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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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, ϕ−Laplacian 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 p−Laplacian-like operators.
In this paper we use the continuation methods to prove the existence of solutions to the ϕ−Laplacian problem for differential, equations with impulse effects and periodic boundary conditions
(ϕ(y′(t)))′=f(t,y(t),y′(t)),t∈J:=[0,b],t≠tk,k=1,…,m, | (1.1) |
y(t+k)−y(t−k)=Ik(y(t−k)),t=tk,k=1,…,m, | (1.2) |
y′(t+k)−y′(t−k)=¯Ik(y(t−k)),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,¯Ik∈C(Rn,Rn) and ϕ:Rn→Rn is a suitable monotone homeomorphism.
Because the ϕ−Laplacian 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:Rn→Rn be a continuous function which satisfies the following two conditions:
(H1) For any x1,x2∈Rn, x1≠x2,
⟨ϕ(x1)−ϕ(x2),x1−x2⟩>0. |
(H2) There exists a function α:[0,∞)→[0,∞), α(s)→+∞ as s→+∞, such that
⟨ϕ(x),x⟩≥α(‖x‖)‖x‖, for all x∈Rn. |
It is well-known that under these two conditions ϕ is an homeomorphism from Rn onto Rn, and that ‖ϕ−1(y)‖→+∞ as ‖y‖→+∞.
Definition 2.1. A map f:[p,q]×Rn→Rn is said to be L1-Carathéodory if
(ⅰ) t→f(t,y) is measurable for all y∈Rn,
(ⅱ) y→f(t,y) is continuous for almost each t∈[p,q],
(ⅲ) for each r>0, there exists hr∈L1([p,q],R+) such that
‖f(t,y)‖≤hr(t) for almost each t∈[p,q] and for all ‖y‖≤r. |
Let X and Z be two real Banach spaces with norms ‖⋅‖X and ‖⋅‖Z, respectively. A continuous operator M:X∩domM→Z is said to be quasi-linear if
(ⅰ) dimKerM=dimM−1(0)=n<∞;
(ⅱ) ImM=M(X∩domM) is a closed subset of Z.
Let X=X1⊕X2 and Z=Z1⊕Z2, 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:X→X1 and Q:Z→Z1 as the corresponding orthogonal projections such that
ImP=KerM,KerQ=ImM. |
Denote by J:Z1→X1 a homeomorphism with J(0)=0.
Let Ω be a bounded open subset of X, with 0∈Ω such that domM∩Ω≠∅, and consider a parameter family of nonlinear perturbations (generally) Nλ:[0,1]×ˉΩ→Z with N1=N. The continuous operator Nλ is said to be M−compact in ˉΩ with respect to M if there is an operator K:ImM→X2 with K(0)=0 such that for λ∈[0,1],
(I−Q)Nλ(ˉΩ)⊂ImM, |
(I−Q)N0=0,QNλx=0⇔QNx=0,λ∈(0,1), |
KM=I−P,K(I−Q)Nλ:ˉΩ→X2is compact, |
M[P+K(I−Q)Nλ]=(I−Q)Nλ. |
We introduce the intermediate map
F(λ,⋅)=P+K(I−Q)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, Ω⊂X be an open and bounded nonempty set, M be a quasi-linear operator and Nλ be a M−compact operator in ˉΩ. Then (2.2) has a solution x∈ˉΩ [resp x∈∂Ω] if and only if x∈ˉΩ [resp x∈∂Ω] is a fixed point of F(λ,⋅) defined in (2.1).
Theorem 2.2. [12] Let X and Z be two Banach spaces and Ω⊂X be an open and bounded nonempty set. Suppose M:X∩domM→Z is a quasilinear operator and Nλ:¯Ω→Z, λ∈[0,1], is M− compact. In addition, if
i) Mx≠Nλx,λ∈(0,1),x∈∂Ω,
ii) deg(JQN1,Ω∩KerM,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,yk∈C(Jk,Rn),k=0,…,m, such that y(t−k) and y(t+k)exist and satisfy y(t−k)=y(tk) for k=1,…,m}, |
endowed with the norm
‖y‖0=max{‖yk‖∞, k=0,…,m}, ‖yk‖∞=supt∈Jk|y(t)|, |
and let
PCb([0,b],Rn)={y∈PC: y(0)=y(b)} |
be a Banach space with the same norm of PC.
