# Periodic Solutions for a Prescribed Mean Curvature Equation with Multiple Delays

Authors: Shiping Lu; Ming Lu
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We study the existence of periodic solutions for the one-dimensional prescribed mean curvature delay equation(d/dt)(x'(t)/1+x't2)+∑i=1naitgxt-τit=pt. By using Mawhin's continuation theorem, a new result is obtained. Furthermore, the nonexistence of periodic solution for the equation is investigated as well.


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## Body

## 1. Introduction

Prescribed mean curvature equation arises from some problems associated with differential geometry and physics such as combustible gas dynamics [1–3]. In the past years, the one-dimensional mean curvature equation of autonomous type
(1)ddt(x′(t)1+(x′(t))2)=g(x(t))
has been studied by many authors [4–14]. The interesting thing is that most of them focus on the case in which the nonlinearity g(u) is chosen to be various power growth functions. For example, Li and Liu in [4] studied the exact number of solutions for the boundary value problem
(2)-(u′1+(u′)2)′=λ(up+uq),u>0in(0,1),u(0)=u(1)=0.
Clearly, the powers of growth with respect to the variable u of functions on the right side of above two equations are not greater than max⁡{p,q}. Pan in [5] studied the exact multiple solutions of boundary value problem for a one-dimensional prescribed mean curvature equation with exponential nonlinearity
(3)ddt(x′(t)1+(x′(t))2)=λex(t),0<t<L,dasdasdasdasddssfdsfsx(0)=x(L)=0.
Equation (3) can be viewed as a variant of the one-dimensional Liouville-Bratu-Gelfand problem. By using the theory of time map, some results on the existence of multiple solutions are obtained. At the same time, we notice that Pan and Xing in [6] further studied the exact number of solutions for the problem
(4)ddt(x′(t)1+(x′(t))2)=λf(x(t)),-L<t<L,sdfsdfsdfsdfsdfsdfsdfds(-L)=x(L)=0,
where f(u)=eu, f(u)=(1+u)p, and f(u)=eu-1, respectively. For other recent developments and applications on the study of mean curvature equation, we refer the reader to [15–20], while the problem of periodic solution for prescribed mean curvature equation has been rarely studied [21–24]. Considering the delay phenomenon which exists generally in nature, Feng [22] studied the existence of periodic solutions for the one-dimensional mean curvature type equation in the following form:
(5)ddt(x′(t)1+(x′(t))2)+f(x(t))x′(t)+g(t,x(t-τ(t)))=p(t).
By imposing some conditions on functions f and g as follows.(H) there are two constantsl>0 and γ>0 such that
(6)|g(t,x1)-g(t,x2)|≤l|x1-x2|∀(t,x)∈R2;(7)|f(x)|≥γ∀x∈R;
the author obtained that (3) has at least one periodic solution by using Mawhin’s continuation theorem. From [22], we see that assumption (H) is crucial for estimatinga priori bounds of all possible T-periodic solutions.In this paper, we consider the following prescribed mean curvature equation with multiple delays:(8)ddt(x′(t)1+(x′(t))2)+∑i=1n‍ai(t)g(x(t-τi(t)))=p(t),
where g∈C(R,R) and p, ai, and τi are all continuous T-periodic functions, i=1,2,…,n. By using Mawhin’s continuation theorem, some new results are obtained; and the problem of nonexistence of periodic solution for (8) is investigated as well.The significance of this paper lies in the following two respects: firstly, we do not need assumption (7); secondly, the conditions imposed on function g(u) and the methods to estimatea priori bounds of possible T-periodic solutions for the equation Lx=λNx are all essentially different from corresponding ones of [22]. For example, we do not require that the function g satisfies global Lipschitz condition (6). Especially, the function g is allowed to be exponential nonlinearity.


