# Optimal Stable Approximation for the Cauchy Problem for Laplace Equation

Authors: Hongfang Li; Feng Zhou
Journal: Advances in Mathematical Physics
 (2016)

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## Abstract

Cauchy problem for Laplace equation in a strip is considered. The optimal error bounds between the exact solution and its regularized approximation are given, which depend on the noise level either in a Hölder continuous way or in a logarithmic continuous way. We also provide two special regularization methods, that is, the generalized Tikhonov regularization and the generalized singular value decomposition, which realize the optimal error bounds.


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## Body

## 1. Introduction

The Cauchy problem for the Laplace equation in particular, and for other elliptic equations in general, occurs in the study of many practical problems in areas such as plasma physics [1], electrocardiology [2, 3], bioelectric field problems [4], nondestructive testing [5], magnetic recording [6], and the Cauchy problem for elliptic equations [7, 8]. These problems are known to be severely ill-posed [9], in the sense that the solution, if it exists, does not depend continuously on the Cauchy data in some natural norm (see, e.g., [5] and references therein). This is because the Cauchy problem is an initial value problem which represents a transient phenomenon in a time-like variable while elliptic equation describes steady-state processes in physical fields. A small perturbation in the Cauchy data, therefore, can affect the solution largely.In this paper, we will concretely consider the following Cauchy problem for Laplace equation in a strip:(1)uxx+uyy=0,0<x<1,-∞<y<∞,u0,y=gy,-∞<y<∞,ux0,y=0,-∞<y<∞,where we want to determine u(x,y) for 0<x≤1 from the data g(y) with corresponding measured data function gδ(y) [9–11].In [9], the authors constructed a regularization method based on the Meyer wavelet, but the convergence rate of the method was not obtained. In [10, 11], a modification method and a Fourier method for this problem were given, respectively, and some error estimates with satisfactory convergence rates were also proved. However, some very important and difficult problems in the theoretical study, that is, the optimal error bound, are not discussed. The major object of this paper is to give the optimal error bounds in theory for problem (1) by employing a regularized theory based on spectral decomposition. Meanwhile we provide two optimal regularized methods, that is, the generalized Tikhonov regularization and the generalized singular value decomposition method, which realize the optimal error bound.The motivation of this paper is inspired by Tautenhahn, in [12], where he discussed a Cauchy problem for the elliptic equation in a bounded domain and used the eigenvalues of the elliptic operator to express the exact solution of the problem. However, this method does not suit problem (1) in an unbounded strip region. Instead of using eigenvalues we employ the technique of Fourier transform.Letg(y) and gδ(y) denote the exact and measured data, respectively, which satisfy (2)g·-gδ·≤δ,where · denotes the L2-norm and the noise level δ>0 is determined by the accuracy of the instruments. We assume the g(y) and other functions appearing in this paper with respect to variable y belong to L2(R). Let (3)g^ξ≔12π∫-∞∞gye-iξydydenote the Fourier transform of function g(y). We now analyze problem (1) in the frequency space. Taking Fourier transform for problem (1) with respect to the variable y, we get (4)u^xxx,ξ-ξ2u^x,ξ=0,0<x<1,ξ∈R,u^0,ξ=g^ξ,ξ∈R,u^0,ξ=0,ξ∈R.The unique solution of (4) is [9–11] (5)u^x,ξ=g^ξcosh⁡xξ.Due to Parseval formula, u^(x,·)∈L2(R), and therefore (5) implies that g(ξ) must decay rapidly as ξ→∞. However, as the measurement data g^δ(ξ), we can not expect g^δ(ξ) has the same decay in high frequency components; that is, small errors in high frequency components can blow up and completely destroy the solution for 0<x≤1, noting that the factor cosh⁡(xξ) in (5) increases exponentially as ξ→∞, so the problem (1) is severely ill-posed.In order to obtain explicit stability estimate for problem (1), some “source condition” is needed. For this we introduce the Sobolev space Hr|r∈R+ according to H0=L2(R), Hr={v(y):vr<∞}, where (6)vr≔∫-∞∞1+ξ2rv^ξ2dξ1/2is the norm in Hr. We require the a priori smoothness condition for problem (1) concerning the unknown solution u(x,y) according to (7)ux,y∈Mp,E≔ux,·∈L2R∣u1,·p≤Efor  somep≥0.This paper is organized as follows: In Section2 we briefly recount some preliminary results, which are the basis of the discussion for other sections. In Section 3 we give the optimal error bounds between the exact solution and its regularized approximation, which depend on the noise level δ either in a Hölder continuous way or in a logarithmic continuous way. In Section 4 we discuss two concrete regularization methods, that is, the generalized Tikhonov regularization and the generalized singular value decomposition, where both regularization methods realize the optimal error bounds.


