# On Solutions of Fractional Order Boundary Value Problems with Integral Boundary Conditions in Banach Spaces

Authors: Hussein A. H. Salem; Mieczysław Cichoń
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The object of this paper is to investigate the existence of a class of solutions for some boundary value problems of fractional order with integral boundary conditions. The considered problems are very interesting and important from an application point of view. They include two, three, multipoint, and nonlocal boundary value problems as special cases. We stress on single and multivalued problems for which the nonlinear term is assumed only to be Pettis integrable and depends on the fractional derivative of an unknown function. Some investigations on fractional Pettis integrability for functions and multifunctions are also presented. An example illustrating the main result is given.


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## Body

## 1. Introduction

The theory of boundary value problems is one of the most important and useful branches of mathematical analysis. Boundary value problems of various types create a significant subject of several mathematical investigations and appear often in many applications, especially in solving numerous problems in physics and engineering. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can be reduced to nonlocal problems with integral boundary conditions. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to [1–3] and the references therein.The class of boundary value problems with integral boundary conditions considered below contains as special cases numerous two, three, multipoint, and nonlocal boundary value problems. Such problems are mainly investigated when considering functions satisfying some conditions expressed in terms of the strong topology of a Banach spaceE. We will investigate the case, when functions are not strongly continuous and strongly integrable. In this situation we need to introduce more general notion of a solution. We should note that the considered case seems to be a natural case and cover many particular cases considered for both the strong and weak topologies (cf. Lemma 19). A more general notion of solutions allows us to solve the problem under very general assumptions, not so restrictive as before (see our last section).In contrast to the classical approach for the theory of boundary value problems, the theory for fractional order BVP's is still developing one and not satisfactorily described. It is caused by the fact that it is very difficult to find convenient and handy conditions ensuring the existence of solutions of several nonlinear boundary value problems of fractional order. In the considered case of a weak topology onE our results form a relatively new branch of investigations.As a pursuit of this, some sufficient conditions for the existence of solutions are presented for the following nonlinearm-point boundary value problem of fractional type:
(1)Dαu(t)+f(t,u(t),Dβu(t))=0,t∈[0,1],α∈(1,2],β∈(0,1),α>1+β,u(1)+∫01‍ℑ(τ)u(τ)dτ=l,u(0)=0,
where x takes values in a Banach space E and l∈E. Here ℑ∈Lq[0,1] for some q∈[1,∞] and Dα denotes the pseudo fractional differential operator of order α (to be described later). We will assume that f is a vector-valued Pettis integrable function on [0,1]. We remark the following:(1)
for real-valued functions withα,β∈ℕ∶={1,2,3,4,…}, we have problems studied in, for example,  [4, 5],(2)
for real-valued functions withℑ≡0 and α,β∈ℕ, we have problems studied in, for example, [6–10]; see also the references therein,(3)
for real-valued functionsℑ≡0 and when the function f is independent of the fractional derivatives, then we have problems studied in, for example,  [1, 11–13],(4)
in abstract spaces (for vector-valued functions) withℑ≡0 and α,β∈ℕ, we have problems studied in, for example,  [14–19],(5)
in abstract spaces with conditions related to the weak topology onE and when the vector-valued function f is independent of the fractional derivatives, then we have a problems studied in, for example,  [20, 21].In comparison with the existence results in the above list, our assumptions seem to be more natural. In contrast to earlier results, we drop the requirement thatf is a real-valued function independent of the fractional derivatives and we consider the case of vector-valued Pettis, but not necessarily Bochner, integrable functions. As we mentioned above, the assumptions in the existence theorem are expressed in terms of the weak topology. Such a result does not appear in the earlier literature and so it seems to be new. We collect all interesting properties for the fractional Pettis integral. Moreover, we are able also to start some studies for multivalued fractional boundary value problems with Pettis-integral boundary conditions and fractionally Pettis integrable multifunctions.In the paper we stress also on comparison results for Pettis integrals and fractional Pettis integrals. This is also done for the multivalued integrals and seems to be interesting by itself, independently of applicability of our results. The properties of fractional integral operators on the spaces of Pettis integrable functions as well as on some of its subspaces are also investigated.Finally, we remark that, in the Banach spaces, the existence of solutions of some boundary value problems of fractional orders has been considered in terms of Pettis integrals, for the first time, by Salem [20]. In this paper, for clarity of proofs, we restrict ourselves to the case of reflexive spaces, but it is easy to extend our results for nonreflexive spaces by putting contraction hypothesis with respect to some measure of weak noncompactness and by using appropriate fixed point theorem (cf. [22]). Nevertheless, all auxiliary results in this paper are not restricted to reflexive spaces.The question of proving the existence of solutions to the problem (1) reduces to proving the existence of solutions of a Fredholm integral equation. Since the space of all Pettis integrable functions is not complete (in general), we restrict our attention to the case of weakly continuous solution of the Fredholm integral equation (modeled off the problem (1)); hence we are ready to find the so-called pseudo-solutions of the problem (1) (cf. [22, 23]).


## 2. Preliminaries and Auxiliary Results

For the sake of the reader’s convenience here we collect a few facts which will be needed further on. LetI=[0,1]. According to the custom Lp(I), 1≤p≤∞ will denote the Banach space of real-valued measurable functions x defined on I. Let L∞(I) denote the Banach space of real-valued essentially bounded and measurable functions defined on I. Through the paper, E is considered to be a Banach space with norm ∥·∥ and with its dual space E*. Moreover, let Ew=(E,w)=(E,σ(E,E*)) denote the space E with its weak topology. By C[I,E] we will denote the Banach space of strongly continuous functions x:I→E endowed with a standard ∥x∥0=supt∈I∥x(t)∥, while P[I,E] denotes the space of all E-valued Pettis integrable functions in the interval I (see [24, 25] for the definition). Let us also recall that a function h:E→E is said to be weakly-weakly sequentially continuous if h takes each weakly convergent sequence in E into weakly convergent sequence in E. We point out that a bounded weakly measurable function x:I→E need not to be Pettis integrable even if E is reflexive. However, in reflexive Banach spaces, the weakly measurable function x:I→E is Pettis integrable if and only if φ(x(·)) is Lebesgue integrable on I for every φ∈E* [26].Let us recall some basic facts. The following Mazur's lemma can be found in [24, 26].Lemma 1.
