# Strict Monotonicity and Unique Continuation for the Third-Order Spectrum of Biharmonic Operator

Authors: Khalil Ben Haddouch; Zakaria El Allali; El Bekkaye Mermri; Najib Tsouli
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We will study the spectrum for the biharmonic operator involving the laplacian and the gradient of the laplacian with weight, which we call third-order spectrum. We will show that the strict monotonicity of the eigenvalues of the operatorΔ2u+2β·∇(Δu)+|β|2Δu, where β∈ℝN, holds if some unique continuation property is satisfied by the corresponding eigenfunctions.


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## Body

## 1. Introduction

We are concerned here with the eigenvalue problem:(1.1)Find(β,α,u)∈ℝN×ℝ×Hsuch thatΔ2u+2β·∇(Δu)+|β|2Δu=αmuinΩ,u=Δu=0on∂Ω,
where Ω is a bounded domain in ℝN  (N≥1),  H=H2(Ω)∩H01(Ω), Δ2 denotes the biharmonic operator defined by Δ2u=Δ(Δu), and m∈M={m∈L∞(Ω)/meas{x∈Ω/m(x)>0}≠0}.Based on the works of Anane et al. [1, 2], we will determine the spectrum of (1.1), which we call third-order spectrum for the biharmonic operator. This spectrum is defined to be the set of couples (β,α)∈ℝN×ℝ such that the problem
(1.2)Δ2u+2β·∇(Δu)+|β|2Δu=αmuinΩ,u=Δu=0on∂Ω
has a nontrivial solution u∈H. This spectrum, which is denoted by σ3(Δ2,m), is an infinite sequence of eigensurfaces Γ1±,Γ2±,…, see Section 3. When β=0, the zero-order spectrum is defined to be the set of eigenvalues α∈ℝ such that the problem
(1.3)Δ2u=αmuinΩ,u=Δu=0on∂Ω
has a nontrivial solution u∈H. In this case the spectrum is denoted by σ0(Δ2,m). The eigenvalue problem (1.3), which is studied by Courant and Hilbert [3], admits an infinite sequence of real eigenvalues (αn(m))n satisfying
(1.4)1αn(m)=supFn∈ℱn(H)minu∈Fn(∫Ωm|u|2dx)∀n≥1,
where ℱn(H) denotes the class of n-dimensional subspaces Fn of H.Definition 1.1.
We say that solutions of problem (1.1) satisfy the unique continuation property (U.C.P), if the unique solution u∈LLoc2(Ω) which vanishes on a set of positive measure in Ω is u≡0.In the literature there exist several works on unique continuation. We refer to the works of Jerison and Kenig [4] and Garofalo and Lin [5], among others. The unique continuation property as defined above differs from the usual notions of unique continuation, see [6] for more details.Definition 1.2.
We say thatΓk(β,·) is strict monotone with respect to the weight if Γk(β,m)>Γk(β,m^), for all m<≢m^.Here we use the notation<≢ to mean inequality almost everywhere together with strict inequality on a set of positive measure.Since the pioneer works of Carleman [7] in 1939 on the unique continuation, this notion has been the interest of many researchers in partial differential equations, see for instance [4, 5, 8]. In 1992, de Figueiredo and Gossez [6] proved that strict monotonicity holds if and only if some unique continuation property is satisfied by the corresponding eigenfunction of a uniformly elliptic operator of the second order. In 1993, Gossez and Loulit [8] have proved the unique continuation property in the linear case of the laplacian operator. The unique continuation property of the biharmonic operator was proved recently by Cuccu and Porru [9]. Our purpose in the fourth section is to show that strict monotonicity of eigensurfaces for problem (1.1) holds if some unique continuation property is satisfied by the corresponding eigenfunctions.


## 2. Preliminaries

LetH be a finite dimensional separable Hilbert space. We denote by (·,·) and ∥·∥ the inner product and the norm of the space H, respectively. Let T:H→H be a compact operator.Lemma 2.1.
All nonzero eigenvalues of the operatorT are obtained by the following characterizations:
(2.1)μn=supFn∈ℱn(H)min{(T(u),u)suchthat∥u∥=1;u∈Fn},μ-n=infFn∈ℱn(H)Max{(T(u),u)suchthat∥u∥=1;u∈Fn},
where ℱn(H) denotes the class of n-dimensional subspaces Fn of H.Moreover, zero is the only accumulation point of the set of all eigenvalues ofT. Here, the eigenvalues are repeated with its order of multiplicity, and the eigenfunctions are mutually orthogonal [10].