## 2. Preliminary Result

We consider arbitrary ill-posed inverse problem [12–17] (8)Ax=y,where A∈L(X,Y) is a linear injective bounded operator between infinite dimensional Hilbert spaces X and Y with nonclosed range R(A) of A. We assume that yδ∈Y are available noisy data with y-yδ≤δ. Any operator R:Y→X can be considered as a special method for solving (8), and the approximate solution of (8) is given by Ryδ. However, the convergent rate of Ryδ to x can be arbitrarily slow without assuming additional quantitative aprior restrictions on the unknown solution x, which is typical for ill-posed problem.Assume we want to solve (8), we have the a priori information that the exact solution satisfies a source condition; that is, x belongs to the source set (9)Mφ,E≔x∈E∣x=φA∗A1/2v,v≤E,where the operator function φ(A∗A) is well defined via spectral representation [13, 14]: (10)φA∗A=∫0aφλdEλ,where(11)A∗A=∫0aλdEλis the spectral decomposition of A∗A, {Eλ} denotes the spectral family of the operator A∗A, and a is a constant such that A∗A≤a. In the case when A:L2(R)→L2(R) is a multiplication operator, Ax(s)=γ(s)x(s), the operator function φ(A∗A) attains the form (12)φA∗Axs=φγs2xs.Let us assume thatR:Y→X is an arbitrary mapping to approximately recover x from yδ. Then the worst case error for R under the a priori information x∈Mϕ,E is [16, 17] (13)ΔRδ,Mφ,E,A≔sup⁡Ryδ-x∣x∈Mφ,E,Ax-yδ≤δ.This worst case error characterizes the maximal error of the method R if the solution x of problem (8) varies in the set Mϕ,E. The best possible worst case error (or the optimal bound) is defined as (14)Δδ,Mφ,E,A≔infR⁡Δδ,Mφ,E,A,where the minimum is taken over all methods R:Y→X. It can be shown (cf. [13, 18]) that the minimum in (14) is actually obtained and (15)Δδ,Mφ,E,A=ωδ,Mφ,E,Awith the modulus of continuity defined by (16)ωδ,Mφ,E,A≔supx∣x∈Mφ,E,Ax≤δ.In order to derive explicitly the optimal error bounds for the worst case errorΔR(δ,Mφ,E,A) defined in (13) and obtain optimality results for special regularization methods, we assume that the function φ in (9) satisfies the following assumption.Assumption 1 (see [13, 14]).
The functionφ(λ):(0,a]→(0,∞) in (14), where a is a constant such that A∗A≤a, is continuous and has the following properties:(i)
limλ→0φ(λ)=0,(ii)
φ(λ) is strong monotonically increasing on (0,a],(iii)
ρ(λ)≔λφ-1(λ):(0,φ(a)]→(0,aφ(a)] is strong convex.Under Assumption1, the next theorem gives us a formula for the optimal error bound.Theorem 2 (see [13, 14]).
LetMφ,E be given by (14), let Assumption 1 be satisfied, and let δ2/E2∈σ(A∗Aφ(A∗A)), where σ(A∗A) denotes the spectrum of operator A∗A; then (17)ωδ,x=Eρ-1δ2E2.In the following we consider two special methods: the method of generalized Tikhonov regularization and the method of generalized singular value decomposition.For the method of generalized Tikhonov regularization, a regularized approximationxαδ is determined by solving the minimization problem [13, 14, 16, 17] (18)minx∈XJαx,Jαx=Ax-yδ2+αφA∗A-1/2x2,or equivalently, by solving the Euler equation (19)A∗A+αφA∗A-1xαδ=A∗yδ,and the following statement holds.Theorem 3 (see [12, 13]).
LetMφ,E be given by (14), let Assumption 1 be satisfied, let φ(λ):(0,a]→R be two times differentiable, let ρ(λ) be strong convex on (0,φ(a)], and δ2/E2≤aφ(a). If the regularization parameter α is chosen optimally by (20)α=λ0φ-1λ0φ′φ-1λ0δE2withλ0=ρ-1δ2E2,then, for the Tikhonov regularized solution xαδ=Rαyδ defined by (18) or (19), the optimal error estimate (21)Δδ,Rα≤Eρ-1δ2E2holds.The regularized approximationxαδ based on the method of generalized singular value decomposition is given by (22)xαδ=gαA∗AA∗yδwithgαλ=1λ,forλ≥α,1α,forλ≤α.For this method the following result holds [13, 14].Theorem 4.
LetMφ,E be given by (9), let Assumption 1 be satisfied, let φ(λ):(0,a]→R be two times differentiable, let ρ(λ) be strong convex on (0,φ(a)], and δ2/E2≤aφ(a). If the regularization parameter α is chosen optimally by (23)α=φλ0+λ0φ′λ0φ′λ0withλ0φλ0=δE2,then for the regularized solution xαδ=Rαyδ defined by (22) the optimal error estimate (21) holds.


