Citation: Rafael Gotardo, Gustavo A. Piazza, Edson Torres, Vander Kaufmann, Adilson Pinheiro. Soil Loss Vulnerability in an Agricultural Catchment in the Atlantic Forest Biome in Southern Brazil[J]. AIMS Geosciences, 2016, 2(4): 345-365. doi: 10.3934/geosci.2016.4.345
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The article is concerned with the solvability of Dirichlet problems of the fractional p-Laplacian equation with impulsive effects, as follows:
{tDαTϕp(0Dαtu(t))+a(t)ϕp(u(t))=λf(t,u(t)),t≠tj,a.e.t∈[0,T],Δ(tDα−1Tϕp(C0Dαtu))(tj)=μIj(u(tj)),j=1,2,⋯,n,n∈N,u(0)=u(T)=0, | (1.1) |
where C0Dαt is the left Caputo fractional derivative, 0Dαt and tDαT are the left and right Riemann-Liouville fractional derivatives respectively, α∈(1/p,1], p>1, ϕp(x)=|x|p−2x (x≠0), ϕp(0)=0, λ>0, μ∈R, a(t)∈C([0,T],R), f∈C([0,T]×R,R), T>0, 0=t0<t1<t2<⋯<tn<tn+1=T, Ij∈C(R,R), and
Δ(tDα−1Tϕp(C0Dαtu))(tj)=tDα−1Tϕp(C0Dαtu)(t+j)−tDα−1Tϕp(C0Dαtu)(t−j), |
tDα−1Tϕp(C0Dαtu)(t+j)=limt→t+jtDα−1Tϕp(C0Dαtu)(t),tDα−1Tϕp(C0Dαtu)(t−j)=limt→t−jtDα−1Tϕp(C0Dαtu)(t). |
Fractional calculus has experienced a growing focus in recent decades because of its application to real-world problems. This kind of problem has attracted the attention of many scholars and produced a series of excellent works [1,2,3,4,5,6,7,8]. In particular, left and right fractional differential operators have been widely used in the study of physical phenomena of anomalous diffusion, specifically, fractional convection-diffusion equations [9,10]. Recently, the equations containing left and right fractional differential operators have become a new field in the theory of fractional differential equations. For example, the authors of [11] first put forward the following fractional convection-diffusion equation:
{−aD(p0D−βt+qtD−βT)Du(t)+b(t)Du(t)+c(t)u(t)=f, a.e. t∈[0,T], 0≤β<1,u(0)=u(T)=0. |
The authors gained the relevant conclusions about the solution of the above-mentioned problems by using the Lax-Milgram theorem. In [12], the authors discussed the following problem:
{ddt(120D−βt(u′(t))+12tD−βT(u′(t)))+∇F(t,u(t))=0, a.e. t∈[0,T], 0≤β<1,u(0)=u(T)=0. |
By applying the minimization principle and mountain pass theorem, the existence results under the Ambrosetti-Rabinowitz condition were obtained. The following year, in [13], the authors made further research on the following issues:
{tDαT(0Dαtu(t))=∇F(t,u(t)), a.e. t∈[0,T], 12<α≤1,u(0)=u(T)=0. |
Use of impulsive differential equations is an effective method to describe the instantaneous change of the state of things, and it can reflect the changing law of things more deeply and accurately. It has practical significance and application value in many fields of science and technology, such as signal communication, economic regulation, aerospace technology, management science, engineering science, chaos theory, information science, life science and so on. Due to the application of impulsive differential equations to practical problems, more and more attention has been paid to them in recent years, and many scholars at home and abroad have studied such problems. For example, in [14,15], using the three critical points theorem, the authors discussed the impulse problems as follows:
{tDαT(C0Dαtu(t))+a(t)u(t)=λf(t,u(t)),t≠tj,a.e.t∈[0,T],α∈(12,1],Δ(tDα−1T(C0Dαtu))(tj)=μIj(u(tj)),j=1,2,⋯,n,u(0)=u(T)=0, |
where λ,μ>0, Ij∈C(R,R), a∈C([0,T]) and there exist a1 and a2 such that 0<a1≤a(t)≤a2. In addition,
Δ(tDα−1T(C0Dαtu))(tj)=tDα−1T(C0Dαtu)(t+j)−tDα−1T(C0Dαtu)(t−j), |
tDα−1T(C0Dαtu)(t+j)=limt→t+j(tDα−1T(C0Dαtu)(t)),tDα−1T(C0Dαtu)(t−j)=limt→t−j(tDα−1T(C0Dαtu)(t)). |
The p-Laplacian equation originated from the nonlinear diffusion equation proposed by Leibenson in 1983, when he studied the problem of one-dimensional variable turbulence of gas passing through porous media:
ut=∂∂x(∂um∂x|∂um∂x|μ−1),m=n+1. |
When m>1, the above equation is the porous medium equation; When 0<m<1, the above equation is a fast diffusion equation; When m=1, the above equation is a heat equation; However, when m=1,μ≠1, such equations often appear in the study of non-Newtonian fluids. In view of the importance of such equations, the above equation has been abstracted into the p-Laplacian equation:
(ϕp(u′))′=f(t,u), |
where ϕp(x)=|x|p−2x (x≠0),ϕp(0)=0,p>1. When p=2, the p-Laplacian equation is reduced to a classical second-order differential equation. Ledesma and Nyamoradi [16] researched the impulse problem with a p-Laplacian operator as below.
{tDαT(|0Dαtu(t)|p−20Dαtu(t))+a(t)|u(t)|p−2u(t)=f(t,u(t)),t≠tj,a.e.t∈[0,T],Δ(tI1−αT(|0Dαtu(tj)|p−20Dαtu(tj)))=Ij(u(tj)),j=1,2,⋯,n,n∈N,u(0)=u(T)=0, | (1.2) |
where α∈(1p,1], p>1, f∈C([0,T]×R,R), Ij∈C(R,R) and
Δ(tI1−αT(|0Dαtu(tj)|p−20Dαtu(tj)))=tI1−αT(|0Dαtu(t+j)|p−20Dαtu(t+j))−tI1−αT(|0Dαtu(t−j)|p−20Dαtu(t−j)),tI1−αT(|0Dαtu(t+j)|p−20Dαtu(t+j))=limt→t+jtI1−αT(|0Dαtu(t)|p−20Dαtu(t)),tI1−αT(|0Dαtu(t−j)|p−20Dαtu(t−j))=limt→t−jtI1−αT(|0Dαtu(t)|p−20Dαtu(t)). |
By using the mountain pass theorem and the symmetric mountain pass theorem, the authors acquired the related results of Problem (1.2) under the Ambrosetti-Rabinowitz condition. If α=1 and a(t)=0, then Problem (1.2) is reduced to the p-Laplacian equation with impulsive effects, as follows:
{−(|u′|p−2u′)′=f(t,u(t)),t≠tj,a.e.t∈[0,1],u(1)=u(0)=0,u(t+j)=u(t−j),j=1,2,⋯,n,Δ|u′(tj)|p−2u′(tj)=Ij(u(tj)),j=1,2,⋯,n. |
This problem has been studied in [17] and [18]. The main methods used in the above literature are the critical point theory and the topological degree theory. To show the major conclusions of literature [16], the following assumptions are first introduced below:
(F1) There are θ>p and r>0, so that 0<θF(t,ξ)≤ξf(t,ξ),∀t∈[0,T],|ξ|≥r;
(F2) f(t,ξ)=o(|ξ|p−1),ξ→0, for ∀t∈[0,T];
(F3) For ∀j, there are cj>0 and γj∈(p−1,θ−1) so that |Ij(ξ)|≤cj|ξ|γj;
(F4) For u large enough, one has Ij(ξ)ξ≤θ∫u0Ij(ξ)dξ,∀j=1,2,⋯,n.
