# Investigation of the Spectral Properties of a Non-Self-Adjoint Elliptic Differential Operator

Authors: Arezoo Ghaedrahmati; Ali Sameripour
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

---

## Abstract

Non-self-adjoint operators have many applications, including quantum and heat equations. On the other hand, the study of these types of operators is more difficult than that of self-adjoint operators. In this paper, our aim is to study the resolvent and the spectral properties of a class of non-self-adjoint differential operators. So we consider a special non-self-adjoint elliptic differential operator (Au)(x) acting on Hilbert space and first investigate the spectral properties of space H1=L2Ω1. Then, as the application of this new result, the resolvent of the considered operator in ℓ-dimensional space Hilbert Hℓ=L2Ωℓ is obtained utilizing some analytic techniques and diagonalizable way.


---

## Body

## 1. Introduction

Let Ω be a bounded domain inRn with smooth boundary ∂Ωi.e.,∂Ω∈ℂ∞. We introduce the weighted Sobolev space H1=W2,β2Ωℓ as the space of complex value functions u (x) defined on Ω with the finite norm:(1)u+=∑i=1n∫Ωρ2βx∂u∂xiℂℓ2dx+∫Ωuxℂℓ2dx1/2.We denote byH∘1 the closure of ℂ0∞Ωℓ in H with respect to the above norm, i.e., H∘ is the closure of ℂ0∞Ω in H1=W2,β2Ωℓ. The notion ℂ0∞Ω stands for the space of infinitely differentiable functions with compact support in Ω. In this paper, we investigate the spectral properties. In particular, we estimate the resolvent of a non-self-adjoint elliptic differential operator of type(2)Aux=−∑i,j=1n∂∂xjρ2βxαijxqx∂∂xiuxacting on Hilbert space Hℓ=L2Ωℓ with Dirichlet-type boundary conditions. Here, ρx∈ℂ10,1 is a positive function that satisfies the following conditions:(3)c1xα1−xβ≤ρ2x≤M,ρ2′x≤Mxα/2−1+ε11−xβ/2−1+ε2,where α,β≥0,ε1=0 if α≠1and ε1>0 if α=1,and ε2=0if β≠1and ε2>0if β=1., αijx=αjix¯i,j=1,…,n,αijx∈C2Ω¯i,j=1,…,n and the functions αijx satisfy the uniformly elliptic condition, i.e., there exists c > 0 such that(4)cs2≤∑i,j=1nαijx,sisj¯s=s1,…,sn∈ℂn,x∈Ω.Furthermore, suppose thatqx∈ℂ2Ω¯;Endℂℓ such that for each x∈Ω¯, the matrix function q (x) has nonzero simple eigenvalues μjx∈ℂ2Ω¯1≤j≤ℓ arranged in the complex plane in the following way:(5)μ1x,…,μℓx∈ℂ\Φ,where Φ=z∈ℂ:argz≤φ,φ∈0,π is a closed angle with zero vertex (i.e., the eigenvalues μjx of qx lie on the complex plane and outside of the closed angle Φ). For a closed extension of operator A with respect to space H=W2,β2Ωℓ above, we need to extend its domain to the closed domain(6)DA=u∈Hℓ0∩W2,loc20,1ℓ:∂∂xj∑i,j=1nρ2βαi,jq∂∂xiu∈Hℓ,(see [1, 2]), where the local space W2,loe2Ωℓ is the functions uxx∈Ω in this form W2,loc2Ω=ux:∑i=02∫Juix2dx<∞,Jis an open subset ofΩ. Here, and in the sequel, the value of the function arg z∈−π,π and A denotes the norm of the bounded operator A: H⟶H.To get a feeling for the history of the subject under study, refer to our earlier papers [3–5]. Indeed, this paper was written in continuing on our earlier papers. This study is sufficiently more general than our earlier papers; here, we obtain the resolvent estimate of operator A, which satisfies the special and general conditions.


