42.7. AN EVOLUTION INCLUSION 1403

Proof: This is from Theorem 42.6.2. Since x0 ∈ D, it follows that φ (x0)< ∞.Let z ∈ D(Φ) , the effective domain of Φ. Then

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

φ and 42.7.48,

φ (Jλ z(t))≤ e−tλ φ (x0)+

1λ

e−tλ

∫ t

0e

sλ φ (z(s))ds. (42.7.50)

ThenΦ(Jλ z) =∫ T

0φ (Jλ z(t))dt ≤ φ (x0)λ +

∫ T

0

1λ

∫ t

0e−(t−s)/λ

φ (z(s))dsdt

≤ λφ (x0)+1λ

∫ T

0φ (z(s))

∫ T

se−(t−s)/λ dtds

≤ λφ (x0)+

(∫ T

0φ (z(s))ds

)1λ

∫∞

0e−t/λ dt

= φ (x0)λ +∫ T

0φ (z(s))ds

= φ (x0)λ +Φ(z)

The conditions of Theorem 42.6.2 are satisfied. This proves L+∂Φ is maximal mono-tone on L2 (0,T ;H) and consequently there exists a unique solution to the differential in-clusion of the theorem.

Then the main result is the following.

Theorem 42.7.5 Let f ∈ L2 (0,T ;H) and x0 ∈ D. Let φ be as described above, a lowersemicontinuous convex proper function defined on H. Then there exists a unique solutionx ∈ L2 (0,T ;H) ,x′ ∈ L2 (0,T ;H) , to

x′+∂Φ(x) ∋ f in L2 (0,T ;H) , x(0) = x0

This satisfies the pointwise condition

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

Proof: From Theorem 42.7.4, there exists a unique solution to

x′v +∂Φ(xv)+ xv ∋ f + v in L2 (0,T ;H) , xv (0) = x0

whenever v ∈ L2 (0,T ;H). Then a simple argument based on fundamental theorem of cal-culus implies that for a.e. t,

x′v (t)+∂φ (xv (t))+ xv (t) ∋ f (t)+ v(t)

Then for given v,u one can act on xv (t)− xu (t) and integrate. This yields

12|xv (t)− xu (t)|2H +

∫ t

0|xv− xu|2 ds≤

∫ t

0|v(s)−u(s)|2H ds

It follows that a sufficiently high power of the mapping u→ xu is a contraction map onL2 (0,T ;H) and so there exists a unique fixed point v in L2 (0,T ;H). Thus xv = v and so

v′+∂Φ(v) ∋ f in L2 (0,T ;H) , v(0) = x0