# The Existence of Multiple Solutions for Nonhomogeneous Kirchhoff Type Equations inℝ3

Authors: Qi Zhang; Xiaoli Zhu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We are concerned with the existence of multiple solutions to the nonhomogeneous Kirchhoff type equation-(a+b∫ℝ3‍|∇u|2)Δu+u=|u|p-1u+h(x)inℝ3, where a,b are positive constants, p∈(1,5),0⩽h(x)=h(|x|)∈C1(ℝ3)∩L2(ℝ3), we can find a constant mp>0 such that for all p∈(1,5) the equation has at least two radial solutions provided |h|2<mp.


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## Body

## 1. Introduction and Main Result

In this paper, we consider the existence of multiple solutions to the following nonhomogeneous Kirchhoff type equation:(1)-(a+b∫ℝ3‍|∇u|2)Δu+u=|u|p-1u+h(x)inℝ3,
where a,b are positive constants and p∈(1,5)·h∈C1(ℝ3)∩L2(ℝ3) satisfies the following conditions: (h1)
0⩽h(x)=h(|x|)∈L2(ℝ3) and |h|2⩽mp, where
(2)mp=p-12pγ2(p+12pγp+1p+1)1/(p-1),γs is the embedding coefficient of H1(ℝ3)↪Ls(ℝ3)ands∈[2,6];(h2)
(∇h(x),x)∈L2(ℝ3), where (·,·) denotes the usual inner product in ℝ3.Recently, there have been many references about the existence of nontrivial solutions to the following Kirchhoff type equation by using variational method [1–5]:
(3)-(a+b∫ℝ3‍|∇u|2)Δu+V(x)u=f(x,u)inℝN,
where a,b are positive constants. V:ℝN→ℝ,f∈C(ℝN×ℝ,ℝ),N=1,2,3. A main tool to deal with problem (3) is the mountain pass theorem. For this purpose, one usually assumes that f(x,t) is subcritical, superlinear at the origin, and either 4-superlinear at infinity or satisfies the following global Ambrosetti-Rabinowitz type condition (AR in short): (AR)
there existsμ>4 such that 0<μF(x,t)=∫0t‍f(x,s⩽tf(x,t) for all x∈ℝN and t∈ℝ. Under the above assumptions, the mountain pass geometry structure and the boundedness of Palais-Smale sequence or Cerami sequence can be obtained.For example, in [5], when f satisfies above assumptions and the potential V satisfies the following conditions:(V)
V∈C(ℝN,ℝ),infℝNV>0 and for each M>0, meas{x∈ℝN:V(x)⩽M}<∞, where meas denotes the Lebesgue measure,which ensure the compact imbedding ofE={u∈H1(ℝN):∫ℝN‍Vu2<∞}↪Lq(ℝN),q∈[2,2*), the author obtained the existence of a nontrivial solution to problem (3).The existence of infinitely many solutions was considered in [2, 3] respectively, by the fountain theorem and a variant version of the fountain theorem, where f is odd on t∈ℝ and is also subcritical, superlinear at the origin, and either 4-superlinear at infinity or satisfies AR condition or some conditions weaker than AR condition. In [2], N=2,3,V≡1 and in [3], N=3,V∈Lloc∞(ℝ3) satisfies the condition (V).The existence of ground state solutions to problem (3) was also considered in [1, 4]. In [1], the authors studied (3) under the conditions: N=3, a positive potential satisfies V∞=liminf|x|→∞V(x)>V0=infx∈ℝN>0·f(x,t)=f(t)∈C1(ℝ,ℝ) satisfies (AR), limt→0(f(t)/|t|3)=0,limt→∞(f(t)/|t|q)=0 for some q∈(3,5) and f(t)/t3 increases for all t>0. They obtained a positive ground state solution by using the Nehari manifold.Under the same condition ofV in [1], the authors in [4] discussed the existence of multiple ground state solutions, where f(x,t)=λf(t)+|t|4t, which contains a critical growth term.Recently, in [6], the authors studied the existence of a positive solution for the following Kirchhoff equation:
