# Computing theq-Numerical Range of Differential Operators

Authors: Ahmed Muhammad; Faiza Abdullah Shareef
Journal: Journal of Applied Mathematics
 (2020)

---

## Abstract

A linear operator on a Hilbert space may be approximated with finite matrices by choosing an orthonormal basis of thez Hilbert space. In this paper, we establish an approximation of theq-numerical range of bounded and unbounnded operator matrices by variational methods. Application to Schrödinger operator, Stokes operator, and Hain-Lüst operator is given.


---

## Body

## 1. Introduction and Definitions

The simplest concepts which can be used to obtain an enclosure of the spectrum of a linear operatorA in a Hilbert space H are the numerical range.(1)WA=Ax,x,forsomex∈DA,∥x∥=1=,where D. denotes the domain. It is not difficult to see that the point spectrum σpA of A is contained in WA and that the approximate point spectrum σappA of A is contained in the closure of WA; the inclusion σA⊂WA¯ holds if A is closed. Many estimates of eigenvalues of differential operators, for instance, involve calculating estimates of the inner products Ax,x, using partial integration. It is always convex [1]. However, the numerical range often gives a poor localization of the spectrum and cannot reveal the existence of spectral gaps. In [2], the numerical range of a (finite) matrix was approximated by projection methods. This concept was generalized to q-numerical range in [3], as follows:(2)WqA=Ax,y,forsomex,y∈H,∥x∥=1=∥y∥,x,y=q,where 0≤q≤1. It is easy to see that if q=1, then WqA coincides with the numerical range of A. For a closed operator A, the closure of the q-numerical range contains the eigenvalues of A scaled by q, and WqαA+βI=αWqA+βI, for every α,β∈ℂ. Moreover, It is known that WqA is a compact convex subset of ℂ [4]. A review of the properties of the q-numerical range of operator matrices may be found in [3, 5, 6]. The main purpose of this work is the approximation of the q-numerical range of bounded and unbounded operator matrices, which contains both theoretical results and applications to some self-adjoint and non-self-adjoint operator matrices. The paper is organized as follows. In Section 2, we establish an approximation of the q-numerical range of bounded and unbounded operator matrices. In Section 3, we shall apply these results to compute the q-numerical range of differential operators and some new analytic bounds.


## 2. Convergence Theorems

In this section, we will use finite matrices to approximate the numerical range of linear operators. However, the idea of approximating linear operators by finite matrices is an obvious one that must happen again and again, suppose that one wishes to compute theq-numerical range of A by using the following projection method. Let Lkk=1∞ be a nested family of spaces in H given by Lk=spanϕ1,ϕ2,⋯,ϕk, where Lk:k∈ℕ is the orthonormal basis of a Hilbert space H whose element lies in DA and suppose that the corresponding orthogonal projections Pk:H⟶Lk converge strongly to the identity operator I. We identify A with its matrix representation with respect to ϕk:k∈ℕ:(3)A=Aiji,j=1∞,Aij≔Aϕj,ϕii,j∈N.Then, the compression ofA to Lk is denoted by Ak≔PkALk, where(4)Ak=Aϕ1,ϕ1Aϕ1,ϕ2⋯Aϕ1,ϕkAϕ2,ϕ1Aϕ2,ϕ2⋯Aϕ2,ϕk⋮⋮⋮Aϕk,ϕ1Aϕl,ϕ2⋯Aϕk,ϕk.Theorem 1.
LetAbe a bounded operator inHLetLk:k∈ℕbe a nested family of spaces inHgiven byLk=spanϕ1,ϕ2,⋯,ϕkwhereϕk:k∈ℕis orthonormal, andAkbe as in Equation((4)).Then,WqAk⊆WqA ,forq∈0,1.Proof.
Define an isometryi:Lk⟶ℂk by iα1ϕ1+α2ϕ2+⋯+αkϕk≔α1,α2,⋯,αk, and iβ1ϕ1+β2ϕ2+⋯+βkϕk≔β1,β2,⋯,βk. Suppose that λ∈WqAk. Then, for some α,β∈ℂk, with ∥α∥=1=∥β∥,α,β=q such that λ=Akα,β. Choose ξ,η∈Lk, such that iξ=α,iη=β, and ∥ξ∥=1=∥η∥. Then, a direct computation shows that λ=Aξ,η, where ξ=∑j=1kαjϕj, and η=∑j=1kβjϕj. Thus, λ∈WqA.The next inclusion, which will be used in the proof of Theorem3, asserts that WqAk:k≥2 forms an increasing sequence of sets.Lemma 2.
LetLk:k∈ℕandAkbe as in Theorem1. Givenq∈0,1, thenWqAk⊆WqAτ, forr≥k.Proof.
This is an immediate consequence of the fact thatℂk is a subspace of ℂτ. In detail, suppose λ is in WqAk; then, there exists α,β∈ℂk, ∥α∥=1=∥β∥, with α,β=q such that λ=Akα,β. Choose ς,ν∈ℂτ by setting ς=α1,α2,⋯,αj,0,⋯,0T, ν=β1,β2,⋯,βj,0,⋯,0T. A simple calculation shows that ς,ν=α,β=q and Akα,β=Akς,ν and so λ is in WqAτ.Theorem 3.
LetAandAkbe as in Theorem1. LetPk:H→Lkbe orthogonal projection. IfPkconverge strongly to the identity operatorI, thenWqA¯=⋃j=1∞WqAj¯.Proof.
In view of Theorem1, it is sufficient to prove WqA⊆⋃j=1∞WqAj¯. Suppose λ∈WqA. Choose x,y∈H such that ∥x∥=∥y∥=1 and x,y=q, such that λ=Ax,y. We know Pkx⟶x and Pky⟶y as k⟶∞ and if xk≔iPkx,yk≔iPky thus Axk,yk⟶Ax,y as k⟶∞ and xk,yk⟶x,y=q as k⟶∞.
Fixk>0. Let i:spanϕ1,ϕ2,⋯,ϕsk⟶ℂsk be the standard isometries as in the proof of Theorem 1. Define α~k,β~k∈Csk by α~k=iskxk,β~k=iskyk. Consider the sk×sk matrix Ask, where the p,r-element of the sk×sk matrix Ask is equal to Aϕp,ϕr, for p,r=1,2,⋯,sk. A simple calculation shows that Askα~k,β~k=Axk,yk, α~sk,β~sk=xk,yk. Since Axk,yk⟶Ax,y as k⟶∞ and xk,yk⟶x,y=q as k⟶∞, we have Askα~k,β~k⟶Ax,y as k⟶∞. Hence, there exist λk∈WqAsk such that λk⟶λ. In view of Lemma 2, this immediately gives λ∈⋃j=1∞WqAj¯.Remark 4.
The hypothesesPkconverge strongly to the identity operatorIare (in general) necessary. It is easy to construct an example whereλ∈⋃j=1∞WqAj¯is a strict subset ofWqA.Example 1.
LetAbe an operator matrix inH=ℓ2ℕ, whereA=diag1/nn=1∞Lk:k∈ℕbe a nested family of subspace inℓ2ℕ,withLk=spane2,⋯,ek+1,whereej=jthstandard basis vector, andrk=spanψ1,⋯,ψkwhereψkk=1∞is any orthonormal sequence. Then, performing an analysis analogous to Theorem3, we see thatAxk,ykis not convergent toAx,y,unlessxis orthogonal toe1.Remark 5.
We assume readers are familiar with basic notions and results about linear unbounded operators, as well as matrices of nonnecessarily bounded operators. Useful references are [7–9]. We call a few definitions though: a linear operatorAwith a domainDAcontained in a Hilbert spaceHis said to be densely defined ifDA=H. Say that a linear operatorAis closed if its graphΓAis closed inH⊕H. A linear operatorAis called closable if the closureΓA¯of its graph is the graph of some operators. A subspaceD⊂DAis called a core of a closable operatorAifADis closable with closureA. The definition of theq-numerical range for bounded linear operators in Equation((2)) generalizes as follows to unbounded operator matricesAwith dense domainDA.Definition 6.
For a linear operatorAwith domainDA⊂H, we define theq-numerical range ofAfor0≤q≤1by(5)WqA=Ax,y,forsomex,y∈DA,∥x∥=1=∥y∥,x,y=q.Theorem 7.
LetAbe an unbounded operator inH. LetLk:k∈ℕbe a nested family of spaces inDAgiven byLk=spanϕ1,ϕ2,⋯,ϕk, whereϕk:k∈ℕis orthonormal, andAkbe as in Equation((4)). Then,WqAk⊆WqA, forq∈0,1.Proof.
Define an isometryπ:Lk→ℂk by πζ1ϕ1+ζ2ϕ2+⋯+ζkϕk≔ζ1,ζ2,⋯,ζk, and πγ1ϕ1+γ2ϕ2+⋯+γkϕk≔γ1,γ2,⋯,γk. Suppose that λ∈WqAk. Then, for some ζ,γ∈ℂk, with ∥ζ∥=1=∥γ∥,ζ,γ=q such that λ=Akζ,γ. Choose ξ,η∈Lk, such that πξ=ζ,πη=γ, and ∥ξ∥=1=∥η∥. Then, a direct computation shows that λ=Aξ,η, where ξ=∑j=1kζjϕj, and η=∑j=1kγjϕj. Thus, λ∈WqA.The following lemma can be the proof in a similar fashion as Lemma2.Lemma 8.
LetLk:k∈ℕandAkbe as in Theorem7. Givenq∈0,1, thenWqAk⊆WqAτ, forr≥k.In the following result, we describe that the closure of the rangeWqA is approximated by WqAk under the assumption that the linear span of ϕ1,ϕ2,⋯ is a core of A.Theorem 9.
LetAbe an unbounded operator inH. LetLk:k∈ℕbe a nested family of spaces inDAgiven byLk=spanϕ1,ϕ2,⋯,ϕk, whereϕk:k∈ℕis orthonormal, andAkbe as in Equation((4)). Then,WqA¯=⋃k=1∞WqAk¯, forq∈0,1.Proof.
SinceC=spanϕ1,ϕ2,⋯ is a core of A, there exists a sequence xkk=1∞, with each xk∈spanϕ1,ϕ2,⋯,ϕsk for some sk>0 such that ∥x−xk∥⟶0 and ∥Ax−Axk∥⟶0. In a similar way, we may also find a sequence ykk=1∞, with each yk∈spanϕ1,ϕ2,⋯,ϕsk for some sk>0, such that ∥y−yk∥⟶0 and ∥Ay−Ayk∥⟶0, so this means that ∥Axk,yk−Ax,y∥⟶0 as k⟶∞ and ∥xk,yk−q∥⟶0 as k⟶∞. Fix k>0. Let π:spanϕ1,ϕ2,⋯,ϕsk⟶ℂsk be the standard isometries as in the proof of Theorem 7. Define α~k,β~k∈Csk by α~k=πskxk,β~k=πskyk. Consider the sk×sk matrix Ask, where the p,r-element of the sk×sk matrix Ask is equal to Aϕp,ϕr, for p,r=1,2,⋯,sk. A simple calculation shows that Askα~k,β~k=Axk,yk and α~k,β~k=xk,yk. Since ∥Axk,yk−Ax,y∥⟶0 as k⟶∞ and ∥xk,yk−q∥⟶0 as k⟶∞, this implies that ∥Askα~k,β~k−Ax,y∥⟶0 as k⟶∞. Hence, there exists λk∈WqAsk such that λk⟶λ. In view of Lemma 8, this immediately gives λ∈⋃k=1∞WqAk¯.


