Global stability for an SEI epidemiological model with continuous age-structure in the exposed and infectious classes

  • We study a model of disease transmission with continuous age-structure for latently infected individuals and for infectious individuals. The model is very appropriate for tuberculosis. Key theorems, including asymptotic smoothness and uniform persistence, are proven by reformulating the system as a system of Volterra integral equations. The basic reproduction number $\mathcal{R}_{0}$ is calculated. For $\mathcal{R}_{0}<1$, the disease-free equilibrium is globally asymptotically stable. For $\mathcal{R}_{0}>1$, a Lyapunov functional is used to show that the endemic equilibrium is globally stable amongst solutions for which the disease is present. Finally, some special cases are considered.

    Citation: C. Connell McCluskey. Global stability for an SEI epidemiological model with continuous age-structure in the exposed and infectious classes[J]. Mathematical Biosciences and Engineering, 2012, 9(4): 819-841. doi: 10.3934/mbe.2012.9.819

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  • We study a model of disease transmission with continuous age-structure for latently infected individuals and for infectious individuals. The model is very appropriate for tuberculosis. Key theorems, including asymptotic smoothness and uniform persistence, are proven by reformulating the system as a system of Volterra integral equations. The basic reproduction number $\mathcal{R}_{0}$ is calculated. For $\mathcal{R}_{0}<1$, the disease-free equilibrium is globally asymptotically stable. For $\mathcal{R}_{0}>1$, a Lyapunov functional is used to show that the endemic equilibrium is globally stable amongst solutions for which the disease is present. Finally, some special cases are considered.


    The Newton kernel $ \Phi_N \colon \mathbb{R}^N \to \mathbb{R} $ is defined by

    $ ΦN(x)={Γ(N22)4πN/2|x|N2if N312πlog1|x|if N=2.
    $

    The non-local partial differential equation

    $ Δu+au=[ΦN|u|2]uin RN,
    $
    (1.1)

    where $ a $ is a positive function, was proposed in the study of quantum physics of electrons in a ionic crystal (the so-called Pekar polaron model) for $ N = 3 $. The same equation can also be seen as a coupling of quantum physics with Newtonian gravitation: indeed, the system

    $ {iψtΔψ+E(x)ψ+γwψ=0Δw=|ψ|2
    $

    in the unknown $ \psi \colon \mathbb{R}^N \times \mathbb{R} \to \mathbb{R} $, $ \psi = \psi(x, t) $, reduces to the single equation

    $ Δu+au+γ[ΦN|u|2]u=0
    $

    via the ansatz $ \psi(x, t) = \mathrm{e}^{-\mathrm{i} \lambda t} u(x) $ with $ \lambda \in \mathbb{R} $ and $ a(x) = E(x)+\lambda $. E.H. Lieb proved in [11] that (1.1) possesses, in dimension $ N = 3 $, a unique ground state solution which is positive and radially symmetric. E. Lenzmann proved in [10] that this solution is also non-degenerate. The analysis of (1.1) in dimension $ N = 3 $ is heavily based on the algebraic properties of the kernel $ \Phi_3 $, in particular its homogeneity. Lieb's proof of existence carries over to $ N = 4 $ and $ N = 5 $, while no solution with finite energy can exist in dimension $ N \geq 6 $, see [6].

    In this note we consider (1.1) in the plane, i.e., when $ N = 2 $. The appearance of the logarithm in $ \Phi_2 $ changes drastically the setting of the problem, which has been an open field of study for several years. One of the main obstructions to a straightforward analysis in the planar case is the lack of positivity of the kernel $ \Phi_2 $.

    Some preliminary numerical results contained in [9] encouraged Ph. Choquard, J. Stubbe and M. Vuffray to prove the existence of a unique positive radially symmetric solution by applying a shooting method, see [7]. But only in very recent years have variational methods been used to solve (1.1) for $ N = 2 $: the formal definition of a Euler functional

    $ I(u)=12R2(|u|2+a|u|2)+18πR2×R2log|xy||u(x)|2|u(y)|2dxdy
    $

    is not consistent with the metric structure of the Sobolev space $ H^1(\mathbb{R}^2) $.

    Stubbe proposed in [14] a variational setting for (1.1) in dimension two within the closed subspace

    $ X={uH1(R2)R2log(1+|x|)|u(x)|2dx<+}
    $

    endowed with the norm

    $ u2X=R2(|u|2+a|u|2)+R2log(1+|x|)|u(x)|2dx.
    $

    Although this space permits to use variational methods, several difficulties arise from the logarithmic term.

    Using this functional approach, S. Cingolani and T. Weth (see [8]) proved some existence results for (1.1) under either a periodicity assumption on the potential $ a $, or the action of a suitable group of transformations. Uniqueness and monotonicity of positive solutions are also proved.

    Later on, D. Bonheure, S. Cingolani and J. Van Schaftingen (see [6]) proved that the positive solution $ u $ of (1.1) with $ a > 0 $ is non-degenerate, in the sense that the only solutions of the linearized equation associated to (1.1) are the (linear combinations of) the two partial derivatives of $u$.

    Motivated by these results, we consider the following perturbed equation, based on (1.1):

    $ Δu+au12π[log1||u2]u=εh(x)|u|p1uin R2.
    $
    (1.2)

    The quantity $ \varepsilon $ plays the rôle of a "small" perturbation, and the function $ h $ is a "weight" for the local nonlinearity $ |u|^{p-1}u $. We refer to the next Sections for the precise assumptions we make.

    We will face the problem of constructing solutions to (1.2) by means of a general technique in Critical Point Theory, introduced by A. Ambrosetti and M. Badiale in [1,2,3]. We refer to [4] for a presentation in book form. For the reader's convenience, we summarize here the main ideas of this method.

