Bifurcation results for a class of prescribed mean curvature equations in bounded domains

Bifurcation results for a class of prescribed mean curvature equations in bounded domains

Nonlinear Analysis 171 (2018) 21–31 Contents lists available at ScienceDirect Nonlinear Analysis www.elsevier.com/locate/na Bifurcation results for...

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Nonlinear Analysis 171 (2018) 21–31

Contents lists available at ScienceDirect

Nonlinear Analysis www.elsevier.com/locate/na

Bifurcation results for a class of prescribed mean curvature equations in bounded domains Hongjing Pana , Ruixiang Xingb, * a b

School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China School of Mathematics, Sun Yat-sen University, Guangzhou 510275, China

article

info

Article history: Received 4 September 2017 Accepted 16 January 2018 Communicated by Enzo Mitidieri MSC: 35J93 47J15 35B32 34B15 Keywords: Prescribed mean curvature equation Direction of bifurcation Nonlinear eigenvalue problem Transcritical bifurcation Pitchfork bifurcation Exponential nonlinearity Logistic nonlinearity Pendulum Perturbed Gelfand problem

abstract Bifurcation for prescribed mean curvature equations in one dimension has been intensively investigated in recent years, and a striking phenomenon discovered is that the length of the interval may affect the shapes of bifurcation curves. However, to our best knowledge, no such results are known in higher dimensions. In this paper, we study a class of prescribed mean curvature equations in bounded domains of RN with general nonlinearities f satisfying f (0) = 0 and f ′ (0) > 0. We establish some formulas for directions of bifurcation at simple eigenvalues, which lead to a sufficient and necessary condition to ensure that the directions depend on the size of the domain. In contrast, this phenomenon does not occur for the semilinear case. Some interesting examples, including logistic and perturbed exponential nonlinearities, are also investigated. © 2018 Published by Elsevier Ltd.

1. Introduction Consider the prescribed mean curvature problem ⎧ ∇u ⎪ ) = λf (u), ⎨ −div( √ 2 1 + |∇u| ⎪ ⎩ u = 0,

in Ω , (1.1) on ∂Ω ,

where Ω is a bounded smooth domain in RN (N ⩾ 1) and λ is a positive parameter. We are concerned with the effect of the size of the domain on the shape of bifurcation curve.

*

Corresponding author. E-mail addresses: [email protected] (H. Pan), [email protected] (R. Xing).

https://doi.org/10.1016/j.na.2018.01.010 0362-546X/© 2018 Published by Elsevier Ltd.

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

22

Bifurcation for the semilinear problem { −∆u = λf (u),

in Ω ,

u = 0,

on ∂Ω ,

(1.2)

has been well studied; see e.g., the nice surveys in [18,21] and the recent monograph [13]. Different from ∆, the mean curvature operator u ↦→ div( √ ∇u 2 ) is nonlinear and nonhomogeneous. This leads to that the 1+|∇u|

quasilinear problem (1.1) is more difficult to handle than the semilinear problem (1.2). Bifurcation for the prescribed mean curvature equation is still far from being well understood; see [19] for f (u) = u, [6,30] for Ω = RN , and [15,16] for solutions in the bounded variation space. Recently, one-dimensional case of (1.1) has received considerable attention and various results on existence and exact multiplicity have been obtained (see e.g., [1–5,7–9,11,12,14,17,19,20,22–28,31,32]). Among them, a striking phenomenon discovered is that the length of the interval may affect on the shapes of bifurcation curves [3,4,7–9,12,14,22–28]. However, to our best knowledge, no such results are known in higher-dimension cases of (1.1). Our research is inspired by [4,8]. In [4], the authors investigate the case Ω = (0, L) and f (u) = u − u3 with Neumann boundary conditions, and apply the Liapunov–Schmidt reduction to prove nπ 2 √ that at each eigenvalue λn = ( nπ and a subcritical pitchfork L ) , a supercritical pitchfork occurs if L > 2 nπ occurs if L < √2 . In [8], the authors use a time-map method to establish bifurcation diagrams for positive au

