On matrix equations X ± A X 2 A = I

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1 Linear Algebra and its Applications On matrix equations X ± A X 2 A = I I.G. Ivanov,V.I.Hasanov,B.V.Minchev Faculty of Mathematics and Informatics, Shoumen University, Shoumen 9712, Bulgaria Received 7 March 1999; accepted 11 September 2 Submitted by F. Uhlig Abstract The two matrix equations X + A X 2 A = I and X A X 2 A = I are studied. We construct iterative methods for obtaining positive definite solutions of these equations. Sufficient conditions for the existence of two different solutions of the equation X + A X 2 A = I are derived. Sufficient conditions for the existence of positive definite solutions of the equation X A X 2 A = I are given. Numerical experiments are discussed. 21 Elsevier Science Inc. All rights reserved. AMS classification: 65F1 Keywords: Matrix equation; Positive definite solution; Iterative method 1. Introduction We consider the two matrix equations X + A X 2 A = I 1 and X A X 2 A = I, 2 where I is the n n unit matrix and A is an n n invertible matrix. Several authors [2 5] have studied the matrix equation X ± A X 1 A = I and they have obtained theoretical properties of these equations. Nonlinear matrix equations of type 1 and 2 arise in dynamic programming, stochastic filtering, control theory and statistics [4]. Corresponding author. address: i.gantchev@fmi.shu-bg.net I.G. Ivanov, vejdi@dir.bg V.I. Hasanov /1/$ - see front matter 21 Elsevier Science Inc. All rights reserved. PII:S

2 28 I.G. Ivanov et al. / Linear Algebra and its Applications In many applications one must solve a system of linear equations [1] Mx = f, where the positive definite matrix M arises from a finite differenceapproximationto an elliptic partial differential operator. As an example, let I A M = A. I Solving this linear system leads us to computing a solution of one of the equations X + A X 1 A = I, X + A X 2 A = I, or X A X 2 A = I. This approach was considered in [6]. The system Mx = f can be solved by Woodbury s formula [7]. In this paper we show that there are two positive definite solutions of Eq. 1. We study the existence question and how to find a Hermitian positive definite solution X of 2. We propose two iterative methods which converge to a positive definite solution of 2. When norm of A is small enough the considered iteration methods converge. The rate of convergence of these methods depends on two parameters. Numerical examples are discussed and some results of the experiments are given. We start with some notations which we use throughout this paper. We shall use A to denote the spectral norm of the matrix A, i.e., A = max i λ i,wheretheλ i are the eigenvalues of AA. Let matrices P and Q be Hermitian. The notation P> QP Q means that P Q is positive definite semidefinite. The assumption P Q>implies P 1 Q 1 and P Q. 2. The matrix equation X + A X 2 A = I In this section we generalize theorems which were proved in [6]. We shall assume that A 2. In [6] two iterative methods converging to a positive definite 27 solution were investigated. We consider the matrix sequence X = γi, X k+1 = AI X k 1 A, k =, 1, 2,... 3 For this iterative method we have proved [6, Theorem 2]. Theorem 1. If there exist the numbers α, β with <α<β<1 for which the inequalities α 2 1 αi < AA <β 2 1 βi are satisfied, then Eq. 1 has a positive definite solution. We can change this theorem slightly. Consider the scalar function ϕx = x 2 1 x.wehave 2 max ϕx = ϕ. [,1] 3

3 I.G. Ivanov et al. / Linear Algebra and its Applications This function is monotonically increasing where x, 2 3 and monotonically decreasing where x 2 3, 1. We obtain the following. Theorem 2. If there exist numbers α, β with <α β 2 3 for which the inequalities α 2 1 αi AA β 2 1 βi are satisfied, then Eq. 1 has a positive definite solution X such that αi X βi. We are proving the new theorem: Theorem 3. Let α and β be solutions of scalar equations α 2 1 α = min λ i AA and β 2 1 β = max λ i AA, respectively. Assume < α β 2 3. Consider {X k } defined by 3. Then: i If γ [, α], then {X k } is monotonically increasing and converges to a positive definite solution X α. ii If γ [ β, 2 3 ], then {X k} is monotonically decreasing and converges to a positive definite solution X β. iii If γ α, β and β 2 /2 α1 β < 1, then {X k } converges to a positive definite solution X γ. Proof. Since the function ϕx = x 2 1 x is monotonically increasing where x [, 2 3 ] we have <α α β β 2 3 and the inequalities α 2 1 αi AA β 2 1 βi are satisfied. i Assume γ [, α]. Hence X = γi βi.wehave X 1 = AI γi 1 A α 2 1 α I X, 1 γ β X 1 = AI γi 1 A 2 1 β I βi. 1 γ Thus X X 1 βi. Assume that X k 1 X k βi.forx k+1 compute X k+1 = AI X k 1 A AI X k 1 1 A = X k, X k+1 = AI X k 1 A β AI βi 1 A 2 1 β 1 β I = βi. Hence {X k } is monotonically increasing and bounded from above by the matrix βi. Consequently the sequence {X k } converges to a positive definite solution X α.

