The Hille-Yosida Theorem
We have seen that when the abstract IVP is uniquely solvable then the solution operator
defines a semigroup of bounded operators. We have not yet discussed the conditions under
which the IVP is uniquely solvable. However, it is clear that S(t) is some sort of generalized
version of exptA where A is an unbounded operator. To see this connection between S(t)
and the exponential of A, consider the scalar equation
u ′ t = −a ut,
u0 = u 0 .
ut = Stu 0  = e −at u 0 .
Then
In this simple situation, we have also
∞
ûs = Lut = ∫ e −st e −at u 0 dt = s u+0 a ;
0
i.e.,
∞
| ûs|= |∫ e −st Stu 0 dt|= s + a −1 u 0 ≤ 1s | u 0 |.
0
We will see this result appear again in a more general setting. An example more general
than this is provided by the system of linear ODE’s
⃗ ′ t = −AX
⃗ t
X
⃗ 0 = X
⃗ 0,
X
where A denotes an n by n symmetric matrix of constants. Then the solution to this system
is given by
n
⃗ t = ∑ X
⃗ 0, V
⃗ k e −λ k t V
⃗k
X
k=1
where
⃗k
λk, V
denote the eigenvalues and normalized eigenvectors of A. Then
n
⃗ t = ∑ X
⃗ 0, V
⃗k
X
∞
k=1
m=0
n
∞
⃗ 0, V
⃗k
=∑ X
k=1
n
−λ k t m ⃗
⃗ 0, V
⃗k
Vk = ∑ X
m!
k=1
∑
∞
t m − λ k  m V
⃗k
∑ m!
m=0
m
∞
t m −A m V
⃗k = ∑ X
⃗ 0, V
⃗ k ∑ −At V
⃗k
∑ m!
m!
n
m=0
k=1
n
n
k=1
k=1
m=0
⃗ 0, V
⃗ k e −At  V
⃗ k = e −At  ∑ X
⃗ 0, V
⃗k V
⃗ k = e −At X
⃗ 0.
=∑ X
∞
−At m
and, in fact, when A is a bounded linear operator on a
m!
m=0
Hilbert space H (as is the case in this example for H = R 1 ) then we can expect that
We observe that e −At  = ∑
M
St = lim M→∞ ∑
m=0
−At m
. = lim M→∞ S M t.
m!
For any fixed value of t, we have
M
||S M t − S N t|| LH ≤ ∑
m=N
|t| m
|| A|| mLH → 0 as M, N → ∞
m!
so the meaning of St = lim M→∞ S M t is to be understood as a limit in the complete normed
1
linear space, LH. Clearly the limit St satisfies
S0 = I
St + τ = StSτ
and
since these equalities hold for every S M t. Also, an easy calculation with the series shows
that
|| St − I|| LH ≤ |t| ||A|| LH e − |t| ||A|| Lh
and
St − I
+A
t
LH
≤ ||A|| LH ||St − I|| LH .
We say, in this case, that the semigroup St is uniformly continuous on H. Of course, all
this is true under the assumption that A is bounded.
On the other hand when A is unbounded, since DA n+1  ⊂ DA n , it may be that the
domain in the infinite sum shrinks to zero. We could make use of the fact that
Stu 0  solves
u ′ t = −Aut,
u0 = u 0 ,
ut + h  ut − h Aut = I − h Aut
to write
from which we find
ut =
I − nt A
n
u0
t = nh, n = 1, 2, . . . .
Since we are again iterating an unbounded operator, the difficulty of the shrinking
domain has not disappeared. However, if we write
ut + h  ut − h Aut + h or
I + h Aut + h = ut,
then
ut + h  I + h A −1 ut
and
ut 
I + nt A
−n
u0
so in this case we are iterating a bounded operator, I + h A −1 . The point of this discussion
is just that it is plausible that we will be able to find a meaningful definition for e At even
when A is unbounded but we must be careful how we do it. Of course, there must also be
some restrictions on A. We will now state and prove a theorem giving a set of necessary
and sufficient conditions on A in order that −A generates a C 0 −semigroup.
Theorem (Hille-Yosida) The following statements are equivalent:
1.
2.
−A : D A  H generates a C 0 −semigroup of contractions on H
a) A is closed and densely defined
b) ∀λ > 0
with
λ + A : D A  H is one to one and onto
λ + A −1
LH
≤ 1
λ
Proof- We already showed in a previous lemma that if B = −A, generates a C 0 −semigroup,
then B = −A is closed and densely defined. We also found that
2
t
i
Stu 0 − u 0 = ∫ Sτ Bu 0 dτ
ii
Stx − x = ∫ B Sτ x dτ
∀u 0 ∈ D A = D B
0
t
∀x ∈ H.
