Research article

Stationary Kirchhoff equations and systems with reaction terms

  • Received: 19 December 2021 Revised: 07 June 2022 Accepted: 08 June 2022 Published: 17 June 2022
  • MSC : 35J65, 47J25

  • In this paper, the operator approach based on the fixed point principles of Banach and Schaefer is used to establish the existence of solutions to stationary Kirchhoff equations with reaction terms. Next, for a coupled system of Kirchhoff equations, it is proved that under suitable assumptions, there exists a unique solution which is a Nash equilibrium with respect to the energy functionals associated to the equations of the system. Both global Nash equilibrium, in the whole space, and local Nash equilibrium, in balls are established. The solution is obtained by using an iterative process based on Ekeland's variational principle and whose development simulates a noncooperative game.

    Citation: Radu Precup, Andrei Stan. Stationary Kirchhoff equations and systems with reaction terms[J]. AIMS Mathematics, 2022, 7(8): 15258-15281. doi: 10.3934/math.2022836

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  • In this paper, the operator approach based on the fixed point principles of Banach and Schaefer is used to establish the existence of solutions to stationary Kirchhoff equations with reaction terms. Next, for a coupled system of Kirchhoff equations, it is proved that under suitable assumptions, there exists a unique solution which is a Nash equilibrium with respect to the energy functionals associated to the equations of the system. Both global Nash equilibrium, in the whole space, and local Nash equilibrium, in balls are established. The solution is obtained by using an iterative process based on Ekeland's variational principle and whose development simulates a noncooperative game.



    The famous Kirchhoff equation [1]

    utt(a+bΩu2dx)u=h(t,x) 

    (a,b>0) is an extension of the classical D'Alembert's wave equation for vibrations of elastic strings, which takes into account the changes in mass density and/or tension force of the string produced by transverse vibrations. In higher dimensions, the equation reads as follows

    utt(a+bΩ|u|2dx)Δu=h(t,x).

    One can also consider the parabolic type equation

    ut(a+bΩ|u|2dx)Δu=h(t,x)

    which models diffusion processes with a diffusion coefficient globally dependent on gradient.

    Several authors (see, e.g., [2,3,4,5,6,7,8]) have considered a more general Kirchhoff type equation, by replacing the integral factor a+b|u|2L2 with an expression of the form η(|u|L2), where η is an increasing and nonnegative function.

    Kirchhoff type problems are referred to be nonlocal due to the presence of the integral over the entire Ω, and due to this specificity, some difficulties arise in their investigation.

    The study of such equations and systems have been made using variational and topological methods, as well as upper and lower solution techniques (see, e.g., [9,10,11,12,13,14,15,16,17,18,19,20] and the references therein).

    In this paper, we first study the Dirichlet problem for a stationary integro-differential equation of Kirchhoff type with a reaction external force term, on a bounded domain ΩRn,

    {(a+bΩ|u|2dx)Δu=f+g(x,u,u)   in Ωu|Ω=0,

    and next we focus on the Dirichlet problem for a coupled system of Kirchhoff equations

    {(a+b|u|2H10)Δu=f1+g1(x,u,v)(a+b|v|2H10)Δv=f2+g2(x,u,v)u|Ω=v|Ω=0 (1.1)

    for which the solution is a Nash type equilibrium.

    To our knowledge, Nash equilibria of system (1.1) have not been considered so far, and our objective is to provide sufficient conditions for such solutions to exist. To this aim we use the approach initiated in [21] (see also [22,23,24,25,26,27]). The idea is to put system (1.1) in an operator form, as a fixed point system,

    {N1(u,v)=uN2(u,v)=v, (1.2)

    where the operators N1 and N2 admit a variational structure, i.e., there exist (energy) functionals E1(u,v) and E2(u,v) such that system (1.2) is equivalent with

    {E11(u,v)=0E22(u,v)=0. (1.3)

    where E11 is the partial Fréchet derivative of E1 with respect to the first variable and E22 is the partial Fréchet derivative of E2 with respect to the second variable. A solution (u,v) of (1.1) is a Nash equilibrium if

    {E1(u,v)=infE1(,v)E2(u,v)=infE2(u,) .

    The notion of a Nash equilibrium originated in game theory and economics, where a number of players or traders with their own costing criteria are in competition and each aims to optimize its cost in relation to the others. When no one can further improve his criterion, it means that the system has reached a Nash equilibrium state. Such kind of situations also hold for systems modeling real processes from physics, biology etc., when stationary states are Nash equilibria for the associated energy functionals.

    Non-cooperative games in which the players move alternately suggest an iterative method based on Ekeland's variational principle for finding and approximating Nash equilibria. The convergence of the iterative process is established by using unilateral Lipschitz conditions on the reaction terms and working techniques with inverse-positive matrices.

    The outline of this paper is as follows: Section 3 provides a comprehensive picture of the theoretical aspects of the Kirchhoff solution operator for the Dirichlet problem. Section 4 is dedicated to the Dirichlet problem for the stationary Kirchhoff equation with a reaction force term; the existence of solutions is established via Banach contraction principle and Schaefer's fixed point theorem. Finally in Section 5 there are provided sufficient conditions for a system of two Kirchhoff equations to admit a Nash equilibrium.

    In this section we collect a number of notions and results that will be used in the following.

    First we recall the weak form of Ekeland's variational principle (see, e.g., [28,Corollary 8.1]).

    Theorem 1 (Ekeland). Let (X,d) be a complete metric space and E:XR a lower semicontinuous functional bounded from below. For each ε>0, there is an element xX such that the following twoproperties hold:

    E(x)infyXE(y)+ε,E(x)E(y)+εd(x,y)   forallyX.

