The same class of stationary solutions to some multidimensional kinetic systems with extensive background density

The same class of stationary solutions to some multidimensional kinetic systems with extensive background density

J. Math. Anal. Appl. 466 (2018) 1–17 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com/lo...

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J. Math. Anal. Appl. 466 (2018) 1–17

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

The same class of stationary solutions to some multidimensional kinetic systems with extensive background density Limei Zhu School of Mathematics, South China University of Technology, Guangzhou 510640, PR China

a r t i c l e

i n f o

Article history: Received 18 July 2017 Available online 12 April 2018 Submitted by H.K. Jenssen Keywords: Extensive background density Multidimensional kinetic system Stationary solution

a b s t r a c t In this study, we investigate the existence and uniqueness of the same class of stationary solutions for some kinetic systems comprising the Vlasov–Poisson– Fokker–Planck system, Vlasov–Poisson–Boltzmann system, and Vlasov–Maxwell– Boltzmann system in arbitrary space dimensions N (N ≥ 3). Essentially, the problem can be reduced to solving the problem of a second order elliptic equation with exponential nonlinearity. This result had been proved in three spatial dimensions but the extension to a higher-dimensional setting makes the existence proof nontrivial. In particular, it necessary to highlight that the requirement regarding the background density function is relaxed compared with previous studies in the case of three spatial dimensions. © 2018 Published by Elsevier Inc.

Contents 1. Introduction . . . . . . . . . . . . . . . . . . . . . . 2. Proof of the main theorem . . . . . . . . . . . . . 3. Precise description of the decay based on the 4. Appendix . . . . . . . . . . . . . . . . . . . . . . . . Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1. Introduction In this study, we investigate the existence and uniqueness of the same class of stationary solutions for some kinetic systems comprising the Vlasov–Poisson–Fokker–Planck (VPFP) system, Vlasov–Poisson–Boltzmann (VPB) system, and Vlasov–Maxwell–Boltzmann (VMB) system in arbitrary space dimensions N (N ≥ 3). Many studies have investigated this problem for the three-dimensional case, but the requirement regarding the background density function is relaxed compared with these previous investigations. The extension is E-mail address: [email protected]. https://doi.org/10.1016/j.jmaa.2018.03.051 0022-247X/© 2018 Published by Elsevier Inc.

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L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

nontrivial and it can be achieved successfully by seeking appropriate functions h(σ) and a(σ), as stated in Assumptions (A1 )–(A3 ) in the following. Essentially, the problem can be reduced to solving the following problem for a second order elliptic equation with exponential nonlinearity: Δu + e−u = ρ(x), u(x) → 0,

x ∈ RN ,

as |x| → ∞.

(1.1) (1.2)

In fact, Equation (1.1) can be recovered from the VPFP system, VPB system, and VMB system, and the solution u to the (1.1) and (1.2) can be used to construct the stationary solution to the three systems mentioned above. The rigorous and detailed derivation in the case of three dimensions was given by [11,25, 27,28] and it can be summarized in the following manner. Fact 1. Equation (1.1) is recovered from the three-dimensional VPFP system (see [28]). The VPFP system with all physical coefficients taken as 1 is written as: ∂t F + v · ∇x F + ∇x Φ · ∇v F = ∇v · (vF ) + Δv F,  Δx Φ = F dv − ρ(x).

(1.3) (1.4)

R3

From [6,16,17,28], the stationary solutions to the VPFP system (1.3) and (1.4) are given by: F = α(x) · μ(v),

(1.5)

where α(x) is determined in the following and   1 1 2 μ(v) = − |v| . 3 exp 2 (2π) 2

(1.6)

By substituting (1.5) into (1.3), we have (v · ∇x α)μ − α∇x Φ · vμ = 0, which will be satisfied if and only if: ∇x α = α∇x Φ.

(1.7)

F is a density so the condition that α(x) ≥ 0 must be satisfied. For the set where α(x) > 0, we can write α = e−u and recast (1.7) in the form of: ∇x u = −∇x Φ,

(1.8)

which together with (1.4) means that:  Δx u = −Δx Φ = −α(x)

μ(v)dv + ρ(x) = −α(x) + ρ(x).

R3

By entering α = e−u into (1.9), we obtain Equation (1.1) exactly.