## 3. Optimal Error Bounds for Problem (1)

Let us formulate the problem (1) for identifying u(x,y) from unperturbed data u(0,y) as operator equation (24)Axux,y=u0,y,with a linear operator A(x)∈L(L2(R),L2(R)); then the equivalent operator equation of (24) in the Fourier domain is given by (25)A^xu^x,ξ=u^0,ξ,A^x=FAxF-1,where F:L2(R)→L2(R) is the (unitary) Fourier operator that maps any L2(R) function v(y) into its Fourier transform v^(ξ). From (5) and (25), we have (26)A^xu^x,ξ=1cosh⁡xξu^x,ξ,where A^(x):L2(R)→L2(R) is a multiplication operator. It is easy to know that the operator A^(x) is self-adjoint, so A^∗(x)A^(x):L2(R)→L2(R) is given by (27)A^∗xA^x=1cosh2⁡xξ.Due to Parseval formula,u(x,y)∈Mp,E is equivalent to u^(x,ξ)∈M^p,E, where (28)u^x,ξ∈M^p,E≔u^x,·∈L2R∣u^1,·p≤Efor  somep≥0with(29)u^1,·p≔∫-∞+∞1+ξ2u^1,ξ2dξ1/2.Note that the source condition (9) for problem (1) can be written as (30)Mφ,E=ux,·∈L2R∣φA∗xAx-1/2ux,·≤E,and then its equivalent form in Fourier frequency space is given by (31)M^φ,E=u^x,·∈L2R∣φA^∗xA^x-1/2u^x,·≤E.Due to the equivalence of conditions (30) and (7), we know the conditions (28) and (31) are equivalent and we have the following result.Proposition 5.
For operator equation (25), the set Mp,E given by (7) is equivalent to the general source set Mφ,E given by (30) provided φ=φ(λ) is given (in parameter representation) by (32)λξ=1cosh2⁡xξ,φξ=1+ξ2-pcosh2⁡xξcosh2⁡ξ,ξ∈R,0<x≤1.Proof.
From (5) we have (33)u^1,ξ=g^ξcosh⁡ξ,which gives (34)u^1,ξ=cosh⁡ξcosh⁡xξu^x,ξ,and then the inequality u(1,y)p≤E is equivalent to (35)1+ξ2p/2u^1,ξ=1+ξ2p/2cosh⁡ξcosh⁡xξu^x,ξ≤E.Note that the set M^p,E given by (28) is equivalent to the set M^φ,E given in (31), we know (35) is equivalent to (36)φA^∗xA^x-1/2u^x,ξ≤E.So, we obtain that the operator function φ=φ(λ) in (30) has the representation (37)φA^∗xA^x=1+ξ2-pcosh2⁡xξcosh2⁡ξ.Together with (27), φ is given by (32) in its parameter representation. The proof is complete.The functionφ=φ(λ) defined by (32) possesses the following properties.Proposition 6.
The functionφ=φ(λ) defined by (32) is continuous and satisfies the properties:
(i)limλ→0φ(λ)=0.
(ii)φ(λ) is strong monotonically increasing.
(iii)ρ(λ)=λφ-1(λ) is strong monotonically increasing and possesses the parameter representation: (38)λξ=1+ξ2-pcosh2⁡xξcosh2⁡ξ⁡,ρξ=1+ξ2-p1cosh2⁡ξ,ξ∈R,0<x≤1.
(iv)ρ-1(λ) is strong monotonically increasing and possesses the parameter representation: (39)λξ=1+ξ2-p1cosh2ξ,ρ-1ξ=1+ξ-pcosh2xξcosh2ξ,ξ∈R,0<x≤1.
(v) For the inverse functionρ-1(λ) of ρ(λ) the following holds: (40)ρ-1λ=λ41-xln⁡1λ-2px1+o1,forλ⟶0for any fixed x∈(0,1].
(vi) The functionρ defined by (38) is strong convex for p=0, x∈(0,1) and p>0, x∈(0,1].Proof.
