Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 467985, 5 pages http://dx.doi.org/10.1155/2013/467985 Research Article Nonexistence Results of Semilinear Elliptic Equations Coupled with the Chern-Simons Gauge Field Hyungjin Huh Department of Mathematics, Chung-Ang University, Seoul 156-756, Republic of Korea Correspondence should be addressed to Hyungjin Huh; huh@cau.ac.kr Received 1 November 2012; Accepted 15 January 2013 Academic Editor: Khalil Ezzinbi Copyright © 2013 Hyungjin Huh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We discuss the nonexistence of nontrivial solutions for the Chern-Simons-Higgs and Chern-Simons-SchroΜdinger equations. The Derrick-Pohozaev type identities are derived to prove it. 1. Introduction and Main Results In this paper, we are concerned with the nonexistence of nontrivial solutions to some elliptic equations coupled with Chern-Simons gauge field. More precisely, let us first consider the following system: 2 σ΅¨ σ΅¨2 −(π + π΄ 0 ) π − π·1 π·1 π − π·2 π·2 π + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π = 0, 2 (1) (2) π2 π΄ 0 = Im (ππ·1 π) , (3) σ΅¨ σ΅¨2 π1 π΄ 2 − π2 π΄ 1 = − (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ , (4) which is derived from the system (5) with stationary solution ansatz π(π‘, π₯) = ππππ‘ π(π₯), π ∈ C, and π΄ π (π‘, π₯) = π΄ π (π₯) for π = 0, 1, 2. Consider σ΅¨ σ΅¨2 π·0 π·0 π − (π·1 π·1 + π·2 π·2 ) π + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π = 0, π0 π΄ 2 − π2 π΄ 0 = − Im (ππ·1 π) , σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ πΈ (π‘) = ∫ ∑ σ΅¨σ΅¨σ΅¨π·πΌ π (π‘, π₯)σ΅¨σ΅¨σ΅¨ + π (σ΅¨σ΅¨σ΅¨π (π‘, π₯)σ΅¨σ΅¨σ΅¨ ) ππ₯ = πΈ (0) . R2 πΌ=0 π1 π΄ 0 = − Im (ππ·2 π) , π0 π΄ 1 − π1 π΄ 0 = Im (ππ·2 π) , The Chern-Simons-Higgs system in (5) was introduced in [1, 2] to deal with the electromagnetic phenomena in planar domain such as fractional quantum Hall effect or high temperature superconductivity. The system in (5) has the conservation of the total energy (6) The special case with a self-dual potential π(|π|2 ) = (1/4) |π| (|π|2 − 1)2 has received much attention and has been studied by several authors, where one can derive the following system of first-order equations called self-dual equations (see [1, 2]) 2 π·1 π − ππ·2 π = 0, 1 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 π1 π΄ 2 − π2 π΄ 1 + σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − 1) = 0, 2 (5) π1 π΄ 2 − π2 π΄ 1 = − Im (ππ·0 π) , where π0 = π/ππ‘, π1 = π/ππ₯1 , π2 = π/ππ₯2 for (π‘, π₯1 , π₯2 ) ∈ R1+2 , π : R1+2 → C is the complex scalar field, π΄ π : R1+2 → R is the gauge field, π·π = ππ + ππ΄ π is the covariant derivative for π = 0, 1, 2, and π denotes the imaginary unit. π + π΄0 − (7) π σ΅¨σ΅¨ σ΅¨σ΅¨2 (σ΅¨πσ΅¨ − 1) π = 0. 2 σ΅¨ σ΅¨ We note that solutions to the self-dual equations (7) provide solutions to (1)–(4). For the self-dual potential π(|π|2 ) = (1/4)|π|2 (|π|2 − 1)2 , there are two possible boundary conditions to make the energy finite; either |π| → 1 or |π| → 0 as |π₯| → ∞. The former boundary condition is called “topological” while the latter “non-topological.” A lot 2 Abstract and Applied Analysis of works have been done for the existence of solutions to the self-dual system [3–7]. Some existence results for the nonselfdual Chern-Simons-Higgs equations with the topological boundary condtion have been proved in [8–10]. From the mathematical