(1.9)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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The VPFP system is a kinetic model used in plasma physics for describing the motion of charged particles (e.g., electrons and ions) under the influence of the self-consistent electrostatic force and when collisions are governed by the Fokker–Planck operator. The existence, uniqueness, and regularity of this system have been studied widely, e.g., both the hypoelliptic [4,5] and hypocoercive [6,13,14,18,23] properties, as well as the existence and large-time behavior of solutions in different settings [2,3,8–10,15,20,41,42,45]. Fact 2. Equation (1.1) is recovered from the three-dimensional VPB system (see [25,27]). The VPB system can be written as: ∂t F + v · ∇x F + ∇x Φ · ∇v F = Q(F, F ),  Δx Φ = F dv − ρ(x).

(1.10) (1.11)

R3

Similar to [6,16,17,27], stationary solutions to system (1.10) and (1.11) are introduced in the form of: F = eΨ(x) M (v),

(1.12)

where Ψ(x) needs to be determined below and M (v) is a global Maxwellian distribution with zero bulk velocity, unit density, and temperature, i.e.,   1 1 2 M (v) = − |v| . 3 exp 2 (2π) 2

(1.13)

It is well known that the Boltzmann collision operator Q(F, F ) satisfies the basic property of: Q(M, M )(v) = 0.

(1.14)

By substituting (1.12) and (1.13) into (1.10) and using (1.14), we obtain the following nonlinear elliptic equation for Ψ(x): −ΔΨ + eΨ = ρ(x).

(1.15)

Then, Equation (1.15) with u = −Ψ becomes Equation (1.1) exactly. The VPB system is a physical model that describes the mutual interactions of dilute charged particles (e.g., electrons) via collisions in the self-consistent electronic field. Extensive investigations of the VPB system with different background charges were presented by [1,12,21,25,26,30–33,40,46–49]. Fact 3. Equation (1.1) is recovered from the three-dimensional VMB system (see [11]). The VMB system where the physical constants are normalized to one has the following form: ∂t F + v · ∇x F + (E + v × B) · ∇v F = Q(F, F ),  ∂t E − ∇x × B = vF dv,

(1.16) (1.17)

R3

∂t B + ∇x × E = 0,  ∇x · B = 0, ∇x · E = F dv − ρ(x). R3

(1.18) (1.19)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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Similar to the VPFP system and VPB system, Equation (1.1) can be recovered from the VMB system. The details are omitted but they were given by [11]. Recently, this system has attracted the attention of mathematicians and some important insights have been obtained (e.g., see [19,22,24,34–36,38,39,43]). The facts given above show that Equation (1.1) can actually be recovered from the VPFP system, VPB system, and VMB system. In addition, after problem (1.1) and (1.2) have been solved, we can then construct the stationary solution to the VPFP system, VPB system, and VMB system by obtaining the solution u for problem (1.1) and (1.2). Thus, investigating (1.1) and (1.2) has a key role in studying the stationary solution to such of the kinetic systems mentioned above. In fact, many previous studies have investigated the three-dimensional case (e.g., Cheng [11], Duan– Yang [25], Duan–Yang–Zhu [27], and Glassey–Schaeffer–Zheng [28]). The existence of solutions to (1.1) and (1.2) was proved by [28] when ρ(x) satisfies |ρ(x) − 1| ≤ C (1 + |x|)

− 12

,

x ∈ R3 .

(1.20)

Subsequently, [27] generalized the condition given above to |ρ(x) − 1| ≤ C (ln (e + |x|))

−α

,

α > 0, x ∈ R3 .

(1.21)

In [25], it was assumed that ρ(x) − 1Wkm,∞ is sufficiently small. Then, the problem comprising (1.1) and (1.2) was proved in three dimensions by contraction mapping theorem, which admits a unique solution u = u(x) that satisfies uWkm,∞ ≤ Cρ(x) − 1Wkm,∞ for some constant C and the weighted norm gWkm,∞ = sup (1 + |x|)k x∈R3

(1.22)  |α|≤m

|∂xα g(x)| with suitable g = g(x)

and integers m ≥ 0, k ≥ 0. We also note that the nonlinear elliptic problem Equation (1.1) arises naturally in applications. On a bounded domain, the two-dimensional case and three-dimensional case were studied by [7] and [29], respectively. This equation was also formulated by [37] for the equilibrium of a charged gas in a container. In addition, there is great interest in these types of equations in geometry [44]. The main result obtained in the present study involves showing that under a weaker assumption on the decay rate on the background density ρ(x) than those made by [27,28], the existence and the uniqueness of the solution to the problem comprising (1.1) and (1.2) still hold. In particular, our result also holds for arbitrary dimensions N (N ≥ 3). We make the following assumptions regarding η(x) := ρ(x) − 1, which describe the more precise behavior of the background density function as |x| → ∞. (A1 ) A function h(σ) ∈ C 2 ([1, ∞)) exists that satisfies: 0 < h(σ) ≤ 1,

lim h(σ) = 0.

