Research article

Positive radial solutions for a boundary value problem associated to a system of elliptic equations with semipositone nonlinearities

  • Received: 31 August 2022 Revised: 30 September 2022 Accepted: 12 October 2022 Published: 17 October 2022
  • MSC : 34B16, 34B25, 35J57

  • In this paper we use the fixed point index theory to study the existence of positive radial solutions for a system of boundary value problems with semipositone second order elliptic equations. Some appropriate concave and convex functions are utilized to characterize coupling behaviors of our nonlinearities.

    Citation: Limin Guo, Jiafa Xu, Donal O'Regan. Positive radial solutions for a boundary value problem associated to a system of elliptic equations with semipositone nonlinearities[J]. AIMS Mathematics, 2023, 8(1): 1072-1089. doi: 10.3934/math.2023053

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  • In this paper we use the fixed point index theory to study the existence of positive radial solutions for a system of boundary value problems with semipositone second order elliptic equations. Some appropriate concave and convex functions are utilized to characterize coupling behaviors of our nonlinearities.



    Let Ω={zRn:R1<|z|<R2,R1,R2>0}. In this work we study the existence of positive radial solutions for the following system of boundary value problems with semipositone second order elliptic equations:

    {Δφ+k(|z|)f(φ,ϕ)=0, zΩ,Δϕ+k(|z|)g(φ,ϕ)=0, zΩ,αφ+βφn=0, αϕ+βϕn=0, |z|=R1,γφ+δφn=0, γϕ+δϕn=0, |z|=R2, (1.1)

    where α,β,γ,δ,k,f,g satisfy the conditions:

    (H1) α,β,γ,δ0 with ργβ+αγ+αδ>0;

    (H2) kC([R1,R2],R+), and k is not vanishing on [R1,R2];

    (H3) f,gC(R+×R+,R), and there is a positive constant M such that

    f(u,v),g(u,v)M, u,vR+.

    Elliptic equations have attracted a lot of attention in the literature since they are closely related to many mathematical and physical problems, for instance, incineration theory of gases, solid state physics, electrostatic field problems, variational methods and optimal control. The existence of solutions for this type of equation in annular domains has been discussed in the literature, see for example, [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18] and the references therein. In [1] the authors used the fixed point index to study positive solutions for the elliptic system:

    {Δu+a(|x|)f(u,v)=0,Δv+b(|x|)g(u,v)=0,

    with one of the following boundary conditions

    u=v=0,|x|=R1,|x|=R2,u=v=0,|x|=R1,ur=vr=0,|x|=R2,ur=vr=0,|x|=R1,u=v=0,|x|=R2.

    In [2] the authors used the method of upper and lower solutions to establish the existence of positive radial solutions for the elliptic equation

    {Δu=f(|x|,u,|u|), xΩ,u|Ω=0,

    where Ω={xRN: |x|<1},N2, and f:[0,1]×R+×R+R is a continuous function.

    However, we note that in most of the papers on nonlinear differential equations the nonlinear term is usually assumed to be nonnegative. In recent years boundary value problems for semipositone equations (f(t,x)M,M>0) has received some attention (see [19,20,21,22,23,24,25,26,27,28,29,30,31,32]), and these equations describe and solve many natural phenomena in engineering and technical problems in real life, for example in mechanical systems, suspension bridge design, astrophysics and combustion theoretical models. In [19] the authors used a fixed point theorem to study the system for HIV-1 population dynamics in the fractional sense

    {Dα0+u(t)+λf(t,u(t),Dβ0+u(t),v(t))=0,t(0,1),Dγ0+v(t)+λg(t,u(t))=0,t(0,1),Dβ0+u(0)=Dβ+10+u(0)=0,Dβ0+u(1)=10Dβ0+u(s)dA(s),v(0)=v(0)=0,v(1)=10v(s)dB(s),

    where Dα0+,Dβ0+,Dγ0+ are the standard Riemann-Liouville derivatives, and f, g are two semipositone nonlinearities. In [28] the authors used the nonlinear alternative of Leray-Schauder type and the Guo-Krasnosel'skii fixed point theorem to study the existence of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations

    {Dα0+u(t)+λf(t,v(t))=0,0<t<1, λ>0,Dα0+v(t)+λg(t,u(t))=0,0<t<1, λ>0,u(j)(0)=v(j)(0)=0,0jn2,u(1)=μ10u(s)ds,v(1)=μ10v(s)ds,

    where f,g satisfy some superlinear or sublinear conditions:

    (HZ)1 There exist M>0 such that lim supz0g(t,z)z<M uniformly for t[0,1] (sublinear growth condition).

