r < t - Department of Mathematics

422 R. E. Showalter
problem is well posed.
Proceeding similarly for the hyper-parabolic equation
(2) U'(t)+Au(t)- Bu(t) = 0,
we see that if u E C1 is a solution on the interval [7,tl, then
_rl_ exp(-A(t- s))exp(-E(s- 7))u(s) = 0, 7 < s < t,
ds
(3) t(t,T) = exp(-A(t- s))exp(-B(s-T))u(s),
is independent of s. Thus, we are led to define a weak solution of (2) on [r,tl
as a continuous H-valued funct-ion for which (3) holds, i.e., the right side of
(3) is independent of s E [7,t].
Lemma. If u is a weak solution on 17,t.I then it is a weak solution on each
[71't11 c k,tl and then E(t,r) = exP(-A(t- t,))exp(-B(7,-.))E(t1,Tl), and
u(t) = E(t,t-).
If u is a continuous H-valued function on [0,11, then u is a weak solution
iff exp(-Bt)u(t) = exp(-At)u(O), 0 _< t 5 1. Thus the initial-value problem of
finding a weak solution of (2) on [0,11 with u(0) = f given in H is equiva-
-
lent to
0
u E C ([0,11 ,H) with exp(-Bt)u(t) = exp(-At)f, 0 5 t 5 1.
Since each exp(-Bt) is one-to-one, there is at most one solution of the initialvalue
problem. Also, the representation via unbounded operators as u(t)
= exp(Bt)exp(-At)f shows the initial-value problem is not well posed. Considering
existence, we see that if f E RgIexp(-B)} = domIexp(B)l, then
u(t) = exp(-At). exp(-B(l- t)). exp(B)f
defines a strong solution (C") of the initial-value problem. More generally we
have the following
Pro osition 1. There exists a weak solution of the initial-value problem if and
only if exp -A)f = exp(-B)g for some g E H, i.e., f E domIexp(B- A)}.
QUASI -REVERSIBI L ITV METHOD
Since the lack of well-posedness of the initial-value problem for (2) is due to
the unboundedness of B, we use a Q- R method [3,5,13,141 to obtain an approximate
solution. First replace B by its bounded Vosida approximation
so
7 - < s 5 t
BE = B(I+ EB)-', E > 0, and solve the equation (2) for exp((BE- A)t)f, O