σ→∞

(1.23)

(A2 ) A positive constant B and a function p(z) > 0 for z ∈ (0, 1] exist such that: −B ≤ a(σ) ≤ −p(h(σ)),

(1.24)

where a(σ) is defined by a(σ) :=

4(σ − 1)h (σ) + 2N h (σ) . h(σ)

(1.25)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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(A3 ) A constant C > 1 exists such that: |η(x)| ≤ Ch(σ)

(1.26)

with

σ = 1 + r2 ,

r = |x| =

N 

 12 x2i

,

x = (x1 , x2 , · · · , xN ).

(1.27)

i=1

The main result obtained in this study can be stated as follows. Theorem 1.1. Suppose that η(x) ∈ L∞ (RN ) satisfies Assumptions (A1 )–(A3 ). Then, the problem comprising

(1.1) and (1.2) has a unique smooth solution u ∈ C ∞ RN that satisfies |u(x) − 1| ≤ Ch(σ)

(1.28)

for some constant C, i.e., the VPFP system, VPB system, and VMP system have a unique stationary solution. Remark 1.1. The existence on h(σ) is shown in the final section. Our result extends those obtained by [27, 28] to arbitrary space dimensions N (N ≥ 3). In particular, the decay required is weaker than those stated by [27,28]. It should be noted that the construction on a(σ) in (1.25) is nontrivial in order to relax the requirement in terms of the background density. In addition, the result obtained by [25] for the case of three dimensions can be generalized to the arbitrary dimension N (N ≥ 3), as summarized in the Appendix. The remainder of this paper is organized as follows. In the next section, we prove Theorem 1.1 by using the method of lower and upper solutions. In Section 3, we give a precise description of the decay based on the background density, which involves previous results. In the Appendix, we summarize a natural generalization of the three-dimensional case in the functional space Wkm,∞ . 2. Proof of the main theorem In this section, we prove Theorem 1.1 by using the method of lower and upper solutions. As described by [27,28], we define a function g by: g(y) = e−y + A2 y − 1,

y ∈ R,

(2.1)

where the positive constant A is determined in the following. Some basic observations regarding the function g were given by [27,28]. Lemma 2.1. The function g satisfies the following properties. (i) g  (y) = −e−y + A2 > 0 if y > − ln A2 . Thus, g is a strictly increasing smooth function on [− ln A2 , ∞) such that the inverse function g −1 : [gmin , ∞) is well defined, where:

gmin = g(− ln A2 ) = −A2 ln A2 − 1 − 1.

(2.2)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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(ii) For any x ∈ R,

g  (y) = e−y ≥ 0 and g(y) ≥ g  (0)y = A2 − 1 y. To apply the method of lower and upper solutions, we rewrite (1.1) as a linear equation by considering a solution operator L that satisfies:

Δ − A2 Lu = η(x) − g(u), x ∈ RN , (2.3) Lu(x) → 0, as |x| → ∞. In the following, the property of the operator Δ − A2 is explained. Lemma 2.2. Denote σ by (1.27) and then for any function h(σ) ∈ C 2 ([1, ∞)) that satisfies Assumption (A1 ), we have:



Δ − A2 h(σ) = − A2 − a(σ) h(σ),

(2.4)

where a(σ) is defined by (1.25). Proof. Direct calculations yield:

d2 N −1 d h(σ) − A2 h(σ) Δ − A2 h(σ) = 2 h(σ) + dr r dr = 4r2 h (σ) + 2N h (σ) − A2 h(σ)   4(σ − 1)h (σ) + 2N h (σ) 2 = −h(σ) A − h(σ)

2 = − A − a(σ) h(σ).