    (HZ)2 There exists [θ1,θ2](0,1) such that lim infz+f(t,z)z=+ and lim infz+g(t,z)z=+ uniformly for t[θ1,θ2] (superlinear growth condition).

    Inspired by the aforementioned work, in particular [31,32,33,34], we study positive radial solutions for (1.1) when the nonlinearities f,g satisfy the semipositone condition (H3). Moreover, some appropriate concave and convex functions are utilized to characterize coupling behaviors of our nonlinearities. Note that our conditions (H4) and (H6) (see Section 3) are more general than that in (HZ)1 and (HZ)2.

    Using the methods in [1,4], we transform (1.1) into a system of ordinary differential equations involving Sturm-Liouville boundary conditions. Let φ=φ(r),ϕ=ϕ(r),r=|z|=ni=1z2i. Then (1.1) can be expressed by the following system of ordinary differential equations:

    {φ(r)+n1rφ(r)+k(r)f(φ(r),ϕ(r))=0, R1<r<R2,ϕ(r)+n1rϕ(r)+k(r)g(φ(r),ϕ(r))=0, R1<r<R2,αφ(R1)βφ(R1)=0, γφ(R2)+δφ(R2)=0,αϕ(R1)βϕ(R1)=0, γϕ(R2)+δϕ(R2)=0. (2.1)

    Then if we let s=R2r(1/tn1)dt,t=(ms)/m,m=R2R1(1/tn1)dt, (2.1) can be transformed into the system

    {φ(t)+h(t)f(φ(t),ϕ(t))=0,0<t<1,ϕ(t)+h(t)g(φ(t),ϕ(t))=0,0<t<1,αφ(0)βφ(0)=0,γφ(1)+δφ(1)=0,αϕ(0)βϕ(0)=0,γϕ(1)+δϕ(1)=0, (2.2)

    where h(t)=m2r2(n1)(m(1t))k(r(m(1t))). Consequently, (2.2) is equivalent to the following system of integral equations

    {φ(t)=10G(t,s)h(s)f(φ(s),ϕ(s))ds,ϕ(t)=10G(t,s)h(s)g(φ(s),ϕ(s))ds, (2.3)

    where

    G(t,s)=1ρ{(γ+δγt)(β+αs),0st1,(γ+δγs)(β+αt),0ts1, (2.4)

    and ρ is defined in (H1).

    Lemma 2.1. Suppose that (H1) holds. Then

    (i)

    ρ(γ+δ)(β+α)G(t,t)G(s,s)G(t,s)G(s,s), t,s[0,1];

    (ii)

    G(t,s)G(t,t), t,s[0,1].

    Proof. (i) In G(t,s), we fix the second variable s, we have

    G(t,s)=1ρ{(γ+δγt)(β+αs)(γ+δγs)(β+αs),0st1,(γ+δγs)(β+αt)(γ+δγs)(β+αs),0ts1.

    This implies that

    G(t,s)G(s,s),t,s[0,1].

    When ts, we have

    1ρ(γ+δγt)(β+αs)ρ1ρ1ρ(γ+δγt)(β+αt)(γ+δγs)(β+αs)1(β+α)(γ+δ).

    When ts, we have

    1ρ(γ+δγs)(β+αt)ρ1ρ1ρ(γ+δγt)(β+αt)(γ+δγs)(β+αs)1(β+α)(γ+δ).

    Combining the above we obtain

    G(t,s)G(t,t)G(s,s)ρ(β+α)(γ+δ).

    (ii) In G(t,s) we fix the first variable t, and we obtain

    G(t,s)=1ρ{(γ+δγt)(β+αs)(γ+δγt)(β+αt),0st1,(γ+δγs)(β+αt)(γ+δγt)(β+αt),0ts1.

    Thus

    G(t,s)G(t,t),t,s[0,1].

    Lemma 2.2. Suppose that (H1) holds. Let ϑ(t)=G(t,t)h(t),t[0,1]. Then

    κ1ϑ(s)10G(t,s)h(s)ϑ(t)dtκ2ϑ(s),

    where

    κ1=ρ(γ+δ)(β+α)10G(t,t)ϑ(t)dt, κ2=10ϑ(t)dt.

