In this paper, we consider a class of quasilinear Schrödinger equations arising from a model of a self-trapped electrons in quadratic or hexagonal lattices. By variational methods, we prove that this problem admits a positive solution for any positive parameter.

The quasilinear Schrödinger equation
izt=Δz+W(x)z|z|p2zκz2Δ|z|2,(t,x)(0,+)×RN,
(1.1)
where W(x) and ρ are continuous functions and κ > 0 is a parameter, has been introduced in Refs. 1–3 to study a model of a self-trapped electrons in quadratic or hexagonal lattices (see also Ref. 4). In those references numerical and analytical results have been given.
Here, our particular interest is in the existence of standing wave solutions, that is, solutions of the form z(x, t) = exp(−iEt)u(x), where ER and u > 0 is a real function. It is well known that z satisfies (1.1) if and only if the function u solves the following equation of the quasilinear elliptic type
Δu+V(x)u+κu2Δ|u|2=|u|p2u,xRN,
(1.2)
where V(x) = W(x) − E is the new potential function. For the case κ < 0, construction of solutions to (1.2) by variational methods has been a hot topic during the last decade. However, in the mathematical literature, except some one-dimensional results given in Refs. 5 and 6, very little is known in the case κ > 0.
Note that (1.2) is the Euler-Lagrange equation associated with the energy functional
I0(u)=12RN(1κu2)|u|2dx+12RNV(x)u2dx1pRN|u|pdx.
(1.3)
Recently, to overcome two rather essential constraints of functional (1.3) in the use of variational methods, Alves et al. in Ref. 7 introduced a cut off method and proved, by using the Morse iteration, that there exists some κ0 > 0 small, such that for all κ ∈ [0, κ0) the solutions found verify the estimate maxxRN|u|<1/3κ.
The main objective of the present paper is to construct the analytical expression of κ0. To this aim, by using the cut off technique introduced in Ref. 7, we consider the following modified equation:
div(g2(u)u)+g(u)g(u)|u|2+V(x)u=up1,xRN,
(1.4)
where g:[0,+)R given by
g(t)=1κt2,if0t<13κ;132κt+16,if13κt.
(1.5)
Setting g(t) = g(−t) for all t ≤ 0, it follows that gC1R,16,1, g(t) is an even function, g(t) is increasing in (−, 0) and decreasing in [0, +).
We observe that (1.4) is the Euler-Lanrange equation associated with the energy functional
Iκ(u)=12RN[g2(u)|u|2+V(x)u2]dx1pRNu+pdx,
(1.6)
and the previously defined g(t) is obviously bounded. So we can discuss the critical point of (1.6) in H1(R). If we can prove the critical point u of the functional (1.6) satisfies |u|<13κ, then this critical point is good for what we want since g(u)=1κu2 under this situation. That is, in this case, the functional (1.6) is exactly the functional (1.3) and thus u is a solution of (1.2).
For the existence and the L estimate of the critical point of the functional (1.6), we also follow the ideas showed in Refs. 8 and 9 and make the change of variables
v=G(u)=0ug(s)ds,u=G1(v).
(1.7)
Thus the change of variable (1.7) transforms the nonsmooth functional Iκ(u) into a smooth functional
Jκ(v)=12RN[|v|2+V(x)|G1(v)|2]dx1pRNG1(v)+pdx.
(1.8)
Moreover, after this change of variable, the corresponding Euler-Lagrange equation (1.4) is reformulated as the semilinear equation
Δv+V(x)G1(v)g(G1(v))=G1(v)+p1g(G1(v)),xRN.
(1.9)
Then, combining the H1 estimate of the solution v of (1.9), we construct the L estimate |v|. Here, different from the Morse iteration, we use the method of converting integral inequalities into differential inequalities which can be found in Lemma 5.1 on p. 71 in Ladyzhenskaya and Ural’tseva’s book10 and is used to study the L estimate of nonlinear elliptic equations on bounded domains to construct the L estimate of the solution v. We must point out that, in this paper, this L estimate is first used on unbounded domain RN for the quasilinear Schrödinger equations.