0
Now, for any λ > 0, Tt = e −λt St is a C 0 −semigroup of contractions that is generated by
B − λI, D B−λI = D B . Apply the results i, ii to Tt to get
t
Ttu 0 − u 0 = e −λt Stu 0 − u 0 = ∫ e −λτ Sτ B − λIu 0 dτ
0
t
Ttx − x = e −λt Stx − x = ∫ e −λτ B − λI Sτ x dτ
and
0
∀u 0 ∈ D A = D B
∀x ∈ H.
Now let t → ∞ and use the fact that −A = B is closed to conclude
∞
u 0 = ∫ e −λτ Sτ λI − Bu 0 dτ
0
∞
∀u 0 ∈ D A = D B
∞
x = ∫ λI − B e −λτ Sτ x dτ = λI − B ∫ e −λτ Sτ x dτ
0
That is,
0
∞
∀u 0 ∈ D A
u 0 = ∫ e −λτ Sτ λI + Au 0 dτ
∀x ∈ H
x = λI + A ∫ e −λτ Sτ x dτ
0
∀x ∈ H.
i. e. , λI + A is 1-1
∞
0
= λI + A z,
z ∈ DA
i. e. , λI + A is onto
∞
λI + A −1 x = ∫ e −λτ Sτ x dτ
Finally,
0
implies
∞
∞
||λI + A −1 x|| H ≤ ∫ 0 e −λτ dτ|| Sτ x|| H ≤ ∫ 0 e −λτ dτ || x|| H ≤ 1 || x|| H
λ
This proves that 2)a),b) are necessary conditions if −A is to generate a C 0 -semigroup of
contractions. Note that
∞
λI + A −1 x = ∫ e −λτ Sτ x dτ = LSt,
0
and
||λI + A −1 x|| H ≤ 1 || x|| H
λ
which generalizes the analogous result observed earlier for the scalar equation.
Now we suppose A satisfies 2)a),b) and we will show −A generates a C 0 -semigroup of
contractions, S(t). We will accomplish this by approximating A by a bounded linear operator
A λ and showing
A λ ∈ LH
A λ x  Ax
⇓
−tA λ
e
x  St x
in H as λ → ∞
∀x ∈ D A
in LH as λ → ∞
∀x ∈ H.
Now if A satisfies 2a) and 2b) then λI + A : D A → H is bijective for λ > 0 so that λI + A −1
exists. Then we may define A λ := λAλI + A −1 a bounded operator which will be shown to
be an approximation to A.
Lemma 1 Under the conditions 2a) and 2b), the operator A λ ∈ LH, λ > 0 and
3
a.
A λ = λI − λ 2 λI + A −1
b.
||A λ x|| H ≤ || Ax|| H
∀x ∈ D A
c.
A λ x  Ax in H as λ → ∞
∀x ∈ D A
Proof of lemma- For x ∈ D A write
λAλI + A −1 − λI λI + Ax = λAx − λ 2 x − λAx = −λ 2 x.
A λ − λI z = −λ 2 λI + A −1 z
i.e.,
∀z ∈ H
A λ z = λz − λ 2 λI + A −1 z
or
∀z ∈ H.
For x ∈ D A , 2b) implies
||A λ x|| H =
λAλI + A −1 x
H
≤ || Ax|| H .
If we combine these two results, we get
λλI + A −1 z − z
H
= 1 ||A λ z|| H ≤ 1 ||A z|| H
λ
λ
∀z ∈ D A
from which it follows that λλI + A −1 z  z in H as λ → ∞ ∀z ∈ D A . But D A is dense in H
and λλI + A −1 is uniformly bounded on H by 2b) hence, by continuity,
λλI + A −1 z  z
in H
as λ → ∞
∀z ∈ H.
Then A λ x = λλI + A −1 Ax  Ax in H as λ → ∞ ∀x ∈ D B . ■
Since A λ ∈ LH, λ > 0 we can define
∞
S λ t = e −tA λ = ∑
n=0
−t n n
Aλ
n!
for t ≥ 0, λ > 0.
Lemma 2 Under the conditions 2a) and 2b), for each λ > 0, S λ t : t ≥ 0 is a strongly
continuous semigroup of contractions on H with generator equal to −A λ . For each x ∈ D A
S λ tx converges in H as λ → ∞. Moreover, the convergence is uniform in t on all
bounded intervals 0, T.
