Based on properties of Green's function and the some conditions of f(t,u), we found a minimal and a maximal positive solution by the method of sequence approximation. Moreover, based on the properties of Green's function and fixed point index theorem, the existence of multiple positive solutions for a singular p-Laplacian fractional differential equation with infinite-point boundary conditions was obtained and, at last, an example was given to demonstrate the validity of our main results.
Citation: Limin Guo, Weihua Wang, Cheng Li, Jingbo Zhao, Dandan Min. Existence results for a class of nonlinear singular p-Laplacian Hadamard fractional differential equations[J]. Electronic Research Archive, 2024, 32(2): 928-944. doi: 10.3934/era.2024045
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Based on properties of Green's function and the some conditions of f(t,u), we found a minimal and a maximal positive solution by the method of sequence approximation. Moreover, based on the properties of Green's function and fixed point index theorem, the existence of multiple positive solutions for a singular p-Laplacian fractional differential equation with infinite-point boundary conditions was obtained and, at last, an example was given to demonstrate the validity of our main results.
Fractional-order differential operator has become one of the most important tools for mathematical modeling of complex mechanics and physical processes because it can describe mechanical and physical processes with historical memory and spatial global correlation succinctly and accurately. Additionally, and the fractional-order derivative modeling is simple, the physical meaning of parameters is clear and the description is accurate. In recent years, fractional derivative has become an important tool to describe all kinds of complex mechanical and physical behaviors, so the study of a positive solutions of fractional differential equations has attracted much attention. For an extensive collection of such literature, readers can refer to [1,2,2,3,4,5,6,7,8,9,10,11,12]. Alsaedi et al. [13] investigated the following equation
ρDα1−(φP(ρDβ0+u(t)))=v1f(t,u(t),ρDβ0+u(t)))+vρ2Iξ0+g(t,u(t),ρDβ0+u(t))) |
with boundary value condition
u(0)=0,u(1)=λ1u(μ),ρDβ0+u(1)=0,φp(ρDβ0+u(0))=λ2φp(ρDβ0+u(η)) |
where φp(t)=|t|p−2⋅t, 1p+1q=1,p,q>1,1<α,β≤2,ρ>0,ζ>0,0<μ,η<1, 0≤λ1<1μρ(β−1),0≤λ2<1(1−ηρ)α−1, and ρDα0+u and ρDβ0+u denote the right and left fractional derivatives of orders α and β with respect to a power function, respectively. The authors proved the uniqueness of positive solutions for the given problem for the cases 1<p≤2 and p>2 by applying an efficient novel approach together with the Banach contraction mapping principle. Li and Liu [8] considered the fractional differential equation
C0Dαgu(t)+f(t,u)=0,0<t<1 |
with boundary value condition
x(0)=0, C0D1gu(0)=0, C0D1gu(1)=∫10h(t)C0Dvgu(t)g′(t)dt |
where 2<α<3,1<v<2,α−v−1>0, f∈C([0,1]×R+,R+), g′>0, h∈C([0,1],R+), R+=[0,+∞), and the existence of multiple solutions for the following system by the fixed point theorem are on cone.
In recent years, more and more scientists have devoted themselves to the study of Hadamard fractional differential systems. For the part of outstanding results of Hadamard's research on fractional differential systems, please refer to [14,15,16]. Ardjouni[14] studied the following Hadamard fractional differential equations
HDα1+u(t)+ϕ(t,u(t))=HDβ1+φ(t,u(t)),1<t<e |
with integral boundary conditions
u(1)=0,u(e)=1Γ(α−β)∫e1(loges)α−β−1g(s,x(s))dss |
where 1<α≤2,0<β≤α−1, g,f:[1,e]×[0,∞)→[0,∞) are given continuous functions, ϕ is not required for any monotone assumption, and φ is nondecreasing on x. The authors get the existence and uniqueness of the positive solution by the method of upper and lower solutions and Schauder and Banach fixed point theorems. In [15], Berhail and Tabouche studied the following fractional differential equation
HDα1+u(t)=f(t,u(t),u′(t)),1<t<T,T<e |
with integral boundary conditions
u″(1)=0,u(3)(1)=0,u(1)+au′(1)=∫T1g1(s)u(s)ds,u(T)−bu′(T)=∫T1g2(s)u(s)ds |
where 3<α≤4, g1,g2∈C([1,T],[0,+∞)), a,b>0, and HDα1+ denotes the Hadamard factional order of α. Based on the properties of Green's function, the authors get the existence of a positive solution for the equation in [15] by the Avery-Peterson fixed point theorem. In [16], the authors studied the existence of a positive solution and stability analysis of the following equation
HDβ1+(ϕp(Dα1+x))(t)=f(t,x),1<t<e |
with integral boundary conditions
x(1)=HDα1+x(1)=u′(1)=u′(e)=0,ϕp(HDα1+u(e))=μ∫e1ϕp(HDα1+x(t))dtt |
where α,β, and μ are three positive real numbers with α∈(2,3],β∈(1,2], and μ∈[0,β), ϕp(s)=|s|p−2s is the P-Laplacian for p>1,s∈R, f is a continuous function on [1,e]×R, and HDα1+,HDβ1+ is the Hadamard fractional differential equation.The authors get the positive solutions by using the fixed point methods.