## 2. The Resolvent Estimate of Degenerate Elliptic Differential Operators on H in Some Special Cases

Theorem 1.
Let A in (2), i.e., assume that operator A is acting on Hilbert spaceH = L2 (Ω) with Dirichlet-type boundary conditions, and the sector Ω be defined as in Section 1. Let the complex function q (x) satisfy the following conditions:(7)qx∈ℂ1Ω¯,qx∈ℂΦ,∀x∈Ω¯,(8)argqx1q−1x2≤π8,∀x1,x2∈Ω¯.Then, for sufficiently large modulusλ∈Φ, the inverse operator A−λI−1 exists and is continuous in H, and the following estimates are valid:(9)A−λI−1≤MΦλ−1λ∈Φ,λ>ℂΦ,(10)ρβ∂∂xiA−λI−1≤MΦ′λ−1/2λ∈Φ,λ>ℂΦ,fori=1,…,n,where MΦ,ℂΦ>0 are sufficiently large numbers depending on S (Φ set is defined in the previous sections). The symbol . stands for the norm of a bounded arbitrary operator T in H.Proof.
Here, to establish Theorem1, we will first prove the assertion of Theorem 1 together with estimate (9). So, as in Section 1 for a closed extension of operator A (for more explanation, see chapter 6 in [3]), we need to extend its domain to the closed set(11)DA=u∈Hℓ0∩W2,loc20,1ℓ:∂∂xj∑i,j=1nρ2βαi,jq∂∂xiu∈Hℓ.
Let operator A now satisfy (7), (8). Then, there exists a complex number Z∈ℂ (notice that we can take Z = eiy, for a fix real ϒ∈−π,π) such that Z=eiϒ=1, and so (12)c′≤ReZqx,c′λ≤−ReZλ,c′>0∀x∈Ω¯,λ∈Φ.
In view of the uniformly elliptic condition, we have(13)cs2=c∑i=1nsi2≤∑i,j=1nαijxsisj¯,c>0,s=s1,…,sn∈ℂn,x∈Ω,and take si=∂y/∂xi which implies that c∑i=1n∂y/∂xix2≤∑i,j=1nαijx∂y/∂xix∂y/∂xix¯. From this, and according to c′≤ReZqx in (10), we then multiply these two positive relations with each other, implying that(14)c1∑i=1n∂y∂xix2≤ReZqx∑i,j=1nαijx∂y∂xix∂y∂xjx¯fory∈DA.
Multiplying both sides of the latter relation by the positive termρ2αx and then integrating both sides, we will have(15)c1∑i=1n∫Ωρ2β∂y∂xix2dx≤ReZqx∑i,j=1n∫Ωρ2βxαijxqx∂y∂xix∂y∂xjx¯.
Now by applying the integration by parts and using Dirichlet-type condition, then the right sides of the latter relation without multiple ReZ become(16)∑i,j=1n∫Ωρ2βxαijxqx∂y∂xix∂y∂xjx¯dx=−∑i,j=1n∫Ωρ2βxαijxqx∂y∂xixyxdx=−∂y∂xj∑i,j=1nρ2βxαijxqx∂y∂xixyx=Ay,y.
Hence,Ayx=−∂y/∂xj∑i,j=1nρ2βxαijxqx∂y/∂xix
Here, the symbol (,) denotes the inner product in H.
