Let α∈(1,2],β∈(0,1) with α−β>1. This paper focused on the multiplicity of positive solutions for a singular tempered fractional boundary value problem
{ −R0Dtα,λu(t)=p(t)h(eλtu(t),R0Dtβ,λu(t)),t∈(0,1),R0Dtβ,λu(0)=0, R0Dtβ,λu(1)=0,
where h∈C([0,+∞)×[0,+∞),[0,+∞)) and p∈L1([0,1],(0,+∞)). By applying reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the above equation were established. The interesting points were that the nonlinearity contained the lower order tempered fractional derivative and that the weight function can have infinite many singular points in [0,1].
Citation: Xinguang Zhang, Yongsheng Jiang, Lishuang Li, Yonghong Wu, Benchawan Wiwatanapataphee. Multiple positive solutions for a singular tempered fractional equation with lower order tempered fractional derivative[J]. Electronic Research Archive, 2024, 32(3): 1998-2015. doi: 10.3934/era.2024091
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Let α∈(1,2],β∈(0,1) with α−β>1. This paper focused on the multiplicity of positive solutions for a singular tempered fractional boundary value problem
{ −R0Dtα,λu(t)=p(t)h(eλtu(t),R0Dtβ,λu(t)),t∈(0,1),R0Dtβ,λu(0)=0, R0Dtβ,λu(1)=0,
where h∈C([0,+∞)×[0,+∞),[0,+∞)) and p∈L1([0,1],(0,+∞)). By applying reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the above equation were established. The interesting points were that the nonlinearity contained the lower order tempered fractional derivative and that the weight function can have infinite many singular points in [0,1].
In this paper, we consider the multiplicity of positive solutions for the following singular tempered fractional equation with lower order tempered fractional derivative
{ −R0Dtα,λu(t)=p(t)h(eλtu(t),R0Dtβ,λu(t)), t∈(0,1),R0Dtβ,λu(0)=0, R0Dtβ,λu(1)=0, | (1.1) |
where α∈(1,2],β∈(0,1) with α−β>1, h∈C([0,+∞)×[0,+∞),[0,+∞)), p∈L1((0,1),(0,+∞)), which implies that the weight function can have infinite many singular points in [0,1].
The equation (1.1) contains a tempered fractional derivative R0Dtα,λ, which is actually obtained by multiplying an exponential factor in the Riemann-Liouville fractional derivative R0DtDtα, i.e., the following relationship exists between tempered fractional derivative and Riemann-Liouville fractional derivative
R0Dtα,λu(t)=e−λtR0DtDtα(eλtu(t)). | (1.2) |
For the definition of the standard Riemann-Liouville fractional derivative and integral, we refer the reader to [1,2,3,4,5].
As the optimization of the Riemann-Liouville fractional derivative, the tempered fractional derivative has many advantages, which not only overcomes the defect of using the power law of the classical fractional derivative in the mathematical sense, such as the Riemann-Liouville fractional derivatives [6,7], the Caputo fractional derivatives [8], Hadamard fractional derivatives [9,10,11] etc, but also brings many practical applications. It especially describes the anomalous diffusion phenomena in Brownian motion with the semi-heavy tails or semi-long range dependence, such as the limits of random walk with an exponentially tempered jump distribution [12,13], transient super-diffusion [14], anomalous diffusions in heterogeneous systems [15], exponentially tempering Lévy flights with both the α-stable and Gaussian trends [16], and pure jump Lévy process with the fractional derivatives in the risk management of financial derivatives traded over the counter [17].
In recent work [18], an upper and lower solutions technique has been employed to establish the existence of positive solutions for a singular tempered fractional turbulent flow model in a porous medium
{ R0Dtα,λ(φp(R0Dtβ,λu(t)))=f(t,u(t)), t∈(0,1),u(0)=0, R0Dtβ,λu(0)=0, u(1)=∫10e−λ(1−t)u(t)dt, | (1.3) |
with α∈(0,1], β∈(1,2] and the nonlinearity f is decreasing in the second variable. In [19], in view of the monotone iterative method, the iterative properties of positive solutions for a tempered fractional equation
{ R0Dtϑ,μu(t)=f(t,eμtu(t),R0Dtδ,μu(t)), t∈(0,1),R0Dtδ,μu(0)=0, R0Dtδ,μu(1)=∫10e−μ(1−t)R0Dtδ,μu(t)dt, | (1.4) |
were established, where ϑ∈(1,2], δ∈(0,1) with ϑ−δ>1, μ is a positive constant, and f:(0,1)×[0,+∞)×[0,+∞)→[0,+∞) is a continuous and nondecreasing function with respect to the two space variables. Recently, by using spaces theories [20,21,22,23,24], smooth theories [25,26,27], operator method [28,29], the method of moving sphere [30], critical point theories [31,32,33,34] and so on, some other types of fractional equations was also studied [35,36,37,38,39,40,41,42,43,44,45,46].