Motivated by the excellent results above, in this paper, we will devote to considering the following infinite-point singular p-Laplacian Hadamard fractional differential equation:
HDα1+(φp(HDγ1+u))(t)+f(t,u(t))=0,1<t<e | (1.1) |
with infinite-point boundary condition
u(j)(1)=0,j=0,1,2,…,n−2;HDr11+u(e)=∞∑j=1ηHjDr21+u(ξj),HDγ1+u(1)=0;φp(HDγ1+u(e))=∞∑i=1ζjφp(HDγ1+u(ξj)) | (1.2) |
where α,γ∈R+=[0,+∞), 1<α≤2, n−1<γ≤n(n≥3), r1,r2∈[2,n−2], r2≤r1, p-Laplacian operator φp is defined as φp(s)=|s|p−2s, p, q>1, 1p+1q=1, and 0<ηi,ζi<1,1<ξi<e(i=1,2,…,∞), f∈C([1,e]×R1+,R1+))(R1+=[0,+∞), and HDα1+u,HDγ1+u, HDri1+u(i=1,2) are the standard Hadamard fractional-order derivatives.
In this paper, we investigate the existence of positive solutions for a singular infinite-point p-Laplacian boundary value problem. Compared with [15], the equation in this paper is a p-Laplacian fractional differential equation and the method in which we used is a fixed point index and sequence approximation. Compared with [16,17], value at infinite points are involved in the boundary conditions of the boundary value problem (1.1, 1.2) and the minimal positive solution and maximal positive solution are obtained in this paper.
For some basic definitions and lemmas about the theory of Hadamard fractional calculus, the reader can refer to the recent literature such as[6,9,18].
Definition 2.1([18,19]). The Hadamard fractional integral of α(α>0) order of a function ℏ:(0,∞)→R1+ is given by
HIα1+ℏ(t)=1Γ(α)∫t1(lnts)α−1ℏ(s)sds |
Definition 2.2([18,19]). The Hadamard fractional derivative of α(α>0) order of a continuous function ℏ:(0,∞)→R1+ is given by
HDα1+ℏ(t)=1Γ(n−α)(tddt)n∫t1ℏ(s)s(lnts)α−n+1ds |
where n=[α]+1 and [α] denotes the integer part of the number α, provided that the righthand side is pointwise defined on (0,∞).
Lemma 2.1([18,19]). If α,β>0, then
HIα1+(lnx)β−1=Γ(β)Γ(β+α)(lnx)β+α−1,HDα1+(lnx)β−1=Γ(β)Γ(β−α)(lnx)β−α−1 |
Lemma 2.2([19]). Suppose that α>0 and ℏ∈C[0,∞)∩L1[0,∞), then the solution of Hadamard fractional differential equation HDα1+ℏ(t)=0 is
ℏ(t)=c1(lnt)α−1+c2(lnt)α−2+⋯+cn(lnt)α−n,ci∈R(i=0,1,⋯,n),n=[α]+1 |
Lemma 2.3([19]). Suppose that α>0,α is not a natural number and ℏ∈C[1,∞)∩L1[1,∞), then
ℏ(t)=HIα1+HDα1+ℏ(t)+∑nk=1ck(lnt)α−k |
for t∈(1,e] where ck∈R(k=1,2,⋯,n), and n=[α]+1.
