ABSTRACT
The results of this section may be found in KALITVIN [1993, 1993a], ApPELL-DIALLO [1995], and ZABREJKO-KALITVIN [1997].
10.1. Equations with variable operators Since we are now interested in the non-stationary equation (10.1), we have to put
(10.3) A(t)z(s) = c(t,s)z(s) +L11 k(t,8,0')Z(0')dO' and to consider, instead of (9.3), the differential equation
(10.4) dzdt = A(t)z +J(t). We recall that the evolution operator U = U(t, r) of (10.4) has the form
(10.6) e(t, r, 09) = exp {1t c(e, s) de} , and h = h(t, r, 8, 0') is a measurable function on [0, T] X [0, T] X [-1,1] X [-1, 1]. It is therefore not surprising that the results of Subsection 9.2 carryover provided we replace throughout the function
tc(s) by the integral f~ c(e, s) de, and the function h(t, I, 0') by the function h(t, 0, s, 0'). In particular, we may write the solution of (10.4) with initial condition x(o) = Xo in the form
for 0<09 S 1 and -1 S 09 < 0, respectively, where we have put
For obtaining the functions y E X+ and z E X_ (see Subsection 9.2) we therefore have to solve the system
for °< 09 S 1 and -1 S 09 < 0, respectively. Moreover, putting
and
(10.11)
(10.12)
(which is paraJIel to (9.9», solving the system (10.8) is equivalent to solving the Fredholm integral equation
z(s) +1° p(s, u)z(u) du = res). -1
In this way, we arrive at the following generalization of Lemma 9.1:
Theorem 10.1. Suppose that the functions c : Q - lR and k : Q x [-1,1] - lR are measurable, the operator A(t) is regular in the space X = Lp([-l, 1]), the operator function (10.3) is strongly continuous, and -1 ;. 0'(P), where
pz(s) = L: p(s,u)z(u)d(u). Then the problem (10.1)/(10.2) has a unique solution x E Cl(X) for any t/> E X+, ¢ E X_, and f E C(X). This solution may be determined by formula (10.7). Of course, one may also formulate a paraJIel result to Lemma 9.2 in case -1 E u(P).