point of view, it is meaningful to study existence and nonexistence of nontrivial solutions under various conditions on π. In this paper, we are concerned with the nonexistence of the non-trivial solution to (1)–(4) with the non-topological boundary condtion. The following is our first result. Theorem 1. Let (π, π΄ 0 , π΄ 1 , π΄ 2 ) be a classical solution of (1)– (4) such that π ∈ π»1 (R2 ) and π΄ 0 ∈ πΏπ (R2 ), π΄ 1 , π΄ 2 ∈ πΏπ (R2 ) for any 2 < π, π ≤ ∞. Let π : R → R be a πΆ1 function such that π(0) = 0, inf{π₯ > 0 | π(π₯) =ΜΈ 0} = 0 and π(|π|2 ), |π|2 πσΈ (|π|2 ) ∈ πΏ1 (R2 ). Assume that σΈ ππ (π ) π − π (π ) ≥ 0 ∀π ≥ 0, (8) where 0 ≤ π ≤ 1 is a constant. Then, one has π ≡ 0. The proof is based on the following Derrick-Pohozaev type identity for (1)–(4): 2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ π (σ΅¨σ΅¨σ΅¨π·1 πσ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π·2 πσ΅¨σ΅¨σ΅¨ ) + (1 − π) (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ R2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 + ππ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ = 0. (9) σΈ As a typical example, we consider π(|π|2 ) = πΌ|π|6 + π½|π|4 . Then it is easy to check that ππσΈ (π )π − π(π ) = πΌ(3π − 1)π 3 + π½(2π−1)π 2 . If one of the following conditions is satisfied, then we have π ≡ 0. (1) For πΌ > 0, π½ > 0, we take 1/2 < π < 1. (2) For πΌ > 0, π½ < 0, we take 1/3 < π < 1/2. Note that for the self-dual potential π(|π|2 ) = (1/4)|π|2 (|π|2 − 1)2 , we have 3π − 1 3 2π − 1 2 π − 1 π − π + π 4 2 4 (10) which is not nonnegative for π ≥ 0. The following Chern-Simons gauged SchroΜdinger system was proposed in [11] when the second quantized π body anyon problem is considered π0 π΄ 1 − π1 π΄ 0 = − Im (ππ·2 π) , π1 π΄ 2 − π2 π΄ 1 = − 1 σ΅¨σ΅¨ σ΅¨σ΅¨2 σ΅¨πσ΅¨ . 2 σ΅¨ σ΅¨ (12) π1 π΄ 0 = Im (ππ·2 π) , (13) π2 π΄ 0 = − Im (ππ·1 π) , (14) 1 σ΅¨ σ΅¨2 π1 π΄ 2 − π2 π΄ 1 = − σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ . 2 (15) In the special case with the potential π(|π|2 ) = −(1/2)|π|4 , we can derive the following self dual equations [11–13] π·1 π + ππ·2 π = 0, 1 σ΅¨ σ΅¨2 π + π΄ 0 − σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ = 0, 2 (16) 1 σ΅¨ σ΅¨2 π1 π΄ 2 − π2 π΄ 1 + σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ = 0. 2 Note that solutions to the self-dual system (16) provide solutions to (12)–(15). The self-dual equations (16) can be transformed into the Liouville equation, an integrable equation whose solutions are explicitly known. For the nonself-dual potential of the form π(|π|2 ) = −π|π|π (π > 0, π > 2), some existence and nonexistence results have been studied in [14, 15] under the condition of the radially symmetric solution π(π‘, π₯) = ππππ‘ π’(|π₯|) (π’ ∈ R). We prove the following nonexistence result, under various conditions on π, for (12)–(15). Theorem 2. Let (π, π΄ 0 , π΄ 1 , π΄ 2 ) be a classical solution of (1)– (4) such that π ∈ π»1 (R2 ), π΄ 0 ∈ πΏπ (R2 ) and π΄ 1 , π΄ 2 ∈ πΏπ (R2 ) for 1 < π ≤ ∞ and 2 < π ≤ ∞. One also assumes that π : R → R is a πΆ1 function such that π(0) = 0 and π(|π|2 ), |π|2 πσΈ (|π|2 ) ∈ πΏ1 (R2 ). πσΈ (π ) π − π (π ) ≥ 0 ∀π ≥ 0, (17) then one has π ≡ 0. (2) Suppose that π is a real-valued function; that is, π(π₯) = π’(π₯) ∈ R and 1/π + 1/π = 1 for 2 < π < ∞. If the potential π satisfies, for a constant β ≥ 2/3, π (β − 1) π + βπσΈ (π ) π − π (π ) ≥ 0 ∀π ≥ 0, (18) then one has π’ ≡ 0. The proof is based on the Derrick-Pohozaev type identities (40) and (45) for (12)–(15). σ΅¨ σ΅¨2 ππ·0 π + (π·1 π·1 + π·2 π·2 ) π − πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π = 0, π0 π΄ 2 − π2 π΄ 0 = Im (ππ·1 π) , σ΅¨ σ΅¨2 (π + π΄ 0 ) π − π·1 π·1 π − π·2 π·2 π + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π = 0, (1) If the potential V satisfies (3) For πΌ < 0, π½ < 0, we take 0 < π < 1/3. ππσΈ (π ) π − π (π ) = With the stationary solution ansatz π(π‘, π₯) = ππππ‘ π(π₯), π ∈ C and π΄ π (π‘, π₯) = π΄ π (π₯) for π = 0, 1, 2, we arrive at (11) Example 3. For the static solution (π = 0), we consider the potential π(π ) = π 3 − π 2 − π . Then, taking β = 2/3, we can check 2 σΈ 1 1 π (|π’|2 ) |π’|2 − π (|π’|2 ) = |π’|6 − |π’|4 + |π’|2 ≥ 0. (19) 3 3 3 Abstract and Applied Analysis 3 However we have, for the complex solution, σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨6 σ΅¨ σ΅¨4 σ΅¨ σ΅¨2 πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) = 2σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ For II, we have (20) which is not nonnegative. The paper is organized as follows. In Section 2, we prove Theorem 1 by deriving Derrick-Pohozaev type identity. Theorem 2 is proved in Section 3. We conclude this section by giving a few notations. σ΅¨ σ΅¨2 II = ∫ ππ (π₯π π·π ππ·π π) − σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ − π₯π π·π ππ·π π·π πππ₯ π΅π = ∫ π₯π π₯π π ππ΅π π·π ππ·π ππππ σ΅¨ σ΅¨2 − ∫ σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ ππ₯ − ∫ π₯π π·π π (π·π π·π π + ππΉππ π)ππ₯, π΅ π΅ π π (24) (i) π»1 (R2 ) denotes the usual Sobolev space π1,2 (R2 ). (ii) π΅π := {π₯ ∈ R2 | |π₯| ≤ π } and ππ΅π := {π₯ ∈ R2 | |π₯| = π }. where we used the notation πΉππ = ππ π΄ π − ππ π΄ π and the following identity: π·π π·π π = π·π π·π π + ππΉππ π. (iii) πππ := the surface measure on ππ΅π . (25) Taking the real parts and integrating by parts, we obtain 2. Proof of Theorem 1 π₯π π₯π σ΅¨ σ΅¨2 π·π ππ·π π πππ − ∫ σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ ππ₯ π π΅π Re {II} = ∫ ππ΅π We apply Derrick-Pohozaev argument to derive the identity (9) which prove Theorem 1. From now on, we adopt the summation convention for repeated indices. Suppose that (π, π΄ 0 , π΄ 1 , π΄ 2 ) is a solution of (1)–(4). Multiplying (1) by π₯π π·π π and integrating over π΅π , we obtain −∫ π΅π 1 σ΅¨ σ΅¨2 π₯π ππ (σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ ) ππ₯ 2 − ∫ π₯π πΉππ Im (ππ·π π) ππ₯ π΅π 2 − ∫ (π + π΄ 0 ) π π₯π π·π π ππ₯ − ∫ π·π π·π π π₯π π·π π ππ₯ π΅π π΅π σ΅¨ σ΅¨2 + ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯π π·π π ππ₯ = 0. π΅ π₯π π₯π = ∫ π ππ΅π (21) σΈ +∫ π π΅π Now we set π·π ππ·π π πππ − ∫ ππ΅π π σ΅¨σ΅¨ σ΅¨2 σ΅¨σ΅¨π· πσ΅¨σ΅¨σ΅¨ ππ 2σ΅¨ π σ΅¨ π 1 σ΅¨σ΅¨ σ΅¨σ΅¨2 σ΅¨ σ΅¨2 2 σ΅¨πσ΅¨ π₯ π (π΄ ) ππ₯ + ∫ πσ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ π₯π ππ π΄ 0 ππ₯, 2σ΅¨ σ΅¨ π π 0 π΅π (26) where we used (2)–(4) in the following way: 2 I = ∫ (π + π΄ 0 ) ππ₯π π·π πππ₯, − ∫ π₯1 πΉ21 Im (ππ·2 π) + π₯2 πΉ12 Im (ππ·1 π) ππ₯ π΅π π΅π II = ∫ π·π π·π ππ₯π π·π πππ₯, π΅π π σ΅¨ σ΅¨2 III = ∫ πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯π π·π πππ₯. π΅ Combining (23) and (26), we have from the identity (21) π Then, integrating by parts and taking real parts, we have Re {I} = ∫ ππ΅π π΅π Re {III} = ∫ ππ΅π 1 σ΅¨ σ΅¨2 π₯ π (π΄2 ) σ΅¨σ΅¨πσ΅¨σ΅¨ ππ₯, 2 π π 0 σ΅¨ σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨2 σ΅¨ σ΅¨2 π (σ΅¨σ΅¨πσ΅¨σ΅¨ ) πππ₯ − ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯. 