# Existence of Nontrivial Solutions and High Energy Solutions for a Class of Quasilinear Schrödinger Equations via the Dual-Perturbation Method

Authors: Yu Chen; Xian Wu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the quasilinear Schrödinger equation of the form−Δu+Vxu−Δu2u=hx,u, x∈RN. Under appropriate assumptions on Vx and hx,u, existence results of nontrivial solutions and high energy solutions are obtained by the dual-perturbation method.


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## Body

## 1. Introduction and Preliminaries

In this paper we consider the quasilinear Schrödinger equation of the form(1)-Δu+V(x)u-Δ(u2)u=h(x,u),x∈RN,
where h∈C(RN×R,R) and V∈C(RN,R). Solutions of (1) are standing waves of the following quasilinear Schrödinger equation:
(2)iψt+Δψ-V(x)ψ+kΔ(α(|ψ|2))α′(|ψ|2)ψ+g(x,ψ)=0,x∈RN,
where V(x) is a given potential, k is a real constant, and α and g are real functions. The quasilinear Schrödinger equations (2) are derived as models of several physical phenomena; for example, see [1–5]. Several methods can be used to solve (1). For instance, the existence of a positive ground state solution has been proved in [6, 7] by using a constrained minimization argument; the problem is transformed to a semilinear one in [8–11] by a change of variables (dual approach); Nehari method is used to get the existence results of ground state solutions in [12, 13].Recently, some new methods have been applied to these equations. In [14], the authors prove that the critical points are L∞ functions by the Moser’s iteration; then the existence of multibump type solutions is constructed for this class of quasilinear Schrödinger equations. In [15], by analysing the behavior of the solutions for subcritical case, the authors pass to the limit as the exponent approaches to the critical exponent in order to establish the existence of both one-sign and nodal ground state solutions. Another new method which works for these equations is perturbations. In [16] 4-Laplacian perturbations are led into these equations; then high energy solutions are obtained on bounded smooth domain.In this paper, the perturbation, combined with dual approach, is applied to search the existence of nontrivial solution and sequence of high energy solutions of (1) on the whole space RN. For simplicity we call this method the dual-perturbation method.We need the following several notations. LetCc∞(RN) be the collection of smooth functions with compact support. Let
(3)H1(R):={u∈L2(RN):|∇u|∈L2(RN)},
with the inner product
(4)〈u,v〉H1=∫RN‍[∇u·∇v+uv]dx
and the norm
(5)∥u∥H1=〈u,u〉H11/2.
Let the following assumption (V) hold:(V)
V∈C(RN,R) satisfies infx∈RNV(x)≥a0>0 and lim|x|→∞V(x)=+∞.Set(6)E:={u∈H1(RN):∫ℝN‍V(x)u2dx<+∞}
with the inner product
(7)〈u,v〉E=∫RN‍[∇u·∇v+V(x)uv]dx
and the norm
(8)∥u∥E=〈u,u〉E1/2.
Then both H1(RN) and E are Hilbert spaces.By the continuity of theembeddingE↪Ls(ℝN) for s∈[2,2*] we know that, for each s∈[2,2*], there exists constant as>0 such that
(9)∥u∥s≤as∥u∥E,∀u∈E,
where ∥·∥s denotes the Ls-norm. In the following, we use C or Ci to denote various positive constants. Moreover, we need the following assumptions:(h1)
thereexist4<p<2(2*) if N≥3 and 4<p<∞ if N=2 such that
(10)|h(x,s)|≤C(1+|s|p-1),∀s∈R,(h2)
lims→0h(x,s)/s=0 uniformly in x∈RN,(h3)
there existμ>4 and r>0 such that
(11)c0:=infx∈RN,|s|=rH(x,s)>0,μH(x,s)≤h(x,s)sfor all x∈RN and |s|≥r, where H(x,s)=∫0s‍h(x,t)dt.By Lemma 3.4 in [17] we know that, under the assumption (V), the embedding E↪Ls(RN) is compact for each 2≤s<2*.Equation (1) is the Euler-Lagrange equation of the energy functional
(12)J(u)=12∫RN‍[(1+2u2)|∇u|2+V(x)u2]dx-∫RN‍H(x,u)dx,
where H(x,u)=∫0u‍h(x,t)dt. Due to the presence of the term ∫RN‍u2|∇u|2dx, J(u) is not well defined in E. To overcome this difficulty, a dual approach is used in [9, 10]. Following the idea from these papers, let f be defined by
(13)f′(t)=11+2f2(t)
on [0,+∞), f(0)=0 and f(-t)=-f(t) on (-∞,0]. Then f has the following properties:(f1)
f is uniquely defined C∞ function and invertible;(f2)
0<f′(t)≤1 for all t∈R;(f3)
|f(t)|≤|t| for all t∈R;(f4)
limt→0f(t)/t=1;(f5)
limt→+∞f(t)/t=21/4, limt→-∞f(t)/|t|=-21/4;(f6)
(1/2)f(t)≤tf′(t)≤f(t) for all t≥0 and f(t)≤tf′(t)≤(1/2)f(t) for all t≤0;(f7)
|f(t)|≤21/4|t| for all t∈R;(f8)
the functionf2(t) is strictly convex;(f9)
there exists a positiveC such that
(14)|f(t)|≥{C|t|,|t|≤1,C|t|1/2,|t|≥1;(f10)
there exist positive constantsC1 and C2 such that
(15)|t|≤C1|f(t)|+C2|f(t)|2
for all t∈R;(f11)
|f(t)f′(t)|≤1/2 for all t∈R;(f12)
for eachξ>0, there exists C(ξ)>0 such that f2(ξt)≤C(ξ)f2(t).The properties(f1)–(f11) have been proved in [8–11]. It suffices to prove (f12).Indeed, by(f1), (f4), and (f5), there exist δ>0 and M>0 such that, for |t|≤δ,
(16)12t2≤f2(t)≤32t2,
and for |t|≥M,
(17)22|t|≤f2(t)≤322|t|.