    Next we recall Perov's fixed point theorem (see, e.g., [28,pp 151–154]) for mappings defined on the Cartesian product of two metric spaces.

    Theorem 2 (Perov). Let (Xi,di), i=1,2 be complete metric spaces and Ni:X1×X2Xi be two mappings for which thereexists a square matrix M of size two with nonnegative entries and thespectral radius ρ(M)<1 such that the following vectorinequality

    (d1(N1(x,y),N1(u,v))d2(N2(x,y),N2(u,v)))M(d1(x,y)d2(u,v))

    holds for all (x,y),(u,v)X1×X2.Then there exists a unique point (x,y)X1×X2 with x=N1(x,y)and y=N2(x,y). Moreover, the point(x,y) can be obtained by the method ofsuccessive approximations starting from any initial point (x0,y0), and

    (d1(Nk1(x0,y0),x)d2(Nk2(x0,y0),y))Mk(IM)1(d1(x0,N1(x0,y0))d2(y0,N2(x0,y0)))

    for every kN.

    Here I stands for the unit matrix of size two. Note that the property of a square matrix M with nonnegative entries of having the spectral radius ρ(M) less than 1 is equivalent to each one of the properties: (a) Mk tends to the zero matrix as k+; (b) The matrix IM is nonsingular and the entries of its inverse (IM)1 are nonnegative.

    For our Kirchhoff system (1.1) both fixed point and critical point formulations ((1.2) and (1.3)) being available, both Perov approach and Ekeland variational approach can be used. The first approach offers the approximation procedure for the solution given by the method of successive approximations, while by the second approach, an approximation procedure more appropriate to the concept of Nash equilibrium can be established.

    We conclude this preliminary section by some notations and results related to Laplacian. For details we refer the reader to the book [29]. We consider the well-known Sobolev space H10(Ω) whose scalar product and norm are

    (u,v)H10=Ωuvdx,    |u|H10=|u|L2=(Ω|u|2dx)1/2.

    The notation H1(Ω) stands for the dual of H10(Ω) and for any fH1(Ω), uH10(Ω), by (f,u) we mean the value at u of the continuous linear functional f. One has the continuous embeddings H10(Ω)L2(Ω)H1(Ω) and the Poincaré inequalities

    |u|L21λ1|u|H10   (uH10(Ω)),|u|H11λ1|u|L2   (uL2(Ω)),

    where λ1 is the first eigenvalue of the Dirichlet problem for the operator Δ. We use the notation (Δ)1 for the inverse of the Laplacian with respect to the homogeneous Dirichlet boundary condition. More exactly, (Δ)1f=u, where u is the unique function in H10(Ω) satisfying (u,v)H10=(f,v) for all vH10(Ω), i.e., u is the weak solution of the Dirichlet problem Δu=f in Ω; u=0 on Ω. Recall that (Δ)1 is an isometry between H1(Ω) and H10(Ω).

    First we focus on the stationary equation

    (a+bΩ|u|2dx)Δu=h

    under the Dirichlet condition  u=0 on Ω.

    Theorem 3. (The solution operator) For each hH1(Ω), theDirichlet problem has a unique weak solution uH10(Ω), i.e.,

    (a+bΩ|u|2dx)(u,v)H10=(h,v),    vH10(Ω), (3.1)

    and the solution operator S:H1(Ω)H10(Ω), hu, is continuous and

    |S(h)|H101a|h|H1. (3.2)

    Proof. (a) Existence: Let hH1(Ω) be fixed and consider the operator Sh:H10(Ω)H10(Ω) defined by

    Sh(v)=1a+b|v|2H10(Δ)1h.

    Clearly, Sh is completely continuous. In addition,

    |Sh(v)|H101a|h|H1,   vH10(Ω). (3.3)

    Hence, if we denote B={vH10(Ω): |v|H101a|h|H1}, then Sh(B)B and according to Schauder's fixed point theorem, there exists at least one u such that Sh(u)=u. Clearly u is a solution of the Dirichlet problem.

    (b) Uniqueness: Assume that u1,u2 are two solutions of (3.1). Then

    (a+b|u1|2H10)|u1|2H10=(h,u1),(a+b|u2|2H10)(u1,u2)H10=(h,u1).

    It follows that

    (a+b|u1|2H10)|u1|2H10=(a+b|u2|2H10)(u1,u2)H10(a+b|u2|2H10)|u1|H10|u2|H10.

    Hence

    (a+b|u1|2H10)|u1|H10(a+b|u2|2H10)|u2|H10.

    The function (a+bx2)x being strictly increasing on R+, we have that |u1|H10|u2|H10. By symmetry the converse inequality also holds. Thus |u1|H10=|u2|H10. Now the uniqueness of solution for the Dirichlet problem related to Δ yields u1=u2.

    (c) Continuity: Let hkh in H1(Ω) and let uk:=S(hk). Using (3.3) we have that the sequence (uk) is bounded. Hence, passing if necessary to a subsequence, we may assume that the sequence of real numbers (|uk|) is convergent. We now prove that the sequence (uk) is Cauchy. From

    Δuk=1a+b|uk|2H10hk,

    we have

    Δ(ukup)=1a+b|uk|2H10hk1a+b|up|2H10hp

    in the weak sense. Consequently

    |ukup|2H10=(1a+b|uk|2H10hk1a+b|up|2H10hp, ukup)=1a+b|uk|2H10(hkhp, ukup)+(1a+b|uk|2H101a+b|up|2H10)(hp,ukup).

    Furthermore

    |ukup|2H101a|hkhp|H1|ukup|H10+ba2||uk|2H10|up|2H10||hp|H1|ukup|H10,

    whence

    |ukup|H101a|hkhp|H1+ba2||uk|2H10|up|2H10||hp|H1.

