# Existence and Global Asymptotic Behavior of Singular Positive Solutions for Radial Laplacian

Authors: Imed Bachar; Habib Mâagli; Said Mesloub
Journal: Journal of Function Spaces
 (2019)

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## Abstract

The aim of this paper is to establish existence and uniqueness of a positive continuous solution to the following singular nonlinear problem.{-t1-ntn-1u′′=a(t)uσ,t∈(0,1),limt→0⁡tn-1u′(t)=0,u(1)=0}, where n≥3,σ<1, and a denotes a nonnegative continuous function that might have the property of being singular at t=0 and /or t=1 and which satisfies certain condition associated to Karamata class. We emphasize that the nonlinearity might also be singular at u=0, while the solution could blow-up at 0. Our method is based on the global estimates of potential functions and the Schauder fixed point theorem.


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## Body

## 1. Introduction and Main Result

Nonlinear problems of the form(1)-1AAu′′=fx,u,x∈0,1,u>0,in0,1,where A is a positive, differentiable function on (0,1) and satisfying several suitable conditions have been studied by many researchers (see for instance [1–10]). Note that many problems in the boundary layer theory and the theory of pseudoplastic fluids can be modeled by equations of the form (1) (see for example [11, 12]).Equations of the form (1) with A(t)=tn-1(n≥3), appears in a natural manner in those cases when the researcher is looking for radial solutions of Laplace operator.For a multiple results of existence, uniqueness, and asymptotic behavior associated with similar problems, we refer the reader to [13–30] and their bibliographies.Let us first introducing the following functional classK called Karamata class.Definition 1.
Letη>1 and L be a function defined on (0,η]. Then L belongs to the class K if (2)Lt≔cexp⁡∫tηzssds,where c>0 and z∈C([0,η]) with z(0)=0.Here, it is pertinent to note that the functions in the classK are slowly varying, and Karamata developed in [31] the initial theory in this field.Cirstea and Rădulescu have exploited in [32] the Karamata theory to study the asymptotic and qualitative behavior near the boundary of solutions of nonlinear elliptic problems.The aim of this paper is to address the existence, uniqueness and qualitative behavior of positive continuous solution to the following singular nonlinear problem.(3)-t1-ntn-1u′′=atuσ,t∈0,1,limt→0tn-1u′t=0,u1=0,where n≥3, σ<1 and a denotes a nonnegative continuous function on 0,1 that might have the property of being singular at t=0 and /or t=1 and which might satisfies certain condition associated with the Karamata class K. In this situation, the nonlinearity might also have the property of being singular at u=0. Here we emphasizes that the obtained solution may also blow-up at 0, which is not given in the previous works. Our approach relies on Karamata theory and the Schauder fixed point theorem.Notations.(i) B(0,1) (resp. B+0,1), denotes the set of Borel (resp. nonnegative Borel) measurable functions in (0,1).(ii)L1(0,1)≔{q∈B(0,1),∫01qrdr<∞}.(iii)C(X) (resp. C+(X)), is the set of continuous (resp. nonnegative continuous) functions in a metric space X.(iii)C0([0,1])={f∈C((0,1)):limt→0⁡f(t)=limt→1⁡f(t)=0}.(iii) Forn≥3, (4)Cn-20,1=f∈C0,1:t→tn-2ft∈C0,1.(iv) Forf,g∈B+(0,1), we say that ft≈gt, t∈(0,1), if there exists c>0 such that 1/cf(t)≤g(t)≤cf(t), for all t∈(0,1).In what follows, we letn≥3, σ<1 and assume that(H) a∈C+((0,1)) with(5)at≈t-μL1t1-t-λL21-t,t∈0,1,where μ≤n+(2-n)σ, λ≤2 and L1, L2∈K defined on (0,η](η>1) such that(6)∫0ηtn+2-nσ-μ-1L1tdt<∞,∫0ηt1-λL2tdt<∞.We introduce the function θ defined on (0,1] by(7)θt≔tmin0,2-μ/1-σL~1t1/1-σ1-tmin1,2-λ/1-σL~21-t1/1-σ,where (8)L~1t≔1,ifμ<2,∫tηL1ssds,ifμ=2,L1t,if2<μ<n+2-nσ,∫0tL1ssds,ifμ=n+2-nσ,and(9)L~2t≔1,ifλ<1,∫tηL2ssds,ifλ=1,L2t,if1<λ<2,∫0tL2ssds,ifλ=2.Our main result is the following.Theorem 2.
Letσ<1 and assume that a satisfies (H). Then problem (3) has a unique positive solution u∈Cn-2([0,1]) satisfying for t∈(0,1],(10)ut≈θt.Remark 3.
The solution obtained in Theorem2 blow-up at 0 for μ>2.Example 4.
Letn≥3 and assume that μ<n. The unique solution of the linear problem (11)-t1-ntn-1u′′=t-μ,t∈0,1,limt→0tn-1u′t=0,u1=0,is given by (12)uμt=1n-μμ-2t2-μ-1,if2<μ<n,-1n-2ln⁡t,ifμ=2,1n-μ2-μ1-t2-μ,ifμ<2.Clearly the solution blow up at 0 for μ≥2 and also we have (13)uμt≈θt=t2-μ1-t,if2<μ<n,1-tln⁡2-ln⁡t,ifμ=2,1-t,ifμ<2.This implies that the global estimates obtained in Theorem 2 are optimal.


