9
(9th step) For each A e
(0,1],
(P)
x
admits a unique solution on
[0,T].
By the above
uniform estimates, we can extend the solution beyond f
x
if f
x
< T, where [0,7\) is a
maximal existence interval of a solution to (P)
x
. Accordingly, we observe that (P)
x
has
a unique solution on
[0,T].
(10th step) Let A e
(0,1],
T > 0 and {u
x
,6
x
,a
x
} be a solution of (P)
A
on [0,T] ob-
tained in the 9th step. The above uniform estimates with respect to A imply that there
exist a subsequence {A.,} of {A}, with {A
3
} -* 0, and functions u, 0, a and £ on Q(T),
such that u
Xj
-• w weakly* in L°°(0,T; ff
4
(0,1)), weakly in W^O.T; ff
3
(0,1)) and
weakly* in ^^(O.T; #
2
(0,1)), fl
A
. -+ 0 weakly in W^O.T; L
2
(0,1)) and weakly* in
L°°{0,T;
^{0,1)), a
Xj
-> a weakly in W/^TjL^O, 1)), weakly* in L°°(0,T; H\0,1)),
and dI(6
Xj
,£
Xj
;<r
Xj
) -+ £ weakly in L
2
(Q(T)) as j -> oo.
It is easy to see that {u, 9,a} is a solution of (P)(uo, ^o,^o,
o"o)
on
[0,T].
We refer to
[2,
section5] for the detailed proof of this step. •
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