Research article Special Issues

Existence results for ϕ-Laplacian impulsive differential equations with periodic conditions

  • Based on a Manasevich and Mawhin continuation theorem and some analysis skills we obtain sufficient conditions for existence results for φ-Laplacian nonlinear impulsive differential equations with periodic boundary conditions:(ϕ(y))=f(t,y(t),y(t)),a.e. t[0,b], y(t+k)y(tk)=Ik(y(tk)),k=1,,m, y(t+k)y(tk)=¯Ik(y(tk)),k=1,,m, y(0)=y(b),y(0)=y(b), where zhongwenzy <t1<t2<<tm<b, f:[0,b]×Rn×RnRn is a Carathéodory function, Ik,ˉIkC(Rn,Rn) and ϕ:RnRn is a suitable monotone homeomorphism.

    Citation: Johnny Henderson, Abdelghani Ouahab, Samia Youcefi. Existence results for ϕ-Laplacian impulsive differential equations with periodic conditions[J]. AIMS Mathematics, 2019, 4(6): 1610-1633. doi: 10.3934/math.2019.6.1610

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  • Based on a Manasevich and Mawhin continuation theorem and some analysis skills we obtain sufficient conditions for existence results for φ-Laplacian nonlinear impulsive differential equations with periodic boundary conditions:(ϕ(y))=f(t,y(t),y(t)),a.e. t[0,b], y(t+k)y(tk)=Ik(y(tk)),k=1,,m, y(t+k)y(tk)=¯Ik(y(tk)),k=1,,m, y(0)=y(b),y(0)=y(b), where zhongwenzy <t1<t2<<tm<b, f:[0,b]×Rn×RnRn is a Carathéodory function, Ik,ˉIkC(Rn,Rn) and ϕ:RnRn is a suitable monotone homeomorphism.


    Differential equations with impulses were considered for the first time by Milman and Myshkis [19] and then followed by a period of active research which culminated with the monograph by Halanay and Wexler [14]. Many phenomena and evolution processes in the field of physics, chemical technology, population dynamics, and natural sciences may change state abruptly or be subject to short-term perturbations, (see, for instance [1,16] and the references therein). These short perturbations may be seen as impulses. Impulsive problems arise also in various applications in communications, mechanics (jump discontinuities in velocity), electrical engineering, medicine, and biology. For example, in the periodic treatment of some diseases, impulses correspond to the administration of a drug treatment. In environmental sciences, impulses correspond to seasonal changes of the water level of artificial reservoirs. Their models are described by impulsive differential equations and inclusions. To date, a spectrum of mathematical results (such as existence, asymptotic behavior, ) have been obtained (see [2,3,8,9,13,16,28,29] and the references therein).

    There have been many approaches to the study of the existence of solutions of impulsive differential equations, such as fixed point theory, topological degree theory (including continuation methods and coincidence degree theory), comparison methods and monotone iterative methods (see [15,23,27,31,33]).

    Recently in [20,30], the authors studied the existence and multiplicity of solutions of some classes of second order impulsive problems by variational methods.

    In the last few years, ϕLaplacian problems for differential equations such as

    (ϕ(y))=f(t,y,y),
    y(0)=y(b),y(0)=y(b),

    have been investigated by several authors. Existence and multiplicity results for the nonresonance and resonance cases have been presented in [4,5,24,26]. These types of problems with impulse effects have been considered by [25], in which they are based on the lower and upper functions.

    Coincidence degree, introduced by Mawhin in 1972, is a topological tool for the investigation of the semilinear equation Lu+Nu=f, where L is a linear Fredholm operator of index zero (not necessarily invertible) and N is a nonlinear perturbation. Continuation theorems involving these kinds of mappings (L,N) became an effective procedure in proving the existence of solutions of a large variety of boundary value problems. This method is extended in [18] to the case when L is a quasi-linear operator in view of its application to problems involving pLaplacian-like operators.

    In this paper we use the continuation methods to prove the existence of solutions to the ϕLaplacian problem for differential, equations with impulse effects and periodic boundary conditions

    (ϕ(y(t)))=f(t,y(t),y(t)),tJ:=[0,b],ttk,k=1,,m, (1.1)
    y(t+k)y(tk)=Ik(y(tk)),t=tk,k=1,,m, (1.2)
    y(t+k)y(tk)=¯Ik(y(tk)),t=tk,k=1,,m, (1.3)
    y(0)=y(b),y(0)=y(b), (1.4)

    where 0<t1<t2<<tm<b, f:[0,b]×Rn×Rn×Rn is a Carathéodory function, Ik,¯IkC(Rn,Rn) and ϕ:RnRn is a suitable monotone homeomorphism.

    Because the ϕLaplacian is nonlinear, the continuation theorem of Mawhin [11] is not applicable, which leads to difficulty for solving the problems (1.1)–(1.4). By using a Manasevich and Mawhin continuation theorem [18] and some analysis techniques, we establish some sufficient conditions for the existence of periodic solutions of the problems (1.1)–(1.4).

    Let ϕ:RnRn be a continuous function which satisfies the following two conditions:

    (H1) For any x1,x2Rn, x1x2,

    ϕ(x1)ϕ(x2),x1x2>0.