(4)(a+λ∫ℝ3‍[|∇u|2+bu2])(-Δu+bu)=f(u)inℝ3,
where N⩾3,a,b>0,f is subcritical, superlinear at the origin and infinity. In order to construct the mountain pass geometry structure and obtain the bounded PS sequence, they combined a truncation argument with a monotonicity trick introduced by Jeanjean [7], and obtained that there exists λ0>0 such that problem (4) has at least one positive solution for λ∈(0,λ0).Motivated by the aformentioned references, we consider the existence of multiple solutions to the nonhomogeneous Kirchhoff equation (1), where p∈(1,5). By using the variational method, we obtain that the problem has at least two positive radial solutions. Under proper assumptions on h, the problem has a local minimum around the origin with negative energy by Ekeland variational principle. Note that the term |u|p-1u is neither 4-superlinear nor satisfies AR condition for p∈(1,3]. In order to obtain the bounded PS sequence, we also use the indirect method in [7]. Meanwhile, for w∈H1(ℝ3), we take a transform of wt(·)=tw(t-2·) to construct the mountain pass geometry structure. Finally, the combination of Pohozaev identity with the method in [7] obtains the bounded PS sequence. Therefore, we obtain the second solution which has positive energy.LetH1(ℝ3) be the usual Sobolev space equipped with the inner product and norm
(5)(u,v)=∫ℝ3‍[a∇u·∇v+uv],∥u∥=(u,u)1/2.
We denote by |·|s the usual Ls(ℝ3) norm. Then, we have that H1(ℝ3)↪Ls(ℝ3) continuously for s∈[2,6]. Hence, there exists γs such that
(6)|u|s⩽γs∥u∥,u∈H1(ℝ3).
Let H=Hr1(ℝ3) be the subspace of H1(ℝ3) containing only the radial functions. Then the imbeddingH↪Ls(ℝ3) is compact for s∈(2,6) [8, Corollary 1.26, page 18]. Let 𝒟1,2(ℝ3) be the completion of C0∞(ℝ3) with respect to the norm ∥u∥𝒟1,2=(∫ℝ3‍|∇u|2)1/2.Define the energy functionalJ:H1(ℝ3)→ℝ by
(7)J(u)=12∫ℝ3‍[a|∇u|2+u2]+b4(∫ℝ3‍|∇u|2)2-1p+1∫ℝ3‍|u|p+1-∫ℝ3‍hu.
By p∈(1,5),h∈L2(ℝ3), we have J∈C1(H1(ℝ3),ℝ). And, for any u,v∈H1(ℝ3),
(8)(J′(u),v)=∫ℝ3‍[a∇u·∇v+uv]+b(∫ℝ3‍|∇u|2)∫ℝ3‍∇u·∇v-∫ℝ3‍|u|p-1uv-∫ℝ3‍hv.Furthermore, by (h1), h(x)=h(|x|), the functional J is also a C1 functional defined on H. By standard argument, the weak solution of (1) is corresponding to the critical point of the functional J on H.Our main result is as follows.Theorem 1.
Letp∈(1,5) and h satisfy (h1)-(h2). Then, problem (1) has at least two nontrivial radial solutions u0 and v0, satisfying J(u0)<0<J(v0).The paper is organized as follows. In Section2, we give the existence of the negative energy solution u0. The existence of positive energy solution v0 and the proof of Theorem 1 are given in Section 3.


