# On Singular Dissipative Fourth-Order Differential Operator in Lim-4 Case

Authors: Ekin Uğurlu; Elgiz Bairamov
Journal: ISRN Mathematical Analysis
 (2013)

---

## Abstract

A singular dissipative fourth-order differential operator in lim-4 case is considered. To investigate the spectral analysis of this operator, it is passed to the inverse operator with the help of Everitt's method. Finally, using Lidskiĭ's theorem, it is proved that the system of all eigen- and associated functions of this operator (also the boundary value problem) is complete.


---

## Body

## 1. Introduction

In 1910, Weyl showed that [1] the singular second-order differential operators can be fallen into two classes: operators in limit-circle case and operators in limit-point case. The operators in limit-circle case have the solutions that are all in square integrable space. However, in limit-point case, only one linearly independent solution can be in square integrable space. The development of this theory belongs to Titchmarsh [2]. After these fundamental works, second-order singular differential operators have been developed by many authors (e.g., see [3–6]).Following the methods of Weyl and Titchmarsh, the theory for higher order equations and Hamiltonian systems was constructed in [7–19].In [7], a regular self-adjoint fourth-order boundary value problem was investigated. Further Green’s function and the resolvent operator were constructed. In [8, 13], such a construction was done for the singular self-adjoint fourth order boundary value problem. In [9, 10], higher-order differential equations whose coefficients are complex-valued were studied. In 1974, Walker [20] showed that an arbitrary order self-adjoint eigenvalue problem can be represented as an equivalent self-adjoint Hamiltonian systems. Further, the developments in the theory of singular self-adjoint Hamiltonian systems were given in [14–19].On the other hand, an important class of the nonself-adjoint operators is the class of the dissipative operators [21]. It is known that all eigenvalues of the dissipative operators lie in the closed upper half-plane, but this analysis is so weak. There are some methods to complete the analysis of the dissipative operators. Some of them are Livšic’s, Krein’s, and Lidskiĭ’s theorems and functional model [21, 22]. These methods were used in the literature for the second-order differential operators (see [22–26]). In this paper a singular dissipative fourth order differential operator in lim-4 case is investigated. In particular, using Lidskiĭ’s theorem, it is shown that the system of all eigen- and associated functions is complete in L2(Λ).