(2.5)

This completes the proof of Lemma 2.2. 2 Lemma 2.3. If the constants A, B, C, and D1 are chosen so they satisfy the inequality: 1 

A ≥ max e 2 (2+D1 ) , D1 B + C ,

(2.6)

then we have

η − D1 h(σ) A2 − a(σ) ≥ gmin ,

(2.7)

where gmin is defined by (2.2). In the following, C is the same as that in Assumption (A3 ). Proof. In fact, from (A1 )–(A3 ) and (2.2), we easily obtain:

η − D1 h(σ) A2 − a(σ) − gmin



≥ −Ch(σ) − D1 h(σ) A2 − a(σ) + A2 ln A2 − 1 + 1



≥ − C + D1 A2 h(σ) + D1 a(σ)h(σ) + A2 ln A2 − 1 + 1



≥ − C + D1 A2 − D1 B + A2 ln A2 − 1 + 1

= A2 ln A2 − D1 − 1 − (D1 B + C) + 1 ≥ A2 − (D1 B + C) + 1 ≥ 0,

(2.8)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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where we have used the following inequalities: ln A2 − D1 − 1 ≥ 1,

A2 ≥ D1 B + C,

which are naturally satisfied if the inequality (2.6) holds. This completes the proof of Lemma 2.3.

2

Now, we can define U by 

 U = g −1 η − D1 h(σ) A2 − a(σ) ,

(2.9)

which is well defined due to Lemmas 2.1 and 2.3. Lemma 2.4 (Construction of the subsolution). If D1 ≥ C, then we have U ≤ LU = −D1 h(σ).

(2.10)

Proof. First, we prove LU = −D1 h(σ). In fact, by the definition (2.9) for U , we have:

η − g(U ) = D1 h(σ) A2 − a(σ) .

(2.11)



Δ − A2 (−D1 h(σ)) = A2 − a(σ) D1 h(σ),

(2.12)

In addition, Lemma 2.2 implies that:

which together with (2.11) means that:

Δ − A2 (−D1 h(σ)) = η − g(U ).

(2.13)

Thus, by the definition of L and the uniqueness of the solution to problem (2.3), we have: LU = −D1 h(σ).

(2.14)

Next, proving U ≤ LU is equivalent to proving that g(U ) ≤ g(LU ) since g is a strictly increasing function. Thus, based on (2.11) and (2.14), we only need prove that:

η − D1 h(σ) A2 − a(σ) ≤ eD1 h(σ) − A2 D1 h(σ) − 1,

(2.15)

eD1 h(σ) − 1 − η − D1 a(σ)h(σ) ≥ 0.

(2.16)

eD1 h(σ) − 1 − η − D1 a(σ)h(σ) ≥ D1 h(σ) − Ch(σ) = (D1 − C) h(σ) ≥ 0,

(2.17)

i.e.,

In fact, we have:

where we have used (1.24), (1.26), and eD1 h(σ) − 1 ≥ D1 h(σ). Combining (2.16) and (2.15) with (2.17) yields g(U ) ≤ g(LU ). This completes the proof of Lemma 2.4. 2 In the following, we construct the supersolution.

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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Lemma 2.5. If the constants A, B, C, and D2 are chosen so they satisfy the inequality: A≥

D2 B + C,

(2.18)

then we have:

η + D2 h(σ) A2 − a(σ) ≥ gmin .

(2.19)

Proof. Similar to (2.8), from (A1 )–(A3 ) and (2.2), we easily obtain:

η + D2 h(σ) A2 − a(σ) − gmin



≥ −Ch(σ) + D2 h(σ) A2 − a(σ) + A2 ln A2 − 1 + 1



≥ D2 A2 − C h(σ) − D2 a(σ)h(σ) + A2 ln A2 − 1 + 1

≥ D2 A2 h(σ) − (C + D2 B) + A2 ln A2 − 1 + 1

≥ A2 ln A2 − 1 − (D2 B + C) + 1

(2.20)

≥ A2 − (D2 B + C) + 1 ≥ 0. This completes the proof of Lemma 2.5. 2 Now, we can define W by: 

 W = g −1 η + D2 h(σ) A2 − a(σ) ,

(2.21)

which is well defined due to Lemma 2.5. Lemma 2.6 (Construction of the supersolution). If the constants A, C, D2 , and the function p(z) in Assumption (A2 ) are chosen to satisfy the inequalities: ⎧  ⎫ −1 ⎨ ⎬ √  D2 ≥ C min p(y) , C ln 4 , A ≥ max e, C − 1 , 1 ⎩ ⎭ [ 2C ,1]

(2.22)

then we have: D2 h(σ) = LW ≤ W.

