# The Existence of Positive Solution of a Nonlinear Problem on Unit Circle

Authors: Na Wei; Yanli Zhou; Xiangyu Ge
Journal: Journal of Function Spaces
 (2020)

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## Abstract

We consider the following nonlinear problemu″x+λux+fx,u=0,x∈S1, (P) where S1 is a unit circle, λ∈ℝ is a parameter, and fx,u is a nonlinearity. By using the variational method, we obtain the existence of a positive solution to (P) for all λ∈ℝ. This phenomenon is different from the corresponding boundary value problem.


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## Body

## 1. Introduction

We consider the following nonlinear problem(1)u″x+λux+fx,u=0,x∈S1,where S1 is a unit circle, λ is a parameter and the nonlinear term f satisfies.(f1) fx,t is a continuous function on S1×ℝ, and fx,t is odd on t.The second-order differential equation (1) is a nonlinear elliptic equation model arising from studying some physics processes or geometric problems. For example, (1) is the problem of oscillations [1, 2] of a spherical thick shell made of an elastic material when the nonlinearity f satisfies a superlinear condition as t⟶+∞, namely,f2limt→+∞fx,t/t=+∞ uniformly for x∈S1.Forλ=1, fx,t=gxtp−1t with p∈ℝ and gx:S1⟶ℝ, (1) is also used to model the planar Minkowski problem [3–5].As is well known, the strong maximum principle is a powerful tool for studying positive solutions of a nonlinear elliptic equation, see for example [6, 7]. In [8], the existence of positive solutions to the nonlinear elliptic equation can also be obtained by a standard argument, based on transforming the partial differential equation to the equivalent ordinary differential equation. The iterative method can also be used in the existence of a solution to nonlinear elliptic equation (see [9]). If the parameter λ>0 is small, the related Green function of the operator d/dx2+λ on S1 is positive. When the Green function is positive, the upper and lower solution method and Schauder’s fixed point theorem are also suitable to obtain the positive solution to some nonlinear elliptic equations, see for example [10–12]. However, the Green function of operator d/dx2+λ on S1 may be sign-changing for some λ>0.In this paper, we aim at studying the existence of a positive solution of nonlinear equation (1) for all λ∈ℝ and the effect of asymptotic properties of nonlinearity fx,t when t⟶0, to the positivity of solution of (1). Our contribution includes two aspects. Firstly, we establish the following Theorem 1 to show that a nonnegative solution is strictly positive by the analysis based on the Taylor expansion theorem.Theorem 1.
Assumeλ∈ℝ, fsatisfiesf1andfx,t/tis bounded ast⟶0. Letv∈C2S1be a nonnegative solution of (1). Ifv≡0, thenvis strictly positive.Indeed, we can also apply the Green function or the strong maximum principle to study the positive of the solution. However, the direct application of Green function for studying the strictly positive of solution often needs more assumptions on parameterλ and the information of sign of the non-homogeneous term. By using Theorem 1, we could estimate the strictly positive of solution to (1) for all λ∈ℝ as the following applications.Secondly, as an application, we give a strictly positive solution of (1) for all λ∈ℝ by using Theorem 1 and the variational method. To study the existence of positive solution to (1) under a general nonlinearity, we also need the following general assumptions.f3limt→0fx,t/t=0 uniformly for x∈S1;f4 there exists α>2 such that, for every x∈S1 and t∈ℝ(2)αFx,t≤tfx,t;f5 there exists T>0 such that(3)infx∈S1,∣t∣>TFx,t>0;f6 for every x∈S1, fx,t/t is an increasing function of t on 0,+∞.Theorem 2.
Assumef1-f6hold. Forλ<0, (1) has a positive solution.We see that the existence of a positive solution in Theorem2 depends on the assumption that λ<0. It is natural to ask whether there exist nontrivial solutions of (1) for some λ≥0. To shed some light to this problem, we give the existence of positive solutions to (1) with some special nonlinearity as follows.Theorem 3.
Assumefx,u=gxuxp−1uxwithp>1. Letλ≥0. Ifg∈CS1is nonnegative with∫S1gx2/1−pdx<+∞, then (1) has a positive solution.From Theorems2 and 3, we obtain that (1) may own a positive solution for all λ∈ℝ, which is different from the boundary value problem [13–16]. To illustrate that the conclusions above for (1) is a new phenomenon that different from the existence of a positive solution to boundary value problem, we consider the following problem(4)u″x+λux+gxuxp−1ux=0x∈0,2π,u0=u2π=0.Letu∈C20,2π be a solution of (4), we have(5)λ−14∫02πsinx2uxdx+∫02πgxuxp−1sinx2dx=0.Ifλ>1/4, it follows that there is no positive solution to the boundary value problem (4) with gx≥0.Since the proof of Theorem2 depends on f1 and λ<0, it is impossible to deduce Theorem 3 by a similar method for Theorem 2. In the next section, we apply a reverse Hölder’s inequality and constrained variational method to prove Theorem 3.