Since there exists a C0>0 such that f2(2t)≤C0f2(t) (see [10]), we can assume that 0<ξ<1. For |t|≤δ, we have |ξt|≤δ, and hence
(18)f2(ξt)≤32ξ2t2≤3ξ2f2(t);
for |t|≥M/ξ>M, one has |ξt|≥M, and hence
(19)f2(ξt)≤322ξ|t|≤3ξf2(t);
and for δ≤|t|≤M/ξ, there exist m(ξ)>0 and M(ξ)>0 such that f2(ξt)≤M(ξ) and f2(t)≥m(ξ). Then we have
(20)f2(ξt)≤M(ξ)≤M(ξ)m(ξ)f2(t).
Hence f2(ξt)≤C(ξ)f2(t), where C(ξ)=max{3ξ2,M(ξ)/m(ξ)}.After the change of variable,J(u) can be reduced to
(21)I(v):=J(f(v))=12∫RN‍(|∇v|2+V(x)f2(v))dx-∫RN‍H(x,f(v))dx.
From [8, 9, 11] we know that if v∈E is a critical point of I, that is,
(22)〈I′(v),φ〉=∫RN‍∇v∇φdx+∫RN‍V(x)f(v)f′(v)φdx-∫RN‍h(x,f(v))f′(v)φdx=0
for all φ∈E, then u:=f(v) is a weak solution of (1). Particularly, if v∈H1(RN)∩C2(RN) is a critical point of I, then u:=f(v) is a classical solution of (1).A sequence{un}⊂E is called a Cerami sequence of J if {J(un)} is bounded and (1+∥un∥)J′(un)→0 in E*. We say that J satisfies the Cerami condition if every Cerami sequence possesses a convergent subsequence.


## 2. Some Lemmas

Consider the following perturbation functionalIθ defined by
(23)Iθ(v)=I(v)+θ2∫RN‍V(x)v2dx,
where θ∈(0,1]. We have the following lemmas.Lemma 1.
If assumptions(V), (h1), and (h2) hold, then the functional Iθ is well defined on E and Iθ∈C1(E,R).Proof.
By conditions(h1) and (h2), the properties (f2), (f3), (f7), and (f11) imply that there exists δ>0 such that
(24)|h(x,f(v))f′(v)|≤|f(v)|f′(v)≤|v|for|v|<δ,|h(x,f(v))f′(v)|≤C|f(v)|p-1f′(v)≤C|f(v)|p-2≤C|v|(p-2)/2for|v|≥δ.
Hence
(25)|h(x,f(v))f′(v)|≤C(|v|+|v|(p/2)-1),(26)|H(x,f(v))|≤C(|v|2+|v|p/2)
for all v∈R. By (26) and the continuity of the embedding E↪Ls(RN) (s∈[2,2*]),
(27)∫RN‍H(x,f(v))dx<+∞,∀v∈E.
Hence Iθ is well defined in E.
Now, we prove thatIθ∈C1(E,R). It suffices to prove that
(28)Ψ1(v)≔∫RN‍H(x,f(v))dx∈C1(E,R),Ψ2(v):=∫RN‍V(x)f2(v)dx∈C1(E,R).
For anyv,ϕ∈E and 0<|t|<1, by the mean value theorem, (25) and (f2)-(f3), we have
(29)|H(x,f(v+tϕ))-H(x,f(v))||t|≤∫01‍|h(f(x,v+stϕ))f′(v+stϕ)ϕ|ds≤C[|v||ϕ|+|ϕ|2+|v|(p-2)/2|ϕ|+|ϕ|p/2],|V(x)f2(v+tϕ)-V(x)f2(v)||t|≤2∫01‍V(x)|f(v+stϕ)f′(v+stϕ)ϕ|ds≤2V(x)∫01‍|v+stϕ||ϕ|ds≤2V(x)[|v||ϕ|+|ϕ|2].
The Hölder inequality implies that
(30)C[|v||ϕ|+|ϕ|2+|v|(p-2)/2|ϕ|+|ϕ|p/2]∈L1(RN),2V(x)[|v||ϕ|+|ϕ|2]∈L1(RN).
Hence, by the Lebesgue theorem, we have
(31)〈Ψ1′(v),ϕ〉=∫RN‍h(x,f(v))f′(v)ϕdx,〈Ψ2′(v),ϕ〉=2∫RN‍V(x)f(v)f′(v)ϕdx
for all ϕ∈E. Now, we show that Ψi′(·):E→E*, i=1,2, are continuous. Indeed, if vn→v in E, then vn→v in Ls(RN) for all s∈[2,2*].
On the spaceLp1(RN)∩Lp2(RN), we define the norm
(32)∥v∥p1∧p2=∥v∥p1+∥v∥p2.
Then
(33)vn⟶vinL2(RN)∩L(p/2)-1(RN).
Moreover, on the space Lp1(RN)+Lp2(RN), we define the norm
(34)∥v∥p1∨p2=inf{∥u∥p1+∥w∥p2:v=u+w,u∈Lp1(RN),w∈Lp2(RN)∥u∥p1}.