## 2. Karamata Class and Global Estimates

### 2.1. Karamata Class

It is clear to see that for someη>1, the class K is given by (14)K=L:0,η→0,∞,L∈C10,ηandlimt→0+tL′tLt=0.Standard examples of functions which are elements of the class K are presented below (see [33–35])(15)Lt=∏k=1mlogkωtξk,andLt=exp∏k=1mlogkωtνk, where logkt=log∘log∘⋯log⁡t (k times), ξk∈R, νk∈(0,1) and ω is a sufficiently large positive real number such that L is defined and positive on (0,η], for some η>1.Next, we collect several properties of the Karamata functions, which will be useful in the proof of our main result.Lemma 5 (see [34, 35]).
Letγ∈R and L∈K defined on (0,η], η>1. We have
( i ) If γ>-1, then ∫0ηsγL(s)ds<∞ and ∫0tsγL(s)ds~t→0+t1+γL(t)/1+γ.
( i i ) If γ<-1, then ∫0ηsγL(s)ds=∞ and ∫tηsγL(s)ds~t→0+-t1+γL(t)/1+γ.Lemma 6 (see [35, 36]).
( i ) Let L∈K and ϵ>0, then we have (16)limt→0+tϵLt=0.
( i i ) Let L1,L2∈K defined on (0,η], η>1, and p∈R. Then the functions (17)L1+L2,L1L2,andL1pareinK.(iii) Let L∈K defined on (0,η], η>1, then we have (18)limt→0+Lt∫tηLs/sds=0.In particular (19)t→∫tηLssds∈K.If further ∫0ηL(s)/sds<∞, then we have limt→0+⁡L(t)/∫0tL(s)/sds=0.
In particular(20)t→∫0tLssds∈K.


### 2.2. Global Estimates

Letn≥3, then(21)Gt,r≔1n-2rn-11-maxt,rn-2maxt,rn-2,is the Green’s function of the operator u→-t1-ntn-1u′′, with boundary conditions limt→0⁡tn-1u′(t)=u(1)=0.Lemma 7.
( i ) On [0,1]×[0,1], we have(22)Gt,r≈rn-11-maxt,rmaxt,rn-2.
( i i ) There exists a constant c>0 such that for all t,r∈[0,1],(23)1crn-1tn-21-t1-r≤tn-2Gt,r≤crn-1min1-t,1-r.Proof.
( i ) The property follows from (21) and the fact that (24)1-maxt,r≤1-max⁡t,rn-2≤n-21-max⁡t,r.
( i i ) The inequalities follow from (22) and the fact that (25)1-t1-r≤1-maxt,r≤min1-t,1-r.Forf∈B+(0,1), we define the otential of f by (26)Vft≔∫01Gt,rfrdr,t∈0,1.Using Lemma 7, we deduce the following.Corollary 8.
Letf∈B+(0,1), then we have (27)Vft∈Cn-20,1⇔∫01rn-11-rfrdr<∞.Proposition 9.
Letn≥3 and f be a function such that the map r→rn-1(1-r)f(r)∈C(0,1)∩L1(0,1), then Vf belongs to Cn-2([0,1]) and it is the unique solution of the problem(28)-t1-ntn-1u′′=ft,t∈0,1,limt→0tn-1u′t=0,u1=0.Proof.
From Corollary8, the function Vf is in Cn-2([0,1]).
Using (21), we have for t∈(0,1), (29)Vft=∫01Gt,rfrdr=1n-2t2-n-1∫0trn-1frdr+1n-2∫t1r1-rn-2frdr=J1t+J2t.Since the function r→rn-1(1-r)f(r)∈C(0,1)∩L1(0,1), then J1(t) and J2(t) are differentiable on (0,1).
By simple calculation, we obtain(30)Vf′t=-t1-n∫0trn-1frdr.That is(31)tn-1Vf′t=-∫0trn-1frdr.By differentiating (31), we obtain for t∈(0,1), (32)-t1-ntn-1Vf′′t=ft.Using (31) and the fact that Vf∈Cn-2([0,1]), we obtain(33)limt→0tn-1Vf′t=Vf1=0.Finally, we prove the uniqueness. Let u,v∈Cn-2([0,1]) be two solutions of (28) and put w=u-v. Then w∈Cn-20,1 and (tn-1w′)′=0. Since limt→0⁡tn-1w′(t)=0, we deduce that tn-1w′(t)=0 and therefore w(t)=c. Using the fact w(1)=0, we deduce that w=0. That is u=v.Proposition 10.
Letγ≤n, ν≤2 and L3, L4∈K such that(34)∫0ηrn-γ-1L3rdr<∞and∫0ηr1-νL4rdr<∞.Put (35)bt=t-γL3t1-t-νL41-t,fort∈0,1.Then for t∈(0,1], (36)Vbt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t,where (37)L~3t≔1,ifγ<2,∫tηL3rrdr,ifγ=2,L3t,if2<γ<n,∫0tL3rrdr,ifγ=n,and (38)L~4t≔1,ifν<1,∫tηL4rrdr,ifν=1,L4t,if1<ν<2,∫0tL4rrdr,ifν=2.Proof.
Fort∈(0,1], we have (39)Vbt=∫01Gt,rbrdr.Using (22), we obtain that (40)Vbt≈1-tt2-n∫0trn-γ-11-r-νL3rL41-rdr+∫t1r1-γ1-r1-νL3rL41-rdr.In what follows, we distinguish two cases.
Case  1 (0<t≤1/2). In this case 1-t≈1. So we obtain (41)Vbt≈t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γ1-r1-νL3rL41-rdr+∫1/21r1-γ1-r1-νL3rL41-rdr≈t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γL3rdr+∫1/211-r1-νL41-rdr=t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γL3rdr+∫01/2r1-νL4rdr. Since ∫0ηr1-νL4(r)dr<∞, we deduce that (42)Vbt≈t2-n∫0trn-γ-1L3rdr+1+∫t1/2r1-γL3rdr.Using Lemma 5 and (34), we deduce that (43)∫0trn-γ-1L3rdr≈tn-γL3t,ifγ<n,∫0tL3rrdr,ifγ=nand (44)1+∫t1/2r1-γL3rdr≈1,ifγ<2,∫tηL3rrdr,ifγ=2,t2-γL3t,if2<γ≤n.Hence, it follows by Lemmas 5, 6 and (34) that, for 0<t≤1/2, (45)Vbt≈1,ifγ<2,∫tηL3rrdr,ifγ=2,t2-γL3t,if2<γ<n,t2-γ∫0tL3rrdr,ifγ=n. That is(46)Vbt≈tmin0,2-γL~3t.
Case  2 (1/2<t≤1). In this case, t≈1. Therefore, we have (47)Vbt≈t2-n∫01/2rn-γ-11-t1-r-νL3rL41-rdr+∫1/2trn-γ-11-t1-r-νL3rL41-rdr+∫t1r1-γ1-r1-νL3rL41-rdr≈1-t∫01/2rn-γ-1L3rdr+∫1/2t1-r-νL41-rdr+∫t11-r1-νL41-rdr=1-t∫01/2rn-γ-1L3rdr+∫1-t1/2r-νL4rdr+∫01-tr1-νL4rdr. Since ∫0ηrn-γ-1L3(r)dr<∞, we deduce that (48)Vbt≈1-t1+∫1-t1/2r-νL4rdr+∫01-tr1-νL4rdr.Using again Lemma 5 and hypothesis (34), we deduce that (49)∫01-tr1-νL4rdr≈1-t2-νL41-t,ifν<2,∫01-tL4rrdr,ifν=2
and(50)1+∫1-t1/2r-νL4rdr≈1,ifν<1,∫1-tηL4rrdr,ifν=2,1-t1-νL41-t,if1<ν≤2.Hence, it follows by Lemmas 5, 6 and hypothesis (34) that, for 1/2<t≤1, we get (51)Vbt≈1-t,ifν<1,1-t∫1-tηL4rrdr,ifν=1,1-t2-νL41-t,if1<ν<2,∫01-tL4rrdr,ifν=2.That is(52)Vbt≈1-tmin⁡1,2-νL~41-t.Combining (46) and (52), we obtain for t∈(0,1], (53)Vbt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t.This ends the proof.Proposition 11.
Assume that condition(H) is satisfied. Then for t∈(0,1], we have (54)Vpt≈θt,where p(r)≔a(r)θσ(r).Proof.
Leta be a function satisfying H. Using (5) and (7), we obtain (55)pt≈t-γL1tL~1tσ/1-σ1-t-νL21-tL~21-tσ/1-σ,where γ=μ-min0,2-μ/1-σσ and ν=λ-min(1,2-λ/1-σ)σ.
Sinceμ≤n+(2-n)σ and λ≤2, then one can easy check that γ≤n and ν≤2.
Now using Lemmas5, 6 and Proposition 10 with
L 3 = L 1 L ~ 1 σ / 1 - σ ∈ K and L4=L2L~2σ/1-σ∈K, we deduce that for each t∈(0,1], (56)Vpt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t.Since min(0,2-γ)=min(0,2-μ/1-σ) and min(1,2-ν)=min(1,2-λ/1-σ), then we deduce (57)Vpt≈tmin0,2-μ/1-σL~3t1-tmin1,2-λ/1-σL~41-t≈θt.This completes the proof.