2 π΅π π − π −∫ π π₯π π₯π ππ΅π 2 σ΅¨ σ΅¨2 − ∫ (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ π΅ π 2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ − ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ π΅ π΅ =∫ π 2 σ΅¨ σ΅¨2 (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ πππ₯ 2 σ΅¨ σ΅¨2 − ∫ πππ π΄ 0 π₯π σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ π΅ (27) σ΅¨ σ΅¨2 = ∫ π₯π σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ (π + π΄ 0 ) ππ π΄ 0 ππ₯. π΅ (22) (23) π Re (π·π π π·π π) − π σ΅¨σ΅¨ σ΅¨2 σ΅¨σ΅¨π· πσ΅¨σ΅¨σ΅¨ 2σ΅¨ π σ΅¨ (28) π σ΅¨σ΅¨ σ΅¨σ΅¨2 π 2 σ΅¨ σ΅¨2 π (σ΅¨σ΅¨πσ΅¨σ΅¨ ) + (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ πππ . 2 2 Thus we have σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨∫ (π + π΄ 0 )2 σ΅¨σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨σ΅¨2 − π (σ΅¨σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨σ΅¨2 ) ππ₯σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ π΅π σ΅¨ ≤ πΆ∫ ππ΅π σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 π (σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + π2 σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 +σ΅¨σ΅¨σ΅¨π΄ 0 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ + π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ )) πππ , (29) 4 Abstract and Applied Analysis where πΆ is a positive constant. Considering the Sobolev embedding and the condition of Theorem 1, we know that |π·π π|2 , π2 |π|2 , |π΄ 0 |2 |π|2 , π(|π|2 ) ∈ πΏ1 (R2 ). Applying the idea in [16], we know that there exists a sequence {π π } → ∞ such that ∫ ππ΅π π Now we set I = ∫ (π + π΄ 0 ) π π₯π π·π π ππ₯, π΅π II = ∫ π·π π·π π π₯π π·π π ππ₯, (35) π΅π σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 π π (σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + π2 σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π΄ 0 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ + π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ )) πππ π σ΅¨ σ΅¨2 III = ∫ πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π π₯π π·π π ππ₯. π΅π σ³¨→ 0, (30) Re {I} = ∫ and consequently ππ΅π 2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ R2 = lim ∫ π→∞ π΅ π π Then, integrating by parts and taking real parts, we have π΅π (31) σ΅¨ σ΅¨2 0 = ∫ π (−(π + π΄ 0 ) π − π·π π·π π + π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π) ππ₯ 2 R −∫ π΅π (32) σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 + ππσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ = 0, π σ΅¨σ΅¨ σ΅¨σ΅¨2 σ΅¨ σ΅¨2 π (σ΅¨σ΅¨πσ΅¨σ΅¨ ) πππ₯ − ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯. 2 π΅π σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) + (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ π΅π =∫ ππ΅π − π₯π π₯π π Re (π·π π π·π π) + π σ΅¨σ΅¨ σ΅¨2 σ΅¨σ΅¨π· πσ΅¨σ΅¨σ΅¨ 2σ΅¨ π σ΅¨ (37) π σ΅¨σ΅¨ σ΅¨σ΅¨2 π σ΅¨ σ΅¨2 π (σ΅¨σ΅¨πσ΅¨σ΅¨ ) + (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ πππ . 