4 3 I.G. Ivanov et al. / Linear Algebra and its Applications ii Assume γ [ β, 2 3 ]. Hence X = γi αi.wehave β X 1 = AI γi 1 A 2 1 β I γi, 1 γ X 1 = AI γi 1 A α 2 1 α I αi. 1 γ Thus X X 1 αi. Assume that X k 1 X k αi. ForX k+1 compute X k+1 = AI X k 1 A AI X k 1 1 A = X k, X k+1 = AI X k 1 A AI αi 1 A α 2 1 α 1 α I = αi. Hence {X k } is monotonically decreasing and bounded from below by the matrix αi. Consequently the sequence {X k } converges to a positive definite solution X β. iii Assume γ α, β. Hence αi < X = γi < βi. Weprovethat{X k } is a Cauchy sequence. Assume that αi < X k < βi.forx k+1 compute X k+1 = AI X k 1 A AA < 1 β β 2 1 β 1 β I = βi, X k+1 = AA AI X k 1 A > 1 α α 2 1 α 1 α I = αi. Hence αi < X k < βi for k =, 1,... Let us consider the norm X k+p X k = AI X k+p 1 1 A We denote P = AI X k+p 1 1 A and Q = AI X k 1 1 A. AI X k 1 1 A and use the equality P P Q + P Q Q = P Q. Obviously, Y = P Q is a solution of the linear matrix equation PY + Y Q = P Q. According to [8, Theorem 8.5.2] we have Y = e Pt P Qe Qt dt. 4

5 I.G. Ivanov et al. / Linear Algebra and its Applications Thus X k+p X k = e Pt P Qe Qt dt P Q e Pt e Qt dt. Since αi < X k for k =, 1,..., P = X k+p, Q = X k, then P> αi and Q> αi. Hence X k+p X k < = 1 2 α = 1 2 α 1 e 2 αt dt P Q A [ I X k+p 1 1 I X k 1 1] A AI X k 1 1 X k+p 1 X k 1 I X k+p 1 1 A 2 α A 2 I Xk 1 1 I Xk+p 1 1 X k+p 1 X k 1 < β 2 1 β 1 X k+p 1 X k 1 2 α 1 β 2 β 2 = Xk+p 1 X k 1. 2 α1 β Consequently k β 2 X k+p X k X p X. 2 α1 β Since β 2 q = < 1 2 α1 β and X p X X p X p 1 + X p 1 X p X 1 X q p 1 + +q + 1 X 1 X < 1 1 q X 1 X, we have X k+p X k < qk 1 q X 1 X.

6 32 I.G. Ivanov et al. / Linear Algebra and its Applications The sequence {X k } forms a Cauchy sequence considered in the Banach space C n n where X k are n n positive definite matrices. Hence this sequence has a positive definite limit of 1. Theorem 4. Let α 1 and β 1 be real for which the inequalities i α1 2 1 α 1I AA β1 2 1 β 1I, β 2 1 ii 2α 1 1 β 1 < 1 are satisfied. For each two γ 1,γ 2 with α 1 γ 1 γ 2 β the reccurence equation 3 defines two matrix sequences {X k } and {X k } with initial points X = γ 1I and X = γ 2 I. These sequences converge to the same limit X γ which a positive definite solution of 1. Proof. We have α 1 I X k β 1I and α 1 I X k β 1I. We put P = AI X k 1 1 A,Q= AI X k 1 1 A and for X k X k we obtain X k X k AI = X k 1 1 A AI X k 1 1 A = P Q = e Pt P Qe Qt dt [ e α1t dt A I X k 1 1 I X k 1 1] A 1 2α A 2 I X k 1 1 I X k 1 1 X k 1 X k 1 Hence X k X But k β2 1 1 β 1 2α β 1 2 X k 1 X β 2 1 = X 2α 1 1 β 1 k 1 X k 1 β1 2 X k 1 2α 1 1 β 1 X k 1.. k 1 β1 2 2α 1 1 β 1 < 1. Consequently sequences {X k } and {X k } have the common limit.

