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1、精品论文http:/A modified filter method for analytic continuationMin-Hua Zhao Chu-Li Fu School of Mathematics and StatisticsLanzhou University, Lanzhou 730000, China The corresponding author. E-mail: Abstract1The problems of analytic continuation are frequently encountered in many practical ap-plications
2、. These problems are well-known to be severely ill-posed and therefore some regu- larization method is necessary to solve them. In this paper we proposed a modified filter method for stable analytic continuation. The convergence estimate under an appropriate choice of the regularization parameter is
3、 obtained. Some numerical tests show that the pro- posed method is effective and stable.Keywords: Modified filter, Analytic continuation, Regularization, Ill-posed problems.1 IntroductionWe consider the problem of analytic continuation of analytic function. The problem is important in practical appl
4、ications 1,2,3,4,5. Meanwhile, the problem is severely ill-posed.In this paper we consider the problem of analytic continuation of analytic function f (z) =f (x + iy) on a strip domain in the complex plane which appears in integration transform and medical image 5,6: = z = x + iy C|x R, |y| y0 , y0
5、0 is a positive constant,(1.1)where i is the imaginary unit. f (z)|y=0 = f (x) is known approximately and we would extend f analytically from this data to the whole domain . This problem has been considered in 6 by a mollification regularization method.Let g denote the Fourier transform of function
6、g(x) defined by1g() = 2Z +eixg(x)dx, (1.2)1精品论文and k k denotes the norm in Sobolev space H p (R) defined by14kgkp := Z1(1 + 2)p |g()|2 d 2 .(1.3)RWhen p = 0, k k0 := k k denotes the L2 (R) norm.2 Ill-posedness of problem and regularizationWe assume the exact data f (x) and the measured data f (x) al
7、l belong to L2(R) and satisfykf f k , (2.1)where we also assume 0 denotes the noisy level.f ( + iy) L2 (R) f or |y| y0. (2.2) Because the function f (z) is analytic in , the following series converges in :f (z) = f (x + iy) = X f(n)(x) (iy)nnX(iy) =Dn f (x), yy ,(2.3)xnwhere Dn = n .n!n=0n!n=0| | 0I
8、t can be imagined from the ill-posedness of numerical differentiation Dn f (x), n=1,2,., that the calculation of f (z) must possess more severe ill-posedness. In fact, taking the Fourier transform with respect to the variable x in (2.3), we can getf (+ iy)(, y) =X (iy)nn!n=0(i)n f()n=X (y)n!f()n=0=
9、ey f().It is equivalent that1f (x + iy) = 2Z +eixey f()d.(2.4)We know from (2.4) that, ey increases rather quickly for y 0, and y 0, and y 0, +. Small errors in high-frequency components can blow upand completely destroy the solution for |y| y0. However, in practice, the data at y = 0 isoften obtain
10、ed on the basis of reading of physical instrument which is denoted by f (x). In suchcases, we cant assume that its given with absolute accuracy. As the measured data f (x), itsFourier transform f () is merely in L2 (R) and such a decay of the exact data is not likely to精品论文occur in the Fourier trans
11、form f (). So the calculation of problem (2.4) is severely ill-posed and therefore its numerical simulation is very difficult.Due to the ill-posedness of problem (2.4) is caused by the perturbation of high frequencies of the measured data, we stabilize the solution by modifying the factor ey . We re
