# Numerical Algorithms for Computing Eigenvalues of Discontinuous Dirac System Using Sinc-Gaussian Method

Authors: A. H. Bhrawy; M. M. Tharwat; A. Al-Fhaid
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The eigenvalues of a discontinuous regular Dirac systems with transmission conditions at the point of discontinuity are computed using the sinc-Gaussian method. The error analysis of this method for solving discontinuous regular Dirac system is discussed. It shows that the error decays exponentially in terms of the number of involved samples. Therefore, the accuracy of the new method is higher than the classical sinc-method. Numerical results indicating the high accuracy and effectiveness of these algorithms are presented. Comparisons with the classical sinc-method are given.


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## Body

## 1. Introduction

The mathematical modeling of many practical problems in mechanics and other areas of mathematical physics requires solutions of boundary value problems (see, e.g., [1–7]). Boundary value problems with discontinuity conditions arise in different branches of mathematics, mechanics, radio, electronics, geophysics, and other fields of natural science and technology. For example, discontinuous conditions inside an interval are connected with discontinuous or nonsmooth properties of media (see [8–11] and references there in).Sampling theory is one of the most important mathematical tools used in communication engineering since it enables engineers to reconstruct signals from some of their sampled data. A fundamental result in information theory is the Whittaker-Kotel'nikov-Shannon (WKS) sampling theorem [12–14]. It states that any f∈PWσ2, where PWσ2 is the space of all entire functions of exponential type σ>0 which lie in L2(ℝ) when restricted to ℝ, can be reconstructed from its sampled values {f(nπ/σ):n∈ℤ} by the formula
(1.1)f(λ)=∑n∈Z‍f(nπσ)sinc(σλ-nπ),λ∈C,
where
(1.2)sinc(σλ-nπ):={sin⁡(σλ-nπ)(σλ-nπ),λ≠nπσ,1,λ=nπσ.
Series (1.1) converges absolutely and uniformly on compact subsets of ℂ, and uniform on ℝ, cf. [15]. Expansion (1.1) is used in several approximation problems which are known as sinc methods, see, for example, [16–19]. In particular the sinc-method is used to approximate eigenvalues of boundary value problems, see, for examples, [20–24]. The sinc-method has a slow rate of decay at infinity, which is as slow as O(|λ-1|). There are several attempts to improve the rate of decay. One of the interesting ways is to multiply the sinc-function in (1.1) by a kernel function, see, for example, [25–27]. Let h∈(0,π/σ] and γ∈(0,π-hσ). Assume that Φ∈PWγ2 such that Φ(0)=1, then for f∈PWσ2 we have the expansion [28],
(1.3)f(λ)=∑n=-∞∞‍f(nh)sinc(h-1πλ-nπ)Φ(h-1λ-n).
The speed of convergence of the series in (1.3) is determined by the decay of |Φ(λ)|. But the decay of an entire function of exponential type cannot be as fast as e-c|x| as |x|→∞, for some positive c, [28]. In [29], Qian has introduced the following regularized sampling formula. For h∈(0,π/σ], N∈ℕ and r>0,
(1.4)(Gh,Nf)(x)=∑n∈ZN(x)‍f(nh)Sn(h-1πx)G(x-nh2rh),x∈R,
where G(t):=exp(-t2), which is called the Gaussian function, Sn(h-1πx):=sinc(h-1πx-nπ), ZN(x):={n∈ℤ:|[h-1x]-n|≤N} and [x] denotes the integer part of x∈ℝ, see also [30, 31]. Qian also derived the following error bound. If f∈PWσ2, h∈(0,π/σ] and a:=min{r(π-hσ),(N-2)/r}≥1, then
(1.5)|f(x)-(Gh,Nf)(x)|≤2σπ‖f‖2π2a2(2πa+e3/2r2)e-a2/2,x∈R.
