Radial symmetry for a generalized nonlinear fractional p-Laplacian problem

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Fractional-order differential equations are very suitable for describing materials and processes with memory and heritability, and their description of complex systems has the advantages of simple modeling, clear physical meaning of parameters and accurate description. Examples include a fractional differential model for the free dynamic response of viscoelastic single degree of freedom systems [18], a new noninteger model for convective straight fins with temperature-dependent thermal conductivity associated with Caputo–Fabrizio fractional derivative [20] and so on.

In this paper, we are concerned with a generalized nonlinear fractional p-Laplacian equation with negative power

(1)

where

Here 0 < s < 1, 2 < p < ∞, PV means the Cauchy principal value and F is a continuous function. For the purpose of making the integral meaningful, we need that

The operator (−∆)s F,p introduced in this paper includes some special cases. WHen F(·.) is an identity map, (- ∆)SF,P becomes the fractional p-Laplacian (−∆)sp. Based ON this, (-∆)sp will become fractional Laplacian( - ∆)s if p = 2, this is well known. In order to surmount the nonlocality of fractional Laplacian, Caffarelli and Silvestre [4] introduced the extension method that reduced this nonlocal problem into a local one in higher dimensions. This method is briefly described below. Given a function g : Rn → R, let the extension G : Rn × [0, ∞) → Rn that meets the following condition:

They concluded that

The extension method mentioned above has been utilized to discuss equations involving fractional Laplacian; see [3,5,15]. Another way to overcome the nonlocality is the integral equations method. Applications of this method can be found in [7, 12, 13, 23, 25, 26, 36].

However, there are still some operators that cannot be solved by the above methods; see [5]. To overcome the difficulty, a direct method of moving planes is introduced in [11]. Gradually, it is used to tackle a series of problems involving kinds of nonlinear operators. For example, the relevant properties of solutions for nonlinear elliptic equations are obtained, besides, it has been highly applied in studying the properties of fractional Laplacian equations and systems; see [8, 14, 24, 28, 31, 32]. Furthermore, there are some excellent results by using this method to study the radial symmetry and monotonicity of the solutions of fractional p-Laplacian equations and systems; see [9, 16, 22, 27, 33,34,35]. For fully nonlinear nonlocal operators, for example,

this direct method has been further developed by Chen and Li in [10].

Recently, the problems involving negative powers are studied in many fields, for example, in MEMS device, singular minimal surface equations, and described curvature equations in conformal geometry; see [1, 2, 6, 17, 19, 21, 29, 30].

In [17], Davila, Wang and Wei proved sharp Hölder continuity and an estimate of rupture sets for sequences of solutions of the following nonlinear problem with negative exponent:

The above problem arises in modeling an electrostatic microelectromechanical system (MEMS) device. The solution of the singularity in the equation, namely u ≈ 0 in some region, represents the rupture in the device in the physical model.

In [19], Jiang and Ni studied the singular elliptic equation

where Ω Rn, n ≥ 2, is a bounded smooth domain and α > 1. When n = 2 and α = 3, the above equation is used to model steady states of van der Waals force driven thin films of viscous fluids. They also considered the physical problem when total volume

of the fluid is prescribed. Singular elliptic equations modeling steady states of van der Waals force driven thin films have been mathematically rigorously studied with no flux Neumann boundary condition. They gave a complete description of all continuous radially symmetric solutions. In particular, they constructed both nontrivial smooth solutions and singular solutions.

In [6], Ma and Cai studied the following nonlinear fractional Laplacian equation with negative powers by using the direct method of moving planes:

(−∆)α/2v(x) + v−β(x) = 0, x ∈ Rn.

The above results of all encourage us to further study a generalized nonlinear fractional p-Laplacian equation with negative powers by the direct method of moving planes. As far as authors know, up to now, this is a new attempt to study a class of equations (1) combining a generalized fractional p-Laplacian with negative powers. Interestingly, this method can also be analogously applied to a generalized Hénon-type nonlinear fractional p-Laplacian equation with negative power

(2)

where σ < 0 and γ > 0 are constants.

The paper is structured as follows. In Section 2, we mainly present some lemmas used in the following part. In Section 3, we study radial symmetry and monotonicity of two generalized fractional p-Laplacian equations with negative powers by applying the direct method of moving planes.

Let Pκ = {x ϵ Rn │ x1 = κ for some κ ϵ R} be the moving planes, Σκ = {x ϵ Rn│ x1 < κ} be the area to the left of Pκ, and xκ = (2κ - x1, x2, . . . , xn) be the reflection of x about Pκ. Meanwhile, we denote

Throughout the next section, we assume that there exists a constant L > 0 such that F(x) − F(y) ≤ L(x − y) and F(0) = 0.

