1402 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE
which is a contradiction.Define
D(L)≡{
x ∈ L2 (0,T ;H) : such that
x(t) = x0 +∫ t
0x′ (s)ds where x′ ∈ L2 (0,T ;H)
}(42.7.47)
and for x ∈ D(L) ,Lx≡ x′.
Then L is maximal monotone. To see this, consider the equation
λx′+ x = z, x(0) = x0
It clearly has a solution so λL+ I is onto. In fact, the solution is
x = e−tλ x0 +
1λ
e−tλ
∫ t
0e
1λ
sz(s)ds
Also,
(Lx−Ly,x− y)L2(0,T ;H) =∫ T
0
((x′− y′
),x− y
)H dt
=∫ T
0
(x′ (t)− y′ (t) ,
∫ t
0x′ (s)− y′ (s)ds
)dt
=12
∫ T
0
ddt
(∣∣∣∣∫ t
0x′ (s)− y′ (s)ds
∣∣∣∣2)
dt
=
∣∣∣∣∫ T
0x′ (s)− y′ (s)ds
∣∣∣∣2H≥ 0
Thus we have the following lemma.
Lemma 42.7.3 L is maximal monotone and if z ∈ L2 (0,T ;H) , then Jλ z is given by
Jλ [z] (t)≡ (I +λL)−1 ([z]) (t) = e−tλ x0 +
1λ
e−tλ
∫ t
0e
1λ
sz(s)ds. (42.7.48)
The main theorem is the following.
Theorem 42.7.4 Let x0 ∈ D≡ D(φ) . Then L+∂Φ is maximal monotone so there exists aunique solution to
Lx+ x+∂Φ(x) ∋ f (42.7.49)
for every f ∈ L2 (0,T ;H). Thus there exists x ∈ L2 (0,T ;H) such that
x′ ∈ L2 (0,T ;H) ,x(0) = x0 ∈ D(φ) ,
andx′+ x+∂Φ(x) ∋ f , x(0) = x0