The Schrödinger-Maxwell system with Dirac mass

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The Schrödinger–Maxwell system with Dirac mass

Le système de Schrödinger–Maxwell avec masse de Dirac

G.M. Coclite

, H. Holden

aCentre of Mathematics for Applications, University of Oslo, P.O. Box 1053, Blindern, NO-0316 Oslo, Norway bDepartment of Mathematics, University of Bari, Via E. Orabona 4, I-70125 Bari, Italy

cDepartment of Mathematical Sciences, Norwegian University of Science and Technology, NO-7491 Trondheim, Norway Received 9 January 2006; accepted 9 June 2006

Available online 18 December 2006

Abstract

We study a nonrelativistic charged quantum particle moving in a bounded open setΩ⊂R³with smooth boundary under the action of a zero-range potential. In the electrostatic case the standing wave solution takes the formψ (t, x)=u(x)e^−iωtwhereu formally satisfies−u+αϕu−_β¹δ_x0u=ωuand the electric potentialϕis given by−ϕ=u². We give a rigorous definition of this problem and show that it has a weak nontrivial solution.

©2006 Elsevier Masson SAS. All rights reserved.

Résumé

Nous étudions une particule quantique chargée et non relativiste se déplaçant dans un domaine ouvert bornéΩ⊂R³au contour régulier sous l’action d’un potentiel concentré en un point. Dans le cas electrostatique, l’onde solution prends la formeψ (t, x)= u(x)e^−iωt oùusatisfait formellement−u+αϕu−_β¹δx₀u=ωuet le potentiel électriqueϕest donné par−ϕ=u². Nous donnons une définition rigoureuse du problème et montrons qu’il possède une solution faible non triviale.

©2006 Elsevier Masson SAS. All rights reserved.

MSC:primary 35D05, 35Q40; secondary 35J65

Keywords:Schrödinger–Maxwell system; Point interaction

1. Introduction

Consider a nonrelativistic charged quantum particle moving in a bounded open setΩ⊂R³with smooth boundary under the action of a short-range potential centered atx0∈Ω. Denote byψ=ψ (t, x),ϕ=ϕ(t, x)andA=A(t, x) the wave function, the electric potential and the vector potential, respectively, generated by the particle.

✩ Partially supported by the Research Council of Norway.

* Corresponding author.

E-mail addresses:[email protected] (G.M. Coclite), [email protected] (H. Holden).

URL:http://www.math.ntnu.no/~holden/ (H. Holden).

0294-1449/$ – see front matter ©2006 Elsevier Masson SAS. All rights reserved.

doi:10.1016/j.anihpc.2006.06.005

We consider only standing wave solutions in the electrostatic case, i.e., ψ (t, x)=u(x)e^−iωt, ϕ=ϕ(x), A=0,

where

u:Ω−→R, ω∈R.

Arguing as in [12] we have thatu, ϕare solutions of the following system

⎧⎪

⎪⎨

⎪⎪

⎩

−u+αϕu− 1

βδ_x0u=ωu, inΩ,

−ϕ=u², inΩ,

u=ϕ=0, on∂Ω,

(1.1)

whereδ_x0 is Dirac’s delta function located atx₀and

α, β >0 are constants. (1.2)

For the sake of notational simplicity we have scaled all physical constants intoα, β.

Recall that the case of uncharged particles (i.e.,α=0) is treated in [5], while the case of charged particles with β= ∞ is discussed in [8], and a charged particle in the full spaceΩ=R³ andβ= ∞can be found in [10,12].

Furthermore, the case of a nonlinear external vector field,β= ∞, Ω=R³is analyzed in [11]. Under the assumption β= ∞the case of a relativistic particle is considered in [9,14]. A nonlinear version of (1.1) withα=0, β=β(ψ )is studied in [1–3]. Finally, classical papers on the coupling of the Maxwell and the Schrödinger equations are [6,7,13].