solution of (1.1) with f (u) = e a+u − 1 and Ω = (−L, L); the bifurcation diagrams derived imply that when a = 2, the length L may affect the direction of local bifurcation. The motivation of this paper is to generalize these one-dimensional results from special examples to more general f and to higher dimensions. In the present paper, we study (1.1) with general nonlinearities f satisfying f (0) = 0 and f ′ (0) > 0. By establishing and analyzing formulas for directions of bifurcation at simple eigenvalues, we obtain a sufficient and necessary condition to ensure that the direction of bifurcation at the principal eigenvalue depends on the size of the domain. In contrast, this phenomenon does not occur for the corresponding semilinear problem. Some interesting examples are also discussed. A key point is that although the linearization of the mean curvature operator at the trivial solution is the Laplace operator, the higher order terms play a crucial role in determining the directions of bifurcation. These results may lead, if combined with other information, such as suitable a priori bounds, to new multiplicity results. Yet, since the prescribed mean curvature operator is non-uniformly elliptic, it is not easy to establish such a priori bounds, even in specific examples; to obtain multiplicity results far from the local bifurcation solutions is therefore our future research goal. The paper is organized as follows. In Section 2, we give the main theorems. Section 3 provides the proofs. 2. Main results Notations. Denote by λ1,Ω and φ1,Ω the principal eigenvalue and eigenfunction of −∆ with the Dirichlet boundary condition on the domain Ω . We will write λ1 and φ1 for short, unless otherwise indicated. Denote by Λ the set of all simple eigenvalues of −∆ with the Dirichlet boundary condition on the domain Ω . Clearly, λ1 ∈ Λ ̸= ∅. Denote by λs and us the derivatives of λ(s) and u(s) with respect to s, respectively. The main results of this paper are the following theorems. Theorem 2.1. Assume that f ∈ C k (−δ, δ) with k ⩾ 1, f (0) = 0 and f ′ (0) > 0 in (1.1). Let λ0 ∈ Λ and 0 , 0) is a bifurcation point and there exist an interval let φ0 be the corresponding eigenfunction. Then ( f λ′ (0) k 1 (−ε, ε) and C functions λ : (−ε, ε) → R and ψ : (−ε, ε) → Z such that − div( √

∇(sφ0 + sψ(s)) 2

1 + |∇(sφ0 + sψ(s))|

) = λ(s)f (sφ0 + sψ(s)),

(2.1)

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

λ0 , f ′ (0)

λ(0) =

23

ψ(0) = 0,

(2.2)

¯ ) | u|∂Ω = 0}. Moreover, where Z is the orthogonal (in L2 (Ω )) complement of span{φ0 } in X = {u ∈ C 2,α (Ω the following assertions hold: (1) If k ⩾ 2, then ∫ λ0 f ′′ (0) Ω φ30 dx ∫ . (2.3) λs (0) = − 2(f ′ (0))2 Ω φ20 dx (2) If k ⩾ 3 and f ′′ (0) = 0, then λss (0) =

−3f ′ (0)

(3) If k ⩾ 4, f ′′ (0) = 0 and f ′′′ (0) = −

3f ′ (0) λ0

4

∫ Ω

|∇φ0 | dx − λ0 f ′′′ (0) ∫ 3(f ′ (0))2 Ω φ20 dx

∫ Ω

φ40 dx

.

(2.4)

∫ |∇φ |4 dx ∫Ω 4 0 , then Ω

φ0 dx

∫ −λ0 f (4) (0) Ω φ50 dx ∫ λsss (0) = . 4(f ′ (0))2 Ω φ20 dx ∫ 3f ′ (0) |∇φ |4 dx ′′ (4) ′′′ ∫Ω 4 0 , then (4) If k ⩾ 5, f (0) = f (0) = 0 and f (0) = − λ0

λssss (0) =

10f ′ (0)(λ0

∫ Ω

w2 dx −

∫ Ω



(2.5)

φ0 dx

∫ ∫ 2 6 |∇w| dx) + 45f ′ (0) Ω |∇φ0 | dx − λ0 f (5) (0) Ω φ60 dx ∫ , 5(f ′ (0))2 Ω φ20 dx

where w ∈ Z is the unique solution of the problem ⎧ ′′′ ⎨ −∆w − λ w = λ f (0) φ3 − 3div(|∇φ |2 ∇φ ), 0 0 ′ 0 0 f (0) 0 ⎩ w = 0,

in Ω ,

(2.6)

(2.7)

on ∂Ω .

1 When λ = f λ′ (0) , since φ1 > 0 in Ω , we obtain from (2.3) and (2.4) that if f ′′ (0) ̸= 0, then λs (0) ̸= 0, i.e., a transcritical bifurcation occurs; while if f ′′ (0) = 0 and f ′′′ (0) ⩾ 0, then λs (0) = 0 and λss (0) < 0, i.e., a subcritical pitchfork bifurcation occurs. Since the sign of λs (0) or λss (0) remains unchanged in these 1 cases, the bifurcation curve does not change the direction of turning at ( f λ′ (0) , 0) when the size of the domain varies. We next show that the condition that f ′′ (0) = 0 and f ′′′ (0) < 0 is not only a necessary condition, 1 but also a sufficient one to ensure that the direction of bifurcation at ( f λ′ (0) , 0) depends on the size of the domain. Define Ωα = {αx | x ∈ Ω }.