## 2. Existence of Negative Energy Solution

In this section, we give the existence of the negative energy solution. In order to obtain our first solution, we need the following preliminaries.Lemma 2.
Letp∈(1,5) and h satisfy (h1). Then, there exists ρ,α>0 such that J∣∂Bρ⩾α, where Bρ={u∈H:∥u∥<ρ}.Proof.
Foru∈H, by (7), the Hölder inequality and the Sobolev inequality imply that
(9)J(u)⩾12∥u∥2-1p+1γp+1p+1∥u∥p+1-|h|2|u|2⩾∥u∥(12∥u∥-1p+1γp+1p+1∥u∥p-γ2|h|2).
Set
(10)g(t)=12t-1p+1γp+1p+1tp,t⩾0,
since p>1, by calculating directly, we see that maxt⩾0g(t)=g(ρ)>0, where ρ=((p+1)/2pγp+1p+1)1/(p-1),g(ρ)=((p-1)/2p)ρ. Then it follows that, if |h|2γ2<g(ρ), that is, |h|2<γ2-1g(ρ)≜mp, there exists α=ρ(g(ρ)-γ2|h|2)>0 such that J|∂Bρ⩾α, where
(11)mp=p-12pγ2(p+12pγp+1p+1)1/(p-1).Lemma 3.
Letp∈(1,5) and h satisfy (h1). Then c=infB¯ρJ<0, where ρ is given by Lemma 2 and Bρ={u∈H:∥u∥<ρ}.Proof.
By (h1), h∈L2(ℝ3),h≠0, then for ε∈(0,|h|2), there exists ψ∈C0∞(ℝ3) such that |h-ψ|2<ε. Then ∫ℝ3‍(h2-hψ)⩽∫ℝ3‍|h2-hψ|⩽ε|h|2, so ∫ℝ3‍hψ⩾|h|22-ε|h|2>0. Hence, by (7), for t>0 small enough, we have
(12)J(tψ)=t22∥ψ∥2+bt44(∫ℝ3‍|∇ψ|2)2-tp+1p+1∫ℝ3‍|ψ|p+1-t∫ℝ3‍hψ<0.
Then, by the definition of Bρ,c=infB¯ρJ<0.Lemma 4.
Letp∈(1,5) and h satisfy (h1). The bounded PS sequence of the functional J possesses a convergent subsequence.Proof.
Let{un} be a bounded PS sequence of J, that is {un} and {J(un)} are bounded, J′(un)→0 in H′, where H′ is the dual space of H. We may assume that, up to a subsequence,
(13)un⇀uinH,un⟶uinLp+1(ℝ3),un⟶ua.e.onℝ3.
It follows that
(14)∫ℝ3‍(|un|p-1un-|u|p-1u)(un-u)⟶0,n⟶∞.
By (8), we can obtain that
(15)(J′(un)-J′(u),un-u)=∥un-u∥2+b(∫ℝ3‍|∇un|2)∫ℝ3‍∇un·(∇un-∇u)-b(∫ℝ3‍|∇u|2)∫ℝ3‍∇u·(∇un-∇u)-∫ℝ3‍(|un|p-1un-|u|p-1u)(un-u)=∥un-u∥2+b(∫ℝ3‍|∇un|2)∫ℝ3‍|∇un-∇u|2+b(∫ℝ3‍(|∇un|2-|∇u|2))∫ℝ3‍∇u·(∇un-∇u)-∫ℝ3‍(|un|p-1un-|u|p-1u)(un-u).
Since {un} is also bounded in 𝒟1,2(ℝ3), then
(16)b(∫ℝ3‍(|∇un|2-|∇u|2))∫ℝ3‍∇u·(∇un-∇u)⟶0.
Therefore,
(17)0⩽∥un-u∥2⩽∥un-u∥2+b(∫ℝ3|∇un|2)∫ℝ3|∇un-∇u|2⟶0,n⟶∞.
That is, un→u in H.Theorem 5.
Letp∈(1,5) and h satisfy (h1). Then, there exists u0∈H such that
(18)J(u0)=c=infB¯ρJ(u)<0,
where ρ is given by Lemma 2 and Bρ={u∈H:∥u∥<ρ}.Proof.
By Lemma3, c=inf{J(u):u∈B¯ρ}<0, then by Ekeland variational principle [9], there exists {un}⊂B¯ρ such that
(19)c⩽J(un)⩽c+1n,J(ω)⩾J(un)-1n∥ω-un∥∀ω∈B¯ρ.
Then, by Lemma 2  un∈Bρ, then {un} is a bounded PS sequence of J. Therefore, Lemma 4 implies that there exists u0∈H such that un→u0, up to a subsequence. So J(u0)=c<0 and J′(u0)=0.