Throughout this paper, we assume the potential V(x)C(RN,R) satisfies

  • V(x) ≥ V0 > 0;

  • V(x) ≤ V < + ,

and we make use of X to be the completion of the space C0(RN) with respect to the norm
u=RN(|u|2+V(x)u2)dx12.
By (V0) and (V1), X is equivalent to H1(RN). The symbols |u|p and |u| are used for the norms of the space Lp(RN) with 2 < p < + and p = respectively. C denotes positive (possibly different) constants.

The corresponding results are as follows:

Theorem 1.1.
Let
κ0=221a1ap316112aV0aN1+p22apCN2(N2)1a2pp2P(vr)1+p22ap,
where a=1p12*, CN is the best Sobolev constant, vr is a radial solution of
Δv+6Vv=vp1,v>0
and
P(vr)=12RN[|vr|2+6V|vr|2]dx1pRN|vr|pdx.
Then, if κ ∈ (0, κ0), the Eq. (1.2) has a positive solution uκ with maxxRN|uκ(x)|<13κ.

At the beginning of this section, noticing that the inverse function G−1(t) exists, we need the following lemma, which was proved in Ref. 7, to show important properties involving functions g(t) and G−1(t).

Lemma 2.1.

  1. limt0G1(t)t=1;

  2. limtG1(t)t=6;

  3. tG1(t)6t, for all t ≥ 0;

  4. 12tg(t)g(t)0, for all t ≥ 0.

In the following lemma, we establish the existence of the critical point of Jκ.

Lemma 2.2.
The critical point of the functional Jκ(v) is attained by a nonnegative function vκ satisfying
Δvκ+V(x)G1(vκ)g(G1(vκ))=G1(vκ)+p1g(G1(vκ)),xRN.
(2.1)

Proof.

We first show that Jκ verifies the mountain pass theorem geometry.

By Lemma 2.1-(3) and Sobolev embedding,
Jκ(v)=12RN|v|2dx+12RNV(x)|G1(v)|2dx1pRNG1(v)+pdx12RN|v|2dx+12RNV(x)|v|2dx6ppRNv+pdx12v2Cvp.
(2.2)
Thereby, by choosing ρ0 small, we know that
a0=12ρ02Cρ0p>0,
and so,
Jκ(v)a0forv=ρ0.
In order to prove the existence of eH1(RN) such that Jκ(e) < 0, fix ψC0(R,[0,1]) with suppψ=B̄1 and note that by Lemma 2.1-(3),
Jκ(tψ)=t22RN|ψ|2dx+12RNV(x)|G1(tψ)|2dx1pRNG1(tψ)+pdx3t2RN(|ψ|2+Vψ2)dxtppRNψ+pdx.
(2.3)
Since p > 2, it follows that Jκ() → − as t. Then, the result follows considering e = for t large enough.
Thus, we can apply the mountain pass theorem without the (PS)-condition to get a (PS) sequence vn satisfying
Jκ(vn)cκandJκ(vn)0asn,
where cκ is the well known mountain pass level associated with the functional Jκ.

Next, we will show that the sequence vn is bounded.