## 3. Numerical Experiments on Differential Operator

In this section, we study some concrete examples and demonstrate that, in spite of the results obtained in the previous section, practical computation of theq-numerical range of differential operator is very far from being straightforward. We define the inner product u,v to be linear in the first parameter and conjugate linear in the second parameter, and we consider the space of square-integrable functions, L2Ω,dx, where Ω is an interval in ℝ, a Hilbert space with inner product(6)u,v=∫Ωuv¯dx.The computations were performed in Matlab.


### 3.1. Application to Schrödinger Operator

In the Hilbert spaceH≔L20,1, we introduce the Schrödinger operator(7)A=−d2dx2+p(with bounded potential p) and the domain of L is given by(8)DA=u∈H20,1:u0=0=u1.Remark 10.
(i)
BecauseA is self-adjoint and bounded below with purely discrete spectrum, the eigenvalues of A are given by(9)λk≔infF⊂DAdimF=ksupy∈Fy≠0gy,where g is the Rayleigh functional(10)gy≔Ay,yy,y,y∈DA,y≠0(ii)
It is clear the operatorA in L20,1 has a sequence of eigenvalues and normalized eigenfunctions for the operator A in L20,1 are(11)λn=n2π2,(12)ϕnx=2sinnπxforn=1,2,3,⋯under the setting px=0(iii)
Because Equation (7) is a closed operator, then it is not difficult to see that the subspace CA=spanϕ1,ϕ2,⋯⊂DA is a core of A, so in this case, the main Theorem 9 is applicable to this example(iv)
We may use these eigenfunctions in Equation (12) as basis elements for a discretization of the type discussed in Section 2, form the matrix elements −ϕk′′,ϕj, using the inner product in Equation (6) with respect to the orthonormal basis in Equation (12) and consider the (infinite) operator matrix(13)Q≔−ϕk′′,ϕjThe matrixAk in Equation (4) is obtained by taking the leading submatrices of the (infinite) operator matrix in Equation (13), with appropriate dimensions. Observe that −ϕk′′,ϕj=diagπ2,4π2,9π2,⋯ which can be evaluated explicitly. The following figure shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.


