Summary of Semigroup Results
A is closed and densely defined
λI + A −1 exists for all λ > 0
−1
‖λI + A ‖ LH ≤ λ
−1
Hille-Yosida properties
for all λ > 0
H-Y properties  ∀ u ∈ H and ∀ t ≥ 0, S n tu converges strongly in H to Stu.
St = lim
S n t has the following properties:
n
St is stronly continuous in t, t ≥ 0
S0 = I
St is a C 0 − s/g of contractions
St + s = St ∘ Ss s, t ≥ 0
‖St‖ LH ≤ 1
Also
St : D A  D A
AStu = StAu
∀u ∈ D A
′
S tu = −AStu = −StAu
St − I
lim
u = −Au
t
t→0
∀u ∈ D A
Stu − u = − ∫ SτAu dτ
∀u ∈ D A
t
0
∀u ∈ D A
Then
H-Y properties 
− A generates a C 0 − s/g of contractions
Finally, if −A generates the C 0 − s/g of contractions, St, then −λI + A generates the
semigroup, Jt = e −λt St and:
u=
λI + A −1 v =
∞
∫ 0 e −λτ SτλI + Au dτ
∀u ∈ D A
∫ 0 e −λτ Sτv dτ
∀v ∈ H
∞
It follows from these last two results that
−A generates a C 0 − s/g of contractions

H-Y properties
We also have
1
A is accretive
Lumer-Phillips properties
I + A : D A  H is onto
and
L-P properties 
− A generates a C 0 − s/g of contractions
If −A generates a C 0 − s/g of contractions, St, then
∀u 0 ∈ D A and ∀f ∈ C 1 0, T : H
t
ut = Stu 0 + ∫ St − τfτdτ ∈ C0, T : D A  ∩ C 1 0, T : D A 
0
solves
u ′ t + Aut = ft,
0<t<T
u0 = u 0
and
Examples
1.Let H = L 2 0, ∞ = H 0 R + ,
A = −a ∂ x , and D A = H 1 0, ∞ for a > 0.
Note that u ∈ D A implies ux → 0 as x → ∞. Then
∞
Au, u H = −a ∫ 0 u ′ xuxdx = − a2 ux 2 | ∞0 =
a
2
u0 2 ≥ 0
Then A is accretive for a > 0, and for arbitrary v ∈ H
I + Au = ux − au ′ x = vx
has the solution
∞
ux = ∫ e
x
1
a x−y vydy
This shows that I + A is onto (i.e., RI + A = H). Then by the L-P theorem, −A generates a
C 0 − s/g, St and the unique solution of
u ′ t + Aut = ∂ t ux, t − a∂ x ux, t = fx, t,
ux, 0 = u 0 x ∈ D A ;
is given by,
t
t
0
0
ux, t = Stu 0 x + ∫ St − τfx, τdτ = u 0 x + at + ∫ fx + at − τ, τdτ
In order to have this solution it is sufficient to suppose u 0 ∈ D A = H 1 0, ∞ and f is C 1 in t
and L 2 in x.
2. Let H = L 2 0, ∞ = H 0 R + ,
A = a∂ x , and D A =
Note that u ∈ D A implies ux → 0 as x → ∞. Then
∞
Au, u H = a ∫ 0 u ′ xuxdx =
a
2
u ∈ H 1 0, ∞ : u0 = 0 .
ux 2 | ∞0 = 0
2
Then A is accretive (for any real a) and for arbitrary v ∈ H
I + Au = ux + au ′ x = vx
for v ∈ H
has the solution
x
1
ux = ∫ e − a x−y vydy.
0
Note that it is necessary to have a > 0 in order to get u ∈ D A for v ∈ H since a negative
value for a would produce an exponentially growing ux which is not even in H much less in
D A . This shows that RI + A = H for a > 0. .Then −A generates a C 0 − s/g, St and
ut = Stu 0 is the unique solution of
u ′ t + Aut = ∂ t ux, t + a∂ x ux, t = 0, u0, t = 0,
ux, 0 = u 0 x ∈ D A ;
i.e,
ux, t = u 0 x − atHx − at = Stu 0 x.
3. Let H = L 2 R = H 0 R,
ux → 0 as x 2 → ∞. Then
A = ∂ x , and D A = H 1 R. Note that u ∈ D A implies
∞
Au, u H = ∫ −∞ u ′ xuxdx =
1
2
ux 2 | ∞−∞ = 0.
so A is accretive. Note that −A is also accretive in this case. For arbitrary v ∈ H
I ± Au = ux ± u ′ x = vx
for v ∈ H
implies via Fourier transformation that
1 ± iαUα = Vα
Then
Uα = 1 ∓ iα
Vα
= Wα ∓ iαWα
1 + α2
and
ux = wx ∓ w ′ x
wx = T −1
F
Vα
1 + α2
= ∫ e −|x−y| vydy
R
Then the ODE has the solution
x
x
ux = ∫ e −|x−y| vydy ∓ d ∫ e −|x−y| vydy
0
0
dx
which shows that RI ± A = H. Then −A generates a C 0 − s/g, St and ut = Stu 0 is the
unique solution of
u ′ t + Aut = ∂ t ux, t + ∂ x ux, t = 0, ux, 0 = u 0 x ∈ D A ;
i.e,
ux, t = u 0 x + t = Stu 0 x.