Eq. (1.1) is not well-defined as it stands due to the presence of Dirac’s delta function. However, one can rigor- ously define the operator−−_β¹δx₀ as a self-adjoint operator onL²(Ω)by suitably rescaling the coupling constant multiplying the delta function. See [5, Chapter I.1, Appendix K.2.1] for an extensive discussion of this. The actual definition of the self-adjoint operator depends strongly on the dimension of the underlying physical spaceΩ⊂R³. LetLβ,x0 denote this self-adjoint operator. Thus the rigorous version of (1.1) reads

Lβ,x₀u+αϕu=ωu, −ϕ=u², (1.3)

with vanishing Dirichlet boundary conditionsu=ϕ=0 on∂Ω. The operator can be derived in several different ways:

One can defineLβ,x0 as a (strong resolvent) limit of short range Schrödinger operators L= −+λ()

² V

x−x0

as →0 where λ()=1+o(). Here the potential should have short range, for instance, have compact support.

Observe that λ()

² V

x−x₀

=λ()1 ³V

x−x₀

,

thus we see that, formally speaking, the coupling constant multiplying the Dirac delta function is infinitely weak.

Alternatively, one can define the operator Lβ,x0 by studying self-adjoint extensions of the symmetric operator

−|C^∞₀ (Ω\{x₀}) since the two operatorsLβ,x0 and−formally coincide onC₀^∞(Ω\ {x₀}). Further discussions and rigorous definitions can be found in [5].

It turns out that it is considerably easier to work with the resolvent ofLβ,x₀ thanLβ,x₀ itself. Indeed, we have (precise domains are defined in the next section)

Lβ,x₀u= −v if and only if u=v−v(x₀)

β G(·, x₀), (1.4)

whereG=G(x, y)is the Green’s function of−onΩ with homogeneous boundary conditions andv∈H²(Ω)∩ H₀¹(Ω).

Thus we will work with⎧

⎪⎪

⎪⎨

⎪⎪

⎪⎩

−v+αu

Ω

G(·, y)u²(y)dy=ωu, inΩ, u=v−v(x₀)G(·, x₀)β⁻¹, inΩ,

v=0, on∂Ω.

Our main result says that ifω > ω0, whereω0is the lowest eigenvalue of−onΩ, then (1.1) admits a nontrivial weak solution. The result is obtained by construction an iterative approximation that is proved to converge to a weak solution. An additional argument is needed to show that the solution is nontrivial.

2. Mathematical preliminaries

Arguing as in [5, Theorem 1.1.3, Appendix K.2.1], we define the linear operatorLβ,x₀ by¹ D(Lβ,x0):= v−v(x₀)G(·, x₀)β⁻¹|v∈H²(Ω)∩H₀¹(Ω)

, (2.1)

Lβ,x0u= −v, u=v−v(x₀)

β G(·, x₀)∈D(Lβ,x0), (2.2)

whereG=G(x, y)is the Green’s function of−onΩwith homogeneous boundary conditions. Recall that H²(Ω)⊂C

Ω , and observe that

D(Lβ,x0)⊂L^p(Ω)∩W₀^1,q(Ω), 1p <3, 1q <3

2. (2.3)

Definition 2.1.Letu, ϕ:Ω→Rbe maps. We say that(u, ϕ)is a weak solution of (1.1) if u∈D(Lβ,x0), ϕ∈H₀¹(Ω)∩W^2,p(Ω), 1p < 3

2, (2.4)

and

Lβ,x0u+αϕu=ωu, −ϕ=u², inΩ. (2.5)

The main result of this paper is the following:

Theorem 2.1.Assume(1.2). Let

ω > ω₀, (2.6)

whereω₀is the lowest eigenvalue of−onΩ. Then(1.1)admits a nontrivial weak solution in the sense of Defini- tion2.1.