Theorem 2.2. Assume that f ∈ C 3 (−δ, δ), f (0) = 0 and f ′ (0) > 0 in (1.1). Then the direction of bifurcation 1 at ( f λ′ (0) , 0) depends on the size of the domain if and only if f ′′ (0) = 0 and f ′′′ (0) < 0. Furthermore, under this condition and with Ωα instead of Ω in (1.1), there exists an α∗ > 0 such that λ α (1) If α < α∗ , then λss (0) < 0, i.e., subcritical pitchfork bifurcation occurs at λ = f1,Ω ′ (0) . (2) If α > α∗ , then λss (0) > 0, i.e., supercritical pitchfork bifurcation occurs at λ = (3) If α = α∗ , then λss (0) = 0.

λ1,Ωα f ′ (0)

.

Remark∫ 1. For the corresponding semilinear problem (1.2), it is known (e.g., [29]) that λss (0) = −λ0 f ′′′ (0) φ4 dx ∫ Ω 20 , simpler than (2.4). So the sign of λss (0) is completely determined by that of f ′′′ (0), ′ 2 3(f (0))



φ0 dx

independent of Ω .

24

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

The next result is a direct application of Theorem 2.2. Corollary 2.3. Let Ω = BR . Assume that f ∈ C 3 (−δ, δ), f (0) = f ′′ (0) = 0, f ′ (0) > 0 and f ′′′ (0) < 0 in (1.1). Then there exists a R∗ > 0 such that 1 . (1) If R < R∗ , then subcritical pitchfork bifurcation occurs at λ = f λ′ (0) (2) If R > R∗ , then supercritical pitchfork bifurcation occurs at λ = (3) If R = R∗ , then λss (0) = 0.

λ1 f ′ (0) .

To obtain the directions of bifurcation curve for α = α∗ and R = R∗ , we need more information on higher-order derivatives of λ(s) at zero. The formulas (2.5) and (2.6) are very useful; see the examples presented below. When N = 1 and Ω = (−R, R) , since all eigenvalues {λn } are simple, Theorem 2.1 holds true for each n λn := [ (2n−1)π ]2 , i.e., the nth bifurcation from the trivial solution occurs at λ = fλ′ (0) , and the assertions 2R (1)–(4) hold. Furthermore, we can generalize Corollary 2.3 from λ1 to all λn and also provide a simple sign criterion of λssss (0) for the case R = Rn∗ . Theorem 2.4. Let N = 1 and Ω = (−R, R) in (1.1). Assume√that f ∈ C 3 (−δ, δ), f (0) = f ′′ (0) = 0, 3f ′ (0) (2n−1)π f ′ (0) > 0, and f ′′′ (0) < 0. Then for each n, there exists a Rn∗ := −f such that ′′′ (0) 2 (1) If R < Rn∗ , then subcritical pitchfork bifurcation occurs at λ =

λn f ′ (0) . n = fλ′ (0) .

(2) If R > Rn∗ , then supercritical pitchfork bifurcation occurs at λ (3) If R = Rn∗ , then λss (0) = 0. Further assume f ∈ C 5 (−δ, δ) and f (4) (0) = 0, then λssss (0) and n 3(f ′′′ (0))2 − f ′ (0)f (5) (0) have the same signs at λ = fλ′ (0) . We next analyze several interesting and important examples as applications of the above results. au Example 1 (Perturbed Exponential Nonlinearity). Let f (u) = exp( a+u ) − 1 (a > 0) and Ω = BR in au (1.1). The term exp( a+u ) usually arises in the perturbed Gelfand problem. Clearly, f (0) = 0, f ′ (0) = 1, 2

a −6a+6 ′′′ f ′′ (0) = a−2 . Theorem 2.1 implies that (λ1 , 0) is a bifurcation point. Furthermore, a and f (0) = a2 Theorem 2.2 implies that the direction of bifurcation at (λ1 , 0) depends on R if and only if a = 2. Precisely, we have the following results (see Fig. 1):

