# Uniqueness Results for Higher Order Elliptic Equations in Weighted Sobolev Spaces

Authors: Loredana Caso; Patrizia Di Gironimo; Sara Monsurrò; Maria Transirico
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

We prove some uniqueness results for the solution of two kinds of Dirichlet boundary value problems for second- and fourth-order linear elliptic differential equations with discontinuous coefficients in polyhedral angles, in weighted Sobolev spaces.


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## Body

## 1. Introduction

The Dirichlet problem for polyharmonic equations in bounded domains ofRn has been studied, among the first, by Sobolev in [1].The problem was developed in various directions. For instance, Vekua in [2, 3] considers different boundary value problems in not necessarily bounded domains for harmonic, biharmonic, and metaharmonic functions. Successively, analogous problems in more general cases, for what concerns domains and operators, have been studied with different methods by many authors (see, e.g., [4–7]).In particular, in [7], the author obtains a uniqueness result for the Dirichlet problem for polyharmonic operators of order 2m in polyhedral angles of Rn. This result has been later on generalized, in [5], to the case of operators in divergence form of order 2m with discontinuous bounded measurable elliptic coefficients.In [6] the authors study a boundary value problem for biharmonic functions in presence of nonregular points on the boundary of the domain. It is well known that in the neighborhood of these singular points (corners or edges) the solution of the problem presents a singularity that can be characterized by the presence of a suitable weight.Uniqueness results for different Dirichlet problems in weighted Sobolev spaces for different classes of weights can be found in [8–12]. Studies of Dirichlet problems in the framework of weighted Sobolev spaces and in the case of unbounded domains can be found in [13–22].In this paper, we extend the results of [5, 7] to the case of weighted Sobolev spaces. More precisely, we prove some uniqueness results for the solution of two kinds of Dirichlet boundary value problems for second- and fourth-order linear elliptic differential equations with discontinuous coefficients in the polyhedral angle Rln,  0≤l≤n-1,  n≥2, in weighted Sobolev spaces.The first problem we consider is the following:(1)∑i,j=1naijuxixj=f,f∈L-s2Rln,u∈W°s1,2Rln,where, for k∈N0 and s∈R,  Wsk,2(Ω) denotes a weighted Sobolev space where the weight is a power of the distance from the origin, W°sk,2(Ω) is the closure of C0∞(Ω) in Wsk,2(Ω), and Ws0,2(Ω)=Ls2(Ω); see Section 2 for details.The second problem we study is(2)∑i,j=1naijuxixjxixj=f,f∈L-s2Rln,u∈W°s2,2Rln.In both cases the coefficients aij belong to some weighted Sobolev spaces.The main tool in our analysis is a generalization of the Hardy’s inequality proved by Kondrat’ev and Olènik in [23].


## 2. Preliminary Results

LetΩ be an open subset of Rn with n≥2, whose boundary contains x=0. For k∈N0 and s∈R,  Wsk,2(Ω) denotes the space of all functions u:Ω→R such that |x|sDαu∈L2(Ω) for |α|≤k, normed by(3)uWsk,2Ω=∑α≤kxsDαuL2Ω,W°sk,2Ωis  the  closure  ofC0∞ΩinWsk,2Ω,Ws0,2Ω=Ls2Ω.From [24] and Propositions 6.3 and 6.5, we get the following.Proposition 1.
IfG is a bounded open subset in Rn with 0∈∂G, then (4)Wsk,2G↪Wk,2Gfors≤0.Furthermore, for each q∈[1,2[ there exists ϵ0=ϵ0(q)>0 such that (5)Wsk,2G↪Wk,qGfor0<s≤ϵ0.In the present paper we use the following notation:(i)
V⊂Rn is a cone with vertex in the origin of coordinates;(ii)
BR,  R>0, is the open ball of center in the origin and radius R;(iii)
VR=V∩BR;(iv)
for everyl∈{0,…,n-1}, (6)Rln=x=x1,x2,…,xn∈Rn:xi>0,i=n-l,…,n,is the “polyhedral angle” with vertex in the origin;(v)
R+n=R0n is the half-space;(vi)
QR=Rln∩BR.To prove our main results, consisting in two uniqueness theorems, we will use the following inequality. We observe that this is a slightly modified version of a generalized Hardy’s inequality that was proved by Kondrat’ev and Olènik in [23], adapted to our needs (see also [5]).Lemma 2 (generalized Hardy’s inequality).
Letp>1 and r∈R be such that r+n-p≠0. Assume that for a sufficiently smooth function g the following condition is fulfilled:(7)∫VR2∖VR1xr▽gxpdx<+∞, where ▽g=(∂g/∂x1,…,∂g/∂xn) is the gradient of the function g and 0<R1<R2. Then, there exist two constants M,K>0 such that(8)∫VR2∖VR1xr-pgx-Mpdx<K∫VR2∖VR1xr▽gxpdx,where K does not depend on the function g, R1, and R2. If, in addition, g(0)=0 then M=0.Remark 3.
We remark that there are always important restrictions on the dimensionn of the space, the order of “singularity” r, and the summability exponent p (see, e.g., [23, 25–29], where different variants of Hardy or Caffarelli-Kohn-Nirenberg type inequalities are proved).


