# Soliton Solutions for Quasilinear Schrödinger Equations

Authors: Junheng Qu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

By using a change of variables, we get new equations, whose respective associated functionals are well defined inH1(ℝN) and satisfy the geometric hypotheses of the mountain pass theorem. Using this fact, we obtain a nontrivial solution.


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## Body

## 1. Introduction

We study the existence of solutions for the following quasilinear Schrödinger equations:(1)-Δu+V(x)u-[Δ(1+u2)α/2]αu2(1+u2)(2-α)/2=uq+up,-Δu+V(x)u-[Δ(1+u2)α/2]αu2(1+u2)(2-α)/2lllllx∈ℝN,
where V∈C(ℝN,ℝ+) is bounded and periodic in each variable of xi, 1≤i≤N, N≥3, T(α)<q+1<p+1<α2*:=2αN/(N-2), α≥1, and here
(2)T(α):={2α,α0≤α,2α0,1<α<α0,12-46,α=1,
where α0 is defined in Lemma 2. These equations are related to existence of standing wave solutions for quasilinear Schrödinger equations of the form
(3)izt=-Δz+W(x)z-h(|z|2)z-Δg(|z|2)g′(|z|2)z,izt=-Δz+W(x)z-h(|z|2)z-Δg(|z|2)g′x∈ℝN,
where W is a given potential and g and h are real functions. Quasilinear equations such as (3) have been accepted as models of several physical phenomena corresponding to various types of g. The case of g(s)=sα was used for the superfluid film equation in plasma physics [1]. Besides, (3) also appears in plasma physics and fluid mechanics [2], in dissipative quantum mechanics [3], and in the theory of Heisenberg ferromagnetism and magnons [4, 5]. See also [6, 7] for more physical backgrounds. Equations (3) with α=1 have been studied extensively recently; see [8, 9]. When g(s)=(1+s)α/2, then (3) turn into our equations (1) with h(s)=sq+sp. In particular if we let α=1, that is, g(s)=(1+s)1/2, (3) models the self-channeling of a high-power ultrashort laser in matter [10]. In this case, few results are known. In [11], the authors proved global existence and uniqueness of small solutions in transverse space dimensions 2 and 3 and local existence without any smallness condition in transverse space dimension 1. In [12], the authors proved the existence of nontrivial solution. When α>1, although we do not know the physical background of (3), in a mathematical sense, we give the proof of the existence of nontrivial solution.For (1), the main difficulty is that the energy functional associated to (1) is not well defined in H1(ℝN). To overcome this difficulty, enlightened by [8, 9], we give a new change of variables. Then we reduce the quasilinear problem (1) to a semilinear one, which we will prove has a nontrivial solutions.Our main result is the following.Theorem 1.
Assume thatα≥1 and T(α)<q+1<p+1<α2*. Then (1) has a nontrivial solution.In this paper,C denotes positive (possibly different) constant, Lp(ℝN) denotes the usual Lebesgue space with norm |u|p=(∫ℝN‍|u|pdx)1/p, 1≤p<∞, and H1(ℝN) denotes the Sobolev space with norm ∥u∥=(∫ℝN‍(|∇u|2+V(x)u2)dx)1/2.