    Suppose we are given a (real) Hilbert space $ X $ and a functional $ I_ \varepsilon \in C^2(X) $ of the form

    $ Iε(#)=I0(#)+εG(#).
    $

    Here $ I_0 \in C^2 (X) $ is the so-called unperturbed functional, while $ \varepsilon \in \mathbb{R} $ is a (small) perturbation parameter. We will suppose that there exists a (smooth) manifold $ Z $ of dimension $d < \infty$, such that every $ z \in Z $ is a critical point of $ I_0 $.

    Letting $ W = \left(T_z Z\right)^\perp $ for $ z \in Z $, we look for solutions to the equation $I_ \varepsilon'(u) = 0$ of the form $ u = z+w $, where $ z \in Z $ and $ w \in W $. We can split the equation $I_ \varepsilon'(u) = 0$ into two equations by means of the orthogonal projection $ P \colon X \to W $:

    $ {PIε(z+w)=0(IP)Iε(z+w)=0.
    $
    (1.3)

    We will assume that the following conditions hold:

    (ND) for all $ z \in Z $, we have $ T_z Z = \ker I_0''(z) $;

    (Fr) for all $ z \in Z $, we have that the linear operator $ I_0''(z) $ is Fredholm with index zero.

    Remark 1.1. The condition (ND) can be seen as a non-degeneracy assumption, since it is always true that $ T_z Z \subset \ker I_0''(z) $, by definition of $ Z $.

    It is possible to show that the first equation of system (1.3) can be (uniquely) solved with respect to $ w = w(\varepsilon, z) $, with $ z \in Z $ and $ \varepsilon $ sufficiently small. The main result of this perturbation technique can be summarized in the following statement.

    Theorem 1.2 ([4]). Suppose that the function $ \Phi_ \varepsilon \colon Z \to \mathbb{R} $ defined by $ \Phi_ \varepsilon(z) = I_ \varepsilon(z+w(\varepsilon, z)) $ possesses, for $ | \varepsilon| $ sufficiently small, a critical point $ z_ \varepsilon \in Z $. Then $ u_ \varepsilon = z_ \varepsilon + w(\varepsilon, z_ \varepsilon) $ is a critical point of $ I_ \varepsilon = I_0+ \varepsilon G $.

    As it should be clear, the perturbation method of Ambrosetti and Badiale leans on the effect of the function $ h $, which breaks the invariance of $ I_0 $ under translations. As such, the existence of a critical point of the function $ \Phi_ \varepsilon $ depends crucially on the behavior of $ h $.

    We split our existence results into two categories. The first one assumes that the weight function $h$ is not only bounded, but also sufficiently integrable over $ \mathbb{R}^2 $; because of this, we can consider this results as a local existence result.

    Theorem 1.3. Let $ p > 1 $ and $ h \in L^{\infty}(\mathbb{R}^2) \cap L^q(\mathbb{R}^2) $ for some $ q > 1 $. Moreover, suppose that

    (h$ _1 $) $ \int_{ \mathbb{R}^2}h(x)|z_0|^{p+1}\, dx \neq 0 $.

    Then Eq. (1.2) has a solution provided $ | \varepsilon| $ is small enough.

    It is possibile to drop the integrability condition on $ h $, at the cost of a more delicate analysis of the implicit function $ w = w(\varepsilon, z) $ that describes $ \Phi_ \varepsilon $. We have the following global result.

    Theorem 1.4. Let $ p > 2 $ and suppose that $ h $ satisfies

    (h$ _2 $) $ h \in L^{\infty}(\mathbb{R}^2) $ and $ \lim_{|x| \to \infty} h(x) = 0 $.

    Then for all $ | \varepsilon| $ small, Eq. (1.2) has a solution.

    We highlight that our results differ from those appearing in the literature for several reasons. First of all, the non-degeneracy property appearing in Proposition 2.5 can be used as a basis for further investigation. Moreover, the right-hand side of Eq. (1.2) may (and indeed must) depend on $ x $; no symmetry requirement, like radial symmetry, is needed in our proofs.

    The paper is organized as follows. In Section 2 we first give the precise assumptions for our problem, then we recall some known results (classical and not) and we prove some properties for energy functional, such as regularity. In Section 3 we present the proof of the main theorems.

    Consider the equation

    $ Δu+au12π[log1||u2]u=εh(x)|u|p1uin R2,
    $
    (2.1)

    with $ a > 0 $, $ h \in L^{\infty}(\mathbb{R}^2) $ and $ p > 1 $. We introduce the function space

    $ X={uH1(R2)R2|u(x)|2log(1+|x|)dx<},
    $

    endowed with the norm

    $ u2X=u2H1(R2)+|u|2,
    $

    where

    $ u2H1(R2)=R2(|u(x)|2+a|u(x)|2)dx|u|2=R2|u(x)|2log(1+|x|)dx.
    $

    The norm $ \|\cdot \|_X $ is associated naturally to an inner product. Let

    $ Iε(u)=I0(u)+εG(u)
    $
    (2.2)

    be the energy functional associated to the equation, where

    $ I0(u)=12R2(|u(x)|2+au2(x))dx18πR2×R2log1|xy||u(x)|2|u(y)|2dxdy
    $

    and

    $ G(u)=1p+1R2h(x)|u(x)|p+1dx.
    $

    We define (see [8,14]) the symmetric bilinear forms

    $ B1(u,v)=R2×R2log(1+|xy|)u(x)v(y)dxdy,B2(u,v)=R2×R2log(1+1|xy|)u(x)v(y)dxdy,
    $

    and

    $ B(u,v)=B1(u,v)B2(u,v)=R2×R2log(|xy|)u(x)v(y)dxdy,
    $
    (2.3)

    since for all $ r > 0 $ we have

    $ logr=log(1+r)log(1+1r).
    $

    Remark 2.1. The definitions above are restricted to measurable functions $ u $, $ v\colon \mathbb{R}^2 \to \mathbb{R} $ such that the corresponding double integral is well defined in the Lebesgue sense.