Notice that the above equality in (16) is obtained by the well-known theorem of the m-sectorial operators which are closed by extending its domain to the closed domain in H. These operators are associated with the closed sectorial bilinear forms that are densely defined in H (for more explanation of the well-known Theorem 1, see chapter 6 in [2]). This is why we extend the domain of operator A to the closed domain in space H above. Therefore,(17)c1∑i=1n∫Ωρ2βx∂y∂xix2dx≤ReZAy,y.
From (10), we have c′λ≤−ReZλ,c′>0,∀λ∈Φ. Multiply this inequality by ∫Ωyx2dx=y,y=y2>0. It follows that(18)c′λ∫Ωyx2dt≤−ReZλy,y.
From this and the above inequality, we will have(19)c1∑i=1n∫Ωρ2βx∂y∂xix2dx+c′λ∫Ωyx2dx≤ReZAy,y−Zλy,y=ReZA−λIy,y≤ZyA−λIy=yA−λIy′,i.e.,(20)c1∑i=1n∫Ωρ2βx∂y∂xix2dx+c′λ∫Ωyx2dx≤yA−λIy.
Sincec1∑i=1n∫Ωρ2βx∂y/∂xix2dx is positive, we will have either(21)c′λyx2=λ∫Ω∂y∂xix2dx≤yA−λIy,or(22)λyx≤MΦA−λIy.
This inequality ensures that the operatorA−λI is one to one, which implies that ker A−λI=0. Therefore, the inverse operator A−λI−1 exists, and its continuity follows from the proof of estimate (9) of Theorem 1. To prove (9), we set v=A−λI−1f,f∈H in (19), implying that(23)λ∫ΩA−λI−1f2dx≤MΦA−λI−1fA−λIA−λI−1f.
SinceA−λIA−λI−1f=If=f, then(24)λ∫ΩA−λI−1f2dx≤MΦA−λI−1ff,so(25)λA−λI−1f≤MΦA−λI−1ff,which implies that λA−λI−1f≤MΦf. Since λ≠0, then A−λI−1f≤MΦλ−1f;i.e.,A−λI−1≤MΦλ−1. This estimate completes the proof of the assertion of Theorem 1 together with estimate (9). Now, we start to prove estimate (10) of Theorem 1. As in the above argument, we drop the positive term c′λ∫Ωyx2 dx from(26)c1∑i=1n∫Ωρ2βx∂y∂xix2dx+c′λ∫Ωyx2dx≤yA−λIy.
It follows that(27)c1∑i=1n∫Ωρ2βx∂y∂xix2dx≤yA−λIy.
Equivalently(28)c1ρβ∂∂xiA−λI−1f2yA−λIy.
SetA−λI−1f,f∈Hin the latter relation, and proceeding by similar calculation as in the proof of estimate (9), we then obtain(29)c1ρβ∂∂xiA−λI−1f2≤A−λI−1fA−λIA−λI−1f.
SinceA−λIA−λI−1f=If=f, then(30)c1ρβ∂∂xiA−λI−1f2≤A−λI−1f2.
Consequently, by (9), this implies that(31)c1ρβ∂∂xiA−λI−1f2≤MΦ′λ−1λ−1f2.
To this end, we will have(32)ρβ∂∂xiA−λI−1≤MΦ′λ−1/2.
Thus, here, the proof of estimate (10) is finished; i.e., this completes the proof of Theorem 1.
Now let condition (8) not hold. Then we will have the following statement.