Lemma 1 [A simple maximum principle]. Let ℵ be bounded area in Rn. Presume that u ∈ lsp ∩ C1,1loc(ℵ) is lower semicontinuous on ℵ and satisfies

(3)

Then

(4)

If φ(x) = 0 at some point x ϵ ℵ , then φ(x) = 0 holds for almost all points x in Rn. When ℵ is unbounded area, we need further assume that lim │x│ → φ(x) ≥ 0, then the same conclusion still holds.

Proof. Suppose (4) is not true, then there is an x0 such that φ(x0) = minℵ φ < 0. According to the second inequality in (3),

This is a direct contradiction to the first inequality in (3), hence (4) holds. When φ(x0) = 0 at some point x0 ∈ ℵ, then

On the other hand, from (3) we have

so, the integral result is 0. Because u is nonnegative, one could get that φ(x) = 0 almost everywhere in Rn. This completes the lemma.

Lemma 2 [Maximum principle for antisymmetric functions]. Let ℵ be bounded area in Σ and φ ∈ lsp ∩ C1,1 loc (ℵ). If

(5)

then

If dκ(x) = 0 at some point in ℵ , then dκ(x) = 0 almost everywhere in Rn. When ℵ is unbounded area, we need further presume that

then the same conclusions still hold.

Proof. Suppose that dκ(x) ≥ 0 in ℵ is not true, then there is a point xˆ in ℵ such that

To simplify writing, let Q(m) = |m|p−2m, then Q′(m) = (p − 1)|m|p−2 ≥ 0.

(6)

Where

To estimate J1, we notice the fact

Due to

on the basis of strict monotonicity of Q, we have

Q φ(xˆ) − φ(y) − Q φκ(xˆ) − φκ(y) ≤ 0, but ≡ 0.

Therefore,

(7)

To evaluate J2, by using the mean value theorem we obtain

(8)

Combining (6), (7) and (8), one can deduce

This inequality contradicts the first condition in (5), thus dκ(xˆ) ≥ 0.

If dκ(x) = 0 at some point x ϵ ℵ , equivalently, x is a minimum of dκ in , so, J2 = 0. Now, according to the first inequality in (5), we get J1 ≥ 0, which means

Q φ(x) − φ(y) − Q φκ(x) – φκ(y) ≥ 0.

Considering the monotonicity of Q, for almost all y ∈ Σ,

φ(x) − φ(y) − φκ(x) − φκ(y) = dκ(x) − dκ(y) = −dκ(y) ≥ 0.

Consequently, dκ(y) = 0 almost everywhere in Σ. Besides, in the light of the antisymmetry of dκ, we receive dκ(y) = 0 almost everywhere in Rn. If is unbounded area, under this circumstance, in view of assumption lim │x│→∞ dκ(x) ≥ 0, suppose that dκ(x) ≥ 0, x ϵ Σ, is false, then a negative minimum of dκ is obtained at some point x Σ. Being similar to the above argument, one can find a contradiction. The proof is completed.

Lemma 3 [Narrow region principle]. Let ℵ be bounded narrow area in Σ such that it is contained in {x|κ − δ < x1 < κ} with small δ. Presume that c(x) is bounded from below in ℵ and

and there exists y0 Σ satisfying dκ(y0) > 0, then when δ is sufficiently small, one can get

dκ(x) ≥ 0 in ℵ.

Further, if dκ(x) = 0 holds for some point in ℵ, then dκ(x) = 0 holds for almost all points x in Rn. Besides, if ℵ is unbounded region, we need presume that

then above conclusions still hold.

Proof. Suppose the contrary, then for any δ > 0, there exists an xδ ∈ ℵδ such that

Then for δk = 1/k, k = 1, 2, . . . , there exists xδk and ℵδk , let us call them xk and ℵk such that

By inequality (6) we deduce

(9)

where

Similarly to (8), we can get H2 ≤ 0, from [33] we can get H1 ≤ −(C0/2)δxk ,

(10)

Combining (9) with (10), one can get

This contradicts with the equation, hence the proof is completed.

Lemma 4 [Decay at infinity]. Let ℵ be unbound area in Σ, and let φ ∈ lsp ∩ C1,1 loc(ℵ) be a solution of

with

Then there exists a positive constant R0 (depending on c(x) and independent of φ(x) and φκ(x)) such that if dκ(x0) = minℵ dκ(x) < 0, then |x0| ≤ R0.

Proof. By inequality (6) we get

where M is a constant. For each fixed κ, when |x0| ≥ κ, x1 = (3|x0| + x0, (x0)′), then B│x0│ (x1) C ∑. Consequently,

Then we have

this contradicts with the condition of c(x). This completes the proof.