### 3.2. Analytical Estimates for Schrödinger Operator

In order to understand to what extent Figure1 is qualitatively correct, we now analyze the q-numerical range of the Schrödinger operator. Comments:(i)
In Remark10 part (i), it is obvious Ay,y≥π2y,y; thus, the numerical range of A is WA¯=π2,∞(ii)
While for0<q<1, the q-numerical range WqA of A contains the range(14)Wqdiagn2,n2π2for any integer n≥2. Let λ=n2π2>0. Then, the range(15)Wqdiagπ2,λ=z=x+iy:x,y∈ℝ2,x−qπ2+λ22+y21−q2≤λ−π222contains the convex polygon with vertexes(16)−1−qλ2,1+qλ2,−i1−q2λ2,i1−q2λ2.Figure 1
On the left-hand side, the variational method approximation to the Schrödinger operator. The approximationAk is given by a self-adjoint matrix with some real eigenvalues λ1,λ2,⋯,λk with λ1≤λ2≤⋯λk. The q-numerical rangeWqA24 of A24 for q=1 is the line segment λ1,λ24; the red dots are σA24. On the right-hand side, the q-numerical rangeWqA24 of A24 for 0<q<1 is the union of the closed circular discs. The small red circles on the real axis inside the circular disk are σA24 scaled by q.For a sufficiently largeλ, such a convex polygon contains an arbitrary given point in the Gaussian plane. Thus, this differential operator satisfies WqA=ℂ(iii)
If we restrict our attention for the variational approximation to the Schro¨dinger operator, we may have another story. The approximation Ak in Theorem 9 is given by a diagonal matrix with some real eigenvalues λ1,λ2,⋯,λk with λ1≤λ2≤⋯≤λk. The q-numerical range WqAk of Ak for 0<q<1 is given by WqAk=diagλ1,λk. Then, the range Wqdiagλ1,λk is the union of closed circular discs(17)Wqdiagλ1,λk=z=a+ib:a,b∈ℝ2,a−qλ1+λk22+b21−q2≤λk−λ122


### 3.3. Application to Block Differential Operators

In this subsection, we apply Theorem9 to compute the q-numerical range of Stokes-type operators and Hain-Lüst-type operators. First, we study Stokes-type operators.


#### 3.3.1. Application to Stokes-Type Operators

Consider the differential expression in the Hilbert spaceL220,1≔L20,1⊕L20,1, we introduce the matrix differential operator(18)A0≔A0B0C0D0=−d2dx2−ddxddx−32,on the domain(19)DA0≔H20,1∩H10,1⊕H10,1.The operator matrixA0 is not closed but closable [2]. In order to show that its closure is self-adjoint, we need the following:Remark 11.
(i)
Consider the operatorM given by(20)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,such that(21)My1y2=−y′1+y2′y′1−12y2.Lemma 12.
The operatorMin Equation ((21)) coincides with the adjoint of the block operator matrix in Equation ((18)).Proof.
Suppose that(22)g=g1g2∈DA~
with(23)u=u1u2∈L20,12.
Then(24)A0y,g=y,u,y=y1y2∈DA0=H20,1∩H10,1⊕H10,1or, equivalently,(25)−y′1+y2′,g1L20,1+y′1−12y2,g2L20,1=y1,u1L20,1+y2,u2L20,1for all y1,y2∈C0∞0,1 with compact support contained in 0,1. In particular, when y1=0, then Equation (25) becomes(26)−y2′,g1L20,1+−12y2,g2L20,1=y2,u2L20,1.The first part of the left-hand side of Equation (26) is a bounded linear functional of y2, which means that g1∈H10,1, and also, the second part of the left-hand side of equation (26) is a bounded linear functional of y2, which means that g2∈L20,1; again return to Equation (26), and integrating by part, we obtain(27)−y2,g1′−12g2=y2,u2y2∈C0∞0,1.
BecauseC0∞0,1 is a dense in L20,1, this means that u2=g′1−1/2g2. By the same argument, if we set y2=0y1∈H010,1∩H20,1, we find that(28)−y′′1,g1L20,1+y1′,g2L20,1=y1,u1L20,1.Again, return to Equation (28), and integrating by part, we obtain(29)−y1′xg1x¯x=0x=1+−y1,g1′+g2′¯=y1,u1.The first part of the left-hand side of Equation (29) is bounded only when g10=0=g11, and the second part is bounded linear functional of y1, which means that g1′+g2∈H10,1. Then, this implies that g∈DA~, and because C0∞0,1 is a dense in L20,1, this means that u1=−g1′+g2′ on 0,1. It follows that(30)u1u2=A~g1g2.Remark 13.
(i)
By the same argument of Lemma12, it is not difficult to see that the operator A0 in Equation (18) is symmetric in L20,12 with the domain(31)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,My1y2=−y′1+y2′y′1−12y2(ii)
BecauseM is a symmetric operator and has nonempty resolvent, then M self-adjoint. Thus, by [10, Theorem 5.4],M=A¯0The following result shows that the eigenvalues of the Stocks operator coincide with eigenvalues of the operatorM.Proposition 14.
LetA0be as in Equation((18)), thenσpA¯0=σpA0.Proof.
Letλ∈σpA¯0, then there exists an y=y1y2t∈DA¯0, y≠0, such that(32)A0¯y1y2=λy1y2.
We see that Equation (32) is equivalent to the following system of equations:(33)−y1′+y2′=λy1,(34)y′1−12y2=λy2,where y1y2t∈DA0¯ and(35)DA0¯≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1.
Becausey1∈H10,1, then from Equation (33), we get y′1+y2∈H20,1 and Equation (34) can be written as(36)y′1+y2=λ+32y2;this implies that y2∈H20,1 for λ≠−3/2. Since H20,1 is a linear space then y′1∈H20,1, hence y1∈H30,1⊆H20,1, it follows that y=y1,y2t∈DA0, so λ is an eigenvalue of A0.
Conversely, ifλ∈σpA, then there exists an y=y1,y2t∈DA0⊆DA0¯,y≠0, such that(37)A0y1y2=λy1y2.
This means thaty is an eigenfunction of A0¯ since A0¯DA0=A0.Remark 15.
(i)
It may be shown thatA has two series of eigenvalues λk± given by(38)λk±=k2π2−3/2±3/2−k2π22+10k2π22,k=1,2,⋯.
The eigenvalues of the minus series are located in the interval (-5/2,-3/2] and convergent to -5/2 ask⟶∞. The eigenvalues of the plus series are located in the interval 0,∞ and convergent to ∞ as k⟶∞. The essential spectrum of A is the accumulation point of the eigenvalues: σessM=−5/2(ii)
It is not difficult to see that the subspaceC0≔CA⊕CD⊂DA0=DA0∩DC0⊕DB0∩DD0 is a core of A0(iii)
Now form the matrix elements−ϕk′′,ϕj, −ϕk′,ϕj, ϕk′,ϕj, −3/2ϕk,ϕj using the inner product in (6) with respect to the orthonormal basis in Equation (12) and consider the (infinite) block operator matrix(39)G≔−ϕk′′,ϕj−ϕk′,ϕjϕk′,ϕj−32ϕk,ϕj.
The defined matrixAk in (4) is obtained by taking the leading submatrices of the block G, with appropriate dimensions. Observe that −ϕk′′,ϕj=diagπ2,4π2,9π2,⋯,(40)−ϕk′,ϕj=0,ifk=j;−2kπ∫01coskπxsinjπxdx,ifk≠j;ϕk′,ϕj=0,ifk=j;2kπ∫01coskπxsinjπxdx,ifk≠j;and −3/2ϕk,ϕj=−3/2δk,j, which can be evaluated explicitlyFigure2 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 2
On the left-hand side, the variational method approximation to the Stokes operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