But +A also generates a C 0 − s/g, Zt and then ut = Ztu 0 is the unique solution of
u ′ t − Aut = ∂ t ux, t − ∂ x ux, t = 0, ux, 0 = u 0 x ∈ D A ;
i.e,
3
ux, t = u 0 x − t = Ztu 0 x = S−tu 0 x.
So in this case, since both A and −A are accretive and I ± A is onto, there are two
C 0 − s/g ′ s , St and S−t. Moreover, by the semigroup property, St ∘ S−t = S0 = I,
which is to say St −1 = S−t; i.e., St forms a C 0 −group for t ∈ R.
4. Let H = L 2 U = H 0 U,
A = −∇ 2 , and D A = H 10 U ∩ H 2 U. Then
Au, u H = − ∫ u ∇ 2 u = ∫ |∇u| 2 ≥ 0
U
U
u ∈ DA
So A is accretive. For arbitrary f ∈ H, the elliptic problem
λI + Au = λu − ∇ 2 u = f in U,
u = 0 on ∂U
has a unique weak solution u ∈ D A since
bu, v; λ = ∫ ∇u ⋅ ∇v + λuv
U
satisfies
bu, u; λ = ‖∇u‖ 20 + λ‖u‖ 20
u ∈ DA
and, in particular,
bu, u; 1 = ‖u‖ 21
which implies bu, u; 1 is coercive and I + A is then an isomorphism from D A onto H.
Then −A generates a C 0 − s/g, St and ut = Stu 0 is the unique solution of
and
u ′ t + Aut = ∂ t ux, t − ∇ 2 ux, t = 0,
u = 0 on ∂U × 0, T
ux, 0 = u 0 x ∈ D A ;
i.e,
ux, t = ∑ n u 0 , w n  H e −λ n t w n x = Stu 0 x.
where w n  are the family of orthonormal eigenfunctions associated with
−∇ 2 wx = λwx
x ∈ U,
w ∈ DA.
5.Consider the problem
∂ tt ux, t − ∂ xx ux, t = 0,
ux, 0 = fx,
∂ t ux, 0 = gx
0 < x < 1, t > 0
0 < x < 1,
t > 0,
u0, t = u1, t = 0,
Let
Then
u1 = ∂xu
and
0 < x < 1.
u2 = ∂tu
∂ t u 1 = ∂ xt u = ∂ x u 2
∂ t u 2 = ∂ tt u = ∂ x u 1
i.e.,
4
∂t
u1
0 1
−
u2
u1
∂x
1 0
u2
0
=
0
and
u1
u2
i.e.,
f ′ x
x, 0 =
gx
⃗ t + AU
⃗ t = 0,
∂tU
⃗ 0 = U
⃗0
U
where
H = L 2 0, 1 2 ,
0 1
A=−
1 0
∂x
⃗ ∈ H : u 1 ∈ H 1 0, 1, u 2 ∈ H 10 0, 1
D A = U
Then
⃗, U
⃗  H = − ∫ 1 ∂ x u 2 ⋅ u 1 + u 2 ⋅ ∂ x u 1 dx = − ∫ 1 d/dxu 1 u 2 dx
AU
0
0
1
= −u 1 u 2 | x=1
x=0 = 0 (since u 2 ∈ H 0 0, 1
⃗ ∈ H, consider
This proves A is accretive, in fact, conservative. Now for λ ≠ 0, F
λ
u1
u2
+A
u1
u2
=
λu 1 − ∂ x u 2
λu 2 − ∂ x u 1
F1
=
F2
Then
λ∂ x u 1 − ∂ xx u 2 = ∂ x F 1
and
∂ x u 1 = λu 2 − F 2 ,
or
−∂ xx u 2 + λ 2 u 2 = ∂ x F 1 + λF 2
Since ∂ x F 1 + λF 2 ∈ H −1 0, 1, this last equation has a unique weak solution u 2 ∈ H 10 0, 1, by
the previously developed elliptic theory. Then
λu 1 = F 1 + ∂ x u 2 ∈ L 2 0, 1,
∂ x u 1 = λu 2 − F 2 ∈ L 2 0, 1,
⃗ ∈ D A . This shows that λ + A : D A  H is surjective for all λ ≠ 0.
so u 1 ∈ H 1 0, 1 and U
Then A generates a C 0 − s/g, actually both ±A generate C 0 − s/g ′ s so A generates a
group of solution operators, Gt.
This group Gt, (using a previously discovered solution to the wave equation) is seen to be
given by
⃗ x, 0 =
GtU
1
2
1
2
f̃ ′ x + t + ̃f ′ x − t +
f̃ ′ x + t − ̃f ′ x − t +
1
2
1
2
g̃ x + t − g̃ x − t
g̃ x + t + g̃ x − t
where ̃f, g̃ denote the odd 2-periodic extensions of f and g.
5