42.9. AN EVOLUTION INCLUSION 1411

Now taking the inner product of this with x(t)− x(s) it follows from 42.9.59,

|x(s)− x(t)|2H = (z(s)− z(t) ,x(t)− x(s))H

≤ φ (s,x(t))−φ (s,x(s))+φ (t,x(s))−φ (t,x(t))

≤(

K(

φ (t,x(t))+ |x(t)|2H +1)+K

(φ (s,x(s))+ |x(s)|2H +1

))|t− s|

which shows by 42.9.73 that x(·) is Lipschitz continuous and is therefore measurable whichverifies 42.9.70. The assumptions of the lemma include 42.9.71. It follows the conclusionof Lemma 42.9.6 holds.

Remark 42.9.8 Note that if φ (t, ·) has a minimum at ξ (t) and if t → ξ (t) and t →φ (t,ξ (t)) are bounded and measurable, then

ξ (t)+0 = ξ (t)

and 0 ∈ ∂2φ (t,ξ (t)). Therefore, in this case J1 (t)ξ (t) = ξ (t) and so the hypotheses ofLemma 42.9.7 hold.

Corollary 42.9.9 Assume 42.9.57 - 42.9.59 and 42.9.70, 42.9.71. Let x0 ∈ D and let [ f ] ∈L2 (0,T ;H) . Then there exists a unique function, x, satisfying

[x] and[x′]

are in L2 (0,T ;H)

which is a solution to

x′+∂2φ (t,x) ∋ f a.e., x(0) = x0, x(t) = x0 +∫ t

0x′ (s)ds. (42.9.74)

Proof: Let [v] ∈ L2 (0,T ;H) and let [x] be the unique solution to

L [x]+ [x]+∂Φ([x]) ∋ [ f ]+ [v] . (42.9.75)

Letting [xi] be the solution corresponding to 42.9.75 in which v is replaced with vi, andxi ∈ [xi] is such that

xi (t) = x0 +∫ t

0x′i (s)ds, i = 1,2,

from Lemma 42.9.6 and 42.9.75 that for each t ∈ [0,T ] ,

12|x1 (t)− x2 (t)|2H +

12

∫ t

0|x1− x2|2H ds≤ 1

2

∫ t

0|v1 (s)− v2 (s)|2H ds

and so|x1 (t)− x2 (t)|2H ≤

∫ t

0|v1 (s)− v2 (s)|2H ds.

Now define a mapping, Λ : L2 (0,T ;H)→ L2 (0,T ;H) by Λ [v] = [x] where [x] is the solutionto 42.9.75. Then, if [vi] is in L2 (0,T ;H) and [xi] is the corresponding solution to 42.9.75,

||Λ [v1]−Λ [v2]||2L2(0,t;H) ≡