### 3.4. Analytical Estimates for Stokes Operator

In order to understand the results in Figure2, it is useful to find an analytical estimate for WqM. Let ζ→,η→∈DM, where(41)ζ→=ζ1ζ2,η→=η1η2with∥ζ→∥=∥η→∥=1,ζ→,η→=q and let(42)λ=Mζ→,η→=−D2ζ1,η1+−ζ2′,η1+ζ1′,η2+αζ2,η2,whereα=−3/2. Equation (42) gives, as an estimate for the first term on the right-hand side of (42),(43)∫01−ζ1′′η¯1dx=∫01ζ1′η¯1′dx≥qπ2∫01ζ1η¯1dx.For the second and the third term on the right-hand side of (42), the Cauchy-Schwarz inequality and Youngs inequality yield(44)Re∫01−ζ2′η¯1dx=Re∫01ζ2η¯1′dx≥−∫01ζ22+η1′2dx,(45)Re∫01ζ1′η¯2dx≥−∫01ζ1′2+η22dx.Combining Equations (44) and (45), we get that(46)Re∫01ζ2′η¯1dx+Re∫01ζ1′η¯2dx≥−q.The fourth term of the right-hand side of Equation (42) satisfies(47)Re∫01αζ2η¯2dx≥infReαq−∫01ζ1η¯1dx.Hence, from Equations (43), (46), and (47), we get that(48)Reλ≥qπ2∫01ζ1η¯1dx−q+infReαq−∫01ζ1η¯1dx.This simplifies to(49)Reλ≥qπ2∫01ζ1η¯1dx−q+q−∫01ζ1η¯1infReα=qπ2−infRα∫01ζ1η¯1dx+qinfReα−1.This yields(50)Reλ≥qα−1,ifπ2−α≥0;qπ2−1,ifπ2−α<0.For our example, these yieldReλ≥−5/2q.To estimateImλ, observe that(51)Imλ=∫01Imuζ2η¯2dx=0.This completes the estimates onWqM.


### 3.5. The Hain-Lüst Operator

This operator was introduced by Hain and Lüst in application to problems of magnetohydrodynamics [11], and the problems of this type were studied in [7, 8, 12]. Assume that w:0,1→0,∞, w~:0,1→0,∞, and u:0,1→ℂ are such that wx=1, w~x=1, ux=18e2πix−20, for each x∈0,1. We introduce the differential expression(52)τA~≔−d2dx2,τB~≔wx,τC~≔w~x,τD~≔ux.LetA,B,C,D be the operators in the Hilbert space L20,1 induced by the differential expressions τA~, τB~, τC~, and τD~ with domain(53)DA≔H20,1∩H010,1,DB=DC=DD≔L20,1.In the Hilbert spaceL220,1≔L20,1⊕L20,1, we introduce the matrix differential operator(54)A≔ABCD=−d2dx2wxw~xux,on the domain(55)DA≔H20,1∩H010,1⊕L20,1.Remark 16.
(i)
By [13], Corollary VII.2.7, the operator A≔−d2/dx2 with domain DA≔H20,1∩H010,1 is closed. Moreover, because DA⊂DC, then the operator C is A-bounded with relative bound 0. This follows since there is a γ>0 such that, for every ε>0,(56)∥Cf∥2≤γ∥Af,f∥≤γε∥Af∥2+ε−1∥f∥2.
On the other handDD⊂DB, then the operator B is D-bounded, we conclude that the operator matrix DA is diagonally dominant of order 0; it is closed by [14], Corollary 2.2.9 (i)(ii)
In Equation (54), since A is self-adjoint with the purely discrete spectrum, then the linear span CA=spanϕ1,ϕ2,⋯ is a core of A, where ϕk:k∈ℕ is an orthonormal basis in L20,1, and by the same argument because D in Equation (54) is bounded, then the linear span CD=spanϕ1,ϕ2,⋯ is a core of D. Hence, it is not difficult to see that the subspace C≔CA⊕CD⊂DA=DA∩DC⊕DB∩DD is a core of A. So the main Theorem 9 is applicable to this example(iii)
We may use the eigenfunctions in Equation (12) as basis elements for a discretization of the type discussed in Section 2, forming the matrix elements Aϕk,ϕj, wϕk,ϕj, w~ϕk,ϕj, uϕk,ϕj, with respect to the inner product in (6) and considering the infinite block matrix(57)Q≔Aϕk,ϕjwϕk,ϕjw~ϕk,ϕjuϕk,ϕj
The defined matrixAk in (4) is obtained by taking the leading submatrices of the block Q, with appropriate dimensions.
Observe thatAϕk,ϕj=diagπ2,4π2,9π2,⋯,wϕk,ϕj=diag1,1,1,⋯,w~ϕk,ϕj=diag1,1,1,⋯, and uϕk,ϕj=36∫01e2πixsinkπxsinjπxdx−20δk,j, which can be evaluated explicitly. If the operator A included a potential, for instance, then its eigenfunctions would not generally be explicitly computable. We could still use the functions ϕj in (12) as basis functions, but the matrix elements Aϕk,ϕj would have to be computed by quadrature and the corresponding matrix would no longer diagonal. Figure 3 shows attempts to compute the numerical approximation of the boundary of WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 3
On the left-hand side, the variational method approximation to the Hain-Lüst operator for computingq-numerical rangeWqA48 of A48 for q=1. The red dots are σA48. On the right-hand side, for 0<q<1, the variational method approximation for computing WqA48 of A48 for 0<q<1. The red dots are σA48 scaled by q.


### 3.6. Analytical Estimates for Non-Self-Adjoint Hain-Lüst Operator

In order to understand the results in Figure3, it is useful to find an analytical estimate for WqA. Let x→,y→∈DA, where(58)x→=x1x2,y→=y1y2with∥x→∥=∥y→∥=1, x→,y→=q and let(59)λ=Ax→,y→=−D2x1,y1+x2,y1+x1,y2+ux2,y2.The first term of (59) gives, as an estimate,(60)∫01x1′′y¯1dx≥qπ2∫01x1y¯1dx.For the second and the third term on the right-hand side of (59), the Cauchy-Schwarz inequality and Youngs inequality yield(61)R∫01x2y¯1dx≥−∫01x22+y12dx,(62)R∫01x1y¯2dx≥−∫01x12+y22dx.Combining Equations (61) and (62), we get that(63)R∫01x2y¯1dx+R∫01x1y¯2dx≥−q.The fourth term of the right-hand side of Equation (59) satisfies(64)R∫01ux2y¯2dx≥infRuq−∫01x1y¯1dx.Hence, from Equations (60), (63), and (64), we get that(65)Rλ≥qπ2∫01x1y¯1dx−q+infRuq−∫01x1y¯1dx.This simplifies to(66)Rλ≥qπ2∫01x1y¯1−q+q−∫01x1y¯1infRu=qπ2−infRu∫01x1y¯1dx+qinfRu−1.This yields(67)Rλ≥qinfRu−1,ifπ2−infRu≥0;qπ2−1,ifπ2−infRu<0.For our example, these yieldRλ≥−53q.To estimateImλ, observe that(68)Imλ=∫01Imux2y¯2dx≤supx∈0,1Imu∫01x1y¯1dx≤q25,Imλ=∫01Imux2y¯2dx≥infx∈0,1Imz∫01x2y¯2dx≥−q25,hence(69)−q25≤Imλ≤q25.This completes the estimates onWqA.The following result shows that theq-numerical range of Example 1, is unbounded.Remark 17.
Suppose that(70)x→=2sinkπx0,y→=2sinkπx0where x→,y→∈DA, with ∥x→∥=∥y→∥=1, and x→,y→=q then(71)Ax→,y→=22kπ2∫01sin2kπxdx=k2π2.But k∈ℤ can be arbitrarily large. This means that Rλ is unbounded above.


