Special Manásevich-Mawhin Continuation Theorems with Applications

Zhou Kai Zhou Yinggao

(School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, China)

Abstract Avoiding the calculation of any topological degree also means to minimize the processing of practical problems when a continuation theorem of topological degree theory is used. In this paper, a special continuation theorem and several corollaries are given. Compared with the classical Manásevich-Mawhin continuation theorem, we can avoid calculating any topological degree and reduce the conditions of the theorem when using this special continuation theorem and its corollaries in applications. More importantly, the conditions for verifying this special continuation theorem will be easier and more convenient. As an application, we use this special continuation theorem and its corollary to study the existence of periodic solutions and positive periodic solutions for a generalized Rayleigh type p-Laplacian equation with deviating arguments and obtain some new sufficient conditions which generalize and improve the known results in the literatures.

Key words Topological degree Manásevich-Mawhin continuation theorem Rayleigh type p-Laplacian equation Periodic solution Positive periodic solution

1 Introduction

The Manásevich-Mawhin continuation theorem is as follows.

Theorem 1.(Manásevich-Mawhin continuation theorem[17]) Consider the boundary value problem

(φ(x′))′=f(t,x,x′),x(0)=x(T),x′(0)=x′(T),

(1)

wheref:[0,T]×Rn×Rn→Rnis Caratheodory,φ:Rn→Rnis continuous and satisfies the conditions:

(A1)(φ(x1)-φ(x2),x1-x2)>0, for anyx1,x2∈Rn,x1≠x2;

(A2)(φ(x),x)≥a(|x|)|x|,?x∈Rn, wherea:[0,+∞)→[0,+∞) anda(t)→+∞ ast→+∞.

Assume that Ω is a bounded open set inC1Tsuch that the following conditions hold.

(i) For eachλ∈(0,1), the problem (φ(x′))′=λf(t,x,x′),x(0)=x(T),x′(0)=x′(T) has no solution on ?Ω;

(iii) The Brouwer degree deg(F,Ω∩Rn,0)≠0.

To use the Manásevich-Mawhin continuation theorem, we need to calculate the topological degree deg(F,Ω∩Rn,0). But, in practice, it is almost impossible to calculate directly the topological degree. In order to obtain the topological degree, homotopy invariance is often used to transform the problem into the topological degree of a special operator. The process is actually quite complicated. Therefore, it is the goal of researchers to avoid computing any topological degree. Obviously, avoiding the calculation of any topological degree will also greatly simplify the processing of practical problems in applying topological degree theory.

In this paper, we first give a corollary of the above Manásevich-Mawhin continuation theorem and a special continuation theorem with a corollary. Compared with the classical Manásevich-Mawhin continuation theorem, we can avoid calculating any topological degree and reduce the conditions of the theorem when using this special continuation theorem and its corollaries in applications. More importantly, the conditions for verifying this special continuation theorem will be easier and more convenient. As an application, we then use the new continuation theorem and its corollary to study the existence of periodic solutions and positive periodic solutions for a generalized Rayleigh typep-Laplacian equation with deviating arguments.

It is worth pointing out that those proofs through the classical Manásevich-Mawhin continuation theorem in the above mentioned papers can be simplified by using the special Manásevich-Mawhin continuation theorem and its corollary.

2 A Special Manásevich-Mawhin continuation theorem and corollaries

ProofCase I. For (f(x),x)>0, the assertion follows from the Acute Angle Principle in general topological degree theory.

Case II. For(f(x),x)<0, we have (-f(x),x)>0, which means that deg(-f,Ω,0)=1 by the Acute Angle Principle, and so deg(f,Ω,0)=(-1)n≠0. The proof is complete.

Lemma 2Letf(x) be continuous on an open interval (a,b) with closure [a,b], then deg(f,(a,b),0)≠0 if and only iff(a)f(b)<0.

Necessity. If deg(f,(a,b),0)≠0, then 0∈f(?(a,b)) by the definition of Brouwer degree, i.e.,f(a)≠0 andf(b)≠0. Iff(a)f(b)>0, thenf(a)>0,f(b)>0, orf(a)<0,f(b)<0.

(1) deg(g,(a,b),0)=0 by the definition of the Brouwer degree;

(2)fandghave the same the boundary values, i.e.,f(x)=g(x) for allx∈?(a,b).

Therefore, by the Boundary Value Theorem of Brouwer degree,deg(f,(a,b),0)=deg(g,(a,b),0)=0, which is a contradiction. Similarly, we can prove thatf(a)<0,f(b)<0 is also impossible. This completes the proof.

Using Lemma1, we can get a corollary of the Manásevich-Mawhin continuation theorem as follows.

Corollary 1Suppose that the condition (i) of the Manásevich-Mawhin continuation theorem holds. If

then the conclusion of the Manásevich-Mawhin Continuation Theorem holds.

Whenn=1, we can get the following especial Manásevich-Mawhin continuation theorem by using Lemma 2.