In [28], Schmeisser and Stenger extended the operator (1.4) to the complex domain ℂ. For σ>0, h∈(0,π/σ] and ω:=(π-hσ)/2, they defined the operator [28],
(1.6)(Gh,Nf)(λ):=∑n∈ZN(λ)‍f(nh)Sn(πλh)G(ω(λ-nh)Nh),
where ℤN(λ):={n∈ℤ:|[h-1ℜλ+1/2]-n|≤N} and N∈ℕ. Note that the summation limits in (1.6) depend on the real part of λ. Schmeisser and Stenger, [28], proved that if f is an entire function of exponential type σ>0, then for h∈(0,π/σ), ω:=(π-hσ)/2, N∈ℕ, |ℑλ|<N, we have
(1.7)|f(λ)-(Gh,Nf)(λ)|≤2|sin⁡⁡(h-1πλ)|‖f‖∞e-ωNπωNβN(h-1Iλ),λ∈C,
where
(1.8)βN(t):=cosh⁡(2ωt)+2eωt2/NπωN[1-(t/N)2]+12[e2ωte2π(N-t)-1+e-2ωte2π(N+t)-1].The amplitude error arises when the exact valuesf(nh) of (1.6) are replaced by the approximations f~(nh). We assume that f~(nh) are close to f(nh), that is, there is ε>0, sufficiently small such that
(1.9)sup⁡n∈Zn(λ)|f(nh)-f~(nh)|<ε.
Let h∈(0,π/σ), ω:=(π-hσ)/2 and N∈ℕ be fixed numbers. The authors in [9] proved that if (1.9) is held, then for |ℑλ|<N, we have
(1.10)|(Gh,Nf)(λ)-(Gh,Nf~)(λ)|≤Aε,N(Iλ),
where
(1.11)Aε,N(Iλ)=2εe-ω/4N(1+Nωπ)exp⁡((ω+π)h-1|Iλ|).We are concerned with the computation of eigenvalues the Dirac system(1.12)u2′(x,λ)-p1(x)u1(x,λ)=λu1(x,λ),u1′(x,λ)+p2(x)u2(x,λ)=-λu2(x,λ),x∈[-1,0)∪(0,1],
with boundary conditions
(1.13)U1(u):=sin⁡αu1(-1,λ)+cos⁡αu2(-1,λ)=0,(1.14)U2(u):=sin⁡βu1(1,λ)+cos⁡βu2(1,λ)=0,
and transmission conditions
(1.15)U3(u):=u1(0-,λ)-δu1(0+,λ)=0,(1.16)U4(u):=u2(0-,λ)-δu2(0+,λ)=0,
where λ∈ℂ; the real valued function p1(·) and p2(·) are continuous in [-1,0) and (0,1], and have finite limits p1(0±):=limx→0±p1(x), p2(0±):=limx→0±p2(x); δ∈ℝ, α,β∈[0,π) and δ≠0.In this paper we will use the sinc-Gaussian sampling formula (1.6) to compute eigenvalues of the Dirac system (1.12)–(1.16). As expected, the new method reduced the error bounds remarkably, see examples at the end of this paper. Special attention is given to the comparison of the numerical results obtained by the new method with those found by classical sinc-method. We would like to mention that works in direction of computing eigenvalues with the sinc-Gaussian, are few, see for example, [9]. Also papers in computing of eigenvalues with discontinuous are few, see [10, 32]. However, the computing of eigenvalues by sinc-Gaussian technique which has discontinuity conditions, do not exist as for as we know.The paper is organized as follows: Section2 contains some preliminary results and the approximated values of the eigenvalues of the Dirac system with discontinuous. The method with error estimates are contained in Section 3. The last section involves some illustrative examples for showing the high accuracy of the proposed technique.