#### 3.6.1. Application to Self-Adjoint Hain-Lüst Operator

In the Hilbert spaceH=L2−π,π⊕L2−π,π, we introduce the matrix differential operator(72)A~≔LBCV=−d2dx2+411z,where(73)zx=0,for−π≤x<0,2,for0≤x<π,on the domain(74)DA~≔H2−π,π∩H01−π,π⊕L2−π,π.Remark 18.
(i)
It is not difficult to see that the subspaceC≔CA⊕CV⊂DA~=DL∩DC⊕DB∩DV is a core of A~. So the main Theorem 9 is applicable to this example(ii)
The eigenvalues and normalized eigenfunctions of the operator−d2/dx2 in L2−π,pi are(75)λj=j24,ψjx=sinjx+π/2πforj=1,2,3,⋯Now form the matrix elementsAψk,ψj, Bψk,ψj, Cψk,ψj, Dψk,ψj, with respect to orthonormal basis ψjx=1/πsinjx+π/2 where j∈ℕ and consider the infinite block operator matrix(76)P≔Aψk,ψjBψk,ψjCψk,ψjDψk,ψj.The defined matrixAk in (4) is obtained by taking the leading submatrices of the block P, with appropriate dimensions. Observe that Lψk,ψj=diag1/4,1/2,⋯,Bψk,ψj=Cψk,ψj=diag1,1,1,⋯, and(77)Vψk,ψj=1π∫0πsinjx+π2sinkx+π2,ifk≠j;12,ifk=j;which can be evaluated explicitly. Figure 4 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 4
On the left-hand side, the variational method approximation to the Hain-Lüst operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


### 3.7. Analytical Estimates for Self-Adjoint Hain-Lüst Operator

In order to understand the results in Figure4, it is useful to find an analytical estimate for WqA~. By the same arguments in Section 3.6, it is not difficult to see that Reλ≥−3/4q, and(78)Imλ=∫−ππImzx2y2¯dx=0.


## 3.1. Application to Schrödinger Operator

In the Hilbert spaceH≔L20,1, we introduce the Schrödinger operator(7)A=−d2dx2+p(with bounded potential p) and the domain of L is given by(8)DA=u∈H20,1:u0=0=u1.Remark 10.
(i)
BecauseA is self-adjoint and bounded below with purely discrete spectrum, the eigenvalues of A are given by(9)λk≔infF⊂DAdimF=ksupy∈Fy≠0gy,where g is the Rayleigh functional(10)gy≔Ay,yy,y,y∈DA,y≠0(ii)
It is clear the operatorA in L20,1 has a sequence of eigenvalues and normalized eigenfunctions for the operator A in L20,1 are(11)λn=n2π2,(12)ϕnx=2sinnπxforn=1,2,3,⋯under the setting px=0(iii)
Because Equation (7) is a closed operator, then it is not difficult to see that the subspace CA=spanϕ1,ϕ2,⋯⊂DA is a core of A, so in this case, the main Theorem 9 is applicable to this example(iv)
We may use these eigenfunctions in Equation (12) as basis elements for a discretization of the type discussed in Section 2, form the matrix elements −ϕk′′,ϕj, using the inner product in Equation (6) with respect to the orthonormal basis in Equation (12) and consider the (infinite) operator matrix(13)Q≔−ϕk′′,ϕjThe matrixAk in Equation (4) is obtained by taking the leading submatrices of the (infinite) operator matrix in Equation (13), with appropriate dimensions. Observe that −ϕk′′,ϕj=diagπ2,4π2,9π2,⋯ which can be evaluated explicitly. The following figure shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.


## 3.2. Analytical Estimates for Schrödinger Operator

In order to understand to what extent Figure1 is qualitatively correct, we now analyze the q-numerical range of the Schrödinger operator. Comments:(i)
In Remark10 part (i), it is obvious Ay,y≥π2y,y; thus, the numerical range of A is WA¯=π2,∞(ii)
While for0<q<1, the q-numerical range WqA of A contains the range(14)Wqdiagn2,n2π2for any integer n≥2. Let λ=n2π2>0. Then, the range(15)Wqdiagπ2,λ=z=x+iy:x,y∈ℝ2,x−qπ2+λ22+y21−q2≤λ−π222contains the convex polygon with vertexes(16)−1−qλ2,1+qλ2,−i1−q2λ2,i1−q2λ2.Figure 1
On the left-hand side, the variational method approximation to the Schrödinger operator. The approximationAk is given by a self-adjoint matrix with some real eigenvalues λ1,λ2,⋯,λk with λ1≤λ2≤⋯λk. The q-numerical rangeWqA24 of A24 for q=1 is the line segment λ1,λ24; the red dots are σA24. On the right-hand side, the q-numerical rangeWqA24 of A24 for 0<q<1 is the union of the closed circular discs. The small red circles on the real axis inside the circular disk are σA24 scaled by q.For a sufficiently largeλ, such a convex polygon contains an arbitrary given point in the Gaussian plane. Thus, this differential operator satisfies WqA=ℂ(iii)
If we restrict our attention for the variational approximation to the Schro¨dinger operator, we may have another story. The approximation Ak in Theorem 9 is given by a diagonal matrix with some real eigenvalues λ1,λ2,⋯,λk with λ1≤λ2≤⋯≤λk. The q-numerical range WqAk of Ak for 0<q<1 is given by WqAk=diagλ1,λk. Then, the range Wqdiagλ1,λk is the union of closed circular discs(17)Wqdiagλ1,λk=z=a+ib:a,b∈ℝ2,a−qλ1+λk22+b21−q2≤λk−λ122


## 3.3. Application to Block Differential Operators

In this subsection, we apply Theorem9 to compute the q-numerical range of Stokes-type operators and Hain-Lüst-type operators. First, we study Stokes-type operators.