## 2. The Change of Variables

We note that the solutions of (1) are the critical points of the following functional:
(4)I(u)=12∫ℝN‍[1+α2u22(1+u2)2-α]|∇u|2dx+12∫ℝN‍V(x)u2dx-1q+1∫ℝN‍uq+1dx-1p+1∫ℝN‍up+1dx.
Since the functional I(u) may not be well defined in the usual Sobolev spaces H1(ℝN), we make a change of variables as
(5)v=G(u)=∫0u‍g(t)dt,
where g(t)=1+α2t2/2(1+t2)2-α. Since g(t) is monotonous with |t|, the inverse function G-1(t) of G(t) exists. Then after the change of variables, I(u) can be written by
(6)J(v)=12∫ℝN|∇v|2dx+12∫ℝNV(x)|G-1(v)|2dx-1q+1∫ℝN‍|G-1(v)|q+1dx-1p+1∫ℝN‍|G-1(v)|p+1dx.
By Lemma 2 listed below, we have limt→0G-1(t)/t=1 and limt→∞|G-1(t)|α/t=2(α>1) or 2/3(α=1), so J(v) is well defined in H1(ℝN) and J(v)∈C1.Ifu is a nontrivial solution of (1), then for all ϕ∈C0∞(ℝN) it should satisfy
(7)∫ℝN‍[g2(u)∇u∇ϕ+g(u)g′(u)|∇u|2ϕ+V(x)uϕ-uqϕ-upϕg2(u)∇u∇ϕ+g(u)g′(u)|∇u|2ϕ+V(x)uϕ]dx=0.We show that (7) is equivalent to
(8)J′(v)ψ=∫ℝN[|G-1(v)|pg(G-1(v))ψ∇v∇ψ+V(x)G-1(v)g(G-1(v))ψ-|G-1(v)|qg(G-1(v))ψ-|G-1(v)|pg(G-1(v))ψ]dx=0,∀ψ∈C0∞(ℝN).
Indeed, if we choose ϕ=(1/g(u))ψ in (7), then we get (8). On the other hand, since u=G-1(v), if we let ψ=g(u)ϕ in (8), we get (7). Therefore, in order to find the nontrivial solutions of (1), it suffices to study the existence of the nontrivial solutions of the following equations:
(9)-Δv=-V(x)G-1(v)g(G-1(v))+|G-1(v)|qg(G-1(v))+|G-1(v)|pg(G-1(v)),V(x)G-1(v)g(G-1(v))+|G-1(v)|qg(G-1(v))+|G-1(v)|pg(G-1(v)),5x∈ℝN.Before we close this section, we give some properties of the change of variables.Lemma 2.
For allt>0, one has the following: (1)
limt→0(G-1(t)/t)=1,(2)
(i) ifα>1 then limt→∞(|G-1(t)|α/t)=2, and (ii) if α=1 then limt→∞(|G-1(t)|/t)=2/3,(3)
|G-1(t)|≤t,(4)
(i) ifα0≤α then tg′(t)/g(t)≤α-1, (ii) if 1<α≤α0 then tg′(t)/g(t)≤α0-1, and (iii) if α=1 then tg′(t)/g(t)≤5-26, where α0≈1.36 is a real root of the equation α3-4α2+8α-6=0.Proof.
(1) We easily getlimt→0(G-1(t)/t)=(G-1(t))′|t=0=1/g(G-1(0))=1.
For (2) ifα>1, since g(t)=1+α2t2/2(1+t2)2-α=1+(α2t2/2(1+t2))(1+t2)α-1, so g(t)~(α2/2)t2(α-1)=(α/2)tα-1 as t→∞, then G(t)=∫0t‍g(s)ds~(1/2)tα as t→∞. Since G-1(t) is the inverse of G(t), so G-1(t)~(2t)1/α as t→∞, thus we have limt→∞(|G-1(t)|α/t)=2. When α=1, the result is obvious since g(t) is an increasing bounded function.
For (3), since[G-1(t)-(1/g(0))t]′=1/g(G-1(t))-1/g(0)≤0, so G-1(t)≤(1/g(0))t=t, which proves (3).
Now we prove (4), since(t/g(t))g′(t)=t2/2(1+t2)2g2(t)=t2/(2+5t2+3t4)=1/(2/t2+5+3t2)≤5-26, which is (iii). To prove (i), that is,
(10)α2(2-α)t2≤2(α-1)(1+t2)3-α,
we set j(t)=2(α-1)(1+t2)3-α-α2(2-α)t2, so j′(t)=2t[2(α-1)(3-α)(1+t2)2-α-α2(2-α)]:=2tkα(t), where kα(t)=2(α-1)(3-α)(1+t2)2-α-α2(2-α). Then kα′(t)=4(α-1)(3-α)(2-α)t(1+t2)1-α. If α≤2 or α≥3, we get kα′(t)≥0, so kα(t)≥kα(0). We notice that kα(0)=α3-4α2+8α-6 and kα(0) is an increasing function with respect to α. By Cardano's formula for cubic equations, we know that kα(0) has one real root and two complex roots. If we set α0≈1.36 to be the real root of kα(0), then kα(t)≥kα(0)≥0 as α≥α0. So j′(t)=2tkα(t)≥0. That is j(t) is a increasing function, so j(t)≥j(0)=2(α-1)>0 as α>1. If 2<α<3, we get kα′(t)<0, so kα(t) is a decreasing function, but in this case limt→∞kα(t)=(α-2)α2>0, so kα(t)≥0 for all t>0. Thus we have the same result as α≤2 or α≥3, which proves (i). For (ii), by the definition of α0, we have
(11)α02t2(1+(α0-1)t2)2(1+t2)3-α0+α02t2(1+t2)≤α0-1,
so
(12)α02t2(1+(α0-1)t2)≤2(α0-1)(1+t2)3-α0+α02(α0-1)t2(1+t2).
We add (α0-1)(α2-α02)t2(1+t2) to both sides of (12), where α<α0. Then
(13)α02t2+α02(α0-1)t4+(α0-1)(α2-α02)(t2+t4)≤2(α0-1)(1+t2)3-α0+(α0-1)α2t2(1+t2),∴(α02+(α0-1)(α2-α02))t2+(α0-1)α2t4≤(α0-1)[2(1+t2)3-α+α2t2(1+t2)].
We notice that α02+(α0-1)(α2-α02)≥α2. In fact, α02+α0α2-α03-α2+α02≥α2⇔2(α02-α2)+α0(α2-α02)≥0⇔(α02-α2)(2-α0)>0, and the last inequality is obvious. So
(14)α2t2+(α0-1)α2t4≤(α0-1)[2(1+t2)3-α+α2t2(1+t2)](15)∴α2t2+(α-1)α2t4≤(α0-1)[2(1+t2)3-α+α2t2(1+t2)],
which implies that tg′(t)/g(t)≤α0-1.