By (25), we have
(35)|h(x,f(v))f′(v)|≤C(|v|+|v|(p/2)-1)≤C(|v|2/2+|v|q/r),
where q=p/2 and r=p/(p-2). Then Theorem A.4 in [18] implies
(36)h(x,f(vn))f′(vn)-h(x,f(v))f′(v)⟶0inL2(RN)+Lr(RN)
as n→+∞. If h(x,f(vn))f′(vn)-h(x,f(v))f′(v)=yn+zn with yn∈L2(RN) and zn∈Lr(RN), one has
(37)|∫RN‍[h(x,f(vn))f′(vn)-h(x,f(v))f′(v)]ϕdx|≤∫RN‍|yn||ϕ|+|zn||ϕ|dx≤C(∥yn∥2+∥zn∥r)∥ϕ∥E.
Hence
(38)|∫RN‍[h(x,f(vn))f′(vn)-h(x,f(v))f′(v)]ϕdx|≤C∥h(x,f(vn))f′(vn)-h(x,f(v))f′(v)∥2∨r∥ϕ∥E,
and hence
(39)∥Ψ1′(vn)-Ψ1′(v)∥⟶0
as n→∞. Therefore, Ψ1∈C1(E,R).
Define(40)LVs(RN)={∫RN‍u:RN⟶R:uis  measurableand∫RN‍V(x)usdx<∞}
with the norm ∥u∥LVs=(∫RN‍V(x)usdx)1/s. On the space LVp1(RN)∩LVp2(RN), we define the norm
(41)∥v∥p1∧p2=∥v∥LVp1+∥v∥LVp2.
On the space LVp1(RN)+LVp2(RN), we define the norm
(42)∥v∥p1∨p2=inf{∥v∥p1∨p2=∥u∥LVp1+∥w∥LVp2:v=u+w,u∈LVp1(RN),w∈LVp2(RN)∥u∥LVp1}.
From vn→v in E, one has vn,v∈LV2(RN) and
(43)vn⟶vinLV2(RN)∩LV2(RN)
as n→∞. Since |f(v)f′(v)|≤|v|, by the following Lemma 2, we have
(44)f(vn)f′(vn)⟶f(v)f′(v)inLV2(RN)+LV2(RN).
If f(vn)f′(vn)-f(v)f′(v)=yn+zn with yn∈LV2(RN) and zn∈LV2(RN), one has
(45)|∫RN‍V(x)[f(vn)f′(vn)-f(v)f′(v)]ϕdx|≤∫RN‍V(x)|yn||ϕ|+V(x)|zn||ϕ|dx≤(∥yn∥LV2+∥zn∥LV2)∥ϕ∥E.
Hence
(46)|∫RN‍V(x)[f(vn)f′(vn)-f(v)f′(v)]ϕdx|≤∥f(vn)f′(vn)-f(v)f′(v)∥2∨2∥ϕ∥E,
and hence
(47)∥Ψ2′(vn)-Ψ2′(v)∥⟶0
as n→∞. Therefore, Ψ2∈C1(E,R). This completes the proof.Lemma 2.
Assume that1≤p,q,r,s<+∞, g∈C(RN×R) and
(48)|g(x,v)|≤C(|v|p/r+|v|q/s).
Then, for every v∈LVp(RN)∩LVq(RN), g(·,v)∈LVr(RN)+LVs(RN), and the operator
(49)A:LVp(RN)∩LVq(RN)⟶LVr(RN)+LVs(RN):v⟼g(x,v)
is continuous.Proof.
Letη(s) be a smooth cut-off function such that η(s)=1 for |s|≤1 and η(s)=0 for |s|≥2. Define
(50)g1(x,v)≔η(v)g(x,v),g2(x,v):=(1-η(v))g(x,v).
We can assume that p/r≤q/s. Hence
(51)|g1(x,v)|≤C|v|p/r,|g2(x,v)|≤C|v|q/s
for all (x,v)∈RN×R. Assume vn→v in LVp(RN)∩LVq(RN). Then vn→v in LVp(RN) and g(·,vn)→g(·,v) in LVr(RN). As in the proof of Lemma A.1 in [18], there exists a subsequence {wn} of {vn} and α∈LVp(RN) such that wn(x)→v(x) and |v(x)|,|wn(x)|≤α(x) for a.e. x∈RN. Hence, from (51), one has
(52)|g1(x,wn)-g1(x,v)|r≤2rC|α(x)|p
a.e. on RN. It follows from the Lebesgue theorem that g1(·,wn)→g1(·,v) in LVr(RN). Consequently, g1(·,vn)→g1(·,v) in LVr(RN). Similarly, we can prove g2(·,vn)→g2(·,v) in LVs. Since
(53)∥g(·,vn)-g(·,v)∥r∨s≤∥g1(·,vn)-g1(·,v)∥LVr+∥g2(·,vn)-g2(·,v)∥LVs,
it follows that g(·,vn)→g(·,v) in LVr+LVs. This completes the proof.Lemma 3.
Let(V), (h1), and (h2) hold. Then every bounded sequence {vn}⊂E with Iθ′(vn)→0 possesses a convergent subsequence.Proof.