Letu∈D(Lβ,x₀). The unique solution of the Dirichlet problem −ϕ=u², inΩ,

ϕ=0, on∂Ω, (2.7)

is

ϕ(x)=

Ω

G(x, y)u²(y)dy, x∈Ω. (2.8)

Hence, (1.1) is equivalent to

⎧⎨

⎩

Lβ,x0u+αu

Ω

G(·, y)u²(y)dy=ωu, inΩ,

u=0, on∂Ω,

(2.9)

1 In [5] the definition reads(Lβ,x₀−z)u=(−−z)vfor azin the resolvent set of−, and whereu=v−_β−i^v(x√_{z/(4π )}⁰⁾ Gz(·, x₀). HereGzis the resolvent of−−zonL²(Ω)with Dirichlet boundary conditions. Since the lowest eigenvalue of−is positive, we can letz→0.

or from (2.2)

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

−v+αu

Ω

G(·, y)u²(y)dy=ωu, inΩ, u=v−v(x0)G(·, x0)β⁻¹, inΩ,

v=0, on∂Ω.

(2.10)

Finally, we can rewrite the equation in (2.10) only in terms ofv

−v(x)+α

Ω

G(x, y)v²(y)v(x)dy−2v(x0)α β

Ω

G(x, y)G(y, x₀)v(y)v(x)dy +v²(x0)α

β²

Ω

G(x, y)G²(y, x0)v(x)dy−v(x0)α β

Ω

G(x, y)G(x, x0)v²(y)dy +2v²(x₀)α

β²

Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)dy−v³(x₀)α β³

Ω

G(x, y)G²(y, x₀)G(x, x₀)dy

=ωv(x)−v(x₀)ω

βG(x, x₀). (2.11)

We continue with a preliminary regularity result for (2.7).

Lemma 2.1.Assume(1.2). Letu∈D(Lβ,x0). Then ϕ=

Ω

G(·, y)u²(y)dy∈W^2,p(Ω)∩H₀¹(Ω), 1p <3

2. (2.12)

In particular ϕ=

Ω

G(·, y)u²(y)dy∈L^q(Ω), 1q <∞, (2.13)

and

ϕ _H1

0(Ω), ϕ _W2,p(Ω)C0

v ²

H₀¹(Ω)+v(x0)²

, 1p <3

2, (2.14)

whereC₀>0is a constant andvis the unique map inH²(Ω)∩H₀¹(Ω)such thatu=v−v(x₀)G(·, x₀)β⁻¹, see (2.2).

This result is based on the classical Agmon regularity result (see [4, Theorem 8.2]).

Lemma 2.2.Let1< p <∞andf∈L^p(Ω)be a given function. Letφbe the unique solution of the Dirichlet problem −φ=f, inΩ,

φ=0, on∂Ω.

Then

φ∈W^2,p(Ω), (2.15)

and

∇φ _L^p_(Ω),D²φ

L^p(Ω)C₁

f _L^p_(Ω)+ φ _L^p_(Ω)

, (2.16)

for some constantC1>0, whereD²φis the Hessian matrix ofφ.

Proof of Lemma 2.1. Letu∈D(Lβ,x0). From (2.2) and [5, Theorem 1.1.3], we know that there exists a unique v∈H²(Ω)∩H₀¹(Ω)such thatu=v−v(x₀)G(·, x₀)β⁻¹.

We begin by proving that ϕ∈H₀¹(Ω), ϕ _H1

0(Ω)C0

v ²

H₀¹(Ω)+v(x0)²

. (2.17)

Sinceϕsolves (2.7) andG(·, x0)∈L^12/5(Ω), using the Sobolev and Hölder inequalities, we get

Ω

|∇ϕ|²dx=

Ω

ϕu²dx ϕ _L6(Ω) u² _L6/5(Ω)

c1 ∇ϕ _L2(Ω) u ²_L12/5(Ω)

c₂ ∇ϕ _L2(Ω)

v ²_L12/5(Ω)+v(x₀)² c₃ ∇ϕ _L2(Ω)

v ²

H₀¹(Ω)+v(x₀)² , for some constantsc₁, c₂, c₃>0.Clearly, this gives (2.17).