## 3. Third-Order Spectrum of the Biharmonic Operator

We define the third-order eigenvalue problem of the biharmonic operator as follows:(3.1)Find(β,α,u)∈ℝN×ℝ×H∖{0}such  thatΔ2u+2β·∇(Δu)+|β|2Δu=αmuinΩ,u=Δu=0on∂Ω.If(β,α,u) is a solution of (3.1) then (β,α) is called third-order eigenvalue and u is said to be the associated eigenfunction.Lemma 3.1.
Problem (3.1) is equivalent to the following problem:
(3.2)Find(α,u)∈ℝ×H∖{0}suchthatΔ2,βu=αmeβ·xuinΩ,u=Δu=0on∂Ω,
where Δ2,βu=Δ(eβ·xΔu).Proof.
For anyβ∈ℝN, we have
(3.3)Δ(eβ·xΔu)=∇(∇(eβ·xΔu))=∇(βeβ·xΔu+eβ·x∇(Δu))=eβ·x[Δ2u+2(β·∇(Δu))+|β|2Δu].
Hence, problem (3.1) is equivalent to problem (3.2)Remark 3.2.
Letu∈H; we denote by ∂u/∂ν the normal derivative defined by ∂u/∂ν=(∇u|∂Ω)·ν→ where ∇u|∂Ω∈(L2(∂Ω))N and ∂u/∂ν∈L2(∂Ω).Definition 3.3.
A weak solution of (3.2) is a function u in H∖{0} witch satisfies, for (β,α)∈ℝN×ℝ and for all φ∈H,
(3.4)∫Ωeβ·xΔuΔφdx=α∫Ωeβ·xmuφdx.Definition 3.4.
For(β,α)∈ℝN×ℝ, we say that u∈H is a classical solution of problem (3.1) if u∈C4(Ω¯).Proposition 3.5.
Ifu∈H is a weak solution of (3.2) and u∈C4(Ω¯), then u is a classical solution of (3.2).Proof.
Letu∈C4(Ω¯) be a weak solution of (3.2), then we have
(3.5)∫Ω‍eβ·xΔuΔφdx=α∫Ω‍eβ·xmuφdx∀φ∈H.
Using the Green formula, we obtain(3.6)∫Ω‍Δ(eβ·xΔu)φdx=-∫Ω‍∇(eβ·xΔu)·∇φdx+∫∂Ω‍φ∂∂ν(eβ·xΔu)dx,∫Ω‍eβ·xΔuΔφdx=-∫Ω‍∇(eβ·xΔu)·∇φdx+∫∂Ω‍eβ·xΔu∂φ∂νdx.
Then we have(3.7)∫Ω‍Δ(eβ·xΔu)φdx=∫Ω‍eβ·xΔuΔφdx+∫∂Ω‍φ∂∂ν(eβ·xΔu)dx-∫∂Ω‍eβ·xΔu∂φ∂νdx.
Thus, the prove is complete.Theorem 3.6.
Let𝒮β={u∈H/∥u∥2,2,β2=∫Ω‍eβ·x|Δu|2dx=1}, then we have  (a)
σ3(Δ2,m)=⋃n=1+∞‍G(Γn(m,·)), where the function Γn(m,·): ℝN→ℝ is defined by
(3.8)1Γn(m,β)=supFn∈ℱn(H)minu∈Fn{∫Ω‍eβ·xm|u|2dx,u∈𝒮β∩Fn}∀β∈ℝN,
where ℱn(H) denotes the class of n-dimensional subspaces Fn of H and G(Γn(m,·))⊂ℝN×ℝ is the graph of Γn(m,·).(b)
∫Ω‍eβ·xm|u|2dx≤(Γ1(m,β))-1∥u∥2,2,β2.(c)
For all∈ℝN:limn→+∞Γn(m,β)=+∞.Proof.
Let(β,α,u)∈H∖{0}, then (β,α,u) is a solution of (3.1) if and only if (α,u) is a solution of problem (3.2). We prove that the map
(3.9)l:H2(Ω)∩H01(Ω)→ℝ,(u,v)↦l(u,v)=〈Δ2,βu,v〉
defines a scalar product on H=H2(Ω)∩H01(Ω) equivalent to the usual scalar product ∫ΩΔuΔvdx.
The mapl(·,·) is a continuous symmetric bilinear form. Since Δ2 satisfies the condition of the uniform ellipticity, then we have
(3.10)l(u,u)=〈Δ2,βu,u〉=∫Ω‍eβ·x|Δu|2dx≥c′∥u∥2,22,
where c'=minx∈Ω¯eβ·x. Therefore, the bilinear form l(·,·) is coercive. On the other hand, the operator
(3.11)T2,β:H2(Ω)∩H01(Ω)→H2(Ω)∩H01(Ω),u↦T2,β(u)=(Δ2,β)-1(meβ·xu)
is well defined, linear, symmetric, and compact on H. Then, problem (3.2) can be written as
(3.12)T2,β(u)=1αu,u∈H2(Ω)∩H01(Ω).
Note thatα=0 is not an eigenvalue of (3.2). It follows that (α,β) is an eigenvalue of (1.1) if and only if 1/α is eigenvalue of the operator T2,β. By Lemma 2.1, the eigenvalues are given by the characterizations
(3.13)1αn=supFn∈ℱn(H)min{l(T2,β(u),u)suchthat∥u∥=1;u∈Fn},1α-n=infFn∈ℱn(H)max{l(T2,β(u),u)suchthat∥u∥=1;u∈Fn}.
In addition, we have(3.14)l(T2,βu,u)=〈Δ2,β((Δ2,β)-1meβ·xu),u〉=∫Ω‍meβ·x|u|2dx,
and ∥u∥2=(u,u)=〈Δ2,βu,u〉=∥u∥2,2,β2, then relation (3.8) is satisfied. Since meβ.x∈M, then we have Γn(m,β)>0 for all n∈ℕ*. As zero is the only accumulation point of the sequence (1/αn)n, it follows that Γn(m,β)→+∞ when n→+∞. Therefore, the proof is completed.


