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Existence of ground state solutions for quasilinear Schrödinger equations with super-quadratic condition
Accepted Manuscript Existence of ground state solutions for quasilinear Schr¨odinger equations with super-quadratic condition
Jianhua Chen, Xianhua Tang, Bitao Cheng
PII: DOI: Reference:
S0893-9659(17)30344-0 https://doi.org/10.1016/j.aml.2017.11.007 AML 5369
To appear in:
Applied Mathematics Letters
Received date : 24 September 2017 Revised date : 13 November 2017 Accepted date : 14 November 2017 Please cite this article as: J. Chen, X. Tang, B. Cheng, Existence of ground state solutions for quasilinear Schr¨odinger equations with super-quadratic condition, Appl. Math. Lett. (2017), https://doi.org/10.1016/j.aml.2017.11.007 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
Existence of ground state solutions for quasilinear Schr¨ odinger equations with super-quadratic condition Jianhua Chen1 Xianhua Tang1,∗ Bitao Cheng1,2 1 School of Mathematics and Statistics, Central South University, Changsha, Hunan, 410083, P. R. China 2 School of Mathematics and Statistics, Qujing Normal University, Qujing,Yunnan, 655011, P. R. China
Abstract. In this paper, we study the following quasilinear Schr¨ odinger equation −∆u + u − ∆(u2 )u = h(u), x ∈ RN , where N ≥ 3, 2∗ =
2N N −2 ,
h is a continuous function. By using a change of variable, we
obtain the existence of ground state solutions. Unlike the condition lim
only need to assume that lim
|u|→∞
Ru 0
h(s)ds |u|2
|u|→∞
Ru 0
h(s)ds |u|4
= ∞, we
= ∞.
Keywords: quasilinear Schr¨odinger equation; ground state solutions; Pohozaev identity; super-quadratic condition; MR(2010) Subject Classification: 35J60, 35J20
1
Introduction This article is concerned with the following quasilinear Schr¨ odinger equation −∆u + u − ∆(u2 )u = h(u), x ∈ RN ,
(1.1)
where N ≥ 3, 2∗ = N2N −2 , h is a continuous function. It is well known that it is a hot problem in nonlinear analysis to study the existence of solitary wave solutions for the following quasi-linear Schr¨ odinger equation i∂t z = −∆z + W (x)z − k(x, |z|) − ∆l(|z|2 )l0 (|z|2 )z
(1.2)
where z : R × RN → C, W : RN → R is a given potential, l : R → R and k : RN × R → R are suitable functions. For various types of l, the quasilinear equation of the form (1.1) has been derived from models of several physical phenomenon. In particular, l(s) = s was used for the superfluid film equation in fluid mechanics by Kurihara [2]. For more physical background, we can refer to [3, 4] and references therein. †
To whom correspondence should be addressed. [email protected](X. Tang), [email protected].
E-mail:
1
[email protected] (J. Chen), tangx-
Set z(t, x) = exp(−iEt)u(x), where E ∈ R and u is a real function, (1.2) can be reduce to the corresponding equation of elliptic type (see [5]): −∆u + V (x)u − ∆l(u2 )l0 (u2 )u = h(x, u) If we take g 2 (u) = 1 +
[(l2 (u))0 ]2 , 2
x ∈ RN .
(1.3)
then (1.3) turns into quasilinear elliptic equations (see [1])
−div(g 2 (u)∇u) + g(u)g 0 (u)|∇u|2 + V (x)u = h(x, u),
x ∈ RN .
(1.4)
For (1.4), there are many papers (see [1]) studying the existence of positive solutions. But in my mind, there are few papers to study ground state solutions for this problem. If we set √ g(s) = 1 + 2s2 , V (x) ≡ 1 and h(x, u) = h(u) in (1.4), then (1.4) reduces to the Equ. (1.1). When h(u) = |u|p−2 u, the condition p > 4 is a necessary condition for mountain pass type theorem and the Nehari technique. However, for 2 < p < 4, there are few papers to study the existence of solutions except for [11]. Very recently, Wu and Wu [11] proved the existence of radial solutions when h(u) = |u|p−2 u and 2 < p ≤ 4. In this paper, we develop super-quadratic condition to prove the existence of ground state solutions for quasilinear Schr¨ odinger equation. To prove our main result, we make the following assumptions: (h1 ) h ∈ C(R, R), and there exist C > 0 and p ∈ (2, 22∗ ) such that |h(s)| ≤ C(1 + |s|p−1 ), ∀ s ∈ R; (h2 ) h(s) = o(|s|) as |s| → 0; H(s) 2 |s|→∞ |s|
(h3 ) lim
= +∞, where H(s) =
(h4 ) there exist two constants θ0 >
Rs 0
2N N −1
h(t)dt; i and ζ ∈ 0, θ0θ−2 such that 0
1 sh(s) − H(s) + ζs2 ≥ 0, ∀ s ∈ R, 2θ0 and sh(s) ≥ 0 for all s ∈ R.
Now, let us recall some basic notions. Let H 1 (RN ) = u ∈ L2 (RN ) : ∇u ∈ L2 (RN ) with 1 R the norm kuk = RN (|∇u|2 + u2 )dx 2 , and we denote by Lp (RN ) the usual Lebesgue space 1 R with norms kukp = RN |u|p dx p , where 1 ≤ p < ∞. The embedding H 1 (RN ) ,→ Ls (RN ) is continuous for s ∈ [2, 2∗ ]. In (1.1), we can deduce formally that the Euler-Lagrange functional associated with the equation (1.1) is Z Z 1 2 2 2 J(u) = [(1 + 2u )|∇u| + u ] − H(u). (1.5) 2 RN RN R For (1.1), due to the appearance of the nonlocal term RN u2 |∇u|2 , J may be not well defined. To overcome this difficulty, we apply an argument developed by Liu et al. [7] and Colin and Jeanjean [8]. We make the change of variables by v = f −1 (u), where f is defined by f 0 (t) =
1 1
(1 + 2f 2 (t)) 2
on [0, ∞) and f (t) = −f (−t) on(−∞, 0],
and then the functional (1.5) in form can be transformed into Z Z 1 2 2 I(v) = (|∇v| + f (v)) − H(f (v)), 2 RN RN 2
x ∈ RN .