    In order to find estimates for $ B_1 $ and $ B_2 $ we recall a classical result of Measure Theory.

    Theorem 2.2 (Hardy-Littlewood-Sobolev's inequality [12]). Let $ p > 1 $, $ q > 1 $ and $ 0 < \lambda < N $ with $ \frac{1}{p} + \frac{\lambda}{N} + \frac{1}{q} = 2. $ If $ f \in L^p(\mathbb{R}^N) $ and $ g \in L^q(\mathbb{R}^N) $, then there exists a sharp constant $ C(N, \lambda, p) $, independent of $ f $ and $ g $, such that

    $ RN×RN|f(x)g(y)||xy|λdxdyC(N,λ,p)fLp(RN)gLq(RN).
    $
    (2.4)

    The sharp constant satisfies

    $ C(N,λ,p)N(Nλ)(|SN1|N)λN1pq((λ/N11p)λN+(λ/N11q)λN).
    $

    If $ p = q = \frac{2N}{2N - \lambda} $, then

    $ C(N,λ,p)=C(N,λ)=πλ2Γ(N/2λ/2)Γ(Nλ/2)(Γ(N/2)Γ(N))1+λ/N.
    $

    In this case there is equality in (2.4) if and only if $ g \equiv cf $ with $ c $ constant and

    $ f(x)=A(γ2+|xα|2)(2Nλ)/2
    $

    for some $ A \in \mathbb{R} $, $ \gamma \in \mathbb{R} \setminus \{0\} $ and $ \alpha \in \mathbb{R}^N. $

    We note that, since

    $ log(1+|xy|)log(1+|x|+|y|)log(1+|x|)+log(1+|y|),
    $

    we have by Schwarz's inequality

    $ |B1(uv,wz)|R2×R2[log(1+|x|)+log(1+|y|)]|u(x)v(x)||w(y)z(y)|dxdy|u||v|wL2(R2)zL2(R2)+uL2(R2)vL2(R2)|w||z|
    $
    (2.5)

    for $ u $, $ v $, $ w $, $ z \in X $. Next, since $ 0 \le \log(1+r) \le r $ for all $ r > 0 $, we have by Hardy-Littlewood-Sobolev's inequality

    $ |B2(u,v)|R2×R21|xy|u(x)v(y)dxdyCuL43(R2)vL43(R2),
    $
    (2.6)

    for $ u, v \in L^\frac{4}{3}(\mathbb{R}^2) $, for some constant $ C > 0 $. In particular, from (2.5) we have

    $ B1(u2,u2)2|u|2u2L2(R2)
    $
    (2.7)

    for all $ u \in X $ and from (2.6) we have

    $ B2(u2,u2)Cu4L83(R2)
    $
    (2.8)

    for all $ u \in L^\frac{8}{3}(\mathbb{R}^2) $.

    Proposition 2.3. The functional $ I_{ \varepsilon} $ is of class $ C^2(X) $.

    Proof. The proof is similar to [8,Lemma 2.2], so we just sketch the main ideas. Recalling (2.7), (2.8), the assumption $ h \in L^{\infty}(\mathbb{R}^2) $ and the fact that $ X $ is compactly embedded into $ L^s(\mathbb{R}^2) $, $ s \in [2, +\infty) $ (see [5,6,8] for a proof) we have

    $ |Iε(u)|12uH1(R2)+18πB(u2,u2)+1p+1εhup+1Lp+1(R2)12uH1(R2)+14π|u|2u2L2(R2)+18πCu4L43(R2)+hup+1X<+.
    $

    The first Gâteaux derivative of $ I_{ \varepsilon} $ along $ v $ is

    $ Iε(u)v=R2(uv+auv)dx12πR2×R2log1|xy|u2(x)u(y)v(y)dxdyεR2h(x)|u|p1uvdx.
    $
    (2.9)

    We add and subtract $ \int\limits_{ \mathbb{R}^2} u(x)v(x)\log(1+|x|) \, dx $ to recover the scalar product of $ X $, so we obtain

    $ Iε(u)v=(u|v)X12πB(u2,uv)R2u(x)v(x)log(1+|x|)dxεR2h(x)|u|p1uvdx.
    $

    Now,

    $ |Iε(u)v|(u|v)X+12πB(u2,uv)+R2u(x)v(x)log(1+|x|)dx+εR2|h(x)||u|p|v|dx(u|v)X+12π|u|2uvL2(R2)+12πu2L2(R2)|uv|+12πCu2L83(R2)uvL43(R2)+εhupppvp
    $

    The second Gâteaux derivative of $ I_{ \varepsilon}(u) $ along $ (v, w) $ is

    $ Iε(u)(v,w)=R2(vw+avw)dx12πR2×R2log1|xy|u2(x)v(y)w(y)dxdy1πR2×R2log1|xy|u(x)v(x)u(y)w(y)dxdy(p1)εR2h(x)|u|p1wdxεR2h(x)|u|p1vwdx.
    $

    In this case, we add and subtract $ \int_{ \mathbb{R}^2} v(x)w(x)\log(1+|x|) \, dx $, hence

    $ Iε(u)(v,w)=(v|w)XR2v(x)w(x)log(1+|x|)dx12πB(u2,vw)1πB(uv,uw)(p1)εR2h(x)|u|p1wdxεR2h(x)|u|p1vwdx.
    $

    Finally,

    $ |Iε(u)(v,w)|(v|w)X+R2v(x)w(x)log(1+|x|)dx+12πB(u2,vw)+1πB(uv,uw)+(p1)εR2|h(x)||u|p1|w|dx+εR2|h(x)||u|p1|v||w|dx(v|w)X+R2v(x)w(x)log(1+|x|)dx+12π|u|2vwL2(R2)+12πu2L2(R2)|v||w|+12πCu2L83(R2)vwL43(R2)+1π|u||v|uL2(R2)wL2(R2)+1πuL2(R2)vL2(R2)|u||w|+1πCuL2(R2)vL2(R2)uL2(R2)wL2(R2)+(p1)εhup1(p1)pwp+εhup1(p1)pvwp.
    $

    It is now standard to conclude that the first and the second Gâteaux derivatives are continuous (with respect to $ u \in X $), so that $ I_ \varepsilon \in C^2(X) $.