Lemma 2.4. Let y∈L1(1,e)∩C(1,e), then the equation of the BVPs
−HDγ1+u(t)=y(t), 1<t<e | (2.1) |
with boundary condition u(j)(1)=0(j=0,1,2,…,n−2),HDr11+u(e)=∑∞j=1ηHjDr21+u(ξj) has integral representation
u(t)=∫e1Ψ(t,s)y(s)dss | (2.2) |
where
Ψ(t,s)=1ΔΓ(γ){Γ(γ)(lnt)γ−1Ξ(s)(lne−lns)γ−r1−1−Δ(lnt−lns)γ−1,1≤s≤t≤e,Γ(γ)(lnt)γ−1Ξ(s)(lne−lns)γ−r1−1, 1≤t≤s≤e | (2.3) |
in which
Ξ(s)=1Γ(γ−r1)−1Γ(γ−r2)∑s≤ξjηj(lnξj−lnslne−lns)γ−r2−1(lne−lns)r1−r2, |
Δ=Γ(γ)Γ(γ−r1)−Γ(γ)Γ(γ−r2)∞∑j=1ηjlnξγ−r2−1j≠0 | (2.4) |
Proof.By means of the Lemma 2.3, we reduce (2.1) to an equivalent integral equation
u(t)=−HIγ1+y(t)+C1(lnt)γ−1+C2(lnt)γ−2+⋯+Cn(lnt)γ−n |
for Ci(i=1,2,⋯,n)∈R. From u(0)=0, we have Cn=0, then taking the first derivative, we have
u′(t)=−HIγ−11+y(t)+C1(γ−1)(lnt)γ−11t+C2(γ−2)(lnt)γ−31t+⋯+… |
By u′(1)=0, we have Cn−1=0, and taking the derivative step by step and combining u(i)(1)=0,(i=2,⋯,n−2), we have Ci=0(i=2,3,⋯,n−2). Consequently, we get
u(t)=C1(lnt)γ−1−HIγ1+y(t) |
By some properties of the fractional integrals and fractional derivatives, we have
HDr11+u(t)=C1Γ(γ)Γ(γ−r1)(lnt)γ−r1−1−HIγ−r11+y(t),HDr21+u(t)=C1Γ(γ)Γ(γ−r2)(lnt)γ−r2−1−HIγ−r21+y(t) | (2.5) |
On the other hand, HDr11+u(1)=∑∞j=1ηHjDr21+u(ξj), and combining with (2.5), we get
C1=∫e1(lne−lns)γ−r1−1Γ(γ−r1)Δy(s)dss−∞∑j=1ηj∫ξj1(lnξj−lns)γ−r2−1Γ(β−r2)Δy(s)dss=∫e1(lne−lns)γ−r1−1Ξ(s)Δy(s)dss |
where Ξ(s),Δ are as (2.3), then,
u(t)=C1(lnt)γ−1−HIγ1+y(t)=−∫t1Δ(lnt−lns)γ−1Γ(γ)Δy(s)dss+∫e1(lne−lns)γ−r1−1(lnt)γ−1Ξ(s)Δy(s)dss=∫e1Ψ(t,s)y(s)dss |
Therefore, Ψ(t,s) is as (2.3). ◻
Lemma 2.5. The Green functions (2.3) have the following properties:
(i)Ψ(t,s)>0,∂∂tΨ(t,s)>0,1<t,s<e;
(ii)maxt∈[1,e]Ψ(t,s)=Ψ(e,s)=1ΔΓ(γ)[Γ(γ)Ξ(s)(lne−lns)γ−r1−1−Δ(lne−lns)γ−1];
(iii)Ψ(t,s)≥(lnt)α−1Ψ(e,s),1≤t,s≤e
Proof. (i) By simple calculation, we have Ξ′(s)>0. For s∈[1,e], we have Ξ(s)≥Ξ(1), then by the expression of Ξ(s) and Δ, we have
Γ(γ)Ξ(s)≥Δ=Γ(γ)Ξ(1) |
For 1<s≤t<e by the preceding formula, we have