Finally, we have to prove

ϕ∈W^2,p(Ω), ϕ _W2,p(Ω)C0

v ²

H₀¹(Ω)+v(x0)²

, 1p <3

2. (2.18)

Due to (2.3), (2.2) and the Sobolev inequality we know that u²∈L^p(Ω), u²

L^p(Ω)c₄ v ²

H₀¹(Ω)+v(x₀)²

, 1p <3 2,

for some constantc₄>0. Then (2.18) is consequence of Lemma 2.2 and of the Sobolev embedding theorem. 2 Our argument is based on the following recursive approximation of (2.11). We begin by fixing a mapv0∈H²(Ω)∩ H₀¹(Ω). Then, for eachn∈N, we definevn∈H²(Ω)∩H₀¹(Ω)as one of theminimal energy solutionsof the following nonlinear problem:

−v_n(x)+α

Ω

G(x, y)v_n²(y)v_n(x)dy−2vn−1(x₀)α β

Ω

G(x, y)G(y, x₀)v_n(y)v_n(x)dy +v²_n−1(x₀)α

β²

Ω

G(x, y)G²(y, x₀)v_n(x)dy−v_n−1(x₀)α β

Ω

G(x, y)G(x, x₀)v_n²(y)dy +2v_n²₋₁(x0)α

β²

Ω

G(x, y)G(y, x0)G(x, x0)vn(y)dy−v_n³₋₁(x0)α β³

Ω

G(x, y)G²(y, x0)G(x, x0)dy

=ωvn(x)−vn−1(x0)ω

βG(x, x0). (2.19)

Observe that, denoting u_n:=v_n−vn−1(x0)

β G(·, x₀), n∈N, (2.20)

we can rewrite (2.19) in the following form

−v_n(x)+αu_n(x)

Ω

G(x, y)u²_n(y)dy=ωu_n(x). (2.21)

3. Approximate solutions

In this section we prove that the sequence{v_n}n∈Nexists. Arguing by induction, it is enough to fixn∈Nand show that we can findv_nusingv_n−1.

In other words we have to prove the existence of aminimal energy solutionfor the equation

−v(x)+α

Ω

G(x, y)v²(y)v(x)dy−2vn−1(x₀)α β

Ω

G(x, y)G(y, x₀)v(y)v(x)dy +v_n²₋₁(x0)α

β²

Ω

G(x, y)G²(y, x0)v(x)dy−vn−1(x0)α β

Ω

G(x, y)G(x, x0)v²(y)dy +2v²_n−1(x₀)α

β²

Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)dy−v³_n−1(x₀)α β³

Ω

G(x, y)G²(y, x₀)G(x, x₀)dy

=ωv(x)−v_n−1(x₀)ω

βG(x, x₀), (3.1)

endowed with the boundary condition

v=0, on∂Ω. (3.2)

3.1. The variational approach Consider the functional

J_n:H₀¹(Ω)−→R defined as follows

Jn(v)=1 2

Ω

∇v(x)²dx+α 4

Ω×Ω

G(x, y)v²(y)v²(x)dxdy

−v_n−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)v(y)v²(x)dxdy +v_n²₋₁(x₀)α

β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)v(x)dxdy +v_n²₋₁(x0) α

2β²

Ω×Ω

G(x, y)G²(y, x0)v²(x)dxdy

−v_n³₋₁(x₀)α β³

Ω×Ω

G(x, y)G²(y, x₀)G(x, x₀)v(x)dxdy

−ω 2

Ω

v²(x)dx−v_n−1(x₀)ω β

Ω

G(x, x₀)v(x)dx.

Lemma 3.1.Assume (1.2). ThenJ_nis of classC¹onH₀¹(Ω)and its critical points are the distributional solutions of (3.1)–(3.2).