However, when the nonlinearity contains lower order tempered fractional derivative, the results of multiplicity of positive solutions have not yet been obtained. In order to fill this gap, by applying the reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the above equation are established in this paper. The interesting points are that the nonlinearity contains the lower order tempered fractional derivative and the weight function can have infinitely many singular points in [0,1].
In this section, we first recall definitions and some useful properties of the Riemann-Liouville fractional derivative and integral.
Definition 2.1 ((2.1.1) on page 69 in [1]). The Riemann-Liouville fractional integral of order α>0 of a function u:(0,+∞)→R is given by
Iαu(t)=1Γ(α)∫t0(t−s)α−1u(s)ds |
provided that the righthand side is pointwise and defined on (0,+∞).
Definition 2.2 ((2.1.5) on page 70 in [1]). The Riemann-Liouville fractional derivative of order α>0 of a function u:(0,+∞)→R is given by
DDαu(t)=1Γ(n−α)(ddt)n∫t0(t−s)n−α−1u(s)ds, |
where n=[α]+1, [α] denotes the greatest integer part of the number α, provided that the righthand side is pointwise and defined on (0,+∞).
The following properties of the Riemann-Liouville fractional derivative and integral can be found on pages 73–75 (Lemmas 2.3–2.5) in [1].
Lemma 2.1. Suppose u(t)∈C[0,1]∩L1[0,1] and α>β>0. Let n=[α]+1, then
(i)
IαR0DtDtαu(t)=u(t)+c1tα−1+c2tα−2+⋅⋅⋅+cntα−n, | (2.1) |
where ci∈R,i=1,2,3,...,n.
(ii)
IαIβu(t)=Iα+βu(t), R0DtDtβIαu(t)=Iα−βu(t), R0DtDtβIβu(t)=u(t). |
Lemma 2.2. Let p∈L1([0,1],(0,+∞)), then the singular linear tempered fractional equation
{ −R0Dtα−β,λu(t)=p(t),u(0)=0, u(1)=0, | (2.2) |
has a unique positive solution u(t) provided that 1<α−β≤2, which can be expressed by
u(t)=∫10H(t,s)p(s)ds, | (2.3) |
where
H(t,s)={ tα−β−1(1−s)α−β−1−(t−s)α−β−1Γ(α−β)e−λteλs,0≤s≤t≤1;tα−β−1(1−s)α−β−1Γ(α−β)e−λteλs, 0≤t≤s≤1. | (2.4) |
is the Green function of (2.2).
Proof. In fact, it follows from 1<α−β≤2, (1.2) and (2.1) that
eλtu(t)=−1Γ(α−β)∫t0(t−s)α−β−1eλsp(s)ds+b1tα−β−1+b2tα−β−2,t∈[0,1]. |
Since u(0)=0 and u(1)=0, we have b2=0 and
b1=1Γ(α−β)∫10(1−s)α−β−1eλsp(s)ds. |
Thus
u(t)=1Γ(α−β)[∫10(1−s)α−β−1e−λteλsp(s)dstα−β−1−∫t0(t−s)α−β−1e−λteλsp(s)ds]=1Γ(α−β)∫t0[tα−β−1(1−s)α−β−1−(t−s)α−β−1]e−λteλsp(s)ds+1Γ(α−β)∫1ttα−β−1(1−s)α−β−1e−λteλsp(s)ds=∫10H(t,s)p(s)ds, t∈[0,1]. |
The following Lemma has been proven (see Lemma 2.3 of [6] or Lemma 3 of [4]).