Also, consider
PC1([0,b],Rn)={y∈PC:y′k∈C(Jk,Rn),k=0,…,m, such thaty′(t−k) and y′(t+k) exist and satisfy y′(t−k)=y′(tk),for k=1,…,m}, |
endowed with the norm
‖y‖1=max(‖y‖0,‖y′‖0),or‖y‖1=‖y‖0+‖y′‖0, |
where yk=y|Jk and Jk=(tk,tk+1], and let
PC1b={y∈PC1: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
‖z‖Z=max{‖x‖L1,‖c1‖,‖c2‖,‖c3‖}. |
Let
M:domM⊆X→Z,y↦((ϕ(y′))′,△y(t1),…,△y(tm),△y′(t1),…,△y′(tm),0). |
Let h∈PC(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+∫t0ϕ−1[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))+∫tt1ϕ−1[ϕ(L′0(h)(t1)+ˉI1(L0(h)(t1)))+∫st1h(τ)dτ]ds,t∈(t1,t2],L2(h)(t)=L1(h)(t2)+I2(L1(h)(t2))+∫tt2ϕ−1[ϕ(L′1(h)(t2)+ˉI2(L1(h)(t2)))+∫st2h(τ)dτ]ds,t∈(t2,t3],…Lm(h)(t)=Lm−1(h)(tm)+Im(Lm−1(h)(tm))+∫ttmϕ−1[ϕ(L′m−1(h)(tm)+ˉIm(Lm−1(h)(tm)))+∫stmh(τ)dτ]ds,t∈(tm,b]. |
We will prove that M is a quasi-linear operator.
KerM={y∈domM: y(t)=y(0), t∈J}. |
For y∈KerM, we have
(ϕ(y′(t))′=0⇒ϕ(y′(t))=c:=(c0c1…cm−1), |
and then
y(t)=(L0(h)(t)L1(h)(t)…Lm−1(h)(t))=ϕ−1(c)t+y(0)=t(ϕ−1(c0)ϕ−1(c1)…ϕ−1(cm−1))+(L0(h)(0)L1(h)(0)…Lm−1(h)(0)). |
The boundary conditions imply that
(ϕ−1(c0)ϕ−1(c1)…ϕ−1(cm−1))=(00…0)⇒(L0(h)(t)L1(h)(t)…Lm−1(h)(t))=(L0(h)(0)L1(h)(0)…Lm−1(h)(0)). |
Thus
y(t)=y(0)=c∈Rn. |
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=Lm−1(x)(tm)+am(Lm−1(x)(tm))+∫btmϕ−1[ϕ(L′m−1(x)(tm)+ˉam(Lm−1(x)(tm)))+∫stmx(τ)dτ]ds, |
c=Lm−2(x)(tm−1)+am−1(Lm−2(x)(tm))+∫tmtm−1ϕ−1[ϕ(L′m−2(x)(tm)+ˉam−1(Lm−2(x)(tm)))+∫stm−1x(τ)dτ]ds+am(Lm−1(x)(tm))+∫btmϕ−1[ϕ(L′m−1(x)(tm)+ˉam(Lm−1(x)(tm)))+∫stmx(τ)dτ]ds, |
… |
c=c+m∑k=1ak+∫t10ϕ−1[ϕ(d)+∫s0x(τ)dτ]ds+∫t2t1ϕ−1[ϕ(L′0(x)(t1)+ˉa1(L0(x)(t1)))+∫st1x(τ)dτ]ds+…+∫tmtm−1ϕ−1[ϕ(L′m−2(x)(tm−1)+ˉam−1(Lm−2(x)(tm−1)))+∫stm−1x(τ)dτ]ds+∫btmϕ−1[ϕ(L′m−1(x)(tm)+ˉam(Lm−1(x)(tm)))+∫stmx(τ)dτ]ds, |
so that
m∑k=1ak+∫b0ϕ−1[ϕ(d+m∑k=1¯ak)+∫t0x(s)ds]dt=0, |
and then
ImM={(x,a1,…,am,¯a1,…,¯am,d)∈Z:m∑k=1ak+∫b0ϕ−1[ϕ(d+m∑k=1ˉak)+∫t0x(s)ds]dt=0}. |
Using the projections,
P: PC1b→PC1b, y↦y(0), |
Q: Z→Z, z→Q(z), |
z=(x,a1,…,am,¯a1,…,¯am,d)↦(0,…,0,1b(m∑k=1ak+∫b0ϕ−1[ϕ(d+m∑k=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 Ω⊂domM, define the family
Nλ:ˉΩ→Z,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 M−compact in ˉΩ. It is easy to show that
(I−Q)Nλ(ˉΩ)⊂ImM and QNλ(ˉΩ)=0. |
Let
K:ImM→domM∩KerP,z↦∑0<tk<tak+∫t0ϕ−1[ϕ(d+∑0<tk<s¯ak)+∫s0x(τ)dτ]ds, |
and the homeomorphism
J:ImQ→KerM |
is given by
(0,d)→J(0,d)=d. |
We supposed that f(t,y,y′) is a Carathéodory function and ϕ−1 is continuous. Then we have the following result.