    Since |hp|H1 is bounded, (hk) and (|uk|2H10) are convergent, one immediately obtain that (uk) is Cauchy. Hence there is u with uku and passing to the limit we see that u=S(h). Finally the uniqueness of the solution implies that the whole sequence (uk) converges to S(h), that is S(hk)S(h).

    Theorem 4. (Monotonicity) If 0h1h2, then |S(h1)|H10|S(h2)|H10.

    Proof. Denote u:=S(h1) and v=S(h2). Since h1,h20, one has u,v0. Then

    (1+|u|2H10)|u|2H10=(h1,u)(h2,u)=(1+|v|2H10)(u,v)(1+|v|2H10)|u|H10|v|H10

    which gives

    (1+|u|2H10)|u|H10(1+|v|2H10)|v|H10,

    whence |u|H10|v|H10.

    Theorem 5. (The energy functional) A function uH10(Ω) isthe weak solution of the Dirichlet problem if and only if it is a criticalpoint of the C1 functional E:H10(Ω)R,

    E(v)=14(2a+b|v|2H10)|v|2H10(h,v). (3.4)

    Proof. One has

    |v+λw|2H10|v|2H10=2λ(v,w)H10+λ2|w|2H10,|v+λw|4H10|v|4H10=(|v+λw|2H10|v|2H10)(|v+λw|2H10+|v|2H10)=(2λ(v,w)H10+λ2|w|2H10)(|v+λw|2H10+|v|2H10).

    Consequently

    limλ0E(v+λw)E(v)λ=(a+b|v|2H10)(v,w)H10(h,w).

    Hence

    (E(v),w)=(a+b|v|2H10)(v,w)H10(h,w). (3.5)

    Theorem 6. Function uH10(Ω) solves the Dirichlet problemif and only if it minimizes the energy functional (3.4).

    Proof. If u is a minimum point of E, then E(u)=0 and according to (3.5) it solves the problem. Assume now that u is a solution. Then for every v, by direct computation, we have

    E(u+v)=E(u)+(a+b|u|2H10)(u,v)H10(h,v)+a2|v|2+b4(|v|4+2|u|2|v|2+4(u,v)2+4|v|2(u,v)2)=E(u)+a2|v|2+b4(|v|4+2|u|2|v|2+4(u,v)2+4|v|2(u,v)2)E(u)+a2|v|2+b4((|v|2+2(u,v))2+2|u|2|v|2)>0

    for every v0. Hence u is the unique minimum point of E.

    Consider the Dirichlet problem

    {(a+bΩ|u|2dx)Δu=f+g(x,u,u)  inΩu=0   onΩ. (4.1)

    Here ΩRn is open bounded, fH1(Ω), g:Ω×R×RnR satisfies the Carathéodory conditions and g(,0,0)=0.

    We look for weak solutions to (4.1), namely uH10(Ω) with g(,u,u)H1(Ω) and

    (a+bΩ|u|2dx)(u,v)H10=(f+g(,u,u), v)  forallvH10(Ω).

    A function uH10(Ω) is a weak solution of (4.1) if

    u=1a+bΩ|u|2dx(Δ)1(f+g(,u,u)),

    that is u is a fixed point of the operator

    A(u)=S(f+g(,u,u)).

    We apply Banach contraction principle. Assume the Lipschitz condition

    (HL)

    |g(x,u,v)g(x,¯u,¯v)|L1|u¯u|+L2|v¯v|

    for all u,¯uR, v,¯vRn and a.e. xΩ, where

    θ:=1a(L1λ1+L2λ1)<1. (4.2)

    Step 1: Invariance of a ball.

    We prove that if L1, L2 are small, then for any large enough number R, one has |A(u)|H10R for all uH10(Ω) with |u|H10R. According with (3.2), using (HL) and Poincaré 's inequality, one has

    |A(u)|H10=|S(f+g(,u,u))|H101a|f+g(,u,u)|H11a(|f|H1+|g(,u,u)|H1)1a(|f|H1+1λ1|g(,u,u)|L2)1a(|f|H1+1λ1(L1|u|L2+L2|u|L2))1a|f|H1+1a(L1λ1+L2λ1)|u|H10.

    Hence in virtue of (4.2), the invariance condition holds for any number R |f|H1/(a(1θ)).

    Step 2: Contraction condition.

    Fix any number R as guaranteed at the previous step. Let u,vH10(Ω) with |u|H10, |v|H10R be arbitrary and let w=S(f+g(,u,u)) and z=S(f+g(,v,v)). Assume without loss of generality that |w|H10|z|H10. Then

    (a+b|w|2H10)|w|2H10=(f+g(.,u,u),w),(a+b|z|2H10)(w,z)H10=(f+g(.,v,v),w),

    whence

    (a+b|w|2H10)|w|2H10(a+b|z|2H10)(w,z)H10=(g(.,u,u)g(.,v,v), w).

    For the left side, one has

    (a+b|w|2H10)|w|2H10(a+b|z|2H10)(w,z)H10(a+b|w|2H10)|w|2H10(a+b|z|2H10)|w|H10|z|H10

    and for the right side

    (g(.,u,u)g(.,v,v), w)|g(.,u,u)g(.,v,v)|L2|w|L2(L1λ1+L2λ1)|uv|H10|w|H10.

    Hence

    (a+b|w|2H10)|w|2H10(a+b|z|2H10)|w|H10|z|H10(L1λ1+L2λ1)|uv|H10|w|H10

    and since |w|H10|z|H10,

    0a(|w|H10|z|H10)(a+b|w|2H10)|w|H10(a+b|z|2H10)|z|H10(L1λ1+L2λ1)|uv|H10.