## 2.1. Karamata Class

It is clear to see that for someη>1, the class K is given by (14)K=L:0,η→0,∞,L∈C10,ηandlimt→0+tL′tLt=0.Standard examples of functions which are elements of the class K are presented below (see [33–35])(15)Lt=∏k=1mlogkωtξk,andLt=exp∏k=1mlogkωtνk, where logkt=log∘log∘⋯log⁡t (k times), ξk∈R, νk∈(0,1) and ω is a sufficiently large positive real number such that L is defined and positive on (0,η], for some η>1.Next, we collect several properties of the Karamata functions, which will be useful in the proof of our main result.Lemma 5 (see [34, 35]).
Letγ∈R and L∈K defined on (0,η], η>1. We have
( i ) If γ>-1, then ∫0ηsγL(s)ds<∞ and ∫0tsγL(s)ds~t→0+t1+γL(t)/1+γ.
( i i ) If γ<-1, then ∫0ηsγL(s)ds=∞ and ∫tηsγL(s)ds~t→0+-t1+γL(t)/1+γ.Lemma 6 (see [35, 36]).
( i ) Let L∈K and ϵ>0, then we have (16)limt→0+tϵLt=0.
( i i ) Let L1,L2∈K defined on (0,η], η>1, and p∈R. Then the functions (17)L1+L2,L1L2,andL1pareinK.(iii) Let L∈K defined on (0,η], η>1, then we have (18)limt→0+Lt∫tηLs/sds=0.In particular (19)t→∫tηLssds∈K.If further ∫0ηL(s)/sds<∞, then we have limt→0+⁡L(t)/∫0tL(s)/sds=0.
In particular(20)t→∫0tLssds∈K.