Theorem 1. Assume that φ ∈ lsp ∩ C1,1 loc (Rn) is the positive solution of equation (1) with

φ(x) = Q|x|t + o(1) as |x| → ∞,

where sp/(γ+1) < t < 1 and Q > 0 are constants. Then φ(x) must be radially symmetric and monotone increasing about some point in Rn.

Proof. Step 1. In the first step, we indicate that when κ is sufficiently negative,

(11)

According to equation (1), for x ∈ Σκ−, we can derive

where ξκ(x) values between φκ(x) and φ(x), that means

here c(x) = −γφ−γ−1(x). Consider φ(x) = Q|x|t + o(1) near infinity, so

near infinity. Therefore, c(x) satisfies the condition of Lemma 4. Since φ(x) and φκ(x) satisfy

with c(x) ~ 1/ │x│ t(γ+1) for │x│ large, using Lemma 4 to dκ(x), if dκ(x) acquires the negative minimum at point x˜ in Σκ, then │x˜│ ≤ R0, that is, when κ is sufficiently negative, (11) holds.

Step 2. In this step, we will move Pκ to the limiting position, this moving process could give the symmetry about the positive solution φ(x). (11) means that there is an initial point to move Pκ, we could move Pκ so long as (11) holds. Let

We show

(12)

If (12) does not hold, then by Lemma 2 we get

dκ0 (x) > 0 ∀x ∈ Σκ0 ,

therefore, there exists a small δ > 0 and a constant bδ, which satisfy

According to the continuity of dκ about κ, there exists 0 < α < δ such that

(13)

In Lemma 3, ℵ = Σκ− \ Σκ0 −δ is the narrow region, then

dκ(x) ≥ 0 ∀x ∈ Σκ \ Σκ0 −δ.

This, combining with (13), indicates

This violates the definition of κ0. Thus, (12) holds. The proof is completed.

Next, we discuss the radial symmetry of a generalized nonlinear Hénon-type fractional p-Laplacian equation with negative power.

Theorem 2. Presume φ ∈ lsp ∩ C1,1(Rn) is a positive solution of (2) with

φ(x) = Q|x|t + o(1) for |x| large,

where γ, Q > 0 and σ < 0 are constants, 0 < t < 1 and t > (sp + σ)/(γ + 1). Then φ(x) is radially symmetric about origin.

Proof. Before we prove Theorem 2, we first consider the singularity of equation (2) at the origin. For κ < 0 and x ∈ Σκ− \ {0κ}, φ(x) and φκ(x) satisfy

where ξκ(x) is between φ(x) and φκ(x). This implies,

with c(x) = −γ|x|σφ−γ−1(x). Seeing that φ(x) = Q|x|t + o(1) near infinity, so

By our assumption t > (sp + σ)/(γ + 1) we get c(x) ~ o(1/ │x│ sp). Next, we still prove the conclusion of Theorem 2 in two steps.

Step 1. We demonstrate that

(14)

holds when κ is sufficiently negative.

Near the singular point 0κ of φ(x) and φκ(x), we manifest that Σκ− has no intersection with Bє(0κ) for certain small ε > 0. In fact, we consider that when κ is sufficiently negative, obviously, 0κ is a sufficiently negative point, we have

Since c(x) ~ o(1/ │x│ sp), by applying Lemma 4 we could work out that there is an R0 > 0, when dκ(x) gets the negative minimum at x∗ in Σκ, then the following relation is true:

(15)

That is to say, (14) holds when κ sufficiently negative.

Step 2. So long as (14) holds, we could move Pκ from left to right up to its critical position. Let

We prove that

Suppose κ0 < 0, we can use Lemmas 3 and 4 to state that Pκ can be moved further right, this contradicts with κ0. By condition of u(x),

That is, there exist ε, h0 > 0 such that

From (15) the situation where the negative minimum of dκ (x) is obtained in Bc(0) does not exist. We also show that it cannot be gained in the internal of BR0 (0). That is,

when κ is close enough to κ0,

When κ0 < 0, by Lemma 2 one can get

dκ0 (x) > 0 ∀x ∈ Σκ0

There is a positive constant j0, which satisfies

Since dκ(x) is continuous with respect to κ, there is 0 < ϵ < δ,

holds for κ ∈ (κ0, κ0 + ϵ).

Considering that c(x) is bounded from below, for narrow region Σκ−\Σκ0 −δ , using

Lemma 3 to dκ(x), we receive

From all above we get

This goes against the definition of κ0. Hence, κ0 = 0, dκ0 (x) = 0, x ∈ Σκ0 \{0κ0 } hold.

Therefore, the proof is completed.