    Consequently

    0|w|H10|z|H10θ|uv|H10.

    On the other hand, from

    (a+b|w|2H10)(w,wz)H10=(f+g(.,u,u), wz),(a+b|z|2H10)(z,wz)H10=(f+g(.,v,v), wz),

    we deduce that

    |wz|2H10=(f+g(.,u,u)a+b|w|2H10f+g(.,v,v)a+b|z|2H10, wz)=1a+b|w|2H10(g(.,u,u)g(.,v,v), wz)+(1a+b|w|2H101a+b|z|2H10)(f+g(.,v,v), wz).

    We have

    (g(.,u,u)g(.,v,v), wz)(L1λ1+L2λ1)|uv|H10|wz|H10

    and

    (1a+b|w|2H101a+b|z|2H10)(f+g(.,v,v), wz)b(|w|H10|z|H10)|w|H10+|z|H10(a+b|w|2H10)(a+b|z|2H10)|f+g(.,v,v)|H1|wz|H10.

    Since

    |w|H10+|z|H10(a+b|w|2H10)(a+b|z|2H10)3381aab

    and

    |g(.,v,v)|H1aθR,

    we obtain

    |wz|H10θ(1+θ338aba(|f|H1+aθR))|uv|H10.

    Hence if

    (HC)

    θ(1+θ338aba(|f|H1+aθR))<1,

    then the operator A is a contraction on the ball of H10(Ω) centered at the origin and of radius R. Notice that condition (HC) is fulfilled for example if θ<1 (invariance condition for the ball of radius R) and b is small enough.

    Thus Banach's contraction principle applied to operator A in the ball of radius R yields the following existence and uniqueness result.

    Theorem 7. Assume that conditions (HL) and (HC)hold. Then problem (4.1) has a unique solution u such that

    |u|H10|f|H1/(a(1θ)).

    Example 8. Consider the Dirichlet problem,

    {(4+B|u|2dx)Δu=2|x|+λ1u+λ1sin|u|  onBu|B=0, (4.3)

    where Ω=B and B is the open ball centered at the origin of Rn and of radius ρ whose measure equals 1. Here

    a=4,b=1,f(x)=2|x| and g(x,u,v)=λ1u+λ1sin|v|,

    foruRandvRn. Note that fH1(B) with |f|H1=1. Indeed, the function u0(x)=|x|1 is the weak solution of Dirichlet problem Δu=f in B, u|B=0 and consequently

    |f|H1=|u0|H10=|u0|L2=|x|x||L2=1.

    Clearly, g is a Carathéodory function, g(,0,0)=0 and satisfies condition (HL) with L1=λ1  and  L2=λ1 and θ=2/a=1/2.

    For R=|f|H1/(a(1θ))=1/2, the condition (HC) is fulfilled, since

    θ(1+θ338aba(|f|H1+aθR))=12(1+3364)<1.

    Therefore, the problem (4.3) has a unique solution uH10(B) with |u|H101/2.

    Step 1: Complete continuity of the operator A:H10(Ω)H10(Ω).

    Recall that (Δ)1:H1(Ω)H10(Ω) is an isometry between H1(Ω) and H10(Ω). This implies that the operator A is completely continuous if the operator

    uB(u):=g(,u,u)

    is well-defined and completely continuous from H10(Ω) to H1(Ω).

    Assume that n3. Then the embedding H10(Ω)Lp(Ω) is continuous for 1p2=2n/(n2), and compact for 1p<2, and consequently the embedding Lq(Ω)H1(Ω) holds and is compact for q>(2)=2n/(n+2).

    We would like to represent B as a composition of three operators: B=JNP, where

    P:H10(Ω)L2(Ω)×L2(Ω;Rn),   P(u)=(u,u),N:L2(Ω)×L2(Ω;Rn)Lq(Ω),   N(w1,w2)=g(,w1,w2),J:Lq(Ω)H1(Ω),   J(v)=v.

    Clearly, since the embedding H10(Ω)L2(Ω) is continuous, P is a bounded linear operator. Also, if q>(2), then J is completely continuous. It remains to clarify the case of Nemytskii's operator N. It suffices that N is well-defined, continuous and bounded (maps bounded sets into bounded sets). To this aim, recall the main result about Nemytskii's operator (see, e.g., [29,Section 9.1]). According to this result, we need a growth condition on g, namely

    |g(x,w1,w2)|c1|w1|2q+c2|w2|2q+h(x)   (w1R, w2Rn, a.a.xΩ)

    where c1,c2R+ are constants and hLq(Ω). Notice that instead of the exponents 2/q, 2/q one may have smaller exponents, let they be α and β, hence a growth condition like

    |g(x,w1,w2)|c1|w1|α+c2|w2|β+h(x) (4.4)

    with 1α2q, 1β2q. These give some conditions on q:

    q2α,   q2β.

    Thus we can take

    q=min{2α, 2β}.

    Finally, the condition  q>(2)  holds if

    α<2(2),   β<2(2).

    Note that

    2(2)=n+2n2,   2(2)=n+2n.

    Therefore, the operator N is as desired provided that g satisfies the growth condition (4.4) for

    1α<n+2n2,    1β<n+2n 

    and hL2(Ω)

    Step 2: A priori boundedness of solutions.

    Let uH10(Ω) be any solution of the equation λA(u)=u for some λ(0,1). Then  u is a weak solution of the problem

    {(a+bλ2Ω|u|2dx)Δu=λf+λg(x,u,u)inΩu=0onΩ.

    Hence

    (a+bλ2Ω|u|2dx)(u,v)H10=(λf+λg(,u,u), v),   vH10(Ω).

    Letting v=u gives

    (a+bλ2|u|2H10)|u|2H10=λ(f,u)+λ(g(,u,u), u).

