Topics In Analysis

33.7 An Evolution Inclusion

In this section is a theorem on existence and uniqueness for the initial value problem

x′ + ∂2ϕ(t,x) ∋ f,x(0) = x0.

Suppose

t∈

[0,T]

is a family of functions mapping H to which satisfy the following axioms.

ϕ (t,⋅) is convex and lower semicontinuous, (33.7.56)

(33.7.56)

D (ϕ (t,⋅)) = D, independent of t ∈ [0,T], (33.7.57)

(33.7.57)

There exists a constant, K, such that for all x ∈ D,

( 2 ) |ϕ (t,x)− ϕ (s,x)| ≤ K ϕ (r,x)+ |x| + 1 |t− s| (33.7.58)

(33.7.58)

for all r ∈

.

Proof: Let

→ and let λ ≡ liminf _n→∞ϕ. Is

ϕ (t,x) ≤ λ?

It suffices to assume λ < ∞ and by taking a subsequence, x_n ∈ D for all n and

ϕ (t ,x ) → λ. n n

Then

lim inf ϕ(t ,x ) = lim inf [ϕ(t ,x )− ϕ(t,x )+ ϕ(t,x )]. (33.7.59) n→ ∞ n n n→∞ n n n n

(33.7.59)

Now

limns→up∞ |ϕ (tn,xn)− ϕ (t,xn)| ≤

( ) lim sup K ϕ (tn,xn)+ |xn|2 +1 |tn − t| = 0. n→∞

Therefore, from 33.7.59

λ = limni→nf∞ ϕ(tn,xn) = lim nin→f∞ ϕ (t,xn) ≥ ϕ (t,x)

because of the assumption that ϕ

is lower semicontinuous. This proves the lemma.

In all that follows

is an element of L². Thus is the equivalence class of measurable square integrable functions which equal x a.e. This seems a little fussy but since the existence results are based on surjectivity theorems and the Hilbert space they apply to is L², it seems best to emphasize the equivalence classes of functions by using this notation, at least while proving theorems on existence and uniqueness.

Proof: This follows because, due to Lemma 33.7.1, ϕ is Borel measurable and so ϕ ∘ x is also measurable.

Now define the following function, Φ, on the Hilbert space, L²

.

{ ∫T Φ([x]) ≡ 0 ϕ(t,x(t))dt if x (t) ∈ D for all t for some x(⋅) ∈ [x] (33.7.60) +∞ otherwise

(33.7.60)

Note that since the functions ϕ

are proper, the top condition is equivalent to the condition

∫ T ϕ(t,x(t))dt if x (t) ∈ D a.e.for all x(⋅) ∈ [x]. 0

Proof: Since each ϕ

is nonnegative and convex, it follows that Φ is also nonnegative and convex. It remains to verify lower semicontinuity. Suppose, → in L² and let

λ = lim inf Φ ([xn]). n→∞

Is λ ≥ Φ

? It suffices to assume λ < ∞, x_n ∈ D for all t, and x_n → x a.e. say for tN where N has measure zero. Let

{ x (t) if t ∕∈ N ^x (t) = x (t) if t ∈ N 1

Then

= and ∈ D for all t. Then by pointwise convergence and Fatou’s lemma,

∫ ∫ Φ([x]) = Φ([^x]) = T ϕ(t,^x(t))dt ≤ T lim inf ϕ (t,x (t))dt 0 0 n→∞ n

∫ T ≤ limni→nf∞ 0 ϕ (t,xn(t))dt = lim nin→f∞ Φ ([xn]) ≡ λ.

This proves the lemma.

Define

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

∫ t x (t) = x0 + x′(s)ds where [x′] ∈ L2(0,T;H )} (33.7.61) 0

(33.7.61)

and for

∈ D,

L [x ] ≡ [x′].

The following lemma is easily obtained.

The main theorem is the following.

Proof: This is from Theorem 33.6.2. Since x₀ ∈ D, it follows from 33.7.58 that ϕ

is bounded.

Let

∈ D, the effective domain of Φ. Then there exists z ∈ such that z ∈ D for all t, and ∫ ₀^Tϕdt < ∞, so by convexity of ϕ and 33.7.62,

∫ t ϕ (t,Jλ[z](t)) ≤ e−tλ ϕ (t,x0)+ 1e −λt esλϕ(t,z(s))ds. (33.7.64) λ 0

(33.7.64)

Now the first term in 33.7.64 is bounded so consider the second. The integral in this term is of the form

∫ t ∫ t esλϕ (s,z (s))ds+ esλ (ϕ(t,z (s))− ϕ(s,z(s)))ds. (33.7.65) 0 0

(33.7.65)

Since

∈ D, ϕ < ∞ for all s and also the first integral in 33.7.65 is finite. By 33.7.58, the second term in 33.7.65 is dominated by

∫ t ( 2) Cλ K 1 + ϕ(s,z(s))+ |z(s)| |t− s|ds < ∞. 0

This shows ϕ

< ∞ for all t and so Φ is given by the top line of 33.7.60. Therefore, by convexity of ϕ and Jensen’s inequality,

∫ ( ∫ ) T −λt 1- −λt t sλ Φ([Jλ[z]]) = 0 ϕ t,e x0 + λ e 0 e z(s)ds dt

∫ T ( ∫ t ) ≤ e−tλ ϕ (t,x0)+ 1-e−λt esλϕ(t,z(s))ds dt 0 λ 0

∫ T −t ∫ T 1 −t ∫ t s = eλ ϕ(t,x0)dt+ -e λ eλϕ(s,z(s))dsdt 0 0 λ 0

∫ T ∫ t + 1e−tλ e sλ (ϕ(t,z (s))− ϕ (s,z(s)))dsdt. (33.7.66) 0 λ 0

(33.7.66)

By 33.7.58, the last term is dominated by

∫ T∫ t 1-−(t−s) ( 2) 0 0 λe λ K 1+ ϕ(s,z(s))+ |z (s)| |t− s|dsdt =

∫ ∫ T Te−(t−λs)t−-sdtK (1+ ϕ (s,z (s))+ |z (s)|2)ds 0 s λ

( 2) ≤ Cλ + Cλ Φ ([z]) + |[z]| . (33.7.67)

(33.7.67)

for some constant, C. From 33.7.58, ϕ

is bounded and so the first term in 33.7.66 is dominated by an expression of the form Cλ. Now consider the middle term of 33.7.66. Since ϕ is nonnegative,

∫ ∫ ∫ ∫ T 1-−tλ t sλ T T 1-−(t−λs) 0 λe 0 e ϕ(s,z(s))dsdt = 0 s λe dtϕ (s,z(s))ds

∫ T ∫ ∞ −u ≤ e duϕ(s,z(s))ds = Φ ([z]). (33.7.68) 0 0

(33.7.68)

It follows

( ) Φ ([Jλ[z]]) ≤ Φ ([z])+ Cλ + Cλ Φ([z])+ |[z]|2 .

The conditions of Theorem 33.6.2 are satisfied with K₁ = K₂ = C. This proves L + ∂Φ is maximal monotone on L²

and consequently there exists a unique solution to the differential inclusion of the theorem.

Of course it is desirable to be able to say that

∈ ∂Φ if and only if y ∈ ∂₂ϕ for some x ∈. To obtain this, here are two more assumptions. For all x ∈ H,

t → J1(t)x is measurable, (33.7.69)

(33.7.69)

where J₁

x is the solution, y, to y + ∂₂ϕ ∋ x, and there exists ∈ L² such that

2 [J1 (⋅)[ξ]] ∈ L (0,T;H ). (33.7.70)

(33.7.70)