## 2. Preliminaries

In order to investigate the existence of periodic solutions for (8), we give some definitions and lemmas in this section.In this paper, unless otherwise specified, we use the following notation. LetPT={φ∈C(R,R):φ(t+T)≡φ(t)} with the norm defined by |φ|∞=max⁡t∈[0,T]|φ(t)|; CT={x=(x1,x2)Τ:xi∈C(R,R),xi(t)≡xi(t+T),i=1,2} and the norm defined by ∥x∥=max⁡{|x1|∞,|x2|∞}. Clearly, PT and CT are two Banach spaces. Furthermore, define |φ|p=(∫0T‍|φ(s)|pds)1/p for all φ∈PT, where p>1 is a constant.Lemma 1 (see [25]).
Supposeτ∈PT and τ′(t)<1,∀t∈[0,T]. Then the function t-τ(t) has its inverse μ(t) satisfying μ∈C(R,R) with μ(t+T)≡μ(t)+T,∀t∈R. Furthermore, if α,β∈PT, then
(9)∫0Tα(t)β(t-τ(t))dt=∫0Tα(μ(t))1-τ′(μ(t))β(t)dt,(10)∫0Tα(μ(t))1-τ′(μ(t))dt=∫0Tα(t)dt.Now, let us recall Mawhin’s continuation theorem. LetX and Y be real Banach spaces and let L:D(L)⊂X→Y be a Fredholm operator with index zero; here D(L) denotes the domain of L. This means that Im⁡L is closed in Y and dim⁡ker⁡L=codimIm⁡L<+∞. If L is a Fredholm operator with index zero, then there exist continuous projectors P:X→X,Q:Y→Y such that Im⁡P=ker⁡L,Im⁡L=ker⁡Q and LD(L)∩ker⁡P:(I-P)X→Im⁡L is invertible. Denote by Kp the inverse of LP.LetΩ be an open bounded subset of X; a continuous map N:Ω-⊂X→Y is said to be L-compact in Ω¯ if QN(Ω¯) is bounded and the operator Kp(I-Q)N(Ω¯) is relatively compact.Lemma 2 (see [26]).
Suppose thatX and Y are two Banach spaces, and L:D(L)⊂X→Y is a Fredholm operator with index zero. Furthermore, Ω⊂X is an open bounded set and N:Ω¯→Y is L-compact on Ω¯. If all the following conditions hold:(1)
Lx≠λNx, ∀x∈∂Ω∩D(L), λ∈(0,1),(2)
Nx∉Im⁡L, ∀x∈∂Ω∩ker⁡L,(3)
deg⁡{JQN,Ω∩ker⁡L,0}≠0, where J:Im⁡Q→ker⁡L is an isomorphism,then equation Lx=Nx has a solution on Ω¯∩D(L).Throughout this paper, for eachi∈{1,2,…,n}, besides τi∈PT, we suppose in addition τi∈C1(R,R) with τ′(t)<1, ∀t∈[0,T].Remark 3.
From above assumption, one can find from Lemma1 that, for each i∈{1,2,…,n}, the function t-τi(t) has its inverse denoted by μ(t). Define
(11)Γ(t)∶=∑i=1n‍ai(μi(t))1-τi′(μi(t)),Γ1(t)∶=∑i=1n‍|ai(μi(t))1-τi′(μi(t))|.
Since ai∈PT, it follows from Lemma 1 again that
(12)∫0TΓ(t)dt=∑i=1n‍∫0Tai(t)dt,∫0TΓ1(t)dt=∑i=1n‍∫0T|ai(t)|dt.For the sake of convenience, we list the following assumptions which will be used for us to study the existence of periodic solutions to (8) in Section 3.(A1) The functionsΓ, Γ1, and p satisfy Γ(t)>0 for all t∈[0,T] and
(13)σ:=|Γ1Γ|∞∫0Tp(s)ds+∫0T|p(s)|ds∈(0,1),
where Γ and Γ1 are all continuous functions determined by Remark 3.(A2) The functiong satisfies g(x)>0 for all x∈R, and there is a constant M>0 such that, for all x∈(M,+∞),
(14)-g(x)(∑i=1n‍∫0Tai(s)ds)+∫0Tp(s)ds<0,-g(-x)(∑i=1n∫0T‍ai(s)ds)+∫0Tp(s)ds>0.Remark 4.
Since (8) does not contain the term f(x(t))x′(t), condition (7) in assumption (H) of [22] does not hold. Furthermore, in our paper, the function g is not required to satisfy the global Lipschitz condition (6). So the conditions in our paper are all essentially different from corresponding ones of [22].
Since the differential term ofx′(t)/1+(x′(t))2 is nonlinear with respect to x(t), the differential operator L:(Lx)(t)=(d/dt)(x′(t)/1+(x′(t))2) associated with Mawhin’s continuation Theorem is not Fredholm type. So we need to convert (8) to the following two-dimensional system:
(15)x1′(t)=x2(t)1-(x2(t))2,x2′(t)=-∑i=1nai(t)g(x1(t-τi(t)))+p(t).
Clearly, if (x1(t),x2(t))Τ is a T-periodic solution to (15), then x1(t) must be a T-periodic solution to (8). From this, we see that, in order to investigate the existence of T-periodic solution for (8), it suffices for us to prove that (15) has a T-periodic solution.
For using Mawhin’s continuation theorem, let(16)L:D(L)⟶CT,[Lx](t)=(x1′(t),x2′(t))Τ,t∈R,N:Σ⊂CT⟶CT,[Nx](t)=(ψ(x2(t))-∑i=1nai(t)g(x1(t-τi(t)))+p(t)),t∈R,
where Σ={x:x=(x1,x2)Τ∈CTwith|x2|∞<δ,|x1|∞<+∞}, where δ∈(σ,1) is a constant and σ∈(0,1) is determined in assumption (A1).