    (H2) There exists a function α:[0,)[0,), α(s)+ as s+, such that

    ϕ(x),xα(x)x,   for all  xRn.

    It is well-known that under these two conditions ϕ is an homeomorphism from Rn onto Rn, and that ϕ1(y)+ as y+.

    Definition 2.1. A map f:[p,q]×RnRn is said to be L1-Carathéodory if

    (ⅰ) tf(t,y) is measurable for all yRn,

    (ⅱ) yf(t,y) is continuous for almost each t[p,q],

    (ⅲ) for each r>0, there exists hrL1([p,q],R+) such that

    f(t,y)hr(t)  for almost each t[p,q] and for all yr.

    Let X and Z be two real Banach spaces with norms X and Z, respectively. A continuous operator M:XdomMZ is said to be quasi-linear if

    (ⅰ) dimKerM=dimM1(0)=n<;

    (ⅱ) ImM=M(XdomM) is a closed subset of Z.

    Let X=X1X2 and Z=Z1Z2, where X1=KerM, Z2=ImM and X2,Z1 are respectively the complementary spaces of X1 in X, Z2 in Z. Assume that dimX1=dimZ1; then we can define P:XX1 and Q:ZZ1 as the corresponding orthogonal projections such that

    ImP=KerM,KerQ=ImM.

    Denote by J:Z1X1 a homeomorphism with J(0)=0.

    Let Ω be a bounded open subset of X, with 0Ω such that domMΩ, and consider a parameter family of nonlinear perturbations (generally) Nλ:[0,1]×ˉΩZ with N1=N. The continuous operator Nλ is said to be Mcompact in ˉΩ with respect to M if there is an operator K:ImMX2 with K(0)=0 such that for λ[0,1],

    (IQ)Nλ(ˉΩ)ImM,
    (IQ)N0=0,QNλx=0QNx=0,λ(0,1),
    KM=IP,K(IQ)Nλ:ˉΩX2is compact,
    M[P+K(IQ)Nλ]=(IQ)Nλ.

    We introduce the intermediate map

    F(λ,)=P+K(IQ)N+JQN, (2.1)

    which is clearly compact under the above assumptions.

    Consider the abstract equation

    Mx=Nλx,λ(0,1]. (2.2)

    Now, we consider the continuation theorem for a quasilinear operator due to Ge and Ren [12].

    Lemma 2.1. [12] Let X and Z be Banach spaces, ΩX be an open and bounded nonempty set, M be a quasi-linear operator and Nλ be a Mcompact operator in ˉΩ. Then (2.2) has a solution xˉΩ [resp xΩ] if and only if xˉΩ [resp xΩ] is a fixed point of F(λ,) defined in (2.1).

    Theorem 2.2. [12] Let X and Z be two Banach spaces and ΩX be an open and bounded nonempty set. Suppose M:XdomMZ is a quasilinear operator and Nλ:¯ΩZ, λ[0,1], is M compact. In addition, if

    i) MxNλx,λ(0,1),xΩ,

    ii) deg(JQN1,ΩKerM,0)0,

    then the abstract equation Mx=N1x has at least one solution in ¯Ω.

    In order to define a solution for Problems (1.1)–(1.2), consider the space of piece-wise continuous functions:

    PC([0,b],Rn)={y:[0,b]Rn,ykC(Jk,Rn),k=0,,m, such that y(tk) and y(t+k)exist and satisfy y(tk)=y(tk) for k=1,,m},

    endowed with the norm

    y0=max{yk,  k=0,,m}, yk=suptJk|y(t)|,

    and let

    PCb([0,b],Rn)={yPC: y(0)=y(b)}

    be a Banach space with the same norm of PC.

    Also, consider

    PC1([0,b],Rn)={yPC:ykC(Jk,Rn),k=0,,m, such thaty(tk) and y(t+k) exist and satisfy y(tk)=y(tk),for k=1,,m},

    endowed with the norm

    y1=max(y0,y0),ory1=y0+y0,

    where yk=y|Jk and Jk=(tk,tk+1], and let

    PC1b={yPC1:y(0)=y(b),y(0)=y(b)}

    be a Banach space with the same norm of PC1. Define X=PC1b, Z=L1([0,b],Rn)×Rnm×Rnm×Rn, and for z=(x,c1,c2,c3)Z, we have

    zZ=max{xL1,c1,c2,c3}.

    Let

    M:domMXZ,y((ϕ(y)),y(t1),,y(tm),y(t1),,y(tm),0).

    Let hPC(J,Rn). Then we define y by:

    y(t)={L0(h)(t),ift[0,t1],L1(h)(t),ift(t1,t2],Lm(h)(t),ift(tm,b], (3.1)

    where

    L0(h)(t)=c+t0ϕ1[ϕ(d)+s0h(τ)dτ]ds,t[0,t1], c=y(0),and d=y(0),L1(h)(t)=L0(h)(t1)+I1(L0(h)(t1))+tt1ϕ1[ϕ(L0(h)(t1)+ˉI1(L0(h)(t1)))+st1h(τ)dτ]ds,t(t1,t2],L2(h)(t)=L1(h)(t2)+I2(L1(h)(t2))+tt2ϕ1[ϕ(L1(h)(t2)+ˉI2(L1(h)(t2)))+st2h(τ)dτ]ds,t(t2,t3],Lm(h)(t)=Lm1(h)(tm)+Im(Lm1(h)(tm))+ttmϕ1[ϕ(Lm1(h)(tm)+ˉIm(Lm1(h)(tm)))+stmh(τ)dτ]ds,t(tm,b].

