# Sign-Changing Solutions for Biharmonic Equations with Navier Boundary Conditions

Authors: Ruiting Jiang
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

This work is concerned with sign-changing solutions to the biharmonic equation with Navier boundary conditions. By using the method of invariant sets of descending flow and symmetric mountain pass lemma, we obtain the existence of multiple sign-changing solutions when the nonlinear term is odd.


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## Body

## 1. Introduction

In this paper, we deal with the following biharmonic equation with Navier boundary conditions(1)Δ2u=fx,u,inΩ,u=Δu=0,on∂Ω,where Ω is a smooth bounded domain in ℝN.The weak form of (1) is to look for u∈X:=H2Ω∩H02Ω satisfying(2)∫ΩΔuΔφdx=∫Ωfx,uφdx,∀φ∈X,where X is equipped with the norm(3)u=∫ΩΔu2dx1/2.The equations involving the biharmonic operator have received special attention from many researchers, in part because they describe the traveling waves in a suspension bridge. Moreover, we cannot apply the maximum principle for the biharmonic operator for a general smooth bounded domain involving the Dirichlet boundary conditions, and this fact makes some problems involving the biharmonic operator very interesting from a mathematical point of view.In the last decades, many authors have attached their attention to the existence and multiplicity of nontrivial solutions for biharmonic equations [1–7]. Using the dual method [2], establish the existence of a nodal ground state solution with fx,u=fu for (1). When fx,u satisfies the asymptotically linear condition, An and Liu [1] proved the existence of nontrivial solutions for (1), their tool is the mountain pass theorem, which is usually used to get the nontrivial solution, see, for example, [8]. For the results of multiple nontrivial solutions to problem (1), see [9–12]. If fx,u is odd in u and satisfies some additional conditions, by using the sign-changing critical theorems, the authors in [13] obtained that problem (1) admitted infinitely many sign-changing solutions [14] and established the existence and multiplicity of solutions without the AR condition.In the case of the Laplacian operator, the study of the existence of nodal solutions has rich literature. However, we cannot use or adapt some techniques developed for the Laplacian, because in general, the authors prove the existence of sign-changing solutions by minimizing the energy functional on the set:(4)M=u∈H01Ω:I′u±,u±=0.In problems involving the biharmonic operator, we cannot even ensure that givenu∈H2Ω, one also has u+,u−∈H2Ω. In this work, the sign-changing solutions for (1) were obtained by the method of invariant sets of descending flows, and one of the key points is to construct a vector field A, which keeps the positive and negative cones invariant. Consequently, constructing operator A (see Section 3 for details) and verifying its properties need special handling.In what follows, we assume thatf satisfies the following conditions:f1f∈CΩ¯×ℝ,ℝ, fx,tt>0, ∀t≠0;f2limt⟶0fx,t/t=0 uniformly in x∈Ω¯;f3limt⟶0fx,t/t2∗∗−1=0 uniformly in x∈Ω¯, where 2∗∗=2N/N−4;f4 there exist R>0 and η>2 such that 0<ηFx,t≤fx,tt for a.e. x∈Ω,t≥R, where Fx,t=∫0tfx,sds. Here are the main results of the paper.Theorem 1.
Assumef1−f4 hold. Then, the problem (1) has a sign-changing solution. Moreover, if fx,t is odd in t, then, the problem (1) has a sequence of sign-changing solutions.