Since{vn}⊂E is bounded, then, by the compactness of the embedding E↪Ls(RN) (2≤s<2*), passing to a subsequence, one has vn⇀v in E, vn→v in Ls(RN) for all 2≤s<2*, and vn(x)→v(x) for a.e. x∈RN. By (25)
(54)|∫RN‍h(x,f(vn))f′(vn)(v-vn)dx|≤∫RN‍C(|vn|+|vn|(p/2)-1)|vn-v|dx≤C(∥vn∥2∥vn-v∥2+∥vn∥p/2(p/2)-1∥vn-v∥p/2)≤C(∥vn∥E∥vn-v∥2+∥vn∥E(p/2)-1∥vn-v∥p/2)⟶0
as n→∞. Similarly, ∫RN‍h(x,f(v))f′(v)(v-vn)dx→0 as n→∞. Hence, by the property of (f8), we have
(55)〈Iθ′(vn)-Iθ′(v),vn-v〉=∫RN‍|∇(vn-v)|2dx+θ∫RN‍V(x)|vn-v|2dx+∫RN‍V(x)[f(vn)f′(vn)-f(v)f′(v)]×(vn-v)dx-∫RN‍[h(x,f(vn))f′(vn)-h(x,f(v))f′(v)](vn-v)dx≥θ∥vn-v∥E2-on(1),
where on(1)→0 as n→∞. This shows that ∥vn-v∥E2→0 as n→∞. This completes the proof.The following Lemma4 has been proved in [10] (see Proposition 2.1(3) in [10]).Lemma 4.
Ifvn(x)→v(x) a.e. in RN and limn→∞∫RN‍V(x)f2(vn)dx=∫RN‍V(x)f2(v)dx, then ∫RN‍V(x)f2(vn-v)dx→0 as n→∞.


## 3. Main Results

Theorem 5.
Assume conditions(V), (h1)–(h3) hold. Let {θn}⊂(0,1] be such that θn→0. Let vn∈E be a critical point of Iθn with Iθn(vn)≤c for some constant c independent of n. Then, up to subsequence, one has vn→v in E, Iθn(vn)→I(v) and v is a critical point of I.Proof.
By(h2), for 0<ε0<(1/4)(1/2-1/μ)a0, there exists δ0>0 such that
(56)|1μsh(x,s)-H(x,s)|≤ε0s2,∀s∈[-δ0,δ0].
By (h1), for δ0≤|s|≤r (r is the constant appearing in condition (h3)), we have
(57)|1μsh(x,s)-H(x,s)|≤2C(1δ02+rp-2)s2,
where C is the constant appearing in condition (h1). Hence
(58)|1μsh(x,s)-H(x,s)|≤ε0s2+2C(1δ02+rp-2)s2,∀s∈[-r,r].
Since lim|x|→∞V(x)=+∞, there exists ρ0>0 such that
(59)14(12-1μ)V(x)>2C(1δ02+rp-2)
for all |x|≥ρ0. Hence
(60)(12-1μ)∫RN‍V(x)f2(v)dx+∫{x:|f(v)|≤r}‍[1μf(v)h(x,f(v))-H(x,f(v))1μ]dx≥12(12-1μ)∫RN‍V(x)f2(v)dx-2C(1δ02+rp-2)r2|Bρ0|.
Since vn is a critical point of Iθn,
(61)〈Iθn′(vn),ϕ〉=∫RN‍∇vn∇ϕdx+∫RN‍V(x)f(vn)f′(vn)ϕdx+θn∫RN‍V(x)vnϕdx-∫RN‍h(x,f(vn))f′(vn)ϕdx=0
for all ϕ∈E. Consequently, taking ϕ=f(vn)/f′(vn)∈E, by (h3) and (f6) we have
(62)c≥Iθn(vn)=Iθn(vn)-1μ〈Iθn′(vn),f(vn)f′(vn)〉=∫RN‍[12-1μ(1+2f2(vn)1+2f2(vn))]|∇vn|2dx+(12-1μ)∫RN‍V(x)f2(vn)dx+θn2∫RN‍V(x)vn2dx-θnμ∫RN‍V(x)vnf(vn)f′(vn)dx+∫RN‍[1μh(x,f(vn))f(vn)-H(x,f(vn))]dx≥(12-2μ)∫RN‍|∇vn|2dx+(12-1μ)∫RN‍V(x)f2(vn)dx+(12-2μ)θn∫RN‍V(x)vn2dx+∫{x:|f(vn)|≤r}‍[1μh(x,f(vn))f(vn)-H(x,f(vn))1μ]dx≥(12-2μ)∫RN‍|∇vn|2dx+12(12-1μ)∫RN‍V(x)f2(vn)dx+(12-2μ)θn∫RN‍V(x)vn2dx-2C(1δ02+rp-2)r2|Bρ0|≥(12-2μ)∫RN‍|∇vn|2dx+(14-12μ)∫RN‍V(x)f2(vn)dx+(12-2μ)θn∫RN‍V(x)vn2dx-C1,
and hence
(63)∫RN‍|∇vn|2dx+∫RN‍V(x)f2(vn)dx+θn∫RN‍V(x)vn2dx≤C
for some constant C independent of n. By the boundedness of ∫RN‍|∇vn|2dx, there exists C2>0 such that
(64)2∫ℝN‍f2(vn)|∇f(vn)|2dx≤∫ℝN‍[1+2f2(vn)]|∇f(vn)|2dx=∫ℝN‍|∇vn|2dx≤C2
for all n. Hence, by the Sobolev embedding theorem, one has
(65)∥f(vn)∥2(2*)4=∥f2(vn)∥2*2≤C3∥∇f2(vn)∥22≤C.
Next, we prove thatf(vn)∈L∞(RN) and ∥f(vn)∥L∞≤C, where the positive constant C is independent of n. Setting T>2, r>0, define v~nT=b(vn), where b:R→R is a smooth function satisfying b(s)=s for |s|≤T-1, b(-s)=-b(s); b′(s)=0 for s≥T, and b′(s) is decreasing in [T-1,T].
This means thatv~nT=vn, for |vn|≤T-1; |v~nT|=|b(vn)|≤|vn|, for T-1≤|vn|≤T; |v~nT|=CT>0, for |vn|≥T, where T-1≤CT≤T.