(2.23)

Proof. First, similar to (2.14), we can prove LW = D2 h(σ). In fact, by the definition of (2.21) for W , we have:

η − g(W ) = −D2 h(σ) A2 − a(σ) .

(2.24)

In addition, Lemma 2.2 implies that



Δ − A2 (D2 h(σ)) = − A2 − a(σ) D2 h(σ),

(2.25)

which together with (2.24) means that

Δ − A2 (D2 h(σ)) = η − g(W ).

(2.26)

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Thus, by the definition of L and the uniqueness of the solution to problem (2.3), we have: LW = D2 h(σ).

(2.27)

Next, proving that LW ≤ W is equivalent to proving that g(LW ) ≤ g(W ) since g is a strictly increasing function. Thus, based on (2.24) and (2.27), we only need to prove that:

e−D2 h(σ) + A2 D2 h(σ) − 1 ≤ η + D2 h(σ) A2 − a(σ) ,

(2.28)

1 + η − D2 a(σ)h(σ) − e−D2 h(σ) ≥ 0.

(2.29)

i.e.,

In fact, from (1.24) and (1.26), we have: 1 + η − D2 a(σ)h(σ) − e−D2 h(σ)

(2.30)

≥ 1 − Ch(σ) + D2 p(h(σ))h(σ) − e−D2 h(σ) . We set

Case 1. 0 < z ≤

1 2C

γ(z) = 1 − Cz + D2 p(z)z − e−D2 z .

(2.31)

γ1 (z) = 1 − Cz − e−D2 z ,

(2.32)

< 1.

We consider the function:

1 which satisfies γ1 (0) = 0, γ1 ( 2C ) ≥ 0 and γ1 (z) ≤ 0 for all z ≥ 0. Hence, for 0 ≤ z ≤

Case 2.

1 2C

1 2C ,

it holds that:

γ(z) ≥ γ1 (z) ≥ 0.

(2.33)

γ2 (z) = D2 zp(z) − Cz = z (D2 p(z) − C) .

(2.34)

≤ z ≤ 1.

We consider the function:

 Since D2 ≥ C

−1 min p(z)

1 [ 2C ,1]

, then for

1 2C

≤ z ≤ 1, it holds that: γ(z) ≥ γ2 (z) ≥ 0.

(2.35)

Combining (2.33) with (2.35) gives γ(z) ≥ 0 for 0 < z ≤ 1, which implies that (2.29) holds since 0 < h(σ) ≤ 1. This completes the proof of Lemma 2.6. 2 Now, based on Lemmas 2.4 and 2.6, the existence and decay rate for the solution to the problem comprising (1.1) and (1.2) are obtained by an iterative process in a similar manner to the methods employed by

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

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[27,28]. The uniqueness is determined directly from the maximum principle for the elliptic equation. After the existence and uniqueness of the solution to the problem comprising (1.1) and (1.2) have been proved, the existence and uniqueness of the stationary solutions to the VPFP system, VPB system, and VMB system follow from classical arguments in a similar manner to the three-dimensional case. This completes the proof of Theorem 1.1. 3. Precise description of the decay based on the background density In this section, we specify functions h(σ) to describe the precise decay rate based on the background density ρ(x) = 1 + η(x), cf. (1.26). We also show the admissibility of Assumptions (A1 ) and (A2 ). Case 1. h(σ) = σ −α , σ ∈ [1, ∞), 0 < α ≤

N −2 2

for N ≥ 3.

It is obvious that Assumption (A1 ) is satisfied. Thus, we only need to show that Assumption (A2 ) is satisfied. In fact, we have: h (σ) = −ασ −α−1 ,

h (σ) = α(α + 1)σ −α−2 ,

(3.1)

which by (1.25) means that: a(σ) =

4(σ − 1)α(α + 1)σ −α−2 − 2N ασ −α−1 σ −α (3.2)

= −2N ασ −1 + 4α(α + 1) (σ − 1) σ −2 ≥ −2N α because σ ≥ 1. In addition, we can rewrite a(σ) as follows: a(σ) = −2α(N − 2 − 2α)σ −1 − 4α(α + 1)σ −2 1

2

= −2α(N − 2 − 2α) (h(σ)) α − 4α(α + 1) (h(σ)) α .