The continuity of functionφ(λ) is obvious.
(i) From (32) we have (41)λ˙ξ=-2xsinh⁡xξcosh3⁡xξ=-2λξxtanh⁡xξ.It is easy to know that λ˙(ξ)≤0 for ξ≥0; that is, the function λ(ξ) is decreasing and limξ→+∞λ(ξ)=0, so, (42)limλ→0⁡φλ=limξ→+∞⁡φξ=0.Moreover, when ξ≤0, λ˙(ξ)≥0, we know λ(ξ) is increasing and limξ→-∞λ(ξ)=0, (43)limλ→0⁡φλ=limξ→-∞⁡φξ=0.From (42) and (43), we have limλ→0φ(λ)=0.
(ii) From (32), we know (44)φ˙ξ=φξ-2pξ1+ξ2+2xtanh⁡xξ-2tanh⁡ξ.Together with (41), we have (45)φ′λ=φ˙ξλ˙ξ=φξλξ2pξ/1+ξ2+2tanh⁡ξ-2xtanh⁡xξ2xtanh⁡xξ. It is easy to know that φ(ξ) and λ(ξ) are all even functions about variable ξ; therefore φ′(λ) is also so. Here it is only to consider the case ξ≥0. Note that the function f(t)≔tanh⁡t is strong monotonically increasing; together with (32) we know φ′(ξ)>0 for ξ>0. Moreover, a straightforward computation gives (46)limξ→0+⁡pξ/1+ξ2+tanh⁡ξ-xtanh⁡xξxtanh⁡xξ=p-x2+1x2.So, we can easily see φ′(λ)>0 for p=0, x∈(0,1) and p>0, x∈(0,1]; that is, when p=0, x∈(0,1) and p>0, x∈(0,1], φ(λ) is strong monotonically increasing.
(iii) From (ii) we know thatφ-1(λ) is strong monotonically increasing for p=0, x∈(0,1) and p>0, x∈(0,1]. Therefore, when p=0, x∈(0,1) and p>0, x∈(0,1], ρ(λ)=λφ-1(λ) is also strong monotonically increasing. From (32), it is easy to know that φ-1(λ) has the parameter representation (47)λξ=1+ξ2-pcosh2⁡xξcosh2⁡ξ⁡,φ-1ξ=1cosh2⁡xξ,ξ∈R,0<x≤1.So, the parameter expression (38) of function ρ(λ) holds.
(iv) According to (iii),ρ-1(λ) is also strong monotonically increasing and its parameter representation (39) can be obtained from (38) immediately.
(v) From (39) we have (48)ln⁡1λ=pln⁡1+ξ2+2ξ+ln⁡1+e-4ξ+2e-2ξ2=2ξ1+01forξ⟶∞,and consider(49)e2ξ=4λ1+ξ2-p1+o1forξ⟶∞,thatis,λ⟶0. It is easy to see that (50)ξ=ln⁡1λ1+o1,forλ⟶0. Inserting (49) and (50) in (39), we have (51)ρ-1ξ=1+ξ2-pcosh2⁡xξcosh2⁡ξ=λξe2xξ+e-2xξ+24=λ4e2xξ1+o1forλ⟶0,that  is,forξ⟶∞=λ41-x1+ξ2-px1+o1forλ⟶0,that  is,forξ⟶∞=λ41-xln⁡1λ-2px1+o1forλ⟶0.Indeed, if we denote (52)Fλ≔ρ-1λλ4x-1ln⁡1λ2px,it is easy to prove that (53)limλ→0⁡Fλ=limξ→∞⁡ρ-1ξλξ4x-1ln⁡1λξ2px=1.So, the representation (40) holds.
(vi) We know that the functionρ(λ) is strong convex if and only if ρ′′(λ)>0. Denoting λ(ξ)=ρ(ξ)r(ξ) with r(ξ)=cosh2⁡(xξ), from (38) we have (54)ρ′′λ=ρ¨λ˙-ρ˙λ¨λ˙3=ρ¨ρr˙-2ρ˙2r˙-ρρ˙r¨ρ˙r+ρr˙3,where (55)ρ˙=ρ˙ξ=-2ρξpξ1+ξ2+tanh⁡ξ,ρ¨=ρ¨ξ=-2ρξ-2pξ1+ξ2+tanh⁡ξ2+p1-ξ21+ξ22+1cosh2⁡ξ,r˙=r˙ξ=xsinh⁡xξ,r¨=r¨ξ=2x2cosh⁡2xξ.It is easy to know that ρ′′(λ) is even function about variable ξ; therefore we only need to consider the case ξ≥0.