As vnH1(RN) is a Palais-Smale sequence, we know that
Jκ(vn)=12RN|vn|2dx+12RNV(x)|G1(vn)|2dx1pRNG1(vn)+pdx=cκ+o(1).
(2.4)
Moreover, for any ψH1(RN), Jκ(vn)ψ=o(1)ψ, that is,
RNvnψ+V(x)G1(vn)g(G1(vn))ψG1(vn)+p1g(G1(vn))ψdx=o(1)ψ.
(2.5)
Now, fixing ψ = G−1(vκ)g(G−1(vκ)), it follows from Lemma 2.1-(4),
|(G1(vκ)g(G1(vκ)))|1+G1(vκ)g(G1(vκ))g(G1(vκ))|vκ||vκ|.
(2.6)
On the other hand, by Lemma 2.1-(3),
|G1(vκ)g(G1(vκ))|6|vκ|.
(2.7)
Combining (2.6) and (2.7), we see that ψ=G1(vκ)g(G1(vκ))H1(RN) with ‖ψ‖ ≤ 6‖vκ‖. Thus, using ψ = G−1(vκ)g(G−1(vκ)) as a test function in (2.5), we derive that
o(1)vn=Jκ(vn)G1(vn)g(G1(vn))=RN1+G1(vn)g(G1(vn))g(G1(vn))|vn|2+V(x)|G1(vn)|2G1(vn)+pdxRN|vn|2+V(x)|G1(vn)|2G1(vn)+pdx.
(2.8)
Therefore, combining (2.4), (2.5), and (2.8), we find the inequality
pcκ+o(1)+o(1)vn=pJκ(vn)Jκ(vn)G1(vn)g(G1(vn))p22RN|vn|2+V(x)|G1(vn)|2dxp22vn2,
(2.9)
which shows the boundedness of vn.
Since vn is a bounded sequence in H1(RN), Let us choose a subsequence, still denote by vn, weakly convergent to vκH1(RN). Moreover, vnvκ in Llocq(RN) for q ∈ [2, 2*) and vn(x) → vκ(x) a.e. in RN. We will prove that vκ is the critical point of Jκ. Indeed, it suffices to show that
Jκ(vκ)ψ=0,ψH1(RN),
or equivalently, for all ψH1(RN),
RNvκψ+V(x)G1(vκ)g(G1(vκ))ψG1(vκ)+p1g(G1(vκ))ψdx=0.
(2.10)
As C0(RN) is dense in H1(RN), we will show the last equality only for ψC0(RN).
In what follows, for each R > 0 we consider ψRC0(RN) verifying
0ψR(x)1xRN,ψR(x)=1xBR(0)andψR(x)=0xB2Rc(0).
By Ref. 13, there is zLq(B2R(0)) such that
|vn||z(x)|a.e.inB2R(0).
Consequently,
G1(vn)g(G1(vn))vnG1(vκ)g(G1(vκ))vκa.e.inB2R(0),
and
G1(vn)+p1g(G1(vn))vnG1(vκ)+p1g(G1(vκ))vκa.e.inB2R(0).
Moreover, by Lemma 2.1,
V(x)G1(vn)g(G1(vn))vnψR6V|vn|2|ψR|6V|z(x)|2|ψR|
and
V(x)G1(vn)+p1g(G1(vn))ψR6p+12V|vn|p|ψR|6p+12V|z(x)|p|ψR|.
Hence, by Lebesgue dominated convergence theorem, we have
RNV(x)G1(vn)g(G1(vn))vnψRdxRNV(x)G1(vκ)g(G1(vκ))vκψRdx
(2.11)
and
RNG1(vn)+p1g(G1(vn))vnψRdxRNG1(vκ)+p1g(G1(vκ))vκψRdx.
(2.12)
The same type of argument works to prove the limits below
RNV(x)G1(vn)g(G1(vn))vκψRdxRNV(x)G1(vκ)g(G1(vκ))vκψRdx
(2.13)
and
RNG1(vn)+p1g(G1(vn))vκψRdxRNG1(vκ)+p1g(G1(vκ))vκψRdx.
(2.14)
Now, the above limits combined with Jκ(vn)(vnψ)=on(1) and Jκ(vn)(vκψ)=on(1) give
RN|vnvκ|2ψR(x)dx0,
from where it follows that
BR(0)|vnvκ|2(x)dx0.
Recalling that R is arbitrary and vnvκ in Lloc2(RN), we are able to conclude that vnvκ in Hloc1(RN). Thereby,
Jκ(vn)ψJκ(vκ)ψ,ψC0(RN).
Since Jκ(vn)ψ=on(1), the last limit yields Jκ(vκ)ψ=0 for all ψC0(RN), showing that vκ is a critical point for Jκ.

At last, it follows the arguments used in Refs. 7 and 11, we can prove that vκ ≢ 0.□

This section is to show the L estimate of the solution vκ. To this end, firstly we give two lemmas to find an uniform boundedness of Lp norm of vκ, which is independent of κ > 0.

Lemma 3.1.
The solution vκ satisfies vκ22pdp2, where d is the mountain pass level associated with the functional
P=12RN(|v|2+6V|v|2)dx1pRN(v+)dx.