Lemma 2.3. Suppose α∈(1,2],β∈(0,1) with α−β>1, then H(t,s) is a nonnegative continuous function in [0,1]×[0,1] satisfying, for any (t,s)∈[0,1]×[0,1],
tα−β−1(1−t)e−λtΓ(α−β)(1−s)α−β−1seλs≤H(t,s)≤tα−β−1(1−t)e−λtΓ(α−β) or ((1−s)α−β−1seλsΓ(α−β)). | (2.5) |
Suppose α∈(1,2],β∈(0,1) with α−β>1. In order to use the reducing order technique, we introduce the following integral transformation
u(t)=e−λtIβ(eλty(t)), t∈[0,1] | (2.6) |
and then consider the following reducing order problem
{ −R0Dtα−β,λy(t)=p(t)h(Iβ(eλty(t)),y(t)), t∈(0,1),y(0)=0, y(1)=0. | (2.7) |
Lemma 2.4. Suppose α∈(1,2],β∈(0,1) with α−β>1. The reducing order problem (2.7) is equivalent to the singular tempered fractional equation (1.1). In particular, if y is a positive solution of the problem (2.7), then u(t)=e−λtIβ(eλty(t)) is a positive solution of the singular tempered fractional equation (1.1).
Proof. We first suppose that u is a positive solution of the singular tempered fractional equation (1.1). Let
u(t)=e−λtIβ(eλty(t)). |
Noticing that α∈(1,2],β∈(0,1) with α−β>1, let n=[α]+1, i.e., n is the smallest integer greater than or equal to α, then it follows from (1.2) and Lemma 2.1 that
R0Dtα,λu(t))=e−λtR0DtDtα(eλtu(t))=e−λtdndtnIn−α(eλtu(t))=e−λtdndtnIn−α(Iβ(eλty(t)))=e−λtdndtnIn−α+β(eλty(t)))=e−λtR0DtDtα−β(eλty(t)))=R0Dtα−β,λy(t), | (2.8) |
and
R0Dtβ,λu(t))=e−λtR0DtDtβ(eλtu(t))=e−λtR0DtDtβ(Iβ(eλty(t)))=y(t). | (2.9) |
By (2.8) and (2.9), we have
{ −R0Dtα−β,λy(t)=−R0Dtα,λu(t))=p(t)h(eλtu(t),R0Dtβ,λu(t))=p(t)h(Iβ(eλty(t)),y(t)),t∈(0,1),y(0)=R0Dtβ,λu(0)=0, y(1)=R0Dtβ,λu(1)=0, |
thus, y solves the equation (2.7).
Conversely, suppose that y is any solution of the reducing order problem (2.7), then we have
{ −R0Dtα−β,λy(t)=p(t)h(Iβ(eλty(t)),y(t)), t∈(0,1),y(0)=0, y(1)=0. |
Make integral transformation (2.6), similar to (2.8) and (2.9), we have
−R0Dtα,λu(t))=−R0Dtα−β,λy(t), R0Dtβ,λu(t))=y(t). |
Substituting the above formulas into (2.7), we get
−R0Dtα,λu(t))=p(t)h(Iβ(eλty(t)),y(t))=p(t)h(eλtu(t),R0Dtβ,λu(t)), 0<t<1, |
and
R0Dtβ,λu(0)=0, R0Dtβ,λu(1)=0, |
which implies that u(t)=e−λtIβ(eλty(t)) solves the singular tempered fractional equation (1.1).
Suppose E=C([0,1];R) with the norm
||y||=maxt∈[0,1]|y(t)|. |
Define a cone of E and an operator T, respectively:
P={y∈E:y(t)≥tα−β−1(1−t)e−λt||y||}, |
and
(Ty)(t)=∫10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds. | (2.10) |
In order to obtain the existence of a positive solution of the Eq (1.1), Lemma 2.4 indicates that we only consider the fixed points of the operator T.
Now, we list the following hypotheses, which are used in the rest of this paper.
(C1)h∈C([0,+∞)×[0,+∞),[0,+∞)) and p∈L1((0,1),(0,+∞)).
(C2) There is a constant n>0 such that for any 0≤u+v≤(1+eλΓ(β+1))n,
h(u,v)<μn, |
where
μ=[neλΓ(α−β)∫10p(s)ds]−1. |
(C3) There is a constant ρ>0 such that for any
(Γ(α−β)(14)α−1Γ(α)+(14)α−βe−34λ)ρ≤u+v≤(1+eλΓ(β+1))ρ |
it implies
h(u,v)≥ϱρ, |
where
ϱ=8α−βe14λΓ(α−β)[∫3414p(s)ds]−1. |
Lemma 2.5. Assume α∈(1,2],β∈(0,1) with α−β>1 and (C1) holds, then the operator T:P→P is completely continuous.