    Proof. From (H1) and Lemma 2.1(i) we have

    10G(t,s)h(s)ϑ(t)dt10G(s,s)h(s)ϑ(t)dt=κ2ϑ(s)

    and

    10G(t,s)h(s)ϑ(t)dt10ρ(γ+δ)(β+α)G(t,t)G(s,s)h(s)ϑ(t)dt=κ1ϑ(s).

    Note we study (2.3) to obtain positive solutions for (1.1). However here the nonlinear terms f,g can be sign-changing (see (H3)). Therefore we study the following auxiliary problem:

    u(t)=10G(t,s)h(s)˜f(u(s))ds, (2.5)

    where G is in (2.4) and ˜f satisfies the condition:

    (H2) ˜fC(R+,R), and there exists a positive constant M such that

    ˜f(u)M, uR+.

    Let w(t)=M10G(t,s)h(s)ds,t[0,1]. Then w is a solution of the following boundary value problem:

    {u(t)+h(t)M=0,0<t<1,αu(0)βu(0)=0,γu(1)+δu(1)=0. (2.6)

    Lemma 2.3. (i) If u satisfies (2.5), then u+w is a solution of the equation:

    u(t)=10G(t,s)h(s)˜F(u(s)w(s))ds, (2.7)

    where

    ˜F(u)={˜f(u)+M,u0,˜f(0)+M,u<0. (2.8)

    (ii) If u satisfies (2.7) with u(t)w(t),t[0,1], then uw is a positive solution for (2.5).

    Proof. We omit its proof since it is immediate.

    Let E=C[0,1], u=maxt[0,1]|u(t)|. Then (E,) is a Banach space. Define a set on E as follows:

    P={uE:u(t)0,t[0,1]},

    and note P is a cone on E. Note, E2=E×E is also a Banach space with the norm: (u,v)=u+v, and P2=P×P a cone on E2. In order to obtain positive radial solutions for (1.1), combining with (2.5)–(2.7), we define the following operator equation:

    A(φ,ϕ)=(φ,ϕ), (2.9)

    where A(φ,ϕ)=(A1,A2)(φ,ϕ), Ai(i=1,2) are

    {A1(φ,ϕ)(t)=10G(t,s)h(s)F1(φ(s)w(s),ϕ(s)w(s))ds,A2(φ,ϕ)(t)=10G(t,s)h(s)F2(φ(s)w(s),ϕ(s)w(s))ds, (2.10)

    and

    F1(φ,ϕ)={f(φ,ϕ)+M,φ,ϕ0,f(0,ϕ)+M,φ<0,ϕ0,f(φ,0)+M,φ0,ϕ<0,f(0,0)+M,φ,ϕ<0,
    F2(φ,ϕ)={g(φ,ϕ)+M,φ,ϕ0,g(0,ϕ)+M,φ<0,ϕ0,g(φ,0)+M,φ0,ϕ<0,g(0,0)+M,φ,ϕ<0.

    Lemma 2.4. Define P0={φP:φ(t)ρ(γ+δ)(β+α)G(t,t)φ,t[0,1]}. Then Ai(P×P)P0,i=1,2.

    Proof. We only prove it for A1. If φ,ϕP, note the non-negativity of F1(denoted by F1(,)), from Lemma 2.1(i) we have

    10ρ(γ+δ)(β+α)G(t,t)G(s,s)h(s)F1(,)dsA1(φ,ϕ)(t)10G(s,s)h(s)F1(,)ds.

    This implies that

    A1(φ,ϕ)(t)10ρ(γ+δ)(β+α)G(t,t)G(s,s)h(s)F1(,)dsρ(γ+δ)(β+α)G(t,t)A1(φ,ϕ).

    Remark 2.1. (i) w(t)=M10G(t,s)h(s)dsP0;

    (ii) Note (see Corollary 1.5.1 in [35]):

    If k(x,y,u):˜GטG×RR is continuous (˜G is a bounded closed domain in Rn), then K is a completely continuous operator from C(˜G) into itself, where

    Kψ(x)=˜Gk(x,y,ψ(y))dy.

    Note that G(t,s),h(s),Fi(i=1,2) are continuous, and also Ai, A are completely continuous operators, i=1,2.