1st Case(ξ>0). In this case, λ′(ξ)<0. Then ρ′′(λ)>0 is equivalent to ρ¨λ˙<ρ˙λ¨, that is, (56)ρρ¨-2ρ˙2<ρρ˙r¨r˙.A straightforward computation shows that (56) is equivalent to (57)ξ1+ξ22p2+p2ξ1+ξ2tanh⁡ξ+1-ξ221+ξ22-xξ1+ξ2coth⁡2xξ+tanh2⁡ξ+12cosh2⁡ξ-xtanh⁡ξcoth⁡2xξ>0.From (57) we have (58)tanh2⁡ξ+12cosh2⁡ξ-xtanh⁡ξcoth⁡2xξ=2sinh2⁡ξ+12cosh2⁡ξ-2sinh2⁡xξ+12cosh2⁡xξ·xtanh⁡ξtanh⁡xξ=2sinh2⁡ξ+12cosh2⁡ξ-2sinh2⁡xξ+12cosh2⁡xξ·tanh⁡ξ/ξtanh⁡xξ/xξ. Note that the function f1(t)≔2sinh2⁡t+1/2cosh2⁡t is monotonically increasing and f2(t)≔tanh⁡t/t is monotonically decreasing, and we obtain that 57′tanh2⁡ξ+12cosh2⁡ξ-xtanh⁡ξcoth⁡2xξ>0for all x∈(0,1]. So, when ξ>0, the function ρ(λ) is strong convex if (59)ξ1+ξ22p2+p2ξ1+ξ2tanh⁡ξ+1-ξ221+ξ22-xξ1+ξ2coth⁡2xξ>0,that is, (60)p+21+ξ2ξtanh⁡ξ+1-ξ22ξ2-x1+ξ2ξcoth⁡2xξ>0.Note that the function f3(x)≔x1+ξ2/ξcoth⁡(2xξ) is increasing about x for ξ>0; by numerical computation we know (61)21+ξ2ξtanh⁡ξ+1-ξ22ξ2-x1+ξ2ξcoth⁡2xξ≥13.So, when p≥0, (60) holds naturally. With (57) and 57′ together, we know that ρ(λ) is strong convex for p≥0, ξ>0 and x∈(0,1].
2nd Case (ξ=0). An elementary calculation shows (62)limξ→0+⁡ρ¨ρr˙-2ρ˙2r˙-ρρ˙r¨ρ˙r+ρr˙3=16p-3p+1+2x4-32x2-x46-2p+1+2x2>0for all p≥0 and x∈(0,1). From above, (vi) is proved.
The proof is finished.Now we formulate our main result of this section concerning the best possible worst case errorω(δ,x) defined by (17) for identifying solution u(x,y) of problem (1) from noisy data uδ(0,y) under the condition (2) and u(x,y)∈Mp,E where the set Mp,E is given by (7). We denote (63)ωδ,x≔Δδ,Mp,E,Axwith Δ(δ,Mp,E,A(x)) given as Δ(δ,Mφ,E,A(x)) in (14). Due to Parseval identity and the equivalence of M^p,E which are given by (28) and (31), respectively, we have (64)ωδ,x=Δδ,M^p,E,A^x=Δδ,M^φ,E,A^x=ωδ,M^p,E,A^x.Applying Theorem2 and Proposition 6 we have the following result.Theorem 7.
Letδ2/E2≤1, and then consider the following.
(i) In casep=0 and 0<x<1, the following holds: (65)ωδ,x=Exδ21-x1+o1,forδ⟶0(Hölder stability).
(ii) In casep>0 and x=1, the following holds: (66)ωδ,1=Eln⁡Eδ-p1+o1,forδ⟶0(logarithmic stability).
(iii) In casep>0 and 0<x<1, the following holds: (67)ωδ,x=Exδ21-xlnEδ-px1+o1,forδ⟶0.Proof.
From (32), we have (68)A^x∗A^xφA^x∗A^xξ=1+ξ2-pcosh2⁡ξ.It is easy to know the function f4(ξ)≔1+ξ2-p/cosh2⁡ξ takes maximum at ξ=0, and limξ→∞f4(ξ)=0, so σ(A^(x)∗A^(x)φ(A^(x)∗A^(x)))⊆(0,1]. Due to Theorem 2 and (40) given in Proposition 6, we know if δ2/E2≤1.
(i) Forp=0 and 0<x<1, (69)ωδ,x=E·λ41-xλ=δ2/E21/21+o1=Exδ21-x1+o1,forδ⟶0 (Hölder stability).
(ii) In casep>0 and x=1, (70)ωδ,1=E·ln⁡1λ-2pλ=δ2/E21/21+o1=Eln⁡Eδ-p1+o1,forδ⟶0 (logarithmic stability).
(iii) In casep>0 and 0<x<1, (71)ωδ,x=E·λ41-xln1λ-2pxλ=δ2/E21/21+o1=Exδ21-xln⁡Eδ-px1+o1,forδ⟶0.The proof is complete.