A convex subset of a normed spaceE is closed if and only if it is weakly closed.A simple consequence of the Hahn-Banach theorem is as follows.Proposition 2.
LetE be a normed space with x0≠0. Then there exits φ∈E* with ∥φ∥=1 and φx0=∥x0∥.Now, we are in a position to recall a fixed point theorem being an extension of results from [27].Theorem 3.
LetE be a Banach space with Q a nonempty, closed, convex, and weakly compact subset of C[I,E]. Assume that T:Q→Q is weakly-weakly sequentially continuous. Then T has a fixed point in Q.We need to introduce some subspaces of the spaceP[I,E] of Pettis integrable functions on I which are important in the sequel.Definition 4.
For1≤p≤∞, we define the class ℋp(E) to be the class of all functions x:I→E having φx∈Lp(I) for every φ∈E*. If p=∞, the added condition
(2)sup∥φ∥=1(esssupt∈I|φx(t)|)<∞
must be satisfied by each x∈ℋ∞(E). The class ℋ0p(E) is defined by
(3)ℋ0p(E)∶={x∈P[I,E]:φx∈Lp(I)}.Remark 5.
In a reflexive Banach spaceE the set ℋ01(E) coincides with the space P[I,E]. This is due to the fact that in reflexive Banach spaces, the weakly measurable function x:I→E is Pettis integrable if and only if φ(x(·)) is Lebesgue integrable on I for every φ∈E* [26]. In general, this is the space of Dunford integrable functions.In the remaining part of this paper we letp∈[1,∞] be fixed and q is conjugated with p; that is, 1/p+1/q=1. The following results are due to Pettis (see [25, Theorem 3.4 and Corollary 3.41]).Proposition 6.
In order thatx(·) could be in ℋ0p(E), it is necessary and sufficient that x(·)u(·) be Pettis integrable for every u(·)∈Lq(I).It is worthwhile to recall the following.Definition 7.
Letx:I→E. The (left-sided) fractional Pettis-integral (shortly LS-FPI) of x of order α>0 is defined by
(4)I+αx(t)∶=∫0t‍(t-s)α-1Γ(α)x(s)ds,t>0.In the above definition the sign “∫‍” denotes the Pettis integral. For further purpose, we define the right-sided fractional Pettis-integral (shortly RS-FPI) by
(5)I-αx(t)∶=∫t1‍(s-t)α-1Γ(α)x(s)ds,t<1.We will call a function fractionally Pettis integrable provided this integral exists as an element ofE (for arbitrary t<1).We need to clarify the relations between Pettis integrability and fractional Pettis integrability. Similar results will be proved for classesℋ0p(E). This will be important in our consideration, but it seems to be really interesting in itself. Here we restrict ourselves to the case of left-sided fractional Pettis-integrals.To make the paper more expository, we will consider fractional Pettis integrability for both cases:α<1 and α>1. The last case is more important in our paper, but the first one is necessary to compare our results with some earlier theorems.Let us observe that such an integralI+αx(t)∶=∫0t‍((t-s)α-1/Γ(α))x(s)ds is a convolution of a function h(τ)=τα-1/Γ(α) for τ>0, h(τ)=0 for τ≤0, and the function (x~)(t)=x(t) for t∈I, where (x~)(t)=0 outside the interval I. Note that Pettis integrability of x(t) implies Pettis integrability of x(t+h) (h>0) and x(-t), so the convolution of Pettis integrable function with real-valued function h can be properly defined. We start with an obvious observation that for φ∈E*(6)φ(I+αx)=∫0t‍(t-s)α-1Γ(α)φx(s)ds.As a consequence of some properties of a convolution for the Pettis integral [28, Proposition 9], for arbitrary α, we have the following.Theorem 8.
Ifx:I→E is Pettis integrable, then (a)
I+αx is defined almost a.e. on I,(b)
x is fractionally Pettis integrable on I,(c)
ifx is Pettis integrable and strongly measurable, then I+αx:I→E is bounded, weakly continuous and
(7)sup∥φ∥≤1∫01‍φI+αx(t)dt≤sup∥φ∥≤1∫01‍φx(t)dt·∥h∥1.In the caseE=ℝ, it is a well-known consequence of an inequality of Young that the linear fractional integral operators I±α, send Lq([0,1]) continuously into Lp([0,1]) if p∈[1,∞] satisfy q>1/(α+(1/p)) (see [29]) (a deep result from interpolation theory implies that even q=1/(α+(1/p)) is allowed if 1<p<∞ ). In particular, I±α:Lp([0,1])→Lp([0,1]) is compact for each p∈[1,∞]. Moreover, for p>max{1,(1/α)}, the map I±α:Lp([0,1])→C([0,1]) is compact (see,  e.g., [20, 30]).The following results plays a major rule in our analysis.Lemma 9.
For anyα>0 the operator I±α takes C[I,Ew] into C[I,Ew] and is well defined.Proof.
Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.
It can be easily seen that ifx is weakly continuous, then x∈ℋ0∞(E). Since s→(t-s)α-1∈L1, s<t, in the view of Proposition 6 and Theorem 8, we have that the function s→(t-s)α-1x(s), s<t, is Pettis integrable. Moreover, I+α is well defined. To see this, we define y:I→E by y(t)∶=I+αx(t), t∈[0,1]. From the definition of fractional Pettis integrals and Theorem 8(c), we have for every φ∈E* that
(8)φy(t)=∫0t‍(t-s)α-1Γ(α)φ(x(s))ds.
Since the function φx is continuous, φy is continuous. That is, y is weakly continuous which finishes the proof.Lemma 10.
LetE be a reflexive Banach space. For any α≥1 and arbitrary p∈[1,∞], the operator I±α takes ℋ0p(E) into ℋ0p(E) and is well defined.Proof.
Only the proof in case of the LS-FPI is given since the case of the RS-FPI is very similar.