## 3. Proof of Theorem1

In this section, we will show the existence of the second solution. Note thatp∈(1,5), when p∈(1,3],|u|p-1u neither satisfies (AR) condition nor is 4-superlinear. So, in order to obtain the bounded PS sequence, following the argument in [6], we also use a direct method in [7]. Firstly, we recall the following main result in [7]. The “monotonicity trick” at the core of this theorem was invented by Struwe (see [9]).Theorem 6 (see [7]).
Let(X,∥·∥) be a Banach space and I⊂ℝ+ be an interval. Consider the family of C1 functionals on X(20)Jλ(u)=A(u)-λB(u),λ∈I,
with B nonnegative and either A(u)→∞ or B(u)→∞ as ∥u∥→∞. We assume that there are two points v1,v2 in X such that
(21)cλ=infγ∈Γλmaxt∈[0,1]Jλ(γ(t))>max{Jλ(v1),Jλ(v2)},∀λ∈J,
where
(22)Γ={γ∈C([0,1],X):γ(0)=v1,γ(1)=v2}.
Then, for almost every λ∈I there is a sequence {un(λ)}⊂X such that (i)
{un(λ)} is bounded;(ii)
Jλ(un(λ))→cλ;(iii)
Jλ′(un(λ))→0 in the dual X-1 of X.In our case,X=H,I=[1/2,1], and define Jλ:X→ℝ by
(23)Jλ(u)=A(u)-λB(u),
where λ∈I,
(24)A(u)=12∥u∥2+b4(∫ℝ3‍|∇u|2)2-∫ℝ3hu,B(u)=1p+1∫ℝ3|u|p+1.
Then {Jλ}λ∈I is a family of C1 functionals on H. For any u∈H,B(u)⩾0, and A(u)⩾(1/2)∥u∥2-γ2|h|2∥u∥→∞ as ∥u∥→∞.In the following, we verify that the functionalJλ satisfies the conditions of Theorem 6.Lemma 7.
Letp∈(1,5) and h satisfy (h1)-(h2). Then, the following claims hold:(i)
there existr,a>0 and e∈H such that for all λ∈I(25)Jλ∣∂Bρ(u)⩾a>0,Jλ(e)<0with∥e∥>r;(ii)
for anyλ∈I,
(26)cλ=infγ∈Γmaxt∈[0,1]Jλ(γ(t))>max{Jλ(0),Jλ(e)},where
(27)Γ={γ∈C([0,1],X):γ(0)=0,γ(1)=e}.Proof.
(i) Since for allu∈H and λ∈I=[1/2,1],Jλ(u)⩾J1(u). By Lemma 2, there exist r,a>0 independent of λ∈I such that Jλ(u)⩾a>0 with ∥u∥=r.
We choose a functionw∈H and w≠0. Set wt(·)=tw(t-2·) for t>0. Then, for all λ∈I, by (7) and (h1), we have
(28)Jλ(wt)⩽at42∫ℝ3|∇w|2‍+t82∫ℝ3‍|w|2+bt84(∫ℝ3‍|∇w|2)2-tp+72(p+1)∫ℝ3‍|w|p+1.
Noting p>1, then there exists t0 large enough satisfying ∥wt0∥>r, which is independent of λ∈I, such that for all λ∈I,Jλ(e)<0 with e=wt0.
(ii) Sincecλ is nonincreasing on λ∈I, then by the definition of cλ and (i), for all λ∈I, we have c1/2⩾cλ⩾c1⩾a>0.By Lemma7 and Theorem 6, for almost every λ∈I, there exists a bounded sequence {unλ}⊂H such that Jλ(unλ)→cλ,(Jλ)′(unλ)→0. By Lemma 4, there exists uλ∈H such that unλ→uλ in H. Therefore, (Jλ)′(uλ)=0 and Jλ(uλ)=cλ. It follows from (ii) of Lemma 7 that uλ≠0.Therefore, there exists{λn}⊂I with λn→1- and a nonnegative sequence {uλn} (denoted by {un} for simplicity) satisfying