## 2. Proof of Theorems

We first give our new method to estimate the positivity of the solution of (1) as follows. By assumption, we obtain(6)−v″x−λvx=fx,vx,x∈0,2π.We will prove thatv is strictly positive by the method of contradiction. Assume vθ0=0 for some θ0∈0,2π. We choose θ0=infx∈0,2π:vx=0. Since v≡0, we have two points θr and θl in 0,2π such that θ0≤θr<θl and(7)vx=0forx∈θ0,θr;andvx>0forx∈θr,θl.Since0 is the minimum value of v on 0,2π, we have(8)v′x=0forx∈θ0,θr.Byf1, we see that fx,0=0. It follows from (6) and (7) that(9)v″x=0forx∈θ0,θr.By Taylor’s theorem, we give a formula ofvx for all x∈θr,θl as follows,(10)vx=v″ξx2x−θr2forsomeξx∈θr,x.We claim that there exists a strictly monotone decreasing sequenceθn⊂θr,θl such that(11)θn⟶nθrand∣v″θn∣>0.Otherwise, there exists a smallδ>0 such that ∣v″x∣=0 for all x∈θr,θr+δ. By applying (10), we derive that(12)vx=0forx∈θr,θr+δ,which contradicts to the second part of (7). By (9), we see that v″θr=0. Let θ1 be given by (11), then we have(13)ρ1=supx∈θr,θ1:∣v″x∣=supρ∈θr,θ1∣v″ρ∣>θr.Fork∈ℕ, we can define a sequence of ρk by(14)ρk+1=supx∈θr,ρk−δk:∣v″x∣=supρ∈θr,ρk−δk∣v″ρ∣,where δk=ρk−θr/2. From (11), we see that θn⟶nθr. Fix k∈ℕ, if i∈ℕ is large enough, then θi∈θr,ρk; and it follows from (14) that ∣v″ρk∣≥∣v″θi∣>0. Following this way, we obtain a strictly monotone decreasing sequence ρn⊂θr,θl such that ρn⟶nθr and(15)∣v″ρn∣>0,∣v″x∣≤∣v″ρn∣forx∈θr,ρn.For eachn∈ℕ, via applying (10) with x=ρn, there exists ξn∈θr,ρn such that(16)vρn=vξn2ρn−θr2andv″ξn≤v″ρn.Letx=ρn in (6), we have(17)−v″ρn−λvρn=fρn,vρn.Substitutingvρn in (17) by using the first part of (16), and then multiplying (17) by 1/v″ρn, we deduce that(18)−1−λv″ξn2v″ρnρn−θr2=fρn,vρnvρnv″ξn2v″ρnρn−θr2.By using the second part of (16) and the assumption that fθ,t/t is bounded, we derive that(19)fρn,vρnvρn=O1andv″ξn2v″ρnρn−θr2=o1asn⟶+∞.From (18), we get a contradiction that 1=o1 as n⟶+∞.To prove Theorem2, we firstly obtain a nonnegative solution of (1) via a minimum of I on manifold N; then, we prove that it is strictly positive by using Theorem 1. The process is standard for studying the existence of nonnegative solutions to nonlinear elliptic equations by the method involving the Nehari manifold. For completeness, we give the main steps. The following deformation lemma is needed.Lemma 1.
(see, for example, Lemma 2.3 in [17]).LetXbe a Banach space,I∈C1X,ℝ, S⊂X, c∈ℝ, ε,δ>0such that(20)infϕ=0I′uϕ∥ϕ∥≥8ε/δforallu∈u:∥u−v∥≤2δ,c−2≤εIu≤c+2ε.
We denote byId the set u∈X:Iu≤d for any d∈ℝ. Then, there exists η∈C0,1×X,X such that(i)
ηt,u=u, if t=0 or if u∈u∈X:infw∈S∥u−w∥≤2δ,c−2ε≤Iu≤c+2ε,(ii)
η1,Ic+ε∩S⊂Ic−ε(iii)
ηt,· is an homeomorphism of X for any t∈0,1(iv)