## 2. Preliminaries

LetL denote the linear nonself-adjoint operator in the Hilbert space H with the domain D(L). The element y∈D(L),  y≠0, is called a root function of the operator L corresponding to the eigenvalue λ0, if all powers of L are defined on this element and (L-λ0I)ny=0 for some n>0.The functionsy1,y2,…,yk are called the associated functions of the eigenfunction y0 if they belong to D(L) and the equalities Lyj=λ0yj+yj-1,  j=1,2,…,k, hold.The completeness of the system of all eigen- and associated functions ofL is equivalent to the completeness of the system of all root functions of this operator.If, for the operatorL with dense domain D(L) in H, the inequality ℑ(Ly,y)≥0 (y∈D(L)) holds, then L is called dissipative.Theorem 1 (see [26]).
LetL be an invertible operator. Then, -L is dissipative if and only if the inverse operator L-1 of L is dissipative.It is known that a kernelA(t,s) of an integral operator 𝒜 defned by
(1)𝒜f=∫abA(t,s)f(s)ds,f∈L2(a,b),
where -∞≤a<b≤∞, is called a Hilbert-Schmidt kernel if |A(t,s)|2 is integrable on (a,b)2(2)∫ab∫ab|A(t,s)|2dtds<∞.A kernel satisfying the propertyA(t,s)=A(s,t)- is called a Hermitian kernel. Properties of Hermitian kernels and related integral operators can be found in [27] (further see [28]). Now we shall remind some results. Let us consider the equation which is related with (1) as
(3)μ𝒜f=f.The functionf which differs from zero is called a characteristic function of 𝒜 which corresponds to the characteristic value μ. If a Hermitian Hilbert-Schmidt kernel is not null then it posseses at least one characteristic value and this characteristic value (every characteristic value) is real. Further there is a finite orthonormal base of characteristic functions for each characteristic value μ of 𝒜. Using the union of these bases one obtains an orthonormal system of characteristic functions of the kernel A(t,s). If φ1,φ2,…,φn are distinct members of such a system belonging respectively to the characteristic value μ1,μ2,…,μn (not necessarily all different) then for 1≤v≤n, φv⊗φv=φv(t)φv(s)- is an orthonormal system of two variables. For such a system the equivalence
(4)A(t,s)≗∑n=1∞φn⊗φnμn
holds. Moreover if g=𝒜f, f∈L2, then the equivalence
(5)g≗∑n=1∞(g,φn)φn=∑n=1∞(f,φn)μnφn
holds. Further the series in (5) are relatively uniformly absolutely convergent. However, equivalence can be replaced by equality. Before showing this, let us consider the equality
(6)g(t)=λ∫A(t,s)f(s)ds.
Following the same idea of [27, pages 22] if A(t,s) is an L2-kernel such that A(t,s) is continuous and f is an L2-function then g(t) is continuous.Hence ifA(t,s) is a continuous Hermitian L2-kernel (see [27, pages 127]) and g=𝒜f, f∈L2, then the equality
(7)g(t)=∑n=1∞(g,φn)φn(t)=∑n=1∞(f,φn)μnφn(t)
holds and the series are uniformly absolutely convergent. In this case taking g(t)=(f,A(t,·)-) the series given in (4) converge to the continuous kernel A(t,s) uniformly in the variable t for every s. Hence from (4) one can get
(8)A(t,t)=∑n=1∞|φn(t)|2μn.
Integrating both side from a to b we obtain
(9)∫abA(t,t)dt=∑n=1∞1μn.
Consequently if A(t,t) is integrable on (a,b) then the series converge. This implies that 𝒜 is of trace-class (nuclear). For the definition and properties of trace-class operators see, for example, [21].Above arguments given in [27] hold for Hermitian kernels. However from arbitrary L2-kernels one can pass to the Hermitian kernels. For example K1(t,s)=A(t,s)+A(s,t)- and K2(t,s)=i(A(t,s)-A(s,t)-) are Hermitian kernels. These Hermitian kernels are continuous and L2-kernels if A(t,s) is so. Hence above arguments hold for K1(t,s) and K2(t,s) if A(t,s) is continuous L2-kernel such that A(t,t) is integrable on (a,b). Further
(10)A=12(K1-iK2).Hence ifK1 and K2 are trace-class kernels then so is A. However this result has been given in [29, pages 526] as a definition.Lidskiĭ’s Theorem (see [21, page 231]). If the dissipative operatorLis the trace class operator, then its system of root functions is complete in the Hilbert spaceH.