Theorem 1. ([16]). If the conditions (F1)–(F4) hold, then the impulsive problem (1.2) possesses one weak solution.
The research work of this paper is to further study the impulse problem (1.1) on the basis of the above work. To compare with Theorem 1, the supposed conditions and main results are given as below.
(H0) a(t)∈C([0,T],R) satisfies essinft∈[0,T]a(t)>−λ1, where λ1=infu∈Eα,p0∖{0}∫T0|0Dαtu(t)|pdt∫T0|u(t)|pdt>0;
(H1) For ∀t∈R, j=1,2,⋯,m,m∈N, Ij(t) satisfies ∫t0Ij(s)ds≥0;
(H2) There are aj,dj>0 and γj∈[0,p−1) so that |Ij(t)|≤aj+dj|t|γj,∀t∈R;
(H3) The map s→Ij(s)/Ij(s)|s|p−1|s|p−1 is strictly monotonically decreasing on R∖{0};
(H4) The map s→f(t,s)/|s|p−1 is strictly monotonically increasing on R∖{0}, for ∀t∈[0,T];
(H5) f(t,u)=o(|u|p−1)(|u|→0), uniformly for ∀t∈[0,T];
(H6) There are M>0, L>0 and θ>p so that
uf(t,u)−θF(t,u)≥−M|u|p,∀t∈[0,T],|u|≥L, |
where F(t,u)=∫u0f(t,s)ds;
(H7) lim|u|→∞F(t,u)|u|θ=∞, uniformly for ∀t∈[0,T].
Theorem 2. Let f∈C1([0,T]×R,R) and Ij∈C1(R,R). Assume that the conditions (H0)–(H7) hold. Then, Problem (1.1) with λ=μ=1 has at least one nontrivial ground-state solution.
Remark 1. Obviously, the conditions (H6) and (H7) are weaker than (F1) of Theorem 1. In addition, for this kind of problem, the existence of solutions has been discussed in the past, while the ground-state solutions have been rarely studied. Therefore, our finding extends and enriches Theorem 1 in [16].
Next, further research Problem (1.1) with the concave-convex nonlinearity. The function f∈C([0,T]×R,R) studied here satisfies the following conditions:
f(t,u)=f1(t,u)+f2(t,u), | (1.3) |
where f1(t,u) is p-suplinear as |u|→∞ and f2(t,u) denotes p-sublinear growth at infinity. Below, some supposed conditions are given on f1 and f2, as below:
(H8) f1(t,u)=o(|u|p−1)(|u|→0), uniformly for ∀t∈[0,T];
(H9) There are M>0,L>0 and θ>p so that
uf1(t,u)−θF1(t,u)≥−M|u|p,∀t∈[0,T],|u|≥L, |
where F1(t,u)=∫u0f1(t,s)ds;
(H10) lim|u|→∞F1(t,u)|u|θ=∞, uniformly for ∀t∈[0,T];
(H11) There are 1<r<p and b∈C([0,T],R+), R+=(0,∞), so that
F2(t,u)≥b(t)|u|r,∀(t,u)∈[0,T]×R, |
where F2(t,u)=∫u0f2(t,s)ds;
(H12) There is b1∈L1([0,T],R+) so that |f2(t,u)|≤b1(t)|u|r−1,∀(t,u)∈[0,T]×R;
(H13) There are aj,dj>0 and γj∈[0,θ−1) so that |Ij(t)|≤aj+dj|t|γj,∀t∈R;
(H14) For t large enough, Ij(t) satisfies θ∫t0Ij(s)ds≥Ij(t)t;
(H15) For ∀t∈R, Ij(t) satisfies ∫t0Ij(s)ds≥0.
Theorem 3. Assume that the conditions (H0) and (H8)–(H15) hold. Then, the impulse problem (1.1) with λ=μ=1 possesses at least two non-trivial weak solutions.
Remark 2. Obviously, the conditions (H9) and (H10) are weaker than (F1) of Theorem 1. And, the condition (H13) is weaker than the condition (F3) of Theorem 1. Further, the function f studied in Theorem 3 contains both p-suplinear and p-sublinear terms, which is more general. Thus, our finding extends Theorem 1 in [16].
Finally, the existence results of the three solutions of the impulse problem (1.1) in the case of the parameter μ≥0 or μ<0 are considered respectively. We need the following supposed conditions.
(H16) There are L,L1,⋯,Ln>0, 0<β≤p, 0<dj<p and j=1,⋯,n so that
F(t,x)≤L(1+|x|β),−Jj(x)≤Lj(1+|x|dj),∀(t,x)∈[0,T]×R, | (1.4) |
where F(t,x)=∫x0f(t,s)ds and Jj(x)=∫x0Ij(t)dt;
(H17) There are r>0, and ω∈Eα,p0 so that 1p‖ω‖p>r, ∫T0F(t,ω(t))dt>0,n∑j=1Jj(ω(tj))>0 and
Al:=1p‖ω‖p∫T0F(t,ω(t))dt<Ar:=r∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dt. | (1.5) |
Theorem 4. Assume that the conditions (H0) and (H16)–(H17) hold. Then, for every λ∈Λr=(Al,Ar), there is
γ:=min{r−λ∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dtmax|x|≤Λ∞(pr)1pn∑j=1(−Jj(x)),λ∫T0F(t,ω)dt−1p‖ω‖pn∑j=1Jj(ω(tj))} | (1.6) |
so that, for each μ∈[0,γ), the impulse problem (1.1) possesses at least three weak solutions.
(H18) There are L,L1,⋯,Ln>0, 0<β≤p, 0<dj<p and j=1,⋯,n so that
F(t,x)≤L(1+|x|β),Jj(x)≤Lj(1+|x|dj); | (1.7) |
(H19) There are r>0 and ω∈Eα,p0 so that 1p‖ω‖p>r, ∫T0F(t,ω(t))dt>0,n∑j=1Jj(ω(tj))<0 and (1.5) hold.
Theorem 5. Assume that the conditions (H0) and (H18)-(H19) hold. Then, for every λ∈Λr=(Al,Ar), there is
γ∗:=max{λ∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dt−rmax|x|≤Λ∞(pr)1pn∑j=1Jj(x),λ∫T0F(t,ω)dt−1p‖ω‖pn∑j=1Jj(ω(tj))} |
so that, for each μ∈(γ∗,0], the impulse problem (1.1) possesses at least three weak solutions.
Remark 3. The assumptions (H16) and (H18) study both 0<β<p and β=p. When p=2, the assumptions (H16) and (H18) contain the condition 0<β<2 in [14,15]. In addition, this paper allows a(t) to have a negative lower bound, satisfying essinft∈[0,T]a(t)>−λ1, where λ1=infu∈Eα,p0∖{0}∫T0|0Dαtu(t)|pdt∫T0|u(t)|pdt>0, and a(t) in [14,15] has a positive lower bound satisfying 0<a1≤a(t)≤a2. Thus, our conclusions extend the existing results.
This paper studies Dirichlet boundary-value problems of the fractional p-Laplacian equation with impulsive effects. By using the Nehari manifold method, the existence theorem of the ground-state solution of the above impulsive problem is given. At the same time, the p-suplinear condition required for the proof is weakened. This is the research motivation for this paper. There is no relevant research work on this result. In addition, the existence and multiplicity theorems of nontrivial weak solutions to the impulsive problem are given by means of a variational method. In the process of building the proof, the conditions of nonlinear functions with the concave-convex terms are weakened and the conditions of impulsive terms and variable coefficient terms are weakened. Our work extends and enriches the existing results in [14,15,16], which is the innovation of this paper.