    From Lemma 2.3 if there exists (φ,ϕ)P2{(0,0)} such that (2.9) holds with (φ,ϕ)(w,w), then φ(t),ϕ(t)w(t),t[0,1], and (φw,ϕw) is a positive solution for (2.3), i.e., we obtain positive radial solutions for (1.1). Note that φ,ϕP0, and from Lemma 2.1(ii) we have

    φ(t)w(t)ρ(γ+δ)(β+α)G(t,t)φM10G(t,t)h(s)ds,
    ϕ(t)w(t)ρ(γ+δ)(β+α)G(t,t)ϕM10G(t,t)h(s)ds.

    Hence, if

    φ,ϕM(γ+δ)(β+α)ρ10h(s)ds,

    we have (φ,ϕ)(w,w). As a result, we only need to seek fixed points of (2.9), when their norms are greater than M(γ+δ)(β+α)ρ10h(s)ds.

    Let E be a real Banach space. A subset XE is called a retract of E if there exists a continuous mapping r:EX such that r(x)=x, xX. Note that every cone in E is a retract of E. Let X be a retract of real Banach space E. Then, for every relatively bounded open subset U of X and every completely continuous operator A:¯UX which has no fixed points on U, there exists an integer i(A,U,X) satisfying the following conditions:

    (i) Normality: i(A,U,X)=1 if Axy0U for any x¯U.

    (ii) Additivity: i(A,U,X)=i(A,U1,X)+i(A,U2,X) whenever U1 and U2 are disjoint open subsets of U such that A has no fixed points on ¯U(U1U2).

    (iii) Homotopy invariance: i(H(t,),U,X) is independent of t (0t1) whenever H:[0,1]ׯUX is completely continuous and H(t,x)x for any (t,x)[0,1]×U.

    (iv) Permanence: i(A,U,X)=i(A,UY,Y) if Y is a retract of X and A(¯U)Y.

    Then i(A,U,X) is called the fixed point index of A on U with respect to X.

    Lemma 2.5. (see [35,36]). Let E be a real Banach space and P a cone on E. Suppose that ΩE is a bounded open set and that A:¯ΩPP is a continuous compact operator. If there exists ω0P{0} such that

    ωAωλω0,λ0,ωΩP,

    then i(A,ΩP,P)=0, where i denotes the fixed point index on P.

    Lemma 2.6. (see [35,36]). Let E be a real Banach space and P a cone on E. Suppose that ΩE is a bounded open set with 0Ω and that A:¯ΩPP is a continuous compact operator. If

    ωλAω0,λ[0,1],ωΩP,

    then i(A,ΩP,P)=1.

    Denote OM,h=M(γ+δ)(β+α)ρ10h(s)ds, Bζ={uE:u<ζ},ζ>0,B2ζ=Bζ×Bζ. We list our assumptions as follows:

    (H4) There exist p,qC(R+,R+) such that

    (i) p is a strictly increasing concave function on R+;

    (ii) lim infvf(u,v)p(v)1, lim infug(u,v)q(u)1;

    (iii) there exists e1(κ21,) such that lim infzp(LG,hq(z))ze1LG,h, where LG,h=maxt,s[0,1]G(t,s)h(s).

    (H5) There exists Qi(0,OM,hκ2) such that

    Fi(uw,vw)Qi,u,v[0,OM,h],i=1,2.

    (H6) There exist ζ,ηC(R+,R+) such that

    (i) ζ is a strictly increasing convex function on R+;

    (ii) lim supvf(u,v)ζ(v)1, lim supug(u,v)η(u)1;

    (iii) there exists e2(0,κ22) such that lim supzζ(LG,hη(z))ze2LG,h.

    (H7) There exists ˜Qi(OM,hκ2LG,) such that

    Fi(uw,vw)˜Qi,u,v[0,OM,h],i=1,2,

    where LG=maxt[0,1]ρ(γ+δ)(β+α)G(t,t).

    Remark 3.1. Condition (H4) implies that f grows p(v)-superlinearly at uniformly on uR+, g grows q(u)-superlinearly at uniformly on vR+; condition (H6) implies that f grows ζ(v)-sublinearly at uniformly on uR+, g grows η(u)-sublinearly at uniformly on vR+.

    Theorem 3.1. Suppose that (H1)–(H5) hold. Then (1.1) has at least one positive radial solution.

    Proof. Step 1. When φ,ϕBOM,hP, we have

    (φ,ϕ)λA(φ,ϕ),λ[0,1]. (3.1)

    Suppose the contrary i.e., if (3.1) is false, then there exist φ0,ϕ0BOM,hP and λ0[0,1] such that

    (φ0,ϕ0)=λ0A(φ0,ϕ0).

