
402
The Abstract Cauchy Problem for the Time Dependent Equations
Clearly, part (b) will be fulfilled as well if we can prove that any
solution (as defined in (A)) of (7.2.1) must perforce be of the form S(t, s) u.
This will be achieved by essentially the same means as in the time-invariant
case (see the end of the proof of Theorem 2.1.1). In the present situation,
this would involve differentiation of the function s - S(t, s)u(s) for an
arbitrary solution of (7.2.1). However, we lack information on the behavior
of S as a function of s. Taking a hint from Example 7.1.3, we may expect
that
DSS(t,s)=-S(t,s)A(s) (0<s<t<T)
(7.2.46)
for each t. To prove this directly would not be simple, thus we use a trick
similar to that in Example 7.1.3; namely, we construct an operator-valued
solution S of equation (7.2.46) and show that S = S. The construction of S
is based on the integral equation (7.2.4), instead of (7.2.6), which was the
starting point for the construction of S. As we did in relation to (7.2.6), we
seek an S of the form
S(t, s) = S(t - s; A(s))+ f tf (t, a)S(a - s; A(s)) do. (7.2.47)
S
Formally, we have
DDS(t, s) = D5S(t - s; A(s))-A(t, s)
+ f tA(t, a) DsS(o - s; A(s)) do,
s
S(t, s)A(s) = D,S(t - s; A(s))
+ f tk(t,a)D0S(u -s; A(s)) do.
S
Therefore, if (7.2.46) is to be satisfied,
0 = DsS(t, s)+S(t, s)A(s)
= (D1 + DD)S(t - s; A(s))- A(t, s)
+ f tA(t, o)(DQ + DS)S(o - s; A(s)) do.
S
so that A must be a solution of the integral equation
A(t,s)- f'A(t,a)A, (a,s)do=AI(t,s)
(7.2.48)
S
with A t (t, s) = (Dt + DD)S(t - s; A(s)). The construction of S now proceeds
in a way exactly analogous to that of S, thus we limit ourselves to state the
main steps and leave the details to the reader.
7.2.7 Lemma.
A, (t, s) is continuous in 0 < s <t < T in the norm of
(E) and satisfies there the inequality
IIRI(t,s)II <C/(t-s)P.
(7.2.49)