## 2. Verification of the PS Condition

Now, we assume thatf satisfies f1–f4 and define a functional on X by(5)Iu=12∫ΩΔu2dx−∫ΩFx,udx.Clearly,I is of class C1 and(6)I′u,φ=∫ΩΔuΔφdx−∫Ωfx,uφdx,∀φ∈X.Lemma 1.
I satisfies the (PS) condition.Proof.
Letun be a (PS) sequence, then(7)c+o1un≥Iun−1ηI′un,un=12−1η∫ΩΔun2dx+∫Ω1ηfx,unun−Fx,undx≥12−1ηun2−M0Ω,where M0=maxx∈Ω¯,t≤RFx,t−1/ηfx,tt. Hence, un is bounded in X. Assume un⇀u in X. By Sobolev’s imbedding theorem, one has un⟶u in LrΩ,1≤r<2∗∗.
LetBε0 be a ball with a radius ε and centered at 0. Define a cutoff function φε, such that φεx=1 if x≥ε; φε=0 if x≤ε/2; ∇φε≤c/ε, Δφε≤c/ε2. Take un−umφε as a test function, combining with Hölder inequality, we obtain that(8)o1=I′un−I′um,un−umφε=∫ΩΔun−ΔumΔun−Δumφεdx+∫ΩΔun−Δumun−umΔφεdx+2∫ΩΔun−Δum∇un−∇um∇φεdx−∫Ωfx,un−fx,umun−umφεdx=∫ΩΔun−Δum2φεdx+o1.
Therefore, for allε>0, one has(9)∫Ω/Bε0Δun−Δum2dx⟶0,n,m⟶∞,(10)∫Ω/Bε0Δun−Δu2dx⟶0,n⟶∞.
First, we verify that the weak limitu satisfies equation (2), that is, u is a weak solution.
Letψ∈C∞Ω¯∩C0Ω¯ and take ψ as a test function, we obtain(11)∫ΩΔunΔψdx=∫Ωfx,unψdx+o1,∀ψ∈X.
Without losing generality, we assumeψx=0 for x<ε. By the local convergence (8), then(12)∫ΩΔuΔψdx=∫Ωfx,uψdx.
For a general functionalψ∈C∞Ω¯∩C0Ω¯, take ψφε as a test function, where φε is the cutoff function defined before, then(13)∫ΩΔuΔψφεdx=∫Ωfx,uψφεdx.
Letε⟶0; thus, the Lebesgue dominant convergence theorem implies that(14)∫Ωfx,uψφεdx⟶∫Ωfx,uψdx.
Clearly,(15)∫ΩΔuΔψφεdx=∫ΩΔuΔψφεdx+2∫ΩΔu∇ψ∇φεdx+∫ΩΔuψΔφεdx.
Once again, the Lebesgue dominant convergence theorem implies that(16)∫ΩΔuΔψφεdx⟶∫ΩΔuΔψdx,ε⟶0,∫ΩΔuψΔφεdx≤Cε−2∫BεΔu2dx1/2∫Bεdx1/2≤CεN/2−2⟶0,ε⟶0.
Similarly,(17)∫ΩΔu∇ψ∇φεdx≤Cε−1∫ΩΔudx≤CεN/2−1⟶0,ε⟶0.
Altogether by lettingε⟶0 in equation (13), equation (2) holds for all ψ∈C∞Ω¯∩C0Ω¯. By a further approximation, equation (2) holds for all φ∈X, that is, u is a weak solution. In particular, we have(18)∫ΩΔu2dx=∫Ωfx,uudx.
Also, one obtains that(19)∫ΩΔun2dx=∫Ωfx,unundx+o1=∫Ωfx,uudx+o1.
Combining equation (18) with equation (19), then(20)∫ΩΔun2dx=∫ΩΔu2dx+o1.
Next, we show thatun⟶u. In order to get the results, we need the equality below.(21)∫ΩΔun2dx=∫ΩΔu2dx+∫ΩΔun−Δu2dx+o1.
In fact, forε>0, by the local convergence (5), we obtain(22)∫Ω/BεΔun2dx−∫Ω/BεΔu2dx−∫Ω/BεΔun−Δu2dx⟶0,n⟶∞.
On the ballBε,(23)∫BεΔun2dx−∫BεΔun−Δu2dx≤C∫BεΔun+ΔuΔudx≤C∫BεΔun2+Δu2dx1/2∫BεΔu2dx1/2≤C∫BεΔu2dx1/20,ε⟶0.
Hence,(24)∫BεΔun2dx−∫BεΔu2dx−∫BεΔun−Δu2dx≤C∫BεΔu2dx1/2⟶0,ε⟶0,
(21) follows from (22) and (24). Now, from (20) and (21), we get(25)∫ΩΔun−Δu2=o1,that is, un⟶u.