Proof. Define J_n,1(v)=1

2

Ω

∇v(x)²dx+v_n²₋₁(x₀) α 2β²

Ω×Ω

G(x, y)G²(y, x₀)v²(x)dxdy

−v_n³₋₁(x0)α β³

Ω×Ω

G(x, y)G²(y, x0)G(x, x0)v(x)dxdy

−ω 2

Ω

v²(x)dx−v_n−1(x₀)ω β

Ω

G(x, x₀)v(x)dx,

J_n,2(v)=α 4

Ω×Ω

G(x, y)v²(y)v²(x)dxdy,

Jn,3(v)= −vn−1(x0)α β

Ω×Ω

G(x, y)G(y, x0)v(y)v²(x)dxdy,

J_n,4(v)=v²_n−1(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)v(x)dxdy, and observe thatJ_n=J_n,1+J_n,2+J_n,3+J_n,4.

Letε∈R, f, v∈H₀¹(Ω). We begin by computing the derivative ofJ_n,1: J_n,1(v+εf )=1

2

Ω

∇

v(x)+εf (x)²dx+v_n²₋₁(x0) α 2β²

Ω×Ω

G(x, y)G²(y, x0)

v(x)+εf (x)2

dxdy

−v_n³₋₁(x₀)α β³

Ω×Ω

G(x, y)G²(y, x₀)G(x, x₀)

v(x)+εf (x) dxdy

−ω 2

Ω

v(x)+εf (x)2

dx−vn−1(x0)ω β

Ω

G(x, x0)

v(x)+εf (x) dx.

Since

J_n,1 (v)[f] =dJn,1

dε (v+εf ) ε=0

=

Ω

∇v(x)· ∇f (x)dx+v²_n−1(x₀)α β²

Ω×Ω

G(x, y)G²(y, x₀)v(x)f (x)dxdy

−v_n³₋₁(x₀)α β³

Ω×Ω

G(x, y)G²(y, x₀)G(x, x₀)f (x)dxdy

−ω

Ω

v(x)f (x)dx−v_n−1(x₀)ω β

Ω

G(x, x₀)f (x)dx, we conclude

J_n,1 (v)= −v(x)+v_n²₋₁(x0)α β²

Ω

G(x, y)G²(y, x0)v(x)dy

−v_n³₋₁(x₀)α β³

Ω

G(x, y)G²(y, x₀)G(x, x₀)dy−ωv(x)−v_n−1(x₀)ω

βG(x, x₀). (3.3) We continue by computing the derivative ofJn,2:

J_n,2(v+εf )=α 4

Ω×Ω

G(x, y)

v(y)+εf (y)2

v(x)+εf (x)2

dxdy, dJn,2

dε (v+εf )=α 2

Ω×Ω

G(x, y)

v(y)+εf (y) f (y)

v(x)+εf (x)2

dxdy

+α 2

Ω×Ω

G(x, y)

v(y)+εf (y)2

v(x)+εf (x)

f (x)dxdy.

Using the symmetry ofG(i.e.,G(x, y)=G(y, x))

J_n,2 (v)[f] =dJn,2

dε (v+εf ) ε=0

=α 2

Ω×Ω

G(x, y)v(y)f (y)v²(x)dxdy+α 2

Ω×Ω

G(x, y)v²(y)v(x)f (x)dxdy

=α

Ω×Ω

G(x, y)v²(y)v(x)f (x)dxdy, namely

J_n,2 (v)=α

Ω

G(x, y)v²(y)v(x)dy. (3.4)

We pass to the derivative ofJ_n,3: J_n,3(v+εf )= −v_n−1(x₀)α

β

Ω×Ω

G(x, y)G(y, x₀)v(y)

v(x)+εf (x)2

dxdy

−εv_n−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)f (y)

v(x)+εf (x)2

dxdy,

dJn,3

dε (v+εf )= −2vn−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)v(y)

v(x)+εf (x)

f (x)dxdy

−v_n−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)f (y)

v(x)+εf (x)2

dxdy

−2εvn−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)f (y)

v(x)+εf (x)

f (x)dxdy.