    We will prove that M is a quasi-linear operator.

    KerM={ydomM: y(t)=y(0), tJ}.

    For yKerM, we have

    (ϕ(y(t))=0ϕ(y(t))=c:=(c0c1cm1),

    and then

    y(t)=(L0(h)(t)L1(h)(t)Lm1(h)(t))=ϕ1(c)t+y(0)=t(ϕ1(c0)ϕ1(c1)ϕ1(cm1))+(L0(h)(0)L1(h)(0)Lm1(h)(0)).

    The boundary conditions imply that

    (ϕ1(c0)ϕ1(c1)ϕ1(cm1))=(000)(L0(h)(t)L1(h)(t)Lm1(h)(t))=(L0(h)(0)L1(h)(0)Lm1(h)(0)).

    Thus

    y(t)=y(0)=cRn.

    Next, ImM={(x,a1,,am,ˉa1,,ˉam,d)Z:(ϕ(y(t))=x(t) , a. e. t[0,b],y(tk)=ak, y(tk)=ˉak,k=1, ,m,x(t)domM}.

    From the following problem

    {(ϕ(y(t)))=x(t),a.e. t[0,b],y(tk)=ak,k=1,,m,y(tk)=¯ak,k=1,,m, (3.2)

    we have

    y(t)={L0(x)(t),ift[0,t1],L1(x)(t),ift(t1,t2],Lm(x)(t),ift(tm,b].

    Since y(0)=y(b), we also have

    L0(x)(0)=Lm(x)(b),
    c=Lm1(x)(tm)+am(Lm1(x)(tm))+btmϕ1[ϕ(Lm1(x)(tm)+ˉam(Lm1(x)(tm)))+stmx(τ)dτ]ds,
    c=Lm2(x)(tm1)+am1(Lm2(x)(tm))+tmtm1ϕ1[ϕ(Lm2(x)(tm)+ˉam1(Lm2(x)(tm)))+stm1x(τ)dτ]ds+am(Lm1(x)(tm))+btmϕ1[ϕ(Lm1(x)(tm)+ˉam(Lm1(x)(tm)))+stmx(τ)dτ]ds,
    c=c+mk=1ak+t10ϕ1[ϕ(d)+s0x(τ)dτ]ds+t2t1ϕ1[ϕ(L0(x)(t1)+ˉa1(L0(x)(t1)))+st1x(τ)dτ]ds++tmtm1ϕ1[ϕ(Lm2(x)(tm1)+ˉam1(Lm2(x)(tm1)))+stm1x(τ)dτ]ds+btmϕ1[ϕ(Lm1(x)(tm)+ˉam(Lm1(x)(tm)))+stmx(τ)dτ]ds,

    so that

    mk=1ak+b0ϕ1[ϕ(d+mk=1¯ak)+t0x(s)ds]dt=0,

    and then

    ImM={(x,a1,,am,¯a1,,¯am,d)Z:mk=1ak+b0ϕ1[ϕ(d+mk=1ˉak)+t0x(s)ds]dt=0}.

    Using the projections,

    P: PC1bPC1b, yy(0),
    Q: ZZ, zQ(z),
    z=(x,a1,,am,¯a1,,¯am,d)(0,,0,1b(mk=1ak+b0ϕ1[ϕ(d+mk=1ˉak)+t0x(s)ds]dt)),

    we obtain that

    ImP=KerM, KerQ=ImM.

    Thus

    dimKerM=n=dim(Z/ImM),

    and moreover, ImM is a closed subspace of Z. Therefore M is a quasi-linear.

    For any ΩdomM, define the family

    Nλ:ˉΩZ,y(λf(t,y,y),I1(y(t1)),,Im(y(tm)),¯I1(y(t1)),,¯Im(y(tm)),0).

    The problems (1.1)–(1.4) is equivalent to the operator equation

    My=N1y. (3.3)

    We will prove that Nλ is Mcompact in ˉΩ. It is easy to show that

    (IQ)Nλ(ˉΩ)ImM and QNλ(ˉΩ)=0.

    Let

    K:ImMdomMKerP,z0<tk<tak+t0ϕ1[ϕ(d+0<tk<s¯ak)+s0x(τ)dτ]ds,

    and the homeomorphism

    J:ImQKerM

    is given by

    (0,d)J(0,d)=d.

    We supposed that f(t,y,y) is a Carathéodory function and ϕ1 is continuous. Then we have the following result.

    Lemma 3.1. Suppose ΩPC1([0,b],Rn) is a bounded open set. Then Nλ is M-compact in ˉΩ.

    Proof. We will prove that

    QNλ:XZ is continuous and sends bounded sets into bounded sets.

    K(IQ)Nλ:ˉΩX is completely continuous.

    Step 1: QNλ sends bounded sets into bounded sets and it is continuous.