## 3. Statement of the Problem

The fourth order differential expression is considered as(11)η(y)=yiv+q(x)y,x∈Λ:=[a,b),
where -∞<a<b≤∞,  a is the regular point and b is the singular point for η, and q is real-valued, Lebesgue measurable, and locally integrable function on Λ.LetL2(Λ) be the Hilbert space consisting of all functions y such that ∫ab|y(x)|2dx<∞ with the inner product (y,χ)=∫aby(x)χ¯(x)dx.Let(12)Ω={y∈L2(Λ):y,y′,y′′,y′′′∈ACloc(Λ),η(y)∈L2(Λ)},
where ACloc(Λ) denotes the set of all locally absolutely continuous functions on Λ. For arbitrary y,χ∈Ω, Green’s formula is obtained as
(13)∫x1x2‍η(y)χ(x)dx-∫x1x2‍y(x)η(χ)dx=[y,χ](x2)-[y,χ](x1),
where a≤x1<x2≤b and [y,χ](x)=y′′′(x)χ(x)-y(x)χ′′′(x)+y′(x)χ′′(x)-y′′(x)χ′(x). Green’s formula implies the fact that if y(x,λ) and χ(x,λ) both satisfy the equation η(y)=λy for the same value of λ, then [y,χ](x) is independent of x and depends only on λ. Further for arbitrary y,χ∈Ω, at singular point b, the limits [y,χ](b)=limx→b[y,χ](x) and [y,χ¯](b)=limx→b[y,χ¯](x) exist and are finite. The latter also follows from Green’s formula. In fact, it is sufficient to get the second factors with their complex conjugates on the left-hand side in Green’s formula.In this paper it is assumed thatq(x) satisfies the lim-4 case conditions at b (see [30] and references therein). Lim-4 case is also known as Weyl’s limit-circle case [8].Let us consider the solutionsφ1(x,λ),  φ2(x,λ),  ψ1(x,λ), and ψ2(x,λ) of the equation
(14)yiv+q(x)y=λy,x∈Λ,
where λ is some complex parameter, satisfying the conditions
(15)[φ1(a,λ)φ2(a,λ)ψ1(a,λ)ψ2(a,λ)φ1′(a,λ)φ2′(a,λ)ψ1′(a,λ)ψ2′(a,λ)φ1′′(a,λ)φ2′′(a,λ)ψ1′′(a,λ)ψ2′′(a,λ)φ1′′′(a,λ)φ2′′′(a,λ)ψ1′′′(a,λ)ψ2′′′(a,λ)]=[α20γ100β20θ10-β10-θ2-α10-γ20],
where all αi,  βi,  γi, and θi(i=1,2) are real numbers such that α2γ2-α1γ1=1 and β1θ1-β2θ2=1. For the existence of these solutions, see [7, 8, 13]. Since lim-4 case holds for η, all solutions φi(x,λ) and ψi(x,λ) belong to L2(Λ). It is clear that [φr,ψs](a)=δrs(1≤r,s≤2), where δrs is the Kronocker delta, [φr,φs](a)=0 and [ψr,ψs](a)=0.Let us setui(x)=φi(x,0) and zi(x)=ψi(x,0)(i=1,2),  with x∈Λ. Then ui and zi become the real solutions of η(y)=0(x∈Λ). Further, they belong to Ω. Hence for arbitrary y∈Ω, the values [y,u1](b),  [y,u2](b),  [y,z1](b), and [y,z2](b) exist and are finite.LetD(N) be the set of all functions y∈Ω satisfying the boundary conditions
(16)α1y(a)+α2y′′′(a)=0,(17)β1y′(a)+β2y′′(a)=0,(18)[y,u1](b)-h1[y,z1](b)=0,(19)[y,u2](b)-h2[y,z2](b)=0,
where αi and βi(i=1,2) are real numbers given as previously stated and h1 and h2 are some complex numbers such that h1=ℜh1+iℑh1 and h2=ℜh2+iℑh2 with ℑhi>0,  i=1,2.It should be noted that for any solutionsy(x,λ) of (14), conditions (16) and (17), respectively, can be written as
(20)[y,φ1](a)=0,(21)[y,φ2](a)=0.The operatorN is defined on D(N) as Ny=η(y), y∈D(N),  x∈Λ. The main aim of this paper is to investigate the spectral analysis of the operator N (the boundary value problem (14)–(19)).