## 3. Mountain Pass Geometry

In this section, we establish the geometric hypotheses of the mountain pass theorem.Lemma 3.
There existρ0,a0>0 such that J(v)≥a0 for all ∥v∥=ρ0.Proof.
Let(16)Q(x,t):=-12V(x)|G-1(t)|2+1q+1|G-1(t)|q+1+1p+1|G-1(t)|p+1.
Then, by Lemma 2 and p+1<α2*, we have
(17)limt→0Q(x,t)t2=limt→0[-12V(x)(G-1(t)t)2limt→0Q(x,t)t2=limt→0555+1q+1(G-1(t)t)2|G-1(t)|q-1limt→0Q(x,t)t2=limt→0555+1p+1(G-1(t)t)2|G-1(t)|p-1]limt→0Q(x,t)t2=-12V(x),limt→∞Q(x,t)t2*=limt→∞[-12V(x)(|G-1(t)|αt)2/α1t2*-2/αlimt→∞Q(x,t)t2*=limt→∞555+1q+1(|G-1(t)|αt)(q+1)/α1t2*-(q+1)/αlimt→∞Q(x,t)t2*=limt→∞555+1p+1(|G-1(t)|αt)(p+1)/α1t2*-(p+1)/α]limt→∞Q(x,t)t2*=0.
Thus, for ϵ>0 sufficiently small, there exists a constant Cϵ>0 such that
(18)Q(x,t)≤(-12V(x)+ϵ)t2+Cϵ|t|2*.
Then, we have
(19)J(v)=12∫ℝN‍|∇v|2dx+12∫ℝN‍V(x)|G-1(v)|2dx-1q+1∫ℝN‍|G-1(v)|q+1dx-1p+1×∫ℝN‍|G-1(v)|p+1dx≥12∫ℝN‍|∇v|2dx+12∫ℝN‍V(x)v2dx-ϵ×∫ℝN‍v2dx-Cϵ∫ℝN‍v2*dx≥C∥v∥2-C∥v∥2*.
Thus, by choosing ρ0 small, we get the result when ∥v∥=ρ0.Lemma 4.
There existsv∈H1(ℝN) such that J(v)<0.Proof.
Givenϕ∈C0∞(ℝN,[0,1]) with suppϕ:=B¯1, we will prove that J(sϕ)→-∞ as s→∞, which will prove the result if we take v=sϕ with s large enough. By the proof of Lemma 2, we have G-1(t)≥Ct1/α as t≥1, so
(20)J(sϕ)≤12s2∫ℝN‍|∇ϕ|2dx+12s2×∫ℝN‍V(x)ϕ2dx-s(p+1)/α×∫{|sϕ|≥1}‍ϕ(p+1)/αdx⟶-∞,
as s→∞. Thus, we get the result.


