SOLVING A CLASS OF GENERALIZED LYAPUNOV OPERATOR DIFFERENTIAL EQUATIONS WITHOUT THE EXPONENTIAL OPERATOR FUNCTION

In this paper a method for solving operator differential equations of the type X' = A + BX + XD; X(0) = C0, avoiding the operator exponential function, is given. Results are applied to solve initial value problems related to Riccati type operator differential equations whose associated algebraic equation is solvable.


SOLVING A CLASS OF GENERALIZED LYAPUNOV OPERATOR DIFFERENTIAL EQUATIONS WITHOUT THE EXPONENTIAL OPERATOR FUNCTION A bstract LUCAS JÓDAR
In this paper a method for solving operator differential equations of the type X ' = A -}-BX + X D ; X (O) = Co, avoiding the operator exponential function is given .Results are applied to solve initial value problems related to Riccati type operator differential equations whose associated algebraic equation is solvable .

. Introduction
It is well-known that the solution of the matrix differential equation (1 .1)X0)(t) = A + BX(t) + X(t)D; X(o) = Co where A, B, Co, D and X (t) are non complex matrices, and D* denotes the adjoint matrix of D, is given by the expression (1 .2) X(t) = exp(tB)Co exp(tD*) + lo t exp(B(t -s))A exp(D*(t -s»ds see [1, p. 28] for details .It is easy to show that the expression (1.2) defines the solution of problem (1 .1)when A, B, Co and D are bounded linear operators defined on a Hilbert space H .Although the exponential matrix function has been widely studied ( [13], [17], [18]), its computation presents some inconvenients ( [13]) so, thinking of applications, an expression of the solution of (1 .1)avoiding the use of the exponential matrix function is interesting.The aim of this paper is to present an alternative method for solving (1 .1)avoiding the exponential matrix function and the computation of integrals involving exponentials of matrices .Let us denote by L(H) the algebra of all bounded linear operators defined on the Hilbert space H, and for T in L(H) let us denote its spectrum by o(T) .We recall that an operator T in L(H) is said to be algebraic if there exists a polynomial p(z) such that p(T) = 0.It is cleax that a finite-dimensional operator is algebraic and from [4, p. 569], an algebraic operator in L(H) has a finite spectrum, but there are operators with a finite spectrum that are not algebraic operators in L(H), [14] .In [5], P.R. Halmos observed that an operator in L(H) that is annihilated by an entire analytic function, is algebraic.An account of the properties of algebraic operators may be found in [7], [14].
Let T be an operator in L(H) and let zoea(T) an isolated point of u(T), then zo is said to be a pole of T if the resolvent function R(z,T) = (zI -T)-1 has a pole at zo .By the order w(zo) of a pole zo is meant the order of zo as a pole of R(z,T) .
(ii) B -E L(H) has a finite spectrum and each z,O(B) is a pole of B. For the finite-dimensional case the condition (ii) is always satisfied, and the condition (i) means that D is similar to a normal operator, [7, p. 14] .Section 2 concerns with the resolution problem (1 .1)and section 3 provides an explicit solution for a class of generalized Riccati operator differential equations in terms of a solution of certain generalized Lyapunov equation associated to the problem .

. Solving generalized Lypunov differential operator differential equations without the exponential operator function
We begin this section with an algebraic result that provides a finite algebraic expression of the solution of generalized algebraic Lyapunov operator equation, under certain uniqueness hypothesis.For the finite-dimensional case, an analogous result is given in [9] .Lemma 1.Let A1 , B1 and Dl be operators in L(H) such that Dl is algebraic and satisfies the condition Proof.Under the hypothesis (2.1), the equation (2.2) has only one solution, [16], [3], and from [3], corollary 2, if X is the only solution of such equation one gets Also, considering the powers Vi, for j = 0, 1, . . ., n, it follows that the (i, 2) block entry of the operator Vi, denoted by Vgj2, for j = 1, 2, . . . .n and i = 1, 2, satisfy and V °2 = 0, V20,2 = I.
Considering the polynomial calculus and computing it follows that for certain operator M one has From (2.5) and (2.7), one gets M = -p(B1 )X, and from the spectral mapping theorem, [4, p. 569], and (2.1), the operator p(B 1 ) is invertible in L(H).Thus, we have By multiplying the operator V1j,2 by the coefficient aj, for j = 0, 1, . . ., n, and by addition it follows that the block entry (1,2) of the operator matrix p(V), is given by the expression (2.9) From (2.8) and (2 .9)one gets (2.3) .