(3.3)

If we take: B = 2N α

(3.4)

and 1

2

p(z) = −2α(N − 2 − 2α)z α − 4α(α + 1)z α ,

(3.5)

then (3.2) and (3.3) mean that Assumption (A2 ) is satisfied. Remark 3.1. When N = 3, h(σ) is the exact same function described by [28]. Case 2. h(σ) = (1 + ln σ)−α , σ ∈ [1, ∞), 0 < α ≤ min

1

2 (3N

 − 8), N − 2 for N ≥ 3.

In fact, we have: h (σ) = −α(1 + ln σ)−α−1 σ −1 , h (σ) = α(α + 1)(1 + ln σ)−α−2 σ −2 + α(1 + ln σ)−α−1 σ −2 ,

(3.6)

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which by (1.25) means that: a(σ) = 4α(α + 1)(1 + ln σ)−2 σ −1 − 4α(α + 1)(1 + ln σ)−2 σ −2 + 4α(1 + ln σ)−1 σ −1 − 4α(1 + ln σ)−1 σ −2 − 2N α(1 + ln σ)−1 σ −1 ≥ −4α(α + 1)(1 + ln σ)−2 σ −2 − 2α(N − 2)(1 + ln σ)−1 σ −1 − 4α(1 + ln σ)−1 σ −2

(3.7)

≥ −4α(α + 1) − 2α(N − 2) − 4α

= − 4α2 + 4α + 2N α because σ ≥ 1. In addition, we can rewrite a(σ) as follows: a(σ) = 2ασ −2 (1 + ln σ)−2 a1 (σ),

(3.8)

a1 (σ) = 2(α + 1)σ − 2(α + 1) − (N − 2)σ(1 + ln σ) − 2(1 + ln σ).

(3.9)

where

Direct calculations yield: a1 (1) = −N

(3.10)

and a1 (σ) = (2α − 2N + 6) − (N − 2) ln σ − 2σ −1 = σ −1 {(2α − 2N + 6)σ − (N − 2)σ ln σ − 2} .

(3.11)

We set: a2 (σ) = (2α − 2N + 6)σ − (N − 2)σ ln σ − 2.

(3.12)

Then, direct calculations yield: a2 (1) = 2α − 2N + 4 ≤ 0

(3.13)

a2 (σ) = 2α − 3N + 8 − (N − 2) ln σ ≤ 2α − 3N + 8 ≤ 0,

(3.14)

and

where we have used α ≤ N − 2 and α ≤ 12 (3N − 8) in (3.13) and (3.14), respectively. Combining (3.13) with (3.14) yields: a2 (σ) ≤ a2 (1) ≤ 0,

(3.15)

which together with (3.10), (3.11) means that: a1 (σ) ≤ a1 (1) = −N.

(3.16)

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Thus, from (3.8), we have: a(σ) ≤ −2ασ −2 (1 + ln σ)−2 N = −2N αe2−2(h(σ))

−1 α

2

(3.17)

(h(σ)) α .

If we take: B = 4α2 + 4α + 2N α

(3.18)

and p(z) = 2N αe2−2z

−1 α

2

zα,

(3.19)

then (3.7) and (3.17) mean that Assumption (A2 ) is satisfied. Remark 3.2. When N = 3, h(σ) is the exact same function described by [27]. Case 3. h(σ) = (1 + ln(1 + ln σ))−α , σ ∈ [1, ∞). In this case, we require that the constant α satisfies   0 < α ≤ min 32 (N − 2), 14 (7N − 18), 16 (15N − 44) for N ≥ 3. In fact, we have: h (σ) = − α(1 + ln(1 + ln σ))−α−1 (1 + ln σ)−1 σ −1 , h (σ) = α(α + 1)(1 + ln(1 + ln σ))−α−2 (1 + ln σ)−2 σ −2 + α(1 + ln(1 + ln σ))−α−1 (1 + ln σ)−2 σ −2

(3.20)