## 3. Properties of the Operator A

In this section, we define an operatoru∈X↦v=Au∈X, which is important to construct descending flow for the functional I. v=Au is defined by(26)Δ2v=fx,u.The weak form of equation (26) is(27)∫ΩΔvΔφdx=∫Ωfx,uφdx,∀φ∈X,Then,(28)I′u,φ=∫ΩΔu−vΔφdx.Lemma 2.
The operatorA is well-defined and continuous.Proof.
Givenu∈X, define(29)Jv=12∫ΩΔv2dx−∫Ωfx,uvdx.then J is coercive and strictly convex, hence has a unique minimizer v=Au. The operator A maps bounded set into bounded set. Let v1=Au1, v2=Au2, by equation (27), we have(30)∫ΩΔv2−Δv1Δv2−v1dx=∫Ωfx,u2−fx,u1v2−v1dx,
Asu1⟶u2, the right side of (30) tends to 0, and the left side of (30) also tends to 0, that is,(31)∫ΩΔv2−Δv12dx=o1,u1⟶u2.Lemma 3.
It holds that(32)I′u,u−Au=u−Au2,I′u≤u−Au.Proof.
(33)I′u,u−Au=∫ΩΔu−v2dx=u−Au2,I′u=supφ=1I′u,φ=supφ=1∫ΩΔu−vΔφdx≤∫ΩΔu−v2dx1/2=u−Au.
Now, for0<δ<1, we define two convex cones P+ and P− by(34)P±=u|u∈X:1−δλ12∫Ωu∓2dx+δS∫Ωu∓2∗∗dx2/2∗∗<ε,where u±=max±u,0, ε>0 is a parameter, λ1 is the positive first eigenvalue of the operator −Δ, and S is the imbedding constant for X to L2∗∗Ω:(35)S=infu∈X∫ΩΔu2dx∫Ωu2∗∗dx2/2∗∗.Lemma 4.
There existsε0>0 such that if ε<ε0, then(36)A∂P+⊂P+,A∂P−⊂P−Proof.
Givenu∈X, define w∈X by(37)∫ΩΔwΔφdx=∫Ωfx,u+φdx,∀φ∈X.
First, we provew≥0. For ψ∈C∞Ω¯, ψ≥0, then the problem(38)−Δφ=ψ,inΩ,φ=0,on∂Ω,then, φ∈C∞Ω¯∩C0Ω¯ and φ≥0 in Ω. Substitute φ into equation (37), we have(39)∫ΩΔwΔψdx=−∫Ωfx,u+φdx≤0,for all ψ∈C∞Ω¯, ψ≥0. Hence, Δw≤0. Consequently, w≥0.
Next, we provew≥v. The function v=Au satisfies(40)∫ΩΔvΔφdx=∫Ωfx,uφdx.
SubstitutingΔφ by −ψ, then(41)∫ΩΔvψdx=−∫Ωfx,uφdx,then, for all ψ∈C∞Ω¯, ψ≥0, it holds(42)∫ΩΔwψdx≤∫ΩΔvψdx.
It follows thatΔw≤Δv; therefore, w≥v. We have w≥maxv,0=v+. Take φ=w as a test function in (37), then(43)∫ΩΔw2dx=∫Ωfx,u+wdx,∫ΩΔw2dx≥1−δλ12∫Ωw2dx+δS∫Ωw2∗∗dx2/2∗∗.
Byf2 and f3, we may assume that fx,t≤1−δλ12t+Ct2∗∗−1 for some C>0, then(44)∫Ωfx,u+wdx≤1−δλ12∫Ωu+wdx+C∫Ωu+2∗∗−1wdx,1−δλ12∫Ωu+wdx≤1−δλ122∫Ωu+2dx+1−δλ122∫Ωw2dx,C∫Ωu+2∗∗−1wdx≤C∫Ωu+2∗∗dx2∗∗−1/2∗∗∫Ωw2∗∗dx1/2∗∗≤C22δS2∗∗δS∫Ωu+2∗∗dx2/2∗∗+12δS∫Ωw2∗∗dx2/2∗∗.
Now, we chooseε0>0, such that(45)C2δS2∗∗ε02∗∗−2<1.
Then, ifε<ε0 and u∈∂P−, we have δS∫Ωu+2∗∗dx2/2∗∗<ε0. By equation (44), we(46)C∫Ωu+2∗∗−1wdx≤C22δS2∗∗ε02∗∗−2δS∫Ωu+2∗∗dx2/2∗∗+12δS∫Ωw2∗∗dx2/2∗∗≤12δS∫Ωu+2∗∗dx2/2∗∗+12δS∫Ωw2∗∗dx2/2∗∗.
With equation (43) and all the above estimates, we have for u∈∂P_,(47)1−δλ12∫Ωw2dx+δS∫Ωw2∗∗dx2/2∗∗<1−δλ12∫Ωu+2dx+δS∫Ωu+2∗∗dx2/2∗∗=ε.
Notice thatw≥v+, we obtain(48)1−δλ12∫Ωv+2dx+δS∫Ωv+2∗∗dx2/2∗∗<ε.
Hence,v=Au∈P− for u∈∂P−. Similarly, A∂P+⊂P+.