    Since g(,u,u)Lq(Ω), one has (g(,u,u), u)=Ωug(x,u,u). Assume that g satisfies the sign condition

    ug(x,u,v)0   foralluR, vRn, a.a.xΩ. (4.5)

    Then (g(,u,u), u)0 and so

    (a+bλ2|u|2H10)|u|2H10λ(f,u)|f|H1|u|H10.

    Thus

    a|u|H10(a+bλ2|u|2H10)|u|H10|f|H1,

    that is the solutions are bounded independently of λ, namely |u|H10|f|H1/a.

    Therefore, based on Schaefer's fixed point theorem, we have the following existence result.

    Theorem 9. Assume that g satisfies the growthcondition (4.4) for some numbers  1α<(n+2)/(n2), 1β<(n+2)/n  andfunction  hL2(Ω). Also assume that g has thesign property (4.5). Then problem (4.1) has atleast one weak solution uH10(Ω) with |u|H10|f|H1/a.

    Example 10. Consider the Dirichlet problem,

    {(1+B|u|2dx)Δu=2|x|u3u2+1uu2+1|u|  onBu|B=0, (4.6)

    where B is as in Example 8. We apply Theorem 9. Here

     f(x)=2|x|,   g(,u,v)=u3u2+1uu2+1|v|

    for uR and vRn. Similarly to Example 8, one has fH1(Ω) and |f|H1=1. Moreover, g satisfies the growth condition (4.4) with α=β=1 and the sign condition (4.5) since

    |g(x,u,v)||u|+12|v|

    and

    ug(x,u,v)=u4u2+1u2u2+1|v|0,

    for all uR and vRn. Consequently, problem (4.6) has at least one weak solution in H10(B) with |u|H101.

    In this section our focus is on system (1.1), where we look for a solution which is a Nash equilibrium.

    We start by an existence and uniqueness result in the whole space H10(Ω)×H10(Ω).

    Each equation of system (1.1) has a variational structure given respectively by the energy functionals  E1,E2:H10(Ω)×H10(Ω)R,

    E1(u,v)=14(2a+b|u|2H10)|u|2H10(f1,u)ΩG1(x,u(x),v(x))dx,E2(u,v)=14(2a+b|v|2H10)|v|2H10(f2,v)ΩG2(x,u(x),v(x))dx,

    where G1(x,u,v)=u0g1(x,s,v)ds and G2(x,u,v)=v0g2(x,u,s)ds. Using (3.5), we easily see that

    E11(u,v)=(a+b|u|2H10)u(Δ)1(f1+g1(,u,v)),E22(u,v)=(a+b|v|2H10)v(Δ)1(f2+g2(,u,v)),

    for every u,vH10(Ω).

    Before stating the main result of this section we introduce the following notion: A function H:Ω×RR is said to be of coercive-type if the functional ϕ:H10(Ω)R,

    ϕ(v)=14(2a+b|v|2H10)|v|2H10(f2,v)ΩH(x,v)dx (5.1)

    is coercive, i.e., ϕ(v)+ as |v|H10+.

    We have the following result on the existence of a Nash equilibrium under unilateral Lipschitz (monotonicity type) conditions.

    Theorem 11. Assume that for i=1,2, fiH1(Ω), gi: Ω×R2R is a Carathéodory function and gi(,0,0)=0. In addition assume that the following conditions aresatisfied:

    (h1) There exist constants aijR+ (i,j=1,2) such that

    aii<λ1a,   i=1,2,
    a12a21<(λ1aa11)(λ1aa22) (5.2)

    and

    (g1(x,u,v)g1(x,¯u,¯v))(u¯u)a11|u¯u|2+a12|u¯u||v¯v|,(g2(x,u,v)g2(x,¯u,¯v))(v¯v)a21|u¯u||v¯v|+a22|v¯v|2 (5.3)

    for all u,v,¯u,¯vR and a.e. xΩ.

    (h2) There exist  two functions H1, H2:Ω×RR of coercive-type such that

    H1(x,v)G2(x,u,v)H2(x,v)

    for all u,vR, a.e. xΩ.

    Then system (1.1) has a unique solution which is aNash equilibrium for the pair of functionals (E1,E2).

    Proof. The proof follows the idea from [22]. For a clear comprehending, westructure our proof in six steps.

    Step 1:The functionals E1(,v) and E2(u,) arebounded from below. First let us remark that from (5.3), for every u,vR, there exist θ(0,1) such that

    G1(x,u,v)=u0g1(x,s,v)ds=ug1(x,θu,v)=1θg1(x,θu,v)θu1θ(a11|θu|2+a12|θu||v|)=a11θu2+a12|u||v|a11u2+a12|u||v|.

    Similarly

    G2(x,u,v)a21|u||v|+a22v2.

    Now let vH10(Ω) be fixed. For any uH10(Ω), one has

    E1(u,v)=14(2a+b|u|2H10)|u|2H10(f1,u)ΩG1(x,u(x),v(x))dx14(2a+b|u|2H10)|u|2H10|f1|H1|u|H10(a11|u|2L2+a12|u|L2|v|L2)14(2a+b|u|2H10)|u|2H10a111λ1|u|2H10a121λ1|u|H10|v|H10|f1|H1|u|H10b4|u|4H10+(a2a11λ1)|u|2H10(|f1|H1+a12λ1|v|H10)|u|H10,

    which is bounded from below since the coefficient of the term of forthdegree of the quartic expression in |u|H10is positive. Similarly the functional E2(u,) isbounded from below for each u.

    Step 2:Construction of an approximation sequence (uk,vk).