## 2.2. Global Estimates

Letn≥3, then(21)Gt,r≔1n-2rn-11-maxt,rn-2maxt,rn-2,is the Green’s function of the operator u→-t1-ntn-1u′′, with boundary conditions limt→0⁡tn-1u′(t)=u(1)=0.Lemma 7.
( i ) On [0,1]×[0,1], we have(22)Gt,r≈rn-11-maxt,rmaxt,rn-2.
( i i ) There exists a constant c>0 such that for all t,r∈[0,1],(23)1crn-1tn-21-t1-r≤tn-2Gt,r≤crn-1min1-t,1-r.Proof.
( i ) The property follows from (21) and the fact that (24)1-maxt,r≤1-max⁡t,rn-2≤n-21-max⁡t,r.
( i i ) The inequalities follow from (22) and the fact that (25)1-t1-r≤1-maxt,r≤min1-t,1-r.Forf∈B+(0,1), we define the otential of f by (26)Vft≔∫01Gt,rfrdr,t∈0,1.Using Lemma 7, we deduce the following.Corollary 8.
Letf∈B+(0,1), then we have (27)Vft∈Cn-20,1⇔∫01rn-11-rfrdr<∞.Proposition 9.
Letn≥3 and f be a function such that the map r→rn-1(1-r)f(r)∈C(0,1)∩L1(0,1), then Vf belongs to Cn-2([0,1]) and it is the unique solution of the problem(28)-t1-ntn-1u′′=ft,t∈0,1,limt→0tn-1u′t=0,u1=0.Proof.
From Corollary8, the function Vf is in Cn-2([0,1]).
Using (21), we have for t∈(0,1), (29)Vft=∫01Gt,rfrdr=1n-2t2-n-1∫0trn-1frdr+1n-2∫t1r1-rn-2frdr=J1t+J2t.Since the function r→rn-1(1-r)f(r)∈C(0,1)∩L1(0,1), then J1(t) and J2(t) are differentiable on (0,1).
By simple calculation, we obtain(30)Vf′t=-t1-n∫0trn-1frdr.That is(31)tn-1Vf′t=-∫0trn-1frdr.By differentiating (31), we obtain for t∈(0,1), (32)-t1-ntn-1Vf′′t=ft.Using (31) and the fact that Vf∈Cn-2([0,1]), we obtain(33)limt→0tn-1Vf′t=Vf1=0.Finally, we prove the uniqueness. Let u,v∈Cn-2([0,1]) be two solutions of (28) and put w=u-v. Then w∈Cn-20,1 and (tn-1w′)′=0. Since limt→0⁡tn-1w′(t)=0, we deduce that tn-1w′(t)=0 and therefore w(t)=c. Using the fact w(1)=0, we deduce that w=0. That is u=v.Proposition 10.
Letγ≤n, ν≤2 and L3, L4∈K such that(34)∫0ηrn-γ-1L3rdr<∞and∫0ηr1-νL4rdr<∞.Put (35)bt=t-γL3t1-t-νL41-t,fort∈0,1.Then for t∈(0,1], (36)Vbt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t,where (37)L~3t≔1,ifγ<2,∫tηL3rrdr,ifγ=2,L3t,if2<γ<n,∫0tL3rrdr,ifγ=n,and (38)L~4t≔1,ifν<1,∫tηL4rrdr,ifν=1,L4t,if1<ν<2,∫0tL4rrdr,ifν=2.Proof.
Fort∈(0,1], we have (39)Vbt=∫01Gt,rbrdr.Using (22), we obtain that (40)Vbt≈1-tt2-n∫0trn-γ-11-r-νL3rL41-rdr+∫t1r1-γ1-r1-νL3rL41-rdr.In what follows, we distinguish two cases.
Case  1 (0<t≤1/2). In this case 1-t≈1. So we obtain (41)Vbt≈t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γ1-r1-νL3rL41-rdr+∫1/21r1-γ1-r1-νL3rL41-rdr≈t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γL3rdr+∫1/211-r1-νL41-rdr=t2-n∫0trn-γ-1L3rdr+∫t1/2r1-γL3rdr+∫01/2r1-νL4rdr. Since ∫0ηr1-νL4(r)dr<∞, we deduce that (42)Vbt≈t2-n∫0trn-γ-1L3rdr+1+∫t1/2r1-γL3rdr.Using Lemma 5 and (34), we deduce that (43)∫0trn-γ-1L3rdr≈tn-γL3t,ifγ<n,∫0tL3rrdr,ifγ=nand (44)1+∫t1/2r1-γL3rdr≈1,ifγ<2,∫tηL3rrdr,ifγ=2,t2-γL3t,if2<γ≤n.Hence, it follows by Lemmas 5, 6 and (34) that, for 0<t≤1/2, (45)Vbt≈1,ifγ<2,∫tηL3rrdr,ifγ=2,t2-γL3t,if2<γ<n,t2-γ∫0tL3rrdr,ifγ=n. That is(46)Vbt≈tmin0,2-γL~3t.
Case  2 (1/2<t≤1). In this case, t≈1. Therefore, we have (47)Vbt≈t2-n∫01/2rn-γ-11-t1-r-νL3rL41-rdr+∫1/2trn-γ-11-t1-r-νL3rL41-rdr+∫t1r1-γ1-r1-νL3rL41-rdr≈1-t∫01/2rn-γ-1L3rdr+∫1/2t1-r-νL41-rdr+∫t11-r1-νL41-rdr=1-t∫01/2rn-γ-1L3rdr+∫1-t1/2r-νL4rdr+∫01-tr1-νL4rdr. Since ∫0ηrn-γ-1L3(r)dr<∞, we deduce that (48)Vbt≈1-t1+∫1-t1/2r-νL4rdr+∫01-tr1-νL4rdr.Using again Lemma 5 and hypothesis (34), we deduce that (49)∫01-tr1-νL4rdr≈1-t2-νL41-t,ifν<2,∫01-tL4rrdr,ifν=2
and(50)1+∫1-t1/2r-νL4rdr≈1,ifν<1,∫1-tηL4rrdr,ifν=2,1-t1-νL41-t,if1<ν≤2.Hence, it follows by Lemmas 5, 6 and hypothesis (34) that, for 1/2<t≤1, we get (51)Vbt≈1-t,ifν<1,1-t∫1-tηL4rrdr,ifν=1,1-t2-νL41-t,if1<ν<2,∫01-tL4rrdr,ifν=2.That is(52)Vbt≈1-tmin⁡1,2-νL~41-t.Combining (46) and (52), we obtain for t∈(0,1], (53)Vbt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t.This ends the proof.Proposition 11.
Assume that condition(H) is satisfied. Then for t∈(0,1], we have (54)Vpt≈θt,where p(r)≔a(r)θσ(r).Proof.
Leta be a function satisfying H. Using (5) and (7), we obtain (55)pt≈t-γL1tL~1tσ/1-σ1-t-νL21-tL~21-tσ/1-σ,where γ=μ-min0,2-μ/1-σσ and ν=λ-min(1,2-λ/1-σ)σ.
Sinceμ≤n+(2-n)σ and λ≤2, then one can easy check that γ≤n and ν≤2.
Now using Lemmas5, 6 and Proposition 10 with
L 3 = L 1 L ~ 1 σ / 1 - σ ∈ K and L4=L2L~2σ/1-σ∈K, we deduce that for each t∈(0,1], (56)Vpt≈tmin0,2-γL~3t1-tmin1,2-νL~41-t.Since min(0,2-γ)=min(0,2-μ/1-σ) and min(1,2-ν)=min(1,2-λ/1-σ), then we deduce (57)Vpt≈tmin0,2-μ/1-σL~3t1-tmin1,2-λ/1-σL~41-t≈θt.This completes the proof.