Here are some definitions and lemmas of fractional calculus. For details, see [19].
Definition 1. ([19]). Let u be a function defined on [a,b]. The left and right Riemann-Liouville fractional derivatives of order 0≤γ<1 for a function u denoted by aDγtu(t) and tDγbu(t), respectively, are defined by
aDγtu(t)=ddtaDγ−1tu(t)=1Γ(1−γ)ddt(∫ta(t−s)−γu(s)ds), |
tDγbu(t)=−ddttDγ−1bu(t)=−1Γ(1−γ)ddt(∫bt(s−t)−γu(s)ds), |
where t∈[a,b].
Definition 2. ([19]). Let 0<γ<1 and u∈AC([a,b]); then, the left and right Caputo fractional derivatives of order γ for a function u denoted by CaDγtu(t) and CtDγbu(t), respectively, exist almost everywhere on [a,b]. CaDγtu(t) and CtDγbu(t) are respectively represented by
CaDγtu(t)=aDγ−1tu′(t)=1Γ(1−γ)∫ta(t−s)−γu′(s)ds, |
CtDγbu(t)=−tDγ−1bu′(t)=−1Γ(1−γ)∫bt(s−t)−γu′(s)ds, |
where t∈[a,b].
Definition 3. ([20]). Let 0<α≤1 and 1<p<∞. Define the fractional derivative space Eα,p as follows:
Eα,p={u∈Lp([0,T],R)|0Dαtu∈Lp([0,T],R)}, |
with the norm
‖u‖Eα,p=(‖u‖pLp+‖0Dαtu‖pLp)1p, | (2.1) |
where ‖u‖Lp=(∫T0|u(t)|pdt)1/1pp is the norm of Lp([0,T],R). Eα,p0 is defined by closure of C∞0([0,T],R) with respect to the norm ‖u‖Eα,p.
Proposition 1 ([19]). Let u be a function defined on [a,b]. If caDγtu(t), ctDγbu(t), aDγtu(t) and tDγbu(t) all exist, then
caDγtu(t)=aDγtu(t)−n−1∑j=0uj(a)Γ(j−γ+1)(t−a)j−γ,t∈[a,b], |
ctDγbu(t)=tDγbu(t)−n−1∑j=0uj(b)Γ(j−γ+1)(b−t)j−γ,t∈[a,b], |
where n∈N and n−1<γ<n.
Remark 4. For any u∈Eα,p0, according to Proposition 1, when 0<α<1 and the boundary conditions u(0)=u(T)=0 are satisfied, we can get c0Dαtu(t)=0Dαtu(t) and ctDαTu(t)=tDαTu(t),t∈[0,T].
Lemma 1. ([20]). Let 0<α≤1 and 1<p<∞. The fractional derivative space Eα,p0 is a reflexive and separable Banach space.
Lemma 2. ([13]). Let 0<α≤1 and 1<p<∞. If u∈Eα,p0, then
‖u‖Lp≤TαΓ(α+1)‖0Dαtu‖Lp. | (2.2) |
If α>1/1pp, then
‖u‖∞≤C∞‖0Dαtu‖Lp, | (2.3) |
where ‖u‖∞=maxt∈[0,T]|u(t)| is the norm of C([0,T],R), C∞=Tα−1pΓ(α)(αq−q+1)1q>0 and q=pp−1>1.
Combined with (2.2), we think over Eα,p0 with the norm as below.
‖u‖Eα,p=(∫T0|0Dαtu(t)|pdt)1p=‖0Dαtu‖Lp,∀u∈Eα,p0. | (2.4) |
Lemma 3. ([13]). If 1/1pp<α≤1 and 1<p<∞, then Eα,p0 is compactly embedded in C([0,T],R).
Lemma 4. ([13]). Let 1/1pp<α≤1 and 1<p<∞. If the sequence {uk} converges weakly to u in Eα,p0, i.e., uk⇀u, then uk→u in C([0,T],R), i.e., ‖uk−u‖∞→0,k→∞.
To investigate Problem (1.1), this article defines a new norm on the space Eα,p0, as follows:
‖u‖=(∫T0|0Dαtu(t)|pdt+∫T0a(t)|u(t)|pdt)1p. |
Lemma 5. ([16]). If essinft∈[0,T]a(t)>−λ1, where λ1=infu∈Eα,p0∖{0}∫T0|0Dαtu(t)|pdt∫T0|u(t)|pdt>0, then ‖u‖ is equivalent to ‖u‖Eα,p, i.e., there are Λ1, Λ2>0, so that Λ1‖u‖Eα,p≤‖u‖≤Λ2‖u‖Eα,p and ∀u∈Eα,p0, where ‖u‖Eα,p is defined in (2.4).
Lemma 6. Let 0<α≤1 and 1<p<∞. For u∈Eα,p0, by Lemmas 2 and 5 and (2.4), we have
‖u‖Lp≤TαΓ(α+1)‖u‖Eα,p≤Λp‖u‖, | (2.5) |
where Λp=TαΛ1Γ(α+1). If α>1/1pp, then
‖u‖∞≤Tα−1pΓ(α)(αq−q+1)1q‖u‖Eα,p≤Λ∞‖u‖, | (2.6) |
where Λ∞=Tα−1pΛ1Γ(α)(αq−q+1)1q,q=pp−1>1.
Lemma 7. ([19]). Let α>0, p≥1, q≥1 and 1/1pp+1/1qq<1+α, or p≠1, q≠1 and 1/1pp+1/1qq=1+α. Assume that the function u∈Lp([a,b],R) and v∈Lq([a,b],R); then,
∫ba[aD−αtu(t)]v(t)dt=∫bau(t)[tD−αbv(t)]dt. | (2.7) |
By multiplying the equation in Problem (1.1) by ∀v∈Eα,p0 and integrating on [0,T], one has
∫T0tDαTϕp(0Dαtu(t))v(t)dt+∫T0a(t)ϕp(u(t))v(t)dt−λ∫T0f(t,u(t))v(t)dt=0. |
According to Lemma 7, we can get
∫T0tDαTϕp(0Dαtu(t))v(t)dt=−n∑j=0∫tj+1tjv(t)d[tDα−1Tϕp(0Dαtu(t))]=−n∑j=0tDα−1Tϕp(0Dαtu(t))v(t)|tj+1tj+n∑j=0∫tj+1tjϕp(0Dαtu(t))0Dαtv(t)dt=n∑j=1[tDα−1Tϕp(0Dαtu(t+j))v(tj)−tDα−1Tϕp(0Dαtu(t−j))v(tj)]+∫T0ϕp(0Dαtu(t))0Dαtv(t)dt = μn∑j=1Ij(u(tj))v(tj)+∫T0ϕp(0Dαtu(t))0Dαtv(t)dt. |
Definition 4. Let u∈Eα,p0 be one weak solution of the impulse problem (1.1), if
∫T0ϕp(0Dαtu(t))0Dαtv(t)dt+∫T0a(t)ϕp(u(t))v(t)dt+μn∑j=1Ij(u(tj))v(tj)−λ∫T0f(t,u(t))v(t)dt=0 |
holds for ∀v∈Eα,p0.