(29)Jλn(un)=cλn,(Jλn)′(un)=0.In order to obtain the boundedness of{un}, we need the following Pohozaev identity. The proof is similar to the argument in [10].Lemma 8.
Under the conditions of (h1) and (h2), if u∈H is a weak solution of (1), the following Pohozaev identity holds:
(30)12(a+b∫ℝ3‍|∇u|2)∫ℝ3‍|∇u|2+32∫ℝ3‍u2=3p+1∫ℝ3‍|u|p+1+∫ℝ3‍(3h+(∇h(x),x))u.Lemma 9.
Consider{un} in (29) is bounded in H.Proof.
Firstly, since(Jλn)′(un)=0, then by Lemma 8, un satisfies the following Pohozaev identity:
(31)a2∫ℝ3‍|∇un|2+b2(∫ℝ3‍|∇un|2)2+32∫ℝ3‍un2-λn3p+1∫ℝ3‍|un|p+1-3∫ℝ3‍h(x)un-∫ℝ3‍(∇h(x),x)un=0.
On the other hand, by ((Jλn)′(un),un)=0 and Jλn(un)=cλn, we have that
(32)a∫ℝ3‍|∇un|2+b(∫ℝ3‍|∇un|2)2+∫ℝ3‍un2-λn∫ℝ3‍|un|p+1-∫ℝ3‍hun=0,(33)a2∫ℝ3‍|∇un|2+b4(∫ℝ3‍|∇un|2)2+12∫ℝ3‍un2-λnp+1∫ℝ3‍|un|p+1-∫ℝ3‍hun=cλn.
Then, combining (33) with (31), we can obtain
(34)a∫ℝ3‍|∇un|2+b4(∫ℝ3‍|∇un|2)2=3cλn-∫ℝ3‍(∇h(x),x)un.
Since cλn⩽c1/2 by Lemma 7, and (∇h(x),x)∈L2(ℝ3), so in order to prove the boundedness of {un} in H, we only need to prove that |un|2 is bounded. By contradiction, we assume that |un|2→∞, up to a subsequence. Let vn=un/|un|2,Xn=a∫ℝ3‍|∇vn|2,Yn=b|un|22(∫ℝ3‍|∇vn|2)2,Zn=λn|un|2p-1∫ℝ3‍|vn|p+1.
Note thatcλn is bounded and h,(∇h(x),x)∈L2(ℝ3). Multiplying (31)–(33) by 1/|un|22, we have that
(35)Xn2+Yn2-3p+1Zn=-32+o(1),Xn+Yn-Zn=-1+o(1),Xn2+Yn4-Znp+1=-12+o(1),
where o(1) denotes the quantity tends to zero as n→∞. By calculating, we obtain that
(36)Xn=1-p5-p+o(1).
Since p∈(1,5) and Xn⩾0 for n∈ℕ, so this is a contradiction for n large enough. Therefore, {un} is bounded in H.Proof of Theorem1.
Since(37)J(un)=Jλn(un)+(λn-1)1p+1∫ℝ3‍|un|p+1,(J′(un),v)=((Jλn)′(un),v)+(λn-1)∫ℝ3‍|un|p-1unv,v∈H.
By Lemma 9 and H↪Lp+1(ℝ3),|un|p+1 is bounded and |∫ℝ3‍|un|p-1unv|⩽γp+1p+1|un|p+1p∥v∥. Thus, when λn→1-, we have that {J(un)} is bounded and J′(un)→0. Therefore {un} is a bounded PS sequence of J. By Lemma 4, {un} has a convergent subsequence. We may assume that un→v0, up to a subsequence. Consequently, J′(v0)=0. According to Lemma 7, we have J(v0)=limn→∞J(un)=limn→∞Jλn(un)⩾a>0. Thus v0 is a positive energy solution to problem (1). Hence, by Theorem 5 problem (1) has two solutions u0 and v0 satisfying J(u0)<0<J(v0).



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