    First,

    QNλ(y)(t)=(001b(mk=1Ik(y(tk))+b0ϕ1[ϕ(mk=1¯Ik(y(tk)))+t0λf(s,y(s),y(s))ds]dt)).

    QNλ sends bounded sets into bounded sets in PC1([0,b],Rn).

    Let yˉΩ={yPC1([0,b],Rn):y1r}.

    QNλ(y)(t)mk=1Ik(y(tk))+b0ϕ1[ϕ(mk=1ˉIk(y(tk)))+t0λf(s,y(s),y(s))ds]dt.

    Clearly y1r, and then, there exists r>0 such that

    ˉIk(y(tk))r and Ik(y(tk))r, k=1,,m.

    Since Ik and ˉIk are continuous and ˉB(0,r) is compact in Rn, we have

    r1=mk=1supxˉB(0,r)Ik(x)< and r2=mk=1supxˉB(0,r)ˉIk(x)<.

    Using the fact that f is a Carathédory function, we obtain

    ϕ(mk=1ˉIk(y(tk)))+t0λf(s,y(s),y(s))dsϕ(mk=1ˉIk(y(tk)))+t0f(s,y(s),y(s))dsϕ(mk=1supxˉB(0,r)ˉIk(x)+t0hq(s)dssupxB(0,r2)ϕ(x)+hqL1:=l.

    Since ϕ1 is continuous, we also have

    supxˉB(0,l)|ϕ1(x)|<.

    Thus

    QNλZr1+bsupxˉB(0,l)|ϕ1(x)|:=r3.

    QNλ is continuous.

    Let (yα)αN be a sequence such that yny in PC1([0,b],Rn). Then there exists r>0 such that

    y1r  for all αN.

    Hence there exists hrL1([0,b],Rn) such that

    f(t,yα(t),yα(t))hr(t)  a.e. tJ.

    From the definition of QNλ, we have

    QNλ(yα)(t)QNλ(y)(t)b0ϕ1[ϕ(mk=1ˉIk(yα(tk)))+t0λf(s,yα(s),yα(s))ds]ϕ1[ϕ(mk=1ˉIk(y)(tk)))+t0λf(s,y(s),y(s))ds]dt.

    By the dominated convergence theorem, and since Ik, ˉIk, ϕ and ϕ1 are continuous functions, we get

    QNλ(yα)QNλ(y)Z1bmk=1Ik(yα(tk))Ik(y(tk))+1bb0ϕ1[ϕ(mk=1ˉI(yα(tk)))+t0λf(s,yα(s),yα(s))ds]ϕ1[ϕ(mk=1ˉIk(y(tk)))+t0λf(s,y(s),y(s))ds]dt.

    Thus

    QNλ(yα)QNλ(y)Z0 as α.

    Hence, QNλ is continuous.

    Step 2: K(IQ)Nλ is completely continuous.

    (KNλy)(t)=0<tk<tIk(u(tk))+t0ϕ1[ϕ(0<tk<s¯Ik(y(tk)))+s0f(τ,y(τ),y(τ))dτ]dt.

    For this we prove that (IQ)Nλ sends bounded sets into bounded sets and KNλ is completely continuous. The first part is immediate. For the second part, as in Step 1, we can prove that KNλ is bounded and continuous.

    It remains to show that KNλ is equicontinuous, then by using the Arzelà-Ascoli theorem KNλis compact. Indeed Let l1, l2[0,b], l1<l2 and Ω be a bounded set in PC1([0,b],Rn), and let yΩ. Then

    (KNλy)(t)=ϕ1[ϕ(0<tk<tˉIk(y(tk)))+t0λf(s,y(s),y(s))ds]supxˉB(0,l)ϕ1(x):=r,

    where l is defined in Step 1. By the mean value theorem, we obtain

    (KNλy)(l2)(KNλy)(l1)=(KNλy)(ξ)(l2l1)r|l2l1|.

    As l2l1 the right hand side of the above inequality tends to zero. Also we have

    ϕ((KNλy)(l2))ϕ((KNλy)(l1))l1tk<l2supxB(0,r)ˉIk(x)+l2l1hr(t)dt0 as l1l2.

    Since ϕ1 is continuous function, we conclude that ¯K(Ω) is compact.

    ● Using the fact that f is a Carathéodory function and that Ik, ˉIk, ϕ and ϕ1 are continuous functions, we can easily prove that KNλ is continuous.

    Let us now consider the abstract differential periodic problem

    My=Nλy, (3.4)

    and define

    F(y,λ):=Py+JQNλy+K(IQ)Nλy, yPC1.

    We obtain that F is a completely continuous operator. Furthermore, the operator equation (3.4) is equivalent to the following fixed point equation:

    y=F(y,λ),yPC1(J,Rn).