Note first that, forx∈ℋ0p(E), we have in the view of Proposition 6 that the function s→(t-s)α-1x(s), s<t, is Pettis integrable. That is, the operator Iα makes sense. Further, I+α is well defined. To see this, define y:I→E by y(t)∶=I+αx(t), t∈[0,1]. From the definition of fractional Pettis integrals, we have for every φ∈E* that
(9)φ(y(t))=∫0t‍(t-s)α-1Γ(α)φ(x(s))ds=I+1I+α-1φ(x(t)).
Since x∈ℋ0p(E), one have φx∈Lp(I) for every φ∈E*. One could see, by the properties of fractional integral operators on the Banach space Lp(I), that φy∈Lp(I) for every φ∈E*. In particular, y is weakly continuous. Since weak continuity implies weak measurability (see [26, page 73]), y(·) is weakly measurable. However, in reflexive Banach spaces, weakly measurable functions y:I→E are Pettis integrable if and only if y is Dunford integrable; that is, φ(y(·)) is Lebesgue integrable on I for every φ∈E*. Thus y(·) is Pettis integrable. That is, I+αx∈ℋ0p(E).Remark 11.
Let us discuss some properties ofℋp(E) and ℋ0p(E). Recall that h(τ)=τα-1/Γ(α) for τ>0 and z(τ)=0 for τ≤0. The case of α≥1 is trivial; that is, h∈L∞(I). When 0<α<1, we see that ∫01‍(z(s))qds=∫01‍(sα-1/Γ(α))qds<∞ whenever q<1/(1-α). This means that z∈L1/(1-α)(I) and by the converse for the Young inequality Iα takes ℋq1(E) into ℋs(E) whenever (1/q1)+(1/q)=1+(1/s). In particular, the space required for this property depends on α. Let us note that Iα need not be continuous as an operator from L1(I) into L1/(1-α)(I) (cf. [29, Remark 4.1.1]).
This means that forα≥1 we have “uniform” estimations for all α, but for 0<α<1 the situation is more complicated (a weakly singular case).Our consideration as well as Theorem 4.1.1 in [29] gives us a new property.Lemma 12.
For any0<α<1 and arbitrary 1≥p≥1/α the operator I±α takes ℋ0p(E) into ℋ0s(E), where s=p/(1-p(α-ε)) with arbitrary ε>0.As a consequence of Lemma10, we are able to prove the following.Lemma 13.
Assume thatα1,α2≥1 and x∈ℋ0p(E). Then,
(10)I+α1I+α2x=I+α2I+α1x=I+α1+α2x.Proof.
As in the proof of Lemma10 it follows that I+α2x and I+α1+α2x exist. By Lemma 10, I+α1I+α2x also exists. Therefore, for any φ∈E* we have
(11)φ(I+α1I+α2x(t))=I+α1φ(I+α2x(t))=I+α1I+α2φ(x(t))=I+α1+α2φ(x(t))=φ(I+α1+α2x(t)),
that is,
(12)φ(I+α1I+α2x(t)-I+α1+α2x(t))=0,foreveryφ∈E*.
Hence I+α1I+α2x=I+α1+α2x(t). Similarly, we are able to show that I+α2I+α1x=I+α1+α2x. This ends the proof.Let us present the caseα≥1.Lemma 14.
Ifα≥1, θ∈Lq[0,1], and y∈ℋ0p(E), then
(13)∫01‍[I-αθ(s)]y(s)ds=∫01‍θ(s)[I+αy(s)]ds.Proof.
Define the real-valued functionh by
(14)h(s)∶=1Γ(α)∫s1‍(τ-s)α-1θ(τ)dτ=I-αθ(s).
Using the properties of fractional calculus in the Banach spaceLq[0,1] (see,  e.g., [20, 30]), we deduce that h∈Lq[0,1]. Now, for y∈ℋ0p(E) we have, in the view of Lemma 10, that I+αy∈ℋ0p(E). Thanks to Proposition 6, the functions t→θ(t)I+αy(t) and t→h(t)y(t) are Pettis integrable on [0,1]. That is, the integrals in both sides of (13) exist. Then there exists J∈E, such that
(15)J=∫01‍θ(s)I+αy(s)ds.
By the definition of the Pettis integral, we have
(16)φJ=∫01‍θ(s)φ(I+αy(s))ds=∫01‍θ(s)I+αφy(s)ds=∫01‍θ(s)∫0s‍(s-τ)α-1Γ(α)φy(τ)dτds,∀φ∈E*.
By changing the order of integration results in(17)φJ=∫01‍(∫τ1‍(s-τ)α-1Γ(α)θ(s)ds)φy(τ)dτ=∫01‍h(τ)φy(τ)dτ=φ(∫01‍h(s)y(s)ds),∀φ∈E*.
Thus
(18)J=∫01‍I-αθ(s)y(s)ds=∫01‍u(s)I+αy(s)ds.Definition 15.
Letx:I→E. We define the fractional pseudo-derivative of x of arbitrary order n+α, α∈[0,1), where n∈ℕ0∶={0,1,2,…} by
(19)Dn+αx(t)∶=Dn+1I+1-αx(t),
where D denote the pseudo-differential operator (cf. [22, 25]).The following Lemma is well known in the caseE=ℝ, but to see that it also holds in the vector-valued case, we provide a proof.Lemma 16.
For0<α≤β we have for every weakly continuous function x:[0,1]→E, D0x=x, and
(20)DβI+αx=Dβ-αx.
In particular, whenα=β, (20) means that the operator DαI+α is defined in C(I,Ew) and that Dα is the left-inverse of I+α.Proof.
The first claim, that is,DI+1x=x, follows from the fact that the integral of weakly continuous function is weakly continuous, then pseudo-differentiable with respect to the right endpoint of the integration interval. Let β=n+γ and β-α=m+δ with n,m∈ℕ0 and γ,δ∈[0,1). Then we have, in view of DI1x=x and Lemmas 9 and 10, that
(21)DβI+αx=Dn+1I+1-γI+αx=Dn+1I+1-γ+αx=Dn+1I+1+n-m-δx=Dm+1Dn-mI+n-mI+1-δx=Dm+1I+1-δx=Dβ-αx.Definition 17.
