Standing wave solution for the generalized Jackiw-Pi model

Hyungjin Huh; Yuanfeng Jin; Youwei Ma; Guanghui Jin

doi:10.1515/anona-2022-0261

Article Open Access

Standing wave solution for the generalized Jackiw-Pi model

Hyungjin Huh , Yuanfeng Jin , Youwei Ma and Guanghui Jin

Published/Copyright: September 13, 2022

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From the journal Advances in Nonlinear Analysis Volume 12 Issue 1

Abstract

We study the existence and nonexistence of the standing wave solution for the generalized Jackiw-Pi model by using variational method. Depending on interaction strength λ , we have three different situations. The existence and nonexistence of the standing wave solution correspond to 1 < λ and 0 < λ < 1 , respectively. We have the explicit solution of self-dual equation for the borderline λ = 1 .

Keywords: generalized Jackiw-Pi model; standing waves; variational method

MSC 2010: 35Q40; 35J20

1 Introduction

In this article, we deal with the following generalized Jackiw-Pi model:

(1.1) i ( m + 1 ) ∣ ϕ ∣ 2 m D 0 ϕ + ( D 1 D 1 + D 2 D 2 ) ϕ + λ 2 ( m + 2 ) ∣ ϕ ∣ 2 m + 2 ϕ = 0 ,

(1.2) ∂ 0 A 1 − ∂ 1 A 0 = − Im ( ϕ ¯ D 2 ϕ ) ,

(1.3) ∂ 0 A 2 − ∂ 2 A 0 = Im ( ϕ ¯ D 1 ϕ ) ,

(1.4) ∂ 1 A 2 − ∂ 2 A 1 = − 1 2 ∣ ϕ ∣ 2 m + 2 ,

where i denotes the imaginary unit, ∂ 0 = ∂ ∂ t , ∂ 1 = ∂ ∂ x 1 , ∂ 2 = ∂ ∂ x 2 for ( t , x 1 , x 2 ) ∈ R 1 + 2 , ϕ : R 1 + 2 → C is the complex scalar field, A μ : R 1 + 2 → R is the gauge field, D μ = ∂ μ + i A μ is the covariant derivative for μ = 0 , 1 , 2 . The strength of interaction potential is represented by a constant λ > 0 and m is a nonnegative integer.

The system (1.1)–(1.4) with m = 0 is called the Jackiw-Pi model and was proposed in [10,11] to deal with electromagnetic phenomena in a planar domain such as the fractional quantum Hall effect or high temperature superconductivity. Several authors [1,2, 3,4,5, 6,9,12, 13,14] have studied the Jackiw-Pi model mathematically. Particularly, the existence and nonexistence of the standing wave solution for the Jackiw-Pi model were shown in [3] by applying variational method to the functional, which is obtained by representing the gauge field A μ in terms of the complex scalar field ϕ . A nontrivial solitary wave solution to the Chern-Simons-Higgs system was recently obtained in [12] by considering the non-relativistic limit to a minimal mass solution of the Jackiw-Pi model.

The generalized Jackiw-Pi model was proposed in [15]. The self-dual equations of the generalized Jackiw-Pi model were studied in [8] where the self-dual system can be transformed into the Liouville-type equation. Here we are interested in the existence and nonexistence of the standing wave solution to the generalized Jackiw-Pi model.

The system of equations (1.1)–(1.4) is invariant under the following gauge transformation:

ϕ → ϕ e i χ , A μ → A μ − ∂ μ χ ,

where χ : R 1 + 2 → R is an arbitrary C ∞ function. Therefore, a solution of the system is formed by a class of gauge equivalent 4-tuples ( ϕ e i χ , A 0 − ∂ 0 χ , A 1 − ∂ 1 χ , A 2 − ∂ 2 χ ) . Here we consider the Coulomb gauge condition

∂ 1 A 1 + ∂ 2 A 2 = 0 .

We are interested in the standing wave solution to (1.1)–(1.4) of the form

(1.5) ϕ ( t , x ) = u ( ∣ x ∣ ) e i ω t , A 0 ( t , x ) = k ( ∣ x ∣ ) , A 1 ( t , x ) = x 2 ∣ x ∣ 2 h ( ∣ x ∣ ) , A 2 ( t , x ) = − x 1 ∣ x ∣ 2 h ( ∣ x ∣ ) ,

where ω > 0 is a given frequency and u , h , k are real valued functions defined on [ 0 , ∞ ) such that h ( 0 ) = 0 . We point out that the ansatz (1.5) satisfies the Coulomb gauge condition. Inserting the ansatz (1.5) into the system of equations (1.1)–(1.4), we obtain the following nonlocal semi-linear elliptic equation for u

(1.6) Δ u − ( m + 1 ) ( ω + ξ ) u 2 m + 1 − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

where ξ ∈ R is a constant and

h ( s ) = ∫ 0 s l 2 u 2 m + 2 ( l ) d l .

We will show that (1.6) is the Euler-Lagrange equation of the functional

(1.7) J ( u ) ≡ 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + ( ω + ξ ) u 2 m + 2 + u 2 ∣ x ∣ 2 h 2 d x − λ 4 ∫ R 2 ∣ u ∣ 2 m + 4 d x ,

where u ∈ H r 1 is a radially symmetric function in H 1 ( R 2 ) . We will show that J ∈ C 1 ( H r 1 ) and a critical point u of J produces a standing wave ( ϕ , A 0 , A 1 , A 2 ) of the form (1.5). The following is our main result.

Theorem 1.1

We have three cases depending on the interaction strength λ .