Lemma 3.1. Suppose Ω⊂PC1([0,b],Rn) is a bounded open set. Then Nλ is M-compact in ˉΩ.
Proof. We will prove that
● QNλ:X→Z is continuous and sends bounded sets into bounded sets.
● K(I−Q)Nλ:ˉΩ→X is completely continuous.
Step 1: QNλ sends bounded sets into bounded sets and it is continuous.
First,
QNλ(y)(t)=(00…1b(∑mk=1Ik(y(tk))+∫b0ϕ−1[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 y∈ˉΩ={y∈PC1([0,b],Rn):‖y‖1≤r}.
‖QNλ(y)(t)‖≤‖∑mk=1Ik(y(tk))‖+∫b0‖ϕ−1[ϕ(∑mk=1ˉIk(y(tk)))+∫t0λf(s,y(s),y′(s))ds]‖dt. |
Clearly ‖y‖1≤r, 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=m∑k=1supx∈ˉB(0,r∗)‖Ik(x)‖<∞ and r2=m∑k=1supx∈ˉB(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))ds‖≤‖ϕ(∑mk=1ˉIk(y(tk)))‖+∫t0‖f(s,y(s),y′(s))‖ds≤‖ϕ(∑mk=1supx∈ˉB(0,r∗)ˉIk(x)‖+∫t0‖hq(s)‖ds≤supx∈B(0,r2)‖ϕ(x)‖+‖hq‖L1:=l. |
Since ϕ−1 is continuous, we also have
supx∈ˉB(0,l)|ϕ−1(x)|<∞. |
Thus
‖QNλ‖Z≤r1+bsupx∈ˉB(0,l)|ϕ−1(x)|:=r3. |
∘ QNλ is continuous.
Let (yα)α∈N be a sequence such that yn→y in PC1([0,b],Rn). Then there exists r>0 such that
‖y‖1≤r for all α∈N. |
Hence there exists hr∈L1([0,b],Rn) such that
‖f(t,yα(t),y′α(t))‖≤hr(t) a.e. t∈J. |
From the definition of QNλ, we have
‖QNλ(yα)(t)−QNλ(y)(t)‖≤‖∫b0ϕ−1[ϕ(∑mk=1ˉIk(yα(tk)))+∫t0λf(s,yα(s),y′α(s))ds]−ϕ−1[ϕ(∑mk=1ˉIk(y)(tk)))+∫t0λf(s,y(s),y′(s))ds]dt‖. |
By the dominated convergence theorem, and since Ik, ˉIk, ϕ and ϕ−1 are continuous functions, we get
‖QNλ(yα)−QNλ(y)‖Z≤1b∑mk=1‖Ik(yα(tk))−Ik(y(tk))‖+1b∫b0‖ϕ−1[ϕ(∑mk=1ˉI(yα(tk)))+∫t0λf(s,yα(s),y′α(s))ds]−ϕ−1[ϕ(∑mk=1ˉIk(y(tk)))+∫t0λf(s,y(s),y′(s))ds]‖dt. |
Thus
‖QNλ(yα)−QNλ(y)‖Z→0 as α→∞. |
Hence, QNλ is continuous.