A functionu:I→E is called pseudo-solution of the problem (1) if u∈C[I,Ew] has fractional pseudo-derivative of order α∈(1,2], u(0)=0, u(1)+∫01‍ℑ(τ)u(τ)dτ=l and satisfies
(22)D2φ(I+2-αu(t))+φ(f(t,u(t),Dβu(t)))=0a.e.on[0,1],foreachφ∈E*.The following auxiliary Lemma will be needed in our techniques.Lemma 18.
Ifv∈C[I,Ew] is a pseudo-solution to the problem
(23)Dα-βv(t)+f(t,I+αv(t),v(t))=0,t∈[0,1],α∈(1,2],β∈(0,1],α>1+β,I+βv(1)+∫01‍ℑ(τ)I+βv(τ)dτ=l,v(0)=0,
then u∶=I+βv is a pseudo-solution for the problem (1).Proof.
Letv∈C[I,Ew] be a pseudo-solution to the problem (23) and φ∈E*. As in the proof of Lemma 9 it follows that I+βv exists and the real function φu is continuous for every φ∈E*; moreover
(24)limt→0+φu(t)=limt→0+(I+βφv)(t)=limt→0+∫0t‍(t-s)β-1Γ(β)φv(s)ds=0.
Thus φu(0)=0 for every φ∈E*; that is, u(0)=0. Further
(25)u(1)+∫01‍ℑ(τ)u(τ)dτ=l.
In the view of Lemma 16 we also have
(26)Dαu(t)=(DαI+βv)(t)=Dα-βv(t).If otherwise is not stated, we will assume from now thatα∈(1,2] and β∈(0,1].To obtain the Hammerstein type integral equation modeled off the problem (23), we keep the boundary value problem (23) in mind and we formally put (cf. [11, Lemma 2.3])
(27)v(t)=-I+α-βf(t,I+βv(t),v(t))+ctα-β-1.
In the view of Lemma 13, we obtain
(28)I+βv(t)=-I+αf(t,I+βv(t),v(t))+c[Γ(α-β)Γ(α)]tα-1.
To facilitate our discussion, let q∈[1,∞] be constant with the conjugate exponents p. Suppose ℑ∈Lq[0,1] be a nonnegative real-valued function and f:[0,1]×E×E→E satisfy the following assumptions:  (1)
for eacht∈I=[0,1], f(t,·,·):(I)×E×E→E is weakly-weakly sequentially continuous,(2)
for eachx,y∈C(I,Ew), f(·,x(·),y(·))∈ℋ0p(E),(3)
for anyr>0 and φ∈E* there exist a Pettis integrable function f~:I→E, function Ψ∈Lp[I,ℝ+], and nondecreasing continuous function Ω:[0,∞)→(0,∞) such that |φ(f(t,x,y))|≤|φ(f~(t))|Ω(r)≤∥φ∥Ψ(t)Ω(r) for a.e. t∈I and all (x,y)∈Br×Br.Let us present two remarks about the above assumptions.(i)
For the interesting discussion about the growth conditions for Pettis integrable functions of the above type see [31]. For differential equations with Caputo fractional integrals (i.e., solutions in the space Cm(0,1), ℝd), where m-1<α<m the problem of dominants for considered functions (Assumption (3)) was considered in [32]. Nevertheless, in the paper by Lin rather strong boundedness conditions are investigated (dominants from L2(I) or L4(I) for α>1/2 or bounded functions for α>0) [32, Remark 2.3].(ii)
For convenience of the readers, let us recall the following lemma describing particular sufficient conditions for Pettis integrability off(·,x(·),y(·)) [23, Lemma 15]. It is obvious that additional growth condition for f allows us to characterize the functions from ℋ0p(E).Lemma 19.
Assume thatx is absolutely continuous and f:I×E×E→E. Thus f(·,x(·),y(·)) is Pettis integrable if at least one of the following cases holds:  (a)
f satisfies Carathédory conditions; that is, f(·,x,y) is measurable, f(t,·,·) is continuous in E×E, and there exists an integrable function h:I→ℝ such that ∥f(t,x,y)∥≤h(t) for all x,y∈E and a.e. t∈I,(b)
f is weakly-weakly continuous and E is a weakly sequentially complete space,(c)
f(·,x,y) is weakly measurable, f(t,·,·) is weakly-weakly continuous in E×E and E is a WCG-space (weakly compactly generated space),(d)
f is strongly measurable and there exists a Young function Γ such that limx→∞Γ(x)/x=+∞ and φf∈LΓ(I),(e)
f is strongly measurable and there exists p>1 such that φf∈Lp for each φ∈E* (here f(·,x(·),y(·))∈ℋ0p(E)),(f)
f(·,x,y) is strongly measurable, f(t,·,·) is weakly-weakly sequentially continuous in E×E, and f is bounded,(g)
f(·,x(·),y(·)) is strongly measurable, E contains no copy of c0, and f is bounded.Now, we would like to pay our attention to solve (27) for c by
(29)I+βv(1)+∫01‍ℑ(τ)I+βv(τ)dτ=l.
It follows that
(30)c[Γ(α-β)Γ(α)]-∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds=l,-∫01‍ℑ(τ)(∫0τ‍(τ-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)c[Γ(α-β)Γ(α)]τα-β-1-∫0τ‍(τ-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds)dτ=l-c[Γ(α-β)Γ(α)]∫01‍ℑ(τ)τα-β-1dτ+∫01‍ℑ(τ)(∫0τ‍(τ-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds)dτ.
Therefore
(31)c[Γ(α-β)Γ(α)](1+γ)=l+∫01‍ℑ(τ)(∫0τ‍(τ-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds)dτ+∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds,
where
(32)γ=∫01‍ℑ(τ)τα-β-1dτ.
Then (in account of Lemma 14), we have
(33)c=Γ(α)Γ(α-β)(1+γ)×[l+∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)dsl+∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds+∫01‍h(s)f(s,I+βv(s),v(s))ds].