## 3. Main Results

In this section, we will apply Lemma2 to study the existence of periodic solutions for (8).Theorem 5.
Suppose that assumptions (A1) and (A2) hold. Then (8) possesses at least one T-periodic solution.Proof.
Suppose thatu∈CT is an arbitrary solution to the equation Lx=λNx for each λ∈(0,1), where L and N are defined by (16), respectively. This implies
(17)u1′(t)=λu2(t)1-(u2(t))2,u2′(t)=-λ∑i=1nai(t)g(u1(t-τi(t)))+λp(t),dsdadsdsdsadsdsadsadsaλ∈(0,1).
From the first formula of (18), we see
(18)u2(t)=(1/λ)u1′(t)1+(u2′(t))2/λ2,λ∈(0,1).
Substituting (18) into the second formula of (17), we have
(19)ddt((1/λ)u1′(t)1+(u1′(t))2/λ2)=-λ∑i=1nai(t)g(u1(t-τi(t)))+λp(t).
Integrating both sides of (19) on the interval [0,T], we obtain
(20)∫0T∑i=1nai(t)g(u1(t-τi(t)))dt=∫0Tp(t)dt.
Since τi′(t)<1 for all t∈[0,T], by using Lemma 1, we see that the function t-τi(t) has its inverse μi(t)(i=1,2,…,n). So by applying (9) to (20), we have
(21)∫0TΓ(t)g(u1(t))dt=∫0Tp(t)dt,
where Γ(t)=∑i=1n‍(ai(μi(t))/(1-τi′(μi(t)))) is determined by Remark 3. Similarly, from (19), we have
(22)∫0T|((1/λ)u1′(t)1+(u1′(t))2/λ2)′|dt≤λ∫0TΓ1(t)|g(u1(t))|dt+λ∫0T|p(t)|dt,
which together with the condition of g(x)>0 for all x∈R in assumption (A2) yields that
(23)∫0T|((1/λ)u1′(t)1+(u1′(t))2/λ2)′|dt≤∫0TΓ1(t)g(u1(t))dt+∫0T|p(t)|dt.
It follows from the condition of Γ(t)>0 for all t∈[0,T] in assumption (A1) that
(24)∫0T‍|((1/λ)u1′(t)1+(u1′(t))2/λ2)′|dt≤∫0TΓ1(t)Γ(t)Γ(t)g(u1(t))dt+∫0T|p(t)|dt≤|Γ1Γ|∞∫0TΓ(t)g(u1(t))dt+∫0T|p(t)|dt.
Substituting (21) into the above formula, we obtain that
(25)∫0T‍|((1/λ)u1′(t)1+(u1′(t))2/λ2)′|dt≤|Γ1Γ|∞∫0Tp(t)dt+∫0T|p(t)|dt.
On the other hand, since∫0T‍u1′(s)ds=u1(T)-u1(0)=0, there must be a point t*∈[0,T] such that u1′(t*)=0; that is, (1/λ)u1′(t*)/1+(u1′(t*))2/λ2=0. So
(26)|(1/λ)u1′(t)1+(u1′(t))2/λ2|≤∫t*t|((1/λ)u1′(s)1+(u1′(s))2/λ2)′|ds≤∫t*t*+T‍|((1/λ)u1′(s)1+(u1′(s))2/λ2)′|ds=∫0T‍|((1/λ)u1′(s)1+(u1′(s))2/λ2)′|ds,fdsfdsfdsfdsfdt∈[t*,t*+T].
This together with (25) implies that
(27)|(1/λ)u1′(t)1+(u1′(t))2/λ2|≤|Γ1Γ|∞∫0Tp(t)dt+∫0T|p(t)|dt,fdfdsfdsffdsft∈[t*,t*+T].
From assumption (A1), we know σ=|Γ1/Γ|∞∫0Tp(t)dt+∫0T|p(t)|dt<1; it follows from (18) and (27) that
(28)|u2|∞=max⁡t∈[t*,t*+T]|(1/λ)u1′(t)1+(u1′(t))2/λ2|≤σ<1.
Furthermore, from the fact of the function x/1+x2 being strongly increasing for x∈R, it follows from (27) that
(29)|u1′(t)λ|≤σ1-σ2,t∈[t*,t*+T];
that is,
(30)|u1′|∞=max⁡t∈[t*,t*+T]|u1′(t)|≤λσ1-σ2<σ1-σ2:=M0.
By using (28), we have
