33.2 Evolution Inclusions
One of the interesting things about maximal monotone operators is the concept of
evolution inclusions. To facilitate this, here is a little lemma.
Lemma 33.2.1 Let f :
→ ℝ be continuous and suppose
+ f (t+ h)− f (t)
D f (t) ≡ lim sup------h-------< g(t)
h→0+
where g is a continuous function. Then
∫ t
f (t)− f (0) ≤ g(s)ds.
0
Proof: Suppose this is not so. Then let
{ ∫ t }
S ≡ t ∈ [0,T ] : f (t)− f (0) > g(s)ds
0
and it would follow that S ≠ ∅ . Let a = inf S. Then there exists a decreasing sequence
h _{n} → 0 such that
∫ a+hn
f (a+ hn)− f (0) > g(s)ds (33.2.5)
0
(33.2.5)
First suppose a = 0. Then dividing by h _{n} and taking the limit,
a contradiction. Therefore, assume a > 0. Then by continuity
∫ a
f (a)− f (0) ≥ g(s)ds
0
If strict inequality holds, then a ≠ inf S . It follows
∫
f (a)− f (0) = ag(s)ds
0
and so from 33.2.5
f (a+-hn-)−-f (a) 1-∫ a+hn
hn > hn a g (s)ds.
Then doing limsup_{n→∞} to both sides,
the same sort of contradiction obtained earlier. Thus S = ∅ and this proves the
lemma.
The following is the main result.
Theorem 33.2.2 Let H be a Hilbert space and let A be a maximal monotone operator
as described above. Let f :
→ H be continuous such that f ^{′} ∈ L ^{2} . Then
there exists a unique solution to the evolution inclusion
y′ + Ay ∋ f,y(0) = y0 ∈ D (A)
Here y ^{′} exists a.e., y
∈ D a.e., y is continuous.
Proof: Let y _{λ} be the solution to
y′λ +A λyλ = f,yλ(0) = y0
I will base the entire proof on estimating the solutions to the corresponding integral
equation
∫ t ∫ t
yλ(t)− y0 + 0 Aλyλ(s)ds = 0 f (s) ds (33.2.6)
(33.2.6)
Let h,k be small positive numbers. Then
∫ t+h ∫ t+h
y (t+ h)− y (t)+ A y (s)ds = f (s)ds (33.2.7)
λ λ t λ λ t
(33.2.7)
Next consider the difference operator
g(t+-k)−-g-(t)
Dkg (t) ≡ k
Do this D _{k} to both sides of 33.2.7 where k < h . This gives
(∫ ∫ )
1- t+h+k t+k
Dk (yλ(t+ h)− yλ(t))+ k t+h Aλyλ (s)ds − t A λyλ(s)ds
( )
1 ∫ t+h+k ∫ t+k
= k- t+h f (s)ds− t f (s)ds (33.2.8)
(33.2.8)
Now multiply both sides by y _{λ}
− y _{λ} . Consider the first term. To
simplify the ideas consider instead
( )
(Dkg (t),g (t+ k)) = 1-|g(t+ k)|2 − (g(t),g(t+ k))
k( )
≥ 1-|g(t+ k)|2 − |g(t)||g(t+ h)|
k( )
≥ 1- 1|g(t+ k)|2 − 1|g(t)|2 (33.2.9)
k 2 2
Then applying this simple observation to
33.2.8 ,
( )
11- |yλ (t +h + k)− yλ(t+ k)|2 − |yλ(t+ h)− yλ(t)|2 +
2k
( ( ) )
1 ∫ t+h+k ∫ t+k
k- A λyλ(s)ds− Aλyλ(s)ds ,yλ(t+ h+ k)− yλ (t+ k) +
t+h t
( ( ∫ ∫ ) )
1- t+h+k t+k
≤ k t+h f (s)ds− t f (s)ds ,yλ(t+ h+ k)− yλ (t +k)
Taking limsup_{k→0} of both sides yields
( )
1D+ |yλ(t+ h)− yλ (t)|2 + (A λyλ(t+ h)− Aλyλ (t),yλ(t+ h)− yλ(t))
2
≤ (f (t+ h)− f (t),yλ(t+ h)− yλ(t))
Now recall that A _{λ} is monotone. Therefore,
( )
D+ |yλ(t+ h)− yλ (t)|2 ≤ |f (t+ h)− f (t)|2 + |yλ (t+ h)− yλ(t)|2
From Lemma 33.2.1 it follows that for all ε > 0,
|yλ(t+ h)− yλ(t)|2 − |yλ(h)− y0|2
∫ t 2 ∫ t 2
≤ |f (s+ h) − f (s)| ds+ |yλ(s+ h)− yλ(s)|ds + εt
0 0
and so since
ε is arbitrary, the term
εt can be eliminated. By Gronwall’s inequality,
( ∫ t )
|yλ(t+ h)− yλ(t)|2 ≤ et |yλ (h) − y0|2 + |f (s+ h)− f (s)|2ds . (33.2.10)
0
(33.2.10)
The last integral equals
∫ |∫ |2 ∫ ∫
t|| s+h ′ || t s+h ′ 2
0 ||s f (r)dr|| ds ≤ 0 h s |f (r)|drds
[∫ h∫ r ∫ t∫ r ∫ t+h∫ t ]2
= h |f ′(r)|2 dsdr + |f′(r)|2dsdr + |f′(r)| dsdr
0 0 h r− h t r−h
∫
2 t+h ′ 2
≤ h 0 |f (r)| dr
and now it follows that for all t + h < T,
|| ||2 ( || ||2 )
||yλ(t+-h)−-yλ(t)|| ≤ eT ||yλ(h)−-y0|| + ||f′||2L2(0,T;H) . (33.2.11)
h h
(33.2.11)
Now return to 33.2.7 .
| | | | | |
||yλ-(h-)−-y0|| ||1∫ h || ||1∫ h ||
| h | ≤ ||h 0 Aλyλ(s)ds||+ ||h 0 f (s)ds||
Then taking limsup_{h→0} of both sides
|| ||
lim sup||yλ(h)−-y0|| ≤ |A λy0|+ |f (0)|
h→0 h
From Lemma 33.1.3 ,
≤ for all
a ∈ Ay _{0} . This is where
y _{0} ∈ D is used.