Define a functional φ:Eα,p0→R as below:
φ(u)=1p‖u‖p+μm∑j=1∫u(tj)0Ij(t)dt−λ∫T0F(t,u(t))dt, | (2.8) |
where F(t,u)=∫u0f(t,s)ds. According to the continuity of the functions f and Ij, it is easy to prove that φ∈C1(Eα,p0,R). In addition,
⟨φ′(u),v⟩=∫T0ϕp(0Dαtu(t))0Dαtv(t)dt+∫T0a(t)ϕp(u(t))v(t)dt+μn∑j=1Ij(u(tj))v(tj)−λ∫T0f(t,u(t))v(t)dt,∀u,v∈Eα,p0. | (2.9) |
Thus, the critical point of φ(u) corresponds to a weak solution of the impulse problem (1.1). The ground-state solution here refers to the minimum energy solution of the functional φ.
Definition 5. ([21]). Let X be a real Banach space, φ∈C1(X,R). For ∀{un}n∈N⊂X, {un}n∈N possesses one convergent subsequence if φ(un)→c(n→∞) and φ′(un)→0 (n→∞). Then, φ(u) satisfies the (PS)c condition.
Lemma 8. ([21]). Let X be a real Banach space and φ∈C1(X,R) satisfy the (PS)c condition. Assume that φ(0)=0 and
(i) there exist ρ, η>0 such that φ|∂Bρ≥η>0;
(ii) there exists an e∈X/B¯Bρ¯Bρ such that φ(e)≤0.
Then, φ has one critical value c≥η. Moreover, c can be described as c=infg∈Γmaxs∈[0,1]φ(g(s)), where Γ={g∈C([0,1],X):g(0)=0,g(1)=e}.
Lemma 9. ([22]). Let X be one reflexive real Banach space, Φ:X→R be one sequentially weakly lower semi-continuous, coercive and continuously Gâteaux differentiable functional whose Gâteaux derivative admits one continuous inverse on X∗ and Ψ:X→R be one continuously Gâteaux differentiable functional whose Gâteaux derivative is compact such that infx∈XΦ(x)=Φ(0)=Ψ(0)=0. Suppose there are r>0 and ¯x∈X with r<Φ(¯x) so that
(i) sup{Ψ(x):Φ(x)≤r}<rΨ(¯x)Φ(¯x),
(ii) for each λ∈Λr=(Φ(¯x)Ψ(¯x),rsup{Ψ(x):Φ(x)≤r}), the functional Φ−λΨ is coercive.
Then, for each λ∈Λr, the functional Φ−λΨ possesses at least three distinct critical points in X.
Define N={u∈Eα,p0∖{0}|G(u)=0}, where G(u)=⟨φ′(u),u⟩=‖u‖p+m∑j=1Ij(u(tj))u(tj)−∫T0f(t,u(t))u(t)dt. Then, any non-zero critical point of φ must be on N. For j=1,2,⋯,m and t∈[0,T], by (H3) and (H4), one has
I′j(u(tj))u2(tj)<(p−1)Ij(u(tj))u(tj),(p−1)f(t,u(t))u(t)<∂f(t,u(t))∂uu2(t). | (3.1) |
So, for u∈N, by (3.1), we get
⟨G′(u),u⟩=p‖u‖p+m∑j=1(I′j(u(tj))u2(tj)+Ij(u(tj))u(tj))−∫T0(∂f(t,u(t))∂u⋅u2(t)+f(t,u(t))u(t))dt=m∑j=1(I′j(u(tj))u2(tj)−(p−1)Ij(u(tj))u(tj))+∫T0((p−1)f(t,u(t))u(t)−∂f(t,u(t))∂u⋅u2(t))dt<0. | (3.2) |
The formula indicates that N has one C1 structure, which is a Nehari manifold. Here are some necessary lemmas to verify Theorem 2.
Lemma 10. Let the assumptions given in (H3) and (H4) be satisfied. Additionally, we assume that u∈N is one critical point of φ|N; then, φ′(u)=0. In other words, N is one natural constraint on φ(u).
Proof. If u∈N is one critical point of φ|N, there is one Lagrange multiplier λ∈R such that φ′(u)=λG′(u). Therefore, ⟨φ′(u),u⟩=λ⟨G′(u),u⟩=0. Combining with (3.2), we know that λ=0, so φ′(u)=0.
To discuss the critical point of φ|N, let us examine the structure of N.
Lemma 11. Let the assumptions given in (H0) and (H7) be satisfied. For ∀u∈Eα,p0∖{0}, there is one unique y=y(u)>0 so that yu∈N.
Proof. The first step is to show that there are ρ,σ>0 such that
φ(u)>0,∀u∈Bρ(0)∖{0},φ(u)≥σ,∀u∈∂Bρ(0). | (3.3) |
It is easy to know that 0 is one strict local minimizer of φ. By (H5), for ∀ε>0, there is δ>0 so that F(t,u)≤ε|u|p,|u|≤δ. So, for u∈Eα,p0,‖u‖=ρ, ‖u‖∞≤Λ∞‖u‖=δ, by (H1), one has
φ(u)=1p‖u‖p+m∑j=1∫u(tj)0Ij(t)dt−∫T0F(t,u(t))dt≥1p‖u‖p−∫T0F(t,u(t))dt≥1p‖u‖p−εTΛp∞‖u‖p. |
Select ε=12pTΛp∞; one has φ(u)≥12p‖u‖p. Let ρ=δΛ∞ and σ=δp2pΛp∞. Therefore, we can conclude that there are ρ,σ>0 so that, for ∀u∈Bρ∖{0}, one has φ(u)>0, and for ∀u∈∂Bρ, one has φ(u)≥σ.
Second, we prove that φ(yu)→−∞ as y→∞. In fact, by (H7), there exist c1,c2>0 so that
F(t,u)≥c1|u|θ−c2,(t,u)∈[0,T]×R. |
By (H2), we have that φ(yu)≤ypp‖u‖p+m∑j=1ajC∞y‖u‖+m∑j=1djyγj+1Cγj+1∞γj+1‖u‖γj+1−c1yθ‖u‖θLθ+Tc2. Because γj∈[0,p−1),p>1,θ>p, φ(yu)→−∞, as y→∞. Let g(y):=φ(yu), where y>0. From the above proof, it can be seen that there exists at least one yu=y(u)>0 so that g(yu)=maxy≥0g(y)=maxy≥0φ(yu)=φ(yuu). Next, we show that, when y>0, g(y) possesses one unique critical point, which must be the global maximum point. In fact, if y is the critical point of g, then
g′(y)=⟨φ′(yu),u⟩=yp−1‖u‖p+m∑j=1Ij(yu(tj))u(tj)−∫T0f(t,yu(t))u(t)dt=0. |
By (3.1), we obtain
g″(y)=(p−1)yp−2‖u‖p+m∑j=1I′j(yu(tj))u2(tj)−∫T0∂f(t,yu(t))∂(yu)⋅u2(t)dt=1y2m∑j=1(I′j(yu(tj))(yu(tj))2−(p−1)Ij(yu(tj))yu(tj))+1y2∫T0((p−1)f(t,yu(t))yu(t)−∂f(t,yu(t))∂yu⋅(yu(t))2)dt<0. | (3.4) |
Therefore, if y is one critical point of g, then it must be one strictly local maximum point, and the critical point is unique. In addition, according to
g′(y)=⟨φ′(yu),u⟩=1y⟨φ′(yu),yu⟩, | (3.5) |
if yu∈N, then y is one critical point of g. Define m=infNφ. By (3.3), we have that m≥inf∂Bρφ≥σ>0.
Lemma 12. Assume that the conditions (H0) and (H7) hold; then, there is u∈N so that φ(u)=m.