For λ ∈ ( 0 , 1 ) and ω > 0 , there exists no nontrivial standing wave solution ( ϕ , A 0 , A 1 , A 2 ) of the form (1.5) satisfying equations (1.1)–(1.4).
For λ = 1 and ω > 0 , any standing wave solution ( ϕ , A 0 , A 1 , A 2 ) of the form (1.5) satisfying equations (1.1)–(1.4) and ∣ ϕ ∣ > 0 in R 2 has the following form:
ϕ = 8 l 1 + ( m + 1 ) ∣ l x ∣ 2 1 m + 1 e i ω t , A 0 = 1 2 8 l 1 + ( m + 1 ) ∣ l x ∣ 2 2 m + 1 − ω , A 1 = 2 l 2 x 2 1 + ( m + 1 ) ∣ l x ∣ 2 , A 2 = − 2 l 2 x 1 1 + ( m + 1 ) ∣ l x ∣ 2 .
For λ > 1 and ω > 0 , there exists a standing wave solution ( ϕ , A 0 , A 1 , A 2 ) of form (1.5) satisfying equations (1.1)–(1.4) and ∣ ϕ ∣ > 0 in R 2 . Moreover, if their frequencies ω are different, then any two standing waves of the form (1.5) are not gauge equivalent to each other.

Theorem 1.1 is a natural extension of the result in [3] where the case of m = 0 was studied. We will follow the basic idea in [3] with some modifications to prove Theorem 1.1. Making use of (1.2)–(1.4) and the ansatz (1.5), we have the integral representation of the gauge fields in terms of u . Plugging the representation into equation (1.1), we obtain a nonlinear equation (1.6), which has the nonlocal terms associated with h ( ∣ x ∣ ) . Then we apply a variational method to the energy functional (1.7) from which (1.6) is derived as the Euler-Lagrange equation.

The rest of the article is organized as follows. In Section 2, we introduce a mathematical setting and some preliminary propositions to obtain our main result. Theorem 1.1 is proved in Section 3.

2 Preliminaries

In this section, we introduce some known facts and present a mathematical setting for our result.

Lemma 2.1

We consider the following Gagliardo-Nirenberg inequality:

(2.1) ‖ u ‖ L 2 m + 2 ( R 2 ) ≤ C ‖ ∇ u ‖ L 2 ( R 2 ) m m + 1 ‖ u ‖ L 2 ( R 2 ) 1 m + 1 ,

which plays an important role.

We also note that, for any u ∈ H r 1 , the following well-known inequality called Strauss inequality holds

(2.2) ∣ x ∣ 1 2 ∣ u ( ∣ x ∣ ) ∣ ≤ C ‖ u ‖ , ∣ x ∣ > 0 ,

where ‖ u ‖ 2 = ∫ R 2 ∣ ∇ u ∣ 2 + ∣ u 2 ∣ d x denotes the Sobolev norm on H r 1 ( R 2 ) .

Now we plug our ansatz (1.5) in (1.1)–(1.4) to obtain

(2.3) Δ u − ( m + 1 ) ( ω + A 0 ) u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

(2.4) 1 s h ′ ( s ) = 1 2 u 2 m + 2 ( s ) ,

(2.5) k ′ ( s ) = − 1 s h ( s ) u 2 ( s ) ,

where ′ = d d s . From the condition h ( 0 ) = 0 , we obtain from (2.4)

h ( r ) = ∫ 0 r s 2 u 2 m + 2 ( s ) d s .

Thus, we have

A 1 = x 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s , A 2 = − x 1 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s .

Since u is in H r 1 ( R 2 ) , A 1 and A 2 are well defined and continuous on R 2 \ { 0 } . Also, by Lemma 2.1 and Hölder’s inequality, we obtain the following estimates:

(2.6) h ( r ) = ∫ 0 r s 2 u 2 m + 2 ( s ) d s = 1 4 π ∫ B ∣ x ∣ u 2 m + 2 ( ∣ x ∣ ) d x ≤ 1 4 π ‖ u ‖ L 2 m + 2 2 m + 2 ≤ C ,

(2.7) h ( r ) = 1 4 π ∫ B ∣ x ∣ u 2 m + 2 ( ∣ x ∣ ) d x ≤ C r ‖ u ‖ L 4 m + 4 2 m + 2 ≤ C r ‖ ∇ u ‖ L 2 2 m + 1 ‖ u ‖ L 2 ≤ C r .

It follows that

∣ A 1 ∣ = x 2 ∣ x ∣ 2 h ( ∣ x ∣ ) ≤ C ∣ x 2 ∣ ∣ x ∣ < C , ∣ A 2 ∣ = x 1 ∣ x ∣ 2 h ( ∣ x ∣ ) < C , ∣ x ∣ > 0 .

Therefore, A 1 , A 2 are L ∞ functions on R 2 ⧹ { 0 } .

We obtain by integrating both sides of (2.5) from ∣ x ∣ to ∞

k ( ∣ x ∣ ) = ξ + ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s ,

where ξ is an arbitrary constant which is just the value of k at infinity. Using inequalities (2.2) and (2.6), we obtain the following estimate:

k ( ∣ x ∣ ) − ξ = ∫ ∣ x ∣ ∞ h ( s ) s u 2 d s ≤ C ∫ ∣ x ∣ ∞ u 2 ( s ) s d s ≤ C ∫ ∣ x ∣ ∞ ‖ u ‖ 2 s 2 d s .

This means that A 0 is well defined on R 2 ⧹ { 0 } , but at this point it is not certain whether or not A 0 belongs to L ∞ . In fact, we need a condition u ∈ L loc ∞ to obtain A 0 ∈ L ∞ as we see in the following proposition.

Proposition 2.2

If u is in H r 1 ( R 2 ) ∩ L loc ∞ ( R 2 ) , then A 0 is in L ∞ ( R 2 ) . Furthermore, if u is in H r 1 ( R 2 ) ∩ C ( R 2 ) , then A 0 , A 1 , and A 2 are in L ∞ ( R 2 ) ∩ C 1 ( R 2 ) .