    Critical points of the unperturbed functional $ I'_0(u) = 0 $ are solutions of the equation

    $ Δu+au12π[log1||u2]u=0in R2,
    $
    (2.10)

    which admits for every $ a > 0 $ a unique — up to translations — radially symmetric solution $ u \in X $ (see [8], Theorem 1.3).

    Since (2.10) is invariant under translations, we can consider $ z_{\xi}(x) = u(x-\xi) $, with $ \xi \in \mathbb{R}^2 $. The manifold

    $ Z={zξξR2}
    $

    is therefore a critical manifold for $ I_0 $. We want to show that the manifold $ Z $ satisfies the properties (ND) and (Fr).

    The (ND) property follows from the following theorem, proved in [6].

    Theorem 2.4. If $ a > 0 $ and $ u \in X $ is a radial solution of (2.10) then there exists $ \mu \in (0, \infty) $ such that

    $ u(x)=(μ+o(1))|x|(log|x|)1/4exp(MeaMeaM|x|1(logsds)
    $

    with

    $ M=12πR2|u|2.
    $

    To prove that (Fr) holds, we actually show that $ I''_0(z_{\xi}) $ is a compact perturbation of the identity operator.

    Proposition 2.5. $ I''_0(z_{\xi}) = L-K $, where $ L $ is a continuous invertible operator and $ K $ is a continuous linear compact operator in $ X $.

    Proof. We recall that

    $ I0(zξ)(v,w)=R2(vw+avw)dx12πR2×R2log1|xy|z2ξ(x)v(y)w(y)dxdy1πR2×R2log1|xy|zξ(x)v(x)zξ(y)w(y)dxdy
    $

    We add and subtract $ \int_{ \mathbb{R}^2} \log(1+|x|)v(x)w(x) \, dx $, hence

    $ I0(zξ)(v,w)=R2(vw+avw)dx+R2log(1+|x|)v(x)w(x)dxR2log(1+|x|)v(x)w(x)dx12πR2×R2log1|xy|z2ξ(x)v(y)w(y)dxdy1πR2×R2log1|xy|zξ(x)v(x)zξ(y)w(y)dxdy,
    $

    and so

    $ I0(zξ)(v,w)=(v|w)XR2log(1+|x|)v(x)w(x)dx12πR2×R2log1|xy|z2ξ(x)v(y)w(y)dxdy1πR2×R2log1|xy|zξ(x)v(x)zξ(y)w(y)dxdy.
    $

    The second derivative is therefore the linear operator defined by

    $ L(zξ):φΔφ+(aw)φ+2zξ(log2π(zξφ)),
    $

    where

    $ w(x)=12πR2log|xy||zξ(y)|2dy,xR2.
    $

    Following the proof in [13], Lemma 15, let $ \{v_n\}_n $, $ \{w_n\}_n $ be two sequences in $ X $ such that $ \|v_n\| \leq 1 $, $ \|w_n\| \leq 1 $, $ v_n \rightharpoonup v_0 $ and $ w_n \rightharpoonup w_0 $. Without loss of generality we can assume $ v_0 = w_0 = 0 $, so that

    $ vn0,wn0.
    $

    From the compact embedding of $ X $ in $ L^s(\mathbb{R}^2) $ for $ s \geq 2 $, we can say that

    $ vn0,wn0
    $
    (2.11)

    in $ L^s(\mathbb{R}^2) $, for every $ s \geq 2 $. We compute, using Theorem 2.4,

    $ R2×R2log1|xy|zξ(x)vn(x)zξ(y)wn(y)dxdyCzξvnL43(R2)zξwnL43(R2).
    $

    Now, by Hölder's inequality,

    $ lim supn+zξvn43L43(R2)lim supn+(R2|zξ(x)|4dx)13(R2|vn(x)|2dx)23=0
    $

    thanks to Theorem 2.4 and (2.11).

    Similarly,

    $ lim supn|R2×R2log1|xy|zξ(x)vn(x)zξwn(y)dxdy|=0.
    $
    (2.12)

    This proves that the linear operator

    $ φ2zξ(log2π(zξφ))
    $

    is compact. At this point, we should notice that the linear operator

    $ φΔφ+(aw)φ
    $

    is not invertible on $ X $. To overcome this difficulty, we set

    $ c2=12πR2|zξ(x)|2dx
    $

    and rewrite $ \mathcal{L}(z_\xi) $ as follows:

    $ L(zξ):φΔφ+(a+c2log(1+|x|))φ(c2log(1+|x|)+w)φ+2zξ(log2π(zξφ)).
    $

    Since

    $ lim|x|+(w(x)+c2log(1+|x|))=0
    $

    by [8,Proposition 2.3], the multiplication operator

    $ φ(c2log(1+|x|)+w)φ
    $

    is compact. We may conclude that the functional $ I''_0(z_{\xi}) $ is of the form $ L - K $ where

    $ Lφ=Δφ+(a+c2log(1+|x|))φ
    $

    is a linear, continuous, invertible operator and $K$ is a linear, continuous, compact operator. A different proof, based on a direct computation, appears in [?].

    Since the properties (ND) and (Fr) hold we can say that, for $ | \varepsilon| $ small, the reduced functional has the following form:

    $ Φε(zξ)=Iε(zξ+wε(ξ))=c0+εG(zξ)+o(ε),
    $

    with $ c_0 = I_0(z_{\xi}) $.