7 I.G. Ivanov et al. / Linear Algebra and its Applications Next consider the matrix sequence X = αi, X s+1 = I A Xs 2 A, s =, 1, 2,... 5 Theorem 5. Suppose there is a matrix A and numbers η, γ for which the conditions i 2 3 <γ η 1, ii η 2 1 ηi A A γ 2 1 γi are satisfied. Then {X s } defined by 5 for α [γ,η] converges to a positive definite solution X of 1 at least linearly and γi X ηi. Proof. Since X = αi we have γi X ηi. Suppose γi X s ηi. Then 1 I X 2 η2 s 1 γ 2 I and 1 η 2 A A A Xs 2 A 1 γ 2 A A. According to condition ii we obtain 1 ηi 1 1 η 2 A A and γ 2 A A 1 γi. Hence γi X s+1 = I A Xs 2 A ηi. 6 We have X s+p X s =A Xs 1 2 X 2 s+p 1 A =A X 2 s+p 1 Xs+p 1 2 X2 s 1 X 2 s 1 A =A Xs+p 1 2 [ Xs+p 1 Xs+p 1 X s 1 + ] X s+p 1 X s 1 Xs 1 X 2 s 1 A =A Xs+p 1 1 Xs+p 1 X s 1 X 2 s 1 A Then +A X 2 s+p 1 Xs+p 1 X s 1 X 1 s 1 A. X s+p X s A X 1 s+p 1 X 2 s 1 A + A X 2 s+p 1 X 1 s 1 A X s+p 1 X s 1. But for all k we have X 1 k 1 γ have A = ρa A γ 2 1 γ. Therefore γ A Xk 1 A Xk γ, γ I and thus X 1 k γ 1, X 2 k 1. Using ii we γ 2

8 34 I.G. Ivanov et al. / Linear Algebra and its Applications γ A Xk 2 A Xk γ γ 2. Hence for all integer p>and arbitrary s we have 21 γ X s+p X s X s+p 1 X s 1. γ We also have q = 21 γ/γ <1since 2 3 <γ. We obtain X s+p X s q s X p X. Since X p X < 1 1 q X 1 X, we have X s+p X s qs 1 q X 1 X. The matrices {X s } form a Cauchy sequence considered in the Banach space C n n where X s are n n positive definite matrices. Hence this sequence has a limit X. Using 6 we obtain γi X ηi. We have X s+1 X A Xs 1 X 2 A + A Xs 2 X 1 A X s X. Similarly, X s+1 X 21 γ X s X. γ Remark 1. Condition ii of Theorem 5 implies that A Theorem 6. Suppose the iterative method 5 converges to a positive definite solution X. Then for any ε> X s+1 X < 2 X 1 A X 2 A +ε X s X for all s large enough. Proof. Since X s X for any ε>, there is an integer s so that for s>s Xs 1 A X 2 A < X 1 A X 2 A +ε

9 and Then I.G. Ivanov et al. / Linear Algebra and its Applications X 1 A Xs 2 A < X 1 A X 2 A +ε. X s+1 X A Xs 1 X 2 A + A Xs 2 X 1 A X s X <2 X 1 A X 2 A +ε X s X. Theorem 7. If A < 2 27, then X 1 A X 2 A < 1 2, where X is the limit of sequence 5. Proof. From the condition A < 2 we obtain that sequence 5 is convergent to 27 X, with 2 3 I<X I. Then X 1 < 3 2 and X 2 < 9 4. We obtain X 1 A X 2 A X 1 X 2 A 2 < = 1 2. Remark 2. If A < 2, then algorithms 3 and 5 converge to two different 27 positive definite solutions X and X and X >X see Example Solutions of the matrix equation X A X 2 A = I We will describe an iterative method which compute a positive definite solution of Eq. 2. Consider the following sequence of matrices: X = αi, X s+1 = AX s I 1 A, s =, 1, 2,... 7 We have the following. Theorem 8. If A is nonsingular and for some real α>, we have i AA <α 2 α 1I, AA ii α 1 1 α 2 A A>I, 2 1 α 1 α 1 iii 2 A 2 < 1, µ µ α 1 where µ is the smallest eigenvalue of the matrix AA, then Eq. 2 has a positive definite solution.