## 3. Dirichlet Problem for Second-Order Elliptic Equations

We consider the following differential operator in divergence form in the polyhedral angleRln,  0≤l≤n-1:(9)∑i,j=1naijuxixj,where the coefficients aij are measurable functions such that there exist two positive constants λ and μ such that (10)λx2sξ2≤∑i,j=1naijxξiξj≤μx2sξ2a.e.  inRln,∀ξ∈Rn.We study the Dirichlet problem(11)∑i,j=1naijuxixj=f,a.e.  inRln,u∈W°s1,2Rln,where f∈L-s2(Rln).Definition 4.
We say that a functionu is a generalized solution of problem (11) if it satisfies the integral identity(12)∫QR∑i,j=1naijuxivxjdx=-∫QRfvdx,for any R>0 and any function v∈W°s1,2(QR).Now we prove our first uniqueness result.Theorem 5.
Letu∈W°s1,2(Rln) be a generalized solution of problem (11), with f=0. Then there exists ϵ0>0 such that if s≤ϵ0/2 and s≠2-n/2 one has u≡0 in Rln.Proof.
LetΘ(t) be an auxiliary function in C0∞([0,∞[) defined by(13)Θt≡10≤t≤1,θt1≤t≤2,0t≥2,where θ(t) is such that 0≤θ(t)≤1. Let us also assume that there exists a positive constant K0 such that(14)Θ′t2≤K0Θt.Set, for any R>0,(15)ΘRx=ΘxR.Note that the function ΘR is such that, for any j=1,…,n, one has(16)ΘRxjx=Θ′xRxjRx.
Letu∈W°s1,2(Rln) be a generalized solution of problem (11), with f=0. We put (17)vRx=uxΘRx. Clearly, by definition of ΘR and as a consequence of our boundary condition, one has that vR∈W°s1,2(Q2R).
Thus, usingvR as test function in (12), we get(18)∫Q2R∑i,j=1naijuxiuxjΘRxdx+∫Q2R∖QR∑i,j=1naijuxiuΘRxjxdx=0.From (10), (16), and (18) we deduce that there exists a positive constant K1=K1(n,μ) such that(19)∫Q2R∑i,j=1naijuxiuxjΘRxdx=∫Q2R∖QR∑i,j=1naijuxiuΘ′xRxjRxdx≤K1∫Q2R∖QRx2suxΘ′xRuxdx=K1∫Q2R∖QRxsuxΘ′xRxs-1udx, where ux denotes the modulus of the gradient of u.
By applying Young’s inequality one gets that for anyϵ>0(20)∫Q2R∑i,j=1naijuxiuxjΘRxdx≤ε2K1∫Q2R∖QRx2sux2Θ′2xRdx+K12ε∫Q2R∖QRx2s-2u2dx.Thus, taking into account (14) and applying the generalized Hardy’s inequality (8) (with p=2 and r=2s) to the second term in the right-hand side of (20), we deduce that if s≠2-n/2,(21)∫Q2R∑i,j=1naijuxiuxjΘRxdx≤ε2K1K0∫Q2R∖QRx2sux2ΘRdx+K12ε∫Q2R∖QRx2s-2u2dx≤ε2K1K0∫Q2R∖QRx2sux2ΘRdx+K1K2ε∫Q2R∖QRx2sux2dx. From the ellipticity condition in (10) and for ϵ=λ/K1K0, we have (22)∫Q2Rx2sux2ΘRdx≤K2∫Q2R∖QRx2sux2dx, where the constant K2=K2(n,λ,μ,K0,K).
Thus for anyP>0 and for any R>P we obtain(23)∫QPx2sux2dx≤K2∫Q2R∖QRx2sux2dx.Since u is a generalized solution of problem (11), with f=0, and the constant K2 does not depend on the radius R and on the solution u, the right-hand side of (23) tends to zero when R→+∞ and then (24)∫QPx2sux2dx=0∀P>0. This implies that (25)x2sux2=0a.e.  inQP∀P>0; therefore(26)ux=0a.e.  inQP∀P>0.By Proposition 1 we deduce that if the solution u∈Ws1,2(QP) with s≤0, then u∈W1,2(QP), for any P>0. On the other hand, if s>0 for any q∈[1,2[ there exists ϵ0=ϵ0(q)>0 such that if 0<s≤ϵ0/2, then u∈W1,q(QP) for any P>0. Thus, by (26) the function u(x) is a constant in Rln, and since u∈W°s1,2(Rln) one concludes that u=0 in Rln.