Here h=I-α-βℑ. Substituting c into (27), one has
(34)v(t)=-I+α-βf(t,I+βv(t),v(t))+lΓ(α)tα-β-1Γ(α-β)(1+γ)+Γ(α)tα-β-1Γ(α-β)(1+γ)×∫01‍[(1-s)α-β-1Γ(α-β)+h(s)]f(s,I+βv(s),v(s))ds=lΓ(α)tα-β-1Γ(α-β)(1+γ)+∫0t‍[Γ(α)(t(1-s))α-β-1(1+γ)(Γ(α-β))2-(t-s)α-β-1Γ(α-β)]f(s,I+βv(s),v(s))ds+∫t1‍Γ(α)(t(1-s))α-β-1(1+γ)(Γ(α-β))2×f(s,I+βv(s),v(s))ds+Γ(α)tα-β-1Γ(α-β)(1+γ)∫01‍h(s)f(s,I+βv(s),v(s))ds.
Thus
(35)v(t)=p(t)+∫01‍G(t,s)f(s,I+βv(s),v(s))ds,t∈[0,1],
where p(t)=lΓ(α)tα-β-1/Γ(α-β)(1+γ). The Green function G is given by G(t,s)=G1(t,s)+G2(t,s) with
(36)G1(t,s)∶={Γ(α)Γ(α-β)[(t(1-s))α-β-1(1+γ)Γ(α-β)-(t-s)α-β-1Γ(α)],0≤s≤t≤1,Γ(α)Γ(α-β)[(t(1-s))α-β-1(1+γ)Γ(α-β)],0≤t≤s≤1,G2(t,s)∶=Γ(α)tα-β-1Γ(α-β)(1+γ)h(s),t,s∈[0,1].
Since α-β>1, the following can be easily seen.Lemma 20.
The mapt→G(t,·) is continuous from [0,1] to Lq[0,1].Remark 21.
We point out that ifE is reflexive, it is not necessary to assume any compactness conditions on the nonlinearity of f. This will be due to [33, Lemma 2] and the fact that a subset of reflexive Banach spaces is weakly compact if and only if it is weakly closed and norm bounded.


## 3. Weak Solutions of the Hammerstein Integral Equation

In this section, in the light of the Assumptions (1)–(3) imposed onf, we proceed to obtain a result which relies on the fixed point Theorem 3 to ensure the existence of weak solution to the integral equation (35). For the sake of convenience, we introduce the following.Definition 22.
By a solution to (35) we mean a function v∈C(I,E) which satisfies the integral equation (35). This is equivalent to the finding v∈C(I,E) with
(37)φ(v(t))=φ(p(t)+∫01‍G(t,s)f(s,I+βv(s),v(s))ds),t∈I∀φ∈E*.We need to explain why we consider continuous solutions. By the properties of the Pettis integral this should be weakly continuous function. Sincep is continuous and we impose (local) boundedness hypothesis for f, our solutions are strongly continuous (cf. [22]). We restrict our attention to the space C(I,E) and then our integral operators will be defined on this space. In contrast to the case of weakly-weakly continuous functions f, we need to replace the space C(I,Ew) endowed with its topology of weak uniform convergence by the space of (strongly) continuous functions C(I,E) with its weak topology. We will utilize in our proofs some characterization of its weak topology.Now, we are in the position to state and prove the first existence result.Theorem 23.
Assume thatα∈(1,2], β∈(0,1) with α>1+β and ℑ∈Lq([0,1]) be a nonnegative real-valued function. If the Assumptions (1)–(3) hold along with
(38)(supt∈[0,1]∫01‍|G(t,s)|Ψ(s)ds)limsupr→∞Ω(r)r<Γ(1+β),
then the integral equation (35) has at least one solution v∈C[I,E].Proof.
First of all, observe the expression ofG and note that the following implications:
(39)ℑ∈Lq[0,1]⟹h∈Lq[0,1]⟹G(t,·)∈Lq[0,1],t∈[0,1],
hold. Consequently G(t,·)Ψ(·)∈L1[0,1] for any t∈[0,1]. Let
(40)ρ=(supt∈[0,1]∫01‍|G(t,s)|Ψ(s)ds)limsupr→∞Ω(r)r.
Consider the set S of real numbers r≥0 which satisfy the inequality
(41)rΓ(1+β)≤∥p∥0+Ω(r)(supt∈[0,1]∫01‍|G(t,s)|Ψ(s)ds).
Then S is bounded above; that is, there exists a constant R0 with
(42)r≤R0∀r∈S.
To see this, suppose (42) is false. Then there exists a sequence 0≠rn∈S with rn→∞ as n→∞ and
(43)Γ(1+β)≤∥p∥0rn+Ω(rn)rn(supt∈[0,1]∫01‍|G(t,s)|Ψ(s)ds).
Since limsup(sn+tn)≤limsupsn+limsuptn for any sequences sn≥0,tn≥0, we have ρ≥Γ(1+β). This contradicts (38). Then, for every R>R0 the inequality
(44)∥p∥0+Ω(R)(supt∈[0,1]∫01‍|G(t,s)|Ψ(s)ds)<Γ(1+β)R
holds, which is in contradiction with R∈S and then contradicts (42).
Now, define the operatorT:C[I,E]→C[I,E] by
(45)Tv(t)∶=p(t)+∫01‍G(t,s)f(s,I+βv(s),v(s))ds,t∈[0,1].
We remark that for v∈C[I,E] we have that, by Lemma 9, I+βv is weakly continuous, and consequently, f(·,I+βv(·),v(·))∈ℋ0p(E) (Assumption (2)). Since s↦G(t,s)∈Lq(I), for all t∈[0,1], G(t,·)f(·,I-βv(·),v(·)) is Pettis integrable for all t∈[0,1] (thanks to Proposition 6), and thus the operator T makes sense. Note that T is well defined. To see this, let t1,t2∈[0,1] with t2>t1. Since β∈(0,1), we deduce that if ∥v∥≤σ1, then ∥v∥<σ and ∥I+βv∥≤σ, where σ=σ1/Γ(1+β). Without loss of generality, assume Tv(t2)-Tv(t1)≠0. Then there exists (as a consequence of Proposition 2 ) φ∈E* with ∥φ∥=1 and ∥Tv(t2)-Tv(t1)∥=φ(Tv(t1)-Tv(t1)).