Step 2: K(I−Q)Nλ is completely continuous.
(KNλy)(t)=∑0<tk<tIk(u(tk))+∫t0ϕ−1[ϕ(∑0<tk<s¯Ik(y(tk)))+∫s0f(τ,y(τ),y′(τ))dτ]dt. |
For this we prove that (I−Q)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 y∈Ω. Then
‖(KNλy)′(t)‖=‖ϕ−1[ϕ(∑0<tk<tˉIk(y(tk)))+∫t0λf(s,y(s),y(s))ds]‖≤supx∈ˉB(0,l)‖ϕ−1(x)‖:=r′, |
where l is defined in Step 1. By the mean value theorem, we obtain
‖(KNλy)(l2)−(KNλy)(l1)‖=‖(KNλy)′(ξ)(l2−l1)‖≤r′|l2−l1|. |
As l2→l1 the right hand side of the above inequality tends to zero. Also we have
‖ϕ((KNλy)′(l2))−ϕ((KNλy)′(l1))‖≤∑l1≤tk<l2supx∈B(0,r∗)‖ˉIk(x)‖+∫l2l1hr(t)dt→0 as l1→l2. |
Since ϕ−1 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 ϕ−1 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(I−Q)Nλy, y∈PC1. |
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),y∈PC1(J,Rn). |
Let us now consider the simple periodic boundary value problem
(ϕ(y′(t)))′=h(t),t∈[0,b],t≠tk,k=1,…,m, | (3.5) |
y(t+k)−y(t−k)=Ik(y(t−k)),t=tk,k=1,…,m, | (3.6) |
y′(t+k)−y′(t−k)=¯Ik(y(t−k)),t=tk,k=1,…,m, | (3.7) |
y(0)=y(b),y′(0)=y′(b), | (3.8) |
where h∈L1 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+∫t0ϕ−1[ϕ(d)+H0(h)(s)]ds,t∈[0,t1],L1(h)(t)=L0(h)(t1)+I1(L0(h)(t1))+∫tt1ϕ−1[ϕ(L′0(h)(t1)+ˉI1(L0(h)(t1)))+H1(h)(s)]ds,t∈(t1,t2],L2(h)(t)=L1(h)(t2)+I2(L1(h)(t2))+∫tt2ϕ−1[ϕ(L′1(h)(t2)+ˉI2(L1(h)(t2)))+H2(h)(s)]ds,t∈(t2,t3],…Lm(h)(t)=Lm−1(h)(tm)+Im(Lm−1(h)(tm))+∫ttmϕ−1[ϕ(L′m−1(h)(tm)+ˉIm(Lm−1(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(Lk−1(h)(tk))+∫b0ϕ−1[a+H(h)(t)]dt=0, |
with
a=ϕ(y′(0)+∑0<tk<b¯Ik(Lk−1(h)(tk))), H(h)(t)=∫t0h(s)ds. |
For fixed l∈PC([0,b],Rn) and c∈Rn, let us define
Gc,l(a)=∑0<tk<bIk(Lk−1(h)(tk))+∫b0ϕ−1[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 l∈PC([0,b],Rn) and c∈Rn, the equation
Gc,l(a)=0 | (3.11) |
has unique solution ˜a(l).