(1.6)
It is easy to check to that I ∈ C 1 . We also know that if v is a critical point of the functional I, then u = f (v) is a critical point of the functional I, i.e. u = f (v) is a solution of problem (1.1). Next, let us recall some properties of the change of variables f : R → R which are proved in [7, 8, 9] as follows: Lemma 1.1 [7, 8, 9] The function f (t) and its derivative satisfy the following properties: (1) f (t)/t → 1 as t → 0; √ 1 (2) f (t)/ t → 2 4 as t → +∞; (3) f 2 (t)/2 ≤ tf (t)f 0 (t) ≤ f 2 (t) for all t ∈ R; (4) there exists a positive constant C such that ( C|t|, if |t| ≤ 1, |f (t)| ≥ 1 2 C|t| , if |t| ≥ 1; Now, we are ready to state the main result of this paper as follows. Theorem 1.2. solution.
2
Assume that (h1 )-(h4 ) are satisfied. Then problem (1.1) has a ground state
Proof of Theorem 1.1
In this section, we prove Theorem 1.1. In the proof of this paper, we will use the following critical point theorem. Theorem 2.1. [10] Let (X, k · k) be a Banach space and I ⊂ R+ an interval. Consider the following family of C 1 -functionals on X: Iµ (v) = A(v) − µB(v),
µ∈I
with B nonnegative and neither A(v) → +∞ or B(v) → +∞ as kvk → ∞. We assume there are two points v1 , v2 in X such that cµ = inf max Iµ (γ(t)) > max{Iµ (v1 ), Iµ (v2 )} for all µ ∈ I γ∈Γµ t∈[0,1]
where Γµ = {γ ∈ C([0, 1], X) : γ(0) = v1 , γ(1) = v2 }. Then for almost every µ ∈ I there is a sequence {vn } ⊂ X such that (i) {vn } is bounded, (ii) Iµ (vn ) → cµ , (iii) Iµ0 (vn ) → 0 in the dual X −1 of X. Moreover, the map µ 7→ cµ is non-increasing and continuous from the left. Next, let I = [λ, 1], where λ ∈ (0, 1). We define the following energy functional Z Z 1 2 2 Iµ (v) = H(f (v)), (|∇v| + f (v)) − µ 2 RN RN where µ ∈ [λ, 1]. Lemma 2.2. Assume that (h1 )-(h3 ) are satisfied. Then there holds: 3
(i) there exists v ∈ H 1 (RN )\{0} independent of λ such that Iµ (v) ≤ 0 for all µ ∈ [λ, 1]. (ii) cµ = inf γ∈Γ maxt∈[0,1] Iµ (γ(t)) > max{Iµ (0), Iµ (v)} for all µ ∈ [λ, 1], where Γ = {γ ∈ C([0, 1], H 1 (RN )) : γ(0) = 0,
γ(1) = v};
(iii) for any v ∈ H 1 (RN )\{0}, there exists a constant C > 0 independent of µ such that cµ ≤ C for all µ ∈ [λ, 1]. Proof. (i) Let v ∈ H 1 (RN )\{0} be fixed. For any µ ∈ [λ, 1], by (f4 ), one has Z Z 1 2 2 H(f (v)). Iµ (v) ≤ Iλ (v) = (|∇v| + f (v)) − λ 2 RN RN
Set vt = f −1 (tf (v(t−1 x))), where t > 0. It follows that Z Z Z 1 N 1 + 2t2 f 2 (v) 1 N +2 2 N 2 Iλ (vt ) = t H(tf (v)) f (v) − λt |∇v| + t 2 1 + 2f 2 (v) 2 N RN RN R Z Z Z H(tf (v)) 2 1 + 2t2 f 2 (v) 1 N +2 1 −2 2 2 =t f (v) − λ t |∇v| + f (v) 2 2 2 2 2 RN RN t f (v) RN 1 + 2f (v) H(s) 2 |s|→∞ |s|
Since lim
H(tf (v)) 2 |t|→∞ |tf (v)|
= +∞, then lim
= +∞, which implies that Iλ (vt ) → −∞ as t → +∞.