Ψ(t,s)=1ΔΓ(γ)[Γ(γ)(lnt)γ−1Ξ(s)(lne−lns)γ−r1−1−Δ(lnt−lns)γ−1]=1ΔΓ(γ)(lnt)γ−1[Γ(γ)Ξ(s)−(lne−lns)γ−r1−1−Δ(1−lnslnt)γ−1]≥1Γ(γ)(lnt)γ−1[(lne−lns)γ−1−(1−lnslnt)γ−1]≥0 |
For 1<t≤s<e, obviously, Ψ(t,s)>0. Furthermore, by direct calculation, we get
∂∂tΨ(t,s)=1ΔΓ(γ){(γ−1)Γ(γ)(lnt)γ−2Ξ(s)(lne−lns)γ−r1−1−Δ(lnt−lns)γ−1t,1≤s≤t≤e,(γ−1)Γ(γ)(lnt)γ−2Ξ(s)(lne−lns)γ−r1−1t, 1≤t≤s≤e | (2.6) |
Clearly, ∂∂tΨ(t,s) is continuous on [1,e]×[1,e]. By the similar method, for 1<s≤t<e, we get
∂∂tΨ(t,s)=1ΔΓ(γ)t[(γ−1)Γ(γ)(lnt)γ−2Ξ(s)(lne−lns)γ−r1−1−Δ(γ−1)(lnt−lns)γ−2]=γ−1ΔΓ(γ)(lnt)γ−2[Γ(γ)Ξ(s)(lne−lns)γ−r1−1−Δ(1−lnslnt)γ−2]≥γ−1Γ(γ)t(lnt)γ−2[(lne−lns)γ−r1−1−(1−lnslnt)γ−2]≥γ−1Γ(γ)t(lnt)γ−2[(lne−lns)γ−2−(1−lnslnt)γ−2]≥0 |
(ii) By (i), we can easily get Ψ(t,s) is increasing on t; hence, we get
maxt∈[1,e]Ψ(t,s)=Ψ(e,s)=1ΔΓ(γ)[Γ(γ)Ξ(s)(lne−lns)γ−r1−1−Δ(lne−lns)γ−1],1≤s≤e |
(iii) For 1≤s≤t≤e, we have
Ψ(t,s)=1ΔΓ(γ)[Γ(γ)(lnt)γ−1Ξ(s)(lne−lns)γ−r1−1−Δ(lnt−lns)γ−1]=1ΔΓ(γ)(lnt)γ−1[Γ(γ)Ξ(s)−(lne−lns)γ−r1−1−Δ(1−lnslnt)γ−1]≥1ΔΓ(γ)(lnt)γ−1[Γ(γ)Ξ(s)(lne−lns)γ−r1−1−Δ(1−lns)γ−1]=(lnt)γ−1Ψ(e,s) |
For 1≤t≤s≤e, we have
Ψ(t,s)=1Δ(lnt)γ−1Ξ(s)(lne−lns)γ−r1−1=(lnt)γ−1⋅1ΔΞ(s)(lne−lns)γ−r1−1≥(lnt)γ−1Ψ(e,s) |
Lemma 2.6. Let f∈C([1,e]×(0,+∞),[0,+∞)), then the BVP (1.1,1.2) has a solution
u(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss | (2.7) |
where
H(t,s)=1¯ΔΓ(α){(lnt)α−1Γ(α)¯Ξ(s)(lne−lns)α−1−¯Δ(lnt−lns)α−1,1≤s≤t≤e,(lnt)α−1Γ(α)¯Ξ(s)(lne−lns)α−1, 1≤t≤s≤e, | (2.8) |
in which
¯Ξ(s)=1Γ(α)−1Γ(α)∑s≤ξjζj(lnξj−lnslne−lns)α−1, |
¯Δ=1−∞∑j=1ζjlnξα−1j≠0 |
Proof. Let v=φp(HDγ1+u),¯h(t)∈C[1,e], then the Eqs (1.1),(1.2) can be changed into the following equation
HDα1+v(t)+¯h(t)=0,1<t<e,v(1)=0,v(e)=∞∑j=1ζjv(ξj) |
then by the similar method with Lemma 2.4, we get v(t)=∫e1H(t,s)dss, and H(t,s) is as (2.8). Let u(t) be the solution of BVP (1.1,1.2) and let κ(t)=HDγ1+u(t). By Lemma 2.4, we have
u(t)=∫e1Ψ(t,s)κ(s)dss | (2.9) |
Putting v(t)=φp(κ(t)), we have
v(t)=∫e1H(t,s)f(s,u(s))dss | (2.10) |
Combining (2.9) and (2.10), we have (2.7). The proof of Lemma 2.6 is completed.
