K_{P}y\Vert \leq C\Vert y\Vert _{1}, $$
Proof

\(\Vert \phi (y)\Vert _{\mathcal{H}} \leq \frac{1}{\Gamma (\alpha)} ( 1 + \Vert A\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\),

\(\Vert K_{P}y\Vert _{\infty }\leq \frac{1}{\Gamma (\alpha)} ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} + \Vert A\Vert _{ \mathcal{L(H)}}\Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\),

\(\Vert D^{\alpha 1}(K_{P}y)\Vert _{\infty }\leq ( 1 + \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} + \Vert A\Vert _{\mathcal{L(H)}} \Vert T^{\dagger }\Vert _{\mathcal{L(H)}} ) \Vert y\Vert _{1}\).
2.3.2 Lcomplete continuity of the nonlinear operator N
Lemma 2.12
The operator N is (2.2) is Lcompact.
Proof
On the other hand, in view of (A4), the set \(\{ Nx(t): x\in \Omega \} \) is relatively compact in \(\mathcal{H}\) almost all \(t\in [0,1]\). Then, due to the Lebesgue dominated convergence theorem, we can prove that \(\{ K_{P,Q}Nx(t): x\in \Omega \} \) and \(\{ D^{\alpha 1}K_{P,Q}Nx(t): x\in \Omega \} \) are relatively compact in \(\mathcal{H}\) for every \(t\in [0,1]\). Therefore, Lemma 2.6 guarantees that \(K_{P,Q}N(\Omega)\) is relatively compact in X. This means that the operator \(K_{P,Q}N\) is compact. The lemma is then proved. □
3 Main result
The main result of this paper is the following theorem.
Theorem 3.1
 (B1)There exist positive real functions \(a, b, c \in L^{1}[0,1]\) with \(( \frac{1}{\Gamma (\alpha)} +C ) ( \Vert a\Vert _{L^{1}} + \Vert b\Vert _{L^{1}} ) < 1 \) such thatfor almost every \(t\in [0,1]\), for every \((x, y) \in H\times H\), with constant C given in Lemma 2.11.$$ \bigl\Vert f(t, x, y)\bigr\Vert _{\mathcal{H}} \leq a(t)\Vert x\Vert _{ \mathcal{H}} + b(t)\Vert y\Vert _{\mathcal{H}} + c(t) $$(3.1)
 (B2)There is a constant \(\Lambda_{1}>0\) such that if \(x \in \operatorname {dom}(L)\) and \(\min_{t \in [0, 1]}\Vert D^{\alpha 1}x(t)\Vert _{\mathcal{H}} > \Lambda_{1}\), then$$ \int_{\eta }^{1}f \bigl( t, x(t), D^{\alpha 1}x(t) \bigr)\,dt \notin \operatorname {Im}T. $$(3.2)
 (B3)There is a constant \(\Lambda_{2}>0\) together with an isomorphism \(J: \operatorname {Im}Q \rightarrow \ker L \) satisfying eitheror$$ \bigl\langle \mathbf{c}, \mathrm{JQN}\bigl( \mathbf{c} t^{\alpha 1} \bigr) \bigr\rangle \geq 0, $$(3.3)for every \(\mathbf{c} \in \ker T\) satisfying \(\Vert \mathbf{c}\Vert _{\mathcal{H}} > \Lambda_{2}\), and for some \(t\in (0,1]\).$$ \bigl\langle \mathbf{c}, \mathrm{JQN}\bigl( \mathbf{c} t^{\alpha 1} \bigr) \bigr\rangle \leq 0 $$(3.4)
The proof of the theorem needs several auxiliary results. We present them first, in the next three lemmas.
Lemma 3.2
Let \(\Omega_{1} = \{ x \in \operatorname {dom}(L) \setminus \ker L : Lx = \lambda Nx, 0<\lambda \leq 1 \} \). Then \(\Omega_{1}\) is a bounded subset in X.
Proof
Lemma 3.3
Let \(\Omega_{2} = \{ x \in \ker L : Nx \in \operatorname {Im}L \} \). Then \(\Omega_{2}\) is a bounded subset in X.
Proof
Lemma 3.4
Proof
(ii) Similarly, the boundedness of \(\Omega_{3}^{+}\) is established. The lemma is proved. □
Now we focus on proving Theorem 3.1.
Proof of Theorem 3.1
Remark 3.5
4 Example and discussion
This section is to provide one illustrative example of Theorem 3.1. Before presenting the example, we should note that giving such a significant example is a challenging task in the case that \(\mathcal{H}\) is infinite dimensional. It is caused by two major reasons. The first one is that it is not trivial to give a suitable nonlinear function f 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 MoorePenrose inverse of a general bounded linear operator on Hilbert spaces or of a Fredholm operator with zeroindex. For instance, some recent works [26, 32] 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 [26], Section 4, and also in [32], Section 4, for the case \(\mathcal{H}=l^{2}\), there is a gap in setting the expression of f. Namely, the function \(f_{1}(t, u, v)\) in [26], Section 4, is not continuous with respect to the third variable at points which \(\Vert v\Vert _{l^{2}}=\Vert (y_{i})_{i=1}^{\infty }\Vert _{l^{2}}\geq 1\) such that \(y_{1}=0\). This means that f is not a Carathéodory function, one assumption must be satisfied. Moreover, in [32], Section 4, finding the MoorePenrose inverse of \(\mathcal{M}\) is not much helpful because the given operator \(\mathcal{M}\) does not need to be Fredholm with zeroindex as a prerequisite. This is derived from another gap in [32], Lemma 2.6, that the condition on A is not sufficient to conclude that \(\operatorname {Im}Q=\ker L\).
In what follows, we give an example for the case that \(\mathcal{H}\) is finite dimensional. The example in an infinite dimensional space would be presented in our future research with more careful consideration.
Example 4.1
Remark 4.2
We should comment on the computation of the MoorePenrose inverse matrix \(T^{\dagger }\) in the above example since this step is crucial. Basically, the common method for computing the MoorePenrose inverse of matrix is the singular value decomposition (SVD) method (see [31]). This method is accurate but timeintensive since it requires a large amount of computational resources, especially in the case of large scale matrix. Another approach for computing the MoorePenrose inverse of matrix has been proposed recently, i.e., the method based on Tensorproduct matrix (see [33]). The new method can be applied for rectangular matrix with fullrank and square matrix with rankdeficient only. However, the method works well compared to SVD method in some aspects: larger dimension matrix and less timeconsuming.
5 Conclusions
In this work, we have established an existence result for the class of fractional boundary value problems with a general resonant condition in Hilbert spaces. We have provided a sufficient and necessary condition for which some RiemannLiouville fractional differential operators associated with threepoint boundary condition are Fredholm with zeroindex. This means that we have handled the problem with a wide range of resonant conditions in which Mawhin’s continuation theorem can be applied. Our result is a natural generalization of some recent ones [24, 25, 26, 27].
Notes
Acknowledgements
We are thankful to the editor and the anonymous reviewers for many valuable suggestions to improve this paper.
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