Due to the symmetry ofG J_n,3 (v)[f] =dJn,3

dε (v+εf ) ε=0

= −2vn−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)v(y)v(x)f (x)dxdy

−v_n−1(x₀)α β

Ω×Ω

G(x, y)G(x, x₀)f (x)v²(y)dxdy, we get

J_n,3 (v)= −2vn−1(x₀)α β

Ω

G(x, y)G(y, x₀)v(y)v(x)dy−v_n−1(x₀)α β

Ω

G(x, y)G(x, x₀)v²(y)dy. (3.5) Finally, we look at the derivative ofJ_n,4:

Jn,4(v+εf )=v_n²₋₁(x0)α β²

Ω×Ω

G(x, y)G(y, x0)G(x, x0)v(y)

v(x)+εf (x) dxdy

+εv_n²₋₁(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)f (y)

v(x)+εf (x) dxdy,

dJn,4

dε (v+εf )=v_n²₋₁(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)f (x)dxdy +v_n²₋₁(x₀)α

β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)f (y)

v(x)+εf (x) dxdy

+εv²_n−1(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)f (y)f (x)dxdy.

Again by the symmetry ofGwe find J_n,4 (v)[f] =dJn,4

dε (v+εf ) ε=0

=2v_n²₋₁(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)f (x)dxdy, and thus we obtain

J_n,4 (v)=2v_n²₋₁(x₀)α β²

Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)dy. (3.6)

The claim is direct consequence of (3.3)–(3.6). 2

The next step in the analysis of (3.1) is a discussion of the topological properties of the functionalJn.

Lemma 3.2(Weak lower semicontinuity). Assume(1.2). The functionalJnis weakly lower semicontinuous onH₀¹(Ω), namely

vk vweakly inH₀¹(Ω)⇒lim inf

k Jn(vk)Jn(v). (3.7)

In particular, the functionals I_1,n(v)=

Ω×Ω

G(x, y)v²(y)v²(x)dxdy,

I_2,n(v)=

Ω×Ω

G(x, y)G(y, x₀)v(y)v²(x)dxdy,

I_3,n(v)=

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(y)v(x)dxdy,

I_4,n(v)=

Ω×Ω

G(x, y)G²(y, x₀)v²(x)dxdy,

I_5,n(v)=

Ω×Ω

G(x, y)G²(y, x₀)G(x, x₀)v(x)dxdy,

I_6,n(v)=

Ω

v²(x)dx,

I_7,n(v)=

Ω

G(x, x₀)v(x)dx, are all weakly continuous, i.e.,

v_k vweakly inH₀¹(Ω)⇒lim

k I_i,n(v_k)=I_i,n(v),i∈ {1, . . . ,7}. (3.8)

Proof. Let{vk}k∈N⊂H₀¹(Ω)andv∈H₀¹(Ω)such that

v_k v weakly inH₀¹(Ω). (3.9)

We know that lim inf

k

Ω

∇v_k(x)²dx

Ω

∇v(x)²dx. (3.10)

The compact embeddings ofH₀¹(Ω)

v_k−→v strongly inL^p(Ω), 1p <6, (3.11)

imply

limk I_6,n(v_k)=I_6,n(v). (3.12)

Moreover, sinceG(·, x₀)∈L²(Ω)we have also

limk I7,n(vk)=I7,n(v). (3.13)

Observe that I_5,n(v)=

Ω

f₁(x)v(x)dx, where

f₁:=

Ω

G(·, y)G²(y, x₀)G(·, x₀)dy∈L^3/2(Ω), indeed from (2.14) we have that

f₂:=

Ω

G(·, y)G²(y, x₀)dy∈H₀¹(Ω), (3.14)

so, using the Tonelli theorem and the Hölder inequality, f₁ ^3/2

L^3/2(Ω)=

Ω Ω

G(x, y)G²(y, x₀)G(x, x₀)dy 3/2

dx

=

Ω

G^3/2(x, x₀)

Ω

G(x, y)G²(y, x₀)dy 3/2

dx

=

Ω

G^3/2(x, x₀)f₂^3/2(x)dx G^3/2(·, x₀)

L^4/3(Ω)f^3/2

2

L⁴(Ω)

=G(·, x₀)^2/3

L²(Ω) f₂ ^2/3

L⁶(Ω)<∞. (3.15)