Proof. First by (C1), T is continuous on [0,1]. For any y∈P, there exists a constant M>0 such that ||y||≤M, then
0≤Iβ(eλsy(s))=∫t0(t−s)β−1eλsy(s)Γ(β)ds≤MeλΓ(β). | (2.11) |
Let
ℵ=max(u,v)∈[0,MeλΓ(β)]×[0,M]h(u,v), |
then by (2.11), for any y∈P, we have
||Ty||=maxt∈[0,1]∫10H(t,s)p(s)h(Iβ(eλty(s)),y(s))ds≤∫10(1−s)α−β−1seλsΓ(α−β)p(s)h(Iβ(eλty(s)),y(s))ds≤ℵ∫10(1−s)α−β−1seλsΓ(α−β)p(s)ds≤ℵeλΓ(α−β)∫10p(s)ds<+∞, | (2.12) |
which implies that T:P→E is well-defined. In addition, by (2.5) and (2.12), we have
(Ty)(t)≥∫10(1−s)α−β−1seλsΓ(α−β)p(s)h(Iβ(eλty(s)),y(s))ds×tα−β−1(1−t)e−λt≥||Ty||tα−β−1(1−t)e−λt, |
which implies that T(P)⊂P.
In the end, by using the standard arguments and combining the Ascoli-Arzela theorem, we know T(P)⊂P is completely continuous.
Our proof of main results depends on the fixed point theorem of cone expansion and compression (see Theorem 2.3.3 on page 93 of [47]).
Lemma 2.6. [47] Suppose P is a cone of real Banach space E, the bounded open subsets Ω1,Ω2 of E satisfy θ∈Ω1,¯Ω1⊂Ω2. Let T:P∩(¯Ω2∖Ω1)→P be a completely continuous operator such that either
(1) ‖Tz‖≤‖z‖,z∈P∩∂Ω1 and ‖Tz‖≥‖z‖,z∈P∩∂Ω2, or
(2) ‖Tz‖≥‖z‖,z∈P∩∂Ω1 and ‖Tz‖≤‖z‖,z∈P∩∂Ω2;
then T has a fixed point in P∩(¯Ω2∖Ω1).
For the convenience of the proof, we first introduce the following notations whenever the limits exist or not:
h0=limu+v→0h(u,v)u+v, h∞=limu+v→+∞h(u,v)u+v, |
and then state our main results as follows.
Theorem 3.1. Assume that (C1) and (C2) hold, and
h0=+∞,h∞=+∞, |
then the tempered fractional equation (1.1) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that
A1tα−1e−λt≤u1(t)≤B1eλ(1−t), A2tα−1e−λt≤u2(t)≤B2eλ(1−t). |
Proof. It follows from h0=+∞ that there exist 0<m<n and a sufficiently large constant
N>32α−βeλΓ(α−β)∫3414p(s)ds |
such that for any 0<u+v≤(1+eλΓ(β+1))m,
h(u,v)≥N(u+v). |
Take Ωm={y∈E:||y||<m}, and ∂Ωm={y∈E:||y||=m}, then for any y∈P∩∂Ωm, one has
Iβ(eλsy(s))=∫t0(t−s)β−1Γ(β)eλsy(s)ds≤eλmtβΓ(β+1)≤eλmΓ(β+1), | (3.1) |
then
Iβ(eλsy(s))+y(s)≤(1+eλΓ(β+1))m. | (3.2) |
Thus,
||Ty||≥(Ty)(12)=∫10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds≥2−α+βe−12λΓ(α−β)∫10(1−s)α−β−1seλsp(s)N(Iβ(eλsy(s))+y(s))ds≥2−α+βe−12λΓ(α−β)∫10(1−s)α−β−1seλsp(s)Ny(s)ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)Ny(s)ds≥N8α−βe54λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)ds||y||≥N32α−βeλΓ(α−β)∫3414p(s)ds||y||≥||y||. | (3.3) |
So, for any y∈P∩∂Ωm, we have ||Ty||≥||y||.