Thus taking v = vt for t > 0 large enough, we have Iλ (v) < 0 for all µ ∈ [λ, 1]. (ii) Let K(s) := − 21 f 2 (s) + H(f (s)), then by Lemma 2.1-(1) and (2), we have " # K(s) 1 f (s) 2 H(f (s)) 1 lim = lim − + =− 2 2 + s→0 s 2 s s 2 s→0 and
" # 1 f (s) 2 K(s) H(f (s)) 1 √ lim lim − + = 0. ∗ = s→+∞ |s|2 s→+∞ 2 |s|2∗ −1 |s|2∗ s
(2.1)
(2.2)
It follows from (2.1) and (2.2) that for any ε > 0, there exists Cε > 0 such that K(s) ≤ ∗ − 12 s2 + εs2 + Cε s2 . Thus we have Z Z 1 1 ∗ ∗ Iµ (v) ≥ (|∇v|2 + v 2 ) − µ ε|v|2 + Cε |v|2 ≥ kvk2 − Ckvk2 , 2 RN 4 N R
where ε is small enough. Since 2∗ > 2, we deduce that Iµ has a strict local minimun at 0 and hence cµ > 0. (iii) For any v ∈ H 1 (RN )\{0}, cµ ≤ maxt>0 Iµ (vt ) ≤ maxt>0 Iλ (vt ) for µ ∈ [λ, 1]. Thus we can choose C > maxt>0 Iλ (vt ) ≥ 0 such that cµ ≤ C. This completes the proof. 2 By Theorem 2.1, it is easy to know that for any µ ∈ [λ, 1], there exists a bounded sequence {vn } ⊂ H 1 (RN ) such that Iµ (vn ) → cµ and Iµ0 (vn ) → 0, which is called (P S)-sequence. Lemma 2.3. Assume that {vn } ⊂ H 1 (RN ) is the sequence obtained above. Then there exits vµ 6= 0 such that Iµ0 (vµ ) = 0. Proof. By the boundedness of {vn }, up to a subsequence, there exists vµ ∈ H 1 (RN ) such that vn * vµ , vn → vµ in Lrloc (RN ) for all 2 ≤ r < 2∗ and vn → vµ a.e. in RN . If δ := R lim sup sup B1 (y) |vn |2 dx = 0, by Lions’ Lemma in [6], we can know that vn → 0 in Lr (RN ) for n→∞ y∈RN
˜ = [t − f (t)f 0 (t)] + h(f (t))f 0 (t) and all 2 < r < 2∗ . Let h(t) Z t 1 ˜ ˜ H(t) = h(s)ds = [t2 − f 2 (t)] + H(f (t)). 2 0 4
It is easy to prove that ˜ ˜ h(t)t h(t)t = 0, lim 2 = 0, ∗ 2 t→+∞ t t→0 t
˜ ˜ H(t) H(t) = 0, lim 2 = 0. ∗ 2 t→+∞ t t→0 t
lim
lim
Hence for any ε > 0, there exists a constant Cε > 0 such that ∗ ˜ H(v) ≤ ε(|v|2 + |v|2 ) + Cε |v|r ,
(2.3)
where 2 < r < 2∗ . For ε = cµ /4C 2 > 0, it follows from (2.3) that Z 3cµ 1˜ 3 ˜ lim sup h(vn )vn − H(vn ) ≤ 3εC 2 + Cε lim sup kvn kr = . 2 4 n→∞ n→∞ RN 2
(2.4)
Therefore, by (2.4), we have 1 0 < cµ = Iµ (vn ) − hIµ0 (vn ), vn i + on (1) = 2
Z
RN
1˜ ˜ n ) + on (1) ≤ 3cµ + on (1), h(vn )vn − H(v 2 4
which is a contradiction. Thus δ > 0. Going if necessary to a subsequence, we may assume the R existence of kn ∈ ZN such that B √ (kn ) |vn |2 > 2δ . By vn → vµ in L2loc (RN ), we know that 1+
N
vµ 6= 0. Now, it is a standard to prove that Iµ0 (vµ ) = 0. This completes the proof. 2
By Lemma 2.3, we can choose a sequence {µn } ⊂ [λ, 1] and vn ∈ H 1 (RN )\{0} such that µn → 1 and Iµ0 n (vn ) = 0. Next, we prove our results. By an argument as Theorm B.1 in [6], we can get the following Pohozaev identity. Lemma 2.4. If v ∈ H 1 (RN ) be a critical point of Iµ (v), then v satisfies Z Z Z N −2 N 2 2 |∇v| + f (v) = µN H(f (v)). 2 2 RN RN RN Lemma 2.5. Suppose that (h1 )-(h4 ) hold. Then there exist α > 0 and % > 0 independent of µ such that kvk ≥ % and Iµ (v) ≥ α for any nontrivial critical point v ∈ H 1 (RN ) of Iµ (v). Proof. By Sobolev inequality and Lemma 1.1-(4), we can get Z Z Z Z Z ∗ (|∇v|2 + v 2 ) ≤ |∇v|2 + |v|2 + v2 ≤ C RN
RN
|v|>1
|v|≤1
RN
(|∇v|2 + f 2 (v)).