Proof. According to the continuity of Ij and f and Lemma 4, it is easy to verify that φ is weakly lower semi-continuous. Let {uk}⊂N be the minimization sequence of φ that satisfies φ(uk)→infNφ=m, so
φ(uk)=m+o(1),G(uk)=0. | (3.6) |
Now, we show that {uk} is bounded in Eα,p0. Otherwise, ‖uk‖→∞ as k→∞. For u∈Eα,p0∖{0}, choose vk=uk‖uk‖; then, ‖vk‖=1. Since Eα,p0 is one reflexive Banach space, there is one subsequence of {vk} (still denoted as {vk}) such that vk⇀v in Eα,p0; then, vk→v in C([0,T],R). On the one hand, combining (2.8) and (H2), one has
∫T0F(t,uk)dt=1p‖uk‖p+m∑j=1∫uk(tj)0Ij(t)dt−φ(uk)≤1p‖uk‖p+m∑j=1ajΛ∞‖uk‖+m∑j=1djΛγj+1∞γj+1‖u‖γj+1+M1, |
where M1>0. Because γj∈[0,p−1),p>1,θ>p, we have that
∫T0F(t,uk)‖uk‖θdt≤o(1),k→∞. | (3.7) |
On the other side, according to the continuity of f, there is M2>0 so that
|uf(t,u)−θF(t,u)|≤M2,∀|u|≤L,t∈[0,T]. |
Combining the condition (H6), we have
uf(t,u)−θF(t,u)≥−M|u|p−M2,∀|u|∈R,t∈[0,T]. | (3.8) |
Combining the conditions (H1) and (H2), we get
m+o(1)=φ(uk)=1p‖uk‖p+m∑j=1∫uk(tj)0Ij(t)dt−∫T0F(t,uk(t))dt≥1p‖uk‖p−1θ∫T0uk(t)f(t,uk(t))dt−Mθ∫T0|u(t)|pdt−M2Tθ≥(1p−1θ)‖uk‖p−1θm∑j=1Ij(uk(tj))uk(tj)−MTθ‖u‖p∞−M2Tθ≥(1p−1θ)‖uk‖p−1θm∑j=1aj‖uk‖∞−1θm∑j=1dj‖uk‖γj+1∞−MTθ‖u‖p∞−M2Tθ. |
This means that there is M3>0 so that limk→∞‖vk‖∞=limk→∞‖uk‖∞‖uk‖≥M3≥0. Therefore, v≠0. Let Ω1={t∈[0,T]:v≠0} and Ω2=[0,T]∖Ω1. According to the condition (H7), there exists M4>0 so that F(t,u)≥0,∀t∈[0,T] and |u|≥M4. Combining with the condition (H5), there exist M5,M6>0 so that F(t,u)≥−M5up−M6,∀t∈[0,T],u∈R. According to the Fatou lemma, one has lim infk→∞∫Ω2F(t,uk)‖uk‖θdt>−∞. Combining with the condition (H7), for t∈[0,T], one has
lim infk→∞∫T0F(t,uk)‖uk‖θdt=liminfk→∞∫Ω1F(t,uk)|uk|θ|vk|θdt+liminfk→∞∫Ω2F(t,uk)|uk|θ|vk|θdt→∞. |
This contradicts (3.7). So, the sequence {uk}k∈N is bounded. Assume that {uk}k∈N possesses one subsequence, still recorded as {uk}k∈N; there exists u∈Eα,p0 so that uk⇀u in Eα,p0, so uk→u in C([0,T],R). For the last step, we show that u≠0. According to the condition (H5), for ∀ε>0, there exists δ>0 so that
f(t,u)u≤ε|u|p,∀(t,u)∈[0,T]×[−δ,δ]. | (3.9) |
Suppose that ‖uk‖∞≤δ; for uk∈N, by (H2) and (3.9), we obtain
Λ−p∞‖uk‖p∞≤‖uk‖p=∫T0f(t,uk(t))uk(t)dt−m∑j=1Ij(uk(tj))uk(tj)≤εT‖uk‖p∞−m∑j=1aj‖uk‖∞−m∑j=1dj‖uk‖γj+1∞. |
There is one contradiction in the above formula, so the hypothesis is not valid, namely, ‖u‖∞=limk→∞‖uk‖∞≥δ>0, so u≠0. According to Lemma 11, there is one unique y>0 so that yu∈N. Because φ is weakly lower semi-continuous,
m≤φ(yu)≤lim_k→∞φ(yuk)≤limk→∞φ(yuk). | (3.10) |
For ∀uk∈N, by (3.4) and (3.5), we get that yk=1 is one global maximum point of g, so φ(yuk)≤φ(uk). Combined with (3.10), one has m≤φ(yu)≤limk→∞φ(uk)=m. Therefore, m is obtained at yu∈N.
The proof process of Theorem 2 is given below.
Proof of Theorem 2. By Lemmas 11 and 12, we know that there exists u∈N so that φ(u)=m=infNφ>0, i.e., u is the non-zero critical point of φ|N. By Lemma 10, one has φ′(u)=0; thus, u is the non-trivial ground-state solution of Problem (1.1).
Lemma 13. Let f∈C([0,T]×R,R) Ij∈C(R,R). Assume that the conditions (H0) and (H8)–(H15) hold. Then, φ satisfies the (PS)c condition.
Proof. Assume that there is the sequence {un}n∈N⊂Eα,p0 so that φ(un)→c and φ′(un)→0 (n→∞); then, there is c1>0 so that, for n∈N, we have
|φ(un)|≤c1,‖φ′(un)‖(Eα,p0)∗≤c1, | (3.11) |
where (Eα,p0)∗ is the conjugate space of Eα,p0. Next, let us verify that {un}n∈N is bounded in Eα,p0. If not, we assume that ‖un‖→+∞ as (n→∞). Let vn=un‖un‖; then, ‖vn‖=1. Since Eα,p0 is one reflexive Banach space, there is one subsequence of {vn} (still denoted as {vn}), so that vn⇀v (n→∞) in Eα,p0; then, vn→v in C([0,T],R). By (H11) and (H12), we get
|f2(t,u)⋅u|≤b1(t)|u|r,|F2(t,u)|≤1rb1(t)|u|r. | (3.12) |
Two cases are discussed below.
Case 1: v≠0. Let Ω={t∈[0,T]||v(t)|>0}; then, meas(Ω)>0. Because ‖un‖→+∞ (n→∞) and |un(t)|=|vn(t)|⋅‖un‖, so for t∈Ω, one has |un(t)|→+∞ (n→∞). On the one side, by (2.6), (2.8), (3.11), (3.12) and (H13), one has
∫T0F1(t,un)dt=1p‖un‖p+m∑j=1∫un(tj)0Ij(t)dt−∫T0F2(t,un)dt−φ(un)≤1p‖un‖p+m∑j=1ajΛ∞‖un‖+m∑j=1djΛγj+1∞‖un‖γj+1+TrΛr∞‖b1‖∞‖un‖r+c1. |
Since γj∈[0,θ−1),θ>p>r>1,
limn→∞∫T0F1(t,un)‖un‖θdt≤o(1),n→∞. | (3.13) |
On the other side, Fatou's lemma combines with the properties of Ω and (H10), so we get
limn→∞∫T0F1(t,un)‖un‖θdt≥limn→∞∫ΩF1(t,un)‖un‖θdt=limn→∞∫ΩF1(t,un)|un(t)|θ|vn(t)|θdt=+∞. |
This contradicts (3.13).