Proof

To prove A 0 ∈ L ∞ ( R 2 ) , it is enough to show k ( 0 ) is finite since the function k is non-increasing. Since u ∈ H r 1 ( R 2 ) ∩ L loc ∞ ( R 2 ) , we have

(2.8) h ( ∣ x ∣ ) = ∫ 0 ∣ x ∣ s 2 u 2 m + 2 d s = 1 4 π ∫ B ∣ x ∣ u 2 m + 2 d x ≤ 1 4 ‖ u ‖ L ∞ 2 m + 2 ∣ x ∣ 2 .

Thus, we have

∫ 0 ∞ h ( s ) s u 2 ( s ) d s = 1 2 π ∫ R 2 h ( ∣ x ∣ ) ∣ x ∣ 2 u 2 ( ∣ x ∣ ) d x ≤ C ∫ R 2 u 2 ( ∣ x ∣ ) d x < ∞ .

This proves that A 0 ∈ L ∞ ( R 2 ) .

For the second claim A 0 , A 1 , A 2 ∈ C 1 ( R 2 ) , suppose that u ∈ H r 1 ( R 2 ) is continuous. Then we see that A 0 , A 1 , A 2 ∈ C 1 ( R 2 ⧹ { 0 } ) . Now, we see from the continuity of u that

h ( s ) s 2 = 1 s 2 ∫ 0 s r 2 u 2 m + 2 ( r ) d r = 1 π s 2 ∫ B s 1 4 u 2 m + 2 ( r ) d r → 1 4 u 2 m + 2 ( 0 ) as s → 0 .

This means that

lim r → 0 k ( r ) − k ( 0 ) r = − lim r → 0 1 r ∫ 0 r u 2 ( s ) s h ( s ) d s = − lim r → 0 h ( r ) r u 2 ( r ) = − lim r → 0 r h ( r ) r 2 u 2 ( r ) = 0 .

Note that k ′ ( r ) = − u 2 ( r ) r h ( r ) for r > 0 and that lim r → 0 k ′ ( r ) = 0 . This implies A 0 ∈ C 1 ( R 2 ) .

Now we consider A 1 . By inequality (2.8), we have

A 1 ( 0 ) = lim x → 0 x 2 ∣ x ∣ 2 h ( ∣ x ∣ ) ≤ lim x → 0 C x 2 = 0 .

Note that

∂ 1 A 1 ( 0 ) = lim h → 0 A 1 ( h , 0 ) − A 1 ( 0 , 0 ) h = 0

and

∂ 2 A 1 ( 0 ) = lim h → 0 A 1 ( 0 , h ) − A 1 ( 0 , 0 ) h = lim h → 0 1 h 2 ∫ 0 h r 2 u 2 m + 2 ( r ) d r = lim h → 0 1 π h 2 ∫ B h 1 4 u 2 m + 2 ( r ) d r = 1 4 u 2 m + 2 ( 0 ) .

Moreover, we see that

∂ 1 A 1 ( x ) = − 2 x 1 x 2 ∣ x ∣ 4 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s + x 1 x 2 2 ∣ x ∣ 2 u 2 m + 2 ( s ) = x 1 x 2 2 ∣ x ∣ 2 u 2 m + 2 ( ∣ x ∣ ) − 1 π ∣ x ∣ 2 ∫ B ∣ x ∣ u 2 m + 2 ( ∣ x ∣ ) d x → 0 as ∣ x ∣ → 0

and

∂ 2 A 1 ( x ) = ∣ x ∣ 2 − 2 x 2 2 ∣ x ∣ 4 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s + x 2 2 2 ∣ x ∣ 2 u 2 m + 2 ( ∣ x ∣ ) = 1 4 π ∣ x ∣ 2 ∫ B ∣ x ∣ u 2 m + 2 ( ∣ x ∣ ) d x + x 2 2 2 ∣ x ∣ 2 u 2 m + 2 ( ∣ x ∣ ) − 1 π ∣ x ∣ 2 ∫ B ∣ x ∣ u 2 m + 2 ( ∣ x ∣ ) d x → 1 4 u 2 m + 2 ( 0 ) as ∣ x ∣ → 0 .

Thus A 1 belongs to C 1 . By a similar procedure, we obtain A 2 ∈ C 1 ( R 2 ) . This completes the proof.□

Now, we apply a variational argument to obtain a solution of equation (1.6). In fact, (1.6) is the Euler-Lagrange equation of the following functional:

J ( u ) ≡ 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + ( ω + ξ ) u 2 m + 2 + u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x − λ 4 ∫ R 2 ∣ u ∣ 2 m + 4 d x ,

for u ∈ H r 1 ( R 2 ) .

Proposition 2.3

The functional J is continuously differentiable on H r 1 and its critical point u is a weak solution of (1.6). Furthermore, a critical point u of J belongs to C 2 ( R 2 ) , so the weak solution u is a classical solution of (1.6).

Proof

We define

J ˆ ( u ) = ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x = ∫ R 2 u 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s 2 d x ,

and a map L from H r 1 to the space of linear functionals on H r 1 by

L ( u ) ϕ = ∫ R 2 ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 d s u 2 m + 1 ϕ + h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u ϕ d x , ϕ ∈ H r 1 ( R 2 ) .

First, we claim that L is bounded. By inequalities (2.1) and (2.7), we have

(2.9) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s = ∫ ∣ x ∣ 1 h ( s ) s u 2 ( s ) d s + ∫ 1 ∞ h ( s ) s u 2 ( s ) d s ≤ C ‖ ∇ u ‖ L 2 m + 1 ‖ u ‖ L 2 m + 2 m + 1 ∫ ∣ x ∣ 1 s − 1 2 2 3 ∫ ∣ x ∣ 1 s u 6 ( s ) d s 1 3 + ‖ ∇ u ‖ L 2 m + 1 ‖ u ‖ L 2 m + 2 m + 1 ‖ u ‖ L 2 2 ≤ C ‖ ∇ u ‖ L 2 2 m + 5 3 ‖ u ‖ L 2 7 3 + C ‖ ∇ u ‖ L 2 2 m + 1 ‖ u ‖ L 2 3 .