## 3. The Resolvent Estimate of Some Classes of Degenerate Elliptic Differential Operators on H

In this section, we will derive a new general theorem by dropping the assumption (8) from Theorem 1 in Section 2.Theorem 2.
As inSection 1, let Φ be some closed sector with vertex at 0 in the complex plane (for more explanation, see [3]), and let the complex function q (x) satisfy the following equations:(33)qx∈ℂ1Ω¯,qx∈ℂΦ;∀x∈Ω¯.Then, for sufficiently large modulusλ∈Φ, the inverse operator A−λI−1 exists and is continuous in H, and the following estimates hold:(34)A−λI−1f≤MΦ′λ−1,λ∈Φ,λ>ℂΦ,where MΦ′,ℂΦ>0 are sufficiently large numbers depending on Φ.Proof.
Let us (9) not satisfy. To prove the assertion of Theorem 2 together with (34), we construct the functions φ1x,…,φmx,q1x,…,qmx so that each one of the functionsq1x,…,qmxx∈Ω¯ as the function q (x) in Theorem 1 satisfies (8).
Therefore, let(35)φ1x,…,φmx,q1x,…,qmx∈C0∞Ωsatisfy(36)0≤φrx,r=1,…,m,φ12x+⋯+φm2x≡1,x∈Ω¯ddtφrx∈ℂ0∞Ω,qrx=qx,∀x∈suppφr,qrx∈ℂ\Φ,∀x∈Ω¯,r=1,…,m,argqrx1qr−1x2≤π8,∀x1,x2∈suppφr,r=1,…,m.
In view of Theorem1 and by (9) and (10), set Ar = A in the definition of the differential operator, which implies that(37)Arux=−∑i,j=1nρ2βxαijxqxuxi′xxj′is acting on H where(38)DAr=u∈Ho∩W2,loc2Ω;∂u∂xj∑i,j=1nρ2βαijqr∂u∂xi∈H.
Due to the assertion of Theorem1, for 0≠λ∈Φ, the inverse operator A−λI−1 exists and is continuous in space H=L2Ω and satisfies(39)A−λI−1≤MΦλ−1,ρβ∂∂xiA−λI−1≤MΦ′λ−1/2,λ∈Φ,λ>ℂΦ,0≠λ∈Φ.
Let us introduce(40)Gλ=∑r=1mφrAr−λI−1φr.
Hereφr is the multiplication operator in H by the function φrx. Consequently, it is easily verified that(41)Ar−λIGλ=I+ρ2β−1x∑r=1mηrxAr−λI−1φr+ρβx∑i=1n∑r=1mϒirx∂∂xiAr−λI−1φr,where ηr,ϒir∈L∞Ω; supp βr and supp ϒir are contained in supp φr. Let us take the right side of (41) equal to I + T (λ). Thus, we will have(42)A−λIGλ=I+Tλ.
Now according toSection 2, if we put A = Ar for r = 1,…,m in (8), we will have(43)Ar−λI−1≤M1sλ−1,ρβ∂∂xiAr−λI−1≤MΦ′λ−1/2.
Owing to the definition of T (λ) in (41) easily, it follows that(44)Tλ≤MΦ′λ−1/2,λ∈Φ,λ>1.
Sinceλ is a sufficiently large number, it easily implies that Tλ<1/2<1. From this and using the well-known theorem in the operator theory, we conclude that I + T (λ) and so (A − λI) G (λ) are invertible. Hence, ((A − λI) G (λ))−1 exists and is equal to(45)Gλ−1A−λI−1=I+Tλ−1.
By adding + I and –I to the right side of (44), it follows that(46)Gλ−1A−λI−1=I+Tλ−1−I+I.
We now set(47)Fλ=I+Tλ−1−I.
Then(48)Gλ−1A−λI−1=I+Fλ.
In view of kT (λ) < 1 and (44), we now estimate F (λ) by the following geometric series:(49)Fλ≤∑i=2+∞Tkλ≤Tλ21+Tλ+Tλ2+⋯≤Tλ2MΦ1+12+⋯≤2MΦMΦ′λ−1/22,i.e., Fλ≤2M1Φλ−1.ByAr−λI−1≤M1Φλ−1, for we will have(50)Gλ=∑r=1mφrAr−λI−1φr≤MΦ″Ar−λI−1≤MΦ″M1Φλ−1,i.e., Gλ≤M2Φλ−1. Now from (45), we have(51)A−λI−1=GλI+Tλ−1=GλI+Fλ.
Therefore(52)A−λI−1=GλI+Fλ≤M2Φλ−11+2M1Φλ−1,i.e., here the assertion of Theorem 2 is proved. Therefore, to complete the proof Theorem 2, we must prove the estimate (34). To the end, according to the latter inequality, we have (53)A−λI−1≤M2Φλ−1+2M2ΦM1Φλ−1λ−1,and since λ−1λ−1=λ−2≤λ−1, it follows that(54)A−λI−1≤MΦλ−1,λ≥ℂ,λ∈Φ.
This completes the proof of Theorem2.