Iη·,u is non increasing for any u∈XLetS be parameterized by angle x, H≔H1S be the Sobolev space equipped with the usual normal(21)∥u∥=∫Su′2+u2dx1/2.DefineFx,u=∫0ufx,tdt. Under the assumptions f1−f3, we have the following C1 variational function of (1).(22)Iu=12∫S1u′2dx−λ2∫S1u2dx−∫S1Fx,udx.The related Nehari manifold is defined by(23)N≔u∈H1S1\0:∫S1u′2dx−λ∫S1u2dx−∫S1fx,uudx=0.Similar to Lemma 4.1 in [17], we have the following lemma.Lemma 2.
Assumef1−f6hold. Letu∈Hbe nontrivial andht=Itufort>0. Ifλ<0, there existstu>0such thattuu∈N, htu=maxt>0ht. The mapu⟶tuis continuous, andu⟶tuudefines a homeomorphism of the unit sphere ofHwithN.Proof.
Ifλ<0, we see that 1/2∫S1u′2dx−λ/2∫S1u2dx>0 for the nontrivial u∈H. For any u∈N, let gt=1/t∫S1fx,tudx. Then g is an increasing function of t by f6. By the definition of ht=Itu, we see that h0=0, and tuu∈N if and only if ∫S1u′2dx−λ∫S1u2dx=1/tu∫S1fx,tuuudx, which is equivalent to h′tu=0. Under the assumptions f3−f5, we see that ht>0 for small t>0 and ht<0 for large t. Therefore, there exists a unique tu such that h′tu=0 and tuu∈N. By f4, there exists a constant C0>0 such that(24)C0uα−1≤ufx,u,where α>2 is given by f4.
Assume that there exists a nontrivial sequenceun⟶nu in H1S1 with u=0, we have a sequence tun that tunun∈N. This and f4 imply that(25)Ctun∥un∥2≥tun∫S1un′2dx−λ∫S1un2dx=∫S1fx,tununundx≥∫S1C0tunαunα−1dx.It follows thattun is bounded. And a subsequence of tun converges to a number t0. By using (25) and the uniqueness of map tu=tu, we see that t0=tu. That is, tun converge to tu. The inverse map of u⟶tuu is the retraction u⟶u/∥u∥.By using the properties ofI in Lemma 2. We obtain the following Lemma 3.Lemma 3.
Assumef1−f6hold andλ<0. Letc1=infu∈NIu, thenc1>0. Ifv∈NandIv=c1, thenvis a critical point ofIonH.Proof.
Letc2≔infu∈H,u=0maxt≥0Itu and c≔infγ∈Γmaxt∈0,1Iγt, where Γ≔γ∈C0,1,H:Iγ1<0,γ0=0. We prove this Lemma by the following two steps.Step 1.
c1=c2=c>0.
By Lemma2, we see that Ituu=maxt≥0Itu if and only if tuu∈N. This and the definitions of c1,c2 imply that c1=c2. By f1−f2, we see that Itu<0 for u=0 and t>0 large. Hence, c≤c2. From Lemma 2, we see that the Nehari manifold N separates H into two parts. By assumptions f1−f3, we deduce that infu∈N∥u∥>0 and there exists r>0 such that(26)min∥u∥≤rIu=0,inf∥u∥=rIu>0.
It follows that the passγ∈Γ has to cross N and c≥inf∥u∥=rIu>0. These facts imply that c1≤c.Step 2.
v is a critical point of I.
Assume thatI′v=0. Then there exists δ>0 and ε>0 such that(27)infϕ=0I′uϕ∥ϕ∥≥εforall∥u−v∥≤3δ.
Forε≔c/2,εδ/8 and S≔Bv,δ, Lemma 1 yields a deformation η such that(i)
η1,u=u if u∈u∈H:c−ε≤Iu≤c+2ε,(ii)
η1,Ic+ε∩Bv,δ⊂Ic−ε(iii)
Iη1,u≤Iu, u∈H