(1) For eachλ∈(0,1), the problem (known as auxiliary equation)

(φ(x′))′=λf(t,x,x′),x(0)=x(T),x′(0)=x′(T)

has no solution on ?Ω;

Remark 1Notice that the condition (3) in Theorem 2 is equivalent to the two conditions (ii) and (iii) in Theorem 1 whenn=1. So Theorem 2 is an equivalent one of Theorem 1 in the casen=1.

Corollary 2Assume that the conditions (1) and (2) in Theorem 2 hold. If, for almost everyt∈[0,T],

f(t,a,0)f(t,b,0)<0,

3 Some applications

Consider the following Rayleigh typep-Laplacian equation with deviating arguments

(φp(x′))′+f(t,x′(t-σ(t)))+g(t,x(t-τ(t)))=e(t),

(2)

wherep>1 is a constant,φp:R→Ris given byφp(y)=|y|p-2yfory≠0 andφp(0)=0,f,g∈C(R2,R) areT- periodic in its first argument andf(t,0)=0,σ,τ, ande∈C(R,R) areT- periodic (T>0).

Whenp=2, the Eq.(2) is the Rayleigh equation with deviating arguments. In the last few years, the existence of periodic solutions for a kind of Rayleigh equation with deviating arguments has received a lot of attention, see [13,14,19,30,31,33] and the references therein.

More recently, periodic solutions under special cases for the Eq.(2) have been studied by some researchers, for example, the case:σ=τ≡0 is studied by the authors in [6,21,24], and the case:σ≡0,τ≠0 is investigated by the authors in [3,8,10,20,22,32]. Whenσ≠0,τ≠0, Zong and Liang [27] discussed the existence of periodic solutions for the Eq.(2) and obtained some sufficient conditions under some special cases.

In this section, we discuss the existence ofT- periodic solutions of the Eq.(2). By using the above special Manásevich-Mawhin continuation theorem and its corollaries, we obtain some new sufficient conditions for the existence ofT-periodic solutions of the Eq.(2). These results generalize and improve those in [27].

Lemma 3[29]Letx(t) be a continuous derivableT-periodic function. Then for everyt*∈(-∞,+∞),

3.1 Existence of periodic solutions for the Laplacian equation

Theorem 3Assume that there exist constantsK1,K2>0,d>0,α:0<α≤p-1 and non-negative continuous functionsr1(t),r2(t) such that

(H1) |f(t,x)|≤r1(t)|x|α+K1, for (t,x)∈R2;

(H2)x(g(t,x)-e(t))>0 and |g(t,x)-e(t)|>r1(t)|x|α+K1, fort∈R,|x|>d;

(H3)g(t,x)-e(t)≥-r2(t)|x|α-K2, fort∈R,x<-d.

ProofConsider the auxiliary equation

(φp(x′(t)))′+λf(t,x′(t-σ(t)))+λg(t,x(t-τ(t)))=λe(t),λ∈(0,1).

(3)

Letx=x(t) be anyT- periodic solution of the Eq.(3). Integrating both sides of the Eq.(3) on [0,T], we have

(4)

It follows that there exists at1∈[0,T] such that

f(t1,x′(t1-σ(t1)))+g(t1,x(t1-τ(t1)))-e(t1)=0.

(5)

We assert that there exists at*∈[0,T] such that

(6)

Case 1:r1(t1)=0. From (5) and (H1), we have

|g(t1,x(t1-τ(t1)))-e(t1)|≤K1,

which, together with (H2), implies that

|x(t1-τ(t1))|≤d.

(7)

Case 2:r1(t1)≠0. If |x(t1-τ(t1))|>d, then it follows from (5), (H1) and (H2) that

r1(t1)|x(t1-τ(t1))|α+K1<|g(t1,x(t1-τ(t1)))-e(t1)|

≤r1(t1)|x′(t1-σ(t1))|α+K1.

Thus, we have

(8)

Combining (7) and (8), we see that

Note thatx(t) is periodic. Then there exists at*∈[0,T] such that (6) holds. By Lemma 3, we have

(9)

LetE1={t:t∈[0,T],x(t-τ(t))>d},E2={t:t∈[0,T],x(t-τ(t))<-d},E3={t:t∈[0,T],|x(t-τ(t))|≤d}. From (4) and (H2), we obtain

Asx(0)=x(T), there exists at*∈[0,T] such thatx′(t*)=0. By Lemma 3, we get

It follows from (9) that

LetM>max{M0,M1} and

Ω={x∈X:|x(t)|

(f(t,0)+g(t,M)-e(t))(f(t,0)+g(t,-M)-e(t))

=(g(t,M)-e(t))(g(t,-M)-e(t))<0.

Therefore, from corollary 2, the Eq. (2) has at least a periodic solution. This completes the proof.

Similarly, we have the following theorem.

Theorem 4Assume that (H1) and (H2) hold and the following condition

(H4)g(t,x)-e(t)≤r2(t)xα+K2, fort∈R,x>d

(H2)′xg(t,x)>0 and |g(t,x)|>r1(t)|x|α+K1, fort∈R,|x|>d;

(H3)′g(t,x)≥-r2(t)|x|α-K2, fort∈R,x<-d

and (5) is reduced to

f(t1,x′(t1-σ(t1)))+g(t1,x(t1-τ(t1)))=0.