(2.5)
If v 6= 0 is a critical point of Iµ (v), then by Lemma 1.1-(3), it follows from hIµ0 (v), vi = 0 that Z Z Z Z 1 2 2 2 0 (|∇v| + v ) ≤ |∇v| + f (v)f (v)v = µ h(f (v))f 0 (v)v 2C RN N N N R R ZR (2.6) ≤µ ε|v|2 + Cε |v|p , RN
which implies that there exists % > 0 such that kvk ≥ %, where the positive constant % is
5
independent of µ. Since v satisfies Pohozaev identity, it follows from (2.6) and (h4 ) that Z N −1 1 2 θ0 − N h(f (v))f 0 (v)v − H(f (v)) + |∇v|2 θ (N − 1)θ − N N N 0 0 R R Z Z N −1 N −1 2 θ0 + f 2 (v) − f (v)f 0 (v)v (N − 1)θ0 − N RN (N − 1)θ0 − N RN Z Z N −1 (N − 1)θ0 1 2 2 θ0 − N ≥ µ h(f (v))f (v) − H(f (v)) + ζf (v) + |∇v|2 (N − 1)θ0 − N RN 2θ0 (N − 1)θ0 − N RN Z Z Z N −1 N −1 (N − 1)θ0 µζ 2 2 2 θ0 + f (v) − f (v) − f 2 (v) (N − 1)θ0 − N RN (N − 1)θ0 − N RN (N − 1)θ0 − N RN Z Z N −1 N −1 2 2 θ0 − N 2 θ0 − (N − 1) − (N − 1)θ0 µζ ≥ |∇v| + f 2 (v) (N − 1)θ0 − N RN (N − 1)θ0 − N N R Z [|∇v|2 + f 2 (v)]. ≥C
Iµ (v) =
(N − 1)θ0 (N − 1)θ0 − N
Z
µ
RN
(2.7)
Thus by (2.5) and (2.7), we have Iµ (v) ≥ C%2 := α > 0. This completes the proof 2 Lemma 2.6. Suppose that (h1 )-(h4 ) hold. Then there exists a sequence {µn } ⊂ [λ, 1] and vµn ∈ H 1 (RN )\{0} such that µn → 1, Iµ0 n (vµn ) = 0 and α ≤ Iµn (vµn ) ≤ cµn . Moreover, the sequence {vµn } is bounded in H 1 (RN ). Proof. The first part of this lemma is obvious. We only need to prove that {vµn } is bounded in H 1 (RN ). By Lemma 2.2, we know that there exists a nontrivial critical point vµ ∈ H 1 (RN ) of Iµ for a.e. µ ∈ [λ, 1], such that Iµ0 (vµ ) = 0 and Iµ (vµ ) = cµ . We can choose a sequence µn ∈ [λ, 1] satisfying µn → 1, then there exists a sequence of nontrivial critical points Iµ0 n (vµn ) = 0 and Iµn (vµn ) = cµn Similar to (2.7), one has cµn + on (1) ≥ Ckvµn k2 . By Lemma 2.2-(iii), we can conclude that the sequence {vµn } is bounded in H 1 (RN ). 2 Proof of Theorem 1.1. From Lemma 2.6, we obtain a bounded sequence of nontrivial critical points {vµn } of Iµn such that µn → 1 and α ≤ Iµn (vµn ) ≤ cµn . For convenience, we denote vµn by vn . By Lemma 2.2, we know that the mapping µ 7→ cµ is left continuous. Thus cµn → c1 = c ≥ α > 0. Moreover, for any ψ ∈ C0∞ (RN ), we can infer that I(vµn ) = Iµn (vµn ) + on (1) and hI 0 (vµn ), ψi = hIµ0 n (vµn ), ψi + on (1). Since {vn } is bounded in H 1 (RN ), up to a subsequence, there exists v0 ∈ H 1 (RN ) such that vn * v0 in H 1 (RN ). Similar to the proof in Lemma 2.3, we can show that v0 ∈ H 1 (RN ) is a nontrivial critical point of I. Similar to the prove in [12], we need to set m = inf{I(v) : v ∈ H 1 (RN )\{0}, I 0 (v) = 0}. By Lemma 2.5, we know that α ≤ m ≤ I(v0 ), where v0 is the nontrivial critical point obtained above. Now, we can choose {vn } ∈ H 1 (RN )\{0} such that I(vn ) → m and I 0 (vn ) = 0. Similar to the proof in Lemma 2.6, we know that {vn } is bounded in H 1 (RN ). With the same approach used in Lemma 2.3, we can infer that there exists a nontrivial critical point v of I. Obviously, I(v) ≥ m. In order to complete the proof, we need to prove that I(v) ≤ m. In fact, since I 0 (vn ) = 0 and I 0 (v) = 0, as Lemma 2.6, by the weakly lower semi-continuity of norm, (i) of
6
Lemma 2.2 and Fatou’s Lemma, one has m = lim inf I(vn ) n→∞ Z 1 (N − 1)θ0 µ h(f (vn ))f 0 (vn )vn − H(f (vn )) + ζf 2 (vn ) = lim inf n→∞ (N − 1)θ0 − N RN θ0 Z Z N −1 N −1 θ0 − (N − 1)θ0 µζ 2 θ0 − N + |∇vn |2 + 2 f 2 (vn ) (N − 1)θ0 − N RN (N − 1)θ0 − N N R Z N −1 − f (v)f 0 (vn )vn (N − 1)θ0 − N RN Z Z N −1 (N − 1)θ0 1 0 2 2 θ0 − N ≥ µ h(f (v))f (v)v − H(f (v)) + ζf (v) + |∇v|2 (N − 1)θ0 − N RN 2θ0 (N − 1)θ0 − N RN Z Z N −1 N −1 2 2 θ0 − (N − 1)θ0 µζ + f (v) − f (v)f 0 (v)v = I(v). (N − 1)θ0 − N (N − 1)θ0 − N RN RN Thus I(v) ≤ m. This completes the proof. 2
Acknowledgements This work was supported by National Natural Science Foundation of China (Grant No. 11461043, 11571370 and 11601525), and supported partly by the Provincial Natural Science Foundation of Jiangxi, China (20161BAB201009) and the outstanding youth scientist foundation plan of jiangxi (20171BCB23004) and Hunan Provincial Innovation Foundation For Postgraduate (Grant No. CX2016B037).
References [1] Y. Shen, Y. Wang, Soliton solutions for generalized quasilinear Schr¨ odinger equations, Nonlinear Anal., Theory Methods Appl. 80 (2013) 194-201. [2] S. Kurihara, Large-amplitude quasi-solitons in superfluid films, J. Phys. Soc. Japan 50 (1981) 3262-3267. [3] F. G. Bass, N. N. Nasanov, Nonlinear electromagnetic-spin waves. Phys. Rep. 189 (1990) 165-223. [4] V. G. Makhankov, V. K. Fedyanin, Nonlinear effects in quasi-one-dimensional models and condensed matter theory. Phys. Rep. 104 (1984) 1-86. [5] S. Cuccagna, On instability of excited states of the nonlinear Schr¨ odinger equation, Phys. D 238 (2009) 38-54. [6] M. Willem, Minimax theorems. Birkhauser, Berlin (1996). [7] J. Q. Liu, Y. Wang, Z. Q. Wang, Solutions for quasilinear Schr¨ odinger equations, II, J. Differential Equations 187 (2003) 473-793. [8] M. Colin, L. Jeanjean, Solutions for a quasilinear Schr¨ odinger equation: a dual approach, Nonlinear Anal. 56 (2004) 213-226. ´ U. Severo, Solitary waves for a class of quasilinear Schr¨ [9] J. Marcos do O, odinger equations in dimension two, Calc. Var. Partial Differential Equations 38 (2010) 275-315.