Letϕ=(f(vn)/f′(vn))|f(v~nT)|2r; then ϕ∈E. By (61) 〈I′(vn),ϕ〉=0. Hence
(66)I1+I2+I3+I4+I5=∫ℝN‍h(x,f(vn))f(vn)|f(v~nT)|2rdx,
where
(67)I5:=∫RN‍V(x)f2(vn)|f(v~nT)|2rdx+θn∫RN‍V(x)vnf(vn)f′(vn)|f(v~nT)|2rdx≥∫RN‍V(x)f2(vn)|f(v~nT)|2rdx,I1≔∫{x:|vn|≥T}‍[1+2f2(vn)1+2f2(vn)]|f(v~nT)|2r|∇vn|2dx≥∫{x:|vn|≥T}‍|f(v~nT)|2r|∇vn|2dx=∫{x:|vn|≥T}‍[1+2f2(vn)]×|∇f(vn)|2|f(v~nT)|2rdx≥2∫{x:|vn|≥T}‍f2(vn)|∇f(vn)|2|f(v~nT)|2rdx=12∫{x:|vn|≥T}‍|∇[f2(vn)fr(v~nT)]|2dx,(68)I2≔∫{x:|vn|≤T-1}‍[2r+1+2f2(vn)1+2f2(vn)]×|f(v~nT)|2r|∇vn|2dx≥∫{x:|vn|≤T-1}‍2f2(vn)1+2f2(vn)|f(v~nT)|2r|∇vn|2dx≥∫{x:|vn|≤T-1}‍|f(vn)|2r+2|∇f(vn)|2dx=1(r+2)2∫{x:|vn|≤T-1}‍|∇fr+2(vn)|2dx=1(r+2)2∫{x:|vn|≤T-1}‍|∇[f2(vn)fr(v~nT)]|2dx,I3:=∫{x:T-1≤|vn|≤T}‍[1+2f2(vn)1+2f2(vn)]f2r(v~nT)|∇vn|2dx≥∫{x:T-1≤|vn|≤T}‍2f2(vn)1+2f2(vn)f2r(v~nT)|∇vn|2dx=12∫{x:T-1≤|vn|≤T}‍[fr(v~nT)∇f2(vn)]2dx,I4:=2r∫{x:T-1≤|vn|≤T}‍f2r-1(v~nT)f′(v~nT)×b′(vn)f(vn)f′(vn)|∇vn|2dx.
For T-1≤|vn|≤T, |v~nT|=|b(vn)|≤|vn|. By the properties of f and b, the mean value theorem implies
(69)|f(b(vn))|≥f′(b(vn))b′(vn)|vn|≥12f′(b(vn))b′(vn)f2(vn).
Hence
(70)I4=2r∫{x:T-1≤|vn|≤T}‍f2r-1(v~nT)f′(v~nT)×b′(vn)f(vn)f′(vn)|∇vn|2dx=2r∫{x:T-1≤|vn|≤T}‍f2r-1(b(vn))f′(b(vn))×b′(vn)f(vn)×1+2f2(vn)|∇vn|2dx≥2r∫{x:T-1≤|vn|≤T}‍[fr-1(b(vn))f′(b(vn))b′(vn)]2×f(vn)vn2f2(vn)|∇vn|2dx≥2r∫{x:T-1≤|vn|≤T}‍[fr-1(b(vn))f′(b(vn))b′(vn)]2×f4(vn)|∇vn|2dx=2r∫{x:T-1≤|vn|≤T}‍f4(vn)×[fr-1(b(vn))∇f(b(vn))]2dx=2r∫{x:T-1≤|vn|≤T}‍[f2(vn)∇fr(v~nT)]2dx.
Consequently,
(71)I3+I4=12∫{x:T-1≤|vn|≤T}‍[fr(v~nT)∇f2(vn)]2dx+2r∫{x:T-1≤|vn|≤T}‍[f2(vn)∇fr(v~nT)]2dx≥1(r+2)2∫{x:T-1≤|vn|≤T}‍2[fr(v~nT)∇f2(vn)]2+2[f2(vn)∇fr(v~nT)]2dx≥1(r+2)2∫{x:T-1≤|vn|≤T}‍|∇[f2(vn)fr(v~nT)]|2dx.
Combining (67) and (68), we have
(72)I1+I2+I3+I4≥1(r+2)2∫ℝN‍|∇[f2(vn)fr(v~nT)]|2dx.
For any ε>0, by (h1) and (h2), there exists C(ε)>0 such that
(73)|h(x,s)|≤ε|s|+C(ε)|s|p-1.
Combining (66), (72), and (73), one has
(74)1(r+2)2∫RN‍|∇[f2(vn)fr(v~nT)]|2dx≤C(ε)∫RN‍|f(vn)|p|f(v~nT)|2rdx.
By the Hölder inequality and (65),
(75)∫RN|f(vn)|p|f(v~nT)|2rdx=∫RN‍|f(vn)|p-4|f(v~nT)|2rf4(vn)dx≤(∫RN‍|f(vn)|(p-4)·(4N/(p-4)(N-2))dx)(p-4)(N-2)/4N·(∫RN‍[|f(v~nT)|2r×f4(vn)]4N/(4N-(p-4)(N-2))dx∫RN‍)(4N-(p-4)(N-2))/4N=(∫RN‍|f(vn)|22*dx)((p-4)(N-2))/4N·(∫RN‍[|f(v~nT)|r×f2(vn)]8N/(4N-(p-4)(N-2))dx∫RN‍)(4N-(p-4)(N-2))/4N≤C(∫RN‍[|f(v~nT)|r×f2(vn)]8N/(4N-(p-4)(N-2))dx∫RN‍)(4N-(p-4)(N-2))/4N.