Then, applying the Gargliardo-Nirenberg inequality to L ( u ) ϕ , we see that L ( u ) ∈ ( H r 1 ) ∗ .

Next we show that L is continuous. Let { u n } be a sequence converging to some u in H r 1 as n → ∞ . The estimates (2.7) and (2.9) imply that there are sequences { a n } and { b n } of uniformly bounded functions, which converge pointwise to a and b on R 2 as n → ∞ satisfying

L ( u n ) ϕ = ∫ R 2 a n ( x ) u n 2 m + 1 ϕ + b n ( x ) u n ϕ d x

and

L ( u ) ϕ = ∫ R 2 a u 2 m + 1 ϕ + b u ϕ d x .

Then, we obtain

∣ L ( u n ) ϕ − L ( u ) ϕ ∣ ≤ ∫ R 2 a n ( x ) u n 2 m + 1 ϕ − a u 2 m + 1 ϕ d x + ∫ R 2 b n ( x ) u n ϕ − b u ϕ d x ≤ ∫ R 2 ∣ a n ( u n 2 m + 1 − u 2 m + 1 ) ϕ ∣ d x + ∫ R 2 ∣ a n − a ∣ u 2 m + 1 ϕ d x + ∫ R 2 ∣ b n ( u n − u ) ϕ ∣ d x + ∫ R 2 ∣ b n − b ∣ ∣ u ϕ ∣ d x ≤ ∫ R 2 ∣ a n ( u n 2 m + 1 − u 2 m + 1 ) ϕ ∣ d x + C ( ∥ ∣ a n − a ∣ u 2 m + 1 ∥ L 2 + ‖ u n − u ‖ L 2 + ∥ ∣ b n − b ∣ u ∥ L 2 ) ‖ ϕ ‖ .

For the first term of the above inequality, it suffices to show the general estimate:

∫ R 2 ∣ a n ( u n 2 m + 1 − u 2 m + 1 ) ϕ ∣ d x ≤ C ∫ R 2 ∣ u n − u ∣ ∣ u n 2 m + u 2 m ∣ ∣ ϕ ∣ d x ≤ C ‖ u n − u ‖ L 2 1 2 ‖ ∇ ( u n − u ) ‖ L 2 1 2 ‖ ϕ ‖ ,

where we use ‖ u ‖ L 4 ( R 2 ) ≤ C ‖ ∇ u ‖ L 2 1 2 ‖ u ‖ L 2 1 2 and ‖ u 2 m ‖ L 4 ( R 2 ) ≤ C ‖ ∇ u ‖ L 2 4 m − 1 2 ‖ u ‖ L 2 1 2 .

Using the dominated convergence theorem and strong convergence of u n to u in H r 1 , we see that lim n → ∞ ∣ L ( u n ) ϕ − L ( u ) ϕ ∣ ‖ ϕ ‖ = 0 . This proves the continuity of L . Now it is easy to see that

J ˆ ( u + ϕ ) − J ˆ ( u ) − 2 L ( u ) ϕ = o ( ‖ ϕ ‖ ) ,

which implies J ˆ ∈ C 1 .

Finally, suppose that u ∈ H r 1 ( R 2 ) is a critical of J ( u ) . Then u weakly solves

Δ u − ( m + 1 ) ( ω + ξ ) u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u = ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u .

Considering estimates (2.7) and (2.9), we see that h 2 ( ∣ x ∣ ) ∣ x ∣ 2 , ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s ∈ L ∞ . Then the standard elliptic estimates [7] imply that u ∈ C loc 1 , γ for some γ > 0 . Then, we obtain h 2 ( ∣ x ∣ ) ∣ x ∣ 2 , ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s ∈ C ( R 2 ) . Since u is radially symmetric, we obtain u ∈ C 2 . This completes the proof.□

Proposition 2.4

(Pohozaev identity) Let b, c, and d be real constants and u ∈ H 1 r ( R 2 ) be a weak solution of the equation:

Δ u + b ( m + 1 ) u 2 m + 1 + c ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 + c h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + d ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 in R 2 ,

where h ( s ) = ∫ 0 s r 2 u 2 m + 2 d r . Then, there holds the following integral identity:

b ∫ R 2 u 2 m + 2 d x + 2 c ∫ R 2 h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u 2 d x + d ∫ R 2 ∣ u ∣ 2 m + 4 d x = 0 .

Proof

Multiplying by ∇ u ⋅ x and integrating by parts on B R , we obtain the following identities:

(1) ∫ B R Δ u ( ∇ u ⋅ x ) d x = R 2 ∫ ∂ B R ∣ ∇ u ∣ 2 d S x ≡ ( i ) , (2) ( m + 1 ) ∫ B R u 2 m + 1 ( ∇ u ⋅ x ) d x = − ∫ B R u 2 m + 2 d x + R 2 ∫ ∂ B R u 2 m + 2 d S x ≡ − ∫ B R u 2 m + 2 d x + ( ii ) , (3) ( m + 2 ) ∫ B R ∣ u ∣ 2 m + 2 u ( ∇ u ⋅ x ) d x = − ∫ B R ∣ u ∣ 2 m + 4 d x + R 2 ∫ ∂ B R ∣ u ∣ 2 m + 4 d S x ≡ − ∫ B R ∣ u ∣ 2 m + 4 d x + ( iii ) , (4) ( m + 1 ) ∫ B R ∫ ∣ x ∣ ∞ h ( s ) s u 2 d s u 2 m + 1 ( ∇ u ⋅ x ) d x + ∫ B R h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u ( ∇ u ⋅ x ) d x = 1 2 d d t t = 1 ∫ B R u 2 ( t x ) ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( t s ) d s 2 d x − ∫ B R u 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s × ∫ 0 ∣ x ∣ ( m + 1 ) s 2 u 2 m + 1 ( s ) u ′ ( s ) d s d x + ( m + 1 ) ∫ B R ∫ ∣ x ∣ ∞ h ( s ) s u 2 d s u 2 m + 1 ( ∇ u ⋅ x ) d x ≡ R 2 ∫ ∂ B R u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d S x − 2 ∫ B R u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x + ( iv ) ≡ − 2 ∫ B R u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x + ( iv ) + ( v ) ,