(ii) If ⟨∑mk=1Ik(x),˜a(l)⟩>0, x∈Rn, 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),a1−a2⟩>0, for a1≠a2, |
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(Lk−1(h)(tk))+∫b0ϕ−1(a+l(t))dt,a⟩=⟨∑0<tk<bIk(Lk−1(h)(tk)),a⟩+∫b0⟨ϕ−1(a+l(t)),a⟩dt=⟨∑0<tk<bIk(Lk−1(h)(tk)),a⟩+∫b0⟨ϕ−1(a+l(t)),a+l(t)⟩dt−∫b0⟨ϕ−1(a+l(t)),l(t)⟩dt, |
and thus
⟨Gc,l(a),a⟩≥⟨∑0<tk<bIk(Lk−1(h)(tk)),a⟩+∫b0⟨ϕ−1(a+l(t)),a+l(t)⟩dt−‖l‖0∫b0‖ϕ−1(a+l(t))‖dt. | (3.12) |
From (H2), ∀y∈Rn, we have that
⟨ϕ−1(y),y⟩≥α(‖ϕ−1(y)‖)‖ϕ−1(y)‖. | (3.13) |
Thus from (3.12) and (3.13),
⟨Gc,l(a),a⟩≥⟨∑0<tk<bIk(Lk−1(h)(tk)),a⟩+∫b0α(‖ϕ−1(a+l(t))‖)‖ϕ−1(a+l(t))‖−‖l‖0∫b0‖ϕ−1(a+l(t))‖dt≥⟨∑0<tk<bIk(Lk−1(h)(tk)),a⟩+∫b0(α(‖ϕ−1(a+l(t))‖)−‖l‖0)‖ϕ−1(a+l(t))‖dt. | (3.14) |
Since ‖a‖→∞ implies that ‖ϕ−1(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 a∈Rn with ‖a‖=r. |
It follows by an elementary topological degree argument that the equation Gc,l(a)=0 has a solution for each l∈PC, which by our previous argument is unique. In this way we define a function ˜a:PC→Rn which satisfies
∑0<tk<bIk(Lk−1(h)(tk))+∫b0ϕ−1[˜a+l(t)]dt=0. | (3.15) |
To prove (ⅱ) let B be a bounded subset of PC and let l∈B. From (3.15)
⟨∑0<tk<bIk(Lk−1(h)(tk))+∫b0ϕ−1[˜a(l)+l(t)]dt,˜a(l)⟩=0. |
Then
⟨∑0<tk<bIk(Lk−1(h)(tk)),˜a(l)⟩+∫b0⟨ϕ−1[˜a(l)+l(t)]dt,˜a(l)⟩=0, |
and hence
⟨∑0<tk<bIk(Lk−1(h)(tk)),˜a(l)⟩+∫b0⟨ϕ−1[˜a(l)+l(t)]dt,˜a(l)+l(t)⟩=∫b0⟨ϕ−1[˜a(l)+l(t)]dt,l(t)⟩. | (3.16) |
Assume that {˜a(l),l∈B} is not bounded. Then, for an arbitrary A>0, there is an l∈B, with ‖l‖0 sufficiently large, so that
A≤α(|ϕ−1(˜a(l)+l(t))|). |
Hence by using (3.16) and (3.13), we find that
A∫b0‖ϕ−1(˜a(l)+l(t))‖≤∫b0α(‖ϕ−1(˜a(l)+l(t))‖)‖ϕ−1(˜a(l)+l(t))‖dt≤∫b0⟨ϕ−1(˜a(l)+l(t)),˜a(l)+l(t)⟩dt=−⟨∑mk=1Ik(Lk−1(h)(tk)),˜a(l)⟩+∫b0⟨ϕ−1(˜a(l)+l(t)),l(t)⟩dt≤∫b0⟨ϕ−1(˜a(l)+l(t)),l(t)⟩≤‖l‖0∫b0‖ϕ−1(˜a(l)+l(t))‖. |
Thus A≤‖l‖0, 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 lα→l, 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
∫b0ϕ−1(˜a(lαj)+lαj(t))dt=0, |
we find that
∫b0ϕ−1(ˆa+l(t))dt=0, |
and hence ˜a(l)=ˆa, which shows the continuity of ˜a.
We will assume in this section that ϕ:Rn→Rn 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(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(d)+∑mk=1¯Ik(lk−1(h)(tk)))+∫t0f(s,a,0)ds]dt]=0, | (4.2) |
where h=f(t,a,0) and L0(h)(t)=c+∫t0ϕ−1[(ϕ(d)+∫s0f(τ,a,0)dτ]ds, t∈[0,t1] has no solution on ∂Ω∩Rn.