12、place ey y by e 1+| and define1f, (x + iy) = 2Z +eix e y 1+| f ()d. (2.5) y Note that if is chosen small, then for bounded |, e 1+| in (2.5) is close to ey in (2.4). y Further, as | +, e 1+| tends to be bounded. In the next sections we will study theproperty of function f, (x + iy) which is consider
13、ed as an approximation of f (x + iy). However,before studying such a property we need to impose an a priori bound which is necessary by 7on the exact solution of problem (2.4) at y = y0, i.e.,kf ( + iy0)k E, (2.6)andwhere E is a finite positive constant.3 Error estimatekf ( iy0)k E, (2.7)We now stud
14、y the properties of (2.5) considered as an approximation to (2.4). Because the results and the proof are analogous for the case 0 y y0 and y0 y 0, so we onlyconsider the domainWe can get fromand therefore+ = z = x + iy C|x R, 0 y y0.(3.1)f (+ iy0)(, y0) = ey0 f(),(3.2)f (+ iy)(, y) = ey f(),f() = ey
15、0 f (+ iy0 )(, y0). (3.3)In the following, we first consider the case 0 y y0 and then discuss the case y = y0.Theorem 3.1 Suppose f (z) = f (x + iy) is an analytic function on the domain + and it is known only on the line y = 0. Let f (x) and f (x) denote the exact and measured data of f (z) at y =
16、0, respectively, and they satisfy noisy level (2.1). Moreover, we assume the a prioribounds (2.7) and (2.8) hold. Let denotes the regularization parameter and define the Fourierregularization approximate f, (x + iy) of f (z) as:R =for 0 y y0, where1f, (x + iy) = 2Z +eixR f ()d (3.4)ey ,y 0,(3.5)Deno
17、tese 1+| , 0. Then, we canrewrite A() asytyt22A(t) = et(yy0 )(1 e 1+t ) et(yy0 ) et(yy0 )yt2 (3.11)Let h(t) := et(yy0 )t2 , then h(t) attains its maximum0 4hmax = e2(y1 + ty)2 ,(3.12)for t = 2 . Therefore we can estimate(note that is given in (3.7)y0 y4y0 A(t) e2 (yy)2 Cln(E/)(3.13)where C = 4yy0 .
18、The theorem now follows by combining (3.9), (3.10) and (3.13).e2 (y0 y)2Notice that the convergence estimate in Theorem 3.1 does not give any useful information on the continuous dependence of the solution at y = y0 . This is common in the theory of ill-posed problems, if we do not have additional c
19、onditions on the smoothness of the solution. To retain the continuous dependence of the solution at y = y0 , instead of (2.6), one has to introduce a stronger a priori assumptionkf ( + iy0 )kp E, p 0,(3.14)where k kp denotes the norm in Sobolev space H p (R) defined as (1.3).Theorem 3.2 Suppose f (x
20、 + iy) is the exact solution and f, (x + iy) is the regularization approximate defined as (3.4). Let f (x) and f (x) denote the exact and measured data of f (z) at y = 0, respectively, and they satisfy noisy level (2.1). Moreover, we assume the a priori bound (3.14) holds.If we take the regularizati
21、on parameter as below: =y0ln E (ln E )2p .(3.15)Then for p 0 we have1kf ( + iy0) f, ( + iy0)k where = maxp/2, y0p/2, y0 .(ln E )2p+ E. (3.16)Proof . Taking the similar procedure of the proof of Theorem (3.1), and using the a prioriassumption (3.14), we getkf ( + iy0 ) f, ( + iy0)k kf (+ iy0)(, y0) f
22、 (+ iy0)(, y0 )k+ kf (+ iy0 )(, y0) f,(+ iy0)(, y0 )k= k(ey0 R )ey0 f (+ iy0 )(, y0 )k+ kR f() R f ()kpp= k(1 R ey0 )(1 + 2 ) 2 (1 + 2 ) 2 f (+ iy0)(, y0 )k+ kR f() f ()k sup A()E + sup B ()RR sup A()E + sup B ()wherey0 0 01Case 1. s s0 := 2 . We can easily estimate A() in (3.17) asppA() (1 + s2 ) 2