L . JÓDAR
For the sake of clarity in the presentation of the next results we recall some concepts and properties concerned with the Riesz-Dunford functional calculus, [4], and the Laplace transform of operátor valued functions, [8] .
Let zo be an isolated point in the spectrum u(T) of an operator TeL(H), then the Laurent expansion of R(z,T) = (zI -T) -1 in a neighborhood 0 < Iz -zo 1 < 5, of zo , is given by 00 (2.10) where E(z o ;T) denotes the spectral projection corresponding to the spectral set {zo} see [4, p. 573], for details .If zo an isolated point in o(T), it follows that zo is a pole of order p, if and only if, (2.11) (zoI -T)PE(zo;T) = 0 and (zoI -T)P-1 E(zo;T) qÉ 0 We recall that a L(H) valued operator function t -> V(t), is said to be an original function if, V(t) = 0, for t < 0, V is locally integrable and there exist a real number so and a positive number M, such that 11V(t)jj _< Mexp(sot), for t > 0. Under these hypotheses the Laplace transform of V, represented by V, is defined in the usual way, see [8] for details and related properties .In particular, if V(1) is an original function, it follows that V(') (s) = sV(s)-V(0).Finally, if z -> f(z), is a L(H) valued meromorphic function and zo is a pole of f, we represent by Res(f; zo) the residue of f in the pole zo .
Let us consider the problem (1 .1)where A and Co are arbitrary operators in L(H) and B, D are operators in L(H) satisfying the properties (i) and (ii) given in page 1.Let X be the function defined by the expression (1.2) for t >_ 0, and X(t) = 0, for t < 0. From (1 .2) it follows that X(t) and X(1)(t) are original functions, where X1)(0) means the right lateral derivative of X at t = 0. Let X(s) be Laplace transform of X. Taking into account the properties of the Laplace transform, as X satisfies the problem (1.1), by application of the Laplace transform to the differential equation arising in (1 .1), it follows that there exists a positive number a such that if Re(s) > a one gets From the spectral mapping theorem, [4], it follows that p(sI -B) is invertible in L(H) for values of s enough advanced in module, and Let us suppose that v(B) = {b,, . . ., b.}, then taking into account (2 .17), the set of poles of (sI -Ri ) -1 is the set of points sij = bi + zj, where 1 _< j < m, and as q1(s) and q2(s) are holomorphic functions, by aplication of the Laplace inversion formula, for t > 0 we have + Res(X2(s) exp(st) ; 0) if sij :~0, 1 < i < n, 1 < j < m L. JÓDAR   and if there exist some sjj = 0, then n m (2.19) X (t) = EE(Res(Xj (s) exp(st) ; sij) -1-Res(XZ (s) exp(st) ; sij) %-1 j-1 In order to compute the residues of Xj(s) exp(st), for i = 1, 2, we need the order of each singularity sij and 0 for such functions .Note that the spectral projection E(sij ; Rj) = E(bi + zj ; B -{-zj) = E(b% ; B), and the order of sjj as a singular point of (zI -Rj) -1 coincides with the order of b% as a singular point of (zI -B)-' .Also, considering the decomposition (2.20) Xj (s)exp(st) = (sI -Rj) -1 (R 1 (sI-R%) 1 )gl(s)exp(st) _ (sI-Rj)-1Qj(s) .
¡# .i The Taylor expansion of Qj(s) at the point sjj takes the form and the Laurent expansion of (sI -Rj)-i at the point sjj is given by the expression (2 .28) In order to compute the residue of X(s) exp(st) at s = 0, note that Res(Xi (s) exp(st) ; 0) is the operator 0 because Xl(s) exp(st) is holomorphic at s=0and (2.29) X2(s) exp(st) = -s-1(p(sI -B)) -l g2(s) exp(st) .