where (iv) is given by

− ∫ B R u 2 h ( ∣ x ∣ ) ∣ x ∣ 2 ∫ 0 ∣ x ∣ ( m + 1 ) s 2 u 2 m + 1 u ′ ( s ) d s d x + ( m + 1 ) ∫ B R ∫ ∣ x ∣ ∞ u 2 h ( s ) s d s u 2 m + 1 ( ∇ u ⋅ x ) d x .

Therefore, we deduce that

(2.10) ( i ) + b ( ii ) + d ( iii ) + c ( ( iv ) + ( v ) ) = b ∫ B R u 2 m + 2 d x + d ∫ B R ∣ u ∣ 2 m + 4 d x + 2 c ∫ B R u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x .

We note that if f ( x ) ≥ 0 is integrable on R 2 , then lim inf R → ∞ R ∫ ∂ B R f d S = 0 . In fact, if there exist N > 0 and ε > 0 such that inf R R ∫ ∂ B R f d S ∣ R > N = ε > 0 , then we obtain

∫ R 2 f ( x ) d x = ∫ 0 ∞ ∫ ∂ B R f ( s ) d s d R ≥ ∫ N ∞ ∫ ∂ B R f d s d R ≥ ∫ N ∞ ε R d R = ∞ ,

which contradicts the integrability of f . Since the integrands in the terms (i), (ii), (iii), and (v) are all nonnegative and contained in L 1 ( R 2 ) , we can take a sequence { R j } such that the terms (i), (ii), (iii), and (v) with R j replacing R converge to 0 as j → ∞ . Now it suffices to show that (iv) with R j replacing R converges to 0 as j → ∞ . Since

∫ 0 R s 2 u 2 m + 1 ( s ) u ′ ( s ) d s = 1 2 ( m + 1 ) π R 2 ∫ ∂ B R u 2 m + 2 d S x − ∫ B R u 2 m + 2 d x ,

it follows that the sequence ∫ 0 R j s 2 u 2 m + 1 ( s ) u ′ ( s ) d s converges. Applying Fubini’s theorem to the first identity of (iv), we have

∫ B R j u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) ∫ 0 ∣ x ∣ s 2 u 2 m + 1 ( s ) u ′ ( s ) d s d x = ∫ R 2 u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) χ { ∣ x ∣ ≤ R j } ∫ 0 ∞ s 2 u 2 m + 1 ( s ) u ′ ( s ) χ s ≤ ∣ x ∣ d s d x = ∫ 0 ∞ s 2 u 2 m + 1 ( s ) u ′ ( s ) χ s ≤ ∣ x ∣ ∫ R 2 u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) χ { ∣ x ∣ ≤ R j } d x d s = ∫ 0 R j s 2 u 2 m + 1 ( s ) u ′ ( s ) ∫ R 2 ⧹ B s u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) d x d s − ∫ 0 R j s 2 u 2 m + 1 ( s ) u ′ ( s ) ∫ R 2 ⧹ B R j u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) d x d s = ∫ B R j x u 2 m + 1 ( ∣ x ∣ ) ∇ u ∫ ∣ x ∣ ∞ u 2 ( s ) s h ( s ) d s d x − ∫ R 2 ⧹ B R j u 2 ( ∣ x ∣ ) ∣ x ∣ 2 h ( ∣ x ∣ ) d x ∫ 0 R j s 2 u 2 m + 1 ( s ) u ′ ( s ) d s .

Then we obtain

( iv ) = − ( m + 1 ) ∫ B R j u 2 ( ∣ x ∣ ) ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s ∫ 0 ∣ x ∣ s 2 u 2 m + 1 ( s ) u ′ ( s ) d s d x + ( m + 1 ) ∫ B R j ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 ( ∇ u ⋅ x ) d x = ( m + 1 ) ∫ R 2 ⧹ B R j u 2 ∣ x ∣ 2 h ( ∣ x ∣ ) d x ∫ 0 R j s 2 u 2 m + 1 ( s ) u ′ ( s ) d s → o ( 1 ) as j → ∞ .

Taking a limit for (2.10) with R j replacing R as j → ∞ , we obtain

b ∫ R 2 u 2 m + 2 d x + d ∫ R 2 ∣ u ∣ 2 m + 4 d x + 2 c ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x = 0 .

This completes the proof.□

Proposition 2.5

For u ∈ H r 1 ( R 2 ) , the following inequality holds:

∫ R 2 ∣ u ( ∣ x ∣ ) ∣ 2 m + 4 d x ≤ 4 ∫ R 2 ∣ ∇ u ∣ 2 d x 1 / 2 ∫ R 2 ∣ u ( ∣ x ∣ ) ∣ 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s 2 d x 1 / 2 .

Furthermore, the equality is attained by a continuum of functions

u l = ( 8 l ) 1 m + 1 ( 1 + ( m + 1 ) l 2 ∣ x ∣ 2 ) 1 m + 1 l ∈ ( 0 , + ∞ ) ,

and we have

1 4 ∫ R 2 ∣ u l ∣ 2 m + 4 d x = ∫ R 2 ∣ ∇ u l ∣ 2 d x = ∫ R 2 ∣ u l ∣ 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u l 2 m + 2 ( s ) d s 2 d x = 2 π m + 3 ( 8 l 2 ) 1 m + 1 .