Therefore, employing (3.11) and (3.15)

limk I5,n(v_k)=I5,n(v). (3.16)

From (3.11)

v_k²−→v² strongly inL^p(Ω), 1p <3, (3.17)

so using I_4,n(v)=

Ω

f₂(x)v(x)dx and (3.14) we conclude

limk I_4,n(v_k)=I_4,n(v). (3.18)

We continue withI3,n. Observe that I_3,n(v_k)−I_3,n(v)=

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)

v_k(y)v_k(x)−v(y)v(x) dxdy

=

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v_k(y)

v_k(x)−v(x) dxdy

+

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v(x)

v_k(y)−v(y) dxdy

=

Ω

f_3,k(x)G(x, x₀)

v_k(x)−v(x) dx+

Ω

f₄(y)G(y, x₀)

v_k(y)−v(y) dy, where

f_3,k:=

Ω

G(·, y)G(y, x₀)v_k(y)dy, f₄:=

Ω

G(x,·)G(x, x₀)v(x)dx.

Using the symmetry of G, (3.11) and (2.14) we have thatf₄∈H₀¹(Ω)and the sequence {f_3,k}k∈N is bounded in H₀¹(Ω). Moreover, using the Hölder inequality

Ω

f_3,k(x)G(x, x₀)

v_k(x)−v(x) dx

f_{3,k L}6(Ω)G(·, x₀)(v_k−v)

L^6/5(Ω)

f_{3,k L}6(Ω)G^6/5(·, x₀)^5/6

L²(Ω)|v_k−v|^6/55/6

L²(Ω)

= f3,k L⁶(Ω)G(·, x0)

L^12/5(Ω) vk−v _L12/5(Ω),

Ω

f4(y)G(y, x0)

vk(y)−v(y) dy

f4 L⁶(Ω)G(·, x0)(vk−v)

L^6/5(Ω)

f_{4 L}6(Ω)G^6/5(·, x₀)^5/6

L²(Ω)|v_k−v|^6/55/6

L²(Ω)

= f_{4 L}6(Ω)G(·, x₀)

L^12/5(Ω) v_k−v _L12/5(Ω), hence, (3.11) implies

limk I3,n(v_k)=I3,n(v). (3.19)

The next step is the weak semicontinuity ofI_2,n. Observe that I_2,n(v_k)−I_2,n(v)=

Ω×Ω

G(x, y)G(y, x₀)

v_k(y)v_k²(x)−v(y)v²(x) dxdy

=

Ω×Ω

G(x, y)G(y, x₀)v_k(y)

v_k²(x)−v²(x) dxdy

+

Ω×Ω

G(x, y)G(y, x₀)v²(x)

v_k(y)−v(y) dxdy

=

Ω

f_5,k(x)

v²_k(x)−v²(x) dx+

Ω

f₆(y)

v_k(y)−v(y) dy, where

f_5,k:=

Ω

G(·, y)G(y, x₀)v_k(y)dy, f₆:=

Ω

G(x,·)G(·, x₀)v²(x)dx.

From (3.11) and (2.14) we have that the sequence{f_5,k}k∈Nis bounded inH₀¹(Ω). Moreoverf₆∈L^3/2(Ω),indeed f₆:=G(·, x₀)f₇, wheref₇:=

Ω

G(x,·)v²(x)dx.

Due to (2.14) we know thatf₇∈H₀¹(Ω), so using the Hölder inequality f₆ ^3/2

L^3/2(Ω)=

Ω

G^3/2(y, x₀)f₇(y)^3/2dy G^3/2(·, x₀)

L^4/3(Ω)f₇^3/2

L⁴(Ω)

=G(·, x0)^2/3

L²(Ω)f₇^3/22/3

L⁴(Ω)<∞. Hence, (3.11) and (3.17) imply

limk I_2,n(v_k)=I_2,n(v). (3.20)

Finally, we considerI1,n. Observe that I_1,n(v_k)−I_1,n(v)=

Ω×Ω

G(x, y)

v_k²(y)v²_k(x)−v²(y)v²(x) dxdy

=

Ω×Ω

G(x, y)v²_k(y)

v²_k(x)−v²(x) dxdy+

Ω×Ω

G(x, y)v²(x)

v²_k(y)−v²(y) dxdy.