## 4. Strict Monotonicity and Unique Continuation

In this section, we will show that strict monotonicity of eigensurfaces for problem (3.1) holds if some unique continuation property is satisfied by the corresponding eigenfunctions.Theorem 4.1.
Letm and m^ be two weights with m<≢m^ and k∈ℕ. If the eigenfunctions φk associated to Γk(β,m) satisfy the (U.C.P) then Γk(β,m)>Γk(β,m^).Theorem 4.2.
Letm be a weight and k∈ℕ. If the eigenfunctions φk associated to Γk(β,m) do not satisfy the (U.C.P) then there exists a weight m^ with m<≢m^, such that, for some i∈ℕ with Γi(β,m)=Γk(β,m), one has Γi(β,m)=Γi(β,m^).As a consequence of Theorems4.1 and 4.2 we have the following result.Corollary 4.3.
Letm∈L∞(Ω) and k∈ℕ. If Γk(β,1)<≢m<≢Γk+1(β,1), then the only solution of the problem
(4.1)(Pm){Δ2u+2β·∇(Δu)+|β|2Δu=muinΩ,u=Δu=0onΩ
is u≡0.Proof of Theorem4.1.
Letk∈ℕ; we define the space
(4.2)Fk=〈φ1,φ2,…,φk〉,
spanned by the eigenfunctions φi associated to Γi(β,m) with
(4.3)∫Ω‍eβ·x|φi|2dx=1,fori=1,…,k.
We have(4.4)1Γk(β,m)=minu∈Fk{∫Ω‍meβ·x|u|2dxsuchthat∫Ω‍eβ·x|Δu|2dx=1}=∫Ω‍meβ·xφk2dx.
Letu∈Fk, with ∫Ω‍eβ·x|Δu|2dx=1. We have either u achieves the infimum in (4.4) or not. In the case u is an eigenfunction associated to Γk(β,m), then by the (U.C.P) and since u2(x)>0 a.e. x∈Ω, we have
(4.5)1Γk(β,m)=∫Ω‍meβ·xu2dx<∫Ω‍m^eβ·xu2dx=1Γk(β,m^).
Thus,Γk(β,m)>Γk(β,m^). In the other case, we have
(4.6)1Γk(β,m)<∫Ω‍meβ·x|u|2dx<∫Ω‍m^eβ·x|u|2dx=1Γk(β,m^).
Thus, in both cases we have(4.7)1Γk(β,m)<∫Ω‍m^eβ·xu2dx.
It follows that(4.8)1Γk(β,m)<infu∈Fk{∫Ω‍m^eβ·xu2dx;∫Ω‍eβ·x|Δu|2dx=1}.
This yields the desired inequality(1/Γk(β,m))<(1/Γk(β,m^)). Hence, we have Γk(β,m)>Γk(β,m^).Proof of Theorem4.2.
Denote byu an eigenfunction associated to Γk(β,m) which vanishes on a set of positive measure. Take i such that Γk(β,m)=Γi(β,m)<Γi+1(β,m) and define
(4.9)m^(x)={m(x)ifu(x)≠0,m(x)+ϵifu(x)=0,
where ϵ>0 is chosen such that Γi(β,m)<Γi+1(β,m^), which is possible by the continuous dependence of the eigenvalues with respect to the weight. We have
(4.10)Δ2u+2β·∇(Δu)+|β|2Δu=Γi(β,m)mu=Γi(β,m)m^u,
which shows that Γi(β,m) is an eigenvalue for the weight m^, that is, Γi(β,m)=Γl(β,m^) for some l∈ℕ. Let us choose the largest l such that this equality holds. It follows from Γi(β,m)<Γi+1(β,m^) that l<i+1. Moreover, the monotone dependence, Γi(β,m^)≤Γi(β,m), implies l≥i. Then we conclude that l=i. Hence, we have Γi(β,m^)=Γi(β,m).Proof of Corollary4.3.
Suppose that(Pm) has nontrivial solution, that is, 1∈σ3(Δ2,m). From the inequality Γk(β,1)<≢m<≢Γk+1(β,1) and the strict monotonicity, we deduce
(4.11)Γk(β,Γk(β,1))>Γk(β,m),Γk+1(β,m)>Γk+1(β,Γk+1(β,1)).
Since(4.12)Γk+1(β,Γk+1(β,1))=Γk(β,Γk(β,1))=1
we deduce that(4.13)Γk(β,m)<1<Γk+1(β,m)
which is a contradiction. Hence, the proof is complete.



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