Lemma 2.7. The Green functions (2.8) have the following properties:
(i)H(t,s)>0,∂∂tH(t,s)>0,1<t,s<e;
(ii)maxt∈[1,e]H(t,s)=H(e,s)=1¯ΔΓ(α)[Γ(α)¯Ξ(s)(lne−lns)α−1−¯Δ(lne−lns)α−1];
(iii)H(t,s)≥(lnt)α−1H(e,s),1≤t,s≤e
Proof. The proof is similar to Lemma 2.5 of this paper and we omit it here.
Lemma 2.8[20]. Let E Be a real Banach space, P⊂E be a cone, and Ωr=u∈P:‖u‖≤r. Let the operator T:P∩Ωr→P be completely continuous and satisfy Tx≠x,∀x∈∂Ωr, then
(i) If ‖Tx‖≤‖x‖,∀x∈∂Ωr, then i(T,Ωr,P)=1;
(ii) If ‖Tx‖≥‖x‖,∀x∈∂Ωr, then i(T,Ωr,P)=0
Now, we define the Banach space E=C([1,e],R), which is assigned a maximum norm, that is, ‖u‖=sup1≤t≤e|u(t)|. Let P={u∈E|u(t)≥0}, then P is a cone in E. Define an operator T:P→P by
(Tu)(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss | (3.1) |
then equation (1.1,1.2) has a solution if, and only if, the operator T has a fixed point.
Lemma 3.1 If f∈C([1,e]×[0,+∞),[0,+∞)), then the operator T:P→P is completely continuous.
Proof. From the continuity and nonnegativeness of Ψ(t,s) and f(t,u(t)), we know that T:P→P is continuous. Let Ω⊂P be bounded, then for all t∈[1,e] and u∈Ω, there exists a positive constant M such that |f(t,u(t))|≤M. Thus,
|(Tu)(t)|=|∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss|≤|∫e1Ψ(e,s)φq(∫e1H(e,τ)dττ)dss×Mq−1|=ωMq−1L |
where
L=∫e1Ψ(e,s)dss, ω=φq(∫e1H(e,τ)dττ) |
which means that T(Ω) is uniformly bounded.
On the other hand, from the continuity of Ψ(t,s) on [1,e]×[1,e], we have Ψ(t,s) is uniformly continuous on [1,e]×[1,e]. Hence, for fixed s∈[1,e] and for any ε>0, there exists a constant δ>0 such that t1,t2∈[1,e] and |t1−t2|<δ, |Ψ(t1,s)−Ψ(t2,s)|<1ωMq−1ε, then for all u∈Ω, we have
|(Tu)(t2)−(Tu)(t1)|≤∫e1|Ψ(t2,s)−Ψ(t1,s)|φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss≤ωMq−1∫e1|Ψ(t2,s)−Ψ(t1,s)|dss<ε |
which implies that T(Ω) is equicontinuous. By the Arzela-Ascoli theorem, we have that T:P→P is completely continuous. The proof is complete.
Theorem 3.1 If f∈C([1,e]×[0,+∞),[0,+∞)), f(t,u) is nondecreasing in u, and λ∈(0,+∞), then BVP(1.1, 1.2) has a minimal positive solution ¯v in Br and a maximal positive solution ¯ϱ in Br. Moreover, vm(t)→¯v(t), ϱm(t)→¯ϱ(t) as m→∞ uniformly on [1,e], where
vm(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,vm−1(τ))dττ)dss | (3.2) |
and
ϱm(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,ϱm−1(τ))dττ)dss | (3.3) |
Proof. Let
Br={u∈P:‖u‖≤r}, |
r≥ωMq−11∫e1Ψ(e,s)dss |
For u∈Br, there exists a positive constant M1 such that |f(t,u(t))|≤M1,
|(Tu)(t)|=|∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss|≤|∫e1Ψ(e,s)φq(∫e1H(e,τ)dττ)dss×Mq−11|=ωMq−11∫e1G(e,s)dss≤r |
hence,
T:Br→Br |
By Lemma 3.1, we get that T:Br→Br is completely continuous. Therefore, by the Schauder fixed point theorem, the operator T has at least one fixed point, and so BVP(1.1, 1.2) has at least one solution in Br. Next, we show that BVP(1.1, 1.2) has a positive solution in Br, which is a minimal positive solution.