Proof.
As vκ is a critical point of Jκ, it follows that
Jκ(vκ)=12RN|vκ|2dx+12RNV(x)|G1(vκ)|2dx1pRNG1(vκ)+pdx=cκ
(3.1)
and, for any ψH1(RN),
Jκ(vκ)ψ=RNvκψ+V(x)G1(vκ)g(G1(vκ))ψG1(vκ)+p1g(G1(vκ))ψdx=0.
(3.2)
Using ψ = G−1(vκ)g(G−1(vκ)) as a test function in (3.2), we derive that
pcκ=pJ(vκ)J(vκ)G1(vκ)g(G1(vκ))p22RN|vκ|2dx+p22RNV(x)|G1(vκ)|2dx.
Then, by Lemma 2.1,
pcκp22RN|vκ|2dx+p22RNV(x)|vκ|2dx,
implying that
vκ22pcκp2.
(3.3)
We define the functional
P(v)12RN(|v|2+6V|v|2)dx1pRN(v+)pdx
and denote by d as the mountain pass level associated with the functional P, which is independent of κ. Since Jκ(v) ≤ P(v), we deduce that cκd. Consequently, combining (3.3), the solution vκ must satisfy the estimate
vκ22pdp2.

Remark 3.1.
Following the arguments given in Ref. 11,, we know that d = P(v0), where v0 is the least energy solution of
Δv+6Vv=|v|p2v,xRN.
(3.4)
Moreover, the result of Ref. 12  indicates the solution v0 is spherically symmetric. So for a given radial solution vr of (3.4), there holds P(v0) ≤ P(vr).

Lemma 3.2.

The solution vκ also satisfies |vκ|pV0aN2CN1aNvκ, where a=1p12* and CN is the best Sobolev constant.

Proof.
Using Hölder inequality and Sobolev inequality, we achieve that
|vκ|p|vκ|2aN|vκ|2*1aN|vκ|2aNCN|vκ|21aNRNV(x)V0vκ2dxaN2CN1aNvκ1aNV0aN2CN1aNvκ.

Now, we construct the estimate of |vκ|.

Lemma 3.3.
The solution vκ has the following estimate:
|vκ|21+12a+12ap614aCN1ap|vκ|p1+p22ap.