### 3.3.1. Application to Stokes-Type Operators

Consider the differential expression in the Hilbert spaceL220,1≔L20,1⊕L20,1, we introduce the matrix differential operator(18)A0≔A0B0C0D0=−d2dx2−ddxddx−32,on the domain(19)DA0≔H20,1∩H10,1⊕H10,1.The operator matrixA0 is not closed but closable [2]. In order to show that its closure is self-adjoint, we need the following:Remark 11.
(i)
Consider the operatorM given by(20)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,such that(21)My1y2=−y′1+y2′y′1−12y2.Lemma 12.
The operatorMin Equation ((21)) coincides with the adjoint of the block operator matrix in Equation ((18)).Proof.
Suppose that(22)g=g1g2∈DA~
with(23)u=u1u2∈L20,12.
Then(24)A0y,g=y,u,y=y1y2∈DA0=H20,1∩H10,1⊕H10,1or, equivalently,(25)−y′1+y2′,g1L20,1+y′1−12y2,g2L20,1=y1,u1L20,1+y2,u2L20,1for all y1,y2∈C0∞0,1 with compact support contained in 0,1. In particular, when y1=0, then Equation (25) becomes(26)−y2′,g1L20,1+−12y2,g2L20,1=y2,u2L20,1.The first part of the left-hand side of Equation (26) is a bounded linear functional of y2, which means that g1∈H10,1, and also, the second part of the left-hand side of equation (26) is a bounded linear functional of y2, which means that g2∈L20,1; again return to Equation (26), and integrating by part, we obtain(27)−y2,g1′−12g2=y2,u2y2∈C0∞0,1.
BecauseC0∞0,1 is a dense in L20,1, this means that u2=g′1−1/2g2. By the same argument, if we set y2=0y1∈H010,1∩H20,1, we find that(28)−y′′1,g1L20,1+y1′,g2L20,1=y1,u1L20,1.Again, return to Equation (28), and integrating by part, we obtain(29)−y1′xg1x¯x=0x=1+−y1,g1′+g2′¯=y1,u1.The first part of the left-hand side of Equation (29) is bounded only when g10=0=g11, and the second part is bounded linear functional of y1, which means that g1′+g2∈H10,1. Then, this implies that g∈DA~, and because C0∞0,1 is a dense in L20,1, this means that u1=−g1′+g2′ on 0,1. It follows that(30)u1u2=A~g1g2.Remark 13.
(i)
By the same argument of Lemma12, it is not difficult to see that the operator A0 in Equation (18) is symmetric in L20,12 with the domain(31)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,My1y2=−y′1+y2′y′1−12y2(ii)
BecauseM is a symmetric operator and has nonempty resolvent, then M self-adjoint. Thus, by [10, Theorem 5.4],M=A¯0The following result shows that the eigenvalues of the Stocks operator coincide with eigenvalues of the operatorM.Proposition 14.
LetA0be as in Equation((18)), thenσpA¯0=σpA0.Proof.
Letλ∈σpA¯0, then there exists an y=y1y2t∈DA¯0, y≠0, such that(32)A0¯y1y2=λy1y2.
We see that Equation (32) is equivalent to the following system of equations:(33)−y1′+y2′=λy1,(34)y′1−12y2=λy2,where y1y2t∈DA0¯ and(35)DA0¯≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1.
Becausey1∈H10,1, then from Equation (33), we get y′1+y2∈H20,1 and Equation (34) can be written as(36)y′1+y2=λ+32y2;this implies that y2∈H20,1 for λ≠−3/2. Since H20,1 is a linear space then y′1∈H20,1, hence y1∈H30,1⊆H20,1, it follows that y=y1,y2t∈DA0, so λ is an eigenvalue of A0.
Conversely, ifλ∈σpA, then there exists an y=y1,y2t∈DA0⊆DA0¯,y≠0, such that(37)A0y1y2=λy1y2.
This means thaty is an eigenfunction of A0¯ since A0¯DA0=A0.Remark 15.
(i)
It may be shown thatA has two series of eigenvalues λk± given by(38)λk±=k2π2−3/2±3/2−k2π22+10k2π22,k=1,2,⋯.
The eigenvalues of the minus series are located in the interval (-5/2,-3/2] and convergent to -5/2 ask⟶∞. The eigenvalues of the plus series are located in the interval 0,∞ and convergent to ∞ as k⟶∞. The essential spectrum of A is the accumulation point of the eigenvalues: σessM=−5/2(ii)
It is not difficult to see that the subspaceC0≔CA⊕CD⊂DA0=DA0∩DC0⊕DB0∩DD0 is a core of A0(iii)
Now form the matrix elements−ϕk′′,ϕj, −ϕk′,ϕj, ϕk′,ϕj, −3/2ϕk,ϕj using the inner product in (6) with respect to the orthonormal basis in Equation (12) and consider the (infinite) block operator matrix(39)G≔−ϕk′′,ϕj−ϕk′,ϕjϕk′,ϕj−32ϕk,ϕj.
The defined matrixAk in (4) is obtained by taking the leading submatrices of the block G, with appropriate dimensions. Observe that −ϕk′′,ϕj=diagπ2,4π2,9π2,⋯,(40)−ϕk′,ϕj=0,ifk=j;−2kπ∫01coskπxsinjπxdx,ifk≠j;ϕk′,ϕj=0,ifk=j;2kπ∫01coskπxsinjπxdx,ifk≠j;and −3/2ϕk,ϕj=−3/2δk,j, which can be evaluated explicitlyFigure2 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 2
On the left-hand side, the variational method approximation to the Stokes operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


## 3.3.1. Application to Stokes-Type Operators

Consider the differential expression in the Hilbert spaceL220,1≔L20,1⊕L20,1, we introduce the matrix differential operator(18)A0≔A0B0C0D0=−d2dx2−ddxddx−32,on the domain(19)DA0≔H20,1∩H10,1⊕H10,1.The operator matrixA0 is not closed but closable [2]. In order to show that its closure is self-adjoint, we need the following:Remark 11.
(i)
Consider the operatorM given by(20)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,such that(21)My1y2=−y′1+y2′y′1−12y2.Lemma 12.
The operatorMin Equation ((21)) coincides with the adjoint of the block operator matrix in Equation ((18)).Proof.
Suppose that(22)g=g1g2∈DA~
with(23)u=u1u2∈L20,12.
Then(24)A0y,g=y,u,y=y1y2∈DA0=H20,1∩H10,1⊕H10,1or, equivalently,(25)−y′1+y2′,g1L20,1+y′1−12y2,g2L20,1=y1,u1L20,1+y2,u2L20,1for all y1,y2∈C0∞0,1 with compact support contained in 0,1. In particular, when y1=0, then Equation (25) becomes(26)−y2′,g1L20,1+−12y2,g2L20,1=y2,u2L20,1.The first part of the left-hand side of Equation (26) is a bounded linear functional of y2, which means that g1∈H10,1, and also, the second part of the left-hand side of equation (26) is a bounded linear functional of y2, which means that g2∈L20,1; again return to Equation (26), and integrating by part, we obtain(27)−y2,g1′−12g2=y2,u2y2∈C0∞0,1.
BecauseC0∞0,1 is a dense in L20,1, this means that u2=g′1−1/2g2. By the same argument, if we set y2=0y1∈H010,1∩H20,1, we find that(28)−y′′1,g1L20,1+y1′,g2L20,1=y1,u1L20,1.Again, return to Equation (28), and integrating by part, we obtain(29)−y1′xg1x¯x=0x=1+−y1,g1′+g2′¯=y1,u1.The first part of the left-hand side of Equation (29) is bounded only when g10=0=g11, and the second part is bounded linear functional of y1, which means that g1′+g2∈H10,1. Then, this implies that g∈DA~, and because C0∞0,1 is a dense in L20,1, this means that u1=−g1′+g2′ on 0,1. It follows that(30)u1u2=A~g1g2.Remark 13.
(i)
By the same argument of Lemma12, it is not difficult to see that the operator A0 in Equation (18) is symmetric in L20,12 with the domain(31)DM≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1,My1y2=−y′1+y2′y′1−12y2(ii)
BecauseM is a symmetric operator and has nonempty resolvent, then M self-adjoint. Thus, by [10, Theorem 5.4],M=A¯0The following result shows that the eigenvalues of the Stocks operator coincide with eigenvalues of the operatorM.Proposition 14.
LetA0be as in Equation((18)), thenσpA¯0=σpA0.Proof.
Letλ∈σpA¯0, then there exists an y=y1y2t∈DA¯0, y≠0, such that(32)A0¯y1y2=λy1y2.
We see that Equation (32) is equivalent to the following system of equations:(33)−y1′+y2′=λy1,(34)y′1−12y2=λy2,where y1y2t∈DA0¯ and(35)DA0¯≔y1y2:y1∈H010,1,y1′+y2∈H10,1,andy2∈L20,1.
Becausey1∈H10,1, then from Equation (33), we get y′1+y2∈H20,1 and Equation (34) can be written as(36)y′1+y2=λ+32y2;this implies that y2∈H20,1 for λ≠−3/2. Since H20,1 is a linear space then y′1∈H20,1, hence y1∈H30,1⊆H20,1, it follows that y=y1,y2t∈DA0, so λ is an eigenvalue of A0.
Conversely, ifλ∈σpA, then there exists an y=y1,y2t∈DA0⊆DA0¯,y≠0, such that(37)A0y1y2=λy1y2.
This means thaty is an eigenfunction of A0¯ since A0¯DA0=A0.Remark 15.
(i)
It may be shown thatA has two series of eigenvalues λk± given by(38)λk±=k2π2−3/2±3/2−k2π22+10k2π22,k=1,2,⋯.
The eigenvalues of the minus series are located in the interval (-5/2,-3/2] and convergent to -5/2 ask⟶∞. The eigenvalues of the plus series are located in the interval 0,∞ and convergent to ∞ as k⟶∞. The essential spectrum of A is the accumulation point of the eigenvalues: σessM=−5/2(ii)
It is not difficult to see that the subspaceC0≔CA⊕CD⊂DA0=DA0∩DC0⊕DB0∩DD0 is a core of A0(iii)
Now form the matrix elements−ϕk′′,ϕj, −ϕk′,ϕj, ϕk′,ϕj, −3/2ϕk,ϕj using the inner product in (6) with respect to the orthonormal basis in Equation (12) and consider the (infinite) block operator matrix(39)G≔−ϕk′′,ϕj−ϕk′,ϕjϕk′,ϕj−32ϕk,ϕj.
The defined matrixAk in (4) is obtained by taking the leading submatrices of the block G, with appropriate dimensions. Observe that −ϕk′′,ϕj=diagπ2,4π2,9π2,⋯,(40)−ϕk′,ϕj=0,ifk=j;−2kπ∫01coskπxsinjπxdx,ifk≠j;ϕk′,ϕj=0,ifk=j;2kπ∫01coskπxsinjπxdx,ifk≠j;and −3/2ϕk,ϕj=−3/2δk,j, which can be evaluated explicitlyFigure2 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 2
On the left-hand side, the variational method approximation to the Stokes operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