    Now, similarly to [21], starting with an arbitrary v0 and usingEkeland's variational principle, we recursively construct a sequence (uk,vk)H10(Ω)×H10(Ω) such that

    E1(uk,vk1)infH10(Ω)E1(,vk1)+1k,   E2(uk,vk)infH10(Ω)E2(uk,)+1k,|E11(uk,vk1)|H101k,|E22(uk,vk)|H101k. (5.4)

    Step 3: Boundedness of the sequence (vk).

    Let ϕ1,ϕ2 be the functionals of type (5.1) with ϕreplaced by ϕ1 and ϕ2, respectively. As coercivefunctionals they are bounded from below.

    Obviously, for every u,v, one has

    ϕ1(v)E2(u,v)ϕ2(v).

    The coerciveness of ϕ2 implies that there is R>0 with

    ϕ2(v)infH10(Ω)ϕ1+1,    |v|H10>R.

    Since infH10(Ω)ϕ1infH10(Ω)E2(u,) for all u, we obtain

    E2(u,v)infH10(Ω)E2(u,)+1   forallu,vH10(Ω), |v|H10>R. (5.5)

    Since for k2,

    E2(uk,vk)infH10(Ω)E2(uk,)+1k<infH10(Ω)E2(uk,)+1,

    in view of (5.5) we must have |vk|H10R, that is the boundedness of the sequence (vk).

    Step 4:Convergence of the sequences (uk) and (vk).

    For every u,¯u,v,¯vH10(Ω), we have

    (E11(u,v)E11(¯u,¯v),u¯u)H10=((a+b|u|2H10)u(a+b|¯u|2H10)¯u,u¯u)H10(g1(,u,v)g1(,¯u,¯v), u¯u)L2=a|u¯u|2H10+b(|u|2H10u|¯u|2H10¯u, u¯u)H10(g1(,u,v)g1(,¯u,¯v), u¯u)L2.

    Since

    (|u|2H10u|¯u|2H10¯u, u¯u)H10=|u|4H10+|¯u|4H10(|u|2H10+|¯u|2H10)(u,¯u)H10|u|4H10+|¯u|4H10(|u|2H10+|¯u|2H10)|u|H10|¯u|H10=(|u|2H10+|¯u|2H10+|u|H10|¯u|H10)(|u|H10|¯u|H10)20

    we obtain

    (E11(u,v)E11(¯u,¯v),u¯u)H10a|u¯u|2H10(g1(,u,v)g1(,¯u,¯v), u¯u)L2a|u¯u|2H10a11|u¯u|2L2a12|u¯u|L2|v¯v|L2(aa11λ1)|u¯u|2H10a12λ1|u¯u|H10|v¯v|H10. (5.6)

    Similarly

    (E22(u,v)E22(¯u,¯v),v¯v)H10(aa22λ1)|v¯v|2H10a21λ1|u¯u|H10|v¯v|H10. (5.7)

    On the other hand, from (5.4) we obtain

    (E11(uk+p,vk+p1)E11(uk,vk1),uk+puk)H10(1k+p+1k)|uk+puk|H10,
    (E22(uk+p,vk+p)E11(uk,vk),vk+pvk)H10(1k+p+1k)|vk+pvk|H10.

    Consequently, if we denote mii=aaiiλ1 (i=1,2), m12=a12λ1 and m21=a21λ1, then

    m11|uk+puk|H10m12|vk+p1vk1|H102k,m21|uk+puk|H10+m22|vk+pvk|H102k. (5.8)

    Under the notations xk,p:=|uk+puk|H10 and yk,p=|vk+pvk|H10, relations (5.8) can be put underthe matrix form

    M[xk,pyk,p]2k[11](MM)[xk1,pyk1,p],

    where

    M=[m11m12m21m22],   M=[m110m21m22].

    Since M is invertible and its inverse

    M1=[1m110m21m11m221m22]

    is nonnegative, we obtain

    [xk,pyk,p]M12k[11]M1[0m1200][xk1,pyk1,p]=[2k1m112k(m21m11m22+1m22)]+[m12m11yk1,pm21m12m11m22yn1,p]

    and therefore

    xk,p2km11+m12m11yk1,p, (5.9)
    yk,p2k(m12m11m22+1m22)+m12m21m11m22yk1,p

    From (5.2) one has α:=m12m21m11m22<1 andhence

    yk,pαyk1,p+2k(m12m11m22+1m22).

    Now we use the following lemma provided in [21].

    Lemma 12. Let (yk,p),(zk,p) be two sequences of real numbersdepending on a parameter p, such that

    (yk,p)isboundeduniformlywithrespecttop

    and

    0yk,pαyk1,p+zk,pforsomeα(0,1).

    If zk,p0 as k+ uniformly with respectto p, then yk,p0 as k+ uniformlywith respect to p.

    According to this result, since (vk) is bounded and then (yk,p) is bounded uniformly with respect to p, weconclude that yk,p0 as k+ uniformlywith respect to p. It follows that (vk) is a Cauchy sequence. Next, the inequality (5.9) implies that (uk) is also a Cauchy sequence.Denote by  u,v their limits.

    Step 5:Transition to the limit.

    If we pass to the limit in (5.4) we obtain

    E1(u,v)=infH10(Ω)E1(,v),  E2(u,v)=infH10(Ω)E2(u,),E11(u,v)=E22(u,v)=0,

    i.e., (u,v) is a solution of (1.1) andalso is a Nash equilibrium for the pair of functional (E1,E2).

    Step 6: Uniqueness.

    Assume there are two different solutions of the system (1.1), denoted with (u,v) and (¯u,¯v). Then

    E11(u,v)=0 , E22(u,v)=0,E11(¯u,¯v)=0 , E22(¯u,¯v)=0.