+ α(1 + ln(1 + ln σ))−α−1 (1 + ln σ)−1 σ −2 , which by (1.25) means that: a(σ) = 4α(α + 1)(σ − 1)(1 + ln(1 + ln σ))−2 (1 + ln σ)−2 σ −2 + 4α(σ − 1)(1 + ln(1 + ln σ))−1 (1 + ln σ)−2 σ −2 + 4α(σ − 1)(1 + ln(1 + ln σ))−1 (1 + ln σ)−1 σ −2

(3.21)

− 2N α(1 + ln(1 + ln σ))−1 (1 + ln σ)−1 σ −1 . We can easily obtain: 1 + ln σ ≥ 1,

1 + ln(1 + ln σ) ≥ 1

(3.22)

because σ ≥ 1. Thus, we have: a(σ) ≥ −2N α. In addition, we can denote a(σ) as:

(3.23)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

13

a(σ) = 2ασ −2 (1 + ln σ)−2 (1 + ln(1 + ln σ))−2 {2(α + 1)(σ − 1) + 2(σ − 1)(1 + ln(1 + ln σ)) + 2(σ − 1)(1 + ln σ)(1 + ln(1 + ln σ)) − N σ(1 + ln σ)(1 + ln(1 + ln σ))} = 2ασ −2 (1 + ln σ)−2 (1 + ln(1 + ln σ))−2 {(2α − N + 6)σ − (2α + 6)

(3.24)

+ (4 − N )σ ln(1 + ln σ) − 4 ln(1 + ln σ) − (N − 2)σ ln σ − 2 ln σ − (N − 2)σ ln σ ln(1 + ln σ) − 2 ln σ ln(1 + ln σ)} ¯1 (σ), =: 2ασ −2 (1 + ln σ)−2 (1 + ln(1 + ln σ))−2 a where a ¯1 (σ) = (2α − N + 6)σ − (2α + 6) + (4 − N )σ ln(1 + ln σ) − 4 ln(1 + ln σ) − (N − 2)σ ln σ − 2 ln σ

(3.25)

− (N − 2)σ ln σ ln(1 + ln σ) − 2 ln σ ln(1 + ln σ). Direct calculations yield: a ¯1 (1) = −N

(3.26)

and a ¯1 (σ) = (2α − 2N + 8) + 2(3 − N ) ln(1 + ln σ) + (4 − N )(1 + ln σ)−1 − 4σ −1 (1 + ln σ)−1 − (N − 2) ln σ − 2σ −1 − (N − 2) ln σ ln(1 + ln σ) − (N − 2) ln σ ln(1 + ln σ)−1 − 2σ −1 ln(1 + ln σ) − 2σ −1 ln σ(1 + ln σ)−1 = σ −1 (1 + ln σ)−1 {(2α − 2N + 8)σ(1 + ln σ) + 2(3 − N )σ(1 + ln σ) ln(1 + ln σ) + (4 − N )σ − 4

(3.27)

− (N − 2)σ ln σ(1 + ln σ) − 2(1 + ln σ) − (N − 2)σ ln σ(1 + ln σ) ln(1 + ln σ) − (N − 2)σ ln σ − 2(1 + ln σ) ln(1 + ln σ) − 2 ln σ} =: σ −1 (1 + ln σ)−1 {a2 (σ) + 2(3 − N )σ(1 + ln σ) ln(1 + ln σ)} ≤ σ −1 (1 + ln σ)−1 a ¯2 (σ), where a ¯2 (σ) = (2α − 2N + 8)σ(1 + ln σ) + (4 − N )σ − 4 − (N − 2)σ ln σ(1 + ln σ) − 2(1 + ln σ) − (N − 2)σ ln σ(1 + ln σ) ln(1 + ln σ)

(3.28)

− (N − 2)σ ln σ − 2(1 + ln σ) ln(1 + ln σ) − 2 ln σ. Direct calculations yield: a ¯2 (1) = 2α − 3N + 6 ≤ 0 due to α ≤ 32 (N − 2), and

(3.29)

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

14

a ¯2 (σ) = σ −1 {(4α − 7N + 24)σ + (2α − 7N + 18)σ ln σ − 6 − (N − 2)σ(ln σ)2 − (N − 2)σ(1 + ln σ) ln(1 + ln σ) − (N − 2)σ ln σ(1 + ln σ) ln(1 + ln σ) − (N − 2)σ ln σ ln(1 + ln σ) − 2 ln(1 + ln σ)}  =: σ −1 a3 (σ) − (N − 2)σ(ln σ)2

(3.30)

− (N − 2)σ ln σ(1 + ln σ) ln(1 + ln σ) − (N − 2)σ ln σ ln(1 + ln σ) − 2 ln(1 + ln σ)} ≤ σ −1 a3 (σ), where a3 (σ) = (4α − 7N + 24)σ + (2α − 7N + 18)σ ln σ − 6 − (N − 2)σ(1 + ln σ) ln(1 + ln σ).