• If a < 2, then f ′′ (0) < 0. So λs (0) > 0, i.e., a transcritical bifurcation occurs. • If a > 2, then f ′′ (0) > 0. So λs (0) < 0, i.e., a transcritical bifurcation occurs. • If a = 2, then f ′′ (0) = 0, f ′′′ (0) < 0 and f (4) (0) > 0. So λs (0) = 0. By Corollary 2.3, there exists a R∗ > 0 such that at λ = λ1 , a subcritical pitchfork bifurcation occurs if R < R∗ , and a supercritical pitchfork bifurcation occurs if R > R∗ . For R = R∗ , since λss (0) = 0, we obtain from (2.5) that λsss (0) < 0, i.e., a transcritical bifurcation occurs. Notice that the directions of bifurcation for positive solutions obtained in [8, Theorems 2.1–2.3] by the time-map method, are consistent with those in the above results. Remark 2. In Example 1, for N = 1 and different λn , one can have all of transcritical bifurcation, subcritical and supercritical pitchforks. For example, when a = 2 and R = R2∗ , a transcritical bifurcation occurs at λ2 ; by Theorem 2.4, a supercritical pitchfork occurs at λ1 due to R > R1∗ , and a supercritical pitchfork occurs at λ3 due to R < R3∗ . Example 2 (Logistic Nonlinearity). Let f (u) = u − up (p ⩾ 2 is an integer) and Ω = BR in (1.1). Clearly, f (0) = 0 and f ′ (0) = 1. Theorem 2.1 implies that (λ1 , 0) is a bifurcation point. Furthermore, Theorem 2.2

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

25

au ) − 1 and Ω = BR . Fig. 1. Directions of Bifurcation at λ1 for (1.1) with f (u) = exp( a+u

implies that the direction of bifurcation at (λ1 , 0) depends on R if and only if p = 3. Precisely, we have the following results: • If p = 2, then f ′′ (0) < 0. So λs (0) > 0, i.e., a transcritical bifurcation occurs. • If p > 3, then f ′′ (0) = f ′′′ (0) = 0. So λs (0) = 0 and λss (0) < 0, i.e., a subcritical pitchfork bifurcation occurs. • If p = 3, then f ′′ (0) = 0 and f ′′′ (0) = −6 < 0. So λs (0) = 0. By Corollary 2.3, there exists a R∗ > 0 such that at λ = λ1 , a subcritical pitchfork bifurcation occurs if R < R∗ , and a supercritical pitchfork bifurcation occurs if R > R∗ . For R = R∗ , the situation is more complicated: since f (4) (0) = 0 and λss (0) = λsss (0) = 0, we have to analyze the sign of λssss (0). By Theorem 2.4, we can provide the answer for N = 1: since f (5) (0) = 0 and 3(f ′′′ (0))2 − f ′ (0)f (5) (0) = 3(−6)2 − 0 > 0, we get λssss (0) > 0, i.e., a supercritical pitchfork bifurcation occurs for R = R∗ . Example 3 (Pendulum Nonlinearity). Let f (u) = sin u and let Ω = BR in (1.1). Clearly, f (0) = f ′′ (0) = f (4) = 0, f ′ (0) = f (5) = 1, and f ′′′ (0) = −1. Theorem 2.1 implies that (λ1 , 0) is a bifurcation point. Furthermore, Theorem 2.2 implies that the direction of bifurcation at (λ1 , 0) depends on R. By Corollary 2.3, there exists a R∗ > 0 such that a subcritical pitchfork bifurcation occurs if R < R∗ , and a supercritical pitchfork bifurcation occurs if R > R∗ . For R = R∗ , the situation is more complicated and similar to Example 2: since f (4) (0) = 0 and λss (0) = λsss (0) = 0, we need to calculate the sign of λssss (0). By Theorem 2.4, we can provide the answer for N = 1: since 3(f ′′′ (0))2 − f ′ (0)f (5) (0) = 3 − 1 > 0, we get λssss (0) > 0, i.e., a supercritical pitchfork bifurcation occurs for R = R∗ . The proofs of the main results will be given in the next section. 3. Proofs ¯ ) | u|∂Ω = 0}, Y = C α (Ω ¯ ), and consider Proof of Theorem 2.1. Take H¨ older spaces X = {u ∈ C 2,α (Ω the function F : R × X → Y , ∇u ) − λf (u). F (λ, u) = −div( √ 2 1 + |∇u| Clearly, F (λ, 0) = 0. If f ∈ C k (−δ, δ), k ⩾ 1, it is easy to see that F : R × X → Y is also C k and Fu (λ, u)w = −div

[

∇w

] 2 1 2

(1 + |∇u| )

+ div

[ ∇u(∇u·∇w) ] 2 3

(1 + |∇u| ) 2

− λf ′ (u)w.