Next, let Ωn={y∈E:||y||<n} and ∂Ωn={y∈E:||y||=n}, then it follows from (3.2) and (C2) that for any y∈P∩∂Ωn, one has
(Ty)(t)=∫10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds≤∫10(1−s)α−β−1seλsΓ(α−β)p(s)h(Iβ(eλty(s),y(s))ds≤eλΓ(α−β)∫10p(s)h(Iβ(eλty(s),y(s))ds≤μneλΓ(α−β)∫10p(s)ds≤n. | (3.4) |
Therefore for any y∈P∩∂Ωn, we have ||Ty||≤||y||.
On the other hand, it follows from h∞=+∞ that there exists M>0 and
η≥32α−βeλΓ(α−β)(∫3414p(s)ds)−1 |
such that for any u+v>M, we have
h(u,v)≥η(u+v). |
Take R=n+4α−βe34λM and let ΩR={y∈E:||y||<R} and ∂ΩR={y∈E:||y||=R}, then for any y∈P∩∂ΩR and [14,34], one has
y(t)≥(14)α−βe−34λ||y||≥(14)α−βe−34λ(n+4α−βe34λM)≥M. |
Thus, for any y∈P∩∂ΩR, we have
||Ty||≥(Ty)(12)=∫10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)η(Iβ(eλsy(s))+y(s))ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)ηy(s)ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)ηy(s)ds≥η8α−βe54λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)ds||y||≥η32α−βeλΓ(α−β)∫3414p(s)ds||y||≥||y||. | (3.5) |
Therefore, for any y∈P∩∂ΩR, we have ||Ty||≤||y||.
According to Lemma 2.6, T has two fixed points y1∈P∩(¯Ωm∖Ωn) and y2∈P∩(¯ΩR∖Ωn) with m≤||y1||≤n≤||y2||≤R. Thus, it follows from Lemma 2.4 that the tempered fractional equation (1.1) has at least two positive solutions satisfying
mΓ(α−β)tα−1e−λtΓ(α)≤u1(t)=e−λtIβ(eλty1(t))≤neλ(1−t)Γ(β+1), |
and
nΓ(α−β)tα−1e−λtΓ(α)≤u2(t)=e−λtIβ(eλty2(t))≤Reλ(1−t)Γ(β+1). |
Theorem 3.2. Assume that (C1) and (C3) hold and
h0=0,h∞=0, |
then the tempered fractional equation (1.1) has at least two positive solutions u3,u4; moreover, there exist four constants A3,B3,A4,B4>0 such that
A3tα−1e−λt≤u3(t)≤B3eλ(1−t), A4tα−1e−λt≤u4(t)≤B4eλ(1−t). |
Proof. First, notice that h0=0, for any ϵ>0. Let us select 0<κ<ρ such that for any 0<u+v<(1+eλΓ(β+1))κ,
h(u,v)≤ϵ(u+v). |
Choose ϵ such that
ϵeλΓ(α−β)(1+eλΓ(β+1))∫10p(s)ds≤1. |
Now, let Ωκ={y∈E:||y||<κ} and ∂Ωκ={y∈E:||y||=κ}, then for any y∈P∩∂Ωκ, the same as (3.2), we also have
Iβ(eλsy(s))+y(s)≤(1+eλΓ(β+1))κ, | (3.6) |
which implies that for any y∈P∩∂Ωκ, the following estimation is valid:
h(Iβ(eλty(s)),y(s)≤ϵ(Iβ(eλty(s))+y(s)). |
Consequently, for any y∈P∩∂Ωκ, one gets
(Ty)(t)=∫10H(t,s)p(s)h(Iβ(eλty(s)),y(s))ds≤∫10(1−s)α−β−1seλsΓ(α−β)p(s)h(Iβ(eλty(s)),y(s))ds≤eλΓ(α−β)∫10p(s)h(Iβ(eλty(s)),y(s))ds≤eλΓ(α−β)∫10p(s)ϵ(Iβ(eλty(s))+y(s))ds≤eλΓ(α−β)∫10p(s)ϵ(1+eλΓ(β+1))κds≤ϵeλΓ(α−β)(1+eλΓ(β+1))∫10p(s)dsκ≤κ. | (3.7) |
(3.7) implies that ||Ty||≤||y||, y∈P∩∂Ωκ.
