# Blow-Up Solutions for a Singular Nonlinear Hadamard Fractional Boundary Value Problem

Authors: Imed Bachar; Said Mesloub
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We consider singular nonlinear Hadamard fractional boundary value problems. Using properties of Green’s function and a fixed point theorem, we show that the problem has positive solutions which blow up. Finally, some examples are provided to explain the applications of the results.


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## Body

## 1. Introduction

Fractional-order differential equations appear extensively in a variety of applications in science and engineering; see, for instance, [1–13] and the references therein.In [14], Hadamard introduced a new definition of fractional derivatives which differs from the Riemann-Liouville and Caputo fractional derivatives in the sense that its kernel integral contains the logarithmic function of an arbitrary exponent. Hadamard fractional derivatives are viewed as a generalization of the operator δ=xd/dx. For further details, properties, and generalizations of this type of derivative, we refer the reader to [5, 15–21] and the references therein.The study of existence, uniqueness, and global asymptotic behavior of a continuous solution of fractional differential equations involving Hadamard fractional derivatives has been investigated by several researchers; see, for example, [22–32].In [23], Ahmad and Ntouyas studied the following problem:
(1)HDαur−gr,ur=0,r∈1,e,u1=0,ue=1Γβ∫1elnesβ−1ussds,β>0,where HDα is the Hadamard fractional derivative order α∈1,2 and g:1,e×ℝ⟶ℝ is a continuous function satisfying
(2)gr,x−gr,y≤Lx−y,foreachr∈1,eandx,y∈ℝ,where L>0 denotes a convenient Lipschitz constant.The authors used the classical Banach fixed point theorem to obtain the existence and uniqueness of a solution for the abovementioned problem.In [24], the authors studied the existence of solutions for a fractional boundary value problem involving Hadamard-type fractional differential inclusions and integral boundary conditions. Their approach was based on standard fixed point theorems for multivalued maps. In [25], the authors used some classical ideas of fixed point theory to investigate the existence and uniqueness of solutions of a boundary value problem comprising nonlinear Hadamard fractional differential equations and nonlocal nonconserved boundary conditions in terms of the Hadamard integral. In [15], the authors studied a Cauchy problem for a differential equation with a left Caputo-Hadamard fractional derivative. By using Banach’s fixed point theorem, they proved the existence and uniqueness of the solution in the space of continuously differentiable functions.The primary objective of this paper is to address the existence and qualitative properties of a solution for the following problem:(3)HDαur−λfr,ur=0,r∈1,e,u>0,in1,e,limr→1+lnr2−αur=a>0,ue=b>0,where HDα is the Hadamard fractional derivative of order α∈1,2,λ≥0, and f satisfies(H1) f:1,e×0,∞⟶0,∞) is continuous, such that for each each fixed r∈1,e,s⟶fr,s is nondecreasing on 0,∞.(H2) For all c>0,∫1elnr1−lnrα−1fr,clnrα−2dr<∞.The following are some examples of functions that satisfy hypotheses (H1) and (H2).
(i)
fr,s=s, which is not a Lipschitz function on 0,∞(ii)
fr,s=lnrγ2−α1−lnr−βsγ, where 0<β<α and γ≥0. It is to be noted that fr,s is singular at r=1(iii)
fr,s=prsγ, where p is any nonnegative continuous function on 1,e and γ∈0,2/2−α.Before stating our main result, we explain some notations.


### 1.1. Notations

(i)
B+1,e≔gg:1,e⟶0,∞isameasurablefunction. If α∈1,2, then(ii)
Cα,ln1,e≔g:lnr2−αgr∈C1,e(iii)
Gαr,s is Green’s function of the operator u⟶−HDαu on 1,e with limr→1+lnr2−αur=0 and ue=0.(iv)
u0r≔alnrα−2−lnrα−1+blnrα−1 is the unique solution of the problem
(4)HDαur=0,r∈1,e,u>0,in1,e,limr→1+lnr2−αur=a>0,ue=b>0(v)
AssumingH1 and H2, we define
(5)λ0≔infr∈1,eu0r∫1eGαr,sfs,u0s/sdsIt will be proven thatλ0>0.The main result of this paper can be stated as follows.Theorem 1.
Letα∈1,2 and assume that hypotheses (H2) and (H2) are satisfied. Then, for λ∈0,λ0, problem (3) has a solution uλ∈Cα,ln1,e satisfying for all r∈1,e,
(6)1−λλ0mina,blnrα−2≤uλr≤maxa,blnrα−2.Remark 2.
(i)
Forα∈1,2, we have limr→1+uλr=∞.(ii)
Ifλ=0, then u0 satisfies (6).The remainder of this paper is organized as follows. In Section2, some relevant properties of Hadamard fractional calculus are presented. Additionally, we construct Green’s function and establish certain interesting inequalities. Theorem 1 is proven in Section 3. To illustrate our existence results, some examples are provided at the end of Section 3.