    This implies that

    φ0,ϕ0P0 (3.2)

    and

    φ0A1(φ0,ϕ0), ϕ0A2(φ0,ϕ0). (3.3)

    From (H5) we have

    Ai(φ0,ϕ0)(t)=10G(t,s)h(s)Fi(φ0(s)w(s),ϕ0(s)w(s))ds10ϑ(s)Qids<OM,h,i=1,2.

    Thus

    A1(φ0,ϕ0)+A2(φ0,ϕ0)<2OM,h=φ0+ϕ0(φ0,ϕ0BOM,hP),

    which contradicts (3.3), and thus (3.1) holds. From Lemma 2.6 we have

    i(A,B2OM,hP2,P2)=1. (3.4)

    Step 2. There exists a sufficiently large R>OM,h such that

    (φ,ϕ)A(φ,ϕ)+λ(ϱ1,ϱ1),φ,ϕBRP,λ0, (3.5)

    where ϱ1P0 is a given element. Suppose the contrary. Then there are φ1,ϕ1BRP,λ10 such that

    (φ1,ϕ1)=A(φ1,ϕ1)+λ1(ϱ1,ϱ1). (3.6)

    This implies that

    φ1(t)=A1(φ1,ϕ1)(t)+λ1ϱ1(t), ϕ1(t)=A2(φ1,ϕ1)(t)+λ1ϱ1(t),t[0,1].

    From Lemma 2.4 and ϱ1P0 we have

    φ1,ϕ1P0. (3.7)

    Note that φ1=ϕ1=R>OM,h, and thus φ1(t)w(t),ϕ1(t)w(t),t[0,1].

    By (H4)(ii) we obtain

    lim infϕF1(φ,ϕ)p(ϕ)=lim infϕf(φ,ϕ)+Mp(ϕ)1, lim infφF2(φ,ϕ)q(φ)=lim infφg(φ,ϕ)+Mq(φ)1.

    This implies that there exist c1,c2>0 such that

    F1(φ,ϕ)p(ϕ)c1, F2(φ,ϕ)q(φ)c2, φ,ϕR+.

    Therefore, we have

    φ1(t)=A1(φ1,ϕ1)(t)+λ1ϱ1(t)A1(φ1,ϕ1)(t)10G(t,s)h(s)[p(ϕ1(s)w(s))c1]ds10G(t,s)h(s)p(ϕ1(s)w(s))dsc1κ2 (3.8)

    and

    ϕ1(t)=A2(φ1,ϕ1)(t)+λ1ϱ1(t)A2(φ1,ϕ1)(t)10G(t,s)h(s)[q(φ1(s)w(s))c2]ds10G(t,s)h(s)q(φ1(s)w(s))dsc2κ2. (3.9)

    Consequently, we have

    ϕ1(t)w(t)10G(t,s)h(s)q(φ1(s)w(s))dsc2κ2w(t)10G(t,s)h(s)q(φ1(s)w(s))ds(c2+M)κ2.

    From (H4)(iii), there is a c3>0 such that

    p(LG,hq(z))e1LG,hzLG,hc3,zR+.

    Combining with (H4)(i), we have

    p(ϕ1(t)w(t))p(ϕ1(t)w(t)+(c2+M)κ2)p((c2+M)κ2)p(10G(t,s)h(s)q(φ1(s)w(s))ds)p((c2+M)κ2)=p(10G(t,s)h(s)LG,hLG,hq(φ1(s)w(s))ds)p((c2+M)κ2)10p(G(t,s)h(s)LG,hLG,hq(φ1(s)w(s)))dsp((c2+M)κ2)10G(t,s)h(s)LG,hp(LG,hq(φ1(s)w(s)))dsp((c2+M)κ2)10G(t,s)h(s)LG,h(e1LG,h(φ1(s)w(s))LG,hc3)dsp((c2+M)κ2)e110G(t,s)h(s)(φ1(s)w(s))dsp((c2+M)κ2)c3κ2.

    Substituting this inequality into (3.8) we have

    φ1(t)w(t)10G(t,s)h(s)[e110G(s,τ)h(τ)(φ1(τ)w(τ))dτp((c2+M)κ2)c3κ2]ds   (c1+M)κ2e11010G(t,s)h(s)G(s,τ)h(τ)(φ1(τ)w(τ))dτds   p((c2+M)κ2)κ2c3κ22(c1+M)κ2.