Putting the Assumption (3) in mind, one can write the following chain of inequalities:(46)∥Tv(t2)-Tv(t1)∥=φ(Tv(t2)-Tv(t1))≤φ(p(t2)-p(t1))+Ω(σ)∫01‍|G(t2,s)-G(t1,s)|Ψ(s)ds≤∥p(t2)-p(t1)∥+Ω(σ)(∫01‍|G(t2,s)-G(t1,s)|qds)1/q×(∫01‍|Ψ(s)|pds)1/p≤∥p(t2)-p(t1)∥+Ω(σ)∥G(t2,·)-G(t1,·)∥q∥Ψ∥p≤∥p(t2)-p(t1)∥+Ω(σ)(∥G1(t2,·)-G1(t1,·)∥q+∥G2(t2,·)-G2(t1,·)∥q)∥Ψ∥p≤∥p(t2)-p(t1)∥+Ω(σ)(Γ(α)∥(t2α-β-1-t1α-β-1)h(·)∥qΓ(α-β)(1+γ)∥G1(t2,·)-G1(t1,·)∥q+Γ(α)∥(t2α-β-1-t1α-β-1)h(·)∥qΓ(α-β)(1+γ))∥Ψ∥p.
Then
(47)∥Tv(t2)-Tv(t1)∥≤∥p(t2)-p(t1)∥+Ω(σ)(Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ)∥G1(t2,·)-G1(t1,·)∥q+Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ))∥Ψ∥p.
Therefore we deduce, in the view of Lemma20, that T maps C[I,E] into itself.
LetQ∈C[I,E] be the convex, closed and equicontinuous subset (required by Theorem 3). Define this set by(48)Q∶={+Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ)v∈C[I,E]:∥v∥0≤R0,∀t1,t2∈[0,1]wehave∥v(t2)-v(t1)∥≤∥p(t2)-p(t1)∥+Ω(R0Γ(1+β))×(Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ)∥G1(t2,·)-G1(t1,·)∥q+Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ))∥Ψ∥p}.
We claim that T restricted to the set Q maps this set into itself (i.e., T:Q→Q) and is weakly-weakly sequentially continuous. Once the claim is established, Theorem 3 guarantees the existence of a fixed point of T. Hence the integral equation (35) has a solution in C[I,E].
We start by showing thatT:Q→Q. To see this, take v∈Q,t∈[0,1]. Since β∈(0,1), we deduce that ∥v∥≤R0<R0/Γ(1+β), ∥Iβv∥≤R0/Γ(1+β). The monotonicity of Ω and the inequality (47) imply that
(49)∥Tv(t2)-Tv(t1)∥≤∥p(t2)-p(t1)∥+Ω(R0Γ(1+β))×(Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ)∥G1(t2,·)-G1(t1,·)∥q+Γ(α)|t2α-β-1-t1α-β-1|∥h∥qΓ(α-β)(1+γ))∥Ψ∥p.
Now, without loss of generality, assumeTv(t)≠0. Then there exists (consequence of Proposition 2) φ∈E* with ∥φ∥=1 and ∥Tx(t)∥=φ(Tv(t)). By the Assumption (3), we obtain
(50)∥Tv(t)∥≤φ(p(t))+φ(∫01‍G(t,s)f(s,I+βv(s),v(s))ds)≤∥p(t)∥+supt∈[0,1]∫01‍|G(t,s)|·|φf(s,I+βv(s),v(s))|ds≤∥p∥0+Ω(R0Γ(1+β))(supt∈[0,1]∫01‍G(t,s)Ψ(s)ds)<R0Γ(1+β)Γ(1+β)=R0.
Therefore ∥Tv∥0=supt∈[0,1]∥Tv(t)∥≤R0. Hence T:Q→Q.
We need to prove now thatT:Q→Q is weakly-weakly sequentially continuous. Let us recall that the weak convergence in Q⊂C(I,E) is exactly the weak pointwise convergence. Let (vn) be a sequence in Q weakly convergent to v. Then vn(t)→v(t) in Ew for each t∈[0,1]. Since Q is closed, by Lemma 1 we have v∈Q.
Fixt∈I and note, in the view of Lebesgue dominated convergence theorem for the Pettis integral (see [31, 34]), that I+βvn(s)→I+βv(s) in Ew. Let us recall that the topology on C(I,Ew) on equicontinuous subsets coincides with the topology of weak pointwise convergence. Since f satisfies Assumption (1), we have f(t,I-βvn(t),vn(t)) converging weakly to f(t,I-βv(t),v(t)); hence again the Lebesgue dominated convergence theorem for Pettis integral yields Tvn(t) converging weakly to Tv(t) in E, but Q is an equicontinuous subset of C(I,E), and then T:Q→Q is weakly-weakly sequentially continuous. Applying now Theorem 3, we conclude that T has a fixed point in Q, which completes the proof.Let us present a multivalued problem:(51)Dα-βv(t)∈F(t,I+αv(t),v(t)),t∈[0,1],α∈(1,2],β∈(0,1],α>1+β,I+βv(1)+∫01‍ℑ(τ)I+βv(τ)dτ=l,v(0)=0.Some basic results for multivalued boundary value problems with Pettis integrals are due to Maruyama [35], Azzam et al. [36], Azzam-Laouir and Boutana [37], and Satco [38]. However these results are devoted to study the standard case α-β=2 and three-point boundary conditions. Our result is an essential extension for the previous ones.Byck(E) and cwk(E) we denote the family of all nonempty convex compact and nonempty convex weakly compact subsets of E, respectively. For every nonempty convex bounded set C⊂E the support function of C is denoted by s(·,C) and defined on E* by s(φ,C)=supx∈Cφx, for each φ∈E*.Definition 24.
A multifunctionF:E→2E with nonempty, closed values is weakly sequentially upper hemicontinuous if and only if for each φ∈E*s(φ,G(·)):E→ℝ is sequentially upper semicontinuous from (E,w) into ℝ.In the remaining part of the paper a multifunctionF is supposed to be Pettis integrable in the sense of Aumann.Definition 25.
The Aumann-Pettis integral of a multifunctionF:I→E is
(52)IA:=(AP)∫I‍F(s)ds={(P)∫I‍f(s)ds:f∈SFPe},
where SFPe denotes the set of all Pettis integrable selections of F provided that this set is not empty.Let us note that the multivalued Pettis integral can be defined by other methods. The above definition is the best choice for our consideration. This can be deduced from the following theorem.Theorem 26 (see [39]).