## 2. The Approximated Eigenvalues of Dirac System

In this section we derive approximate values of the eigenvalues of problem (1.12)–(1.16). Recall that problem (1.12)–(1.16) has denumerable set of real and simple eigenvalues, compare with [33, 34]. Let
(2.1)ϕ(⋅,λ)=(ϕ1(⋅,λ)ϕ2(⋅,λ)),ϕi(x,λ)={ϕi1(x,λ),x∈[-1,0),ϕi2(x,λ),x∈(0,1],i=1,2,
be the solution of (1.12) satisfying the following initial conditions:
(2.2)(ϕ11(-1,λ)ϕ12(0+,λ)ϕ21(-1,λ)ϕ22(0+,λ))=(cos⁡αδ-1ϕ11(0-,λ)-sin⁡αδ-1ϕ21(0-,λ)).
Since ϕ(·,λ) satisfies (1.13), then the eigenvalues of the problem (1.12)–(1.16) are the zeros of the function
(2.3)Δ(λ)=-δ2(sin⁡βϕ12(1,λ)+cos⁡βϕ22(1,λ)).
Notice that both ϕ(·,λ) and Δ(λ) are entire functions of λ and ϕ(·,λ) satisfies the system of integral equations, see [32, 35],
(2.4)ϕ11(x,λ)=cos⁡(λ(x+1)-α)-T-1,1ϕ11(x,λ)-T~-1,2ϕ21(x,λ),(2.5)ϕ21(x,λ)=sin⁡(λ(x+1)-α)+T~-1,1ϕ11(x,λ)-T-1,2ϕ21(x,λ),(2.6)ϕ12(x,λ)=1δϕ11(0-,λ)cos⁡(λx)-1δϕ21(0-,λ)sin⁡(λx)-T0,1ϕ12(x,λ)-T~0,2ϕ22(x,λ),(2.7)ϕ22(x,λ)=1δϕ11(0-,λ)sin⁡(λx)+1δϕ21(0-,λ)cos⁡(λx)+T~0,1ϕ12(x,λ)-T0,2ϕ22(x,λ),
where 𝒯-1,i, 𝒯~-1,i, 𝒯0,i and 𝒯~0,i, i=1,2, are the Volterra integral operators defined by
(2.8)T-1,1φ(x,λ):=∫-1x‍sin⁡λ(x-t)p1(t)φ(t,λ)dt,T-1,2φ(x,λ):=∫-1x‍sin⁡λ(x-t)p2(t)φ(t,λ)dt,T~-1,1φ(x,λ):=∫-1x‍cos⁡λ(x-t)p1(t)φ(t,λ)dt,T~-1,2φ(x,λ):=∫-1x‍cos⁡λ(x-t)p2(t)φ(t,λ)dt,T0,1φ(x,λ):=∫0x‍sin⁡λ(x-t)p1(t)φ(t,λ)dt,T0,2φ(x,λ):=∫0x‍sin⁡λ(x-t)p2(t)φ(t,λ)dt,T~0,1φ(x,λ):=∫0x‍cos⁡λ(x-t)p1(t)φ(t,λ)dt,T~0,2φ(x,λ):=∫0x‍cos⁡λ(x-t)p2(t)φ(t,λ)dt.For convenience, we define the constants(2.9)c1:=∫-10‍[|p1(t)|+|p2(t)|]dt,c2:=c1exp⁡(c1),c3:=∫01‍[|p1(t)|+|p2(t)|]dt,c4:=c2+2|δ|(1+c2).
Define f-1,i(·,λ) and f0,i(·,λ), i=1,2, to be
(2.10)f-1,1(x,λ):=T-1,1ϕ11(x,λ)+T~-1,2ϕ21(x,λ),f-1,2(x,λ):=T~-1,1ϕ11(x,λ)-T-1,2ϕ21(x,λ),(2.11)f0,1(x,λ):=T0,1ϕ12(x,λ)+T~0,2ϕ22(x,λ),f0,2(x,λ):=T~0,1ϕ12(x,λ)-T0,2ϕ22(x,λ).Lemma 2.1.