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[10] L. Jeanjean, On the existence of bounded Palais-Smale sequences and application to a Landesman-Lazer type problem set on RN . Proc. R. Soc. Edinburgh Sect A. 129 (1999) 787-809. [11] K. Wu, X. Wu, Radial solutions for quasilinear Schr¨ odinger equations without 4-superlinear condition, Appl. Math. Lett. 76 (2018) 53-59. [12] L. Jeanjean, K. Tanaka, A positive solution for an asymptotically linear elliptic problem on RN autonomous at infinity, ESAIM Control Optim. Calc. Var. 7 (2002) 597-614.
8
Jianhua Chen, Xianhua Tang, Bitao Cheng
PII: DOI: Reference:
S0893-9659(17)30344-0 https://doi.org/10.1016/j.aml.2017.11.007 AML 5369
To appear in:
Applied Mathematics Letters
Received date : 24 September 2017 Revised date : 13 November 2017 Accepted date : 14 November 2017 Please cite this article as: J. Chen, X. Tang, B. Cheng, Existence of ground state solutions for quasilinear Schr¨odinger equations with super-quadratic condition, Appl. Math. Lett. (2017), https://doi.org/10.1016/j.aml.2017.11.007 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
*Manuscript Click here to view linked References
Existence of ground state solutions for quasilinear Schr¨ odinger equations with super-quadratic condition Jianhua Chen1 Xianhua Tang1,∗ Bitao Cheng1,2 1 School of Mathematics and Statistics, Central South University, Changsha, Hunan, 410083, P. R. China 2 School of Mathematics and Statistics, Qujing Normal University, Qujing,Yunnan, 655011, P. R. China
Abstract. In this paper, we study the following quasilinear Schr¨ odinger equation −∆u + u − ∆(u2 )u = h(u), x ∈ RN , where N ≥ 3, 2∗ =
2N N −2 ,
h is a continuous function. By using a change of variable, we
obtain the existence of ground state solutions. Unlike the condition lim
only need to assume that lim
|u|→∞
Ru 0
h(s)ds |u|2
|u|→∞
Ru 0
h(s)ds |u|4
= ∞, we
= ∞.
Keywords: quasilinear Schr¨odinger equation; ground state solutions; Pohozaev identity; super-quadratic condition; MR(2010) Subject Classification: 35J60, 35J20
1
Introduction This article is concerned with the following quasilinear Schr¨ odinger equation −∆u + u − ∆(u2 )u = h(u), x ∈ RN ,
(1.1)
where N ≥ 3, 2∗ = N2N −2 , h is a continuous function. It is well known that it is a hot problem in nonlinear analysis to study the existence of solitary wave solutions for the following quasi-linear Schr¨ odinger equation i∂t z = −∆z + W (x)z − k(x, |z|) − ∆l(|z|2 )l0 (|z|2 )z
(1.2)
where z : R × RN → C, W : RN → R is a given potential, l : R → R and k : RN × R → R are suitable functions. For various types of l, the quasilinear equation of the form (1.1) has been derived from models of several physical phenomenon. In particular, l(s) = s was used for the superfluid film equation in fluid mechanics by Kurihara [2]. For more physical background, we can refer to [3, 4] and references therein. †
To whom correspondence should be addressed. [email protected](X. Tang), [email protected].
E-mail:
1
[email protected] (J. Chen), tangx-
Set z(t, x) = exp(−iEt)u(x), where E ∈ R and u is a real function, (1.2) can be reduce to the corresponding equation of elliptic type (see [5]): −∆u + V (x)u − ∆l(u2 )l0 (u2 )u = h(x, u) If we take g 2 (u) = 1 +
[(l2 (u))0 ]2 , 2
x ∈ RN .
(1.3)
then (1.3) turns into quasilinear elliptic equations (see [1])
−div(g 2 (u)∇u) + g(u)g 0 (u)|∇u|2 + V (x)u = h(x, u),
x ∈ RN .
(1.4)
For (1.4), there are many papers (see [1]) studying the existence of positive solutions. But in my mind, there are few papers to study ground state solutions for this problem. If we set √ g(s) = 1 + 2s2 , V (x) ≡ 1 and h(x, u) = h(u) in (1.4), then (1.4) reduces to the Equ. (1.1). When h(u) = |u|p−2 u, the condition p > 4 is a necessary condition for mountain pass type theorem and the Nehari technique. However, for 2 < p < 4, there are few papers to study the existence of solutions except for [11]. Very recently, Wu and Wu [11] proved the existence of radial solutions when h(u) = |u|p−2 u and 2 < p ≤ 4. In this paper, we develop super-quadratic condition to prove the existence of ground state solutions for quasilinear Schr¨ odinger equation. To prove our main result, we make the following assumptions: (h1 ) h ∈ C(R, R), and there exist C > 0 and p ∈ (2, 22∗ ) such that |h(s)| ≤ C(1 + |s|p−1 ), ∀ s ∈ R; (h2 ) h(s) = o(|s|) as |s| → 0; H(s) 2 |s|→∞ |s|
(h3 ) lim
= +∞, where H(s) =
(h4 ) there exist two constants θ0 >
Rs 0
2N N −1
h(t)dt; i and ζ ∈ 0, θ0θ−2 such that 0
1 sh(s) − H(s) + ζs2 ≥ 0, ∀ s ∈ R, 2θ0 and sh(s) ≥ 0 for all s ∈ R.