## 4. Completeness Theorems

Let{yj(x,λ);1≤j≤r} be any r(1≤r≤4) solutions of η(y)=λy. The notation W{y1,y2,…,yr}=W{y1(x,λ),y2(x,λ),…,yr(x,λ)} denotes the Wronskian of order r of this set of functions [7, 8, 31].It is known that the following equality holds [7, 8, 13]:
(22)W{y1,y2,y3,y4}=-[y1,y2](x)[y3,y4](x)W{y1,y2,y3,y4}=+[y1,y3](x)[y2,y4](x)W{y1,y2,y3,y4}=-[y1,y4](x)[y2,y3](x).
This equation also shows that W{y1,y2,y3,y4} of any four solutions of η(y)=λy is independent of x and depends only on λ [7, 8].Now consider the solutionsϑ1(x,λ) and ϑ2(x,λ) of the equation η(y)=λy,  x∈Λ, satisfying the conditions
(23)[[ϑ1,u1](b)[ϑ2,u1](b)[ϑ1,z1](b)[ϑ2,z1](b)[ϑ1,u2](b)[ϑ2,u2](b)[ϑ1,z2](b)[ϑ2,z2](b)]=[h10100h201].
For the existence of these solutions given with the previous conditions, see [13]. Clearly these solutions satisfy conditions (18) and (19), respectively. Now let us set
(24)W(λ)=W{φ1,φ2,ϑ1,ϑ2}.
Then W(λ) becomes an entire function, and the zeros of W(λ) coincide with the eigenvalues of the operator N [13]. This implies that all zeros of W (all eigenvalues of N) are discrete and that possible limit points of these zeros (eigenvalues of N) can only occur at infinity.Using (22), we immediately have
(25)W{u1,u2,z1,z2}=1,[[u1,u1](x)[u2,u1](x)[z1,u1](x)[z2,u1](x)[u1,u2](x)[u2,u2](x)[z1,u2](x)[z2,u2](x)[u1,z1](x)[u2,z1](x)[z1,z1](x)[z2,z1](x)[u1,z2](x)[u2,z2](x)[z1,z2](x)[z2,z2](x)]=[00-10000-110000100].
Hence, the Plücker identity for the fourth-order case is obtained (see [13, p. 435]):
(26)[y,χ](x)=[y,u1](x)[χ,z1](x)-[y,z1](x)[χ,u1](x)+[y,u2](x)[χ,z2](x)-[y,z2](x)[χ,u2](x),
where y,χ∈D(N).Theorem 2.
The operatorN is dissipative in L2(Λ).Proof.
Fory∈D(N), we have
(27)(Ny,y)-(y,Ny)=[y,y¯](b)-[y,y¯](a).
Since y∈D(N), a direct calculation shows that
(28)[y,y¯](a)=0.
Further, using (26) and conditions (18) and (19) one obtains
(29)[y,y¯](b)=2iℑh1|[y,z1](b)|2+2iℑh2|[y,z2](b)|2.
Substituting (28) and (29) into (27) it is obtained that
(30)ℑ(Ny,y)=ℑh1|[y,z1](b)|2+ℑh2|[y,z2](b)|2,
and this completes the proof.Theorem2 shows that all eigenvalues of N lie in the closed upper half-plane.Theorem 3.
The operatorN has no real eigenvalue.Proof.
Fory∈D(N), a direct calculation shows that
(31)ℑ(Ny,y)=ℑ(λ∥y∥2).
Now let λ=λ0 be a real eigenvalue of N, and let φ1(x,λ0) be the corresponding eigenfunction. Then (31) and (30) gives [φ1,z1](b)=[φ1,z2](b)=0. Using these equalities in (18) and (19) one gets that [φ1,u1](b)=[φ1,u2](b)=0.
Let us consider the solutionψ1(x,λ0). Hence, using (26) it is obtained that
(32)1=[φ1,ψ1](a)=[φ1,ψ1](b)=[φ1,u1](b)[ψ1,z1](b)-[φ1,z1](b)[ψ1,u1](b)+[φ1,u2](b)[ψ1,z2](b)-[φ1,z2](b)[ψ1,u2](b)=0.
This contradiction completes the proof.From Theorem3 it is obtained that all eigenvalues of N lie in the open upper half-plane. In particular zero is not an eigenvalue of N. Hence, the operator N-1 exists.Consider the solutionsu1(x),  u2(x),  v1(x) and v2(x), where v1(x)=u1(x)-h1z1(x) and v2(x)=u2(x)-h2z2(x). u1(x) and u2(x) satisfy conditions (16) and (17), respectively and v1(x) and v2(x) satisfy conditions (18) and (19), respectively.The equationNy=f(x),  y∈D(N),  x∈Λ, is equivalent to the nonhomogeneous differential equation
(33)yiv+q(x)y=f(x),y∈D(N),x∈Λ,
subject to the boundary conditions
(34)[y,u1](a)=0,[y,u2](a)=0,[y,u1](b)-h1[y,z1](b)=0,[y,u2](b)-h2[y,z2](b)=0
(compare the boundary conditions at a with (20) and (21)). Using Everitt’s method [7] (further see [8, 13]) the general solution is obtained as
(35)y(x)=∫ab‍G(x,ζ)f(ζ)dζ,
where
(36)G(x,ζ)={vT(ζ)u(x),a≤x<ζ,vT(x)u(ζ),ζ<x≤b,v(x)=u(x)-hz(x) and
(37)u(x)=(u1(x)u2(x)),z(x)=(z1(x)z2(x)),h=(h100h2).The operatorK defined by
(38)Kf=∫ab‍G(x,ζ)f(ζ)dζ,
where f∈L2(Λ), is the inverse operator of N. Hence the completeness of the system of all eigen- and associated functions of N is equivalent to the completeness of those of K in L2(Λ).SinceG(x,ζ) is a continuous Hilbert-Schmidt kernel and G(x,x) is integrable on [a,b), the operator K is of trace class.Let us consider the operator-K. Since N is dissipative in L2(Λ),-K is also dissipative in L2(Λ). Thus all conditions are satisfied for Lidskiĭ’s theorem. So we have the following.Theorem 4.
The system of all root functions of-K (also K) is complete in L2(Λ).Since the completeness of the system of root functions (eigen- and associated functions) ofK is equivalent to the completeness of those of N, is obtained that the following.Theorem 5.
All eigenvalues of the problem (14)–(19) lie in the open upper half-plane and they are purely discrete. The system of all eigen- and associated functions of the problem (14)–(19) is complete in L2(Λ).



---

*Source: 549876-2013-09-11.xml*