Moreover,
(76)1(r+2)2∫RN‍|∇[f2(vn)fr(v~nT)]|2dx≥C(r+2)2(∫RN‍[f2(vn)|f(v~nT)|r]2*dx)2/2*.
Hence(77)(∫RN‍[f2(vn)|f(v~nT)|r]2*dx)2/2*≤C(r+2)2×(∫RN‍[|f(v~nT)|r×f2(vn)|f(v~nT)|r]8N/(4N-(p-4)(N-2))dx∫RN‍)(4N-(p-4)(N-2))/4N.Since 4<p<2(2*), d=2*/(8N/(4N-(p-4)(N-2)))=2*/2-p/4+1>1. Set q=8N/(4N-(p-4)(N-2)). Then
(78)(∫RN‍[f2(vn)|f(v~nT)|r]qddx)1/qd(r+2)≤[C(r+2)2]1/2(r+2)×(∫RN‍[|f(v~nT)|rf2(vn)]qdx)1/q(r+2).
Take r=r0 such that (2+r0)q=2(2*). Since |v~nT|=|b(vn)|≤|vn|, |f(v~nT)|≤|f(vn)|. Hence, from (65), we have
(79)∫RN‍[|f(v~nT)|r0f2(vn)]qdx≤∫RN‍|f(vn)|(2+r0)qdx<C.
Since f(v~nT)→f(vn) as T→+∞, taking T→+∞ in (78) with r=r0, we have
(80)(∫RN‍|f(vn)|(2+r0)qddx)1/qd(r0+2)≤[C(r0+2)2]1/2(r0+2)×(∫RN‍|f(vn)|(2+r0)qdx)1/q(r0+2).
Set 2+r1=(2+r0)d. Then
(81)(∫RN‍|f(vn)|(2+r1)qdx)1/q(r1+2)≤[C(r0+2)2]1/2(r0+2)×(∫RN‍|f(vn)|(2+r0)qdx)1/q(r0+2).
Inductively, we have
(82)(∫RN‍|f(vn)|(2+rk+1)qdx)1/q(rk+1+2)≤[C(rk+2)2]1/2(rk+2)×(∫RN‍|f(vn)|(2+rk)qdx)1/q(rk+2)≤∏i=0k‍[C(ri+2)2]1/2(ri+2)×(∫RN‍|f(vn)|(2+r0)qdx)1/q(r0+2),
where (2+ri)=di(2+r0)  (i=0,1,…,k), and
(83)∏i=0k‍[C(ri+2)2]12(ri+2)=exp{∑i=0k‍lnCdi(r0+2)di(r0+2)}=exp{∑i=0k‍[lnC(r0+2)di(r0+2)+ilnddi(r0+2)]}
is convergent as k→∞. Let Ck=∏i=0k‍[C(ri+2)2]1/2(ri+2). Then Ck→C∞>0 as k→∞. Hence
(84)∥f(vn)∥L(2+r0)qdk+1≤Ck∥f(vn)∥L2(2*).
Let k→∞; by (65), we have
(85)∥f(vn)∥L∞≤C∞∥f(vn)∥L2(2*)≤C,∥f(v)∥L∞≤C.
Hence, by (f9) and (85), we have
(86)∫RN‍V(x)vn2dx=∫{x:|vn(x)|≤1}‍V(x)vn2dx+∫{x:|vn(x)|>1}‍V(x)vn2dx≤1C∫{x:|vn(x)|≤1}‍V(x)f2(vn)dx+1C∫{x:|vn(x)|>1}‍V(x)f4(vn)dx≤1C∫{x:|vn(x)|≤1}‍V(x)f2(vn)dx+C∫{x:|vn(x)|≥1}‍V(x)f2(vn)dx≤C∫RN‍V(x)f2(vn)dx.
By (63) we know that ∫RN‍V(x)vn2dx is bounded, and hence {vn} is bounded in E. Up to subsequence, one has vn⇀v in E, vn→v in Ls(RN) for s∈[2,2*), and vn(x)→v(x) a.e. x∈RN.
Now, we show thatv is a critical point of I. For any ψ∈C0∞(RN) with ψ≥0, by (85), we know that ϕ=ψexp(-f(vn))∈E. Take ϕ=ψexp(-f(vn)) as the test function in (61); we have
(87)0=∫RN‍exp(-f(vn))∇vn·∇ψdx-∫RN‍|∇vn|2ψexp(-f(vn))f′(vn)dx+θn∫RN‍V(x)vnψexp(-f(vn))dx+∫RN‍V(x)f(vn)f′(vn)ψexp(-f(vn))dx-∫RN‍h(x,f(vn))f′(vn)ψexp(-f(vn))dx.
By |∇(vn-v)|2ψexp(-f(vn))f′(vn)≥0, one has
(88)limsupn→∞∫RN‍-|∇vn|2ψexp(-f(vn))f′(vn)dx≤-∫RN‍|∇v|2ψexp(-f(v))f′(v)dx.
Since θn→0, by (63)
(89)θn∫RN‍V(x)vnψexp(-f(vn))dx⟶0
as n→∞. Moreover, notice that vn⇀v in E, vn→v in Ls(RN) for s∈[2,2*), and vn(x)→v(x) a.e. x∈RN; by Hölder inequality and Lebesgue theorem, we have
(90)∫RN‍exp(-f(vn))∇vn·∇ψdx⟶∫RN‍exp(-f(v))∇v·∇ψdx,∫RN‍V(x)f(vn)f′(vn)ψexp(-f(vn))dx⟶∫RN‍V(x)f(v)f′(v)ψexp(-f(v))dx,∫RN‍h(x,f(vn))f′(vn)ψexp(-f(vn))dx⟶∫RN‍h(x,f(v))f′(v)ψexp(-f(v))dx.