Proof

Since C 0 ∞ ( R 2 ) is dense in H r 1 ( R 2 ) , it suffices to prove the inequality for u ∈ C 0 ∞ ( R 2 ) . From the Fubini theorem and Hölder’s inequality, we see that

∫ R 2 ∣ u ( ∣ x ∣ ) ∣ 2 m + 4 d x = ∫ 0 ∞ 2 π r ∣ u ( r ) ∣ 2 m + 4 d r = ∫ 0 ∞ 2 π r u 2 m + 2 ( r ) − ∫ r ∞ ( u 2 ( s ) ) ′ d s d r ≤ ∫ 0 ∞ 4 π r u 2 m + 2 ( r ) ∫ 0 ∞ ∣ u ( s ) ∣ ∣ u ′ ( s ) ∣ χ { s ≥ r } d s d r = ∫ 0 ∞ 4 π ∣ u ( s ) ∣ ∣ u ′ ( s ) ∣ ∫ 0 ∞ r u 2 m + 2 ( r ) χ { s ≥ r } d r d s = 4 ∫ R 2 ∣ ∇ u ∣ ∣ u ( ∣ x ∣ ) ∣ ∣ x ∣ ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s d x ≤ 4 ∫ R 2 ∣ ∇ u ∣ 2 d x 1 / 2 ∫ R 2 ∣ u ( ∣ x ∣ ) ∣ 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u 2 m + 2 ( s ) d s 2 d x 1 / 2 .

This shows that the inequality holds. On the other hand, by the elementary calculation, we have

∫ R 2 ∣ u l ∣ 2 m + 4 d x = ∫ 0 ∞ ∣ u l ∣ 2 m + 4 2 π r d r = 8 π m + 3 ( 8 l 2 ) 1 m + 1

and

∫ R 2 ∣ ∇ u l ∣ 2 d x = ∫ 0 ∞ ∣ ∇ u ∣ 2 2 π r d r = 2 π ( 8 l 2 ) 1 m + 1 m + 3 .

Hence, we can conclude that

1 4 ∫ R 2 ∣ u l ∣ 2 m + 4 d x = ∫ R 2 ∣ ∇ u l ∣ 2 d x = ∫ R 2 ∣ u l ∣ 2 ∣ x ∣ 2 ∫ 0 ∣ x ∣ s 2 u l 2 m + 2 ( s ) d s 2 d x = 2 π m + 3 ( 8 l 2 ) 1 m + 1 .

This completes the proof.□

3 Proof of Theorem 1.1

We consider the three cases separately.

3.1 The case of λ ∈ ( 0 , 1 )

Suppose that for given λ ∈ ( 0 , 1 ] and ω > 0 , a 4-tuple of function ( ϕ = u e i ω t , A 0 , A 1 , A 2 ) is a nontrivial standing wave solution of the form (1.5). Then, the real valued function u is a solution of the equation

(3.1) Δ u − ( m + 1 ) ( ω + ξ ) u 2 m + 1 − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

for some real ξ . From the Pohozaev identity in Proposition 2.4, it follows that

( ω + ξ ) ∫ R 2 u 2 m + 2 d x + 2 ∫ R 2 h 2 ( ∣ x ∣ ) ∣ x ∣ 2 d x − λ 2 ∫ R 2 ∣ u ∣ 2 m + 4 d x = 0 .

We multiply (3.1) by u and integrate by parts to obtain

∫ R 2 ∣ ∇ u ∣ 2 + ( m + 1 ) ( ω + ξ ) u 2 m + 2 + ( 2 m + 3 ) h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u 2 − λ 2 ( m + 2 ) ∣ u ∣ 2 m + 4 d x = 0 .

Combining the above two identities, we obtain

(3.2) J ˆ ( u ) = 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) − λ 2 ∣ u ∣ 2 m + 4 d x = 0 .

However, if λ < 1 , Proposition 2.5 implies that

J ˆ ( u ) = 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) − λ 2 ∣ u ∣ 2 m + 4 d x ≥ 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x − λ ∫ R 2 ∣ ∇ u ∣ 2 d x 1 2 ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x 1 2 > 1 2 ∫ R 2 ∣ ∇ u ∣ 2 + u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x − ∫ R 2 ∣ ∇ u ∣ 2 d x 1 2 ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x 1 2 = 1 2 ∫ R 2 ∣ ∇ u ∣ 2 d x 1 2 − ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x 1 2 2 ≥ 0

for every u ≠ 0 . This contradicts the existence of a nontrivial solution u of (3.1) when λ < 1 and proves the first assertion of our theorem.

3.2 The case of λ = 1

Proposition 2.5 tells that for each l > 0 , the function u l is a minimizer of J ˆ . Note that the Euler-Lagrange equation of J ˆ is

Δ u − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

and using (3.1), we obtain ξ = − ω . Next, we show that for λ = 1 , any standing wave solution ( ϕ , A 0 , A 1 , A 2 ) of the form (1.5) with ∣ ϕ ∣ > 0 is exactly given by the following explicit formula:

(3.3) ( ϕ , A 0 , A 1 , A 2 ) = 8 l 1 + ( m + 1 ) ∣ l x ∣ 2 1 m + 1 e i ω t , 1 2 8 l 1 + ( m + 1 ) ∣ l x ∣ 2 2 m + 1 − ω , 2 l 2 x 2 1 + ( m + 1 ) ∣ l x ∣ 2 , − 2 l 2 x 1 1 + ( m + 1 ) ∣ l x ∣ 2

for a constant l > 0 . It is not difficult to see that 4-tuples of functions in (3.3) satisfy (1.1)–(1.4) for all l > 0 . It is also clear that 4-tuples (3.3) are of the form (1.5) and ∣ ϕ ∣ > 0 .

