42.9. AN EVOLUTION INCLUSION 1407

Proof: Since each φ (t, ·) is nonnegative and convex, it follows that Φ is also non-negative and convex. It remains to verify lower semicontinuity. Suppose, [xn]→ [x] inL2 (0,T ;H) and let

λ = lim infn→∞

Φ([xn]) .

Is λ ≥ Φ([x])? It suffices to assume λ < ∞, xn (t) ∈ D for all t, and xn (t)→ x(t) a.e. sayfor t /∈ N where N has measure zero. Let

x̃(t) ={

x(t) if t /∈ Nx1 (t) if t ∈ N

Then [x̃] = [x] and x̃(t) ∈ D for all t. Then by pointwise convergence and Fatou’s lemma,

Φ([x]) = Φ([x̃]) =∫ T

0φ (t, x̃(t))dt ≤

∫ T

0lim inf

n→∞φ (t,xn (t))dt

≤ lim infn→∞

∫ T

0φ (t,xn (t))dt = lim inf

n→∞Φ([xn])≡ λ .

This proves the lemma.Define

D(L)≡{[x] ∈ L2 (0,T ;H) : for some x ∈ [x] such that

x(t) = x0 +∫ t

0x′ (s)ds where

[x′]∈ L2 (0,T ;H)

}(42.9.62)

and for [x] ∈ D(L) ,L [x]≡

[x′].

The following lemma is easily obtained.

Lemma 42.9.4 L is maximal monotone and if [z]∈ L2 (0,T ;H) , then the equivalence class,[Jλ [z]] is determined by the function,

Jλ [z] (t)≡ (I +λL)−1 ([z]) (t) = e−tλ x0 +

1λ

e−tλ

∫ t

0e

1λ

sz(s)ds. (42.9.63)

The main theorem is the following.

Theorem 42.9.5 Let x0 ∈ D. Then L+ ∂Φ is maximal monotone so there exists a uniquesolution to

L [x]+ [x]+∂Φ([x]) ∋ [ f ] (42.9.64)

for every [ f ] ∈ L2 (0,T ;H).

Proof: This is from Theorem 42.8.2. Since x0 ∈D, it follows from 42.9.59 that φ (t,x0)is bounded.

Let [z] ∈ D(Φ) , the effective domain of Φ. Then there exists z ∈ [z] such that z(t) ∈ Dfor all t, and

∫ T0 φ (t,z(t))dt < ∞, so by convexity of φ (t, ·) and 42.9.63,

φ (t,Jλ [z] (t))≤ e−tλ φ (t,x0)+

1λ

e−tλ

∫ t

0e

sλ φ (t,z(s))ds. (42.9.65)