    On the other hand, from (5.6) and (5.7), we have

    0m11|u¯u|2H10m12|u¯u|H10|v¯v|H10,0m22|v¯v|2H10m21|u¯u|H10|v¯v|H10. (5.10)

    If u=¯u or v=¯v then in each case |u¯u|=0 or |v¯v|=0, concluding that u=¯u and v=¯v. In what follows we will work under assumption that u¯u and v¯v. From (5.10) we obtain

    |u¯u|H10m12m11|v¯v|H10,|v¯v|H10m21m22|u¯u|H10,

    whence

    |v¯v|H10m12m21m11m22|v¯v|H10.

    Since from (5.2) one has m12m21m11m22<1, we conclude that

    |v¯v|H10<v¯v|H10,

    which is impossible. Hence u=¯u and v=¯v.

    Remark 13 (Classical Lipschitz conditions). Obviously the unilateral Lipschitz conditions (5.3) are satisfied if g1,g2 fulfill the classical Lipschitz conditions

    |g1(x,u,v)g1(x,¯u,¯v)|a11|u¯u|+a12||v¯v|,|g2(x,u,v)g2(x,¯u,¯v)|a21|u¯u|+a22|v¯v|,

    for all u,v,¯u,¯vR and a.e. xΩ. In this particular case considered in [21] (see also [27]), the required conditions on the coefficients aij make possible to derive the existence and uniqueness of the solution of system (1.2) directly from Perov's fixed point theorem. We note that unilateral Lipschitz conditions for Nash equilibria of systems have been used for the first time in paper [22].

    Example 14 Consider the Dirichlet problem for the system of Kirchhoff type

    {(1+10|u|2)u=usinv(1+10|v|2)v=v+sinuu(0)=v(0)=u(1)=v(1)=0. on (0,1) (5.11)

    We apply Theorem 11, where

    Ω=(0,1),a=b=1,g1(x,u,v)=usinv,g2(x,u,v)=sinu+v.

    Note that condition (5.3) holds with aij=1 (i,j=1,2). The first eigenvalue of the Dirichlet problem u=λu on (0,1), u(0)=u(1)=0 is equal to π2 (see, e.g., [28,p. 72]), whence relation (5.2) is valid since 1<π2 and 1<(π21)2. In order to check (h2) we compute

    G2(x,u,v)=v0(s+sinu)ds=12v2+v sinu.

    Consider the coercive-type functions H1(x,v)=12v2|v| and H2(x,v)=12v2+|v|. Clearly

    H1(x,v)G2(x,u,v)H2(x,v).

    Therefore, the Dirichlet problem (5.11) has a unique solution (u,v)H10(0,1)×H10(0,1) which is a Nash equilibrium for the corresponding energy functionals.

    Let R1,R2>0 and denote by BR1, BR2 two closed balls of H10(Ω), of center the origin and radius R1 and R2, respectively. Now, our interest is focused on an existence and uniqueness result of the system (1.1) on BR1×BR2.

    Here an additional ingredient is given by the Leray-Schauder boundary conditions

    E11(u,v)+μu0 for all (u,v)BR1×BR2 with |u|H10=R1 and all μ>0, E22(u,v)+γv0 for all (u,v)BR1×BR2 with |v|H10=R2 and all γ>0. (5.12)

    Theorem 15. Assume that for i=1,2, fiH1(Ω), gi: Ω×R2R is a Carathéodory function, gi(,0,0)=0, and that condition(h1) holds. In addition assume that

    (h2')

    a11λ1R1+a12λ1R2+|f1|H1aR1+bR31,a21λ1R1+a22λ1R2+|f2|H1aR2+bR32.

    Then system (1.1) has in BR1×BR2 a unique solution which is a Nash equilibrium in BR1×BR2 for the pair of functionals (E1,E2).

    Proof. Step 1: As at Step 1 from the proof of Theorem 11, thefunctionals E1 and E2 are bounded from below on BR1×BR2.

    Step 2: E1 and E2 satisfy the Leray-Schauder boundaryconditions (5.12).

    Assume that there exist (u,v)BR1×BR2 with |u|H10=R1 and μ>0 such that

    E11(u,v)+μu=0.

    Then

    (a+b|u|2H10)|u|2H10+μ|u|2H10((Δ)1(f1+g1(,u,v)),u)H10=0,

    which gives

    (a+bR21)R21+μR21=((Δ)1(f1+g1(,u,v)),u)H10=(f1+g1(,u,v),u)L2R1|f1|H1+a11|u|2L2+a12|u|L2|v|L2R1|f1|H1+a11λ1R21+a12λ1R1R2,

    whence

    (a+bR21)R1+μR1|f1|H1+a11λ1R1+a12λ1R2,

    which contradicts the first relation in (h2'). An analog reasoning appliesfor E2.

    Step 3: Construction of an approximation sequence.

    As in the proof of Lemma 2.1 in [24], starting from an arbitrarilyinitial point v0BR2 and applying recursively Ekeland'svariational principle, we obtain a sequence (uk,vk)BR1×BR2 such that

    E1(uk,vk1)infBR1E1(,vk1)+1k,   E2(uk,vk)infBR2E2(uk,)+1k,|E11(uk,vk1)+μkuk|H101k,|E22(uk,vk)+γkvk|H101k,

    where

    μk={1R21(E11(uk,vk1),uk)H10, if|uk|H10=R1and(E11(uk,vk1),uk)H10<00,otherwise

    and

    γk={1R22(E22(uk,vk),vk)H10, if|vk|H10=R2and(E22(uk,vk),vk)H10<00,otherwise.

    Step 4: Convergence to zero of the sequences (μk) and (γk).