    Let $ \Gamma \colon \mathbb{R}^2 \to \mathbb{R} $ be the function defined as

    $ Γ(ξ)=G(zξ)=1p+1R2h(x)|zξ|p+1dx,ξR2.
    $

    Lemma 3.1. Suppose that $ h \in L^{\infty}(\mathbb{R}^2) \cap L^q(\mathbb{R}^2) $ for some $ q > 1 $. Then

    $ lim|ξ|Γ(ξ)=0.
    $

    Proof. By the Hölder inequality,

    $ |Γ(ξ)|1p+1R2|h(x)||zξ|p+1dx1p+1(R2|h(x)|qdx)1q(R2|zξ|(p+1)qdx)1q
    $

    and since $ z_{\xi} $ decays to zero as $ |\xi| \to \infty $ we have

    $ |Γ(ξ)|C(R2|zξ|(p+1)qdx)1q0as |ξ|+.
    $

    The proof is completed.

    Thanks to the previous Lemma we can now prove the existence of local solutions for (2.1).

    Proof of Theorem 1.3. The hypothesis of the previous Lemma are satisfied, so we have that $ \Gamma(\xi) $ goes to $ 0 $ as $ |\xi| \to \infty $. From (h$ _1 $) follows that $ \Gamma(0) = -\frac{1}{p+1}\int_{ \mathbb{R}^2}h(x)|z_0|^{p+1} \, dx \neq 0 $. Then $ \Gamma $ is not identically zero and follows that $ \Gamma $ has a maximum or a minimum on $ \mathbb{R}^2 $ and the existence of a solution follows from Theorem 2.16 in [4].

    As before, we call $ P = P_{\xi}:X \to W_{\xi} $ the orthogonal projection onto $ W_{\xi} = \left(T_{z_{\xi}}Z\right)^{\perp} $, $ \widetilde{W_{\xi}}: = \langle z_{\xi} \rangle \oplus \left(T_{z_{\xi}}Z\right) $ and $ R_{\xi}(w) = I'_0(z_{\xi}+w) - I''_0(z_{\xi})[w] $.

    Remark 3.2. By the variational characterization of the Mountain-Pass solution $ u $ as in [4,Remark 4.2], the spectrum of $ PI''_0(u) $ has exactly one negative simple eigenvalue with eigenspace spanned by $ u $ itself. Moreover, $ \lambda = 0 $ is an eigenvalue with multiplicity $ N $ and eigenspace spanned by $ D_iu $, $ i = 1, \dotsc, N $ and there exists $ \kappa > 0 $ such that

    $ (PI0(u)[v]|v)κv2,vuTuZ,
    $

    and hence the rest of the spectrum is positive.

    We prove the following

    Theorem 3.3. (i) There is $ C > 0 $ such that $ \|\left(PI''_0(z_{\xi})\right)^{-1}\|_{L(W_{\xi}, W_{\xi})} \leq C $, for every $ \xi \in \mathbb{R}^2 $,

    (ii) $ R_{\xi}(w) = o(\|w\|) $, uniformly with respect to $ \xi \in \mathbb{R}^2 $.

    Proof. Since $ z_{\xi} $ is a Mountain-Pass solution, Remark 3.2 holds, hence it suffices to show that there exists $ \kappa > 0 $ such that

    $ (PI0(zξ)[v]v)κv2,ξR2,v~Wξ.
    $

    For any fixed $ \xi \in \mathbb{R}^2 $, say $ \xi = 0 $, $ PI''_0(z_0) = PI''_0(u) $ is invertible and there exists $ \kappa > 0 $ such that

    $ (PI0(u)[v]v)κv2,v˜W:=uTuZ.
    $

    Let $ v^{\xi}(x) = v(x+\xi) $, then

    $ (PI0(zξ)[v]v)=P[R2(|v|2+av2)dx12πR2×R2log1|xy|z2ξ(x)v2(y)dxdy1πR2×R2log1|xy|zξ(x)v(x)zξ(y)v(y)dxdy]
    $

    and thanks to the change of variables $ x = t+\xi $ and $ y = s+\xi $ we obtain

    $ (PI0(zξ)[v]v)=P[R2(|v(t+ξ)|2+av2(t+ξ))dt12πR2×R2log1|ts|z2ξ(t+ξ)v2(s+xi)dtds1πR2×R2log1|ts|zξ(t+ξ)v(t+ξ)zξ(s+ξ)v(s+ξ)dtds]=P[R2(|vξ|2+a(vξ)2)dt12πR2×R2log1|xy|z20(t)|vξ(s)|2dtds1πR2×R2log1|xy|z0(t)vξ(t)z0(s)vξ(s)dtds]=PI0(u)(vξ,vξ).
    $

    Moreover, $ v^{\xi} \perp \widetilde{W} $ whenever $ v \perp \widetilde{W_{\xi}} $, hence

    $ (PI0(zξ)[(v,v)]v]v)=(PI0(u)[vξ]vξ)κvξ2=κv2,ξR2,v~Wξ
    $

    and so $ (i) $ si true.

    To prove $ (ii) $ we observe that

    $ Rξ(w)=I0(zξ+w)(v)I0(zξ)(w,v)=R2[(zξ+w)v+a(zξ+w)v]dx12πR2×R2log1|xy|(zξ+w)2(x)(zξ+w)(y)v(y)dxdyR2(wv+awv)dx+12πR2×R2log1|xy|z2ξ(x)w(y)v(y)dxdy+1πR2×R2log1|xy|zξ(x)w(x)zξ(y)v(y)dxdy.
    $

    After some computations we obtain

    $ Rξ(w)=R2(zξv+azξv)dx12πR2×R2log1|xy|z2ξ(x)zξ(y)v(y)dxdy12πR2×R2log1|xy|z2ξ(x)w(y)v(y)dxdy1πR2×R2log1|xy|zξ(x)w(x)zξ(y)v(y)dxdy1πR2×R2log1|xy|zξ(x)w(x)w(y)v(y)dxdy12πR2×R2log1|xy|w2(x)zξ(y)v(y)dxdy12πR2×R2log1|xy|w2(x)w(y)v(y)dxdy+12πR2×R2log1|xy|z2ξ(x)w(y)v(y)dxdy+1πR2×R2log1|xy|zξ(x)w(x)zξ(y)v(y)dxdy
    $

    and since $ z_{\xi} $ is critical point we finally have

    $ Rξ(w)=1πR2×R2log1|xy|zξ(x)w(x)w(y)v(y)dxdy12πR2×R2log1|xy|w2(x)zξ(y)v(y)dxdy12πR2×R2log1|xy|w2(x)w(y)v(y)dxdy.
    $