Next, let Ωρ={y∈E:||y||<ρ}, and ∂Ωρ={y∈E:||y||=ρ}, then for any y∈P∩∂Ωρ, we have
Iβ(eλssα−β−1y(s))=∫t0(t−s)β−1Γ(β)sα−β−1eλsy(s)ds≥Γ(α−β)tα−1ρΓ(α). | (3.8) |
Thus, it follows from (3.8) that for any y∈P∩∂Ωρ and [14,34], one has
(Γ(α−β)(14)α−1Γ(α)+(14)α−βe−34λ)ρ=Γ(α−β)(14)α−1Γ(α)ρ+(14)α−βe−34λρ=Iβ(eλssα−β−1y(s))+y(t)≤(1+eλΓ(β+1))ρ. | (3.9) |
So, for any y∈P∩∂Ωρ, we have
||Ty||≥(Ty)(12)=∫10H(12,s)p(s)h(Iβ(eλsy(s)),y(s))ds≥2−α+βe−12λΓ(α−β)∫10(1−s)α−β−1seλsp(s)h(Iβ(eλsy(s)),y(s))ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)h(Iβ(eλsy(s)),y(s))ds≥2−α+βe−12λΓ(α−β)∫3414(1−s)α−β−1seλsp(s)ϱρds≥ϱ8α−βe14λΓ(α−β)∫3414p(s)dsρ≥||y||, | (3.10) |
which implies that ||Ty||≥||y|| holds and y∈P∩∂Ωρ.
On the other hand, since h∞=0, for any ϵ>0, there exists M∗>0 such that for any u+v>M∗
h(u,v)≤ϵ(u+v). |
For the above ϵ>0, choose a sufficiently small one such that
ϵeλΓ(α−β)(1+eλΓ(β+1))∫10p(s)ds<1, |
and take
R∗=max{eλΓ(α−β)max0≤u+v≤M∗h(u,v)∫10p(s)ds1−ϵeλΓ(α−β)(1+eλΓ(β+1))∫10p(s)ds,ρ+M∗}. |
Next, let ΩR∗={y∈E:||y||<R∗}, and ∂ΩR∗={y∈E:||y||=R∗}, then for any y∈P∩∂ΩR∗, one has
(Ty)(t)=∫10H(t,s)p(s)h(Iβ(eλty(s),y(s))ds≤∫10(1−s)α−β−1seλsΓ(α−β)p(s)h(Iβ(eλty(s),y(s))ds≤eλΓ(α−β){∫0≤Iβ(eλty(s)+y(s)≤M∗p(s)h(Iβ(eλty(s),y(s))ds +∫M∗≤Iβ(eλty(s)+y(s)≤(1+eλΓ(β+1))R∗p(s)h(Iβ(eλty(s),y(s))ds}≤eλΓ(α−β){max0≤u+v≤M∗h(u,v)∫10p(s)ds +∫M∗≤Iβ(eλty(s)+y(s)≤(1+eλΓ(β+1))R∗p(s)ϵ(Iβ(eλty(s)+y(s))ds}≤eλΓ(α−β){max0≤u+v≤M∗h(u,v)+ϵ(1+eλΓ(β+1))R∗)∫10p(s)ds≤R∗. | (3.11) |
Therefore, for any y∈P∩∂ΩR, we have ||Ty||≥||y||.
According to Lemma 2.6, T has two fixed points y3∈P∩(¯Ωρ∖Ωκ) and y4∈P∩(¯ΩR∗∖Ωρ) with κ≤||y3||≤ρ≤||y4||≤R∗. Thus, it follows from Lemma 2.4 that the tempered fractional equation (1.1) has at least two positive solutions satisfying
κΓ(α−β)tα−1e−λtΓ(α)≤u3(t)=e−λtIβ(eλty3(t))≤ρeλ(1−t)Γ(β+1), |
and
ρΓ(α−β)tα−1e−λtΓ(α)≤u4(t)=e−λtIβ(eλty4(t))≤R∗eλ(1−t)Γ(β+1). |
The tempered fractional diffusion equation has many important applications, including the tempered fractional Langevin and Vasicek differential equations [48] and the space-time tempered fractional diffusion-wave equation [49]. The new theorems established in the present paper are very useful in the area of tempered fractional calculus. We can find the multiple positive solutions for the singular tempered fractional equations with the lower order tempered fractional derivatives using the proposed theorems.
Let us apply the main results to solve two singular tempered fractional equations with lower order tempered fractional derivative.