The functionsf-1,1(x,λ) and f-1,2(x,λ) are entire in λ for any fixed x∈[-1,0) and satisfy the growth condition
(2.12)|f-1,1(x,λ)|,|f-1,2(x,λ)|≤2c2e|Iλ|(x+1),λ∈C.Proof.
Sincef-1,1(x,λ)=𝒯-1,1ϕ11(x,λ)+𝒯~-1,2ϕ21(x,λ), then from (2.4) and (2.5) we obtain
(2.13)f-1,1(x,λ)=T-1,1cos⁡(λ(x+1)-α)+T~-1,2sin⁡(λ(x+1)-α)-T-1,1f-1,1(x,λ)+T~-1,2f-1,2(x,λ).
Using the inequalities |sinz|≤e|ℑz| and |cosz|≤e|ℑz| for z∈ℂ, leads for λ∈ℂ to
(2.14)|f-1,1(x,λ)|≤|T-1,1cos⁡(λ(x+1)-α)|+|T~-1,2sin⁡(λ(x+1)-α)|+|T-1,1f-1,1(x,λ)|+|T~-1,2f-1,2(x,λ)|≤e|Iλ|(x+1)∫-1x‍[|p1(t)||f-1,1(t,λ)|+|p2(t)||f-1,2(t,λ)|]e-|Iλ|(t+1)dt+2e|Iλ|(x+1)∫-1x‍[|p1(t)|+|p2(t)|]dt≤2c1e|Iλ|(x+1)+e|Iλ|(x+1)∫-1x‍[|p1(t)||f-1,1(t,λ)|+|p2(t)||f-1,2(t,λ)|]e-|Iλ|(t+1)dt.
The above inequality can be reduced to
(2.15)e-|Iλ|(x+1)|f-1,1(x,λ)|≤2c1+∫-1x‍[|p1(t)||f-1,1(t,λ)|+|p2(t)||f-1,2(t,λ)|]e-|Iλ|(t+1)dt.
Similarly, we can prove that
(2.16)e-|Iλ|(x+1)|f-1,2(x,λ)|≤2c1+∫-1x‍[|p1(t)||f-1,1(t,λ)|+|p2(t)||f-1,2(t,λ)|]e-|Iλ|(t+1)dt.
Then from (2.12) and (2.15) and Lemma 3.1 of [34, pp. 204], we obtain (2.12).In a similar manner, we will prove the following lemma forf0,1(·,λ) and f0,2(·,λ).Lemma 2.2.
The functionsf0,1(x,λ) and f0,2(x,λ) are entire in λ for any fixed x∈(0,1] and satisfy the growth condition
(2.17)|f0,1(x,λ)|,|f0,2(x,λ)|≤2c3c4e|Iλ|(x+1),λ∈C.Proof.
Sincef0,1(x,λ)=𝒯0,1ϕ11(x,λ)+𝒯~0,2ϕ21(x,λ), then from (2.6) and (2.7) we obtain
(2.18)f0,1(x,λ)=1δϕ11(0-,λ)T0,1cos⁡(λx)-1δϕ21(0-,λ)T0,1sin⁡(λx)-T0,1f-1,2(x,λ)+1δϕ11(0-,λ)T~0,2sin⁡(λx)+1δϕ21(0-,λ)T~0,2cos⁡(λx)+T~0,2f-1,2(x,λ).
Then from (2.4) and (2.5) and Lemma 2.1, we get
(2.19)f0,1(x,λ)≤1|δ||ϕ11(0-,λ)||T0,1cos⁡(λx)|+1|δ||ϕ21(0-,λ)||T0,1sin⁡(λx)|+|T0,1f-1,2(x,λ)|+1|δ||ϕ11(0-,λ)||T~0,2sin⁡(λx)|+1|δ||ϕ21(0-,λ)||T~0,2cos⁡(λx)|+|T~0,2f-1,2(x,λ)|≤2(c2+2|δ|(1+c2))c3e|Iλ|(x+1)=2c3c4e|Iλ|(x+1).