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

26

Taking u = 0, we have that Fu (λ, 0) = −∆ − λf ′ (0)I. Since λ0 is simple and φ0 ̸≡ 0 on Ω , we obtain that λ0 ker(Fu ( ′ , 0)) = ker(−∆ − λ0 I) = span{φ0 }, f (0) ∫ λ0 Im(Fu ( ′ , 0)) = {v ∈ Y | vφ0 dx = 0}, f (0) Ω λ0 coker(Fu ( ′ , 0)) = span{φ0 }. f (0) 0 0 , 0)) = dim coker(Fu ( f λ′ (0) , 0)) = 1. Thus the dimensions dim ker(Fu ( f λ′ (0)

Moreover, we have Fλu (λ, 0) = −f ′ (0)I. Then Fλu (

λ0 , 0)φ0 = −f ′ (0)φ0 ̸∈ Im(Fu (λ0 , 0)). f ′ (0)

0 0 Applying the Crandall–Rabinowitz theorem ([10, Theorem 1.7]) at ( f λ′ (0) , 0), we obtain that ( f λ′ (0) , 0) is a k 1 bifurcation point and there exist an interval (−ε, ε) and C functions λ : (−ε, ε) → R and ψ : (−ε, ε) → Z such that λ0 F (λ(s), sφ0 + sψ(s)) = 0, λ(0) = ′ , ψ(0) = 0, (3.1) f (0)

where Z is the complement span of {φ0 } in X. That is, the local solution is u(s) = sφ0 + sψ(s) satisfying us (0) = φ0 . We next prove the assertions (1)–(4). (1) Since f ∈ C 2 , we have F ∈ C 2 , λ ∈ C 2 and ψ ∈ C 2 . Differentiating (1.1) once and twice in s, we obtain that [ ] [ ∇u(∇u·∇u ) ] ∇us s − div + div = λs f (u) + λf ′ (u)us , (3.2) 2 1 2 3 2 2 (1 + |∇u| ) (1 + |∇u| ) and [ ] [ ∇u (∇u·∇u ) ] [ ∇u(∇u·∇u ) ] [ ∇u(∇u·∇u )2 ] ∇uss s s s s s −div + 2div + div − 3div 1 3 3 2 2 2 2 5 2 2 2 (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) 2 ′ ′′ 2 ′ = λss f (u) + 2λs f (u)us + λf (u)us + λf (u)uss . (3.3) Taking s = 0 in (3.3), since u(0) = 0, us (0) = φ0 , and f (0) = 0, we have that − ∆uss = 2λs (0)f ′ (0)φ0 + λ0

f ′′ (0) 2 φ + λ0 uss . f ′ (0) 0

(3.4)

Multiplying (3.4) by φ0 , integrating over Ω , and integrating by parts with respect to x, we obtain that ∫ ∫ ∫ ∫ f ′′ (0) ′ 2 3 uss (−∆φ0 ) dx = 2λs (0)f (0) φ0 dx + λ0 ′ φ dx + uss (λ0 φ0 ) dx, f (0) Ω 0 Ω Ω Ω which implies (2.3). (2) Since f ∈ C 3 , we have F ∈ C 3 , λ ∈ C 3 and ψ ∈ C 3 . Differentiating (3.3) in s, we obtain that [ ] [ ∇u (∇u·∇u ) ] [ ∇u (∇u·∇u ) ] [ ∇u (∇u·∇u )2 ] ∇usss ss s s s s s s − div + 3div + 3div − 9div 2 1 2 3 2 3 2 5 2 2 2 (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) 2 [ ∇u(∇u·∇u ) ] [ ∇u(∇u·∇u )(∇u·∇u ) ] [ ∇u(∇u·∇u )3 ] s ss s s s s + div − 9div + 15div 2 3 2 5 2 7 (1 + |∇u| ) 2 (1 + |∇u| ) 2 (1 + |∇u| ) 2 = λsss f (u) + 3λss f ′ (u)us + 3λs f ′′ (u)u2s + 3λs f ′ (u)uss + λf ′′′ (u)u3s + 3λf ′′ (u)us uss + λf ′ (u)usss .

(3.5)

H. Pan, R. Xing / Nonlinear Analysis 171 (2018) 21–31

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2

Notice that (∇u·∇us )s = |∇us | + ∇u·∇uss , f ′′ (0) = 0, and λs (0) = 0. Taking s = 0, we have that ( ) f ′′′ (0) 3 2 − ∆usss + 3div |∇φ0 | ∇φ0 = 3λss (0)f ′ (0)φ0 + λ0 ′ φ + λ0 usss . f (0) 0

(3.6)

Also, we obtain from (3.4) that −∆uss (0) = λ0 uss (0), which implies uss (0) = β0 φ0 .