To prove the opposite direction, suppose that ( ϕ , A 0 , A 1 , A 2 ) is a standing wave solution of the form (1.5) and ∣ ϕ ∣ > 0 . Then, we know that the real function u satisfies (3.1) for ξ = − ω when λ = 1 . We see from elementary calculations that

1 2 ∫ R 2 ∣ ∂ 1 u ∣ 2 + ∣ ∂ 2 u ∣ 2 + ( A 1 2 + A 2 2 ) u 2 d x − 1 4 ∫ R 2 ∣ u ∣ 2 m + 4 d x = 1 2 ∫ R 2 ( ∂ 1 u − A 2 u ) 2 + ( ∂ 2 u + A 1 u ) 2 d x + 1 2 ∫ R 2 ∂ 1 ( A 2 u 2 ) + ∂ 2 ( − A 1 u 2 ) d x + 1 2 ∫ R 2 ( ∂ 2 A 1 − ∂ 1 A 2 ) u 2 d x − 1 4 ∫ R 2 ∣ u ∣ 2 m + 4 d x ,

where we use the divergence theorem and the fact u ∈ L 2 ( R 2 ) , A j ∈ L ∞ ( R 2 ) , and ∫ R 2 ∂ 1 ( A 2 u 2 ) + ∂ 2 ( − A 1 u 2 ) d x = 0 . Also, we recall that ∂ 1 A 2 − ∂ 2 A 1 = − 1 2 ∣ u ∣ 2 m + 2 . Therefore, we obtain

J ˆ ( u ) = 1 2 ∫ R 2 ( ∂ 1 u − A 2 u ) 2 + ( ∂ 2 u + A 1 u ) 2 d x .

Hence, from (3.2) we obtain the following relations:

(3.4) ∂ 1 u = A 2 u , ∂ 2 u = − A 1 u .

Then by a change of variable v = 2 ( m + 1 ) log u , the relation (3.4) is equivalent to

∂ 1 v = 2 ( m + 1 ) A 2 , ∂ 2 v = − 2 ( m + 1 ) A 1 ,

which implies that

Δ v + ( m + 1 ) e v = 0 .

Note that v solves the following initial value problem:

(3.5) v ″ + 1 r v ′ + ( m + 1 ) e v = 0 , v ( 0 ) = 2 ( m + 1 ) log u ( 0 ) , v ′ ( 0 ) = 0 .

We can check that (3.5) is solved by the following solution:

(3.6) v ( r ) = − 2 log ( m + 1 ) u 2 m + 2 ( 0 ) 8 + 1 + 2 ( m + 1 ) log u ( 0 )

and recall that initial value problem (3.5) has a unique solution. This means v is identically same as (3.6). So we arrive at

u ≡ 8 l 1 + ( m + 1 ) ∣ l x ∣ 2 1 m + 1 ,

where l 2 = u 2 m + 2 ( 0 ) 8 . Now, we can directly calculate A 0 , A 1 , and A 2 from u to obtain (3.3).

3.3 The case of λ > 1

We introduce the following constrained minimization problems:

(3.7) I = inf u ∈ M ∫ R 2 u 2 m + 2 d x , M = { u ∈ H r 1 ( R 2 ) \ { 0 } ∣ J ˆ ( u ) = 0 } .

We need to prove the constraint M is nonempty. In fact, taking a minimizer u l in Proposition 2.5, we obtain J ˆ ( u l ) < 0 . We note that

J ˆ ( t u l ) = 1 2 ∫ R 2 ∣ ∇ u l ∣ 2 d x t 2 + ∫ R 2 u l 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x t 4 m + 6 − ( m + 1 ) ∫ R 2 ∣ u l ∣ 2 m + 4 d x t 2 m + 4 .

Since lim t → 0 J ˆ ( t u l ) / t 2 = 1 2 ∫ R 2 ∣ ∇ u l ∣ 2 d x , there exists t 0 ∈ ( 0 , 1 ) such that J ˆ ( t 0 u l ) = 0 . This implies that t 0 u l ∈ M . Thus, M is nonempty.

The minimization problem (3.7) lacks some compactness since both the functional ∫ R 2 u 2 d x and the constraint M are invariant under the transformation

u ↦ u t ( x ) ≔ t 1 m + 1 u ( t x ) , t > 0 ,

since

∫ R 2 u t 2 m + 2 d x = ∫ R 2 t 2 u 2 m + 2 ( t x ) d x = ∫ R 2 u 2 m + 2 d x

and

J ˆ ( u t ) = t 2 m + 1 1 2 ∫ R 2 ∣ ∇ u ∣ 2 d x + 1 2 ∫ R 2 u 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x − λ 4 ∫ R 2 ∣ u ∣ 2 m + 4 d x .

We overcome this by a renormalization argument as follows. Let u n ≥ 0 be a minimizing sequence of (3.7). By the invariance above, we see that for t n = ∫ R 2 ∣ ∇ u n ∣ 2 d x − m + 1 2 , a sequence { v n ( x ) = t n 1 m + 1 u n ( t n x ) } n ∈ M is a minimizing sequence of (3.7) with ∫ R 2 ∣ ∇ v n ∣ 2 d x = 1 . Now we can see { v n } is bounded in H r 1 ( R 2 ) . Taking a subsequence if necessary, we may assume that there exists a nonnegative function u 0 ∈ H r 1 ( R 2 ) such that as n → ∞ , v n ⇀ u 0 in H r 1 ( R 2 ) weakly and v n → u 0 in L q ( R 2 ) strongly for all q > 2 . If u 0 = 0 , then we have lim n → ∞ J ˆ ( v n ) = 1 2 . This contradicts the conditions for the constraint M . Thus, we obtain u 0 ≠ 0 . Define

g ( t ) ≡ J ˆ ( t u 0 ) = 1 2 ∫ R 2 ∣ ∇ u 0 ∣ 2 d x t 2 + ∫ R 2 u 0 2 ∣ x ∣ 2 h 2 ( ∣ x ∣ ) d x t 4 m + 6 − ( m + 1 ) ∫ R 2 ∣ u 0 ∣ 2 m + 4 d x t 2 m + 4 .