From (3.1) and (3.2), we have that
vm(t)=(Tvm−1)(t),t∈[1,e],m=1,2,3,… | (3.4) |
This, together with f(t,u) being nondecreasing in u, yields that
0=v0(t)≤v1(t)≤…≤vm(t)≤…,t∈[1,e] |
Since T is compact, we have that {vm} is a sequentially compact set. Hence, there exists ¯v∈Br such that vm→¯v(m→∞).
Let u(t) be any positive solution of BVP(1.1, 1.2) in Br. Obviously, 0=v0(t)≤u(t)=(Tu)(t); thus,
vm(t)≤u(t),m=0,1,2,…. | (3.5) |
Taking limits as m→∞ in (3.5), we have ¯v(t)≤u(t) for t∈[1,e]. Thus, ¯v is a minimal positive solution, then, we show BVP(1.1, 1.2)has a positive solution in Br, which is a maximal positive solution.
Let ϱ0(t)=r,t∈[1,e] and ϱ1(t)=Tϱ0(t). From T:Br→Br, we get ϱ1∈Br. Therefore,
0≤ϱ1(t)≤r=ϱ0 |
This, together with f(t,u) being nondecreasing in u, yields that
…≤ϱm(t)≤…≤ϱ1(t)≤ϱ0(t),t∈[1,e] |
Using a proof similar to that of the above, we can show that
ϱm(t)→¯ϱ(t)(m→∞) |
and
¯ϱ(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,¯ϱ(τ))dττ)dss |
Let u(t) be any positive solution of BVP(1.1, 1.2) in Br. It is obvious that
u(t)≤ϱ0(t) |
so
u(t)≤ϱm(t) | (3.6) |
Taking limits as m→∞ in (3.6), we have u(t)≤¯ϱ(t) for t∈[1,e]. The proof is complete.
Define
f0=limu→0+supt∈[1,e]f(t,u)φp(l1‖u‖),f0=limu→0+supt∈[1,e]f(t,u)φp(l2‖u‖), |
f∞=limu→+∞supt∈[1,e]f(t,u)φp(l3‖u‖),f∞=limu→+∞supt∈[1,e]f(t,u)φp(l4‖u‖) |
Let
B1=∫e1G(e,s)(lns)(α−1)(q−1)dss |
Theorem 3.2 If f∈C([1,e]×[0,+∞),[0,+∞)) and the following conditions hold:
(H1) f0=f∞=+∞.