Now, let us recall some basic notions. Let H 1 (RN ) = u ∈ L2 (RN ) : ∇u ∈ L2 (RN ) with 1 R the norm kuk = RN (|∇u|2 + u2 )dx 2 , and we denote by Lp (RN ) the usual Lebesgue space 1 R with norms kukp = RN |u|p dx p , where 1 ≤ p < ∞. The embedding H 1 (RN ) ,→ Ls (RN ) is continuous for s ∈ [2, 2∗ ]. In (1.1), we can deduce formally that the Euler-Lagrange functional associated with the equation (1.1) is Z Z 1 2 2 2 J(u) = [(1 + 2u )|∇u| + u ] − H(u). (1.5) 2 RN RN R For (1.1), due to the appearance of the nonlocal term RN u2 |∇u|2 , J may be not well defined. To overcome this difficulty, we apply an argument developed by Liu et al. [7] and Colin and Jeanjean [8]. We make the change of variables by v = f −1 (u), where f is defined by f 0 (t) =
1 1
(1 + 2f 2 (t)) 2
on [0, ∞) and f (t) = −f (−t) on(−∞, 0],
and then the functional (1.5) in form can be transformed into Z Z 1 2 2 I(v) = (|∇v| + f (v)) − H(f (v)), 2 RN RN 2
x ∈ RN .
(1.6)
It is easy to check to that I ∈ C 1 . We also know that if v is a critical point of the functional I, then u = f (v) is a critical point of the functional I, i.e. u = f (v) is a solution of problem (1.1). Next, let us recall some properties of the change of variables f : R → R which are proved in [7, 8, 9] as follows: Lemma 1.1 [7, 8, 9] The function f (t) and its derivative satisfy the following properties: (1) f (t)/t → 1 as t → 0; √ 1 (2) f (t)/ t → 2 4 as t → +∞; (3) f 2 (t)/2 ≤ tf (t)f 0 (t) ≤ f 2 (t) for all t ∈ R; (4) there exists a positive constant C such that ( C|t|, if |t| ≤ 1, |f (t)| ≥ 1 2 C|t| , if |t| ≥ 1; Now, we are ready to state the main result of this paper as follows. Theorem 1.2. solution.
2
Assume that (h1 )-(h4 ) are satisfied. Then problem (1.1) has a ground state
Proof of Theorem 1.1
In this section, we prove Theorem 1.1. In the proof of this paper, we will use the following critical point theorem. Theorem 2.1. [10] Let (X, k · k) be a Banach space and I ⊂ R+ an interval. Consider the following family of C 1 -functionals on X: Iµ (v) = A(v) − µB(v),
µ∈I
with B nonnegative and neither A(v) → +∞ or B(v) → +∞ as kvk → ∞. We assume there are two points v1 , v2 in X such that cµ = inf max Iµ (γ(t)) > max{Iµ (v1 ), Iµ (v2 )} for all µ ∈ I γ∈Γµ t∈[0,1]
where Γµ = {γ ∈ C([0, 1], X) : γ(0) = v1 , γ(1) = v2 }. Then for almost every µ ∈ I there is a sequence {vn } ⊂ X such that (i) {vn } is bounded, (ii) Iµ (vn ) → cµ , (iii) Iµ0 (vn ) → 0 in the dual X −1 of X. Moreover, the map µ 7→ cµ is non-increasing and continuous from the left. Next, let I = [λ, 1], where λ ∈ (0, 1). We define the following energy functional Z Z 1 2 2 Iµ (v) = H(f (v)), (|∇v| + f (v)) − µ 2 RN RN where µ ∈ [λ, 1]. Lemma 2.2. Assume that (h1 )-(h3 ) are satisfied. Then there holds: 3
(i) there exists v ∈ H 1 (RN )\{0} independent of λ such that Iµ (v) ≤ 0 for all µ ∈ [λ, 1]. (ii) cµ = inf γ∈Γ maxt∈[0,1] Iµ (γ(t)) > max{Iµ (0), Iµ (v)} for all µ ∈ [λ, 1], where Γ = {γ ∈ C([0, 1], H 1 (RN )) : γ(0) = 0,
γ(1) = v};
(iii) for any v ∈ H 1 (RN )\{0}, there exists a constant C > 0 independent of µ such that cµ ≤ C for all µ ∈ [λ, 1]. Proof. (i) Let v ∈ H 1 (RN )\{0} be fixed. For any µ ∈ [λ, 1], by (f4 ), one has Z Z 1 2 2 H(f (v)). Iµ (v) ≤ Iλ (v) = (|∇v| + f (v)) − λ 2 RN RN
Set vt = f −1 (tf (v(t−1 x))), where t > 0. It follows that Z Z Z 1 N 1 + 2t2 f 2 (v) 1 N +2 2 N 2 Iλ (vt ) = t H(tf (v)) f (v) − λt |∇v| + t 2 1 + 2f 2 (v) 2 N RN RN R Z Z Z H(tf (v)) 2 1 + 2t2 f 2 (v) 1 N +2 1 −2 2 2 =t f (v) − λ t |∇v| + f (v) 2 2 2 2 2 RN RN t f (v) RN 1 + 2f (v) H(s) 2 |s|→∞ |s|
Since lim
H(tf (v)) 2 |t|→∞ |tf (v)|
= +∞, then lim
= +∞, which implies that Iλ (vt ) → −∞ as t → +∞.