Since J ˆ ( u ) is weakly lower semi-continuous with respect to u , we have g ( 1 ) ≤ 0 . If g ( 1 ) = 0 , u 0 is a minimizer since ∫ R 2 u 0 2 m + 2 ≤ lim n → ∞ ∫ R 2 v n 2 m + 2 d x . Now suppose that g ( 1 ) < 0 . Then we have already seen above that there exists a t 0 ∈ ( 0 , 1 ) such that g ( t 0 ) = 0 . This means that t 0 u 0 ∈ M and

∫ R 2 ( t 0 u 0 ) 2 m + 2 d x < ∫ R 2 u 0 2 m + 2 d x ≤ I ,

which is a contradiction. Therefore, g ( 1 ) = 0 and u 0 is a minimizer.

Now, we have an alternative:

There exists a nonnegative u ∈ M such that J ˆ ′ ( u ) = 0 .
J ˆ ′ ( u ) ≠ 0 for any u ∈ M .

Suppose that the case (A) occurs. Then we have a positive solution u ∈ H r 1 ( R 2 ) of

(3.8) Δ u − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 .

For a transformed function u t ( x ) = t 1 m + 1 u ( t x ) , t > 0 , we can see

Δ u t − ( m + 1 ) ∫ ∣ x ∣ ∞ h t ( s ) s u t 2 d s u t 2 m + 1 − h t 2 ( ∣ x ∣ ) ∣ x ∣ 2 u t + λ 2 ( m + 2 ) ∣ u t ∣ 2 m + 2 u t = t 2 m + 3 m + 1 Δ u − ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

where h t ( r ) = ∫ 0 r r 2 u t 2 m + 2 ( s ) d s . Thus, it follows that u t is also a positive solution of (3.8) for all t > 0 . Then, for any given frequency ω > 0 , a 4-tuple

( ϕ ω , A 0 ω , A 1 ω , A 2 ω ) = u ω e i ω t , ∫ ∣ x ∣ ∞ h ω ( s ) s u ω 2 ( s ) d s − ω , x 2 ∣ x ∣ 2 h ω ( ∣ x ∣ ) , − x 1 ∣ x ∣ 2 h ω ( ∣ x ∣ )

is a standing wave solution of the form (1.5) to equations (1.1)–(1.4). Since u ω is not identically same with u ω ˜ for different ω ≠ ω ˜ , the standing waves ( ϕ ω , A 0 ω , A 1 ω , A 2 ω ) are not gauge equivalent as the standing wave of the form (1.5) for different frequency ω .

Suppose that the case (B) occurs. Then there exists a Lagrange multiplier ω ∗ ∈ R such that

u 2 m + 1 = ω ∗ ( m + 1 ) Δ u − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u .

Since u 0 ≠ 0 , we obtain ω ∗ ≠ 0 . Denoting ω 0 − 1 = ω ∗ , we have

Δ u − ( m + 1 ) ω 0 u 2 m + 1 − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 .

From the fact that the functional ∫ R 2 u 2 m + 2 d x and constraint M are invariant under the transformation u t ( x ) = t 1 m + 1 u ( t x ) , we deduce that u t is also a minimizer of problem (3.7) for all t > 0 and u t satisfies

Δ u t − ( m + 1 ) ω t u t 2 m + 1 − ( m + 1 ) ∫ ∣ x ∣ ∞ h t ( s ) s u t 2 ( s ) d s u t 2 m + 1 − h t 2 ( ∣ x ∣ ) ∣ x ∣ 2 u t + λ 2 ( m + 2 ) ∣ u t ∣ 2 m + 2 u t = 0 ,

for some ω t > 0 . Note that the aforementioned equation is equivalent to

Δ u − ( m + 1 ) ω t t 2 m + 1 u 2 m + 1 − ( m + 1 ) ∫ ∣ x ∣ ∞ h ( s ) s u 2 ( s ) d s u 2 m + 1 − h 2 ( ∣ x ∣ ) ∣ x ∣ 2 u + λ 2 ( m + 2 ) ∣ u ∣ 2 m + 2 u = 0 ,

from which we obtain ω t = t 2 m + 1 ω 0 . Therefore, for any given frequency ω > 0 and σ ( ω ) ≡ ω ω 0 m + 1 2 , a 4-tuple

( ϕ ω , A 0 ω , A 1 ω , A 2 ω ) = u σ ( ω ) e i ω t , ∫ ∣ x ∣ ∞ h σ ( ω ) ( s ) s u σ ( ω ) 2 ( s ) d s , x 2 ∣ x ∣ 2 h σ ( ω ) ( ∣ x ∣ ) , − x 1 ∣ x ∣ 2 h σ ( ω ) ( ∣ x ∣ )

is a standing wave solution of the form (1.5) to equations (1.1)–(1.4). Finally, the standing waves ( ϕ ω , A 0 ω , A 1 ω , A 2 ω ) are not gauge equivalent as the standing wave of the form (1.5) for different frequency ω because u σ ( ω ) is not identically same with u σ ( ω ˜ ) .

Funding information: H. Huh was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (2020R1F1A1A01072197). Y. Jin was supported by the National Natural Science Foundation of China (No. 11761074) and the Jilin Science and Technology Development for Leading Talent of Science and Technology Innovation in Middle and Young and Team Project (No. 20200301053RQ). G. Jin was supported by the Jilin Science and Technology Development Program (No. YDZJ202201ZYTS311) and the National Natural Science Foundation of China (No. 11961073).
Conflict of interest: The authors state no conflict of interest.

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Received: 2021-11-16

Accepted: 2022-07-14

Published Online: 2022-09-13

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https://doi.org/10.1515/anona-2022-0261

Keywords for this article

generalized Jackiw-Pi model; standing waves; variational method

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