    Let us now consider the simple periodic boundary value problem

    (ϕ(y(t)))=h(t),t[0,b],ttk,k=1,,m, (3.5)
    y(t+k)y(tk)=Ik(y(tk)),t=tk,k=1,,m, (3.6)
    y(t+k)y(tk)=¯Ik(y(tk)),t=tk,k=1,,m, (3.7)
    y(0)=y(b),y(0)=y(b), (3.8)

    where hL1 is such that b0h(s)ds=0, and let us recall that y is a continuous solution to (3.5)–(3.8) defined by

    y(t)={L0(h)(t),ift[0,t1],L1(h)(t),ift(t1,t2],Lm(h)(t),ift(tm,b], (3.9)

    where

    L0(h)(t)=c+t0ϕ1[ϕ(d)+H0(h)(s)]ds,t[0,t1],L1(h)(t)=L0(h)(t1)+I1(L0(h)(t1))+tt1ϕ1[ϕ(L0(h)(t1)+ˉI1(L0(h)(t1)))+H1(h)(s)]ds,t(t1,t2],L2(h)(t)=L1(h)(t2)+I2(L1(h)(t2))+tt2ϕ1[ϕ(L1(h)(t2)+ˉI2(L1(h)(t2)))+H2(h)(s)]ds,t(t2,t3],Lm(h)(t)=Lm1(h)(tm)+Im(Lm1(h)(tm))+ttmϕ1[ϕ(Lm1(h)(tm)+ˉIm(Lm1(h)(tm)))+Hm(h)(s)]ds,t(tm,b],

    and

    H0(h)(s)=s0h(τ)dτ,ift[0,t1],H1(h)(s)=st1h(τ)dτ,ift(t1,t2],Hm(h)(s)=stmh(τ)dτ,ift(tm,b].

    The boundary conditions imply that

    0<tk<bIk(Lk1(h)(tk))+b0ϕ1[a+H(h)(t)]dt=0,

    with

    a=ϕ(y(0)+0<tk<b¯Ik(Lk1(h)(tk))),  H(h)(t)=t0h(s)ds.

    For fixed lPC([0,b],Rn) and cRn, let us define

    Gc,l(a)=0<tk<bIk(Lk1(h)(tk))+b0ϕ1[a+l(t)]dt. (3.10)

    Proposition 3.2. If ϕ satisfies conditions (H1) and (H2), then the function Gc,l has the following properties:

    (i) For any fixed lPC([0,b],Rn) and cRn, the equation

    Gc,l(a)=0 (3.11)

    has unique solution ˜a(l).

    (ii) If mk=1Ik(x),˜a(l)>0, xRn, where the function ˜a:PC([0,b],Rn)Rn is defined in (i), then ˜a is continuous and sends bounded sets into bounded sets.

    Proof. (ⅰ) By (H1), it is immediate that

    Gc,l(a1)Gc,l(a2),a1a2>0,   for  a1a2,

    and hence if (3.11) has a solution, it is unique. To prove its existence we will show that Gc,l(a),a>0 for a sufficiently large. Indeed we have

    Gc,l(a),a=0<tk<bIk(Lk1(h)(tk))+b0ϕ1(a+l(t))dt,a=0<tk<bIk(Lk1(h)(tk)),a+b0ϕ1(a+l(t)),adt=0<tk<bIk(Lk1(h)(tk)),a+b0ϕ1(a+l(t)),a+l(t)dtb0ϕ1(a+l(t)),l(t)dt,

    and thus

    Gc,l(a),a0<tk<bIk(Lk1(h)(tk)),a+b0ϕ1(a+l(t)),a+l(t)dtl0b0ϕ1(a+l(t))dt. (3.12)

    From (H2), yRn, we have that

    ϕ1(y),yα(ϕ1(y))ϕ1(y). (3.13)

    Thus from (3.12) and (3.13),

    Gc,l(a),a0<tk<bIk(Lk1(h)(tk)),a+b0α(ϕ1(a+l(t)))ϕ1(a+l(t))l0b0ϕ1(a+l(t))dt0<tk<bIk(Lk1(h)(tk)),a+b0(α(ϕ1(a+l(t)))l0)ϕ1(a+l(t))dt. (3.14)

    Since a implies that ϕ1(a+l(t)), uniformly for t[0,b], we find from (3.14) that there exists an r>0 such that

    Gc,l(a),a>0    for all   aRn   with   a=r.

    It follows by an elementary topological degree argument that the equation Gc,l(a)=0 has a solution for each lPC, which by our previous argument is unique. In this way we define a function ˜a:PCRn which satisfies

    0<tk<bIk(Lk1(h)(tk))+b0ϕ1[˜a+l(t)]dt=0. (3.15)

    To prove (ⅱ) let B be a bounded subset of PC and let lB. From (3.15)

    0<tk<bIk(Lk1(h)(tk))+b0ϕ1[˜a(l)+l(t)]dt,˜a(l)=0.

    Then

    0<tk<bIk(Lk1(h)(tk)),˜a(l)+b0ϕ1[˜a(l)+l(t)]dt,˜a(l)=0,

    and hence

    0<tk<bIk(Lk1(h)(tk)),˜a(l)+b0ϕ1[˜a(l)+l(t)]dt,˜a(l)+l(t)=b0ϕ1[˜a(l)+l(t)]dt,l(t). (3.16)

    Assume that {˜a(l),lB} is not bounded. Then, for an arbitrary A>0, there is an lB, with l0 sufficiently large, so that

    Aα(|ϕ1(˜a(l)+l(t))|).