## 1.1. Notations

(i)
B+1,e≔gg:1,e⟶0,∞isameasurablefunction. If α∈1,2, then(ii)
Cα,ln1,e≔g:lnr2−αgr∈C1,e(iii)
Gαr,s is Green’s function of the operator u⟶−HDαu on 1,e with limr→1+lnr2−αur=0 and ue=0.(iv)
u0r≔alnrα−2−lnrα−1+blnrα−1 is the unique solution of the problem
(4)HDαur=0,r∈1,e,u>0,in1,e,limr→1+lnr2−αur=a>0,ue=b>0(v)
AssumingH1 and H2, we define
(5)λ0≔infr∈1,eu0r∫1eGαr,sfs,u0s/sdsIt will be proven thatλ0>0.The main result of this paper can be stated as follows.Theorem 1.
Letα∈1,2 and assume that hypotheses (H2) and (H2) are satisfied. Then, for λ∈0,λ0, problem (3) has a solution uλ∈Cα,ln1,e satisfying for all r∈1,e,
(6)1−λλ0mina,blnrα−2≤uλr≤maxa,blnrα−2.Remark 2.
(i)
Forα∈1,2, we have limr→1+uλr=∞.(ii)
Ifλ=0, then u0 satisfies (6).The remainder of this paper is organized as follows. In Section2, some relevant properties of Hadamard fractional calculus are presented. Additionally, we construct Green’s function and establish certain interesting inequalities. Theorem 1 is proven in Section 3. To illustrate our existence results, some examples are provided at the end of Section 3.