    Assume the contrary. Then, passing eventually to subsequences, we may assumethat μkμ>0 or γkγ>0.Using the expressions of E11 and E22 and denoting

    αk:=E11(uk,vk1)+μkuk,   βk:=E22(uk,vk)+γkvk, (5.13)

    we have

    uk=S(f1+g1(,uk,vk1))+μka+b|uk|2H10,vk=S(f2+g2(,uk,vk))+γka+b|vk|2H10. (5.14)

    The sequences (uk),(vk) being boundedand the operators S(f1+g1(,u,v)),§(f2+g2(,u,v)) being compact, we havethat the two sequences from the right-hand sides in (5.14) are compact; thus (uk) and (vk) have convergentsubsequences (ukj), (vkj). The samereasoning applied to the second formula in (5.14) with kj1instead of k allows us, passing again to subsequence, to assume that thesequences (ukj), (vkj) and (vkj1) are convergent. Let u,v,¯v be their limits. If we take the limit in (5.13)

    E11(u,¯v)+μu=0,    E22(u,v)+γv=0,

    where |u|H10=R1 if μ>0 and |v|H10=R2 if γ>0. In each case, one of the two Leray-Schauder conditions (5.12) is contradicted.Therefore μk0 and γk0 as k+.

    Step 5: Estimations for (uk) and (vk).

    We can proceed similarly to Theorem 11, Step 4, to obtain inequalities(5.6) and (5.7). Under the notations from Step 4 in the proof ofthe previous theorem, and the additional notations ck,p=|μk+pμk|, dk,p=|γk+pγk|, we arrive to the matrix inequality

    Mk[xk,pyk,p]2k[11](MkMk)[xk1,pyk1,p]+[ck,pR1dk,pR2],

    where now

    Mk=[m11+μkm12m21m22+μk],   Mk=[m11+μk0m21m22+μk].

    Since for any kN, Mk is invertible and

    Mk1=[1m11+μk0m21(m11+μk)(m22+γk)1m22+γk]

    we obtain

    [xk,pyk,p]2kM1k[11]M1k[0m1200][xk1,pyk1,p]+M1k[ck,pR1dk,pR2].

    Thus

    xk,p2k(m11+μk)+m12m11+μkyk1,p+ck,pR11m11+μk2km11+m12m11yk1,p+ck,pR11m11,yk,p2k(1m22+γk)+1(m11+μk)(m22+γk)[2km12+m12m21yk1,p+m21ck,pR1]+1m22+γkdk,pR22k(m12m11m22+1m22)+m12m21m11m12yk1,p+m21m11m22ck,pR1+1m22dk,pR2.

    Hence

    yk,pαyk1,p+2k(m12m11m22+1m22)+m21m11m22ck,pR1+1m22dk,pR2 ,

    where α:=m12m21m11<1 and ck,p, dk,pconverge to zero uniformly with respect to p. Now the conclusion followsas in the proof of Theorem 11 with the limits u and v of the sequences (uk) and (vk)satisfying

    E11(u,v)=0,E22(u,v)=0

    and

    E1(u,v)=infBR1E1(,v),   E22(u,v)=infBR2E2(u,).

    Step 6: Uniqueness.

    Similar to the proof in Theorem 11.

    Example 16. Consider the Dirichlet problem for the system of Kirchhoff type

    {(2+10|u|2)u=u3+usinv+π2sin(πx)(2+10|v|2)v=v3+v+sinuu(0)=v(0)=u(1)=v(1)=0.on (0,1) (5.15)

    For R1=R2=1, we apply Theorem 15, where

    Ω=(0,1),a=2,b=1,f1(x)=π2sin(πx),f20,
    g1(x,u,v)=u3+usinv,g2(x,u,v)=v3+v+sinu.

    Since

    (u3+¯u3)(u¯u)=(u¯u)2(u2+u¯u+¯u2)0,

    one has

    (g1(x,u,v)g1(x,¯u,¯v))(u¯u)(usinv¯u+sin¯v)(u¯u)|u¯u|2+|u¯u||v¯v|.

    Similarly

    (g2(x,u,v)g2(x,¯u,¯v))(v¯v)|u¯u||v¯v|+|v¯v|2.

    Hence, condition (5.3) holds with aij=1 for i,j=1,2. In addition, since λ1=π2, condition (5.2) also holds. Thus assumption (h1) is satisfied. Next we check condition (h2'). We have |f2|H1=0 and that the function u0(x)=sin(πx) is the solution of the Dirichlet problem u=f1 in (0,1), u(0)=u(1)=0. Then

    |f1|H1=|u0|H10=|u0|L2=(10π2cos2(πx))12=π2.

    Now, condition (h2') is verified since both 2 / π2+π /2 and 2 / π2 are less than 3.

    Therefore, the Dirichlet problem (5.15) has a unique solution

    (u,v){uH10(0,1):|u|H101}×{vH10(0,1):|v|H101},

    which is a Nash equilibrium for the corresponding energy functionals.

    In this paper, we have studied the existence, uniqueness, localization and variational properties of solutions for some equations and systems of Kirchhoff type. First we have defined the solution operator associated to nonhomogeneous equations subjected to the Dirichlet boundary condition and we have made the connexion with the corresponding energy functional. Next, we have considered equations with a reaction term and using Banach contraction principle and Schaefer's fixed point theorem we have established sufficient conditions so that a solution exist and be localized in some bounded sets. For a system of two Kirchhoff equations, under appropriate conditions, we have proved the existence of a unique solution which appears as a Nash equilibrium for the associated energy functionals. Both global Nash equilibrium, in the whole space, and local Nash equilibrium, in balls, have been obtained by using an iterative procedure simulating a noncooperative game and based on Ekeland's variational principle.

    The authors declare no conflict of interest.



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