Using the symmetry of G, (3.17) and (2.14) we have that

ΩG(x,·)v²(x)dx ∈ L²(Ω) and the sequence {

ΩG(·, y)v_k²(y)dy}k∈Nis bounded inL²(Ω). Then, employing (3.17), we obtain

limk I1,n(vk)=I1,n(v). (3.21)

Due to (3.10), (3.12), (3.13), (3.16), (3.18)–(3.21) the proof is complete. 2 Lemma 3.3(Coercivity). Assume(1.2). The functionalJnis coercive inH₀¹(Ω), i.e.,

v_{k H}1

0(Ω)−→ ∞ ⇒lim

k J_n(v_k)= ∞. (3.22)

Proof. Let{vk}k∈N⊂H₀¹(Ω)such that v_{k H}1

0(Ω)−→ ∞. (3.23)

We have to prove

limk Jn(vk)= ∞. (3.24)

Define

λ_k:= v_{k H}1

0(Ω), v˜_k:=vk

λk

,

obviously,

v_k=λ_kv˜_k, λ_k−→ ∞, ˜v_{k H}1 0(Ω)=1.

From the definition ofJ_nwe get J_n(v_k)=J_n(λ_kv˜_k)=λ²_k

2 +a_kλ⁴_k+b_kλ³_k+c_kλ²_k+d_kλ_k, (3.25) where

a_k:=α 4

Ω×Ω

G(x, y)v˜²_k(y)v˜²_k(x)dxdy,

bk:= −vn−1(x0)α β

Ω×Ω

G(x, y)G(y, x0)v˜k(y)v˜_k²(x)dxdy,

c_k:=v²_n−1(x₀)α β²

Ω×Ω

G(x, y)G(y, x₀)G(x, x₀)v˜_k(y)v˜_k(x)dxdy +v_n²₋₁(x₀) α

2β²

Ω×Ω

G(x, y)G²(y, x₀)v˜_k²(x)dxdy−ω 2

Ω

˜ v_k²(x)dx,

d_k:= −v_n³₋₁(x0)α β³

Ω×Ω

G(x, y)G²(y, x0)G(x, x0)v˜_k(x)dxdy−v_n−1(x0)ω β

Ω

G(x, x0)v˜_k(x)dx.

Due to the boundedness of{˜vk}kinH₀¹(Ω), there existsv˜∈H₀¹(Ω)such that ˜v _H1

0(Ω)1 and (passing to a subse- quence)

˜

vkv˜ weakly inH₀¹(Ω). (3.26)

We distinguish two cases.

Case1. If

˜ v=0,

then, employing the second part of Lemma 3.2 and (3.26), a_k−→α

4

Ω×Ω

G(x, y)v˜²(y)v˜²(x)dxdy∈(0,∞),

b_k−→ −v_n−1(x₀)α β

Ω×Ω

G(x, y)G(y, x₀)v(y)˜ v˜²(x)dxdy∈R,

ck−→v_n²₋₁(x0)α β²

Ω×Ω

G(x, y)G(y, x0)G(x, x0)v(y)˜ v(x)˜ dxdy +v²_n−1(x₀) α

2β²

Ω×Ω

G(x, y)G²(y, x₀)v˜²(x)dxdy−ω 2

Ω

˜

v²(x)dx∈R,

dk−→ −v³_n−1(x0)α β³

Ω×Ω

G(x, y)G²(y, x0)G(x, x0)v(x)˜ dxdy−vn−1(x0)ω β

Ω

G(x, x0)v(x)˜ dx∈R. Clearly, from (3.25) we get

limk

J_n(v_n) λ⁴_k =lim

k a_k=α 4

Ω×Ω

G(x, y)v˜²(y)v˜²(x)dxdy >0,