## 2. Preliminaries

We recall some relevant properties concerning Hadamard fractional derivative. For more details, the reader can see Section2.7 of [19].Definition 3.
The Hadamard fractional integral of orderγ>0 of the function h is defined as
(7)HIγhr≔1Γγ∫1rlnrsγ−1hssds,1≤r≤e.Forγ=0, we define HI0h=h.Definition 4.
Letγ>0 and γ its integer part. The Hadamard fractional derivative of order γ of the function h is defined as
(8)HDγhr≔δn1Γn−γ∫1rlnrsn−γ−1hssds,1≤r≤e,where n=γ+1 and δ=rd/dr.Example 5.
(Property 2.24 of [19]).Ifγ,σ>0, then
(9)HIγlnsσ−1r=ΓσΓσ+γlnrσ+γ−1,HDγlnsσ−1r=ΓσΓσ−γlnrσ−γ−1.In particular, ifσ=1 and γ∈0,1, then HDγ1r=1/Γ1−γlnr−γ.Lemma 6.
(see [5]).
Letβ>γ>0 and h∈C1,e∩L11,e. Then,
(i)
HDγHIβh=HIβ−γh and HDγHIγh=h(ii)
The equalityHDγhr=0 is valid on 1,e if, and only if,
(10)hr=∑mj=1cjlnrγ−j,where cj∈ℝ,j=1,⋯,m, and m is the smallest integer greater than or equal to γ.(iii)
IfHDγh∈C1,e∩L11,e, then
(11)HIγHDγhr=hr+∑mj=1cjlnrγ−j,where cj∈ℝ,j=1,⋯,m, and m is the smallest integer greater than or equal to γ.Lemma 7.
Letα∈1,2 and h∈C1,e.The unique solution of the problem(12)HDαur+hr=0,1<r<e,limr→1+lnr2−αur=0,ue=0,is given by
(13)ur=∫1eGαr,shssds,where
(14)Gαr,s=1Γαlnr−lnrlnsα−1−lnrsα−1,1≤s≤r≤e,lnr−lnrlnsα−1,1≤r≤s≤e.Proof.
By Lemma6, the solution of problem (12) can be written as
(15)ur=c1lnrα−1+c2lnrα−2−1Γα∫1rlnrsα−1hssds.Sincelimr→1+lnr2−αur=0 and ue=0, we obtain c2=0 and
(16)c1=1Γα∫1elnesα−1hssds.Therefore,(17)ur=∫1elnrα−1Γαlnesα−1hssds−∫1r1Γαlnrsα−1hssds=∫1eGαr,shssds,where Gαr,s is given by (14).In Figure1, we give the representation of the Green function G3/2t,s with the contours and the projections on some coordinate planes. In particular, one can see that G3/2t,s is nonnegative.Figure 1
Gαt,s for α=3/2.
(a)
Gαt,s and contours(b)
Projection ontz(c)
Projection onszLemma 8.
Let1<α≤2. Then,
(i)
Gαr,s∈C1,e×1,e(ii)
On1,e×1,e, one has
(18)α−1Hαr,s≤ΓαGαr,s≤Hαr,s,where Hαr,s≔lnr−lnrlnsα−2minlnr,lns1−maxlnr,lns.In particular,Gαr,s≥0(iii)
On1,e×1,e, one has
(19)α−1lns−lnslnrlnr−lnrlnsα−1≤ΓαGαr,s,ΓαGαr,s≤lns−lnslnslnr−lnrlnsα−2(iv)
For eachr,ξ,s∈1,e, the following holds:
(20)Gαr,ξGαξ,sGαr,s≤1α−1Γαlnξα−11−lnξα−1.Proof.
It is easy to check that (i) holds.To prove (ii), forr,s∈1,e, we have
(21)Gαr,s=1Γαlnr−lnrlnsα−1−lnr−lns+α−1,=1Γαlnr−lnrlnsα−11−lnr−lns+lnr−lnrlnsα−1,where lnr−lns+=maxlnr−lns,0.Therefore, inequalities in (18) follow from the fact that
(22)α−11−ξ≤1−ξα−1≤1−ξ,forξ∈0,1.By using (18) and the fact that lnrlns≤minlnr,lns≤lns and 1−lnr1−lns≤1−maxlnr,lns≤1−lns, we obtain (19).Next, we aim at proving (iv). Let r,ξ,s∈1,e and put
(23)ρr,s≔minlnr,lns1−maxlnr,lns.From (18), we have
(24)Gαr,ξGαξ,sGαr,s≤lnξα−21−lnξα−2α−1Γαρr,ξρξ,sρr,s.By symmetry, one can verify that(25)ρr,ξρξ,sρr,s≤lnξ1−lnξ.Hence, the required results follow from (24) and (25).