10 36 I.G. Ivanov et al. / Linear Algebra and its Applications Proof. We consider sequence 7. According to condition i we have X 1 = AA AX I 1 A = α 1 <αi= X. Hence X 1 <X. Applying condition ii we obtain AA X 1 = α 1 >I+ 1 α 2 A A>I, I<X 1 <X. Moreover, we have X 1 I 1 >X I 1, X 2 = AX 1 I 1 A > AX I 1 A = X 1. Consequently X 1 <X 2. Using condition ii we have X 1 I>1/α 2 A A and X 2 = AX 1 I 1 A <αi= X. Hence I<X 1 <X 2 <X. Analogously one can prove that I<X 1 <X 2s+1 <X 2s+3 <X 2k+2 <X 2k <X = αi for every positive integer s, k. Consequently the subsequences {X 2k }, {X 2s+1 } are convergent to positive definite matrices. These sequences have a common limit. Indeed, we have X 2k X 2k+1 = AX 2k 1 I 1 A AX 2k I 1 A. We denote P = AX 2k 1 I 1 A, Q = AX 2k I 1 A and use the following equality: P P Q + P Q Q = P Q. Since X 2k 1 <X 2k+1 <X 2k for each k = 1, 2,..., the matrix Y = P Q is a positive definite solution of the matrix equation PY + Y Q = P Q. According to [8, Theorem 8.5.2] we have Y = e Pt P Qe Qt dt. 8 Since P, Q are positive definite matrices integral 8 exists and e Pt P Qe Qt ast. Then

11 I.G. Ivanov et al. / Linear Algebra and its Applications X 2k X 2k+1 = e Pt P Qe Qt dt P Q e Pt e Qt dt. We have X 1 <X s for the matrices of sequence 7. Hence X 1 I<X s I and X s I 1 <X 1 I 1.Wehave α 1 X 1 I 1 =. µ α 1 Consequently X s I 1 < Furthermore α 1 µ α 1. X s+1 = AX s I 1 A >X 1 = Thus, we obtain X 2k X 2k+1 P Q Condition iii now implies q = 1 2 A 2 = P Q 1 2 α 1 AA µ α 1 α 1 I. e 2 µ/α 1t dt α 1 µ = 1 2 µ AX 2k 1 I 1 A AX 2k I 1 A = 1 α 1 2 µ AX 2k I 1 X 2k X 2k 1 X 2k 1 I 1 A 1 2 α 1 α 1 2 µ A 2 X 2k X 2k 1. µ α 1 α 1 µ α 1 µ α 1 2 < 1. Consequently the subsequences {X 2k }, {X 2s+1 } are convergent and have a common positive definite limit which is a solution of the matrix equation 2.

12 38 I.G. Ivanov et al. / Linear Algebra and its Applications Similarly we can write the following. Theorem 9. If there is a real α>2for which i AA >α 2 α 1I, ii AA α 1 1 α 2 A A<I, A 2 iii 2αα 1 2 < 1, then Eq. 2 has a positive definite solution. Proof. We consider sequence 7. According to condition i we have X 1 = AA AX I 1 A = α 1 >αi= X. Hence X 1 >X. Moreover, we have X 1 I 1 <X I 1, X 2 = AX 1 I 1 A < AX I 1 A = X 1. Consequently X 2 <X 1. Using condition ii we have X 1 I<1/α 2 A A and X 2 = AX 1 I 1 A >αi= X. Hence X <X 2 <X 1. Analogously one can prove that αi = X <X 2s <X 2s+2 <X 2k+3 <X 2k+1 <X 1. for every positive integer s, k. Consequently the subsequences {X 2k }, {X 2s+1 } are convergent to positive definite matrices. These sequences have a common limit. Indeed, we have X 2k+1 X 2k = AX2k I 1 A AX 2k 1 I 1 A. We denote P = AX 2k I 1 A, Q = AX 2k 1 I 1 A and use the equality P P Q + P Q Q = P Q. Since X 2k <X 2k+1 <X 2k 1 for each k = 1, 2,...the matrix Y = P Q is a positive definite solution of the matrix equation