(H2) There exists constants λ,ℏ>0 such that f(t,u)≤λq−1φp(l5‖u‖) for t∈[1,e],u∈[0,ℏ], then BVP(1.1, 1.2)has at least two positive solutions u1 and u2 such that
0<‖u1‖<ℏ<‖u2‖ |
for
λ∈(1(l2B1)1q−1κ,1(l5L)1q−1κ)∩(1(l4B1)1q−1κ,1(l5l)1q−1κ) | (3.7) |
where
l2B1>l5L and l4B1>l5L |
Proof. Since
f0=limu→0+supt∈[1,e]f(t,u)φp(l2‖u‖)=+∞ |
there is ℏ0∈(0,ℏ) such that
f(t,u)≥φp(l2‖u‖) for t∈[1,e],u∈[0,ℏ0] |
Let
Ωℏ0={u∈P:‖u‖≤ℏ0} |
then, for any u∈∂Ωℏ0, it follows from Lemma 2.4 that
(Tu)(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss≥λq−1∫e1(lnt)γ−1Ψ(e,s)φq(∫e1H(s,τ)φp(l2‖u‖)dττ)dss≥λq−1l2∫e1(lnt)γ−1Ψ(e,s)φq(∫e1(lns)α−1H(e,τ)dττ)dss‖u‖=λq−1l2φq(κ)∫e1(lnt)γ−1Ψ(e,s)(lns)(α−1)(q−1)dss‖u‖=λq−1l2B1φq(κ)‖u‖ | (3.8) |
where e is as (3.8), κ=∫e1H(e,τ)dττ,B1=∫e1Ψ(e,s)(lns)(α−1)(q−1)dss. Thus,
‖Tu‖≥λq−1l2B1φq(κ)‖u‖ |
This, together with (3.7), yields that
‖Tu‖≥‖u‖,∀u∈∂Ωℏ0 |
By Lemma 2.6, we get
i(T,Ωℏ0,P)=0 | (3.9) |
In view of
f∞=limu→+∞supt∈[1,e]f(t,u)φp(l4‖u‖)=+∞ |
there is ℏ⋆0,ℏ⋆0>ℏ, such that
f(t,u)≥λq−1φp(l4‖u‖),for t∈[1,e],u∈[ℏ⋆0,+∞) |
Let
Ωℏ⋆0={u∈P:‖u‖≤ℏ⋆0} |
then, for any u∈∂Ωℏ⋆0, it follows from Lemma 2.4 that
(Tu)(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss≥λq−1∫e1(lnt)γ−1Ψ(e,s)φq(∫e1H(s,τ)φp(l4‖u‖)dττ)dss≥λq−1l4∫e1(lnt)γ−1Ψ(e,s)φq(∫e1(lns)α−1H(e,τ)dττ)dss=λq−1l4φq(κ)∫e1(lnt)γ−1Ψ(e,s)(lns)(α−1)(q−1)dss‖u‖=λq−1l4B1φq(κ)‖u‖ |
Thus, by (3.7), we have
‖Tu‖≥λq−1L4B1φq(κ)‖u‖ |
This, together with (3.7), yields that
‖Tu‖≥‖u‖,∀u∈∂Ωℏ⋆0 |
By Lemma 2.6, we get
i(T,Ωℏ⋆0,P)=0 | (3.10) |
Finally, let Ωℏ={u∈P:‖u‖≤ℏ}. For any u∈∂Ωℏ, it follows from Lemma 2.3 and (H2) that
(Tu)(t)=∫e1Ψ(t,s)φq(∫e1H(s,τ)f(τ,u(τ))dττ)dss≤λq−1∫e1Ψ(e,s)φq(∫e1H(e,τ)φp(l5‖u‖)dττ)dss≤λq−1l5∫e1Ψ(e,s)φq(∫e1H(e,τ)dττ)dss⋅‖u‖=λq−1l5φq(κ)∫e1Ψ(e,s)dss‖u‖=λq−1l5Lφq(κ)‖u‖ |
where e is as (3.8), L=∫e1Ψ(e,s)dss. Thus,
‖Tu‖≤λq−1l5Lφq(κ)‖u‖ |
This, together with (3.7), yields that
‖Tu‖≤‖u‖,∀u∈∂Ωℏ |
By Lemma 2.6, we get
i(T,Ωℏ,P)=1 | (3.11) |
From (3.9)–(3.11) and ℏ0<ℏ<ℏ⋆0, we get
i(T,Ω⋆ℏ0∖¯Ωℏ,P)=−1,i(T,Ω⋆ℏ∖¯Ωℏ0,P)=1 |
Hence, T has a fixed point u1∈Ω⋆ℏ∖¯Ωℏ0 and a fixed point u2∈Ω⋆ℏ0∖¯Ωℏ. Obviously, u1,u2 are both positive solutions of BVP (1.1, 1.2) and 0<‖u1‖<ℏ<‖u2‖. The proof of Theorem 3.2 is completed. In a similar way, we get the following result.
Corollary 3.1 If f∈C([1,e]×[0,+∞),[0,+∞)), and the following conditions hold:
(H3) f0=f∞=0.