Hence, from (87), we have
(91)0≤∫RN‍exp(-f(v))∇v·∇ψdx-∫RN‍|∇v|2ψexp(-f(v))f′(v)dx+θn∫RN‍V(x)vψexp(-f(v))dx+∫RN‍V(x)f(v)f′(v)ψexp(-f(v))dx-∫RN‍h(x,f(v))f′(v)ψexp(-f(v))dx=∫RN‍∇v·∇(ψexp(-f(v)))dx+∫RN‍V(x)f(v)f′(v)ψexp(-f(v))dx-∫RN‍h(x,f(v))f′(v)ψexp(-f(v))dx.
For any φ∈E with φ≥0, by (85) we know that ζ:=φexp(f(v))∈E. By Theorem 2.8 in [19], there exists a sequence {ψn}⊂C0∞(RN) such that ψn≥0 and ψn→ζ and ψn(x)→ζ(x) for a.e. x∈RN. Take ψ=ψn in (91), and let n→∞; we have
(92)0≤∫RN‍∇v·∇φdx+∫RN‍V(x)f(v)f′(v)φdx-∫RN‍h(x,f(v))f′(v)φdx.
The opposite inequality can be obtained by taking ϕ=ψexp(f(vn)) and ζ=φexp(-f(v)). Consequently,
(93)∫RN‍∇v·∇φdx+∫RN‍V(x)f(v)f′(v)φdx-∫RN‍h(x,f(v))f′(v)φdx=0,∀φ∈E.
This shows that v∈E is a critical point of I, and by taking φ=f(v)/f′(v)∈E, one has
(94)∫RN‍[1+2f2(v)1+2f2(v)]|∇v|2dx+∫RN‍V(x)f2(v)dx-∫RN‍h(x,f(v))f(v)dx=0.
Finally, takingϕ=f(vn)/f′(vn) as the test function in (61), we have
(95)∫RN‍[1+2f2(vn)1+2f2(vn)]|∇vn|2dx+θn∫RN‍V(x)vnf(vn)f′(vn)dx+∫RN‍V(x)f2(vn)dx-∫RN‍h(x,f(vn))f(vn)dx=0.
Since
(96)∫RN‍h(x,f(vn))f(vn)dx⟶∫RN‍h(x,f(v))f(v)dx,∫RN‍[1+2f2(vn)1+2f2(vn)]|∇(vn-v)|2dx≥0,
by Fatou’s Lemma, (63), (94), (95), up to subsequence, one has
(97)θn∫RN‍V(x)vnf(vn)f′(vn)dx⟶0,∫RN‍|∇vn|2dx⟶∫RN‍|∇v|2dx,∫RN‍2f2(vn)1+2f2(vn)|∇vn|2dx⟶∫RN‍2f2(v)1+2f2(v)|∇v|2dx,(98)∫RN‍V(x)f2(vn)dx⟶∫RN‍V(x)f2(v)dx.
Hence Iθn(vn)→I(v) as n→∞. Set wn:=vn-v∈E. By (f8), (f12), and (85), one has
(99)f2(wn)=f2(2·wn2)≤C[12f2(vn)+12f2(v)]≤C[f2(vn)+f2(v)]≤C.
Consequently, by (f9), (98), and Lemma 4, one has
(100)∫RN‍V(x)|wn|2dx=∫{x:|wn|≤1}‍V(x)|wn|2dx+∫{x:|wn|≥1}‍V(x)|wn|2dx≤C∫{x:|wn|≤1}‍V(x)f2(wn)dx+C∫{x:|wn|≥1}‍V(x)f4(wn)dx≤C∫{x:|wn|≤1}‍V(x)f2(wn)dx+C∫{x:|wn|≥1}‍V(x)f2(wn)dx=C∫RN‍V(x)f2(wn)dx⟶0
as n→∞. Therefore, vn→v in E. This completes the proof.Theorem 6.
Assume conditions(V), (h1)–(h3) hold; then (1) has a weak solution.Proof.
First, we prove that, for eachθ∈(0,1], Iθ satisfy the Cerami condition. Indeed, let {vn}⊂E be an arbitrary Cerami sequence of Iθ. Set ϕ=f(vn)/f′(vn). Then ∥ϕ∥E≤C∥vn∥E. Similar to the proof of (63), we can prove that {vn} is bounded in E. Hence, by Lemma 3, the sequence {vn} possesses a convergent subsequence in E. This shows that Iθ satisfy the Cerami condition.
Next, for anyε>0, by (h1), (h2), (f3), and (f7), there exists C(ε)>0 such that
(101)H(x,f(t))≤εt2+C(ε)|t|p/2
for all (x,t)∈RN×R. For small 0<ρ≪1, set
(102)Sρ={v∈E:∥v∥E=ρ}.
Then, from (101), for v∈Sρ,
(103)Iθ(v)=12∫RN‍[|∇v|2+V(x)f2(v)]dx+θ2∫RN‍V(x)v2dx-∫RN‍H(x,f(v))dx≥θ2∫RN‍[|∇v|2+V(x)v2]dx-ε∫RN‍v2dx-C(ε)∫RN‍|v|p/2dx≥θ2∥v∥E2-εa22∥v∥E2-C(ε)ap/2p/2∥v∥Ep/2≥ρ2(θ4-Cρ(p-4)/2)≥δ>0
for small ε>0 and ρ>0. Moreover, by (h3), for any (x,z)∈RN×R with |z|≥r, one has
(104)H(x,z)≥c0|z|μ.