Proof of lemma- Lemma 1a) implies S λ t = e −tA λ = e −tλ e λ
result of hypothesis 2b)
|| S λ t|| LH ≤ e −tλ || e λ
2 tλ+A −1
2 tλ+A −1
and then we have as a
|| LH ≤ e −tλ e tλ = 1.
Thus S λ t is a strongly continuous semigroup of contractions on H. Also
d
dt
S λ t = −A λ S λ t
and
S λ t − S μ t = ∫
t d
0 ds
t
S μ t − sS λ s ds = ∫ 0 S μ t − sS λ sA μ − A λ  ds.
Then
|| S λ tx − S μ tx|| H ≤ t|| A μ x − A λ x|| H
∀t ≥ 0, λ, μ > 0, x ∈ D A .
4
S λ tx is a Cauchy sequence in H, uniformly in t∈ 0, T. ■
i.e.,
To complete the proof of the theorem, observe that each S λ t is a contraction and D A is
dense in H, so the limit
S λ tx  Stx in H as λ → ∞,
t≥0
extends to all x in H, and holds uniformly in t ∈ 0, T. Since every S λ t is a strongly
continuous contraction, clearly St ∈ LH is a contraction and in addition, since the
convergence is uniform in t on bounded intervals, 0, T, it follows that St is strongly
continuous. The semigroup identity also holds since S λ t S λ τ  St Sτ etc. Then St is
a strongly continuous semigroup of contractions on H. Now for x ∈ D A and h > 0,
h
S λ hx − x = ∫ S λ t−A λ xdt
0
∀λ > 0,
and as λ → ∞, this becomes
h
Shx − x = ∫ St−Axdt
0
and this implies that the generator B of S(t) is an extension of −A. But we have assumed
that I + A is onto and since B is the generator of S(t), I − B is one to one. Then
I − BD A = I + AD A = H, which is to say, I − B −1 H = D A or B = −A. ■
Now we can prove:
Theorem- Existence and Uniqueness for the IVP
Suppose −A generates St, a strongly continuous semigroup of contractions on H. Then for
any u 0 ∈ D A , ut = Stu 0  satisfies
i
ut ∈ C 0 0, ∞; H ∩ C 1 0, ∞; H,
ii
u ′ t + Aut = 0,
iii
u0 = u 0
t > 0,
and ut is unique.
Proof- We have already shown that for St, a strongly continuous semigroup and
u 0 ∈ D A , Stu 0  satisfies i) and iii), and moreover,
d
dt
Stu 0  = BStu 0  = −AStu 0 
∀u 0 ∈ D A
The Hille-Yosida theorem shows that if −A generates a strongly continuous semigroup,
then
λI + A −1 x
H
≤ 1 || x|| H
λ
λ > 0, x ∈ H,
or
λ 2 || z|| 2H ≤ || λI + Az|| 2H
λ > 0, z ∈ D A .
This implies
2Az, z H ≥ −
||A z|| 2H
λ
λ > 0, z ∈ D A
5
or
Az, z H ≥ 0
∀z ∈ D A .
Then A is accretive and ut = Stu 0  is unique.■
Corollary- Suppose −A generates St, a strongly continuous semigroup of contractions on
H. Then for any u 0 ∈ D A , and every f ∈ C 1 0, T; H, the unique solution of
u ′ t + Aut = ft,
0 < t < T,
u0 = u 0 ,
t
is given by
ut = Stu 0  + ∫ St − sfs ds.
Proof- Let
gt = ∫ St − sfs ds.
0
t
0
Then g0 = 0 and
gt + h − gt = ∫
t+h
0
=∫
t
St + h − sfsds − ∫ St − sfs ds
t+h
0
0
t
Sσft + h − σ ds − ∫ Sσft − σ ds
0
t
t+h
0
t
= ∫ Sσft + h − σ − ft − σ ds + ∫
Sσft + h − σ ds
Dividing by h > 0 and letting h → 0, we find
t
g ′ t = ∫ Sσf ′ t − σ ds + Stf0 ,
0
which shows that g is differentiable.