    Multiply by ϑ(t) on both sides of the above and integrate over [0,1] and use Lemma 2.2 to obtain

    10(φ1(t)w(t))ϑ(t)dte110ϑ(t)1010G(t,s)h(s)G(s,τ)h(τ)(φ1(τ)w(τ))dτdsdt   p((c2+M)κ2)κ22c3κ32(c1+M)κ22e1κ2110(φ1(t)w(t))ϑ(t)dtp((c2+M)κ2)κ22c3κ32(c1+M)κ22.

    From this inequality we have

    10(φ1(t)w(t))ϑ(t)dtp((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211

    and thus

    10φ1(t)ϑ(t)dtp((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+10w(t)ϑ(t)dtp((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22.

    Note that (3.7), φ1P0, and we have

    φ1p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22κ1(e1κ211)+Mκ22κ1.

    On the other hand, multiply by ϑ(t) on both sides of (3.8) and integrate over [0,1] and use Lemma 2.2 to obtain

    κ110ϑ(t)p(ϕ1(t)w(t))dt10φ1(t)ϑ(t)dt+c1κ22p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22+c1κ22.

    From Remark 2.1 we have wP0, note that ϕ1=R>M(γ+δ)(β+α)ρ10h(s)dsw and ϕ1P0, then ϕ1wP0. By the concavity of p we have

    ϕ1wκ1110(ϕ1(t)w(t))ϑ(t)dt=ϕ1wκ1p(ϕ1w)10ϕ1(t)w(t)ϕ1wp(ϕ1w)ϑ(t)dtϕ1wκ1p(ϕ1w)10p(ϕ1(t)w(t)ϕ1wϕ1w)ϑ(t)dtϕ1wκ21p(ϕ1w)[p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22+c1κ22].

    This implies that

    p(ϕ1w)1κ21[p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22+c1κ22].

    From (H4)(i) we have

    p(ϕ1)=p(ϕ1w+w)p(ϕ1w+w)p(ϕ1w)+p(w)1κ21[p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22+c1κ22]+p(w)1κ21[p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22e1κ211+Mκ22+c1κ22]+p(Mκ2)<+.

    Therefore, there exists Oϕ1>0 such that ϕ1Oϕ1.

    We have prove the boundedness of φ1,ϕ1 when (3.6) holds, i.e., when φ1,ϕ1BRP, there exist a positive constant to control the norms of φ1,ϕ1. Now we choose a sufficiently large

    R1>max{OM,h,Oϕ1,p((c2+M)κ2)κ22+c3κ32+(c1+M)κ22κ1(e1κ211)+Mκ22κ1}.

    Then when φ1,ϕ1BR1P, (3.6) is not satisfied, and thus (3.5) holds. From Lemma 2.5 we have

    i(A,B2R1P2,P2)=0. (3.10)

    Combining (3.4) with (3.10) we have

    i(A,(B2R1¯B2OM,h)P2,P2)=i(A,B2R1P2,P2)i(A,B2OM,hP2,P2)=01=1.

    Then the operator A has at least one fixed point (denoted by (φ,ϕ)) on (B2R1¯B2OM,h)P2 with φ(t),ϕ(t)w(t),t[0,1]. Therefore, (φw,ϕw) is a positive solution for (2.2), and (1.1) has at least one positive radial solution.

    Theorem 3.2. Suppose that (H1)–(H3), (H6) and (H7) hold. Then (1.1) has at least one positive radial solution.

    Proof. Step 1. When φ,ϕBOM,hP, we have

    (φ,ϕ)A(φ,ϕ)+λ(ϱ2,ϱ2),λ0, (3.11)

    where ϱ2P is a given element. Suppose the contrary. Then there exist φ2,ϕ2BOM,hP,λ20 such that

    (φ2,ϕ2)=A(φ2,ϕ2)+λ2(ϱ2,ϱ2).

    This implies that

    φ2φ2(t)A1(φ2,ϕ2)(t)+λ2ϱ2(t)A1(φ2,ϕ2)(t),t[0,1],
    ϕ2ϕ2(t)A2(φ2,ϕ2)(t)+λ2ϱ2(t)A2(φ2,ϕ2)(t),t[0,1].