Similarly, we can prove that
(2.20)f0,2(x,λ)≤2c3c4e|Iλ|(x+1).


## 3. The Method and Error Analysis

In this section we derive the method of computing the eigenvalues of problem (1.12)–(1.16) numerically. We aim to approximate Δ(λ) and hence its zeros, that is, the eigenvalues. The idea is to split Δ(λ) into two parts, one is known and the other is unknown, but is an entire function of exponential type. Then we approximate the unknown part using (1.6) to get the approximate Δ(λ) and then compute the approximate zeros. Now, let us split Δ(λ) into
(3.1)Δ(λ):=K(λ)+U(λ),
where 𝒰(λ) is the unknown part involving integral operators
(3.2)U(λ):=δ2sin⁡βf0,1(1,λ)-δ2cos⁡βf0,2(1,λ)-δsin⁡(λ-β)f-1,1(0-,λ)-δcos⁡(λ+β)f-1,2(0-,λ),
and 𝒦(λ) is the known part
(3.3)K(λ):=-δsin⁡(2λ-α+β).
Then, from Lemmas 2.1 and 2.2, we have the following result.Lemma 3.1.
The function𝒰(λ) is entire in λ and the following estimate holds:
(3.4)|U(μ)|≤Me2|Iλ|,
where
(3.5)M:=4|δ|(c2+|δ|c3c4).Proof.
From (3.2), we have
(3.6)|U(λ)|≤|δ|2|sin⁡β||f0,1(1,λ)|+|δ|2|cos⁡β||f0,2(1,λ)|+|δ||sin⁡(λ-β)||f-1,1(0-,λ)|+|δ||cos⁡(λ+β)||f-1,2(0-,λ)|.
Using the inequalities |sinλ|≤e|ℑλ| and |cosλ|≤e|ℑλ| for λ∈ℂ, and Lemmas 2.1 and 2.2 imply (3.4).Then𝒰(λ) is an entire function of exponential type 2. In the following we let λ∈ℝ since all eigenvalues are real. Now we approximate the function 𝒰(λ) using the operator (1.6) where h∈(0,π/2] and ω:=(π-2h)/2 and then we obtain
(3.7)|U(λ)-(Gh,NU)(λ)|≤Th,N(λ),
where
(3.8)Th,N(λ):=2|sin⁡⁡(h-1πλ)|‖U‖∞e-ωNπωNβN(0),λ∈R.
The samples 𝒰(nh)=Δ(nh)-𝒦(nh), n∈ℤN(λ) cannot be computed explicitly in the general case. We approximate these samples numerically by solving the initial-value problems defined by (1.12) and (2.2) to obtain the approximate values 𝒰~(nh), n∈ℤN(λ), that is, 𝒰~(nh)=Δ~(nh)-𝒦(nh). Accordingly we have the explicit expansion
(3.9)(Gh,NU~)(λ):=∑n∈ZN(λ)‍U~(nh)Sn(πλh)G(ω(λ-nh)Nh).
Therefore we get, compare with (1.10),
(3.10)|(Gh,NU)(λ)-(Gh,NU~)(λ)|≤Aε,N(0),λ∈R.
Now let Δ~N(λ):=𝒦(λ)+(𝒢h,NU~)(λ). From (3.7) and (3.10) we obtain
(3.11)|Δ(λ)-Δ~N(λ)|≤Th,N(λ)+Aε,N(0),λ∈R.
Let λ* be an eigenvalue and λN be its desired approximation, that is, Δ(λ*)=0 and Δ~N(λN)=0. From (3.11) we have |Δ~N(λ*)|≤Th,N(λ*)+Aε,N(0). Now we define an enclosure interval Iε,N for λ*. Define the curves
(3.12)a±(λ)=Δ~N(λ)±(Th,N(λ)+Aε,N(0)).