(3.7)

Multiplying (3.6) by φ0 , integrating over Ω , and integrating by parts with respect to x, we obtain that ∫ ∫ 4 usss (−∆φ0 ) dx − 3 |∇φ0 | dx Ω Ω ∫ ∫ ∫ f ′′′ (0) ′ 2 4 φ dx + usss (λ0 φ0 ) dx, = 3λss (0)f (0) φ0 dx + λ0 ′ f (0) Ω 0 Ω Ω which implies (2.4). (3) Since f ∈ C 4 , we have F ∈ C 4 , λ ∈ C 4 and ψ ∈ C 4 . Differentiating (3.5) in s, we obtain that ] [ ∇u (∇u·∇u ) ] [ ∇u (∇u·∇u ) ] [ ∇u sss s ss s s ssss + 4div + 6div − div 2 1 2 3 2 3 (1 + |∇u| ) 2 (1 + |∇u| ) 2 (1 + |∇u| ) 2 [ ∇u (∇u·∇u )2 ] [ ∇u (∇u·∇u ) ] [ ∇u (∇u·∇u )(∇u·∇u ) ] ss s s s ss s s s s − 18div + 4div − 36div 2 5 2 3 2 5 2 2 2 (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) [ ∇u (∇u·∇u )3 ] [ ∇u(∇u·∇u ) ] [ ∇u(∇u·∇u )(∇u·∇u ) ] s s s sss s s ss + 60div + div − 12div 2 7 2 3 2 5 (1 + |∇u| ) 2 (1 + |∇u| ) 2 (1 + |∇u| ) 2 [ ∇u(∇u·∇u )2 ] [ ∇u(∇u·∇u )2 (∇u·∇u ) ] [ ∇u(∇u·∇u )4 ] s s s s s s − 9div + 90div − 105div 2 5 2 7 2 9 2 2 (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) 2 ′ ′′ 2 ′ ′′′ 3 = λssss f (u) + 4λsss f (u)us + 6λss f (u)us + 6λss f (u)uss + 4λs f (u)us + 12λs f ′′ (u)us uss + 4λs f ′ (u)usss + λf (4) (u)u4s + 6λf ′′′ (u)u2s uss + 3λf ′′ (u)u2ss + 4λf ′′ (u)us usss + λf ′ (u)ussss .

(3.8)

Notice that (∇u·∇us )ss = 3∇us ·∇uss + ∇u·∇usss , λss (0) = 0, and uss (0) = β0 φ0 . Taking s = 0, we have that ( ) 2 − ∆ussss + 18β0 div |∇φ0 | ∇φ0 = 4f ′ (0)λsss (0)φ0 + λ0

f (4) (0) 4 f ′′′ (0) 3 φ + 6β λ φ + λ0 ussss . 0 0 f ′ (0) 0 f ′ (0) 0

(3.9)

Multiplying (3.9) by φ0 , integrating over Ω , and integrating by parts with respect to x, we obtain that ∫ ∫ ∫ 4 ussss (−∆φ0 ) dx − 18β0 |∇φ0 | dx = 4f ′ (0)λsss (0) φ20 dx Ω Ω Ω ∫ ∫ ∫ f ′′′ (0) f (4) (0) 5 4 φ dx + 6β0 λ0 ′ φ dx + ussss (λ0 φ0 ) dx, + λ0 ′ f (0) Ω 0 f (0) Ω 0 Ω ∫ 3f ′ (0) |∇φ |4 dx ′′′ ∫Ω 4 0 which, together with f (0) = − , implies (2.5). λ0



φ0 dx

(4) Since f ∈ C 5 , we have F ∈ C 5 , λ ∈ C 5 and ψ ∈ C 5 . Differentiating (3.8) in s, we obtain that − div

[

∇usssss 2 1 2

]

+ 10div

[ ∇u

2 sss |∇us | 3 2

]

+ 30div

[ ∇u (∇u · ∇u ) ] ss s ss 2 3

(1 + |∇u| ) (1 + |∇u| ) 2 (1 + |∇u| ) 2 [ ∇u |∇u |2 ] [ ∇u (∇u · ∇u ) ] [ ∇u |∇u |4 ] s ss s s sss s s + 15div + 20div − 45div + (· · · ) 2 3 2 3 2 5 2 2 (1 + |∇u| ) (1 + |∇u| ) (1 + |∇u| ) 2 = 5λssss f ′ (u)us + λf (5) (u)u5s + 15λf ′′′ (u)us u2ss + 10λf ′′′ (u)u2s usss + λf ′ (u)usssss + ⟨· · · ⟩,

(3.10)

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where the terms in (· · · ) and ⟨· · · ⟩ have no influence on the subsequent analysis and will disappear after taking s = 0 and using f (0) = f ′′ (0) = f (4) (0) = 0 and λs (0) = λss (0) = λsss (0) = 0. 2 Notice that (∇u·∇us )sss = 3|∇uss | + 4∇us ·∇usss + ∇u·∇ussss . Taking s = 0 in (3.10), we get that ( ) ( ) [ ] 2 2 − ∆usssss + 10div ∇usss |∇φ0 | + 45β02 div ∇φ0 |∇φ0 | + 20div ∇φ0 (∇φ0 · ∇usss ) ) ( 4 − 45div ∇φ0 |∇φ0 | = 5λssss (0)f ′ (0)φ0 + λ0

f (5) (0) 5 f ′′′ (0) 3 f ′′′ (0) 2 φ0 + 15β02 λ0 ′ φ0 + 10λ0 ′ φ usss + λ0 usssss . ′ f (0) f (0) f (0) 0