2 2 Applying the same argument in Section 2, the right hand side of the above equality vanishes. Then we conclude that + Repeating the similar argument to the proof of Theorem 1, we derive Derrick-Pohozaev type identities for (12)–(15). Suppose that (π, π΄ 0 , π΄ 1 , π΄ 2 ) is a solution of (12)–(15). Multiplying (12) by π₯π π·π π and integrating over π΅π , we obtain ∫ (π + π΄ 0 ) π π₯π π·π π ππ₯ − ∫ π·π π·π π π₯π π·π π ππ₯ π 1 σ΅¨ σ΅¨2 π₯ π π΄ σ΅¨σ΅¨πσ΅¨σ΅¨ ππ₯, 2 π π 0σ΅¨ σ΅¨ ππ΅π (33) 3. Proof of Theorem 2 σ΅¨ σ΅¨2 + ∫ πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) π π₯π π·π π ππ₯ = 0. π΅ ππ΅π π σ΅¨σ΅¨ σ΅¨2 σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨ πππ σ΅¨ σ΅¨ 2 Then we have from the identity (34) where 0 ≤ π ≤ 1 is a constant. We are ready to prove Theorem 1. (1) For 0 < π ≤ 1, we have π·π π ≡ 0. If π(π₯0 ) =ΜΈ 0, then there exists πΏ > 0 such that π(π₯) =ΜΈ 0 for π΅π₯0 (πΏ) = {π₯ | |π₯ − π₯0 | < πΏ}. In the region π΅π₯0 (πΏ), we have π΄ π = π(ππ π/π). Using (4) we have π΄ 0 (π₯) + π = 0 in π΅π₯0 (πΏ). On the other hand, from (2) and (3), we deduce that π΄ 0 (π₯) = constant = −π. By (1), we obtain πσΈ (|π|2 )π = 0 for all π₯ ∈ R2 . By the condition of πσΈ and π ∈ πΏ2 , we conclude that π ≡ 0. (2) For π = 0, we have π(|π|2 ) = 0. By the condition of π and π ∈ πΏ2 , we have π ≡ 0. π΅π Re {III} = ∫ π·π ππ·π π πππ − ∫ (36) by taking care of the boundary integral terms as before. Combining (31) and (32), we obtain σ΅¨ σ΅¨2 2 σ΅¨ σ΅¨2 ∫ πσ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + (1 − π) (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ 2 R π ππ΅π σΈ σ΅¨ σ΅¨2 2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 = ∫ −(π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯, 2 R π₯π π₯π Re {II} = ∫ On the other hand, we know from (1) that π΅π 1 σ΅¨ σ΅¨2 π₯ π π΄ σ΅¨σ΅¨πσ΅¨σ΅¨ ππ₯, 2 π π 0σ΅¨ σ΅¨ −∫ 2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ = 0. 2 π σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ πππ₯ − ∫ (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ 2 π΅π (34) σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) + (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ = 0. R2 (38) On the other hand, we know from (12) σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ (π + π΄ 0 ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ππ₯ = 0. R2 (39) Combining (38) and (39), we end up with σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 σ΅¨ σ΅¨2 ∫ σ΅¨σ΅¨σ΅¨σ΅¨π·π πσ΅¨σ΅¨σ΅¨σ΅¨ + πσΈ (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ − π (σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨ ) ππ₯ = 0. 2 R (40) Following the reasoning in Theorem 1, we deduce π ≡ 0 from the fact π·π π ≡ 0. For the proof of the second result in Theorem 2, we assume π(π₯) = π’(π₯) ∈ R. Then (13)–(15) can be rewritten by π1 π΄ 0 = π΄ 2 π’2 , π2 π΄ 0 = −π΄ 1 π’2 , 1 π1 π΄ 2 − π2 π΄ 1 = − π’2 . 2 (41) Abstract and Applied Analysis 5 It is easy to check the following identity: 1 π1 (π΄ 2 π΄ 0 ) − π2 (π΄ 1 π΄ 0 ) = (π΄21 + π΄22 − π΄ 0 ) π’2 , 2 (42) from which we derive, with the condition π΄ 0 ∈ πΏπ , π΄ 1 , π΄ 2 ∈ πΏπ for 1/π + 1/π = 1, 2 < π < ∞, ∫ R2 1 π΄ π’2 ππ₯ = ∫ (π΄21 + π΄22 ) π’2 ππ₯. 2 0 π 2 (43) Considering |π·π π’|2 = |∇π’|2 + (π΄21 + π΄22 )π’2 , we have from (38) and (39) ∫ π (|π’|2 ) + π|π’|2 + 2 (π΄21 + π΄22 ) π’2 ππ₯ = 0, R2 ∫ |∇π’|2 + 3 (π΄21 + π΄22 ) π’2 + π|π’|2 + πσΈ (|π’|2 ) |π’|2 ππ₯ = 0. R2 (44) Then we obtain, for a constant β ≥ 2/3, ∫ β|∇π’|2 + (3β − 2) (π΄21 + π΄22 ) π’2 + π (β − 1) |π’|2 R2 σΈ 2 2 (45) 2 + βπ (|π’| ) |π’| − π (|π’| ) ππ₯ = 0, which proves the second result in Theorem 2. 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