    Then we have

    φ2+ϕ2A1(φ2,ϕ2)+A2(φ2,ϕ2). (3.12)

    From (H7) we have

    Ai(φ2,ϕ2)=maxt[0,1]Ai(φ2,ϕ2)(t)maxt[0,1]ρ(γ+δ)(β+α)G(t,t)10G(s,s)h(s)Fi(φ2(s)w(s),ϕ2(s)w(s))dsLG10G(s,s)h(s)˜Qids=˜Qiκ2LG,i=1,2.

    By the condition on ˜Qi we have

    A1(φ2,ϕ2)+A2(φ2,ϕ2)>2OM,h=φ2+ϕ2,

    and this contradicts (3.12), so (3.11) holds. By Lemma 2.5 we have

    i(A,B2OM,hP2,P2)=0. (3.13)

    Step 2. There exists a sufficiently large R>OM,h such that

    (φ,ϕ)λA(φ,ϕ),φ,ϕBRP,λ[0,1]. (3.14)

    Suppose the contrary. Then there exist φ3,ϕ3BRP,λ3[0,1] such that

    (φ3,ϕ3)=λ3A(φ3,ϕ3). (3.15)

    Combining with Lemma 2.4 we have

    φ3,ϕ3P0. (3.16)

    Note that φ3,ϕ3BRP, and then φ3(t)w(t),ϕ3(t)w(t)0,t[0,1]. Hence, from (H6) we have

    lim supϕF1(φ,ϕ)ζ(ϕ)=lim supϕf(φ,ϕ)+Mζ(ϕ)1, lim supφF2(φ,ϕ)η(φ)=lim supφg(φ,ϕ)+Mη(φ)1.

    This implies that there exists ˜M>0 such that

    F1(φ,ϕ)ζ(ϕ), F2(φ,ϕ)η(φ),φ,ϕ˜M. (3.17)

    By similar methods as in Theorem 3.1, choosing R>˜M, and from (3.15) we obtain

    φ3(t)=λ3A1(φ3,ϕ3)(t)10G(t,s)h(s)ζ(ϕ3(s)w(s))ds (3.18)

    and

    ϕ3(t)=λ3A2(φ3,ϕ3)(t)10G(t,s)h(s)η(φ3(s)w(s))ds. (3.19)

    From (H6)(iii), there exists c4>0 such that

    ζ(LG,hη(z))e2LG,hz+c4LG,h,zR+.

    By the convexity of ζ we have

    ζ(ϕ3(t)w(t))ζ(10G(t,s)h(s)η(φ3(s)w(s))ds)10ζ[G(t,s)h(s)η(φ3(s)w(s))]ds=10ζ[G(t,s)h(s)LG,hLG,hη(φ3(s)w(s))]ds10G(t,s)h(s)LG,hζ[LG,hη(φ3(s)w(s))]ds10G(t,s)h(s)LG,h[e2LG,h(φ3(s)w(s))+c4LG,h]ds10G(t,s)h(s)[e2(φ3(s)w(s))+c4]ds. (3.20)

    Substituting this inequality into (3.18) we have

    φ3(t)10G(t,s)h(s)10G(s,τ)h(τ)[e2(φ3(τ)w(τ))+c4]dτdse21010G(t,s)h(s)G(s,τ)h(τ)(φ3(τ)w(τ))dτds+c4κ22. (3.21)

    Consequently, we have

    φ3(t)w(t)10G(t,s)h(s)10G(s,τ)h(τ)[e2(φ3(τ)w(τ))+c4]dτdse21010G(t,s)h(s)G(s,τ)h(τ)(φ3(τ)w(τ))dτds+c4κ22. (3.22)

    Multiply by ϑ(t) on both sides of (3.22) and integrate over [0,1] and use Lemma 2.2 to obtain

    10(φ3(t)w(t))ϑ(t)dte2κ2210(φ3(t)w(t))ϑ(t)dt+c4κ32,

    and we have

    10(φ3(t)w(t))ϑ(t)dtc4κ321e2κ22.

    Note that (3.16), wP0, and

    φ3wc4κ32κ1(1e2κ22).

    By the triangle inequality we have

    φ3=φ3w+wφ3w+wc4κ32κ1(1e2κ22)+Mκ2.

    On the other hand, from (3.20) we have

    ζ(ϕ3(t)w(t))10G(t,s)h(s)[e2(φ3(s)w(s))+c4]ds10ϑ(s)[e2(φ3(s)w(s))+c4]dsc4e2κ321e2κ22+c4κ2.