(3.11)

Multiplying (3.11) by φ0 , integrating over Ω , integrating by parts with respect to x, and using f ′′′ (0) = ∫ 3f ′ (0) |∇φ0 |4 dx ∫Ω 4 − , we obtain that λ0



φ0 dx

∫ − 30



2

6

|∇φ0 | (∇φ0 · ∇usss ) dx + 45 |∇φ0 | dx Ω Ω ∫ ∫ ∫ f ′′′ (0) f (5) (0) φ60 dx + 10λ0 ′ φ30 usss dx. = 5λssss (0)f ′ (0) φ20 dx + λ0 ′ f (0) f (0) Ω Ω Ω

Notice that by (3.6), usss satisfies ⎧ ′′′ ⎨ −∆u − λ u = λ f (0) φ3 − 3div(|∇φ |2 ∇φ ), in Ω , sss 0 sss 0 ′ 0 0 f (0) 0 ⎩ usss = 0, on ∂Ω .

(3.12)

(3.13)

Moreover, usss ∈ Z. Since λ0 is an eigenvalue of −∆, it is not difficult to prove that (3.13) admits a unique solution in Z. Multiplying (3.13) by 10usss , integrating over Ω , integrating by parts with respect to x, we have ∫ ∫ 2 10 |∇usss | dx−10λ0 u2sss dx Ω Ω ∫ ∫ f ′′′ (0) 2 3 = 10λ0 ′ φ usss dx + 30 |∇φ0 | (∇φ0 · ∇usss ) dx, f (0) Ω 0 Ω which, together with (3.12), implies (2.6).



Proof of Theorem 2.2. The necessity of the condition has been given in front of the theorem. It remains to prove the sufficiency of the condition. Let α > 0. Define ∫ ∫ 4 −3f ′ (0) Ωα |∇φ1,Ωα (x)| dx − λ1,Ωα f ′′′ (0) Ωα φ41,Ωα (x) dx ∫ g(α) = . (3.14) 3(f ′ (0))2 Ωα φ21,Ωα (x) dx Setting y =

x α

and writing φ˜1,Ωα (y) = φ1,Ωα (αy), we have that − α−2 ∆y φ˜1,Ωα (y) = −∆φ1,Ωα (x) = λ1,Ωα φ1,Ωα (x) = λ1,Ωα φ˜1,Ωα (y),

and g(α) =

−3f ′ (0)

∫ Ω

4

|∇y φ˜1,Ωα (y)| α−4+n dy − λ1,Ωα f ′′′ (0) ∫ 3(f ′ (0))2 Ω φ˜21,Ωα (y)αn dy

∫ Ω

φ˜41,Ωα (y)αn dy

So α2 λ1,Ωα = λ1,Ω ,

φ˜1,Ωα (y) = βφ1,Ω (y) for some β > 0,

.

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and hence ∫ 4 |∇y φ˜1,Ωα (y)| dy + α4 λ1,Ωα f ′′′ (0) Ω φ˜41,Ωα (y) dy ∫ g(α) = − 3(f ′ (0))2 α4 Ω φ˜21,Ωα (y) dy ∫ ∫ 4 3f ′ (0) Ω |∇φ1,Ω (y)| dy + α2 λ1,Ω f ′′′ (0) Ω φ41,Ω (y) dy ∫ =− . 3(f ′ (0))2 α4 β −2 Ω φ21,Ω (y) dy 3f ′ (0)





Taking   ∫ 4  3f ′ (0) |∇φ1,Ω (y)| dy ∗ Ω √ ∫ α = , −λ1,Ω f ′′′ (0) Ω φ41,Ω (y) dy

(3.15)

we derive that g(α∗ ) = 0, g(α) > 0 for α > α∗ , and g(α) < 0 for α < α∗ . Since λss (0) and g(α) have the same signs, it completes the proof. □ Proof of Theorem 2.4. It is well known that the eigenvalues and eigenfunctions are λn,(−R,R) = [

(2n − 1)π 2 ] , 2R

φn,(−R,R) = cos(

(2n − 1)π x), 2R

n = 1, 2, . . . .