Thus taking v = vt for t > 0 large enough, we have Iλ (v) < 0 for all µ ∈ [λ, 1]. (ii) Let K(s) := − 21 f 2 (s) + H(f (s)), then by Lemma 2.1-(1) and (2), we have " # K(s) 1 f (s) 2 H(f (s)) 1 lim = lim − + =− 2 2 + s→0 s 2 s s 2 s→0 and
" # 1 f (s) 2 K(s) H(f (s)) 1 √ lim lim − + = 0. ∗ = s→+∞ |s|2 s→+∞ 2 |s|2∗ −1 |s|2∗ s
(2.1)
(2.2)
It follows from (2.1) and (2.2) that for any ε > 0, there exists Cε > 0 such that K(s) ≤ ∗ − 12 s2 + εs2 + Cε s2 . Thus we have Z Z 1 1 ∗ ∗ Iµ (v) ≥ (|∇v|2 + v 2 ) − µ ε|v|2 + Cε |v|2 ≥ kvk2 − Ckvk2 , 2 RN 4 N R
where ε is small enough. Since 2∗ > 2, we deduce that Iµ has a strict local minimun at 0 and hence cµ > 0. (iii) For any v ∈ H 1 (RN )\{0}, cµ ≤ maxt>0 Iµ (vt ) ≤ maxt>0 Iλ (vt ) for µ ∈ [λ, 1]. Thus we can choose C > maxt>0 Iλ (vt ) ≥ 0 such that cµ ≤ C. This completes the proof. 2 By Theorem 2.1, it is easy to know that for any µ ∈ [λ, 1], there exists a bounded sequence {vn } ⊂ H 1 (RN ) such that Iµ (vn ) → cµ and Iµ0 (vn ) → 0, which is called (P S)-sequence. Lemma 2.3. Assume that {vn } ⊂ H 1 (RN ) is the sequence obtained above. Then there exits vµ 6= 0 such that Iµ0 (vµ ) = 0. Proof. By the boundedness of {vn }, up to a subsequence, there exists vµ ∈ H 1 (RN ) such that vn * vµ , vn → vµ in Lrloc (RN ) for all 2 ≤ r < 2∗ and vn → vµ a.e. in RN . If δ := R lim sup sup B1 (y) |vn |2 dx = 0, by Lions’ Lemma in [6], we can know that vn → 0 in Lr (RN ) for n→∞ y∈RN
˜ = [t − f (t)f 0 (t)] + h(f (t))f 0 (t) and all 2 < r < 2∗ . Let h(t) Z t 1 ˜ ˜ H(t) = h(s)ds = [t2 − f 2 (t)] + H(f (t)). 2 0 4
It is easy to prove that ˜ ˜ h(t)t h(t)t = 0, lim 2 = 0, ∗ 2 t→+∞ t t→0 t
˜ ˜ H(t) H(t) = 0, lim 2 = 0. ∗ 2 t→+∞ t t→0 t
lim
lim
Hence for any ε > 0, there exists a constant Cε > 0 such that ∗ ˜ H(v) ≤ ε(|v|2 + |v|2 ) + Cε |v|r ,
(2.3)
where 2 < r < 2∗ . For ε = cµ /4C 2 > 0, it follows from (2.3) that Z 3cµ 1˜ 3 ˜ lim sup h(vn )vn − H(vn ) ≤ 3εC 2 + Cε lim sup kvn kr = . 2 4 n→∞ n→∞ RN 2
(2.4)
Therefore, by (2.4), we have 1 0 < cµ = Iµ (vn ) − hIµ0 (vn ), vn i + on (1) = 2
Z
RN
1˜ ˜ n ) + on (1) ≤ 3cµ + on (1), h(vn )vn − H(v 2 4
which is a contradiction. Thus δ > 0. Going if necessary to a subsequence, we may assume the R existence of kn ∈ ZN such that B √ (kn ) |vn |2 > 2δ . By vn → vµ in L2loc (RN ), we know that 1+
N
vµ 6= 0. Now, it is a standard to prove that Iµ0 (vµ ) = 0. This completes the proof. 2
By Lemma 2.3, we can choose a sequence {µn } ⊂ [λ, 1] and vn ∈ H 1 (RN )\{0} such that µn → 1 and Iµ0 n (vn ) = 0. Next, we prove our results. By an argument as Theorm B.1 in [6], we can get the following Pohozaev identity. Lemma 2.4. If v ∈ H 1 (RN ) be a critical point of Iµ (v), then v satisfies Z Z Z N −2 N 2 2 |∇v| + f (v) = µN H(f (v)). 2 2 RN RN RN Lemma 2.5. Suppose that (h1 )-(h4 ) hold. Then there exist α > 0 and % > 0 independent of µ such that kvk ≥ % and Iµ (v) ≥ α for any nontrivial critical point v ∈ H 1 (RN ) of Iµ (v). Proof. By Sobolev inequality and Lemma 1.1-(4), we can get Z Z Z Z Z ∗ (|∇v|2 + v 2 ) ≤ |∇v|2 + |v|2 + v2 ≤ C RN
RN
|v|>1
|v|≤1
RN
(|∇v|2 + f 2 (v)).