Case 2: v≡0. From (H8), for ∀ε>0, there is L0>0, so that |f1(t,u)|≤ε|u|p−1,|u|≤L0. So, for |u|≤L0, there is ε0>0 so that |uf1(t,u)−θF1(t,u)|≤ε0(1+θ)up. For (t,u)∈[0,T]×[L0,L], there is c2>0 so that |uf1(t,u)−θF1(t,u)|≤c2. Combined with the condition (H9), one has
uf1(t,u)−θF1(t,u)≥−ε0(1+θ)up−c2,∀(t,u)∈[0,T]×R. | (3.14) |
By (H14), we obtain that there exists c3>0, such that
θm∑j=1∫un(tj)0Ij(t)dt−m∑j=1Ij(un(tj))un(tj)≥−c3. | (3.15) |
By (2.6), (2.8), (2.9), (3.11), (3.12), (3.14) and (3.15), we get that there exists c4>0 such that
o(1)=θc1+c1‖un‖‖un‖p≥θφ(un)−⟨φ′(un),un⟩‖un‖p=(θp−1)+1‖un‖p[θm∑j=1∫un(tj)0Ij(t)dt−m∑j=1Ij(un(tj))un(tj)]+1‖un‖p∫T0[unf1(t,un)−θF1(t,un)]dt+1‖un‖p∫T0[unf2(t,un)−θF2(t,un)]dt≥(θp−1)+1‖un‖p∫T0[−ε0(1+θ)upn−c2]dt−1‖un‖p(θr+1)∫T0b1(t)|un|rdt−1‖un‖pc3≥(θp−1)−ε0(1+θ)∫T0|un|p‖un‖pdt−Tc2‖un‖p−1‖un‖p(θr+1)‖b1‖L1‖un‖r∞−1‖un‖pc3≥(θp−1)−ε0(1+θ)∫T0|vn|pdt−Tc2‖un‖p−(θr+1)‖b1‖L1Λr∞‖un‖r−p≥(θp−1),n→∞. |
It is a contradiction. Thus, {un}n∈N is bounded in Eα,p0. Assume that the sequence {un}n∈N possesses one subsequence, still recorded as {un}n∈N; there exists u∈Eα,p0 so that un⇀u in Eα,p0; then, un→u in C([0,T],R). Therefore,
{⟨φ′(un)−φ′(u),un−u⟩→0,n→∞,∫T0[f(t,un(t))−f(t,u(t))][un(t)−u(t)]dt→0,n→∞,m∑j=1(Ij(un(tj))−Ij(u(tj)))(un(tj)−u(tj))→0,n→∞,∫T0a(t)(ϕp(un(t))−ϕp(u(t)))(un(t)−u(t))dt→0,n→∞. | (3.16) |
Through (2.9), we can get
⟨φ′(un)−φ′(u),un−u⟩=∫T0(ϕp(0Dαtun(t))−ϕp(0Dαtu(t)))(0Dαtun(t)−0Dαtu(t))dt+∫T0a(t)(ϕp(un(t))−ϕp(u(t)))(un(t)−u(t))dt+m∑j=1(Ij(un(tj))−Ij(u(tj)))(un(tj)−u(tj))−∫T0[f(t,un(t))−f(t,u(t))][un(t)−u(t)]dt. | (3.17) |
From [23], we obtain
∫T0(ϕp(0Dαtun(t))−ϕp(0Dαtu(t)))(0Dαtun(t)−0Dαtu(t))dt≥{c∫T0|0Dαtun(t)−0Dαtu(t)|pdt,p≥2,c∫T0|0Dαtun(t)−0Dαtu(t)|2(|0Dαtun(t)|+|0Dαtu(t)|)2−pdt,1<p<2. | (3.18) |
If p≥2, by (3.16)–(3.18), one has ‖un−u‖→0 (n→∞). If 1<p<2, by the H¨older inequality, one has ∫T0|0Dαtun(t)−0Dαtu(t)|pdt≤c(∫T0|0Dαtun(t)−0Dαtu(t)|2(|0Dαtun(t)|+|0Dαtu(t)|)2−pdt)p2(‖un‖+‖u‖)p(2−p)2. Thus,
∫T0(ϕp(0Dαtun(t))−ϕp(0Dαtu(t)))(0Dαtun(t)−0Dαtu(t))dt≥c(‖un‖+‖u‖)2−p(∫T0|0Dαtun(t)−0Dαtu(t)|pdt)2p. | (3.19) |
By (3.16), (3.17) and (3.19), one has ‖un−u‖→0 (n→∞). Hence, φ satisfies the (PS)c condition.
The proof of Theorem 3.
Step1. Clearly, φ(0)=0. Lemma 13 implies that φ∈C1(Eα,p0,R) satisfies the (PS)c condition.
Step2. For ∀ε1>0, we know from (H8) that there is δ>0 so that
F1(t,u)≤ε1|u|p,∀t∈[0,T],|u|≤δ. | (3.20) |
For ∀u∈Eα,p0, by (2.5), (2.6), (2.8), (3.12) and (H15), we get
φ(u)=1p‖u‖p+m∑j=1∫u(tj)0Ij(t)dt−∫T0F(t,u(t))dt≥1p‖u‖p−∫T0F(t,u(t))dt≥1p‖u‖p−ε1∫T0|u|pdt−1r∫T0b1(t)|u|rdt≥1p‖u‖p−ε1Λpp‖u‖p−1r‖b1‖L1‖u‖r∞≥(1p−ε1Λpp−Λr∞r‖b1‖L1‖u‖r−p)‖u‖p. | (3.21) |
Choose ε1=12pΛpp; one has φ(u)≥(12p−Λr∞r‖b1‖L1‖u‖r−p)‖u‖p. Let ρ=(r4pΛr∞‖b1‖L1)1r−p and η=14pρp; then, for u∈∂Bρ, we obtain φ(u)≥η>0.
Step3. From (H10), for |u|≥L1, there exist ε2,ε3>0 such that
F1(t,u)≥ε2|u|θ−ε3. | (3.22) |
By (H8), for |u|≤L1, there exist ε4,ε5>0 such that
F1(t,u)≥−ε4up−ε5. | (3.23) |
From (3.22) and (3.23), we obtain that there exist ε6,ε7>0 so that
F1(t,u)≥ε2|u|θ−ε6up−ε7,∀t∈[0,T],u∈R, | (3.24) |
where ε6=ε2Lθ−p1+ε4. For ∀u∈Eα,p0∖{0} and ξ∈R+, by (H13), (2.5), (2.6), (2.8), (3.24) and the H¨older inequality, we have
φ(ξu)≤ξpp‖u‖p+m∑j=1ajξΛ∞‖u‖+m∑j=1djξγj+1Λγj+1∞‖u‖γj+1−ε2ξθ∫T0|u|θdt+ε6Λppξp‖u‖p+ε7T≤(1p+ε6Λpp)ξp‖u‖p+m∑j=1ajξΛ∞‖u‖+m∑j=1djξγj+1Λγj+1∞‖u‖γj+1−ε2ξθ(Tp−θθ∫T0|u(t)|pdt)θp+ε7T≤(1p+ε6Λpp)ξp‖u‖p+m∑j=1ajξΛ∞‖u‖+m∑j=1djξγj+1Λγj+1∞‖u‖γj+1−ε2ξθTp−θp‖u‖θLp+ε7T. |
Since θ>p>1 and γj+1∈[1,θ), the above inequality indicates that φ(ξ0u)→−∞ when ξ0 is large enough. Let e=ξ0u; one has φ(e)<0. Thus, the condition (ii) in Lemma 8 holds. Lemma 8 implies that φ possesses one critical value c(1)≥η>0. The specific form is c(1)=infg∈Γmaxs∈[0,1]φ(g(s)), where Γ={g∈C([0,1],Eα,p0):g(0)=0,g(1)=e}. Hence, there is 0≠u(1)∈Eα,p0 so that
φ(u(1))=c(1)≥η>0,φ′(u(1))=0. | (3.25) |
Step4. Equation (3.21) implies that φ is bounded below in ¯Bρ. Choose σ∈Eα,p0 so that σ(t)≠0 in [0,T]. For ∀l∈(0,+∞), by (2.6), (2.8), (H10), (H11) and (H13), we have
φ(lσ)≤lpp‖σ‖p+m∑j=1ajlΛ∞‖σ‖+m∑j=1djlγj+1Λγj+1∞‖σ‖γj+1−∫T0F2(t,lσ(t))dt≤lpp‖σ‖p+m∑j=1ajlΛ∞‖σ‖+m∑j=1djlγj+1Λγj+1∞‖σ‖γj+1−lr∫T0b(t)|σ(t)|rdt. | (3.26) |
Thus, from 1<r<p and γj∈[0,θ−1), we know that for a small enough l0 satisfying ‖l0σ‖≤ρ, one has φ(l0σ)<0. Let u=l0σ; we have that c(2)=infφ(u)<0,‖u‖≤ρ. Ekeland's variational principle shows that there is one minimization sequence {vk}k∈N⊂¯Bρ so that φ(vk)→c(2) and φ′(vk)→0,k→∞, i.e., {vk}k∈N is one (PS)c sequence. Lemma 13 shows that φ satisfies the (PS)c condition. Thus, c(2)<0 is another critical value of φ. So, there exists 0≠u(2)∈Eα,p0 so that φ(u(2))=c(2)<0,‖u(2)‖<ρ.