(3.31)

Direct calculations yield: a3 (1) = 4α − 7N + 18 ≤ 0

(3.32)

and a3 (σ) = 6α − 15N + 44 + (2α − 7N + 18) ln σ − (N − 2)(1 + ln σ) ln(1 + ln σ) − (N − 2) ln(1 + ln σ)

(3.33)

≤ 6α − 15N + 44 ≤ 0, where we have used the facts that α ≤ 14 (7N − 18) in (3.32) and α ≤ 16 (15N − 44) in (3.33). Combining (3.32) and (3.33) yields: a3 (σ) ≤ a3 (1) ≤ 0,

(3.34)

a ¯2 (σ) ≤ 0.

(3.35)

a ¯2 (σ) ≤ a ¯2 (1) ≤ 0,

(3.36)

a ¯1 (σ) ≤ 0.

(3.37)

a ¯1 (σ) ≤ a ¯1 (1) ≤ 0.

(3.38)

a(σ) ≤ −2N ασ −2 (1 + ln σ)−2 (1 + ln(1 + ln σ))−2 .

(3.39)

which together with (3.30) means that:

From (3.35) and (3.29), we also have:

which together with (3.27) means that:

Similarly, (3.26) and (3.37) imply that:

Thus, from (3.24) and (3.38), we have:

L. Zhu / J. Math. Anal. Appl. 466 (2018) 1–17

15

If we take: B = 4α2 + 4α + 2N α

(3.40)

  1    2 1 p(z) = 2N αexp 2 − 2exp z − α − 1 exp 2 − 2z − α z α ,

(3.41)

and

then (3.23) and (3.39) mean that Assumption (A2 ) is satisfied. Remark 3.3. Case 3 shows that the decay based on the background density function is weaker than those in Cases 1 and 2. Thus, our result is a natural extension of the result given by [27,28] in the case of three dimensions. In addition, we note that our result holds for arbitrary dimensions N (N ≥ 3). 4. Appendix In this section, we prove the existence of the problem comprising (1.1) and (1.2) in arbitrary dimensions N (N ≥ 3), which is a natural generalization of the three-dimensional case considered by [25]. However, we use this to construct stationary solutions to VPFPs, VPBs, and VMBs in the functional space Wkm,∞ . Theorem 4.1. Suppose that ρ(x) − 1Wkm,∞ is sufficiently small. Then, the problem comprising (1.1) and (1.2) in N dimensions has been proved by contraction mapping theorem to admit a unique solution u = u(x) that satisfies: uWkm,∞ ≤ Cρ(x) − 1Wkm,∞ for some constant C and the weighted norm gWkm,∞ = sup (1 + |x|)k x∈RN

and integers m ≥ 0, k ≥ 0.

(4.1)  |α|≤m

|∂xα g(x)| with suitable g = g(x)

Proof. As described by [25], Equation (1.1) can be rewritten in integral form as:

u = T(u) =: G ∗ e−u − u + ρ ,

(4.2)

where G = G(x) in N dimensions is given by:

G(x) =

∞

1 (4π)

N 2

s− 2 e−s− N

|x|2 4s

ds,

(4.3)

0

and it is the fundamental solution to the linear elliptic equation ΔxG − G = 0. Thus, (1.1) admits a solution if and only if the nonlinear mapping T has a fixed point. We define:  Bm,k (C) = φ ∈ W m,∞ (RN ); φWkm,∞ ≤ Cρ(x) − 1Wkm,∞ ,

(4.4)

for some constant C that are determined subsequently. Similar to [25], we can prove that if ρ(x) − 1Wkm,∞ is sufficiently small, then a constant C exists such that T : Bm,k (C) → Bm,k (C) is a contraction mapping. The full details are omitted but they were given by [25]. 2

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