## 3. Proof of Theorem 1

We aim at proving Theorem1. First, we need to establish some preliminary results. For 1<α≤2, we denote by
(i)
Jα,ln≔g∈B+1,e:∫1e1−lnξα−1lnξα−1gξdξ<∞(ii)
Forg∈B+1,e,(26)Ag≔supr,s∈1,e∫1eGαr,ξGαξ,sGαr,sgξξdξ(iii)
Forg∈B+1,e,(27)Wgr≔∫1eGαr,sgssds,forr∈1,e(iv)
Fora,b>0, we recall that u0r≔alnrα−2−lnrα−1+blnrα−1 is the unique solution of problem (4).Note that forr∈1,e,
(28)mina,blnrα−2≤u0r≤maxa,blnrα−2.We recall thatCα,ln1,e≔g:1,e⟶ℝ,lnξ2−αg∈C1,e.Proposition 9.
Letα∈1,2 and g∈B+1,e, then
(i)
Wg∈Cα,ln1,e⇔∫1e1−lnsα−1lnsgsds<∞(ii)
Letg be such that the function s⟶1−lnsα−1lnsgs∈C1,e∩L11,e, then Wg∈Cα,ln1,e and it is the unique solution of the problem
(29)HDαur=−gr,1<r<e,limr→1+lnr2−αur=0,ue=0.Proof.
(i)
The property follows from Lemma8 (ii).(ii)
From (i),Wg∈Cα,ln1,e and by using again Lemma 8 (ii), we have
(30)Wgr≤1Γαlnrα−2∫1elns1−lnsα−1gssds,which implies by Example 5 (i) that HI2−αWg is bounded on 1,e.Therefore, we have(31)HI2−αWgr=1Γ2−α∫1rlnrs1−αWgssds=1Γ2−α∫1rlnrs1−α1s∫1eGαs,ξgξξdξds=∫1eKr,ξgξξdξ,where
(32)Kr,ξ=1Γ2−α∫1rlnrs1−αGαs,ξsds.We claim that(33)Kr,ξ=1−lnξα−1lnr−lnr−lnξ+,where lnr−lnξ+=maxlnr−lnξ,0.Indeed, from (32) and (14), we have
(34)Kr,ξ=1ΓαΓ2−α∫1r1−lnξα−1lnrs1−αlnsα−1sds−∫1rlnrs1−αlns−lnξ+α−11sds.Now using the fact that(35)∫ζξξ−κ1−ακ−ζ1−αdκ=ΓαΓ2−αξ−ζ,we deduce that
(36)∫1rlnsα−1lnrs1−α1sds=ΓαΓ2−αlnr.On the other hand,(i)
if1≤ξ≤r, then by using (35), we get
(37)∫1rlns−lnξ+α−1lnrs1−α1sds=∫ξrlns−lnξα−1lnr−lns1−α1sds=ΓαΓ2−αlnr−lnξ(ii)
if1≤r≤ξ, then obviously
(38)∫1rlns−lnξ+α−1lnrs1−α1sds=0Hence,(39)∫1rlns−lnξ+α−1lnrs1−α1sds=ΓαΓ2−αlnr−lnξ+.So, (33) follows from (34), (36), and (39).Next, we claim that(40)HDαWgr≔δ2HI2−αWgr=−gr,forr∈1,e.Indeed, from (33), we have
(41)HI2−αWgr=∫1eKr,sgssds=∫1elnr1−lnsα−1−lnr−lns+gssds=lnr∫1r1−lnsα−1−1gssds+lnr∫re1−lnsα−1gssds+∫1rlnsgssds≔J1r+J2r+J3r.From the hypothesis, the functions⟶1−lnsα−1gs/s is continuous and integrable near e while the function s⟶lnsgs/s becomes continuous and integrable near 1. So J2r and J3r are differentiable on 1,e.On the other hand by observing that(42)1−1−lnsα−1=Olnsnear1,we deduce that J1r is differentiable on 1,e.Therefore,(43)δHI2−αWgr=∫1r1−lnsα−1−1gssds+∫re1−lnsα−1gssds.Applying for the second time theδ-derivative, we obtain
(44)HDαWgr=δ2HI2−αWgr=−gr.By Lemma8 (ii) and (iii), for each s∈1,e, we have
(45)limr→1+lnr2−αGαr,s=0,0≤lnr2−αGαr,s≤1Γαlns1−lnsα−1.This implies by the dominated convergence theorem that(46)limr→1+lnr2−αWgr=0.Similarly, we haveWge=0.Finally, the uniqueness follows from Lemma6 (ii).Remark 10.
The property of the above proposition remains true forα=2.Lemma 11.
Let1<α≤2 and g∈Jα,ln, then
(i)
Ag<∞(ii)
For allr∈1,e,
(47)∫1eGαr,su0sgssds≤Agu0r(iii)
The family(48)Λg=1u0r∫1eGαr,su0shssds,h≤gis relatively compact in C1,e.Proof.
(i)
As consequence of Lemma8 (iv) and definition of Jα,ln, we obtain Ag<∞.(ii)
Observe that for eachr,s∈1,e, we have
(49)limξ→1Gαs,ξGαr,ξ=lnsα−2−lnsα−1lnrα−2−lnrα−1Using this fact, Fatou’s lemma, and (26), we deduce that