## 3. Existence Results

### 3.1. Preliminary Results

Forα≥0, we denote by Pα the following problem (58)Pα-t1-ntn-1u′′=atuσ,in0,1,limt→0tn-1u′t=2-nα,u1=α.Next, we establish several facts to be used within the proof of the main result.Lemma 12.
Letσ<0, 0≤α≤β and uα,uβ∈C(0,1]∩C10,1 be two positive functions satisfying(59)-t1-ntn-1uα′′≤atuασ,in0,1,limt→0tn-1uα′t=2-nα,uα1=α.
and(60)-t1-ntn-1uβ′′≥atuβσ,in0,1,limt→0tn-1uβ′t=2-nβ,uβ1=β.
Thenuα≤uβ in (0,1].Proof.
Letu(t)≔uα(t)-uβ(t) for t∈(0,1). Assume that u(t0)>0 for some t0∈(0,1). Then there exists an interval t1,t2⊂0,1 containing t0 such that u(t)>0, on t1,t2 with u(t2)=0 and u(t1)=0 or t1=0.
On the other hand, sinceσ<0, then we have uβσ(t)>uασ(t) for t∈t1,t2. So (61)t1-ntn-1u′′=t1-ntn-1uα′′-t1-ntn-1uβ′′≥atuβσ-uασ>0ont1,t2.So the function t→tn-1u′ is nondecreasing on t1,t2 with limt→t1⁡tn-1u′(t)≥0.
Hence, the functionu is nondecreasing on t1,t2 with u(t0)>0 and u(t2)=0. This gives a contradiction. Therefore, uα≤uβ.Proposition 13.
Letσ<0, and assume that hypothesis (H) is satisfied. Then for each α>0, problem Pα has a unique positive solution uα∈C(0,1]∩C10,1 satisfying for t∈(0,1)(62)uαt=αt2-n+∫01Gt,raruασrdr.In particular, limt→0⁡tn-2uα(t)=α.Proof.
Letσ<0, α>0 and assume that the function a satisfies hypothesis (H). Note that the function(63)t→ht≔tn-2∫01Gt,rarr2-nσdrisinC00,1.Indeed, since for each r>0, the function t→tn-2G(t,r)∈C0[0,1], it follows from (23), (6) and the convergence dominated theorem that h∈C0[0,1].
Letβ≔α+ασh∞ and Λ be the closed convex set given by (64)Λ=v∈C0,1:α≤v≤β.Define the operator T on Λ by (65)Tvt=α+tn-2∫01Gt,rarr2-nσvσrdr.Since σ<0, then for all v∈Λ, we have (66)α≤Tv≤β.On the other hand, we have for all t1,t2∈[0,1](67)Tvt1-Tvt2≤ασ∫01t1n-2Gt1,r-t2n-2Gt2,rarr2-nσdr.Since, for each r>0, the function t→tn-2G(t,r)∈C0[0,1], then we deduce by (23), (6) and the convergence dominated theorem that TΛ is equicontinuous in [0,1]. In particular, for all v∈Λ, TΛ⊂C[0,1] and so TΛ⊂Λ. Moreover, since the family {Tv(t), v∈Λ} is uniformly bounded in [0,1], then by Ascoli’s theorem that TΛ becomes relatively compact in C[0,1]. Next, we prove the continuity of T in Λ. Let vkk⊂Λ and v∈Λ such that vk-v∞→0 as k→∞. Then we have (68)Tvkt-Tvt≤∫01tn-2Gt,rarr2-nσvkσr-vσrdr.Now, since (69)vkσr-vσr≤2ασ,we deduce by convergence dominated theorem that (70)∀t∈0,1,Tvkt→Tvtask→∞.Since TΛ is relatively compact in C[0,1], we obtain (71)Tvk-Tv∞→0ask→∞.So T is a compact mapping Λ to itself. Therefore, by the Schauder fixed point theorem, there exists vα∈Λ such that for each t∈[0,1](72)vαt=α+tn-2∫01Gt,rarr2-nσvασrdr.Put uα(t)=t2-nvα, for t∈(0,1]. Then uα∈C(0,1] and we have(73)uαt=αt2-n+∫01Gt,raruασrdr,and (74)αt2-n≤uαt≤βt2-n.We have for all r∈(0,1), (75)aruασr≤ασarr2-nσ.Now since by hypothesis (H) the function r→rn-1(1-r)a(r)r2-nσ∈C(0,1)∩L1(0,1), we deduce from (73) and Proposition 9 that uα is a solution of problem Pα. By Lemma 12, we obtain the uniqueness.
Finally, using (73) and the fact that (76)0≤tn-2∫01Gt,raruασrdr≤ασht∈C00,1,we deduce that limt→0⁡tn-2uα(t)=α.From Lemma12 and (73) we deduce the followingCorollary 14.
Letσ<0, and assume that hypothesis (H) is satisfied. For 0<α1≤α2, we denote by uαi∈C(0,1]∩C10,1 the unique positive solution of the problem Pα. Then we have (77)0≤uα2t-uα1t≤α2-α1t2-n,fort∈0,1.Theorem 15.
Letσ<0. Under hypothesis (H), problem (3) has a unique positive solution u∈C(0,1]∩C10,1 satisfying for t∈(0,1)(78)ut=∫01Gt,raruσrdr.Proof.
Letαkk⊂(0,∞) be a sequence that decreases to zero. Let uk be the unique positive continuous solution of the problem Pαk. By Lemma 12 (or Corollary 14), the sequence ukk decreases to a function u, and from (73) the sequence uk-αkt2-nk increases to u. Using this fact and (73), we obtain for each t∈(0,1), (79)ut≥ukt-αkt2-n=∫01Gt,rarukσrdr≥βkσ∫01Gt,rarr2-nσdr>0,where βk≔αk+αkσh∞ and h is given by (63).
By the monotone convergence theorem, we obtain(80)ut=∫01Gt,raruσrdr.Since u=infk⁡uk=supk⁡uk-αkt2-n, then u is upper and lower semi-continuous function on (0,1] and so u∈C((0,1]).
We claim thatu is a solution of problem (3).
From (23), there exists c>0, such that for all t∈[0,1],(81)1c1-t∫01rn-11-raruσrdr≤∫01Gt,raruσrdr=ut.In particular, (82)∫01rn-11-raruσrdr≤2cu12<∞.So the function r→rn-1(1-r)a(r)uσ(r)∈C(0,1]∩L10,1. By Proposition 9, u is a solution of problem (3). Finally, by Lemma 12, we obtain the uniqueness.