    Hence by using (3.16) and (3.13), we find that

    Ab0ϕ1(˜a(l)+l(t))b0α(ϕ1(˜a(l)+l(t)))ϕ1(˜a(l)+l(t))dtb0ϕ1(˜a(l)+l(t)),˜a(l)+l(t)dt=mk=1Ik(Lk1(h)(tk)),˜a(l)+b0ϕ1(˜a(l)+l(t)),l(t)dtb0ϕ1(˜a(l)+l(t)),l(t)l0b0ϕ1(˜a(l)+l(t)).

    Thus Al0, which is a contradiction. Therefore ˜a sends bounded sets in PC into bounded sets in Rn.

    Finally, to show the continuity of ˜a, let (lα) be a convergent sequence in PC, say lαl, as α. Since (˜a(lα)) is a bounded sequence, any subsequence of it contains a convergent subsequence, denoted by (a(lαj)). Let a(lαj)ˆa, as j. By letting j in

    b0ϕ1(˜a(lαj)+lαj(t))dt=0,

    we find that

    b0ϕ1(ˆa+l(t))dt=0,

    and hence ˜a(l)=ˆa, which shows the continuity of ˜a.

    We will assume in this section that ϕ:RnRn is continuous and satisfies conditions (H1)(H2) of Section 2. Our aim in this part is to apply the Manasevich and Mawhin continuation theorem [18] for quasilinear equations to the quasilinear problem with impulse effects (1.1)–(1.4).

    Theorem 4.1. Assume that Ω is an open bounded set in PC([0,b],Rn) such that the following conditions hold:

    (H3) For each λ(0,1) the problem

    {(ϕ(y))=λf(t,y,y),y(tk)=λIk(y(tk)),y(tk)=λ¯Ik(y(tk)),y(0)=y(b),y(0)=y(b), (4.1)

    has no solution on Ω.

    (H4) The equation

    Gc,d(a):=1b[mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(d)+mk=1¯Ik(lk1(h)(tk)))+t0f(s,a,0)ds]dt]=0, (4.2)

    where h=f(t,a,0) and L0(h)(t)=c+t0ϕ1[(ϕ(d)+s0f(τ,a,0)dτ]ds, t[0,t1] has no solution on ΩRn.

    (H5) The Brouwer degree

    dB[G,ΩRn,0]0.

    Then the problems (1.1)–(1.4) has a solution in Ω.

    Proof. We transform the problems (1.1)–(1.4) into the one parameter family of problems

    {(ϕ(y))=λf(t,y,y)+(1λ)(1b[mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1ˉIk(Lk1(h)(tk)))+t0f(s,y(s),y(s))ds]dt]),y(tk)=λIk(y(tk)),y(tk)=λ¯Ik(y(tk)),y(0)=y(b),y(0)=y(b). (4.3)

    For λ(0,1], observe that in both cases, y is a solution of problem (4.1) or y is a solution of problem (4.3). We have necessarily

    1b(mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(Lk1(h)(tk)))+t0f(s,y(s),y(s))ds]dt)=0.

    It follows that, for λ(0,1], problems (4.1) and (4.3) have the same solutions. Furthermore it is easy to see that f Carathéodory implies that N:PC1([0,b],Rn)×[0,1]Z defined by

    N(y,λ)=(λJNf(y)+(1λ)JQNf(y),I1(y(t1)),,Im(y(tm)),¯I1(y(t1)),,¯Im(y(tm)),0),

    where

    Nf(y)=N(y,1), for each yPC1([0,b],Rn),

    is continuous and sends bounded sets into bounded sets. Problem (4.3) can be written in the equivalent form

    y=Ff(y,λ), (4.4)

    with

    Ff(y,λ)=Py+JQNf(y)+K[λJ(IQ)Nf](y).

    We assume that for λ=1, (4.4) does not have a solution on Ω. By hypothesis (H3), it follows that (4.4) has no solution for (y,λ)Ω×(0,1]. For λ=0, (4.3) is equivalent to the problem

    {(ϕ(y))=1b[mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(lk1(h)(tk)))+t0f(s,y(s),y(s))ds]dt],y(0)=y(b),y(0)=y(b). (4.5)

    And thus if y is a solution of this problem, we must have

    mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(Lk1(h)(tk)))+r0f(s,y(s),y(s))ds]dr=0. (4.6)

    Hence

    (ϕ(y(t)))=0,

    which implies that

    y(t)=ϕ1(c),

    where cRn is a constant. Integrating this last equation over [0,b], we obtain ϕ1(c)=0, and thus y(t)=e, a constant, and by (4.6)

    mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(lk1(h)(tk)))+r0f(s,e,0)ds]dr=0,

    which, together with hypothesis (H4), implies that y=eΩ. Thus we prove that (4.4) has no solution (y,λ)Ω×[0,1]. Then we have that for each λ[0,1] the Leray-Schauder degree dLS[IFf(.,λ),Ω,0] is well defined and by properties of that degree, that

    dLS[IFf(,1),Ω,0]=dLS[IFf(,0),Ω,0]. (4.7)

    Now it is clear that the problem

    y=Ff(y,1) (4.8)

    is equivalent to the problem (1.1)-(1.4), and will have a solution, if we can prove that

    dLS[IFf(,0),Ω,0]0.