The curves a+(λ), a-(λ) trap the curve of Δ(λ) for suitably large N. Hence the closure interval is determined by solving a±(λ)=0, gives an interval Iε,N:=[a-,a+]. Next we estimate the error |λ*-λN| for the eigenvalue λ*.Theorem 3.2.
Letλ* be an eigenvalue of (1.12)–(1.16) and λN be its approximation. Then, for λ∈ℝ, one has the following estimate:
(3.13)|λ*-λN|<Th,N(λN)+Aε,N(0)inf⁡ζ∈Iε,N|Δ′(ζ)|,
where the interval Iε,N is defined above.Proof.
Replacingλ by λN in (3.11) we obtain
(3.14)|Δ(λN)-Δ(λ*)|<Th,N(λN)+Aε,N(0),
where we have used Δ~N(λN)=Δ(λ*)=0. Using the mean value theorem yields that for some ζ∈Jε,N:=[min(λ*,λN),max(λ*,λN)],
(3.15)|(λ*-λN)Δ′(ζ)|≤Th,N(λN)+Aε,N(0),ζ∈Jε,N⊂Iε,N.
Since λ* is simple and N is sufficiently large, then infζ∈Iε,N|Δ′(ζ)|>0 and we get (3.13).


## 4. Numerical Examples and Comparisons

This section includes two examples illustrating the sinc-Gaussian method. In the following examples, we considerλk,N being the kth root of 𝒦(λ)+(𝒢h,N𝒰~)=0. Also, it is observed that the approximation λk,N and the exact solution λk are all inside the interval [a-,a+]. We indicate in these two examples the effect of the amplitude error in the proposed method by determining enclosure intervals for different values of ε. All examples are computed in [32] by using the classical sinc method. We see that the sinc-Gaussian method gives remarkably better results.Example 4.1.
Consider the system(4.1)u2′(x)-p(x)u1(x)=λu1(x),u1′(x)+p(x)u2(x)=-λu2(x),x∈[-1,0)∪(0,1],
subject to
(4.2)u1(-1)+u2(-1)=0,u1(1)+u2(1)=0,u1(0-)-2u1(0+)=0,u2(0-)-2u2(0+)=0.
Here
(4.3)p1(x)=p2(x)=p(x)={x2,x∈[-1,0)x,x∈(0,1],α=β=π/4 and δ=2.
Direct calculations give(4.4)K(λ)=-2sin⁡[2λ],Δ(λ)=-2sin⁡[56+2λ],
therefore the eigenvalues areλk=(6kπ-5)/12, k∈ℤ.
LetE and EG denote the absolute errors associated with the results of classical sinc method and sinc-Gaussian method, respectively. In Table 1, we give comparison between the absolute error of sinc-Gaussian and the classical sinc-method.Table 1
The absolute errors of sinc-Gaussian and the classical sinc-method withN=20, h=0.2, ω=1.3708.
λ
k
Sinc
Exactλk
Sinc-Gaussianλk,N
E
E
G
λ
-
2
-
3.5582594249860
-
3.55825932025645
-
3.5582593202564
1.04
×
10
-
7
2.75
×
10
-
14
λ
-
1
-
1.98746291522753
-
1.98746299346156
-
1.98746299346155
7.82
×
10
-
8
3.77
×
10
-
15
λ
0
-
0.41666654116886
-
0.41666666666666
-
0.41666666666666
1.25
×
10
-
7
1.83
×
10
-
15
λ
1
1.15412963731329
1.15412966012822
1.15412966012823
2.28
×
10
-
8
9.77
×
10
-
15
λ
2
2.72492581024868
2.72492598692312
2.72492598692315
1.76
×
10
-
7
3.28
×
10
-
14In Table2, we observe that the approximation λk,N and the exact solution λk are all inside the interval [a-,a+] for different values of ε.Table 2
The approximationλk,N and the exact solution λk are all inside the interval [a-, a+] for different values of ε.