## 4. Sign-Changing Solutions

As we mentioned in the introduction, the operatorA will be used to construct descending flow for the functional I. Since A is merely continuous, we need the following modification of A.Lemma 5.
LetK be the critical point set of I:(49)K=u∈X|I′u=0.
Then, there exists a locally Lipschitz continuous operatorB defined on X0=X/K which inherits the operators of A. More precisely,(50)B∂P+⊂P+,B∂P−⊂P−,12u−Bu≤u−Au≤2u−Bu,u∈X0,I′u,u−Bu≥δu−Bu,u∈X0.Proof.
The proof is similar to Lemma1 of [15], we omit it.
By equations (39), (43) in Lemma 3, and (50), for some constants c, one has(51)I′u≤cu−Bu.Lemma 6.
(1)
Assumeε0 sufficiently small. Then, for ε<ε0 and u∈∂P+∩∂P−, one has(52)Iu≥cε.(2)
LetE be an arbitrary finite-dimensional subspace of X, then for u∈E, one obtains(53)limu⟶∞Iu=−∞.Proof.
(1)
By the conditionsf2 and f3, there exist constants c,c0,c′,c1,c1′ such that for u∈∂P+∩∂P−, one has(54)Iu=12∫ΩΔu2dx−∫ΩFx,udx≥12∫ΩΔu2dx−c1∫Ωu2dxc2∫Ωu2∗∗dx≥c0∫Ωu2dx+∫Ωu2∗∗dx2/2∗∗−c2∫Ωu2∗∗dx≥c0∫Ωu2dx+∫Ωu2∗∗dx2/2∗∗−c′ε2∗∗−2/2∫Ωu2∗∗dx2/2∗∗≥c0−c′ε2∗∗−2/2∫Ωu2dx+∫Ωu2∗∗dx2/2∗∗≥cε.(2)
By the conditionf4, there exists c2,c3 such that Fx,t≥c2tη−c3. Then, for u∈E, we have(55)Iu=12∫ΩΔu2dx−∫ΩFx,udx≤12∫ΩΔu2dx−c2∫Ωuηdx+c4≤c5u2−c6uη+c4⟶−∞,as u⟶+∞, where c4,c5,c6 are constants.
Denote(56)Kc=u∈X|Iu=c,I′u=0,W=P+∪P−,K¯c=KcW.Lemma 7.
LetN be an open neighborhood of K¯c. Then, there exists ε0,ε1,ε2>0 such that for 0<ε1<ε2<ε0, there exists a continuous map σ:0,1×X⟶X satisfying(1)
σ0,u=u for u∈X.(2)
σt,u=0 for t∈0,1,Iu∉c−ε2,c+ε2.(3)
σ1,Ic+ε1/N∪W⊂Ic−ε1.(4)
If the functionfx,t is odd in t, then σt,−u=−σt,u for all t,u∈0,1×X.(5)
σt,P+⊂P+, σt,P−⊂P− for t∈0,1.Proof.
The proof is similar to the proof of Lemma7 of [16], we omit it.Proof of Theorem 1.
LetN be an open neighborhood of K¯c. Let η=σ1,⋅, where σ is the map obtained in Lemma 7. Then, for some ε>0, the map η satisfies.(1)
η|Ic−2ε=Id.(2)
ηIc+ε/N∪W⊂Ic−ε.(3)
ηP¯+⊂P+, ηP¯−⊂P−.(4)
IfI is even, then η−u=−ηu.
That is,I has a deformation property. Moreover, by Lemma 6, I satisfies the symmetric mountain pass geometry. Now, Theorem 1 follows from Theorems 1 and 2 in [15].



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