### 3.2. Proof of Theorem2

Assume that hypothesis(H) is fulfilled. Let θ be the function defined in (7). By Proposition 11, there exists M≥1 such that for each t∈(0,1),(83)1Mθt≤Vpt≤Mθt,where p(r)≔a(r)θσ(r). We will break up the proof in two cases.Case  1 (σ<0). Let u be the solution of problem (3) given by Theorem 15. We claim that u satisfies (10).Letc=M-σ/1-σ and put v≔1/cVp.Using hypothesis(H) and Lemmas 5, 6, we deduce that(84)∫01rn-11-rprdr<∞.So, it follows by Proposition 10 and (83), that v satisfies (85)-t1-ntn-1v′′≤vt,t∈0,1,limt→0tn-1v′t=0,v1=0.Therefore by Lemma 12, we have (86)1cVp≤u.Similarly, we show that (87)u≤cVp.Now since by (83), Vp(t)≈θ(t), we deduce that u(t)≈θ(t).Case  2 (0≤σ<1). We shall use a fixed point argument to construct a solution of problem (3). Let φ(t)=tn-2θ(t), for t∈[0,1]. By (83), we have(88)1Mφt≤tn-2Vpt≤Mφt.Put c=M1/1-σ and consider the closed convex set given by (89)E=v∈C0,1,1cφ≤v≤cφ.Clearly φ∈E.We define the operatorF on E by (90)Fvt≔tn-2∫01Gt,rarr2-nσvσrdr,t∈0,1.Using (88), for v∈E, we have (91)1cφ≤Fv≤cφ.Since for each r>0, the function t→tn-2G(t,r) is in C0([0,1]), it follows by (23), (84) and the convergence dominated theorem that FE⊂E.Letvkk⊂C([0,1]) defined by(92)v0=1cφandvk+1=Fvk,fork∈N.Since the operator F is nondecreasing and FE⊂E, we deduce that (93)v0≤v1≤v2≤⋯≤vk≤vk+1≤cφ.Therefore, the sequence vkk converges by the convergence monotone theorem to a function v satisfying for each t∈[0,1](94)1cφt≤vt≤cφtandvt=tn-2∫01Gt,rarr2-nσvσrdr.Using (23) and (84), we prove that v∈C([0,1]).Putu(t)=t2-nv(t). Then u∈Cn-2([0,1]) and satisfies the equation (95)ut=Vauσt.Now since the function r↦rn-1(1-r)a(t)uσ(r)∈C(0,1]∩L10,1, it follows from Proposition 9 that u is a positive solution of problem (3) satisfying for t∈(0,1], (96)ut≈θt.For the uniqueness, let u and v be two arbitrary solutions of problem (3) in the cone (97)S=u∈Cn-20,1,ut≈θt.There exists a constant m≥1 such that (98)1m≤uv≤m,in0,1.So the set J≔{m≥1:1/m≤u/v≤m} is not empty. Let m0≔inf⁡J. Then m0≥1 and we have uσ≤m0σvσ.Letz≔m0σv-u. Then(99)-t1-ntn-1z′′=atm0σvσ-uσ≥0in0,1,limt→0tn-1z′t=0,z1=0,Which implies by Proposition 9 that m0σv-u=V(a(m0σvσ-uσ))≥0. By symmetry, we obtain that m0σu≥v. Hence, m0σ∈J. Using the fact that σ<1, we get m0=1. Then, we conclude that u=v.This completes the proof of Theorem2.Example 16.
Letσ<1 and a∈C(0,1) such that (100)ar≈r-μlog3r-β1-r-2log31-r-2,where μ<n+2-nσ and β∈R. Then, by Theorem 2, problem (3) has a unique positive solution u in Cn-2([0,1]) satisfying the following estimates:(i) If2<μ<n+2-nσ, then for t∈(0,1), (101)ut≈t2-μ/1-σlog3t-β/1-σlog31-t-1/1-σ.(ii) Ifμ=2 and β>1 or μ<2, then for t∈(0,1), (102)ut≈log31-t-1/1-σ.(iii) Ifμ=2 and β=1, then for t∈(0,1), (103)ut≈log23t1/1-σlog⁡31-t-1/1-σ.(iv) Ifμ=2 and β<1, then for t∈(0,1), (104)ut≈log⁡3t1-β/1-σlog⁡31-t-1/1-σ.