(H5) The Brouwer degree
dB[G,Ω∩Rn,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′)+(1−λ)(1b[∑mk=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+∑mk=1ˉIk(Lk−1(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(m∑k=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+m∑k=1¯Ik(Lk−1(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)+(1−λ)JQNf(y),I1(y(t1)),…,Im(y(tm)),¯I1(y(t1)),…,¯Im(y(tm)),0), |
where
Nf(y)=N(y,1), for each y∈PC1([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(I−Q)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(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+∑mk=1¯Ik(lk−1(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
m∑k=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+m∑k=1¯Ik(Lk−1(h)(tk)))+∫r0f(s,y(s),y′(s))ds]dr=0. | (4.6) |
Hence
(ϕ(y′(t)))′=0, |
which implies that
y′(t)=ϕ−1(c), |
where c∈Rn is a constant. Integrating this last equation over [0,b], we obtain ϕ−1(c)=0, and thus y(t)=e, a constant, and by (4.6)
m∑k=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+m∑k=1¯Ik(lk−1(h)(tk)))+∫r0f(s,e,0)ds]dr=0, |
which, together with hypothesis (H4), implies that y=e∉∂Ω. 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[I−Ff(.,λ),Ω,0] is well defined and by properties of that degree, that
dLS[I−Ff(⋅,1),Ω,0]=dLS[I−Ff(⋅,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[I−Ff(⋅,0),Ω,0]≠0. |
We have that
Ff(y,0)=Py+JQNf(y). |
Thus we obtain
y−Ff(y,0)=y−Py−1b[∑mk=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+∑mk=1¯Ik(lk−1(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(I−Ff(.,0),Ω,0)=(−1)NdB(G,Ω∩Rn,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×Rn→Rn be a Carathéodory function. We will say that f satisfies a generalized Villari condition if there is an ρ0>0 such that for all y∈PC1b, y=(y1,…,yn), with
mint∈J|yj|>ρ0, |
for some j∈{1,…,n}, it holds that
∫b0fi(t,y(t),y′(t))dt≠0, | (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 ν∈C1(Rn,Rn) and h∈L1(J,R+) such that
⟨ϕ(y),ν′(x)y⟩≥0, ⟨x,ν(y)−ν(z)⟩≤0, for any x,y,z∈Rn, |
and
|f(t,x,y)|≤⟨f(t,x,y),ν(x)⟩+h(t), |
for all x,y∈Rn and a.e t∈J.
(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,∀x∈Rn, |
and
∑mk=1⟨ϕ(x+ˉIk(y)),ν(z)−ν(y+Ik(y))⟩≤0,∑mk=1⟨∫tktk−1f(t,x,y)dt,ν(y)⟩≤0, |
for any x,y,z∈Rn.
(K4) There is an R0>0, such that all the possible solutions to the equation
G(a):=1b[∑mk=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+∑mk=1¯Ik(lk−1(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), y∈PC1b, λ∈(0,1), be a solution to (4.1), then using (4.2), we have
0≥−∫b0⟨ϕ(y′(t)),ν′(y(t))y′(t)⟩=∫b0⟨(ϕ(y′(t)))′,ν(y(t))⟩−⟨L′0(k)(0),ν(L0(k)(t1))−ν(L0(k)(0))⟩−m∑k=1⟨ϕ(L′k−1(k)(tk)+ˉIk(Lk−1(k)(tk)))),n(Lk(k)(tk+1))−ν(Lk−1(k)(tk)+Ik(Lk−1(k)(tk)))⟩−⟨∫tktk−1f(t,x,y)dt,ν(Lk−1(k)(tk))⟩=λ∫b0⟨f(t,y(t),y′(t)),ν(y(t))⟩−⟨L′0(k)(0),ν(L0(k)(t1))−ν(L0(k)(0))⟩−m∑k=1⟨ϕ(L′k−1(k)(tk)+ˉIk(Lk−1(k)(tk)))),n(Lk(k)(tk+1))−ν(Lk−1(k)(tk)+Ik(Lk−1(k)(tk)))⟩−⟨∫tktk−1f(t,x,y)dt,ν(Lk−1(k)(tk))⟩. |