λ
k
Exactλk
[a-, a+], ε=10-2
[a-, a+], ε=10-5
λ
k
,
N
λ
-
2
-
3.55825932025645
[-3.59648,-3.52003]
[-3.55829,-3.55822]
-
3.5582593202564
λ
-
1
-
1.98746299346156
[-2.02569,-1.94923]
[-1.98750,-1.98746]
-
1.98746299346155
λ
0
-
0.41666666666666
[-0.454895,-0.37843]
[-0.416704,-0.416628]
-
0.41666666666666
λ
1
1.15412966012822
[1.115901,1.19235]
[1.15409,1.15416]
1.15412966012823
λ
2
2.72492598692312
[2.68669,2.76315]
[2.72488,2.72496]
2.72492598692315
∥
𝒰
∥
∞
=
67.9396, βN(0)=1.2155.Example 4.2.
In this example we consider the system(4.5)u2′(x)-p(x)u1(x)=λu1(x),u1′(x)+p(x)u2(x)=-λu2(x),x∈[-1,0)∪(0,1],
subject to
(4.6)3u1(-1)+u2(-1)=0,u1(1)+3u2(1)=0,u1(0-)-3u1(0+)=0,u2(0-)-3u2(0+)=0,
where
(4.7)p1(x)=p2(x)=p(x)={x,x∈[-1,0)x2+1,x∈(0,1],α=π/3, β=π/6 and δ=3.
Direct calculations give(4.8)K(λ)=3sin⁡[π6-2λ],Δ(λ)=32(cos⁡[56+2λ]-3sin⁡[56+2λ]),
therefore the eigenvalues are λk=((6k+1)π-5)/12, k∈ℤ.
Table3 gives comparison between the absolute error of sinc-Gaussian and the classical sinc-method. Moreover, in Table 4 the approximation λk,N and the exact solution λk are all inside the interval [a-,a+] for different values of ε.Table 3
The absolute errors of sinc-Gaussian and the classical sinc-method withN=20, h=0.3, ω=1.2708.
λ
k
Sinc
Exactλk
Sinc-Gaussianλk,N
E
E
G
λ
-
2
−3.29646018123751
−3.29645993245731
−3.29645993245735
2.48
×
10
-
7
4.21
×
10
-
14
λ
-
1
−1.7256638185270
−1.7256636056624
−1.7256636056621
2.12
×
10
-
7
2.68
×
10
-
13
λ
0
−0.154867300702813
−0.154867278867517
−0.154867278867127
2.18
×
10
-
8
3.89
×
10
-
13
λ
1
1.41592898901685
1.41592904792737
1.41592904792752
5.89
×
10
-
8
1.44
×
10
-
13
λ
2
2.98672509121710
2.9867253747222
2.98672537472225
2.83
×
10
-
7
1.86
×
10
-
14Table 4
The approximationλk,N and the exact solution λk are all inside the interval [a-, a+] for different values of ε.
λ
k
Exactλk
[a-, a+], ε=10-2
[a-, a+], ε=10-5
λ
k
,
N
λ
-
2
−3.29645993245731
[
-
3.322822
,
-
3.270097
]
[
-
3.296486
,
-
3.296433
]
−3.29645993245735
λ
-
1
−1.7256636056624
[
-
1.752025
,
-
1.699301
]
[
-
1.725689
,
-
1.725637
]
−1.7256636056621
λ
0
−0.154867278867517
[
-
0.181229
,
-
0.128505
]
[
-
0.154893
,
-
0.1548409
]
−0.154867278867127
λ
1
1.41592904792737
[
1.389566,1.442291
]
[
1.415902,1.415955
]
1.41592904792752
λ
2
2.9867253747222
[
2.960363,3.013087
]
[
2.986699,3.013087
]
2.98672537472225
∥
𝒰
∥
∞
=
531.544, βN(0)=1.22382.



---

*Source: 925134-2012-07-30.xml*