Under the hypothesis u(B) n a(-D) = 0, the operator p(-B) is invertible, thus the factor (p(sI -B)) -1 q2 (s) is holomorphic at s = 0, and from (2.29) it follows that (2.30) Res(X(s)exp(st and from lemma 1, it follows that ResÑ(s) exp(st) ; 0) = X* , X, being the only solution of the algebraic equation A + BX + X D = 0 .Summarizing we have that under the hypothesis u(B) fl a(-D) _ 0, the solution X(t) of problem (1.1), for t > 0 is given by the expression n m (2 .31) where X* is the only solution of the algebraic equation A + BX + XD = 0, given by (2.30), and Qij(t) and S=j (t) are given by (2.24) and (2.27) respectively .Thus the following result has been proved : Theorem 1 .Let us consider the problem (1.1) where A and Co are operators in L(H) and the operators B and D satisfy the following properties (i) u(B) = {bi ; 1 < i < n}, o,(D) = {z1; 1 < j < m}, and u(B)nu(-D) _ 0 .
(ii) Each bico,(B) is a pole of B .
Then the only solution of problem (1 .1) is given by the expression (2.31), where X* is given by (2 .30),E(bi ; B) are the, spectral projections of B, and Qi1(t) and Sij(t) are given by (2 .2.x)and (2.27) respectively .
Proof.The result is a consequence of the above comments.In fact for t > 0, the expression of the solution coincides with (2.31) .On the other hand, the solution of problem (1 .1),given by (1.2) is an analytic function of the variable t, and coincides with the expression appearing in the right hand side of (2.31), that is also analytic, in consequence they coincide on all the real fine .
Under the hypothesis of theorem 1, note that with the exception of Si;(t) and Qjj (t), all coefficients E(bi; B) and X, given by (2 .30)do not involve the variable t, thus, in order to study the behaviour of the solution when t ---> oo, we have to consider the functions Qjj (t) and Sij(t) .Note that Sij (t) and Qjj (t) are defined by (2.27) and (2.24) in terms of the derivatives (with respect to s) of the functions Qj(s) = (II 1(sI -Ri) -1 )gi (s) exp(st) a9Éj Tj(s) = (n8 1(sI -Ri) -1 )s -1 g2(s)exp(st), i9É7 where q1 (s) and q2 (s) are given by (2.15).Corollary 1 .Let us consider the problem (1.1) under Me hypotheses of theorem 1 .If sis = bi -{-zj, 1 <_ i <_ n, 1 <_ j < m, and all sis are contained in the half plane Re(z) < 0, then all solution of the differential equation arising in (1 .1)converges ío X* when t -> +oo .

An Application to Riccati Operator Differential Equations
The resolution of a Caúchy problem for Riccati operator differenctial equations of the type where A, F, G, E and Co are operators in L(H), is important in control theory, [12], transport theory, [15], and filtering problems, [2] .In a recent paper [10], an explicit expression of the solution of problem (3.1) is given in terms of the block entries of the operator function but an explicit expression of such entries in terms of data is not known .The aim of this section is to obtain an explicit expression of a class of problems of the type (3.1) in terms of a solution of the corresponding algebraic Riccati operator equation and the solution of certain associated generalized Lyapunov operator differential equation .
A resolution method for solving non-symmetric algebraic Riccati operator equation is given in [6] .
Let us consider the problem (3.1) and let us suppose that there exists a solution X, of the algebraic equation (3.3) such that Co -X, is invertible in L(H) .From [111, the problem (3.1) is locally solvable, so, there exists a solution X(t) defined in a neighborhood J of the origin t = 0.As X(0) = Co satisfies Co -X, invertible in L(H), from continuity, it follows that X(t) -X* is invertible in L(H) when t belongs to some neighborhood of t = 0, let us denote this neighborhood by J.