Then, from hypothesis (K1), (K2) and (K3), we get
λ∫b0⟨f(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=1b∫b0ϕ(y′(t))dt. From (4.11) and (4.2), we get
˜b(t)=λf(t,y(t),y′(t)). |
Then,
‖~b′‖L1≤∫b0|f(t,y(t),y′(t))|dt≤∫b0⟨f(t,y(t),y′(t)),ν(y(t))⟩+‖h‖L1≤‖h‖L1, | (4.12) |
which yields
‖˜b‖0≤b‖h‖L1. |
We next find an a priori bound for ¯b. We have that
y′(t)=ϕ−1(˜b(t)+¯b). | (4.13) |
Hence, by integrating on each Jk=(tk−1,tk] and using the boundary conditions, we obtain
y(t)=y(0)+∑0<tk<tIk(Lk−1(h)(tk))+∫t0ϕ−1(˜b(s)+¯b)ds,t∈[0,b]. |
Then,
Ik(Lk−1(h)(tk))+∫b0ϕ−1(˜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 ϕ−1 is continuous, we obtain a positive constant C2 such that
‖y′‖0≤C2. | (4.15) |
Hence for t∈[0,t1], we get
‖∫t0y′(s)ds‖≤∫t10‖y′(t)‖dt≤C2t1:=M1. |
For t∈(t1,t2]
‖∫tt1y′(s)ds‖≤∫t2t1‖y′(t)‖dt≤C2(t2−t1):=M2. |
We continue this process and we obtain that, for t∈(tm,b],
‖∫ttmy′(s)ds‖≤∫btm‖y′(t)‖dt≤C2(b−tm):=Mm. |
Then
‖∫t0y′(s)ds‖≤∫b0‖y′(t)‖dt≤max(M1,M2,…,Mm):=M. | (4.16) |
Next the solution y satisfies
0=1b∫b0f(t,y(t),y′(t))dt=1bf(t,y(0)+∑0<tk<tIk(Lk−1(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)+∫ttj1y′j1(s)ds, |
we get
|yj1(t)|≤|yj1(tj1)|+M1≤ρ1+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))+∫ttj2y′j1(s)ds, |
we get
|yj2(t)|≤|yj2(tj2)|+c1+M2≤ρ2+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(Lm−1(h)(tm))+∫ttjmy′jm(s)ds, |
we get
|yjm(t)|≤ρm+cm+Mm. |
Then for each j∈{1,…,n}, there exists tj∈J such that
|yj(t)|≤maxρ1,…,ρm+m∑k=1ck+M:=¯M. |
Thus there is a constant C3 such that ‖y‖0≤C3. It follows that we can find R0>0 such that, if (y,λ) is a solution to (4.1), then ‖y‖1≤R0. We define next the set Ω⊂PC1b 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 ν∈C1(Rn,Rn) such that conditions (K1) and (K3) hold.
(M2)There exist a function ˜h∈L1(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 t∈J, and all x,y∈Rn.
(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 ‖y′‖0≤C2. We claim that conditions (K3) and (M2) imply that there is a constant C3 such that ‖y‖0≤C3. Indeed, from (4.12), we have that
∫b0|f(t,y(t),y′(t))|dt≤‖h‖L1. |
Then by (4.18)
∫b0η(|y(t)|)dt≤‖h‖L1+‖˜h‖L1. | (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 ‖y‖0.
Now let a∈Rn be such that
m∑k=1Ik(Lk−1(h)(tk))+∫b0ϕ−1[ϕ(y′(0)+m∑k=1¯Ik(lk−1(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)|p−2y′(t))′=−(1+y2(t))+h(t), a.e. t∈J:=[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:J→R defined by
h(t)=cost, t∈[0,π]. |
Let ν=c∈C1(R,R), c∈R+, then
⟨ϕ(y),ν′(x)y⟩≥0, ⟨x,ν(y)−ν(z)⟩≤0, for any x,y,z∈Rn, |
and
|f(t,x,y)|≤⟨f(t,x,y),ν(x)⟩+h(t), |
for all x,y∈R and a.e t∈J.
(ˉK1) There exist positive constants α1,α2, such that
I(x)=α1, ˉI(x)=α2,∀x∈R, |
and
⟨ϕ(x+ˉI(y)),ν(z)−ν(y+I(y))⟩=0, |
and
c∫π40f(t,x,y)dt+c∫ππ4f(t,x,y)dt=c∫π0f(t,x,y)dt=−cπ(1+x2)≤0 |
for any x,y,z∈R.
Define
G(a):=1b[α1+∫b0ϕ−1[d+α2+∫t0f(s,a,0)ds]dt]. |
It is clear that G:R→R 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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