    Note that c4e2κ321e2κ22+c4κ2 is independent to R, and using (H6)(i) there exists Oϕ3>0 such that

    ϕ3wOϕ3,

    and then

    ϕ3=ϕ3w+wϕ3w+wOϕ3+Mκ2.

    Therefore, when φ3,ϕ3BRP, we obtain there is a positive constant to control the norms of φ3,ϕ3. Then if we choose

    R2>{OM,h,Oϕ3+Mκ2,˜M,c4κ32κ1(1e2κ22)+Mκ2},

    then (3.14) holds, and from Lemma 2.6 we have

    i(A,B2R2P2,P2)=1. (3.23)

    From (3.13) and (3.23) we have

    i(A,(B2R2¯B2OM,h)P2,P2)=i(A,B2R2P2,P2)i(A,B2OM,hP2,P2)=10=1.

    Then the operator A has at least one fixed point (denoted by (u,v)) on (B2R2¯B2OM,h)P2 with u(t),v(t)w(t),t[0,1]. Therefore, (uw,vw) is a positive solution for (2.2), and (1.1) has at least one positive radial solution.

    We now provide some examples to illustrate our main results. Let α=β=γ=δ=1, and k(|z|)=e|z|,zRn. Then (H1) and (H2) hold.

    Example 3.1. Let p(ϕ)=ϕ45,q(φ)=φ2,φ,ϕR+. Then lim infzp(LG,hq(z))z=lim infzL45G,hz85z, and (H4)(i), (iii) hold. If we choose

    f(φ,ϕ)=1β1κ2(|sinφ|+1)ϕM, g(φ,ϕ)=O1β3M,hβ2κ2(|cosϕ|+1)φβ3M,β1,β2>1,β3>2,

    then (H3) holds, and when φ,ϕ[0,OM,h], we have

    F1(φ,ϕ)=f(φ,ϕ)+MOM,hβ1κ2:=Q1, F2(φ,ϕ)=g(φ,ϕ)+MO1β3M,hβ2κ2Oβ3M,h=OM,hβ2κ2:=Q2.

    Hence, (H5) holds. Also we have

    lim infϕf(φ,ϕ)p(ϕ)=lim infϕ1β1κ2(|sinφ|+1)ϕMϕ45=, lim infφg(φ,ϕ)q(φ)=lim infφO1β3M,hβ2κ2(|cosϕ|+1)φβ3Mφ2=.

    Then (H4)(ii) holds. As a result, all the conditions in Theorem 3.1 hold, and (1.1) has at least one positive radial solution.

    Example 3.2. Let ζ(ϕ)=ϕ2,η(φ)=φ25, φ,ϕR+. Then lim supzζ(LG,hη(z))z=lim supzL2G,hz45z = 0e2LG,h, and (H7)(i), (iii) hold. If we choose

    f(φ,ϕ)=˜Q1+(ϕ+|cosφ|)α1M, g(φ,ϕ)=˜Q2+(|sinϕ|+φ)α2M,φ,ϕR+,

    where α1(0,2),α2(0,25). Then (H3) holds. Moreover, we have

    F1(φ,ϕ)=f(φ,ϕ)+M˜Q1, F2(φ,ϕ)=g(φ,ϕ)+M˜Q2,

    and

    lim supϕ˜Q1M+(ϕ+|cosφ|)α1ϕ2=0,lim supφ˜Q2M+(|sinϕ|+φ)α2φ25=0.

    Therefore, (H6) and (H7) (ii) hold. As a result, all the conditions in Theorem 3.2 hold, and (1.1) has at least one positive radial solution.

    Remark 3.2. Note that condition (HZ)2 is often used to study various kinds of semipositone boundary value problems (for example, see [19,22,23,26,28,29,30]). However, in Example 3.1 we have

    lim infϕ+f(φ,ϕ)φ=lim infϕ+1β1κ2(|sinφ|+1)ϕMϕ=12β1κ2,φR+.

    Comparing with (HZ)2 we see that our theory gives new results for boundary value problem with semipositone nonlinearities.

    This research was supported by the National Natural Science Foundation of China (12101086), Changzhou Science and Technology Planning Project (CJ20210133), Natural Science Foundation of Chongqing (cstc2020jcyj-msxmX0123), and Technology Research Foundation of Chongqing Educational Committee (KJQN202000528).

    The authors declare no conflict of interest.



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