## 4. Existence

In consequence of Lemmas3  and 4  of the Ambrosetti-Rabinowitz mountain pass Theorem [13], see also [14–16], for the constant
(21)c=infγ∈Γsupt∈[0,1]J(γ(t))>0,
where Γ={γ∈C([0,1],H1(ℝN)):γ(0)=0,γ(1)≠0,J(γ(1))<0}, and there exists a Palais-Smale sequence at level c; that is, J(vn)→c and J′(vn)→0 as n→∞.Lemma 5.
The Palais-Smale sequence{vn} for J is bounded in H1(ℝN).Proof.
Since{vn}⊂H1(ℝN) satisfies
(22)J(vn)=12∫ℝN‍|∇vn|2dx+12∫ℝN‍V(x)|G-1(vn)|2dx-1p+1∫ℝN‍|G-1(vn)|p+1dx-1q+1∫ℝN‍|G-1(vn)|q+1dx=c+o(1)
and for any ψ∈C0∞(ℝN),
(23)J′(vn)ψ=∫ℝN‍[|G-1(vn)|pg(G-1(vn))ψ|G-1(vn)|qg(G-1(vn))∇vn∇ψ+V(x)G-1(vn)g(G-1(vn))ψJ′(vn)ψ=∫ℝN‍55-|G-1(vn)|qg(G-1(vn))ψ-|G-1(vn)|pg(G-1(vn))ψ]dxJ′(vn)ψ=o(1)∥ψ∥.
Now, we consider the function G-1(vn)g(G-1(vn)). Note by Lemma 2 that
(24)|∇(G-1(vn)g(G-1(vn)))|=[1+G-1(vn)g(G-1(vn))g′(G-1(vn))]|∇vn|≤12T(α)|∇vn|.
Combining Lemma 2, we have G-1(vn)g(G-1(vn))∈H1(ℝN). Thus, since C0∞(ℝN) is dense in H1(ℝN), by choosing ψ=G-1(vn)g(G-1(vn)) in (23), we deduce that
(25)o(1)∥vn∥=J′(vn)G-1(vn)g(G-1(vn))=∫ℝN‍[(1+G-1(vn)g(G-1(vn))g′(G-1(vn)))|∇vn|2+V(x)|G-1(vn)|2-|G-1(vn)|q+1-|G-1(vn)|p+1(1+G-1(vn)g(G-1(vn))g′(G-1(vn)))]dx≤∫ℝN‍[12T(α)|∇vn|2+V(x)|G-1(vn)|2-|G-1(vn)|q+1-|G-1(vn)|p+112]dx.
Therefore, by (22) and (25), we have
(26)(q+1)J(vn)-J′(vn)G-1(vn)g(G-1(vn))≥(q+12-12T(α))×∫ℝN‍|∇vn|2dx+q-12∫ℝN‍V(x)|G-1(vn)|2dx+(1-q+1p+1)∫ℝN‍|G-1(vn)|p+1dx≥(q+12-12T(α))∫ℝN‍|∇vn|2dx+q-12×∫ℝN‍V(x)|G-1(vn)|2dx.
Combining (22) and (26), we get ∫ℝN‍|G-1(vn)|p+1dx is bounded. To verify that {vn} is bounded in H1(ℝN) we start splitting
(27)∫ℝN‍V(x)vn2dx=∫{x:|vn(x)|>1}‍V(x)vn2dx+∫{x:|vn(x)|≤1}‍V(x)vn2dx.
By the proof of Lemma 2, we have G-1(t)≥Ct1/α, for all t>1 and G(1)=∫01‍g(t)dt≥∫01‍dt=1. Therefore
(28)∫{x:|vn(x)|>1}‍V(x)vn2dx≤C∫{x:|vn(x)|>1}‍|G-1(vn)|2αdx≤C∫{x:|vn(x)|>1}‍|G-1(vn)|p+1dx≤C.
Since g(t) is increasing and G(t)=∫0t‍g(s)ds≤g(t)t, we have
(29)∫{x:|vn(x)|≤1}‍V(x)|G-1(vn)|2dx≥1g2(G-1(1))∫{x:|vn(x)|≤1}‍V(x)vn2dx.
Hence {vn} is bounded in H1(ℝN), and this proves Lemma 5.Now we give the completion of the proof of Theorem1.Proof.
First, we will prove thatJ′(v)=0. That is, v is a weak solution of (9). To prove this, it suffices to show that