13 I.G. Ivanov et al. / Linear Algebra and its Applications PY+ Y Q = P Q. According to [8, Theorem 8.5.2] we have Y = e Pt P Qe Qt dt. 9 Since P, Q are positive definite matrices, integral 9 exists and e Pt P Qe Qt ast. Then X 2k+1 X 2k = e Pt P Qe Qt dt P Q e Pt e Qt dt. We have αi = X <X s for the matrices of 7. Hence X I<X s I and X s I 1 <X I 1.Wehave X I 1 =1/α 1. Consequently X s I 1 < 1 α 1, P = X2k+1 >αi, Q = X2k >αi. Thus we obtain X 2k+1 X 2k P Q e 2αt dt = P Q 1 2α = 1 2α AX 2k I 1 A AX 2k 1 I 1 A = 1 2α AX 2k 1 I 1 X 2k 1 X 2k X 2k I 1 A A 2 2αα 1 2 X 2k 1 X 2k. Condition iii now implies A 2 q = 2αα 1 2 < 1. Consequently the subsequences {X 2k }, {X 2s+1 } are convergent and have a common positive definite limit which is a solution of the matrix equation 2. Consider the following matrix sequence: X = ηi, X s+1 = I + A Xs 2 A, s =, 1, 2,... 1 Theorem 1. Suppose there is a matrix A and numbers α, β for which the conditions i 1 β<α< β 2 + 1, ii α 2 β 1I A A β 2 α 1I

14 4 I.G. Ivanov et al. / Linear Algebra and its Applications are satisfied. Then {X s } defined by 1 for η [β,α] converges to a positive definite solution X of 2 with at least linear rate of convergence and βi X αi. Proof. Since X = ηi we have βi X αi. Suppose βi X s αi. Then 1 α 2 A A A Xs 2 A 1 β 2 A A. According to condition ii we obtain β 1I 1 α 2 A A and 1 β 2 A A α 1I. Hence βi X s+1 = I + A Xs 2 A αi. 11 We have X s+p X s =A Xs+p 1 2 s 1 A =A Xs 1 2 Xs 1 2 X2 s+p 1 Xs+p 1 2 =A X 2 [ s 1 Xs 1 Xs 1 X s+p 1 + ] X s 1 X s+p 1 Xs+p 1 X 2 =A X 1 s 1 +A X 2 s 1 Xs 1 X s+p 1 X 2 s+p 1 A s+p 1 A Xs 1 X s+p 1 X 1 s+p 1 A. Then X s+p X s A Xs 1 1 X 2 s+p 1 A + A Xs 1 2 X 1 s+p 1 A X s 1 X s+p 1. But for all k we have Xk 1 1/β and Xk 2 1/β 2. Using ii we have A β 2 α 1. Therefore β A Xk 1 2 α 1 β, A Xk 2 β 2 α 1 β 2. Hence for all integer p>and arbitrary s we have 2α 1 X s+p X s X s+p 1 X s 1. β Using i α< β 2 + 1wehave q = 2α 1 β < 1.

15 I.G. Ivanov et al. / Linear Algebra and its Applications We obtain X s+p X s q s X p X. Since X p X < 1 1 q X 1 X, we have X s+p X s < qs 1 q X 1 X. The matrix sequence {X s } is a Cauchy sequence considered in the Banach space C n n where X s are n n positive definite matrices. Hence this sequence has a limit X. From 11 we obtain βi X αi. We have X s+1 X A X 1 X 2 s A + A X 2 Xs 1 A X s X. Similarly, X s+1 X 2α 1 X s X. β Theorem 11. Suppose the iterative method 1 converges to a positive definite solution X. Then for any ε> X s+1 X < 2 X 1 A X 2 A +ε X s X for all s large enough. The proof is similar to that of Theorem Numerical experiments We have made numerical experiments to compute a positive definite solution of Eqs. 1 and 2. The solution was computed for different matrices A and different values of n. Computations were done on a PENTIUM 2 MHz computer. All programs were written in MATLAB. We denote and ε 1 Z = Z + A Z 2 A I ε 2 Z = Z A Z 2 A I.