## 3.4. Analytical Estimates for Stokes Operator

In order to understand the results in Figure2, it is useful to find an analytical estimate for WqM. Let ζ→,η→∈DM, where(41)ζ→=ζ1ζ2,η→=η1η2with∥ζ→∥=∥η→∥=1,ζ→,η→=q and let(42)λ=Mζ→,η→=−D2ζ1,η1+−ζ2′,η1+ζ1′,η2+αζ2,η2,whereα=−3/2. Equation (42) gives, as an estimate for the first term on the right-hand side of (42),(43)∫01−ζ1′′η¯1dx=∫01ζ1′η¯1′dx≥qπ2∫01ζ1η¯1dx.For the second and the third term on the right-hand side of (42), the Cauchy-Schwarz inequality and Youngs inequality yield(44)Re∫01−ζ2′η¯1dx=Re∫01ζ2η¯1′dx≥−∫01ζ22+η1′2dx,(45)Re∫01ζ1′η¯2dx≥−∫01ζ1′2+η22dx.Combining Equations (44) and (45), we get that(46)Re∫01ζ2′η¯1dx+Re∫01ζ1′η¯2dx≥−q.The fourth term of the right-hand side of Equation (42) satisfies(47)Re∫01αζ2η¯2dx≥infReαq−∫01ζ1η¯1dx.Hence, from Equations (43), (46), and (47), we get that(48)Reλ≥qπ2∫01ζ1η¯1dx−q+infReαq−∫01ζ1η¯1dx.This simplifies to(49)Reλ≥qπ2∫01ζ1η¯1dx−q+q−∫01ζ1η¯1infReα=qπ2−infRα∫01ζ1η¯1dx+qinfReα−1.This yields(50)Reλ≥qα−1,ifπ2−α≥0;qπ2−1,ifπ2−α<0.For our example, these yieldReλ≥−5/2q.To estimateImλ, observe that(51)Imλ=∫01Imuζ2η¯2dx=0.This completes the estimates onWqM.


## 3.5. The Hain-Lüst Operator

This operator was introduced by Hain and Lüst in application to problems of magnetohydrodynamics [11], and the problems of this type were studied in [7, 8, 12]. Assume that w:0,1→0,∞, w~:0,1→0,∞, and u:0,1→ℂ are such that wx=1, w~x=1, ux=18e2πix−20, for each x∈0,1. We introduce the differential expression(52)τA~≔−d2dx2,τB~≔wx,τC~≔w~x,τD~≔ux.LetA,B,C,D be the operators in the Hilbert space L20,1 induced by the differential expressions τA~, τB~, τC~, and τD~ with domain(53)DA≔H20,1∩H010,1,DB=DC=DD≔L20,1.In the Hilbert spaceL220,1≔L20,1⊕L20,1, we introduce the matrix differential operator(54)A≔ABCD=−d2dx2wxw~xux,on the domain(55)DA≔H20,1∩H010,1⊕L20,1.Remark 16.
(i)
By [13], Corollary VII.2.7, the operator A≔−d2/dx2 with domain DA≔H20,1∩H010,1 is closed. Moreover, because DA⊂DC, then the operator C is A-bounded with relative bound 0. This follows since there is a γ>0 such that, for every ε>0,(56)∥Cf∥2≤γ∥Af,f∥≤γε∥Af∥2+ε−1∥f∥2.
On the other handDD⊂DB, then the operator B is D-bounded, we conclude that the operator matrix DA is diagonally dominant of order 0; it is closed by [14], Corollary 2.2.9 (i)(ii)
In Equation (54), since A is self-adjoint with the purely discrete spectrum, then the linear span CA=spanϕ1,ϕ2,⋯ is a core of A, where ϕk:k∈ℕ is an orthonormal basis in L20,1, and by the same argument because D in Equation (54) is bounded, then the linear span CD=spanϕ1,ϕ2,⋯ is a core of D. Hence, it is not difficult to see that the subspace C≔CA⊕CD⊂DA=DA∩DC⊕DB∩DD is a core of A. So the main Theorem 9 is applicable to this example(iii)
We may use the eigenfunctions in Equation (12) as basis elements for a discretization of the type discussed in Section 2, forming the matrix elements Aϕk,ϕj, wϕk,ϕj, w~ϕk,ϕj, uϕk,ϕj, with respect to the inner product in (6) and considering the infinite block matrix(57)Q≔Aϕk,ϕjwϕk,ϕjw~ϕk,ϕjuϕk,ϕj
The defined matrixAk in (4) is obtained by taking the leading submatrices of the block Q, with appropriate dimensions.
Observe thatAϕk,ϕj=diagπ2,4π2,9π2,⋯,wϕk,ϕj=diag1,1,1,⋯,w~ϕk,ϕj=diag1,1,1,⋯, and uϕk,ϕj=36∫01e2πixsinkπxsinjπxdx−20δk,j, which can be evaluated explicitly. If the operator A included a potential, for instance, then its eigenfunctions would not generally be explicitly computable. We could still use the functions ϕj in (12) as basis functions, but the matrix elements Aϕk,ϕj would have to be computed by quadrature and the corresponding matrix would no longer diagonal. Figure 3 shows attempts to compute the numerical approximation of the boundary of WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 3
On the left-hand side, the variational method approximation to the Hain-Lüst operator for computingq-numerical rangeWqA48 of A48 for q=1. The red dots are σA48. On the right-hand side, for 0<q<1, the variational method approximation for computing WqA48 of A48 for 0<q<1. The red dots are σA48 scaled by q.