(30)J′(v)ψ=∫ℝN‍[|G-1(v)|pg(G-1(v))∇v∇ψ+V(x)G-1(v)g(G-1(v))ψJ′(v)ψ=∫ℝN‍551-|G-1(v)|qg(G-1(v))ψ-|G-1(v)|pg(G-1(v))ψ]dx=0,(v)-|G-1(v)|qg(G-1(v))ψ-|G-1(v)|pg(G-1(v))ψ]dx=0∀ψ∈C0∞(ℝN).
From Lemma 5, {vn} is a bounded Palais-Smale sequence, and there exists v∈H1(ℝN) such that vn⇀v weakly in H1(ℝN). By the Lebesgue dominated theorem, we have
(31)J′(vn)ψ-J′(v)ψ=∫ℝN‍(∇vn-∇v)∇ψdx+∫ℝN‍V(x)[G-1(vn)g(G-1(vn))-G-1(v)g(G-1(v))]ψdx-∫ℝN‍[|G-1(vn)|qg(G-1(vn))-|G-1(v)|qg(G-1(v))]ψdx-∫ℝN‍[|G-1(vn)|pg(G-1(vn))-|G-1(v)|pg(G-1(v))]ψdx⟶0.
Hence, J′(v)=0. That is, v is a weak solution of (1).
Next, in order to complete the proof of Theorem1, we must show that v is nontrivial. By contradiction, we assume v=0. To prove this, we claim that, for all R>0,
(32)limn→∞supy∈ℝN∫BR(y)‍vn2dx=0
cannot occur. Suppose by contradiction that (32) occurs; that is, {vn} vanishes. Then by the Lions compactness lemma [16], vn→0 in Lr(ℝN) for any r∈(2,2*). By the proof of Lemma 2, we get G-1(t):(2t)1/α as t→∞, so there exists a suitable constant C such that G-1(t)≤Ct1/α. In addition, since G(t)≤g(t)t, we have
(33)limn→∞∫ℝN‍|G-1(vn)|pg(G-1(vn))vndx≤limn→∞∫ℝN‍|G-1(vn)|p+1dx≤limn→∞C∫ℝN‍vn(p+1)/αdx=0,limn→∞∫ℝN‍|G-1(vn)|q+1dx=limn→∞∫ℝN‍|G-1(vn)|p+1dx=0,
and limn→∞∫ℝN‍(|G-1(vn)|q/g(G-1(vn)))vndx=0 is obvious since q<p, which implies that
(34)0=limn→∞J′(vn)vn=limn→∞∫ℝN‍[|G-1(vn)|pg(G-1(vn))|∇vn|2+V(x)G-1(vn)g(G-1(vn))vn=limn→∞∫ℝN‍555-|G-1(vn)|qg(G-1(vn))vn-|G-1(vn)|pg(G-1(vn))vn]dx=limn→∞∫ℝN‍[|∇vn|2+V(x)G-1(vn)g(G-1(vn))vn]dx.
Then,
(35)limn→∞∫ℝN‍|∇vn|2dx=0,(36)limn→∞∫ℝN‍V(x)G-1(vn)g(G-1(vn))vndx=0.
On the other hand, by (25), we have
(37)limn→∞∫ℝN‍V(x)|G-1(vn)|2dx=0.
Combining (35) and (37), we get a contradiction since J(vn)→c>0. Thus, {vn} does not vanish and there exist k,R>0 and {yn}⊂ℝN such that
(38)limn→∞∫BR(yn)‍vn2dx≥k>0.
Define v~n(x)=vn(x+yn). We may assume that the components of {yn} are integer multiples of the periods of V(x). Since {vn} is a Palais-Smale sequence for J and V(x) is periodic in xi, 1≤i≤N, {v~n} is also a Palais-Smale sequence for J with J′(v~)=0 if v~n⇀v~  in H1(ℝN). Since {v~n} does not vanish, we have that v~≠0 is a nontrivial solution of (9).



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