## 3.1. Preliminary Results

Forα≥0, we denote by Pα the following problem (58)Pα-t1-ntn-1u′′=atuσ,in0,1,limt→0tn-1u′t=2-nα,u1=α.Next, we establish several facts to be used within the proof of the main result.Lemma 12.
Letσ<0, 0≤α≤β and uα,uβ∈C(0,1]∩C10,1 be two positive functions satisfying(59)-t1-ntn-1uα′′≤atuασ,in0,1,limt→0tn-1uα′t=2-nα,uα1=α.
and(60)-t1-ntn-1uβ′′≥atuβσ,in0,1,limt→0tn-1uβ′t=2-nβ,uβ1=β.
Thenuα≤uβ in (0,1].Proof.
Letu(t)≔uα(t)-uβ(t) for t∈(0,1). Assume that u(t0)>0 for some t0∈(0,1). Then there exists an interval t1,t2⊂0,1 containing t0 such that u(t)>0, on t1,t2 with u(t2)=0 and u(t1)=0 or t1=0.
On the other hand, sinceσ<0, then we have uβσ(t)>uασ(t) for t∈t1,t2. So (61)t1-ntn-1u′′=t1-ntn-1uα′′-t1-ntn-1uβ′′≥atuβσ-uασ>0ont1,t2.So the function t→tn-1u′ is nondecreasing on t1,t2 with limt→t1⁡tn-1u′(t)≥0.
Hence, the functionu is nondecreasing on t1,t2 with u(t0)>0 and u(t2)=0. This gives a contradiction. Therefore, uα≤uβ.Proposition 13.
Letσ<0, and assume that hypothesis (H) is satisfied. Then for each α>0, problem Pα has a unique positive solution uα∈C(0,1]∩C10,1 satisfying for t∈(0,1)(62)uαt=αt2-n+∫01Gt,raruασrdr.In particular, limt→0⁡tn-2uα(t)=α.Proof.
Letσ<0, α>0 and assume that the function a satisfies hypothesis (H). Note that the function(63)t→ht≔tn-2∫01Gt,rarr2-nσdrisinC00,1.Indeed, since for each r>0, the function t→tn-2G(t,r)∈C0[0,1], it follows from (23), (6) and the convergence dominated theorem that h∈C0[0,1].
Letβ≔α+ασh∞ and Λ be the closed convex set given by (64)Λ=v∈C0,1:α≤v≤β.Define the operator T on Λ by (65)Tvt=α+tn-2∫01Gt,rarr2-nσvσrdr.Since σ<0, then for all v∈Λ, we have (66)α≤Tv≤β.On the other hand, we have for all t1,t2∈[0,1](67)Tvt1-Tvt2≤ασ∫01t1n-2Gt1,r-t2n-2Gt2,rarr2-nσdr.Since, for each r>0, the function t→tn-2G(t,r)∈C0[0,1], then we deduce by (23), (6) and the convergence dominated theorem that TΛ is equicontinuous in [0,1]. In particular, for all v∈Λ, TΛ⊂C[0,1] and so TΛ⊂Λ. Moreover, since the family {Tv(t), v∈Λ} is uniformly bounded in [0,1], then by Ascoli’s theorem that TΛ becomes relatively compact in C[0,1]. Next, we prove the continuity of T in Λ. Let vkk⊂Λ and v∈Λ such that vk-v∞→0 as k→∞. Then we have (68)Tvkt-Tvt≤∫01tn-2Gt,rarr2-nσvkσr-vσrdr.Now, since (69)vkσr-vσr≤2ασ,we deduce by convergence dominated theorem that (70)∀t∈0,1,Tvkt→Tvtask→∞.Since TΛ is relatively compact in C[0,1], we obtain (71)Tvk-Tv∞→0ask→∞.So T is a compact mapping Λ to itself. Therefore, by the Schauder fixed point theorem, there exists vα∈Λ such that for each t∈[0,1](72)vαt=α+tn-2∫01Gt,rarr2-nσvασrdr.Put uα(t)=t2-nvα, for t∈(0,1]. Then uα∈C(0,1] and we have(73)uαt=αt2-n+∫01Gt,raruασrdr,and (74)αt2-n≤uαt≤βt2-n.We have for all r∈(0,1), (75)aruασr≤ασarr2-nσ.Now since by hypothesis (H) the function r→rn-1(1-r)a(r)r2-nσ∈C(0,1)∩L1(0,1), we deduce from (73) and Proposition 9 that uα is a solution of problem Pα. By Lemma 12, we obtain the uniqueness.
Finally, using (73) and the fact that (76)0≤tn-2∫01Gt,raruασrdr≤ασht∈C00,1,we deduce that limt→0⁡tn-2uα(t)=α.From Lemma12 and (73) we deduce the followingCorollary 14.
Letσ<0, and assume that hypothesis (H) is satisfied. For 0<α1≤α2, we denote by uαi∈C(0,1]∩C10,1 the unique positive solution of the problem Pα. Then we have (77)0≤uα2t-uα1t≤α2-α1t2-n,fort∈0,1.Theorem 15.
Letσ<0. Under hypothesis (H), problem (3) has a unique positive solution u∈C(0,1]∩C10,1 satisfying for t∈(0,1)(78)ut=∫01Gt,raruσrdr.Proof.
Letαkk⊂(0,∞) be a sequence that decreases to zero. Let uk be the unique positive continuous solution of the problem Pαk. By Lemma 12 (or Corollary 14), the sequence ukk decreases to a function u, and from (73) the sequence uk-αkt2-nk increases to u. Using this fact and (73), we obtain for each t∈(0,1), (79)ut≥ukt-αkt2-n=∫01Gt,rarukσrdr≥βkσ∫01Gt,rarr2-nσdr>0,where βk≔αk+αkσh∞ and h is given by (63).
By the monotone convergence theorem, we obtain(80)ut=∫01Gt,raruσrdr.Since u=infk⁡uk=supk⁡uk-αkt2-n, then u is upper and lower semi-continuous function on (0,1] and so u∈C((0,1]).
We claim thatu is a solution of problem (3).
From (23), there exists c>0, such that for all t∈[0,1],(81)1c1-t∫01rn-11-raruσrdr≤∫01Gt,raruσrdr=ut.In particular, (82)∫01rn-11-raruσrdr≤2cu12<∞.So the function r→rn-1(1-r)a(r)uσ(r)∈C(0,1]∩L10,1. By Proposition 9, u is a solution of problem (3). Finally, by Lemma 12, we obtain the uniqueness.