Since μ>4, there exists a constant 4<α<min{μ,2(2*)}. Hence, by (f5), we have
(105)lim|t|→∞H(x,f(t))|t|α/2=lim|t|→∞H(x,f(t))|f(t)|α·|f(t)|α|t|α/2=+∞
uniformly in x∈RN. Consequently, there exist constants τ>1 such that
(106)H(x,f(t))≥|t|α/2,∀|t|≥τ,
for all x∈RN. For any finite-dimensional subspace E~⊂E, by the equivalency of all norms in the finite-dimensional space, there is a constant a>0 such that
(107)∥v∥α/2≥a∥v∥E,∀v∈E~.
By (h1), (h2), and (106), there exists a positive constant C>0 such that
(108)H(x,f(t))≥|t|α/2-Ct2,∀(x,t)∈RN×R.
Since 4<α<2(2*), by (f3), (107), and (108), we have
(109)Iθ(v)=12∫RN‍[|∇v|2+V(x)f2(v)]dx+θ2∫RN‍V(x)v2dx-∫RN‍H(x,f(v))dx≤∥v∥E2-∥v∥α/2α/2+C∥v∥22≤C∥v∥E2-aα/2∥v∥Eα/2
for all v∈E~. Hence there exists a large R>0 such that Iθ<0 on E~∖BR. Set a fixed e∈E~ with ∥e∥E=1. For any fixed T>ρ, define the path hT:[0,1]↦E~⊂E by hT(t)=tTe. Then for large T>0, by (109), one has
(110)Iθ(hT(1))≤CT2-aα/2Tα/2<0,∥hT(1)∥E=T>ρ,supt∈[0,1]Iθ(hT(t))≤CT2<+∞.
Hence by Theorem 2.2 with the Cerami condition in [20], Iθ possesses a critical value
(111)cθ≔infγ∈Γmaxt∈[0,1]Iθ(γ(t))≥δ>0,cθ≤supt∈[0,1]Iθ(hT(t))≤CT2,
where
(112)Γ={γ∈C([0,1],E):γ(0)=0,γ(1)=hT(1)}.
Consequently, by Theorem 5, we know that (1) has a weak solution. This completes the proof of Theorem 6.Remark 7.
Letv+=max{v,0} and v-=max{-v,0}. Set
(113)I±(u)=12∫RN‍[|∇v|2+V(x)f2(v)]dx-∫RN‍H(x,f(v±))dx,Iθ±(v)=12θ∫RN‍V(x)v2dx+I±(v)
instead of I(u) and Iθ(u), respectively. Then, under the conditions of Theorem 6, we can obtain the existence of a positive solution and a negative solution for (1).Theorem 8.
Assume conditions(V), (h1)–(h3) hold. If h(x,s) is odd in s, then (1) has a sequence {vm} of solutions such that I(vm)→+∞.Proof.
Consider the eigenvalue of the operateL=-Δ+V. By assumption (V) and the compactness of the embedding E↪L2(RN), we know that the spectrum σ(L)={λ1,λ2,…,λn⋯} of L with
(114)0<λ1<λ2<⋯<λn<⋯
and λn→+∞ as n→+∞ (see page 3820 in [21]). Let φn be the eigenfunction corresponding to λn. By regularity argument we know that φn∈E. Set En=span{φ1,φ2,…,φn}. Then we can decompose the space E as E=En⊕Wn for n=1,2,…, where Wn is orthogonal to En in E. For ρ>0, set
(115)Qρ={v∈E:∫RN‍[|∇v|2+V(x)f2(v)]dx≤ρ2}.
By (109) there exists rn>0 independent of θ such that
(116)Iθ(v)<0,∀v∈En∖Qrn-.
Set
(117)Dn=En∩Qrn,Gn={φ∈C(Dn,E):φis  oddandφ|∂Qrn∩En=id},Γj={φ(Dn∖Qrn¯):φ∈Gn,n≥j,A=-A⊂En∩Qrnis  closed  andγ(A)≤n-j},
where γ(·) is the genus. Let
(118)cj(θ)=infB∈Γjsupv∈BIθ,j=1,2….
We have the following three facts (we refer the reader to [16] for their proofs).Fact (1). For each B∈Γj, if 0<ρ<rn for all n≥j, then B∩∂Qρ∩Wj-1≠∅.Fact (2). There exist constants αj≤βj such that cj(θ)∈[αj,βj] and αj→+∞ as j→+∞.Fact (3).  cj(θ), j=1,2,… are critical values of Iθ.Consequently, Theorem 8 follows from Theorem 5 and the above Facts (2)-(3). This completes the proof.Corollary 9.
If the following conditions(h4) and (h5) are used in place of (h3); then the conclusions of Theorem 5, Theorem 6, and Theorem 8 hold:(h4)
lim|s|→+∞infH(x,s)>0 uniformly in x∈ℝN,(h5)
there existμ>4 and τ>0 such that
(119)μH(x,s)≤h(x,s)sfor all x∈ℝN and |s|≥τ.Proof.
By(h4), there are constants λ>0 and r1>0 such that whenever |s|≥r1, one has
(120)H(x,s)>λ,∀x∈ℝN.
Set r=max{τ,r1}. Then, by (h5),
(121)c0:=infx∈ℝN,|s|=rH(x,s)≥λ>0,μH(x,s)≤h(x,s)s
for all x∈ℝN and |s|≥r. Therefore, condition (h3) holds. This completes the proof.



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*Source: 256324-2013-10-29.xml*