But,
t+h
t
gt + h − gt
= 1 ∫ St + h − sfsds − ∫ St − sfs ds
0
h 0
h
t+h
Sh − I t
∫ 0 St − sfs ds + 1 ∫ t St + h − sfsds
=
h
h
t+h
Sh − I
=
gt + 1 ∫ St + h − sfsds
h
h t
and, as h → 0,
gt + h − gt
 g ′ t
h
and
1 ∫ t+h St + h − sfsds  S0 ft = ft.
h t
Since these limits exist, it follows that the limit
lim h→0
Sh − I
gt
h
exists and equals −Agt. Then we conclude g ′ t − ft = −Agt, and gt satifies
g ′ t + Agt = ft, t ∈ 0, T
and g0 = 0.
Evidently, ut = Stu 0  + gt solves the inhomogeneous IVP. This solution is unique
since if there are two such solutions, their difference satisfies the IVP with
ft = 0, u 0 = 0, and then this difference is zero since A is accretive.■
6
Examples1. Consider the Banach space
X = L 1 0, ∞ with A = − dxd ,
DA =
following initial value problem,
u ∈ X : Au ∈ X . This corresponds to solving the
u ′ t + Aut = ∂ t ux, t − ∂ x ux, t = 0,
ux, 0 = u 0 x ∈ D A
Note that if ux and − u ′ x = Aux both belong to X , then ux → 0 as x → ∞. Then it
follows that for λ ≥ 0, and u ∈ D A
λI + Aux = λux − u ′ x = 0 implies
ux = Ce λx
i.e.,
ux = 0;
and ux → 0 as x → ∞ if and only if C = 0.
Then λI + A −1 exists. In fact,
λI + Aux = vx
x
ux = Ce λx − ∫ e λx−y vy dy,

0
∞
where u ∈ D A if C = ∫ e −λy vy dy. Then we have,
0
∞
∞
ux = ∫ e λx−y vy dy = ∫ e −λz vx + z dz = λI + A −1 vx.
x
0
Note that
λI + A −1 v
X
∞
∞
0
x
= ∫ ∫ e λx−y vy dy dx
∞ ∞
∞ y
0
0
≤ ∫ ∫ e λx−y |vy| dydx ≤ ∫ ∫ e λx−y |vy| dx dy
x
0
∞
≤ 1 ∫ 1 − e −λy |vy| dy ≤ 1 || v|| X
λ 0
λ
If we take A to be a closed extension of the indicated operator then we have an operator
that satisfies the hypotheses of the Hille Yosida theorem and it follows that -A generates a
C 0 −semigroup of contractions on X. Note that the equation at the bottom of page 3 implies
∞
∞
λI + A −1 vx = ∫ 0 e −λτ vx + τ dτ = ∫ 0 e −λτ Sτvx dτ,
from which it follows that for x, τ ≥ 0, Sτvx = vx + τ for v ∈ X. Then the solution of
u ′ t + Aut = ∂ t ux, t − ∂ x ux, t = 0,
is given by
speed one.
ux, 0 = u 0 x ∈ D A
ux, t = Stu 0 x = u 0 x + t. This is a wave travelling from right to left with
2. Consider the Banach space
X = L 1 0, ∞ with A = dxd ,
D A = u ∈ X : Au ∈ X, u0 = 0 . This corresponds to
solving the following initial-boundary value problem,
u ′ t + Aut = ∂ t ux, t + ∂ x ux, t = 0,
ux, 0 = u 0 x ∈ D A
u0, t = 0,
t > 0,
Then it follows that for λ ≥ 0, and u ∈ D A
λI + Aux = λux + u ′ x = 0 implies
ux = 0
7
since
ux = Ce −λx and u0 = 0 implies C = 0.
Then λI + A −1 exists. In fact,
x
ux = ∫ e −λx−y vy dy = λI + A −1 vx.
λI + Aux = vx, u0 = 0 
∞
−1
λI + A vx
Then
X
0
x
= ∫ ∫ e −λx−y vy dy dx ≤ 1 || v|| X
0
0
λ
and
∞
x
λI + A −1 vx = ∫ 0 e −λx−y vy dy = ∫ 0 e −λτ Sτvx dτ.
But
x
x
∫ 0 e −λx−y vy dy = ∫ 0 e −λτ vx − τ dτ
hence
∞
x
∫ 0 e −λτ vx − τ dτ = ∫ 0 e −λτ Sτvx dτ.
Then
Sτvx =
vx − τ
0
if
0≤τ≤x
if
τ>x
,
and the solution of the initial boundary value problem
is given by
ux, t = Stu 0 x =
u 0 x − t
0
if
0≤t≤x
if
t>x
This is a wave travelling from left to right with speed one.
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