23、 2 .(3.20)Case 2. 1 s s0. Using the inequality 1 er r (r 0), we estimate A() in (3.17) as2 ppA(s) y0 s(1 + s2) 2 y s2(1 + s2 ) 2 y s2p .1 + s 0 0If 0 p 2, note that s 1, we getA(s) y0. (3.22)Case 3. s 1. We estimate A(s) in (3.17) as2 ppA(s) y0s(1 + s2) 2 y (1 + s2) 2 y .(3.23)1 + s 0 0Now combining
24、 (3.20)-(3.23), we haveppA() max 2 , y0 2 , y0 := . (3.24)The theorem now follows by combining (3.17)-(3.24).Remark 3.1. Since the regularization parameter 0 as the measured error 0, we caneasily find that, for p 0, 0 ( 0). Thuslim kf ( + iy0) f, ( + iy0 )k = 0,p 0.0Note also that the regularization
25、 parameter in Theorem 3.2 differs from that in Theorem 3.1. Actually, we separately consider the case 0 y y0 (Theorem 3.1) and the case y = y0 (Theo- rem 3.2), in order to distinguish the following facts. For the case 0 y y0, the a priori bound (2.7) is sufficient. However, for the case y = y0 , the
26、 stronger a priori bound (3.14) must be im-posed. However, sometimes, it may be interesting if one uses only one regularization parameter in the both cases. The following remark shows that we can do it by making no more efforts.Remark 3.2. In Theorem 3.1, if instead of (3.7), we let the regularizati
27、on parameter be given as (3.15), and let the a priori bound E be given as (3.14), by following the procedure of the proof of Theorem 3.1, we can easily obtain1 y kf ( + iy0) f, ( + iy0)k E y0 y y0 (lnE 2py) y0 +C E EE ,(3.25)ln (ln )2pwhere p 0, C is the same constant of Theorem 3.1. Note that, for
28、the special case p = 0, this estimate just is that of Theorem 3.1.Combining the case of y0 y 0 and the case of 0 0 denotes the regularization parameter and define the Fourier regularization solution off (z) as:where1f, (x + iy) = 2Z +0eixR f ()d, for 0 y y ,R =ey ,y 0,ande 1+| , 0,1Z +ix wheref, (x
29、+ iy) = e2R f ()d, for y0 y 0,ey , 0, R = ye 1+| , 0,The we have the following result:If we take the regularization parameter =y0ln E (ln E )2p then there holds the estimate in |y|1 | y|E2p|y|CE|f ( + iy) f, ( + iy)| E y0 y0 (ln) y0 + EE ,|y| y0 , p 0. (3.26)wherelnC =4|y|y0 (ln )2p(3.27)4 Numerical
30、 testse2(y0 |y|)2In this section some numerical examples are devised to verify the validity of the regularization method given in section 2. In these numerical experiments we always take y0 = 1 and fix thedomainz = x + iy C|x| 10, 0 y 1.Suppose the vector F represent samples from the function f (x),
31、 then we add a perturbation to each the data and obtain the perturbation dataF = F + Randn(size(F ), (4.1)where the function randn() generates arrays of random numbers whose elements are normally distributed with mean 0, variance 2 = 1, and standard deviation = 1, andv 1u = |F F |l2 := uM +1|F (n) F
32、 (n)|2,(4.2)here we usually choose M = 100.t M + 1Xn=1In these numerical experiments, we computer the regularization parameter according to Theorem 3.1. We computer function f, (x + iy) given by (3.4) by using the Fast Fourier Transform and Inverse Fast Fourier Transform.Example 4.1 The functionf (z
33、) = ez2 = e(x+iy)2 = ey2 x2 (cos 2xy i sin 2xy)is an analytic function in the domain = z = x + iy C|x R, |y| 1with2f (z)|y=0 = ex L2(R)1.21=0.001 =0.0001 exact solution1.41.2=0.001 =0.0001 exact solution10.8Real part at y=0.10.60.4Real part at y=0.50.80.60.40.20.20 00.210 5 0 5100.210 5 0 5 10(a) (b)Figure 1: Example 4.1: (a)-(b) Real parts at y