(3.16)

Define gn (R) =

−3f ′ (0)

∫R −R

4

|∇φn,(−R,R) (x)| dx − λn,(−R,R) f ′′′ (0) ∫R 3(f ′ (0))2 −R φ2n,(−R,R) (x) dx

∫R −R

φ4n,(−R,R) (x) dx

.

(3.17)

Substituting (3.16) into (3.17), similar to the proof of Theorem 2.2, we have that   ∫ (2n−1)π   ′ (2n−1)π 2 ∫  1 4 sin4 (z) dz  3f (0)( 2 )3 (2n−1)π ′ (0)  3f |∇φ (y)| dy − n,(−1,1)  −1 ∗ 2 Rn = √ = ∫1 (2n−1)π −λn,(−1,1) f ′′′ (0) −1 φ4n,(−1,1) (y) dy √ − (2n−1)π f ′′′ (0) ∫ 2 4 (2n−1)π cos (z) dz 2 −

2

(2n − 1)µπ , = 2 where µ =

(3.18)



Let R = become

3f ′ (0) −f ′′′ (0) > 0. Rn∗ = (2n−1)µπ . 2

λn = [

Then for Ω = (− (2n−1)µπ , (2n−1)µπ ), the nth eigenvalue and eigenfunction 2 2

(2n − 1)π 2 1 ] = 2, 2R µ

φn = cos(

x (2n − 1)π x) = cos( ), 2R µ

x ∈ Ω.

So φ′n = − µ1 sin( µx ) and φ′′n = − µ12 cos( µx ). Since λn

f ′′′ (0) 3 3 x 9 x x 3 3x 2 φ − 3(|φ′n | φ′n )′ = − 4 cos3 ( ) + 4 sin2 ( ) cos( ) = − 4 cos( ), f ′ (0) n µ µ µ µ µ µ µ

(2.7) becomes ⎧ 3 3x 1 ⎪ ⎪ ⎨ − w′′ − 2 w = − 4 cos( ), µ µ µ ⎪ (2n − 1)µπ (2n − 1)µπ ⎪ w(− ⎩ ) = 0 = w( ). 2 2

x ∈ (−

(2n − 1)µπ (2n − 1)µπ , ), 2 2

(3.19)

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The problem admits a unique solution w(x) = − ∫ satisfying Ω wφn dx = 0. Moreover, w′ (x) = Recall that (2.6) becomes

9 8µ3

3x 3 cos( ), 8µ2 µ

sin( 3x µ ).

λssss (0) = 10f ′ (0)(λn

∫R −R

w2 dx −

∫R −R

∫R ∫R 2 6 |∇w| dx) + 45f ′ (0) −R |∇φn | dx − λn f (5) (0) −R φ6n dx . ∫R 5(f ′ (0))2 −R φ2n dx

(3.20)

By a direct computation, we have that ∫ (2n−1)µπ 2 45f ′ (0) x 225f ′ (0) sin6 ( ) dx = (2n − 1)π, 6 (2n−1)µπ µ µ 16µ5 −R − 2 ∫ R ∫ (2n−1)µπ 2 5f (5) (0) f (5) (0) (5) 6 x 6 λn f (0) cos ( ) dx = (2n − 1)π, φn dx = (2n−1)µπ µ2 µ 16µ − −R 2 45f ′ (0)



R

6

|∇φn | dx =

and 10f ′ (0)(λn



R

−R

w2 dx −



R

2

|w′ | dx)

−R

∫ (2n−1)µπ ∫ (2n−1)µπ ) 2 2 3 2 1( 3x ′ 2 3x = 10f (0)( 2 ) 2 cos ( ) dx − 9 sin2 ( ) dx (2n−1)µπ (2n−1)µπ 8µ µ µ µ − − 2 2 (2n−1)π (2n−1)π ∫ ∫ ) 2 2 90f ′ (0) ( = 2 5 cos2 (3y) dy − 9 sin2 (3y) dy (2n−1)π (2n−1)π 8 µ − − 2 2 =

90f ′ (0) π 9π 90f ′ (0) (2n − 1)( − )=− (2n − 1)π. 2 5 8 µ 2 2 16µ5

Since [−

5(2n − 1)π 90f ′ (0) 225f ′ (0) 5f (5) (0) + − ](2n − 1)π = [3(f ′′′ (0))2 − f ′ (0)f (5) (0)], 5 5 16µ 16µ 16µ 16µf ′ (0)

we obtain from (3.20) that λssss (0) and 3(f ′′′ (0))2 − f ′ (0)f (5) (0) have the same signs.



Acknowledgments The first author is supported by Pearl River S&T Nova Program of Guangzhou (2014J2200010) and NSF of Guangdong (2014A030313431). The second author is supported by NSF of China (11471339). References

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