We deduce thatmaxt>0Iη1,tv≤c−ε<c=c2, which contradicts the definition of c2.By Lemma3, we see that c1=infu∈NIu>0. There exists a sequence of un⊂N such that Iun⟶c1 as n⟶+∞, that is,(28)12∫S1u′2dx−λ2∫S1u2dx−∫S1Fx,udx=c1+o1,asn⟶+∞,(29)∫S1u′2dx−λ∫S1u2dx−∫S1fx,uudx=0,Letα>2 be given by f4, it follows from (28) and (29) that(30)α−22∫S1u′2dx−λ∫S1u2dx≤2c1forlargen.Sinceλ<0, we have that ∥un∥ is bounded. Then, there exists a weak limitation v of the sequence un in H, i.e., up to a subsequence, un⇀v weakly in H and un⟶v uniformly on S1 as n⟶+∞. This together with the assumptions f1−f3 imply that(31)limn→+∞∫S1fx,unundx=∫S1fx,vvdx.It follows from (29) and (31) that limn→+∞∥un−v∥=0, which implies that v∈N and c1=infu∈NIu=Iv. By Lemma 3, we see that I′v=0. Let v+=max0,v and v−=−max0,−v. Then, we deduce that Iv+v+=Ivv+=Iv−v−=Ivv−=0. If v is a sign-changing function, then v− and v+ are nontrivial. It follows that v−,v+∈N. We thus deduce the following contradiction(32)c1=Iv=Iv++Iv−≥2infu∈NIu=2c1>0.So,v is not a sign-changing function. By the second part of f1, we see that Iv=I−v. Without loss of generality, we assume that v is nonnegative. A standard regularity shows that v is C2 on S1. Since the nonlinear term f satisfies assumptions f1 and f3, we can apply Theorem 1 to get that v is strictly positive.Let us consider the constrained optimal problem(33)c=infFu=1Iu,where(34)Iu=∫Su′2−λu2dx,(35)Fu=∫Sgxup+1dx.Forλ≥0, it follows from reverse Hölder inequality and (34) that(36)infFu=1Iu=infFu=0IuFu2/p+1≥infFu=0−λ∫S1u2dx∫S1gx2/1−pdθ1−p/2∫S1u2dx=−λ∫S1gx2/1−pdθp−1/2>−∞,which implies that Iu is bounded in H1S from below under the constrained condition that Fu=1. By the definition of I,F in (34) and (35), there exists a nonnegative minimizing sequence un⊂H1S such that Fun=1 and(37)Iun⟶infFu=1Iuasn⟶+∞.It follows from2/p−1>p−1/p+1 and the Hölder inequality that(38)∫S1gx−p−1/p+1dx≤C∫S1gx2/1−pdxp−122p+1<+∞.SinceFun=1, by Hölder inequality we see that(39)∫S1un2dx≤∫S1gx−p−1/p+1dxp+1/p+2∫S1gxunp+1dx2/p+1=∫S1gx−p−1/p+1dxp+1/p+2<+∞.This together withIun⟶n→+∞infFu=1Iu imply that unis bounded in H1S. So, there is a subsequence such that(40)un⟶nvweaklyinH1S1.By the compactness of Sobolev imbedding, up to a subsequence, we have(41)un⟶nvstronglyinCαS1forα∈0,1/2.It follows thatF∣v∣=1, and therefore v=0.(42)∫02πgθvp+1dθ=limn→+∞∫02πgθunp+1dθ=1.By the Fatou lemma, we get(43)infFu=1Iu≤I∣v∣≤limn→+∞Iun=infFu=1Iu,So,∣v∣ is a minimizer of I under constrained condition that Fu=1. There exists ω∈ℝ such that(44)I′∣v∣=ωF′vinH−1S.It follows from the regularity theorem that∣v∣∈C2S is a classical nonnegative solution of(45)−u″−λu=ωp+12gxup−1u,x∈S1.Letw=ωp+1/21/p−1∣v∣, we see that w is nonnegative solution of (1) with fx,u=gxup−1u. By applying Theorem 1, we see that w is positive.



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