    By (2.7), (2.8), (2.4), Hölder's inequality we have

    $ |Rξ(w)|1π(|zξ||w|wL2vL2+zξL2wL2|w||v|+C1zξL43w2L43vL43)+12π(|w|2zξL2vL2+w2L2|zξ||v|+C2w2L83zξL43vL43)+12π(|w|2wL2vL2+w2L2|w||v|+C3w2L83wL43vL43),
    $

    by the compact embedding of $ X $ into $ L^s(\mathbb{R}^2) $, $ s \geq 2 $

    $ |Rξ(w)|1π(zξXw2XvX+zξXw2XvX+C1zξXw2XvX)+12π(w2XzξXvX+w2XzξXvX+C2w2XzξXvX)+12π(w3XvX+w3XvX+C3w3XvX)
    $
    (3.1)

    and finally

    $ |Rξ(w)|C4zξXw2XvX+C5w2XzξXvX+C6w3XvX.
    $

    Hence,

    $ |Rξ(w)|wXC4zξXwXvX+C5wXzξXvX+C6w2XvX
    $

    and this goes to $ 0 $ as $ \|w\|_X \to 0 $ uniformly with respect to $ \xi \in \mathbb{R}^2 $.

    This Lemma allows us to use Lemma 2.21 in [4], so there exists $ \varepsilon_0 > 0 $ such that for all $ | \varepsilon| \leq \varepsilon_0 $ and all $ \xi \in \mathbb{R}^2 $ the auxiliary equation $ PI'_ \varepsilon\left(z_{\xi} + w\right) = 0 $ has a unique solution $ w_{ \varepsilon}(z_{\xi}) $ with

    $ limε0wε(zξ)=0,
    $
    (3.2)

    uniformly with respect to $ \xi \in \mathbb{R}^2 $.

    We now prove

    Lemma 3.4. There exists $ \varepsilon_1 > 0 $ such that for all $ | \varepsilon| \leq \varepsilon_1 $, the following result holds:

    $ lim|ξ|wξ=0,strongly in X.
    $

    Proof. We first show two preliminaries results:

    (a) $ w_{\xi} $ weakly converges in $ X $ to some $ w_{ \varepsilon, \infty} \in X $ as $ |\xi| \to \infty $. Moreover, the weak limit $ w_{ \varepsilon, \infty} $ is a weak solution of

    $ Δwε,+awε,12π[log1||w2ε,]wε,=εh(x)|wε,|p1wε,;
    $
    (3.3)

    (b) $ w_{ \varepsilon, \infty} = 0 $.

    As a consequence of (3.2) we have that $ w_{ \varepsilon}(z_{\xi}) $ weakly converges in $ X $ to some $ w_{ \varepsilon, \infty} \in X $, as $ |\xi| \to \infty $. Recall that $ w_{ \varepsilon}(z_{\xi}) $ is a solution of the auxialiary equation $ PI'_{ \varepsilon}\left(z_{\xi}+w_{\xi}(z_{\xi})\right) = 0 $, namely

    $ Δwε,ξ+awε,ξ12π[log1||w2ε,ξ]wε,ξ12π[log1||z2ξ]wε,ξ1π[log1||z2ξ](zξ+wε,ξ)12π[log1||w2ε,ξ]zξ=εh(x)|zξ+wε,ξ|p1(zξ+wε,ξ)zp1ξ2i=1aiDizξ,
    $

    where

    $ ai=R2(εh(x)|zξ+wε,ξ|p1zp1ξ)Dizξdx,
    $

    and $ D_i $ denotes the partial derivative with respect to $ x_i $.

    Let $ v $ be any test function, then

    $ R2(wε,ξv+awε,ξv)dx12πR2×R2log1|xy|w2ε,ξ(x)wε,ξ(y)v(y)dxdy=12πR2×R2log1|xy|z2ξ(x)wε,ξ(y)v(y)dxdy+1πR2×R2log1|xy|z2ξ(x)(zξ+wε,ξ)(y)v(y)dxdy+12πR2×R2log1|xy|w2ε,ξ(x)zξ(y)v(y)dxdy+R2εh(x)|zξ+wε,ξ|p1v(x)dxR2zp1ξ(x)v(x)dx2i=1aiR2Dizξ(x)v(x)dx.
    $

    Following the computations in the proof of Proposition 2.5, in particular (2.12), and by Theorem 2.4 we can say that

    $ 12πR2×R2log1|xy|z2ξ(x)wε,ξ(y)v(y)dxdy0,1πR2×R2log1|xy|z2ξ(x)(zξ+wε,ξ)(y)v(y)dxdy0,12πR2×R2log1|xy|w2ε,ξ(x)zξ(y)v(y)dxdy0,
    $

    as $ |\xi| \to \infty $.

    Now, we need to pass to the limit in

    $ R2εh(x)|zξ+wε,ξ|p1v(x)dx.
    $

    In order to do that, we show that

    $ lim|ξ|0R2zp1kξwkε,ξvdx=0,k[0,p1).
    $
    (3.4)

    We split the integral as

    $ R2zp1kξwkε,ξvdx=|x|ρzp1kξwkε,ξvdx+|x|>ρzp1kξwkε,ξvdx
    $

    where $ \rho > 0 $. Using Hölder's inequality with $ p = k+1 $, so $ p' = \frac{p}{p-1-k} $, we obtain

    $ ||x|ρzp1kξwkε,ξvdx|(|x|ρ|z(p1k)pξdx)1p(|x|ρ|wε,ξ|kp|v|pdx)1pC(|x|ρ|zξ|pdx)1p
    $

    and this goes to $ 0 $ as $ \rho $ goes to $ \infty $. On the other hand,

    $ ||x|>ρzp1kξwkε,ξvdx|(|x|>ρ|z(p1k)pξ|wε,ξ|kpdx)1p(|x|>ρ|v|pdx)1p,
    $

    and since $ v $ is a test function this integral goes to $ 0 $ as $ \rho $ goes to $ \infty $.