## 3.2. Proof of Theorem2

Assume that hypothesis(H) is fulfilled. Let θ be the function defined in (7). By Proposition 11, there exists M≥1 such that for each t∈(0,1),(83)1Mθt≤Vpt≤Mθt,where p(r)≔a(r)θσ(r). We will break up the proof in two cases.Case  1 (σ<0). Let u be the solution of problem (3) given by Theorem 15. We claim that u satisfies (10).Letc=M-σ/1-σ and put v≔1/cVp.Using hypothesis(H) and Lemmas 5, 6, we deduce that(84)∫01rn-11-rprdr<∞.So, it follows by Proposition 10 and (83), that v satisfies (85)-t1-ntn-1v′′≤vt,t∈0,1,limt→0tn-1v′t=0,v1=0.Therefore by Lemma 12, we have (86)1cVp≤u.Similarly, we show that (87)u≤cVp.Now since by (83), Vp(t)≈θ(t), we deduce that u(t)≈θ(t).Case  2 (0≤σ<1). We shall use a fixed point argument to construct a solution of problem (3). Let φ(t)=tn-2θ(t), for t∈[0,1]. By (83), we have(88)1Mφt≤tn-2Vpt≤Mφt.Put c=M1/1-σ and consider the closed convex set given by (89)E=v∈C0,1,1cφ≤v≤cφ.Clearly φ∈E.We define the operatorF on E by (90)Fvt≔tn-2∫01Gt,rarr2-nσvσrdr,t∈0,1.Using (88), for v∈E, we have (91)1cφ≤Fv≤cφ.Since for each r>0, the function t→tn-2G(t,r) is in C0([0,1]), it follows by (23), (84) and the convergence dominated theorem that FE⊂E.Letvkk⊂C([0,1]) defined by(92)v0=1cφandvk+1=Fvk,fork∈N.Since the operator F is nondecreasing and FE⊂E, we deduce that (93)v0≤v1≤v2≤⋯≤vk≤vk+1≤cφ.Therefore, the sequence vkk converges by the convergence monotone theorem to a function v satisfying for each t∈[0,1](94)1cφt≤vt≤cφtandvt=tn-2∫01Gt,rarr2-nσvσrdr.Using (23) and (84), we prove that v∈C([0,1]).Putu(t)=t2-nv(t). Then u∈Cn-2([0,1]) and satisfies the equation (95)ut=Vauσt.Now since the function r↦rn-1(1-r)a(t)uσ(r)∈C(0,1]∩L10,1, it follows from Proposition 9 that u is a positive solution of problem (3) satisfying for t∈(0,1], (96)ut≈θt.For the uniqueness, let u and v be two arbitrary solutions of problem (3) in the cone (97)S=u∈Cn-20,1,ut≈θt.There exists a constant m≥1 such that (98)1m≤uv≤m,in0,1.So the set J≔{m≥1:1/m≤u/v≤m} is not empty. Let m0≔inf⁡J. Then m0≥1 and we have uσ≤m0σvσ.Letz≔m0σv-u. Then(99)-t1-ntn-1z′′=atm0σvσ-uσ≥0in0,1,limt→0tn-1z′t=0,z1=0,Which implies by Proposition 9 that m0σv-u=V(a(m0σvσ-uσ))≥0. By symmetry, we obtain that m0σu≥v. Hence, m0σ∈J. Using the fact that σ<1, we get m0=1. Then, we conclude that u=v.This completes the proof of Theorem2.Example 16.
Letσ<1 and a∈C(0,1) such that (100)ar≈r-μlog3r-β1-r-2log31-r-2,where μ<n+2-nσ and β∈R. Then, by Theorem 2, problem (3) has a unique positive solution u in Cn-2([0,1]) satisfying the following estimates:(i) If2<μ<n+2-nσ, then for t∈(0,1), (101)ut≈t2-μ/1-σlog3t-β/1-σlog31-t-1/1-σ.(ii) Ifμ=2 and β>1 or μ<2, then for t∈(0,1), (102)ut≈log31-t-1/1-σ.(iii) Ifμ=2 and β=1, then for t∈(0,1), (103)ut≈log23t1/1-σlog⁡31-t-1/1-σ.(iv) Ifμ=2 and β<1, then for t∈(0,1), (104)ut≈log⁡3t1-β/1-σlog⁡31-t-1/1-σ.



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*Source: 3572132-2019-02-03.xml*