16 42 I.G. Ivanov et al. / Linear Algebra and its Applications We have tested our iteration processes for solving the equation X + A X 2 A = I on the following n n matrices. We use the stopping criterion ε 1 Z < 1.e 8. Example 1. Consider Eq. 1 with A = The spectral norm of A is.292. Consider the iterative method 3. If X = αi and α =, then 15 iterations are required for computing X. In this case we obtain a monotonically increasing sequence which converges to the solution X. Consider the case X = βi.ifβ = 2 3, then 16 iterations are required for computing X.Ifβ =.368, then 12 iterations are required for computing X. In this case β satisfies conditions of Theorem 2. We obtain a monotonically decreasing sequence which converges to the solution X. There is a value of ββ=.368 for which the iterative method 3 with X = βi is faster than the iterative method 3 with X = αi. Hereβ is a solution of the scalar equation A 2 = β 2 1 β. Consider the iterative method 5. We choose initial value X = αi and 2 3 <α 1 see Theorem 5. In this case the matrix sequence converges to the solution X. If α = 1, then 12 iterations are required for computing X.Ifα = 2/3+1 2,then 11 iterations are required for computing X.Ifα =.892, then nine iterations are required for computing X. We choose α to be a solution of the scalar equation A 2 = α 2 1 α see Theorem 5. Example 2. Consider Eq. 1 with A = For this example the conditions of Theorems 2 and 5 are not satisfied A = > 27. But the iterative method 5 is convergent. We use this method for computing the X.Ifη = 2 3, then 14 iterations are required for computing X.If η = 1, then 13 iterations are required for computing X.Ifη = 2/3+1 2,then13 iterations are required for computing X. We have tested our iteration processes for solving the equation X A X 2 A = I on the following n n matrices. We use the stopping criterion ε 2 Z < 1 8. Example 3. We have A = UÃU 1, where U is a randomly generated matrix and [ Ã = diag n, n,..., ] n 2.

17 I.G. Ivanov et al. / Linear Algebra and its Applications Table 1 Iterative method 7 n A α k Xα The matrix A satisfies the conditions of Theorem 9. Let k Xα be the smallest number s, forwhichε 2 X s <1 8 for the iterative method 7. The results are given in Table 1. Example 4. Consider Eq. 2 with A = Consider the iterative method 1. The matrix A satisfies the conditionsof Theorem 8. We choose β = 1. We compute A =.587. If α = 1.345, then six iterations are required for computing X. 5. Conclusion In this paper we consider special nonlinear matrix equations. We introduce iteration algorithms by which positive definite solutions of the equations can be calculated. It was proved in [4, Theorem 13] that if A < 1 2, then the equation X + A X 1 A = I has a positive definite solution. From Theorem 2 Theorem 5 we can see that if 2 A 2 < max ϕx = x 2 1 x = ϕ = , then the equation X + A X 2 A = I has a positive definite solution. There are matrices A see Examples 3 and 4 for which A > 1 2 and A > 4 27 and for which we can still compute a positive definite solution of the equation X A X 2 A = I.

18 44 I.G. Ivanov et al. / Linear Algebra and its Applications References [1] B.L. Buzbee, G.H. Golub, C.W. Nielson, On direct methods for solving Poisson s equations, SIAM J. Numer. Anal [2] C. Guo, P. Lancaster, iterative solution of two matrix equations, Math. Comput [3] J.C. Engwerda, A.C.M. Ran, A.L. Rijkeboer, Necessary and sufficient conditions for the existence of a positive definite solution of the matrix equation X + A X 1 A = Q, Linear Algebra Appl [4] J.C. Engwerda, On the existence of a positive definite solution of the matrix equation X + A T X 1 A = I, Linear Algebra Appl [5] A. Ferrante, B.C. Levy, Hermitian solutions of the the equation X = Q + NX 1 N, Linear Algebra Appl [6] I.G. Ivanov, S.M. El-Sayed, Properties of positive definite solution of the equation X + A X 2 A = I, Linear Algebra Appl [7] A.S. Householder, The Theory of Matrices in Numerical Analysis, Blaisdell, New York, [8] P. Lancaster, Theory of Matrices, Academic Press, New York, 1969.