## 3.6. Analytical Estimates for Non-Self-Adjoint Hain-Lüst Operator

In order to understand the results in Figure3, it is useful to find an analytical estimate for WqA. Let x→,y→∈DA, where(58)x→=x1x2,y→=y1y2with∥x→∥=∥y→∥=1, x→,y→=q and let(59)λ=Ax→,y→=−D2x1,y1+x2,y1+x1,y2+ux2,y2.The first term of (59) gives, as an estimate,(60)∫01x1′′y¯1dx≥qπ2∫01x1y¯1dx.For the second and the third term on the right-hand side of (59), the Cauchy-Schwarz inequality and Youngs inequality yield(61)R∫01x2y¯1dx≥−∫01x22+y12dx,(62)R∫01x1y¯2dx≥−∫01x12+y22dx.Combining Equations (61) and (62), we get that(63)R∫01x2y¯1dx+R∫01x1y¯2dx≥−q.The fourth term of the right-hand side of Equation (59) satisfies(64)R∫01ux2y¯2dx≥infRuq−∫01x1y¯1dx.Hence, from Equations (60), (63), and (64), we get that(65)Rλ≥qπ2∫01x1y¯1dx−q+infRuq−∫01x1y¯1dx.This simplifies to(66)Rλ≥qπ2∫01x1y¯1−q+q−∫01x1y¯1infRu=qπ2−infRu∫01x1y¯1dx+qinfRu−1.This yields(67)Rλ≥qinfRu−1,ifπ2−infRu≥0;qπ2−1,ifπ2−infRu<0.For our example, these yieldRλ≥−53q.To estimateImλ, observe that(68)Imλ=∫01Imux2y¯2dx≤supx∈0,1Imu∫01x1y¯1dx≤q25,Imλ=∫01Imux2y¯2dx≥infx∈0,1Imz∫01x2y¯2dx≥−q25,hence(69)−q25≤Imλ≤q25.This completes the estimates onWqA.The following result shows that theq-numerical range of Example 1, is unbounded.Remark 17.
Suppose that(70)x→=2sinkπx0,y→=2sinkπx0where x→,y→∈DA, with ∥x→∥=∥y→∥=1, and x→,y→=q then(71)Ax→,y→=22kπ2∫01sin2kπxdx=k2π2.But k∈ℤ can be arbitrarily large. This means that Rλ is unbounded above.


### 3.6.1. Application to Self-Adjoint Hain-Lüst Operator

In the Hilbert spaceH=L2−π,π⊕L2−π,π, we introduce the matrix differential operator(72)A~≔LBCV=−d2dx2+411z,where(73)zx=0,for−π≤x<0,2,for0≤x<π,on the domain(74)DA~≔H2−π,π∩H01−π,π⊕L2−π,π.Remark 18.
(i)
It is not difficult to see that the subspaceC≔CA⊕CV⊂DA~=DL∩DC⊕DB∩DV is a core of A~. So the main Theorem 9 is applicable to this example(ii)
The eigenvalues and normalized eigenfunctions of the operator−d2/dx2 in L2−π,pi are(75)λj=j24,ψjx=sinjx+π/2πforj=1,2,3,⋯Now form the matrix elementsAψk,ψj, Bψk,ψj, Cψk,ψj, Dψk,ψj, with respect to orthonormal basis ψjx=1/πsinjx+π/2 where j∈ℕ and consider the infinite block operator matrix(76)P≔Aψk,ψjBψk,ψjCψk,ψjDψk,ψj.The defined matrixAk in (4) is obtained by taking the leading submatrices of the block P, with appropriate dimensions. Observe that Lψk,ψj=diag1/4,1/2,⋯,Bψk,ψj=Cψk,ψj=diag1,1,1,⋯, and(77)Vψk,ψj=1π∫0πsinjx+π2sinkx+π2,ifk≠j;12,ifk=j;which can be evaluated explicitly. Figure 4 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 4
On the left-hand side, the variational method approximation to the Hain-Lüst operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


## 3.6.1. Application to Self-Adjoint Hain-Lüst Operator

In the Hilbert spaceH=L2−π,π⊕L2−π,π, we introduce the matrix differential operator(72)A~≔LBCV=−d2dx2+411z,where(73)zx=0,for−π≤x<0,2,for0≤x<π,on the domain(74)DA~≔H2−π,π∩H01−π,π⊕L2−π,π.Remark 18.
(i)
It is not difficult to see that the subspaceC≔CA⊕CV⊂DA~=DL∩DC⊕DB∩DV is a core of A~. So the main Theorem 9 is applicable to this example(ii)
The eigenvalues and normalized eigenfunctions of the operator−d2/dx2 in L2−π,pi are(75)λj=j24,ψjx=sinjx+π/2πforj=1,2,3,⋯Now form the matrix elementsAψk,ψj, Bψk,ψj, Cψk,ψj, Dψk,ψj, with respect to orthonormal basis ψjx=1/πsinjx+π/2 where j∈ℕ and consider the infinite block operator matrix(76)P≔Aψk,ψjBψk,ψjCψk,ψjDψk,ψj.The defined matrixAk in (4) is obtained by taking the leading submatrices of the block P, with appropriate dimensions. Observe that Lψk,ψj=diag1/4,1/2,⋯,Bψk,ψj=Cψk,ψj=diag1,1,1,⋯, and(77)Vψk,ψj=1π∫0πsinjx+π2sinkx+π2,ifk≠j;12,ifk=j;which can be evaluated explicitly. Figure 4 shows attempts to calculate WqAk for various k and q and also some attempts to estimate these sets by qualitative means, using existing theorems from the literature as well as the theorems proved above.Figure 4
On the left-hand side, the variational method approximation to the Hain-Lüst operator. The approximationAk is given by a real symmetric matrix with some eigenvalues λ1,λ2,⋯,λ28 with λ1≤λ2≤⋯λ28. The q-numerical rangeWqA28 of A28 for q=1 is the line segment λ1,λ28; the red dots are σA28. On the right-hand side, the q-numerical rangeWqA28 of A28 for 0<q<1 is the union of closed circular disk center at qx∗A28x with radius 1−q2rx where rx=∥A28x∥2−x∗A28x2. The small red circles inside the circular disk are σA28 scaled by q.


## 3.7. Analytical Estimates for Self-Adjoint Hain-Lüst Operator

In order to understand the results in Figure4, it is useful to find an analytical estimate for WqA~. By the same arguments in Section 3.6, it is not difficult to see that Reλ≥−3/4q, and(78)Imλ=∫−ππImzx2y2¯dx=0.


## 4. Conclusions

This paper illustrates the practical difficulties associated with the computation ofq-numerical ranges of operator matrices and block operator matrices of differential operators, even when good theoretical results are available to underpin the approximation procedure.



---

*Source: 6584805-2020-08-28.xml*