Thus from
33.2.11 , there exists a constant
C independent of
t and
h and
λ such
that
||yλ (t +h )− yλ(t)||2
||------h-------|| ≤ C
From the estimate just obtained and 33.2.7 , this implies
∫ t+h ∫ t+h
yλ-(t+-h)-−-yλ-(t) + 1- A y (s)ds = 1 f (s)ds (33.2.12)
h h t λ λ h t
(33.2.12)
Now letting h → 0, it follows that for all t ∈ [0,T ), there exists a constant C independent
of t,λ such that
This is a very nice estimate. The next task is to show uniform convergence of the y _{λ}
as λ → 0. From 33.2.7
(Dh (yλ (t)− yμ(t)) ,yλ (t+ h) − yμ (t+ h)) +
( 1 ∫ t+h )
-- (A λyλ(s)− Aμyμ (s))ds,yλ(t+ h)− yμ(t+ h) = 0
h t
Then from the argument in 33.2.9 ,
( )
1-1 |yλ(t+ h)− yμ(t+ h)|2 − |yλ(t)− yμ(t)|2
h 2
( )
1-∫ t+h
+ h t (A λyλ(s)− Aμyμ(s))ds,yλ(t+ h)− yμ(t+ h) ≤ 0
Now take limsup_{h→0} to obtain
1 2
2D+ |yλ(t)− yμ(t)| + (Aλyλ(t)− Aμyμ (t),yλ(t)− yμ(t)) ≤ 0
Using the definition of A _{λ} this equals
1 2
2 D+ |yλ (t)− yμ(t)|+
(A λyλ (t)− A μyμ(t),λA λyλ(t)+ Jλyλ(t) − (μA μyμ(t) + Jμyμ(t))) ≤ 0
Now this last term splits into the following sum
(A λyλ(t)− A μyμ(t) ,λA λyλ(t)− μAμyμ (t))
+ (Aλyλ (t)− A μyμ(t),Jλyλ(t)− Jμyμ(t))
By Lemma
33.1.3 the second of these terms is nonnegative. Also from the estimate
33.2.13 , the first term converges to 0 uniformly in
t as
λ,μ → 0
. Then by Lemma
33.2.1 it
follows that if
λ is any sequence converging to 0,
y _{λ} is uniformly Cauchy.
Let
Thus y is continuous because it is the uniform limit of continuous functions. Since
A _{λ} y _{λ}
is uniformly bounded, it also follows
y(t) = lim Jλyλ(t) uniformly in t. (33.2.14)
λ→0
(33.2.14)
Taking a further subsequence, you can assume
A λyλ ⇀ z weak ∗ in L∞ (0,T ;H). (33.2.15)
(33.2.15)
Thus z ∈ L ^{∞}
. Recall
A _{λ} y _{λ} ∈ AJ _{λ} y _{λ} .
Now A can be considered a maximal monotone operator on L ^{2}
according to
the rule
where
D (A) ≡ {f ∈ L2 (0,T;H ) : f (t) ∈ D (A) a.e. t}
By Lemma 33.1.5 applied to A considered as a maximal monotone operator on
L ^{2}
and using
33.2.14 and
33.2.15 , it follows
y ∈ D a.e.
t and
z ∈ Ay
a.e.
t. Then passing to the limit in
33.2.6 yields
∫ t ∫ t
y(t)− y0 + z(s)ds = f (s)ds. (33.2.16)
0 0
(33.2.16)
Then by fundamental theorem of calculus, y ^{′}
exists a.e.
t and
where z
∈ Ay a.e.
It remains to verify uniqueness. Suppose
is another pair which works. Then
from
33.2.16 ,
∫ t
y(t) − y1 (t)+ (z(r)− z1(r))dr = 0
∫0
s
y(s)− y1(s)+ 0 (z(r)− z1(r))dr = 0
Therefore for
s < t ,
∫
y(t)− y (t)− (y(s)− y (s)) = t(z(r)− z (r))dr
1 1 s 1
and so
||y (t)− y1(t)|− |y(s)− y1(s)|| ≤ K |s − t|
for some K depending on
_{L∞} , _{L∞} . Since
y,y _{1} are bounded, it follows
that
t → ^{2} is also Lipschitz. Therefore by Corollary
24.4.3 , it is
the integral of its derivative which exists a.e. So what is this derivative? As
before,
(Dh (y (t)− y1(t)) ,y (t +h )− y1(t +h ))
( )
1∫ t+h
+ h-t (z(s)− z1(s))ds,y(t+ h)− y1(t+ h) = 0
and so
( 2 2)
1- |y(t+-h)−-y1(t+-h)| − |y-(t)−-y1(t)|--
h 2 2