    Hence, (3.4) holds and then

    $ R2εh(x)|zξ+wε,ξ|p1v(x)dxR2εh(x)|wε,|p1v(x)dx,.
    $

    Moreover,

    $ R2|wε,ξ|p1v(x)dxR2|w|p1v(x)dx,R2εh(x)|wε,ξ|p1v(x)dxR2εh(x)|w|p1v(x)dx,
    $

    as $ |\xi| \to \infty $ and again by Theorem 2.4,

    $ R2zp1ξ(x)v(x)dx0,R2Dizξ(x)v(x)dx0,
    $

    as $ |\xi| \to \infty $.

    Finally, we obtain

    $ R2(wε,v+awε,v)dx12πR2×R2log1|xy|w2ε,(x)wε,(y)v(y)dxdy=R2εh(x)|wε,|p1v(x)dx,
    $

    thus $ w_{ \varepsilon, \infty} $ is a weak solution of (3.3), namely (a) holds.

    By (3.2) we have that $ \lim\limits_{| \varepsilon| \to 0} w_{ \varepsilon, \infty} = 0 $. Since the unique solution $ w \in X $ of

    $ Δw+aw12π[log1||w2]w=εh(x)|w|p1w
    $

    with small norm is $ w = 0 $. To show that, we need to prove that there exists a constant $ C > 0 $ such that $ \|w\|_X \geq C $.

    Consider the first Gâteaux derivative, computed in (2.9), evaluated at $ w $ along $ w $, namely

    $ 0=Iε(w)w=R2(|w|2+aw2)dx12πR2×R2log1|xy|w2(x)w2(y)dxdyεR2h(x)|w|p+1dx,
    $

    thus

    $ w2H1(R2)=12πR2×R2log1|xy|w2(x)w2(y)dxdy+εR2h(x)|w|p+1dx,
    $

    and thus

    $ w2H1(R2)12πR2×R2log1|xy|w2(x)w2(y)dxdy+εR2|h(x)||w|p+1dx.
    $
    (3.5)

    Now, observe that from (2.3) we have

    $ R2×R2log1|xy||w(x)|2|w(y)|2dxdy=R2×R2log(1+|xy|)|w(x)|2|w(y)|2dxdyR2×R2log(1+1|xy|)w2(x)w2(y)dxdy,
    $

    hence

    $ R2×R2log(1+1|xy|)w2(x)w2(y)dxdy=R2×R2log(1+|xy|)w2(x)w2(y)dxdy+R2×R2log(1|xy|)w2(x)w2(y)dxdy
    $

    and finally

    $ R2×R2log(1+1|xy|)w2(x)w2(y)dxdyR2×R2log(1|xy|)w2(x)w2(y)dxdy.
    $

    Moreover,

    $ εR2|h(x)||w|p+1dxC1wp+1Lp+1(R2)C2wp+1H1(R2).
    $

    By (3.5) we obtain

    $ w2H1(R2)R2×R2log(1|xy|)w2(x)w2(y)dxdy+εR2|h(x)||w|p+1dxCw4H1(R2)+C2wp+1H1(R2)C3wηH1(R2),
    $

    where $ \eta = \max\{p+1, 4\} > 2 $ since $ p > 1 $; recalling that $ \|w\|^{\eta-2}_{H^1(\mathbb{R}^2)} \leq C_4\|w\|^{\eta-2}_{X} $ we obtain

    $ wη2X1C5>0.
    $

    From this we infer that $ w_{ \varepsilon, \infty} = 0 $, provided that $ | \varepsilon| \ll 1 $. Hence (b) is true.

    We recall that $ w_{ \varepsilon, \xi} $ satisfies

    $ wε,ξ=(PI0(zξ))1[εPG(zξ+wε,ξ)PRξ(wε,ξ)],
    $
    (3.6)

    where

    $ G(zξ+wε,ξ)=R2h(x)|zξ+wε,ξ|p1(zξ+wε,ξ)dx
    $

    and

    $ Rξ(wε,ξ)=1πR2×R2log1|xy|zξ(x)wε,ξ(x)wε,ξ(y)v(y)dxdy12πR2×R2log1|xy|w2ε,ξ(x)zξ(y)v(y)dxdy12πR2×R2log1|xy|w2ε,ξ(x)wε,ξ(y)v(y)dxdy.
    $

    From (3.6) and Theorem 3.3 it follows

    $ wε,ξ2C[|ε||(G(zξ+wε,ξ|wε,ξ))|+|(Rξ(wε,ξ)|wε,ξ)|].
    $
    (3.7)

    We infer that

    $ |(G(zξ+wε,ξ|wε,ξ))|R2|h(x)||zξ+wε,ξ|p|wε,ξ(x)|dxh(R2|zξ|p|wε,ξ|dx+R2|wε,ξ|p+1dx)
    $

    and this goes to 0 as $ |\xi| \to \infty $ by the compact embedding of $ X $ in $ L^s(\mathbb{R}^2) $ for all $ s \geq 2 $, (a) and (b) proved above.

