355
To prove this it is enough to take in (2.21) the function v defined as follows: for any
(8, t) E ]0, S[ x ]0, T[ and R6 = [8 - 8,8 + 8[ x ]t - 8, t + 8[ c R,
v = {u in Q\(]a,b[ XR6} ,
w in la, b[ x R6 ,
where w E K is fixed; dividing by 48 2 and letting 8 -+ 0 we obtain (2.27).
We will need the following lemma
Lemma 2.15. Given w E W = {v E L2(Q}: Av E L2(Q}} and a E R,
2 f 1o[ealwAw] ~ eaT 10 Iw(rW - e all 10 Iw(u)12 - a f 1olw2 e a/ ]_ f [[eat Iw(OW] .
(2.28)
Proof: Notice that, given w E W, the trace of won t = 0 belongs to L2(0} and the
trace of w on 8 = 0 belongs to L 2 ([a, b] x [0, TJ). SO we can argue with smooth functions
WE converging to w in W. Using the formula (2.22), we have
2 f foS [[WE AWE eal] = f foS [[AlwEI2 eal] =
= f[[lw EI 2 e at ](S)- f[[lw EI 2 e at ](0)-a flo [lwEl 2 e al ]
+ eaT 10 IwE(rW - e all 10 IwE(uW .•
Define
1
1+1 (
1' +11b
8(t) = I jnl/(r) - 1001 2 + I a lu 1 (r) - u~12 .
(2.29)
Then we have
Theorem 2.16. Let U oo solve the problem (2.5) and U E L1:(O, 00; L2(0, S; HJ(a, b)))
be the solution of (2.21) or (2.27). Suppose that I E LOO(O, +00; L2(0)), 100 E L2(0) and
8(t) --+ 0 when t -+ 00. Then
u(t) --+ U oo in L2(0), when t -+ 00 .
(2.30)
Proof: Taking v = U oo in (2.27) and v = u(t) in (2.5), calling w(t) = u(t) - uoo , we
obtain, for a.e. t ~ S,
10 w(t) Aw(t) + 10 18x w(tW ::; 10 I/(t) - loollw(t)1 .
Multiplying the above inequality by e al , Q E R and then integrate it between ( ~ S
and r > (, we have