## 4. On the Resolvent Estimate of the Differential Operator inHℓ

As inSection 1, let the differential operator(55)Aux=−∑i,j=1n∂∂xjρ2βxαijxqx∂u∂xix,act on Hilbert space Hℓ=Lℓ2Ω with Dirichlet-type boundary conditions, and suppose that qx∈ℂ2Ω¯,Endℂℓ such that for each x∈Ω¯, the matrix function q (x) has nonzero simple eigenvalues μjx∈ℂ2Ω¯,1≤j≤ℓ arranged in the complex plane in the following way:(56)μ1x,…,μℓx∈ℂ\Φ,where(57)Φ=z∈ℂ:argz≤φ,φ∈0,π.Furthermore, suppose that forj=1,…,ℓ, we have(58)μjx∈ℂ1Ω¯,μjx∈ℂ\Φ,∀x∈Ω¯,(59)argμjx1μj−1x2≤π8,∀x1,x2∈Ω¯.Now, according to Theorem1, but here instead of operator A which acts on the space H = L2 (Ω), let operator A act on the space Hℓ=L2Ωℓ. Now by the assumption of Section 1, we will have the following theorem in the general case.Theorem 3.
Let (58) and (59) and the assumptions of Section 1 hold for operator A as in (2), then for sufficiently large modulus λ∈Φ, the inverse operator A−λI−1 exists and is continuous in the space Hℓ=L2Ωℓand the following estimate holds: (60)A−λI−1≤MΦλ−1,ρddxA−λI−1≤M2′λ−1/2,λ∈Φ,λ≥ℂ,where MΦ,CΦ>0 are sufficiently large numbers depending on Φ and λ>CΦ.Proof.
Now by applying the eigenvaluesμ1x,…,μℓx of the matrix function q (x), we define the operators A1,…,Aℓ such that(61)Ajux=−∂u∂xj∑i,j=1nρ2βxαijxμjx∂u∂xix,j=1,…,ℓ,where its extension domains are(62)DAj=y∈Ho∩W2,loc2Ω;∂y∂xj∑i,j=1nρ2βαijμj∂y∂xi∈H,which, as operator A in Theorem 1, the operators Aj, j=1,…,ℓ, acts on space H = L2 (Ω) (notice that here the operators Aj are the same operator A in Section 2, i.e., to define the operators Aj, we just change the function qx in operator A by the eigenvalues functions μj (x), j=1,…,ℓof matrix qx). The conditions which we consider on the eigenvalues μj (x) of the matrix function qx in Section 1 guarantee that one can convert the matrix qx to the diagonal form qx=UxΛxU−1x, where Ux,U−1x∈ℂ20,1;Endℂℓ and Λx=diagμ1x,…,μℓx. Consider space Hℓ=H⊕⋯H⊕ (ℓ times). Put Γλ=UBλU−1 where the operator(63)Bλ=diagA1−λI−1,…,Aℓ−λI−1acts on the direct sum Hℓ=H⊕⋯H⊕ (ℓ times) in which λ∈Φ¯\R+,λ≥C0 and (Uu)(x) = U (x) u (x); (u∈Hℓ). Consequently, it follows that(64)A−λIΓλu=−ddxρ2AxddxUxBλU−1xux=T1+T2+T3,where(65)T1=−ddxρ2AxUxddxBλU−1xux=−ddxρ2UtΛxddxBλU−1xux=−Uddxρ2ΛxddxBλU−1xux−U′xρ2ΛddxBλU−1u=λUBλU−1u−U′xρ2ΛddxBλU−1u+UU−1u,T2=−ddxρ2qU′BλU−1u,T3=−λUxBλU−1u.
Using (9) and (10), we have A−λITλ=I+T10+T20 where T20=ρ2′qU′BλU−1 and T10≤Mλ−1/2. Now by the Hardy-type inequality, we estimate the operator T20 as follows:(66)∫01t−1+ε1′1−t−1+ε2′yt2dt≤Mε1′,ε2′∫01yt2dt+Mε1′,ε2′∫01t1+ε1′1−t1+ε2′y′t2dt,∀y∈Η0,ε1′,ε2′≠0.
SinceqtU′t≤M by (3), we have the following inequality:(67)∫01ρ2t′2Bλutℂℓ2dt≤M2∫01tα−2+2ε1′1−tβ−2+2ε2′Bλutℂℓ2dt≤M3∫01tα1−tβρ−2tρ2Bλutt′ℂℓ2dt+MBλuHℓ2.
Now by (3) and estimate (9), it follows(68)∫01ρ2t′2Bλutℂℓ2dt≤M∫01ρ2Bλutℂℓ2dt+MBλuHℓ2≤M′λ−1uHℓ2,λ∈Φ,λ>ℂ.
Then,T20≤M′λ−1/2 for sufficiently large in modulus of λ∈Φ; consequently,(69)A−λIΓλ=I+Fλ,Fλ≤Mλ−1/2,λ∈Φ,λ>ℂ.
Proceeding as at the end ofSection 2 (e.g., see (43)) from Fλ≤Mλ−1/2, it easily follows that I+Fλ is inversible and then that A−λIΓλ is inversible, that is,(70)A−λIΓλ−1=I+Fλ−1.
Then by adding±I, the last relation we have is (71)I+Fλ−1=I+Fλ−1+I−I.
SinceFλ≤Mλ−1/2, in a calculation as in Section 2, take yλ=I+Fλ−1−I. Then, yλsatisfies(72)yλ≤Mλ−1,λ∈Φ,λ>ℂ.
Consequently,A−λI−1=ΓλI+yλ since(73)Γλ=UBλU−1,Bλ=diagA1−λI−1,…,Aℓ−λI−1.
PutPj=Aj,j=1,…,ℓ as in (39). By (72) and (73), we have Aj−λI1≤Mλ−1,j=1,…,ℓ and it follows that Γλ≤λ−1, so (74)A−λI−1≤ΓλI+yλ≤Mλ−11+Mλ−1≤Mλ−1.
Now we prove estimate (39). Since ρd/dtAj−λI−1≤Mλ−1,j=1,…,ℓfor Γ1λ, we can get the corresponding estimate Γ1λ≤M1λ−1/2, and this implies(75)ρddxA−λI−1≤Γ1λI+y1λ.
Sincey1λ≤M1′λ−1, we have(76)ρddxAj−λI−1ρddxPj−λI−1≤Mλ−1/21+M1′λ−1,which implies ρd/dxAj−λI−1≤Mλ−1/2λ∈Φ,λ≥ℂ so that the proof of the fundamental Theorem 3 in the general case Hℓ=L2Ωℓ is completed.



---

*Source: 5564552-2021-05-15.xml*