(31)11+M02∫0T|u1′(t)|dt=11+M02∫t*t*+T|u1′(t)|dt<∫t*t*+T|u1′(t)λ2+(u1′(t))2|dt=∫t*t*+T|(1/λ)u1′(t)1+(u1′(t))2/λ2|dt≤Tσ,
which implies
(32)∫0T|u1′(t)|dt<Tσ1+M02.
Furthermore, from (21) and the condition of Γ(t)>0 for all t∈[0,T] in assumption (A1), we see that there must be a point t1∈[0,T] such that
(33)g(u1(t1))∫0TΓ(s)ds=∫0Tp(s)ds;
that is,
(34)g(u1(t1))=∫0Tp(s)ds∫0TΓ(s)ds.
By using the conclusion of ∫0TΓ(s)ds=∑i=1n∫0Tai(s)ds in Remark 3, we have
(35)g(u1(t1))=∫0Tp(s)ds∑i=1n∫0Tai(s)ds.
So by using assumption (A2), we see
(36)|u1(t1)|≤M,
which together with (32) yields that
(37)|u1|∞=max⁡t∈[t1,t1+T]|u1(t)|≤|u1(t1)|+∫t1t1+T|u1′(s)|ds≤M+∫0T|u1′(t)|dt<M+Tσ1+M02=:M1.
If set Ω={x∈CT:|x1|∞<M1,|x2|∞<σ}, then, from the above proof, we see that Lu≠λNu for all (λ,u)∈(0,1)×∂Ω. This means that condition (1) of Lemma 2 holds.
Now, supposeu∈∂Ω∩ker⁡L; then u=(u1,u2)∈R2 is a constant vector with |u1|=M1 or |u2|=σ. So
(38)QNu={ψ(u2)=u21-u22-g(u1)(∑i=1n∫0Tai(s)ds)+∫0Tp(s)ds,
which together with assumption (A2) yields that
(39)QNu≠0,∀u∈∂(Ω∩ker⁡L).
From (30), (37), and (39), we see that conditions (1) and (2) in Lemma 2 hold for Ω={x∈CT:∥x1∥∞<M1,∥x2∥∞<σ}.
Below, we will show that condition (3) of Lemma2 also holds. In fact, let
(40)H(μ,x)=μx+(1-μ)JQNx,for(μ,x)∈[0,1]×Ω¯,
where J(x1,x2)Τ=(x1,x2)Τ. Clearly, if u*(t)=(u1*(t),u2*(t))Τ∈Ω¯∩ker⁡L is the solution of equation H(μ,x)=0 for some μ∈[0,1], then (u1*(t),u2*(t))Τ is a constant vector. So
(41)ψ(u2*)=u2*1-(u2*)2=0-g(u1*)(∑i=1n∫0Tai(s)ds)+∫0Tp(s)ds=0.
This together with assumption (A2) results in u2*=0 and |u1*|≤M. By (37), we see that M<M1, and then u*∉∂(Ω¯∩ker⁡L). From this, we conclude that H(μ,x)≠0 for all (μ,x)∈[0,1]×∂(Ω¯∩ker⁡L), which together with the property of homotopy invariance for Brouwer’s topological degree gives that
(42)deg⁡{JQN,Ω∩ker⁡L,0}=deg⁡{H(0,·),Ω∩ker⁡L,0}=deg⁡{H(1,·),Ω∩ker⁡L,0}=deg⁡{I,Ω∩ker⁡L,0}=1≠0.
This proves that condition (3) of Lemma 2 holds. Thus, by using Lemma 2, we have that (15) possesses at least one T-periodic solution u0(t)=(u01(t),u02(t))Τ∈Ω¯={x∈CT:|x1|∞≤M1,|x2|∞≤σ}. Clearly, u01(t) must be a T-periodic solution to (8).Remark 6.
From the proof of Theorem5, it is easy to see that assumption (A2) can be replaced by the following:(A2)′
the functiong satisfies g(x)>0 for all x∈R, and there is a constant M>0 such that, for all x∈(M,+∞),
(43)-g(x)(∑i=1n∫0Tai(s)ds)+∫0Tp(s)ds>0,-g(-x)(∑i=1n∫0Tai(s)ds)+∫0Tp(s)ds<0.Theorem 7.
Suppose thatΓ(t)≥0 and Γ(t)≢0 for t∈[0,T]. If
(44)∫0Tp(s)ds∑i=1n∫0Tai(s)ds∉g(R)∶={g(u):u∈R},
then (8) has no T-periodic solution.
