Comparing the Energy of Convergence of One-Step and Two-Step Iterative Methods
- PDF / 99,776 Bytes
- 5 Pages / 594 x 792 pts Page_size
- 101 Downloads / 193 Views
COMPARING THE ENERGY OF CONVERGENCE OF ONE-STEP AND TWO-STEP ITERATIVE METHODS V. G. Prikazchikov1 and A. N. Khimich2
UDC 519.6
Abstract. The paper establishes the exact error energy compression ratio on each step of the two-step method, which shows a better convergence rate compared with the one-step method with optimal parameter. Keywords: iterative methods, convergence rate estimate, energy norm, characteristic equation, energy equivalence constants. INTRODUCTION The estimates of convergence rates of one-step and two-step iterative methods with constant parameters in energy norm are obtained in [1, pp. 285, 324] for the solution of operator equations in the Hilbert space. In [2, p. 519], the convergence of the two-step method is estimated in special norm with the compressibility parameter smaller than in [1]. In the present paper, we will use spectral decomposition, as well as in the analysis of spectral stability, and will estimate the convergence of the both methods in the energy norm related to the operator of the original problem. The estimate for the one-step method coincides with the estimate obtained in another way in [1]. The estimate for the two-step method coincides with the estimate in [2] obtained in another norm. The obtained estimate in energy norm does not follow from the estimate in [2], and it is more accurate than in [1] with the same norm. ONE-STEP METHOD Let us consider the functional equations in the Hilbert space
Au = f , A : H ® H , A ¹ A (t ) , A = A * > 0.
(1)
The one-step iterative method for the solution of (1) can be written as
B u t + Au = f , u (0) = u 0 , B : H ® H , B ¹ B (t ) , B = B * > 0, where
ut =
(2)
u n +1 - u n u$ - u u ( t + t) - u ( t ) . º º t t t
1
Taras Shevchenko National University of Kyiv, Kyiv, Ukraine, [email protected]. 2V. M. Glushkov Institute of Cybernetics, National Academy of Sciences of Ukraine, Kyiv, Ukraine, [email protected]. Translated from Kibernetika i Sistemnyi Analiz, No. 1, January–February, 2017, pp. 26–30. Original article submitted July 25, 2016. 1060-0396/17/5301-0021 ©2017 Springer Science+Business Media New York
21
For the error z = u - u , we have the equation
Bz t + Az = 0 .
(3)
Let { l k , j k }, k = 1, 2, ..., N , be the eigen pairs in the problem Aj = lBj . Due to the properties of operators A and B , the system of eigen elements {j k } is orthonormal, i.e., ( Bj i , j j ) = d ij . Let us represent the solution of (3) as ¥
å b k (t ) j k ,
z (t ) =
z ( t + t) =
k =1
Substitution into (3) yields
æ b$ k - b k + lk bk t k =1 è ¥
å çç
¥
å b k ( t + t) j k .
k =1
ö ÷ Bj k = 0 . ÷ ø
From here, since j k is basis, we obtain
b$ k = (1 - l k t) b k , k = 1, 2, ..., N . For the value of energy ( Az$ , z$ ) and ( Az, z ) , we obtain ( Az$ , z$ ) = orthonormality of { j k }. Similarly, ( Az, z ) =
¥
¥
k =1
k =1
å b$ 2k l k =å (1 - l k t) 2 b 2k l k . Here we use the
¥
å b 2k l k .
k =1
As a result, we obtain ( Az$ , z$ ) £ max(1 - l k t) 2 ( Az, z ) for "t > 0 . lk
As is known from [2, p. 456],
Data Loading...