Proof. The functionals Φ:Eα,p0→R and Ψ:Eα,p0→R are defined as follows:
Φ(u)=1p‖u‖p,Ψ(u)=∫T0F(t,u(t))dt−μλn∑j=1Jj(u(tj)); |
then, φ(u)=Φ(u)−λΨ(u). We can calculate that
infu∈Eα,p0Φ(u)=Φ(0)=0,Ψ(0)=∫T0F(t,0)dt−μλn∑j=1Jj(0)=0. |
Furthermore, Φ and Ψ are continuous Gâteaux differentiable and
⟨Φ′(u),v⟩=∫T0ϕp(0Dαtu(t))0Dαtv(t)dt+∫T0a(t)ϕp(u(t))v(t)dt, | (3.27) |
⟨Ψ′(u),v⟩=∫T0f(t,u(t))v(t)dt−μλn∑j=1Ij(u(tj))v(tj),∀u,v∈Eα,p0. | (3.28) |
In addition, Φ′:Eα,p0→(Eα,p0)∗ is continuous. It is proved that Ψ′:Eα,p0→(Eα,p0)∗ is a continuous compact operator. Suppose that {un}⊂Eα,p0, un⇀u, n→∞; then, {un} uniformly converges to u on C([0,T]). Owing to f∈C([0,T]×R,R) and Ij∈C(R,R), we have that f(t,un)→f(t,u) and Ij(un(tj))→Ij(u(tj)), n→∞. Thus, Ψ′(un)→Ψ′(u) as n→∞. Then, Ψ′ is strongly continuous. According to Proposition 26.2 in [24], Ψ′ is one compact operator. It is proved that Φ is weakly semi-continuous. Suppose that {un}⊂Eα,p0, {un}⇀u; then, {un}→u on C([0,T]), and liminfn→∞‖un‖≥‖u‖. So, liminfn→∞Φ(un)=liminfn→∞(1p‖un‖p)≥1p‖u‖p=Φ(u). Thus, Φ is weakly semi-continuous. Because Φ(u)=1p‖u‖p→+∞ and ‖u‖→+∞, Φ is coercive. By (3.27), we obtain
⟨Φ′(u)−Φ′(v),u−v⟩=∫T0(ϕp(0Dαtu(t))−ϕp(0Dαtv(t)))(0Dαtu(t)−0Dαtv(t))dt+∫T0a(t)(ϕp(u(t))−ϕp(v(t)))(u(t)−v(t))dt,∀u,v∈Eα,p0. |
From [23], we know that there is c>0 so that
∫T0(ϕp(0Dαtu(t))−ϕp(0Dαtv(t)))(0Dαtu(t)−0Dαtv(t))dt≥{c∫T0|0Dαtu(t)−0Dαtv(t)|pdt,p≥2,c∫T0|0Dαtu(t)−0Dαtv(t)|2(|0Dαtu(t)|+|0Dαtv(t)|)2−pdt,1<p<2. | (3.29) |
If p≥2, then ⟨Φ′(u)−Φ′(v),u−v⟩≥c‖u−v‖p. Thus, Φ′ is uniformly monotonous. When 1<p<2, the H¨older inequality implies
∫T0|0Dαtu(t)−0Dαtv(t)|pdt≤(∫T0|0Dαtu(t)−0Dαtv(t)|2(|0Dαtu(t)|+|0Dαtv(t)|)2−pdt)p2 (∫T0(|0Dαtu(t)|+|0Dαtv(t)|)pdt)2−p2≤c1(∫T0|0Dαtu(t)−0Dαtv(t)|2(|0Dαtu(t)|+|0Dαtv(t)|)2−pdt)p2(‖u‖p+‖v‖p)2−p2, |
where c1=2(p−1)(2−p)2>0. Then,
∫T0|0Dαtu(t)−0Dαtv(t)|2(|0Dαtu(t)|+|0Dαtv(t)|)2−pdt≥c2(‖u‖+‖v‖)2−p(∫T0|0Dαtu(t)−0Dαtv(t)|pdt)2p, |
where c2=1c2p1. Combined with (3.29), we can get
∫T0(ϕp(0Dαtu(t))−ϕp(0Dαtv(t)))(0Dαtu(t)−0Dαtv(t))dt≥c(‖u‖+‖v‖)2−p(∫T0|0Dαtu(t)−0Dαtv(t)|pdt)2p. | (3.30) |
Thus, ⟨Φ′(u)−Φ′(v),u−v⟩≥c‖u−v‖2(‖u‖+‖v‖)2−p. So, Φ′ is strictly monotonous. Theorem 26.A(d) in [24] implies that (Φ′)−1 exists and is continuous. If x∈Eα,p0 satisfies Φ(x)=1p‖x‖p≤r, then, by (2.6), we obtain Φ(x)≥1pΛp∞‖x‖p∞, and
{x∈Eα,p0:Φ(x)≤r}⊆{x:1pΛp∞‖x‖p∞≤r}={x:‖x‖p∞≤prΛp∞}={x:‖x‖∞≤Λ∞(pr)1p}. |
Therefore, from λ>0 and μ≥0, we have
sup{Ψ(x):Φ(x)≤r}=sup{∫T0F(t,x(t))dt−μλn∑j=1Jj(x(tj)):Φ(x)≤r}≤∫T0max|x|≤Λ∞(pr)1pF(t,x)dt+μλmax|x|≤Λ∞(pr)1pn∑j=1(−Jj(x)). |
If max|x|≤Λ∞(pr)1pn∑j=1(−Jj(x))=0, by λ<Ar, we get
sup{Ψ(x):Φ(x)≤r}<rλ. | (3.31) |
If max|x|≤Λ∞(pr)1pn∑j=1(−Jj(x))>0, for μ∈[0,γ), γ=min{r−λ∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dtmax|x|≤Λ∞(pr)1pn∑j=1(−Jj(x)),λ∫T0F(t,ω)dt−1p‖ω‖pn∑j=1Jj(ω(tj))}, we have
sup{Ψ(x):Φ(x)≤r}≤∫T0max|x|≤Λ∞(pr)1pF(t,x)dt+μλmax|x|≤Λ∞(pr)1pn∑j=1(−Jj(x))<∫T0max|x|≤Λ∞(pr)1pF(t,x)dt+r−λ∫T0max|x|≤Λ∞(pr)1pF(t,x)dtmax|x|≤Λ∞(pr)1pn∑j=1(−Jj(x))λ×max|x|≤Λ∞(pr)1pn∑j=1(−Jj(x))<∫T0max|x|≤Λ∞(pr)1pF(t,x)dt+r−λ∫T0max|x|≤Λ∞(pr)1pF(t,x)dtλ=∫T0max|x|≤Λ∞(pr)1pF(t,x)dt+rλ−∫T0max|x|≤Λ∞(pr)1pF(t,x)dt=rλ. |
Thus, (3.31) is also true. On the other side, for μ<γ, one has
Ψ(ω)=∫T0F(t,ω(t))dt−μλn∑j=1Jj(ω(tj))>Φ(ω)λ. | (3.32) |
By combining (3.31) and (3.32), we obtain that Ψ(ω)Φ(ω)>1λ>sup{Ψ(x):Φ(x)≤r}r. This means that the condition (i) of Lemma 9 holds.