Similar to the proof of Theorem 3, we can get (6) and (9). LetE1,E2andE3be defined as in Theorem 3. Then

and

Therefore, by Theorem 2, the Eq.(2) has at least a periodic solution. This completes the proof.

(H4)′g(t,x)≤r2(t)xα+K2, fort∈R,x>d

Remark 2Whenr1=r2≡0 andT=2π, Theorem 5 and Theorem 6 are reduced to Theorem 1 and Theorem 2 in [27], respectively. So our results generalize and improve the corresponding results in [27].

3.2 Existence of positive periodic solutions for the Laplacian equation

Theorem 7Assume that there exist constantsK>0,d>0,α:0<α≤p-1 and non-negative continuous functionr(t) such that

(P1) |f(t,x)|≤r(t)|x|α+K, for (t,x)∈R2;

(P2)g(t,x)-e(t)>r(t)xα+K, fort∈R,x>d;

(P3)g(t,0)-e(t)<0, fort∈R.

Ifα=p-1 andrT<1, orα

ProofLetx=x(t) be any positiveT- periodic solution of the auxiliary equation (3). Integrating both sides of (3) on [0,T], we can obtain at1∈[0,T] such that (5) holds.

We claim that there exists at*∈[0,T] such that

(10)

Case 1:r(t1)=0. From (5) and (P1), we have

g(t1,x(t1-τ(t1)))-e(t1)≤K,

which, together with (P2), implies that

0≤x(t1-τ(t1))≤d.

(11)

Case 2:r(t1)>0. Ifx(t1-τ(t1))>d, then it follows from (5), (P1) and (P2) that

r(t1)x(t1-τ(t1))α+K

Thus, we have

(12)

Combining (11) and (12), we see that

Note thatx(t) is periodic. Then there exists at*∈[0,T] such that (10) holds. By Lemma 3, we have

(13)

LetE1={t:t∈[0,T],0≤x(t-τ(t))≤d},E2={t:t∈[0,T],x(t-τ(t))>d}. From (4) and (P2), we obtain

Asx(0)=x(T), there exists at*∈[0,T] such thatx′(t*)=0. By Lemma 3, we obtain

It follows from (13) that

LetM>max{M0,M1} and

Ω={x∈X:0

then the condition (1) in Corollary 2 holds. Since for anyx∈?Ω∩R,x=M(>d) orx=0, we have, in view of (P2) and (P3),

(f(t,0)+g(t,M)-e(t))(f(t,0)+g(t,0)-e(t))=(g(t,M)-e(t))(g(t,0)-e(t))<0.

Hence, from Corollary 2, the Eq. (2) has at least a positiveT- periodic solution. This completes the proof.

Similarly, we have the following theorem.

Theorem 8Assume that there exist constantsK>0,d>0,α:0<α≤p-1 and non-negative continuous functionr(t) such that (P1) holds, and the following conditions hold:

(P4)g(t,x)-e(t)<-r(t)xα-K, fort∈R,x>d;

(P5)g(t,0)-e(t)>0, fort∈R.

Ifα=p-1 andrT<1, orα

Corollary 3Assume that (P3) holds, and that there exist constantsK>0 andd>0 such that

(C1) |f(t,x)|≤K, for (t,x)∈R2;

(C2)g(t,x)-e(t)>K, fort∈R,x>d.

Then the Eq.(2) has at least a positiveT-periodic solution.

Corollary 4Assume that there exist constantsK>0 andd>0 such that (C1) and (P5) hold. If

(C3)g(t,x)-e(t)<-K, fort∈R,x>d

holds, then the Eq.(2) has at least a positiveT-periodic solution.

Example 1Consider the following equation

(φp(x′(t)))′+costsinx′(t-sint)+g(t,x(t-cost))=sin2t,

(14)

whereg(t,x)=x1/3-1 fort∈R,x≤0, andg(t,x)=2ex-1fort∈R,x>0. Then the Eq. (14) has at least a positive 2π- periodic solution.

ProofFrom (14), we havef(t,x)=costsinxwith |f(t,x)|≤1 andf(t,0)=0,σ(t)=sint,τ(t)=cost,e(t)=sin2tare 2π- periodic, andg(t,x)-e(t)>1 fort∈R,x>1 andg(t,0)-e(t)≤-1 fort∈R. Namely, (C1),(C2) and (P3) hold. By Corollary 3, the Eq.(14) has at least a positive 2π-periodic solution.

4 Acknowledgements

This work is partially supported by the Natural Science Foundation of China(No.11871475), the Natural Science Foundation of Hunan Province (No.2019JJ40354), the Degree and Graduate Education Reform Research Project of Hunan Province(No.2020JGYB031), and the Graduate Education and Teaching Reform Research Project of Central South University(No.2020JGB020).