    We have that

    Ff(y,0)=Py+JQNf(y).

    Thus we obtain

    yFf(y,0)=yPy1b[mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(lk1(h)(tk)))+t0f(s,y(s),y(s))ds]dt].

    Hence by the properties of Leray-Schauder degree (proved in [7]), we have that

    dLS(IFf(.,0),Ω,0)=(1)NdB(G,ΩRn,0),

    where the function G is defined in Proposition 3.2, and dB denotes the Brouwer degree. Since by the Hypothesis (H5) this last degree is different from zero, the theorem is proved.

    Our next theorem is a consequence of Theorem 4.1. We need first the following definition.

    Let f=(f1,,fn):J×Rn×RnRn be a Carathéodory function. We will say that f satisfies a generalized Villari condition if there is an ρ0>0 such that for all yPC1b, y=(y1,,yn), with

    mintJ|yj|>ρ0,

    for some j{1,,n}, it holds that

    b0fi(t,y(t),y(t))dt0, (4.9)

    for some i{1,,n}.

    Let B(R) denote the open ball in Rn with center zero and radius R.

    Theorem 4.2. Assume that the following conditions hold:

    (K1) There exist νC1(Rn,Rn) and hL1(J,R+) such that

    ϕ(y),ν(x)y0, x,ν(y)ν(z)0,     for  any   x,y,zRn,

    and

    |f(t,x,y)|f(t,x,y),ν(x)+h(t),

    for all x,yRn and a.e tJ.

    (K2) f satisfies a generalized Villari condition.

    (K3) There exist positive constants ck, such that for each k=1,,m, we have

    |Ik(x)|ck,xRn,

    and

    mk=1ϕ(x+ˉIk(y)),ν(z)ν(y+Ik(y))0,mk=1tktk1f(t,x,y)dt,ν(y)0,

    for any x,y,zRn.

    (K4) There is an R0>0, such that all the possible solutions to the equation

    G(a):=1b[mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(lk1(h)(tk)))+s0f(s,a,0)ds]dt]=0, (4.10)

    belong B(0,R0).

    (K5) The Brouwer degree

    dB[G,B(0,R0),0]0.

    Then the problems (1.1)–(1.4) has at least one solution.

    Proof. Let (y,λ), yPC1b, λ(0,1), be a solution to (4.1), then using (4.2), we have

    0b0ϕ(y(t)),ν(y(t))y(t)=b0(ϕ(y(t))),ν(y(t))L0(k)(0),ν(L0(k)(t1))ν(L0(k)(0))mk=1ϕ(Lk1(k)(tk)+ˉIk(Lk1(k)(tk)))),n(Lk(k)(tk+1))ν(Lk1(k)(tk)+Ik(Lk1(k)(tk)))tktk1f(t,x,y)dt,ν(Lk1(k)(tk))=λb0f(t,y(t),y(t)),ν(y(t))L0(k)(0),ν(L0(k)(t1))ν(L0(k)(0))mk=1ϕ(Lk1(k)(tk)+ˉIk(Lk1(k)(tk)))),n(Lk(k)(tk+1))ν(Lk1(k)(tk)+Ik(Lk1(k)(tk)))tktk1f(t,x,y)dt,ν(Lk1(k)(tk)).

    Then, from hypothesis (K1), (K2) and (K3), we get

    λb0f(t,y(t),y(t)),ν(y(t))0. (4.11)

    Let us set ϕ(y(t))=˜b(t)+¯b, with b0˜b(t)dt=0, and ¯b=1bb0ϕ(y(t))dt. From (4.11) and (4.2), we get

    ˜b(t)=λf(t,y(t),y(t)).

    Then,

    ~bL1b0|f(t,y(t),y(t))|dtb0f(t,y(t),y(t)),ν(y(t))+hL1hL1, (4.12)

    which yields

    ˜b0bhL1.

    We next find an a priori bound for ¯b. We have that

    y(t)=ϕ1(˜b(t)+¯b). (4.13)

    Hence, by integrating on each Jk=(tk1,tk] and using the boundary conditions, we obtain

    y(t)=y(0)+0<tk<tIk(Lk1(h)(tk))+t0ϕ1(˜b(s)+¯b)ds,t[0,b].

    Then,

    Ik(Lk1(h)(tk))+b0ϕ1(˜b(t)+¯b)dt=0. (4.14)

    By Proposition 3.2, it follows that ¯b=˜a(˜b), where the function ˜a is defined in that proposition. Recalling that ˜a sends bounded sets into bounded sets, we have that there is a positive constant C1 such that

    |¯b|C1.

    Hence, from (4.13) and the fact that ϕ1 is continuous, we obtain a positive constant C2 such that

    y0C2. (4.15)

    Hence for t[0,t1], we get

    t0y(s)dst10y(t)dtC2t1:=M1.

    For t(t1,t2]

    tt1y(s)dst2t1y(t)dtC2(t2t1):=M2.

    We continue this process and we obtain that, for t(tm,b],

    ttmy(s)dsbtmy(t)dtC2(btm):=Mm.