    Then, by (3.1) we find that

    $ |(Rξ(wε,ξ)|wε,ξ)|1π(zξXwε,ξ2Xwε,ξX+zξXwε,ξ2Xwε,ξX+C1zξXwε,ξ2Xwε,ξX)+12π(wε,ξ2XzξXwε,ξX+wε,ξ2XzξXwε,ξX+C2wε,ξ2XzξXwε,ξX)+12π(wε,ξ3Xwε,ξX+wε,ξ3Xwε,ξX+C3wε,ξ3Xwε,ξX)
    $

    thus,

    $ |(Rξ(wε,ξ)|wε,ξ)|C4wε,ξ4X+o(ε).
    $

    Inserting the above inequality in (3.7) we obtain that

    $ wε,ξ2XC4wε,ξ4X+o(ε), as |ξ|
    $

    and passing to the limit we find that

    $ lim|ξ|wε,ξ2XC4lim|ξ|wε,ξ4X.
    $

    Since $ w_{ \varepsilon, \xi} $ is small in $ X $ as $ | \varepsilon| \to 0 $ we conclude that

    $ lim|ξ|wε,ξX=0 provided |ε|1.
    $

    Now we are ready to prove the existence of a global solution for the problem (2.1).

    Proof of Theorem 1.4. Consider the reduced function $ \Phi_{ \varepsilon}(\xi) = I_{ \varepsilon}\left(z_{\xi} + w_{ \varepsilon, \xi}\right) $, namely

    $ Φε(ξ)=12zξ+wε,ξ2H1(R2)18πR2×R2log1|xy|(zξ+wε,ξ)2(x)(zξ+wε,ξ)2(y)dxdyεp+1R2h(x)|zξ+wε,ξ|p+1dx.
    $
    (3.8)

    From $ I_0(z_{\xi}) = \frac{1}{2}\|z_{\xi}\|^2_{H^1(\mathbb{R}^2)} - \frac{1}{8\pi}\int_{ \mathbb{R}^2 \times \mathbb{R}^2}\log\frac{1}{|x-y|}z_{\xi}^2(x)z_{\xi}^2(y) \, dx \, dy $ and setting $ c_0 = I_0(z_{\xi}) $ we have that

    $ 12zξ2H1(R2)=c0+18πR2×R2log1|xy|z2ξ(x)z2ξ(y)dxdy.
    $

    Moreover, $ -\Delta z_{\xi} + az_{\xi} - \frac{1}{2\pi}\left[\log\frac{1}{|\cdot|}*z_{\xi}^2\right]z_{\xi} = 0 $ implies

    $ (zξ|wε,ξ)H1(R2)=12πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy.
    $

    Substituting into (3.8) we obtain

    $ Φε(ξ)=12zξ2H1(R2)+12wε,ξ2H1(R2)+(zξ|wε,ξ)H1(R2)18πR2×R2log1|xy|(zξ+wε,ξ)2(x)(zξ+wε,ξ)2(y)dxdy1p+1R2h(x)|zξ+wε,ξ|p+1dx=c0+18πR2×R2log1|xy|z2ξ(x)z2ξ(y)dxdy+12wε,ξ2H1(R2)+12πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|(zξ+wε,ξ)2(x)(zξ+wε,ξ)2(y)dxdyεp+1R2h(x)|zξ+wε,ξ|p+1dx,
    $

    thus

    $ Φε(ξ)=c0+18πR2×R2log1|xy|z2ξ(x)z2ξ(y)dxdy+12wε,ξ2H1(R2)+12πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|z2ξ(x)z2ξ(y)dxdy14πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|z2ξ(x)w2ε,ξ(y)dxdy14πR2×R2log1|xy|zξ(x)wε,ξ(x)z2ξ(y)dxdy12πR2×R2log1|xy|zξ(x)wε,ξ(x)zξ(y)wε,ξ(y)dxdy14πR2×R2log1|xy|zξ(x)wε,ξ(x)w2ε,ξ(y)dxdy18πR2×R2log1|xy|w2ε,ξ(x)z2ξ(y)dxdy14πR2×R2log1|xy|w2ε,ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|w2ε,ξ(x)w2ε,ξ(y)dxdyεp+1R2h(x)|zξ+wε,ξ|p+1dx,
    $

    and after a short computation

    $ Φε(ξ)=c0+12wε,ξ2H1(R2)+12πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy14πR2×R2log1|xy|z2ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|z2ξ(x)w2ε,ξ(y)dxdy14πR2×R2log1|xy|zξ(x)wε,ξ(x)z2ξ(y)dxdy12πR2×R2log1|xy|zξ(x)wε,ξ(x)zξ(y)wε,ξ(y)dxdy14πR2×R2log1|xy|zξ(x)wε,ξ(x)w2ε,ξ(y)dxdy18πR2×R2log1|xy|w2ε,ξ(x)z2ξ(y)dxdy14πR2×R2log1|xy|w2ε,ξ(x)zξ(y)wε,ξ(y)dxdy18πR2×R2log1|xy|w2ε,ξ(x)w2ε,ξ(y)dxdyεp+1R2h(x)|zξ+wε,ξ|p+1dx.
    $

    Now, by repating the same arguments used in Proposition 2.5 all the integrals over $\mathbb{R}^2 \times \mathbb{R}^2$ converge to $ 0 $.

    Moreover, by Lemma 3.4 we have $ \|w_{ \varepsilon, \xi}\|^2_{H^1(\mathbb{R}^2)} \to 0 $ as $ |\xi| \to +\infty $ and by Minkowski's inequality, Proposition 2.4, Lemma 3.4 and (h$ _\mathbf{2} $):

    $ lim|ξ|+R2h(x)|zξ+wε,ξ|p+1dx=0.
    $

    Hence,

    $ lim|ξ|+Φε(ξ)=c0.
    $

    This means that either $ \Phi_{ \varepsilon} $ is constant, or it has a maximum or minimum. In any case $ \Phi_{ \varepsilon} $ has a critical point and we can apply Theorem 2.23 in [4] to find a solution for problem (2.1).

    The authors are grateful to the anonymous referee for their comments and suggestions about the first version of the paper.

    The authors declare no conflict of interest.

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