Finally, for the third step, we show that, for any λ∈Λr=(Al,Ar), the functional Φ−λΨ is coercive. By (1.4), we obtain
∫T0F(t,x(t))dt≤L∫T0(1+|x(t)|β)dt≤LT+LT‖x‖β∞≤LT+LTΛβ∞‖x‖β,x∈Eα,p0. | (3.33) |
and
−Jj(x(tj))≤Lj(1+|x(tj)|dj)≤Lj(1+‖x‖dj∞)≤Lj(1+Λdj∞‖x‖dj). | (3.34) |
By (3.34), we get
n∑j=1(−Jj(x(tj)))≤n∑j=1Lj(1+Λdj∞‖x‖dj). | (3.35) |
If μλ≥0, for x∈Eα,p0, by (3.33) and (3.35), we have
Ψ(x)≤LT+LTΛβ∞‖x‖β+μλn∑j=1Lj(1+Λdj∞‖x‖dj)=LT+μλn∑j=1Lj+LTΛβ∞‖x‖β+μλn∑j=1LjΛdj∞‖x‖dj. |
Thus, Φ(x)−λΨ(x)≥1p‖x‖p−λ(LT+μλn∑j=1Lj+LTΛβ∞‖x‖β+μλn∑j=1LjΛdj∞‖x‖dj),∀x∈Eα,p0. If 0<β and dj<p, then lim‖x‖→+∞(Φ(x)−λΨ(x))=+∞,λ>0. Thus, Φ−λΨ is coercive. When β=p, Φ(x)−λΨ(x)≥(1p−λLTΛp∞)‖x‖p−λ(LT+μλn∑j=1Lj+μλn∑j=1LjΛdj∞‖x‖dj). Choose L<∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dtprTΛp∞. We have that1p−λLTΛp∞>0, for all λ<Ar. If 0<dj<p, we have thatlim‖x‖→+∞(Φ(x)−λΨ(x))=+∞, for all λ<Ar. Obviously, the functional Φ−λΨ is coercive. Lemma 9 shows that φ=Φ−λΨ possesses at least three different critical points in Eα,p0.
This paper studies the solvability of Dirichlet boundary-value problems of the fractional p-Laplacian equation with impulsive effects. For this kind of problems, the existence of solutions has been discussed in the past, while the ground-state solutions have been rarely studied. By applying the Nehari manifold method, we have obtained the existence result of the ground-state solution (see Theorem 2). At the same time, by the mountain pass theorem and three critical points theorem, some new existence results on this problem were achieved (see Theorems 3–5). In particular, this paper weakens the commonly used p-suplinear and p-sublinear growth conditions, to a certain extent, and expands and enriches the results of [14,15,16]. This theory can provide a solid foundation for studying similar fractional impulsive differential equation problems. For example, one can consider the solvability of Sturm-Liouville boundary-value problems of fractional impulsive equations with the p-Laplacian operator. In addition, the proposed theory can also be used to study the existence of solutions to the periodic boundary-value problems of the fractional p-Laplacian equation with impulsive effects and their corresponding coupling systems.
This research was funded by the Natural Science Foundation of Xinjiang Uygur Autonomous Region (Grant No. 2021D01A65, 2021D01B35), Natural Science Foundation of Colleges and Universities in Xinjiang Uygur Autonomous Region (Grant No. XJEDU2021Y048) and Doctoral Initiation Fund of Xinjiang Institute of Engineering (Grant No. 2020xgy012302).
The authors declare that there is no conflict of interest.
Proof of Theorem 5
Proof. This is similar to the proof process of Theorem 4. Since λ>0, μ∈(γ∗,0], one has
sup{Ψ(x):Φ(x)≤r}=sup{∫T0F(t,x(t))dt−μλn∑j=1Jj(x(tj)):Φ(x)≤r}≤∫T0max|x|≤Λ∞(pr)1pF(t,x)dt−μλmax|x|≤Λ∞(pr)1pn∑j=1Jj(x). |
If max|x|≤Λ∞(pr)1pn∑j=1Jj(x)=0, by λ<Ar, we obtain
sup{Ψ(x):Φ(x)≤r}<rλ. | (8.1) |
For μ∈(γ∗,0], if max|x|≤Λ∞(pr)1pn∑j=1Jj(x)>0, then (A.1) is also true. On the other hand, for μ∈(γ∗,0], we have
Ψ(ω)=∫T0F(t,ω(t))dt−μλn∑j=1Jj(ω(tj))>Φ(ω)λ. | (8.2) |
Combining (A.1) and (A.2), we get Ψ(ω)Φ(ω)>1λ>sup{Ψ(x):Φ(x)≤r}r, which shows that the condition (i) of Lemma 9 holds. Finally, we show that Φ−λΨ is coercive for ∀λ∈Λr=(Al,Ar). For x∈Eα,p0, by (1.7), we get
∫T0F(t,x(t))dt≤LT+LTΛβ∞‖x‖β,Jj(x(tj))≤Lj(1+Λdj∞‖x‖dj). | (8.3) |
So,
n∑j=1Jj(x(tj))≤n∑j=1Lj(1+Λdj∞‖x‖dj). | (8.4) |
For x∈Eα,p0, if −μλ≥0, then, by (A.3) and (A.4), we have
Ψ(x)≤LT+LTΛβ∞‖x‖β−μλn∑j=1Lj(1+Λdj∞‖x‖dj)=LT−μλn∑j=1Lj+LTΛβ∞‖x‖β−μλn∑j=1LjΛdj∞‖x‖dj. |
Thus, for ∀x∈Eα,p0, we get
Φ(x)−λΨ(x)≥1p‖x‖p−λ(LT−μλn∑j=1Lj+LTΛβ∞‖x‖β−μλn∑j=1LjΛdj∞‖x‖dj). |
If 0<β and dj<p, then lim‖x‖→+∞(Φ(x)−λΨ(x))=+∞,λ>0. Thus, Φ−λΨ is coercive. When β=p, Φ(x)−λΨ(x)≥(1p−λLTΛp∞)‖x‖p−λ(LT−μλn∑j=1Lj−μλn∑j=1LjΛdj∞‖x‖dj). Choose L<∫T0max|x|≤Λ∞(pr)1/1ppF(t,x)dtprTΛp∞. We have that 1p−λLTΛp∞>0 for λ<Ar. If 0<dj<p for all λ<Ar, one has lim‖x‖→+∞(Φ(x)−λΨ(x))=+∞. Obviously, the functional Φ−λΨ is coercive. Lemma 9 shows that φ=Φ−λΨ possesses at least three different critical points in Eα,p0.
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