Proof.
For any ϕH1(RN), the solution vκ of (1.9) satisfies
RNvκϕdx+RNV(x)Gκ1(vκ)ϕgκ(Gκ1(vκ))dx=RNGκ1(vκ)+p1ϕgκ(Gκ1(vκ))dx.
(3.5)
By taking ϕ=(vκl)+ as a test function in (3.5) with l > 0, applying the inequality 1g(G1(vκ))6, we have
Al|vκ|2dxAl|Gκ1(vκ)|p1(vκl)+gκ(Gκ1(vκ))dx6Al|Gκ1(vκ)|p1(vκl)dx,
(3.6)
where Al=xRN:vκ(x)>l and |Al| denotes the Lebesgue measure of the set Al. Furthermore, by Hölder inequality, Minkowski inequality and Lemma 2.1-(3), (3.6) implies
Al|vκ|2dx6Al|vκl|pdx1pAl|G1(vκ)|pdx11p6Al|vκl|pdx1pAl|vκ|pdx1pAl|G1(vκ)|pdxp2p6p2Al|vκl|pdx1pAl|vκ|pdx1pAl|vκ|pdxp2p6p2|vκ|pp2Al|vκl|pdx1pAl|vκl|pdx1p+l|Al|1p.
(3.7)
Moreover, since Al is bounded, then there exists a constant R such that AlBR{xRN||x|<R}. Then by Sobolev inequality, we have
|vκl|2*=Al|vκl|2*dx12*=BR|(vκl)+|2*dx12*CNBR|(vκl)+|2dx12=CNAl|vκ|2dx12=CN|vκ|2.
Thus, Using Hölder inequality again, we achieve the following estimate
Al|vκl|pdxAl|vκl|2*dxp2*|Al|1p2*CNpAl|vκ|2dxp2|Al|1p2*.
That is,
Al|vκl|pdx2pCN2|Al|1p2*2pAl|vκ|2dxCN2|Al|2aAl|vκ|2dx,
(3.8)
where a1p12*. Thus, combining (3.7) and (3.8), we have
Al|vκl|pdx2pCN2|Al|2a6p2|vκ|pp2Al|vκl|pdx2p+l|Al|1pAl|vκl|pdx1p.
(3.9)
On the other hand, noticing that
l|Al|AlvκdxAl|vκ|pdx1p|Al|11p|vκ|p|Al|11p,
we obtain l|Al|1p|vκ|p. To go a step further, we have
CN2|Al|2a6p2|vκ|pp2CN26p2|vκ|pp2|vκ|pl2ap.
(3.10)
Moreover, if we take l0=(26p2CN2)12ap|vκ|pp22ap+1, we have
CN26p2|vκ|pp2|vκ|pl02ap=12.
(3.11)
Consequently, combining (3.9) and (3.11), we conclude, if l > l0, that
Al|vκl|pdx1p26p2CN2|vκ|pp2l|Al|2a+1p.
So
Al|vκl|dxAl|vκl|pdx1p|Al|11p26p2CN2|vκ|pp2l|Al|1+2a.
(3.12)
Inspired by Lemma 5.1, which is presented in Chapter two of Ref. 10, we consider the function
f(l)=Al|vκl|dx.
For this function, we have −f′(l) = |Al|. Therefore, (3.12) can be rewritten as
f(l)26p2CN2|vκ|pp2l(f(l))1+2a,
(3.13)
i.e.,
l11+2a26p2CN2|vκ|pp211+2af(l)11+2a(f(l)).
If we integrate this inequality with respect to l from l0 to lmax ≔ |vκ|, we obtain
lmax111+2al0111+2a26p2CN2|vκ|pp211+2a(f(l0))111+2a(f(lmax))111+2a26p2CN2|vκ|pp211+2a(f(l0))111+2a.
(3.14)
Moreover, jointly with (3.13), recalling that l0=(26p2CN2)12ap|vκ|pp22ap+1, we infer that
(f(l0))111+2a26p2CN2|vκ|pp211+2a26p2CN2|vκ|pp2l0|Al0|1+2a2a1+2a26p2CN2|vκ|pp211+2a26p2CN2|vκ|pp2|Al0|2al02a1+2a26p2CN2|vκ|pp2|vκ|p2apl02apl02a1+2a=l02a1+2a.
(3.15)
Therefore, combining (3.14) and (3.15), we have
lmax2a1+2a2l02a1+2a,
which implies the desired inequality
|vκ|=lmax21+12a+12ap614aCN1ap|vκ|pp22ap+1.

Proof of Theorem 1.1.

Recalling that vκ is the solution of (1.9), we know uκG−1(vκ) solves the Eq. (1.4). Then, by the definition of g(t), uκ is exactly the solution of (1.2) if we show that |uκ|<13κ.

A direct consequence of Lemma 3.1, Lemma 3.2 and Remark 3.1 is that vκ has the estimate
|vκ|pV0aN2CN1aNvκV0aN2CN1aN2pp2P(vr)12.
(4.1)
Jointly with Lemma 3.3 and Lemma 2.1-(3), we infer that
|uκ|6|vκ|21+12a+12ap612+14aCN1apV0aN2CN1aN2pp2P(vr)121+p22ap.
(4.2)
Now, to ensure
|uκ|<13κ,
(4.3)
we select
κ0=22+1a+12ap61+12a31V0aN1+p22apCN2(N2)1a2pp2P(vr)1+p22ap.
Thus inequality (4.3) can be satisfied if only κ ∈ (0, κ0) and we complete the proof.□

This work was supported by NSFC (Grant No. 12271179), the Guangdong Basic and Applied Basic Research Foundation (Grant No. 2020A1515010338).

The authors have no conflicts to disclose.

Yongkuan Cheng: Writing – original draft (equal). Yaotian Shen: Conceptualization (equal); Writing – review & editing (equal).

Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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