(2.5)
If v 6= 0 is a critical point of Iµ (v), then by Lemma 1.1-(3), it follows from hIµ0 (v), vi = 0 that Z Z Z Z 1 2 2 2 0 (|∇v| + v ) ≤ |∇v| + f (v)f (v)v = µ h(f (v))f 0 (v)v 2C RN N N N R R ZR (2.6) ≤µ ε|v|2 + Cε |v|p , RN
which implies that there exists % > 0 such that kvk ≥ %, where the positive constant % is
5
independent of µ. Since v satisfies Pohozaev identity, it follows from (2.6) and (h4 ) that Z N −1 1 2 θ0 − N h(f (v))f 0 (v)v − H(f (v)) + |∇v|2 θ (N − 1)θ − N N N 0 0 R R Z Z N −1 N −1 2 θ0 + f 2 (v) − f (v)f 0 (v)v (N − 1)θ0 − N RN (N − 1)θ0 − N RN Z Z N −1 (N − 1)θ0 1 2 2 θ0 − N ≥ µ h(f (v))f (v) − H(f (v)) + ζf (v) + |∇v|2 (N − 1)θ0 − N RN 2θ0 (N − 1)θ0 − N RN Z Z Z N −1 N −1 (N − 1)θ0 µζ 2 2 2 θ0 + f (v) − f (v) − f 2 (v) (N − 1)θ0 − N RN (N − 1)θ0 − N RN (N − 1)θ0 − N RN Z Z N −1 N −1 2 2 θ0 − N 2 θ0 − (N − 1) − (N − 1)θ0 µζ ≥ |∇v| + f 2 (v) (N − 1)θ0 − N RN (N − 1)θ0 − N N R Z [|∇v|2 + f 2 (v)]. ≥C
Iµ (v) =
(N − 1)θ0 (N − 1)θ0 − N
Z
µ
RN
(2.7)
Thus by (2.5) and (2.7), we have Iµ (v) ≥ C%2 := α > 0. This completes the proof 2 Lemma 2.6. Suppose that (h1 )-(h4 ) hold. Then there exists a sequence {µn } ⊂ [λ, 1] and vµn ∈ H 1 (RN )\{0} such that µn → 1, Iµ0 n (vµn ) = 0 and α ≤ Iµn (vµn ) ≤ cµn . Moreover, the sequence {vµn } is bounded in H 1 (RN ). Proof. The first part of this lemma is obvious. We only need to prove that {vµn } is bounded in H 1 (RN ). By Lemma 2.2, we know that there exists a nontrivial critical point vµ ∈ H 1 (RN ) of Iµ for a.e. µ ∈ [λ, 1], such that Iµ0 (vµ ) = 0 and Iµ (vµ ) = cµ . We can choose a sequence µn ∈ [λ, 1] satisfying µn → 1, then there exists a sequence of nontrivial critical points Iµ0 n (vµn ) = 0 and Iµn (vµn ) = cµn Similar to (2.7), one has cµn + on (1) ≥ Ckvµn k2 . By Lemma 2.2-(iii), we can conclude that the sequence {vµn } is bounded in H 1 (RN ). 2 Proof of Theorem 1.1. From Lemma 2.6, we obtain a bounded sequence of nontrivial critical points {vµn } of Iµn such that µn → 1 and α ≤ Iµn (vµn ) ≤ cµn . For convenience, we denote vµn by vn . By Lemma 2.2, we know that the mapping µ 7→ cµ is left continuous. Thus cµn → c1 = c ≥ α > 0. Moreover, for any ψ ∈ C0∞ (RN ), we can infer that I(vµn ) = Iµn (vµn ) + on (1) and hI 0 (vµn ), ψi = hIµ0 n (vµn ), ψi + on (1). Since {vn } is bounded in H 1 (RN ), up to a subsequence, there exists v0 ∈ H 1 (RN ) such that vn * v0 in H 1 (RN ). Similar to the proof in Lemma 2.3, we can show that v0 ∈ H 1 (RN ) is a nontrivial critical point of I. Similar to the prove in [12], we need to set m = inf{I(v) : v ∈ H 1 (RN )\{0}, I 0 (v) = 0}. By Lemma 2.5, we know that α ≤ m ≤ I(v0 ), where v0 is the nontrivial critical point obtained above. Now, we can choose {vn } ∈ H 1 (RN )\{0} such that I(vn ) → m and I 0 (vn ) = 0. Similar to the proof in Lemma 2.6, we know that {vn } is bounded in H 1 (RN ). With the same approach used in Lemma 2.3, we can infer that there exists a nontrivial critical point v of I. Obviously, I(v) ≥ m. In order to complete the proof, we need to prove that I(v) ≤ m. In fact, since I 0 (vn ) = 0 and I 0 (v) = 0, as Lemma 2.6, by the weakly lower semi-continuity of norm, (i) of
6
Lemma 2.2 and Fatou’s Lemma, one has m = lim inf I(vn ) n→∞ Z 1 (N − 1)θ0 µ h(f (vn ))f 0 (vn )vn − H(f (vn )) + ζf 2 (vn ) = lim inf n→∞ (N − 1)θ0 − N RN θ0 Z Z N −1 N −1 θ0 − (N − 1)θ0 µζ 2 θ0 − N + |∇vn |2 + 2 f 2 (vn ) (N − 1)θ0 − N RN (N − 1)θ0 − N N R Z N −1 − f (v)f 0 (vn )vn (N − 1)θ0 − N RN Z Z N −1 (N − 1)θ0 1 0 2 2 θ0 − N ≥ µ h(f (v))f (v)v − H(f (v)) + ζf (v) + |∇v|2 (N − 1)θ0 − N RN 2θ0 (N − 1)θ0 − N RN Z Z N −1 N −1 2 2 θ0 − (N − 1)θ0 µζ + f (v) − f (v)f 0 (v)v = I(v). (N − 1)θ0 − N (N − 1)θ0 − N RN RN Thus I(v) ≤ m. This completes the proof. 2
Acknowledgements This work was supported by National Natural Science Foundation of China (Grant No. 11461043, 11571370 and 11601525), and supported partly by the Provincial Natural Science Foundation of Jiangxi, China (20161BAB201009) and the outstanding youth scientist foundation plan of jiangxi (20171BCB23004) and Hunan Provincial Innovation Foundation For Postgraduate (Grant No. CX2016B037).
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