(H4) There exists constants λ,ℏ2>0 such that f(t,u)≥λq−1φp(l6‖u‖) for t∈[1,e],u∈[0,ℏ2], then BVP(1.1, 1.2)has at least two positive solutions u1 and u2 such that
0<‖u1‖<ℏ2<‖u2‖ |
for
λ∈(1(l6B1)1q−1κ,1(l3L)1q−1κ)∩(1(l6B1)1q−1κ,1(l1L)1q−1κ) | (3.12) |
where
l6B1>l3L and l6B1>l1L |
Example. Consider the following infinite-point p−Laplacian fractional differential equations
{HD321+(φ3(HD521+u))(t)+f(t,u(t))=0, 1<t<e,u(1)=u′(1)=0;HDr11+u(1)=∞∑j=1ηHjDr21+u(ξj),HDα1+u(1)=0;φp(HD521+u(e))=∞∑j=1ζiφp(HD521+u(ηi)) | (4.1) |
where γ=52,α=32,r1=32,r2=12,p=3,q=32, ηj=12j2,ξj=e1j4, ζj=12j2,
f(t,u)=(lnt+2)|u(t)|3+|u(t)| |
Clearly, for f∈C([1,e]×[0,∞),[0,∞)), one can have
|f(t,u)|=|(lnt+2)|u(t)|3+|u(t)||≤3 |
By simple calculation, we have
Δ=Γ(γ)Γ(γ−r1)−Γ(γ)Γ(γ−r2)∞∑j=1ηjlnξγ−r2−1j=Γ(52)Γ(52−32)−Γ(52)Γ(52−12)∞∑j=112j2(1j4)2≈38.18, |
¯Δ=1−∞∑i=1ζilnξα−1i=1−∞∑i=1ζilnξ12i=1−12∞∑i=11j4≈1−0.5411=0.4589, |
P(s)=1Γ(γ−r1)−1Γ(γ−r2)∑s≤ξjηj(lnξj−lnslne−lns)γ−r2−1(lne−lns)r1−r2=1Γ(52−32)−1Γ(52−12)∑s≤1j412j2(1j4−lnslne−lns)52−12−1(lne−lns)32−12=1−12∑s≤ξj(1j4−lnslne−lns)(1−s), |
L=∫e1Ψ(e,s)dss=1Δ∫e1P(s)(lne−lns)γ−r1−1ds−1Γ(γ)∫e1(lne−lns)γ−1dss≈138.18∫e1(1−12∑s≤ξj(1j4−lnslne−lns)(lne−lns))(lne−lns)dss−1Γ(52)∫e1(lne−lns)32dss≈0.5600, |
κ=∫e1H(e,s)dss=1¯Δ∫e1¯P(s)(lne−lns)α−1dss−1Γ(α)∫e1(lne−lns)α−1dss≈10.25∫e1(1Γ(32)−1Γ(32)∑s≤ξjηj(lnξj−lnslne−lns)12)(lne−lns)12dss−1Γ(32)∫e1(lne−lns)12dss≈5.08 |
and
B1=∫e1Ψ(e,s)(lns)(α−1)(q−1)dss=∫e1Ψ(e,s)(lns)14dss≈138.18∫e1(1−12∑s≤ξj(1j4−lnslne−lns)(lne−lns))(lne−lns)(lns)14dss−1Γ(52)∫e1(lne−lns)32(lns)14dss≈3.0235 |
Hence, take l2=10,l5=8,L=0.56,l4=6, and we have
l2B1=10×3.0235>l5L=8×0.56,l4B1=6×3.0235>l5L=8×0.56, |
then there exists λ=0.005 such that
λ∈(1(l2B1)1q−1κ,1(l5L)1q−1κ)∩(1(l4B1)1q−1κ,1(l5l)1q−1κ)=(0.00022,0.01)∩(0.0006,0.01), |
Hence, all the conditions of Theorem 3.2 hold and boundary value problem (4.1) has two positive solutions u1,u2.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the referee for his/her valuable comments and suggestions. This research was supported by the National Natural Science Foundation of China (12101086, 12272064, 62273061), the major project of Basic science (Natural science) research in colleges and universities of Jiangsu Province(22KJA460001, 23KJA580001), Changzhou Science and Technology Plan Project (CE20235049), and Open Project of Applied Mechanics and Structure Safety Key Laboratory of Sichuan Province (SZDKF-202102).
The authors declare there is no conflicts of interest.
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