Example 4.1. Let α=53, β=13, λ=3, and
h(u,v)={ (u+v)12, 0≤u+v≤47,(u+v)24732, u+v>47. |
We consider the multiplicity of positive solutions for the following singular tempered fractional equation with the lower order tempered fractional derivative:
{ −R0Dt53,3u(t)=|12−t|12h(e3tu(t),R0Dt13,3u(t)),0≤t≤1,R0Dt13,3u(0)=0, R0Dt13,3u(1)=0. | (4.1) |
Conclusion. The singular tempered fractional equation (4.1) has at least two positive solutions u1,u2: moreover, there exist four constants A1,B1,A2,B2>0 such that
A1t23e−3t≤u1(t)≤B1e3(1−t), A2t23e−3t≤u2(t)≤B2e3(1−t). |
Proof. Here,
p(t)=|12−t|12, h(u,v)={ (u+v)12600, 0≤u+v≤47,(u+v)2600×4732, u+v>47. |
Clearly, (C1) holds and
h0=limu+v→0h(u,v)u+v=+∞, h∞=limu+v→+∞h(u,v)u+v=+∞. |
In the following, we verify the condition (C2). In fact, take n=2, then
μ=[neλΓ(α−β)∫10p(s)ds]−1=[2e3Γ(43)∫10|12−s|12ds]−1=0.0079, |
and for any 0≤u+v≤(1+eλΓ(β+1))n=2(1+e3Γ(43))=46.9853, we have
h(u,v)<4712600=0.0114<μn=0.0158, |
which implies that (C2) holds.
Consequently, it follows from Theorem 3.1 that the tempered fractional equation (4.1) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that
A1t23e−3t≤u1(t)≤B1e3(1−t), A2t23e−3t≤u2(t)≤B2e3(1−t). |
Example 4.2. Let α=53, β=13, λ=3, and
h(u,v)={ 100√2(u+v)2, 0≤u+v≤12,50(u+v)12, u+v>12. |
We consider the multiplicity of positive solutions for the following singular tempered fractional equation with the lower order tempered fractional derivative
{ −R0Dt53,3u(t)=|12−t|12h(e3tu(t),R0Dt13,3u(t)),0≤t≤1,R0Dt13,3u(0)=0, R0Dt13,3u(1)=0. | (4.2) |
Conclusion. The tempered fractional equation (4.2) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that
A1t23e−3t≤u1(t)≤B1e3(1−t), A2t23e−3t≤u2(t)≤B2e3(1−t). |
Proof. Here
p(t)=|12−t|12, h(u,v)={ 100√2(u+v)2, 0≤u+v≤12,50(u+v)12, u+v>12. |
Clearly, (C1) holds and
h0=limu+v→0h(u,v)u+v=0, h∞=limu+v→+∞h(u,v)u+v=0. |
Take ρ=2, and we have
ϱ=8α−βe14λΓ(α−β)[∫3414p(s)ds]−1=843e34Γ(43)[∫3414=|12−s|12ds]−1=15.1238. |
For any
0.8184=2(Γ(43)(14)23Γ(53)+(14)43e−94)≤u+v≤2(1+e3Γ(43))=46.9844, |
one has
h(u,v)≥50×0.818412=45.232>ϱρ=30.2476, |
which implies that (C3) holds.
Consequently, it follows from Theorem 3.2 that the tempered fractional equation (4.2) has at least two positive solutions u1,u2; moreover, there exist four constants A1,B1,A2,B2>0 such that
A1t23e−3t≤u1(t)≤B1e3(1−t), A2t23e−3t≤u2(t)≤B2e3(1−t). |
This work studies the multiplicity of positive solutions for a class of singular tempered fractional equations with the lower order tempered fractional derivative. By applying reducing order technique and fixed point theorem, some new results of existence of the multiple positive solutions for the equation are established. The interesting points are the nonlinearity contains the lower order tempered fractional derivative and the weight function can have infinitely many singular points in [0,1]. However, in this study, the conditions α∈(1,2],β∈(0,1) with α−β>1 are required; if 0<α−β<1 or h has singularity at space variables, these interesting problems are still worth future studying.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work is granted by the Natural Science Foundation of Shandong Province of China (ZR2022AM015) and an ARC Discovery Project Grant.
Xinguang Zhang is a guest editor of the special issue for ERA and was not involved in the editorial review or the decision to publish this article. All authors declare that there are no competing interests.
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