## 4. Dirichlet Problem for 4th-Order Elliptic Equations

Let us now consider the following differential operator of 4th order in the polyhedral angleRln,  0≤l≤n-1, (27)∑i,j=1naijuxixjxixj, where aij are measurable symmetric coefficients and there exist two positive constants λ and μ such that (28)λx2s≤aijx≤μx2sa.e.  inRln,i,j=1,…n.We want to prove a uniqueness result for the solution of the Dirichlet problem(29)∑i,j=1naijuxixjxixj=f,a.e.  inRln,u∈W°s2,2Rln,where f∈L-s2(Rln).Definition 6.
We say that a functionu is a generalized solution of problem (29) if it satisfies the integral identity(30)∫QR∑i,j=1naijuxixjvxixjdx=∫QRfvdx,for any R>0 and any function v∈W°s2,2(QR).The result is the following.Theorem 7.
Letu∈W°s2,2(Rln) be a generalized solution of problem (29), with f=0. Then there exists ϵ0>0 such that if s≤ϵ0/2 and s≠2-n/2,4-n/2 one has u≡0 in Rln.Proof.
We shall rely on the methods developed in [5, 7]. We consider the function ΘR(x) defined in (13) and satisfying (14). Furthermore, we assume that there exists a positive constant K1 such that(31)Θ′′t2≤K1Θt.
Note that the functionΘR is such that, for any i,j=1,…,n, one has (16) and(32)ΘRxixjx=Θ′′xRxixjR2x2+Θ′xRx2δij-xixjRx3,where δij denotes the Kronecker delta.
Again we put(33)vRx=uxΘRx, where u∈W°s2,2(Rln) is a generalized solution of problem (29), with f=0.
Observe that the definition ofΘR together with the boundary condition satisfied by u gives that vR∈W°s2,2(Q2R). Hence, by the symmetry of aij, if we take vR as test function in (30) we get(34)∫Q2R∑i,j=1naijuxixj2ΘRxdx+2∫Q2R∖QR∑i,j=1naijuxixjuxiΘRxjxdx+∫Q2R∖QR∑i,j=1naijuxixjuΘRxixjxdx=0.From (28) and (34) we deduce that(35)λ∫Q2Rx2s∑i,j=1nuxixj2ΘRxdx≤2μ∫Q2R∖QRx2s∑i,j=1nuxixjuxiΘRxjxdx+μ∫Q2R∖QRx2s∑i,j=1nuxixjuΘRxixjxdx. By applying (16), (32), and Young’s inequality one gets that there exist two positive constants K2=K2(n,λ,μ,K0) and K3=K3(n,λ,μ,K0,K1) such that for any ε,ε1>0(36)∫Q2Rx2s∑i,j=1nuxixj2ΘRxdx≤ε∫Q2R∖QRx2s∑i,j=1nuxixj2ΘRxdx+ε12∫Q2R∖QRx2s∑i,j=1nuxixj2ΘRxdx+K2ε∫Q2R∖QRx2s-2∑i=1nuxi2dx+K32ε1∫Q2R∖QRx2s-4u2dx.Thus, applying repeatedly the generalized Hardy’s inequality (8) (with p=2 and r=2s to the third integral on the right-hand side and with p=2 and r=2s-2 to the last integral on the right-hand side and then again with p=2 and r=2s), we deduce that if s≠2-n/2,4-n/2, (37)∫Q2Rx2s∑i,j=1nuxixj2ΘRxdx≤K4∫Q2R\QRx2s∑i,j=1nuxixj2dx, where the constant K4=K4(n,λ,μ,K0,K1,K).
Thus for anyP>0 and for any R>P we obtain(38)∫QPx2s∑i,j=1nuxixj2dx≤K4∫Q2R∖QRx2s∑i,j=1nuxixj2dx.Now, arguing as in the proof of Theorem 5, since u is a generalized solution of problem (29), with f=0, the right-hand side of (38) tends to zero when R→+∞ and then (39)∫QPx2s∑i,j=1nuxixj2dx=0∀P>0. This implies that (40)x2s∑i,j=1nuxixj2=0a.e.   inQP∀P>0;therefore(41)uxx=0a.e.   inQP∀P>0.
In view of Proposition1 we obtain that if the solution u∈Ws2,2(QP) with s≤0, then u∈W2,2(QP), for any P>0, while if s>0 for any q∈[1,2[ there exists ϵ0=ϵ0(q)>0 such that if 0<s≤ϵ0/2, then u∈W2,q(QP) for any P>0. Thus, by (41) the function ux is constant a.e. in QP, and since u∈W°s2,2(Rln) one concludes that ux=0 a.e. in QP, for any P>0. The thesis follows then as the one of Theorem 5.



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