    Then

    t0y(s)dsb0y(t)dtmax(M1,M2,,Mm):=M. (4.16)

    Next the solution y satisfies

    0=1bb0f(t,y(t),y(t))dt=1bf(t,y(0)+0<tk<tIk(Lk1(h)(tk))+t0y(s)ds,y(t))dt. (4.17)

    By the generalized Villari condition, we have, for each j1{1,,n}, there exists tj1[0,t1] such that

    |yj1(tj1)|ρ1.

    Since

    yj1(t)=yj1(tj1)+ttj1yj1(s)ds,

    we get

    |yj1(t)||yj1(tj1)|+M1ρ1+M1.

    For each j2{1,,n}, there exists a tj2(t1,t2] such that

    |yj2(tj2)|ρ2.

    Since

    yj2(t)=yj2(tj2)+I1(L0(h)(t1))+ttj2yj1(s)ds,

    we get

    |yj2(t)||yj2(tj2)|+c1+M2ρ2+c1+M2.

    We continue this process, and we get that, for each jm{1,,n}, there exists a tjm(tm,b] such that

    |yjm(tjm)|ρm.

    Since

    yjm(t)=yjm(tjm)+Im(Lm1(h)(tm))+ttjmyjm(s)ds,

    we get

    |yjm(t)|ρm+cm+Mm.

    Then for each j{1,,n}, there exists tjJ such that

    |yj(t)|maxρ1,,ρm+mk=1ck+M:=¯M.

    Thus there is a constant C3 such that y0C3. It follows that we can find R0>0 such that, if (y,λ) is a solution to (4.1), then y1R0. We define next the set ΩPC1b that appears in Theorem 4.1 as Ω=B(R0), the open ball in PC1b of center 0 and radius R0. Thus condition (H3) of Theorem 4.1 is satisfied with Ω=B(R0), and since the rest of the conditions of Theorem 4.1 are also satisfied, the proof is complete.

    Corollary 4.3. Suppose that the following conditions are satisfied.

    (M1) There is a mapping νC1(Rn,Rn) such that conditions (K1) and (K3) hold.

    (M2)There exist a function ˜hL1(J,R+) and continuous function η:[0,)[0,), such that

    η(s)   as  s,

    and

    η(|x|)˜h(t)|f(t,x,y)|, (4.18)

    for almost all tJ, and all x,yRn.

    (M3) Condition (K5) holds.

    Then the problems (1.1)–(1.4) has at least one solution.

    Proof. Let (y,λ), λ(0,1) be a solution to (4.1). As in the proof of Theorem 4.2, it follows from conditions (K1) and (K3), that there is a positive constant C2 such that y0C2. We claim that conditions (K3) and (M2) imply that there is a constant C3 such that y0C3. Indeed, from (4.12), we have that

    b0|f(t,y(t),y(t))|dthL1.

    Then by (4.18)

    b0η(|y(t)|)dthL1+˜hL1. (4.19)

    Since (4.15) holds by the reasoning of the previous theorem and η(s) as s, from (4.19) we find the required bound for y0.

    Now let aRn be such that

    mk=1Ik(Lk1(h)(tk))+b0ϕ1[ϕ(y(0)+mk=1¯Ik(lk1(h)(tk)))+s0f(s,a,0)ds]dt=0.

    Then (4.18) implies that η(|a|)C2, and hence |a|C3, for some positive constants C2 and C3. Thus there is R0>0 such that every solutions of (4.10) belongs to B(R0), and since condition (K5) holds, then all condition of Theorem 4.1 are satisfied. Therefore, (1.1)–(1.4) has at least one solution.

    Example 4.1. We consider the following

    (|y(t)|p2y(t))=(1+y2(t))+h(t), a.e. tJ:=[0,π], (4.20)
    y(π4+)y(π4)=I(y(π4)), (4.21)
    y(π4+)y(π4)=ˉI(y(π4)), (4.22)
    y(0)=y(π),y(0)=y(π), (4.23)

    where h:JR defined by

    h(t)=cost, t[0,π].

    Let ν=cC1(R,R), cR+, then

    ϕ(y),ν(x)y0, x,ν(y)ν(z)0,   for any x,y,zRn,

    and

    |f(t,x,y)|f(t,x,y),ν(x)+h(t),

    for all x,yR and a.e tJ.

    (ˉK1) There exist positive constants α1,α2, such that

    I(x)=α1, ˉI(x)=α2,xR,

    and

    ϕ(x+ˉI(y)),ν(z)ν(y+I(y))=0,

    and

    cπ40f(t,x,y)dt+cππ4f(t,x,y)dt=cπ0f(t,x,y)dt=cπ(1+x2)0

    for any x,y,zR.

    Define

    G(a):=1b[α1+b0ϕ1[d+α2+t0f(s,a,0)ds]dt].

    It is clear that G:RR is a continuous function. We can show that there exists R0>0 such that G(B(0,R0))B(0,R0). Then d(G,B(0,R),0)0. So all the conditions of Theorem 4.2 are satisfied. Hence the above problem has at least one solution.

    The authors would like to thank the anonymous referees for their careful reading of the manuscript and pertinent comments; their constructive suggestions substantially improved the quality of the work.

    The authors declare no conflict of interest.



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