(50)∫1eGαr,slnsα−2−lnsα−1lnrα−2−lnrα−1gssds≤liminfξ→1∫1eGαr,ξGαξ,sGαr,sgssds≤Ag.That is,(51)∫1eGαr,slnsα−2−lnsα−1gssds≤Aglnrα−2−lnrα−1,forr∈1,e.Similarly, sincelimξ→eGαs,ξ/Gαr,ξ=lnsα−1/lnrα−1, we obtain
(52)∫1eGαr,slnsα−1gssds≤Aglnrα−1,forr∈1,e.Hence, (47) follows by combining (51) and (52).
(iii)
It follows from (ii) and (i) that the familyΛg is uniformly boundedBy (19) and (28), for r,s∈1,e×1,e, we have
(53)Gαr,su0ru0sgss≤1Γαmaxa,bmina,blns−lnslnsα−1gs.Since the functionr,s⟶Gαr,s/u0r∈C1,e×1,e and g∈Jα,ln, we deduce by (53) that Λg is equicontinuous in 1,e and becomes relatively compact in C1,e by Ascoli’s theorem.Proof of Theorem 1.
We let(54)g0r≔1u0rfr,u0r,forr∈1,e.By hypotheses (H1) and (H2) and (28), we have g0∈Jα,ln.Define(55)λr≔u0r∫1eGαr,sfs,u0s/sds,λ0≔infr∈1,eλr.Using (54) and (47), we obtain
(56)∫1eGαr,sfs,u0ssds=∫1eGαr,su0sg0ssds≤Ag0u0r.Therefore,λ0≥1/Ag0>0.Let0<λ≤λ0 and
(57)S=v∈C1,e:1−λλ0≤v≤1.Forv∈S, define T by
(58)Tvr=1−λu0r∫1eGαr,sfs,vsu0ssds.By usingH1,H2, and Lemma 11 (iii), we prove that TS is relatively compact in C1,e.From (58), H1, and (55), we deduce that TS⊆S.Next, by simple arguments, one can prove thatT is a compact operator.Therefore, it has a fixed pointvλ∈S satisfying
(59)vλr=1−λu0r∫1eGαr,sfs,vλsu0ssds.Letuλr=vλru0r. Then, uλ∈Cα,ln1,e and satisfies
(60)uλr=u0r−λ∫1eGαr,sfs,uλssds.Sincevλ∈S, it follows from (28) that
(61)1−λλ0mina,blnrα−2≤uλr≤maxa,blnrα−2.By usingH1, (61), and H2, we deduce that the function s⟶lns1−lnsα−1fs,uλs∈C1,e∩L11,e.Hence, from (60), Proposition 9 (ii), and (4), we conclude that uλ is a solution of problem (3).Example 12.
Let0<β<1. Then, for some λ0>0 and each λ∈0,λ0, problem
(62)HD4/3ur−λlnr−βu3/2=0,r∈1,e,u>0,in1,e,limr→1+lnr2/3ur=a>0,ue=b>0,has a solution uλ in C4/3,ln1,e satisfying
(63)1−λλ0mina,blnr−2/3≤uλr≤maxa,blnr−2/3.Observe that the nonlinearity considered in this example is singular atr=1.Example 13.
Let1<α<2,γ∈0,2/2−α and p∈C+1,e.Then, there exists a constantλ0>0 such that for λ∈0,λ0, problem
(64)HDαur−λpruγ=0,r∈1,e,u>0,in1,e,limr→1+lnr2−αur=a>0,ue=b>0,admit a solution uλ in Cα,ln1,e satisfying
(65)1−λλ0mina,blnrα−2≤uλr≤maxa,blnrα−2.In particular, for=3/2,γ=2, and a=b=1, we have from (4) and (55),
(66)λr≔lnr−1/2∫1eG3/2r,slns−1ps/sds.Therefore, by choosing some continuous functionspii=1,2,3,4 in (66), we obtain the following graph for λir≔λr with p=pi and a numerical value of the constant λ0≔infr∈1,eλir. In Figure 2, we collect the graph of functions λi, and in Table 1, we summarize the numerical value of λ0.Figure 2
Graph of the functionsλi.Table 1
Value ofλ0.
Functionps in (66)Value ofλ0p1s=s3/2λ0=1.162p2s=esλ0=0.446p3s=lns1/2λ0=3.808p4s=sinsλ0=2.581Example 14.
Let1<α<2 and β∈0,2/2−α. Then, Theorem 1 can be applied forfr,s≔lnrα−2+sβ, where r,s∈1,e×0,∞.


## 4. Conclusion

In this paper, we have considered singular nonlinear Hadamard fractional boundary value problems. By using estimates on Green’s function and the Schauder fixed point theorem, we have proven the existence of a positive solution which blows up.



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