## 4. Optimal Regularization Methods

In this section we consider two special regularization methods, apply them to problem (1), and show how to choose the regularization parameter such that it guarantees the optimal error bounds given by (65)–(67).The method of generalized Tikhonov regularization (18) consists in the determination of a regularized approximation uαδ=uαδ(x,y) by solving the minimization problem:(72)minu^∈M^φ,EJαu,Jαu^=1cosh⁡xξu^x,ξ-u^δ0,ξ2+α1+ξ2p/2cosh⁡ξcosh⁡xξu^x,ξ2,or, equivalently, it is the solution of Euler equation: (73)1cosh2⁡xξ+α1+ξ2pcosh2⁡ξcosh2⁡xξu^αδx,ξ=1cosh⁡xξu^δ0,ξ.Applying Theorem3 and Proposition 5 to problem (1), we obtain the following result.Theorem 8.
Consider the operator equation (24) and assume its unknown solution u(x,y)∈Mp,E given by (7). If δ2/E2≤1 holds, then the method of generalized Tikhonov regularization (72) or (73) is optimal on Mp,E provided that the regularization parameter α is chosen optimally by (74)α0=xtanh⁡xξ0pξ0/1+ξ02+tanh⁡ξ0-xtanh⁡xξ0δE2,where ξ0 is the unique solution of the equation (75)1+ξ2-pcosh2⁡ξ=δ2E2.
(i) In the casex=1 and p>0 the following holds: (76)α0=1pln⁡Eδδ221+o1,forδ⟶0.
(ii) In the casep≥0 and 0<x<1, the following holds: (77)α0=x1-xδ221+o1,forδ⟶0.Furthermore, the optimal error estimate uαδ(x,y)-u(x,y)≤ω(δ,x) holds, where ω(δ,x) is given by (65)–(67), respectively.Proof.
From Theorem3, it follows that the optimal regularization parameter α is given by (20) with φ(λ) given by (32), which is equivalent to (78)α0=φλ0λ0φ′λ0δE2=φξ0λ˙ξ0λξ0φ˙ξ0δE2=xtanh⁡xξ0pξ0/1+ξ02+tanh⁡ξ0-xtanh⁡xξ0δE2,with(79)λ0=ρ-1δ2E2=1+ξ2-pcosh2⁡ξ;that is, ξ0 is the unique solution of (75).
(i) In the casex=1 and p>0, note that ξ0→∞ for δ→0, and tanh⁡x→1 for x→∞, and from (78) and (50), we have(80)α0=xtanh⁡ξ0pξ0/1+ξ02δE2=ξ0pδE21+o1=1pln⁡Eδ1+o1forδ⟶0.