LetF:I→cwk(E)[ck(E)] be measurable and scalarly integrable multifunction (i.e., the support functions are real-valued integrable functions). Then the following statements are equivalent:(a)
the set{φf:φ∈B(E*),f∈SFPe} is uniformly integrable,(b)
every measurable selection ofF is Pettis integrable,(c)
for every measurable subsetA of I the Aumann-Pettis integral IA belongs to cwk(E) [ck(E)] and, for every φ∈E*, one has
(53)s(φ,IA)=∫A‍s(φ,F(s))ds.Taking into account Theorem8, we are able to add one more condition to the above theorem, which seems to be important in our consideration. Since α-β>1, G is continuous, and by taking arbitrary Pettis integrable selection we obtain Pettis integrability of G(t,·)f(·).Theorem 27.
Each of the conditions from Theorem26 implies the following:(d)
for everyt the multifunction F is fractionally Aumann-Pettis integrable; that is, I+αF(t)∶={∫0t‍((t-s)α-1/Γ(α))f(s)ds,f∈SFPe} belongs to cwk(E) [ck(E)].Let us recall that we restrict ourselves to the case of the (left-sided) fractional Pettis-integral.Note that for multivalued mappings we will utilize Kakutani's fixed point theorem (for continuity concepts see [27]).Theorem 28 (see [27]).
IfC is a nonempty weakly compact convex subset of E and F:C→2C is sequentially weakly upper semi-continuous, then there exists a fixed point of F; that is, x∈C with x∈F(x).An immediate consequence of the above theorems as well as our main theorems is the following result.Theorem 29.
Assume thatE is separable. Let F:I×E×E→2E with nonempty convex and weakly compact values satisfy the following: (a)
F(t,·,·):E×E→E is weakly sequentially upper hemicontinuous for each t∈I,(b)
F(·,x,y) has a weakly measurable selection for each (x,y)∈E×E,(c)
F(t,x,y)⊂H(t) a.e. for some cwk-(E-) valued Pettis integrable multifunction G.
Then there exists at least one pseudo-solution of the Cauchy problem  (51) on I.We will follow the idea of the proof for the single-valued problem. Let us only sketch the main steps of the proof.Note that the Assumption (c) implies weak compactness ofSHPe and separability of the space E are sufficient to replace reflexivity of E (as announced in the preliminary part).In this proof we need to define the multifunctionT:C[I,E]→2C[I,E] by
(54)Tv(t)∶=p(t)+∫01‍G(t,s)F(s,I+βv(s),v(s))ds,t∈(I).
By [40, Lemma 3.2] our assumptions (a)–(c) imply that the set SFPe(·,v(·)) is nonempty for arbitrary v∈C(I,E) and the multivalued Nemytskii operator is well defined, so by Theorem 27 the Aumann-Pettis integral of F is nonempty too.LetW={f∈ℋ01(E):f(t)∈H(t)a.e.onI} and U={xf∈C(I,E):xf(t)=p(t)+∫0t‍G(t,s)f(s)ds,t∈I,f∈W}.Forf∈W and φ∈E* we have φf≤s(φ,H). Then, by our assumptions, W is Pettis uniformly integrable. Thus for arbitrary x∈U and t,τ∈I there exists an appropriate f∈W and
(55)∥x(t)-x(τ)∥=sup∥φ∥≤1φ(x(t)-x(τ))=sup∥φ∥≤1(∫0t‍φf(s)ds-∫0τ‍φf(s)ds)=sup∥φ∥≤1∫τt‍φf(s)ds≤sup∥φ∥≤1∫τt‍s(φ,H(s))ds.
By uniform Pettis integrability of W it follows that U is an equicontinuous subset of C(I,E). The property of the multivalued Pettis integral gives us the convexity of  U. Then U is nonempty, convex, bounded, and equicontinuous in C(I,E).AsSHPe is sequentially compact for the topology induced by the tensor product L∞⊗E*, the U¯ is closed. Since U is convex, by Mazur's lemma (Lemma 1) U¯ is weakly closed. Thus by a weak version of Ascoli's theorem U¯ is weakly compact in C(I,E).As the setU is strongly equicontinuous, then for each M>0 there exists α∈I such that for each t∈I and f∈SFPe we have ∥∫0t‍f(s)ds∥≤M.ThenT restricted to a ball with radius R0=M+∥p∥ (as in the previous proof) has nonempty, closed, convex and weakly compact values. As a domain Q for T we put U¯∩BR0. By repeating the proof from [40, Theorem 3.3] we are able to show that T has weakly-weakly sequentially closed graph. Restricted to a weakly compact set Q an operatot T is sequentially weakly upper semi-continuous. This means, that the Kakutani fixed point theorem (Theorem 28) gives us a fixed point of T. The proof is complete.


## 4. Pseudo-Solutions to Fractional Order Boundary Value Problem

In this section, we are looking for sufficient conditions to ensure the existence of pseudo-solution to the boundary value problem (1) under the Pettis integrability assumption imposed on f. In order to obtain the existence of solutions of the problem (1), we can make use of Theorem 23.Theorem 30.
Let the assumptions of Theorem23 be satisfied. Then the boundary value problem (1) has at least one pseudo-solution u∈C(I,Ew).Proof.
Firstly, we remark that, for anyv∈C[I,E], we have (according to Proposition 6) that v(·)ℑ(·)∈P[I,E] for ℑ∈Lq(I),  q∈[1,∞]. Thus the integral boundary condition makes sense.
In account of Theorem23 it can be easily seen that the integral equation (35) has a solution v∈C[I,E]. Let v be a weak solution of (35). Then
(56)v(t)=tα-β-1lΓ(α)Γ(α-β)(1+γ)+∫01‍[G1(t,s)+G2(t,s)]×f(s,I+βv(s),v(s))ds=-∫0t‍(t-s)α-β-1Γ(α-β)f(s,I+βv(s),v(s))ds+ctα-β-1,c=Γ(α)Γ(α-β)(1+γ)×[l+∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)ds+∫01‍h(s)f(s,I+βv(s),v(s))ds∫01‍(1-s)α-β-1f(s,I+βv(s),v(s))Γ(α-β)].