Suppose thatu0(t) is a T-periodic solution to (8). Then
(45)ddt(u0′(t)1+(u0′(t))2)+∑i=1nai(t)g(u0(t-τi(t)))=p(t).
Integrating both sides of (45) on [0,T], we have
(46)∫0TΓ(s)g(u0(s))ds=∫0Tp(s)ds.
Since Γ(t)≥0 and Γ(t)≢0 for t∈[0,T], it follows that there is a t0∈[0,T] such that g(u0(t0))=∫0Tp(s)ds/∫0TΓ(s)ds. In view of the fact of ∫0TΓ(s)ds=∑i=1n‍∫0Tai(s)ds in Remark 3, we see that
(47)g(u0(t0))=∫0Tp(s)ds∑i=1n∫0Tai(s)ds.
So
(48)∫0Tp(s)ds∑i=1n∫0Tai(s)ds∈g(R),
which contradicts the assumption of ∫0Tp(s)ds/∑i=1n∫0Tai(s)ds∉g(R).For illustrating Theorem5, we give the following examples.Example 8.
Consider the problem of existence of periodic solution to the equation(49)(x′(t)1+(x′(t))2)′+14(1-12cos⁡t)ex(t-(1/2)sint)-18(1+12sint)ex(t-(1/2)cos⁡t)=15πsin2t.
Corresponding to (8), we see that g(x)=ex, a1(t)=(1/4)(1-(1/2)cos⁡t), a2(t)=-(1/8)(1+(1/2)sint), τ1(t)=(1/2)sint, τ2(t)=(1/2)cos⁡t, and p(t)=(1/5π)sin2t. Let μi(t) be the inverse of t-τi(t), (i=1,2). Then
(50)Γ(t)=a1(μ1(t))1-τ1′(μ1(t))+a2(μ2(t))1-τ2′(μ2(t))=1-(1/2)cos⁡μ1(t)4(1-(1/2)cos⁡μ1(t))-1+(1/2)sinμ2(t)8(1+(1/2)sinμ2(t))=18>0,∀t∈[0,2π],
and |Γ1/Γ|∞=3. So
(51)σ:=|Γ1Γ|∞∫0Tp(s)ds+∫0T|p(s)|ds=45π∫02πsin2sds=45<1,
which implies that assumption (A1) holds. Furthermore, since g(x)=ex, assumption (A2) also holds. Thus, by using Theorem 5, we have that (49) possesses at least one 2π-periodic solution.Remark 9.
Sinceg(x)=ex, it is easy to see that the function g does not satisfy the global Lipschitz condition (6). So the result of Example 8 cannot be obtained by using the main theorem of [22].Example 10.
Consider the problem of existence of periodic solution to the equation(52)(x′(t)1+(x′(t))2)′+14(1-12cos⁡t)ex(t-(1/2)sint)1+ex(t-(1/2)sint)+18(1+12sint)ex(t-(1/2)cos⁡t)1+ex(t-(1/2)cos⁡t)=θsin2t,
where θ∈R is a constant. Corresponding to (8), we have
(53)Γ(t)≡38,|Γ1Γ|∞=1,g(u)=eu1+eu,∫02π‍p(s)ds∑i=1n∫02π‍ai(s)ds=43θ.
Thus, if θ∈(0,3/4), then there must be a constant M>0 such that
(54)-ex1+ex∫02πΓ(s)ds+θ∫02πsin2sds=-3πex4(1+ex)+θπ<0,(55)-e-x1+e-x∫02πΓ(s)ds+θ∫02πsin2sds=-3πe-x4(1+e-x)+θπ>0
for x>M, and if θ∈(0,1/2π), then
(56)σ∶=|Γ1Γ|∞∫0Tp(s)ds+∫0T|p(s)|ds=2θ∫02πsin2sds=2θπ<1.
Furthermore, since g(R)=(0,1), it follows that ∫02πp(s)ds/∑i=1n∫02π‍ai(s)ds=(4/3)θ∉g(R) for θ∈(-∞,0]∪[3/4,+∞). Therefore, by using Theorems 5 and 7, we obtain the following results: (1)
ifθ∈(0,1/2π), then (52) has at least one 2π-periodic solution;(2)
ifθ∈(-∞,0]∪[3/4,+∞), (52) has no 2π-periodic solution.



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