(ii) In the casep≥0 and 0<x<1, the following holds: (81)α0=xtanh⁡xξ0pξ0/1+ξ02+tanh⁡ξ0-xtanh⁡xξ0δE2=x1-xδE21+o1,forδ⟶0. The proof is finished.Now we consider the method of generalized singular value decomposition. Due to (22), the regularized approximation uαδ(x,y) is given by (82)u^αδx,ξ=A^∗xA^x-1A^∗xu^δ0,ξ,A^∗xA^x≥α,1αA^∗xu^δ0,ξ,A^∗xA^x≤α,=cosh⁡xξu^δ0,ξ,1cosh2⁡xξ≥α,1α1cosh⁡xξu^δ0,ξ,1cosh2⁡xξ≤α.Applying Theorems4 and 7 to problem (1), we have the following.Theorem 9.
Consider the operator equation (24) and assume its unknown solution u(x,y)∈Mp,E given by (7). If δ2/E2≤1 holds, then the method of generalized singular value decomposition (82) is optimal on Mp,E provided that the regularization parameter α is chosen optimally by (83)α0=pξ0/1+ξ02tanh⁡ξ0cosh2⁡xξ0tanh⁡ξ0-xtanh⁡xξ0,where ξ0 is the unique solution of (75).
(i) In the casex=1 and p>0, the following holds: (84)α0=1pln⁡Eδ2p+1δE21+o1forδ⟶0.
(ii) In the case0<x<1,p≥0, the following holds:(85)α0=41-x1-xln⁡Eδ2pxδE2x1+o1,forδ⟶0.Furthermore, the optimal error estimate uαδ(x,y)-u(x,y)≤ω(δ,x) holds, where ω(δ,x) is given by (65)–(67), respectively.Proof.
From Theorem4, we know that the optimal regularization parameter α is given by (23) with φ(λ) given by (32), which is equivalent to (86)α0=φλ0+λ0φ′λ0φ′λ0=φξ0+λξ0φ˙ξ0/λ˙ξ0φ˙ξ0/λ˙ξ0=λ˙ξ0φξ0+λξ0φ˙ξ0φ˙ξ0=pξ0/1+ξ02+tanh⁡ξ0cosh2⁡xξ0pξ0/1+ξ02+tanh⁡ξ0-xtanh⁡xξ0,where ξ0 is the unique solution of (75).
(i) In the casex=1 and p>0, note that ξ0→∞ for δ→0, and tanh⁡x→1 for x→∞, and from (83) and (75), we have(87)α0=pξ0/1+ξ02+tanh⁡ξ0cosh2⁡xξ0pξ0/1+ξ02=pξ0/1+ξ02+11+ξ0p=pξ0/1+ξ02δE21+o1forξ0⟶∞,thatis,δ⟶0=1pξ02p+1δE21+o1=1pln⁡Eδ2p+1δE21+o1forδ⟶0.
(ii) In the casep≥0 and 0<x<1, from (75) and (83), we have(88)α0=tanh⁡ξ0cosh2⁡xξ0tanh⁡ξ0-xtanh⁡xξ0=4e-2xξ011-x1+o1,forξ0⟶∞,thatis,δ⟶0=41-x1-x1+ξ02pxδE2x,forξ0⟶∞,thatis,δ⟶0=41-x1-xln⁡Eδ2pxδE2x1+o1,forδ⟶0. Furthermore, the optimal error estimate uα0δ(x,y)-u(x,y)≤ω(δ,x) holds, where ω(δ,x) is given by (65)–(67), respectively. This completes the proof.



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*Source: 1468634-2016-06-06.xml*