By Lemma14 and using G1(0,s)=G2(0,s)=0, a straightforward estimates show that
(57)v(0)=0,I+βv(1)+∫01‍ℑ(τ)I+βv(τ)dτ=l.
Furthermore, we have
(58)v(t)=-I+α-βf(t,I+βv(t),v(t))+ctα-β-1.
Thus for any φ∈E* we have
(59)φv(t)=-φ(I+α-βf(t,I+βv(t),v(t)))+φctα-β-1=-I+α-βφ(f(t,I+βv(t),v(t)))+φctα-β-1.
Operating by I+2-(α-β) on both sides of (59) and using the properties of fractional calculus in the space L1[0,1] (see,  e.g., [20, 30]) result in
(60)I+2-(α-β)φv(t)=-I+2φ(f(t,I+βv(t),v(t)))+φcΓ(α-β)Γ(2)t.
Therefore
(61)φ(I+2-(α-β)v(t))=-I+2φ(f(t,I+βv(t),v(t)))+φcΓ(α-β)Γ(2)t.
Thus
(62)d2dt2φ(I+2-(α-β)v(t))=-φ(f(t,I+βv(t),v(t)))a.e.on[0,1].
That is,v has the fractional pseudo-derivative of order α-β∈(1,2) and satisfies
(63)Dα-βv(t)=-f(t,I+βv(t),v(t))on[0,1].
Therefore v is a pseudo-solution to the problem (23). This together with Lemma 18 implies that the problem (1) has a pseudo-solution u∈C(I,Ew) which completes the proof.Now, we consider an example to illustrate our result.Example 31.
Letℑ∈L∞(I) and assume that E∶=ℓ2(I) be the space of countably nonzero functions on I∶=[0,1] that are square-summable, under the ℓ2-norm. This is even a reflexive space. For each t∈I we define g(t)∶=et∈ℓ2(I) by
(64)et(s)∶={1,ifs=t,0,ifs≠t.
This function is Pettis, but not Bochner, integrable (it is not even strongly measurable [22, 41]), and for any φ∈ℓ2(I) we have [22, 41]
(65)φet=∑τ∈[0,1]‍φ(τ)et(τ)={0,ifτ≠t,φ(τ),ifτ=t.Since(ℓ2(I))*=ℓ2(I), the function φet is only countably nonzero for each φ∈(ℓ2(I))*. Hence φg=0 a.e. (with respect to the Lebesgue measure).Now, we investigate the existence of pseudo-solutions for the problem(66)Dαu(t)+μDβu(t)+δu(t)=g(t),t∈[0,1],δ,μ∈ℝ,α∈(1,2],β∈(0,1),α>1+β,u(1)+∫01‍ℑ(τ)u(τ)dτ=l,u(0)=0.
Let us define the function f:I×E×E→E by f(t,x,y)=μx+δy-g(t). Remark that for any x,y∈C[I,E], f is Pettis, but not Bochner, integrable and satisfies, by suitable choice of δ, μ, all requirements of Theorem 30 with ψ≡1, Ω(r)=(δ+μ)r, p=1, and q=∞.Therefore one cannot expect the existence of weak or strong solutions to the problem (66). In view of Theorem 30 the existence of pseudo-solutions is guaranteed. Since f satisfies the requirements of Theorem 30 with p=1, q=∞, the problem (66) has a pseudo-solution u∈C(I,Ew) given by u=I+βv, where v denotes the weak solution to the problem
(67)v(t)=lΓ(α)tα-β-1Γ(α-β)(1+γ)+∫01‍G(t,s)[μv(s)+δI+βv(s)-g(s)]ds,t∈(I).
That is, v satisfies the problem
(68)v(t)=lΓ(α)tα-β-1Γ(α-β)(1+γ)+∫01‍G(t,s)[μv(s)+δI+βv(s)]ds-∫01‍G(t,s)g(s)ds,t∈(I).
Since ∫01‍G(t,s)φg(s)ds=0, for each φ∈ℓ2(I), we obtain ∫01‍G(t,s)g(s)ds=0. Hence
(69)v(t)=lΓ(α)tα-β-1Γ(α-β)(1+γ)+∫01‍G(t,s)[μv(s)+δI+βv(s)]ds=lΓ(α)tα-β-1Γ(α-β)(1+γ)-I+α-β[μv(s)+δI+βv(s)]+ctα-β-1,
where
(70)c=Γ(α)Γ(α-β)(1+γ)×[l+∫01‍(1-s)α-β-1[μv(s)+δI+βv(s)]Γ(α-β)ds∫01‍(1-s)α-β-1[μv(s)+δI+βv(s)]Γ(α-β)+∫01‍h(s)[μv(s)+δI+βv(s)]ds].
Therefore
(71)v(t)+μI+α-βv(t)+δI+αv(t)=c*tα-β-1,
where
(72)c*=(c+lΓ(α)Γ(α-β)(1+γ)).
Further
(73)I+βv(t)+μI+α-βI+βv(t)+δI+α+βv(t)=c*Γ(α-β)Γ(α)tα-1.
Now if u=I+βv, we obtain
(74)u(t)+μI+α-βu(t)+δI+αu(t)=c*Γ(α-β)Γ(α)tα-1.
Thus for every φ∈(ℓ2(I))*, we have
(75)φu(t)+μI+α-βφu(t)+δI+αφu(t)=φc*Γ(α-β)Γ(α)tα-1.
Whence
(76)I+2-αφu(t)+μI+2-βφu(t)+δI+2φu(t)=φc*Γ(α-β)Γ(2)t.
Moreover,
(77)d2dt2φ(I+2-αu(t))+d2dt2φI+2-βu(t)+δφu(t)=0.
Since
(78)d2dt2φI+2-βu(t)=d2dt2I+2-βφu(t)=ddt(ddtI+1I+1-βφu(t))=ddtI+1-βφu(t)=ddtφI+1-βu(t),
we arrive, for every φ∈E*, at Dαφu(t)+μDβφu(t)+δφu(t)=0 for all t∈[0,1] while φg=0 a.e.Finally(79)Dαφu(t)+μDβφu(t)+δu(t)=φet,a.e.[0,1].Thenu is a pseudo- (but not a weak or strong) solution to the problem (66).



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*Source: 428094-2013-06-18.xml*