3.1. Preliminary
In order to obtain the explicit solution of
k
(
x
,
y
)
and
Φ
(
x
)
, we will firstly solve the following equation of
Φ
(
x
)
with assumptions
λ
(
x
)
∈
P
[
x
]
n
and
λ
(
x
)
∈
C
[
0
,
l
]
∞
, respectively:
(8)
Φ
′′
(
x
)

Φ
(
x
)
A

∫
0
x
∫
0
x

y
Φ
(
z
)
B
d
z
λ
(
y
)
d
y
+
λ
(
x
)
=
0
Φ
(
0
)
=
K
Φ
′
(
0
)
=
0
.
Lemma 1.
let
Φ
(
x
)
∈
R
1
×
n
,
λ
(
x
)
∈
R
1
×
n
,
A
∈
R
n
×
n
, and
B
∈
R
n
×
1
, let
λ
(
x
)
∈
P
[
x
]
n
, and then (8) about
Φ
(
x
)
has the unique solution.
Proof.
It follows easily from (8) that
(9)
Φ
′′
(
x
)

Φ
(
x
)
A
+
λ
(
x
)
=
∫
0
x
∫
0
x

y
Φ
(
z
)
B
d
z
λ
(
y
)
d
y
.
By taking the derivative on both sides about
x
and applying variable substitution, we have
(10)
Φ
′′′
(
x
)

Φ
′
(
x
)
A
+
λ
′
(
x
)
=
∫
0
x
Φ
(
x

y
)
B
λ
(
y
)
d
y
=
∫
0
x
Φ
(
y
)
B
λ
(
x

y
)
d
y
.
Furthermore, take the derivative repeatedly, and then
(11)
Φ
(
n
+
4
)
(
x
)

Φ
(
n
+
2
)
(
x
)
A
=
Φ
(
n
)
(
x
)
B
λ
(
0
)
+
⋯
+
Φ
(
x
)
B
λ
(
n
)
(
0
)
.
By this equation, we have
(12)
Φ
′
(
x
)
=
Φ
′
(
x
)
Φ
′′
(
x
)
=
Φ
′′
(
x
)
⋮
Φ
(
n
+
3
)
(
x
)
=
Φ
(
n
+
3
)
(
x
)
Φ
(
n
+
4
)
(
x
)
=
Φ
(
n
+
2
)
(
x
)
A
Φ
(
n
+
4
)
(
x
)
+
Φ
(
n
)
(
x
)
B
λ
(
0
)
+
⋯
+
Φ
(
x
)
B
λ
(
n
)
(
0
)
,
and with the initial values of (8), we have following initial values:
(13)
Φ
(
0
)
=
K
Φ
′
(
0
)
=
0
Φ
′′
(
0
)
=
Φ
(
0
)
A

λ
(
0
)
Φ
′′′
(
0
)
=
Φ
′
(
0
)
A

λ
′
(
0
)
⋮
Φ
(
n
+
3
)
(
0
)
=
Φ
(
n
+
1
)
(
0
)
A
+
Φ
(
n

1
)
(
0
)
B
λ
(
0
)
Φ
(
n
+
3
)
(
0
)
+
⋯
+
Φ
(
0
)
B
λ
(
n

1
)
(
0
)
.
Then the solution to the ODEs (12) with the above initial values can be expressed as follows:
(14)
Φ
(
x
)
=
Γ
(
0
)
e
D
x
E
with
Γ
(
0
)
=
(
K
0
Φ
′′
(
0
)
⋯
Φ
(
n
+
2
)
(
0
)
Φ
(
n
+
3
)
(
0
)
)
1
×
[
n
(
n
+
4
)
]
, where the initial values of higherorder derivatives of
Φ
in zero are denoted by (13) and
(15)
D
=
(
0
n
×
n
⋯
⋯
⋯
⋯
0
n
×
n
B
λ
(
n
)
(
0
)
I
n
×
n
B
λ
(
n

1
)
(
0
)
⋱
0
⋮
I
n
×
n
B
λ
(
0
)
I
n
×
n
0
n
×
n
0
I
n
×
n
A
I
n
×
n
0
n
×
n
)
[
n
×
(
n
+
4
)
]
×
[
n
×
(
n
+
4
)
]
,
E
=
(
I
n
×
n
0
n
×
n
⋮
0
n
×
n
)
[
n
×
(
n
+
4
)
]
×
n
.
Lemma 2.
Let
Φ
(
x
)
∈
R
1
×
n
,
λ
(
x
)
∈
R
1
×
n
,
A
∈
R
n
×
n
,
B
∈
R
n
×
1
, and
λ
(
x
)
∈
C
[
0
,
l
]
∞
, and then (8) has the following unique solution:
(16)
Φ
(
x
)
=
∑
n
=
0
+
∞
Δ
Φ
n
(
x
)
,
with
Δ
Φ
0
(
x
)
=
K

∫
0
x
∫
0
τ
λ
(
ξ
)
d
ξ
d
τ
,
Δ
Φ
n
+
1
(
x
)
=
F
[
Δ
Φ
n
]
(
x
)
,
n
=
0,1
,
2
,
…
where
(17)
F
[
Δ
Φ
n
]
(
x
)
=
∫
0
x
∫
0
τ
{
[
∫
0
ξ
(
∫
0
ξ

y
Δ
Φ
n
(
z
)
B
d
z
)
λ
(
y
)
d
y
]
Δ
Φ
n
(
ξ
)
A
f
f
f
f
f
i
f
f
i
f
f
i
+
[
∫
0
ξ
(
∫
0
ξ

y
Δ
Φ
n
(
z
)
B
d
z
)
λ
(
y
)
d
y
]
}
d
ξ
d
τ
.
Proof.
By integrating into (8) over
[
0
,
x
]
, we have
(18)
Φ
′
(
x
)
=
∫
0
x
{
Φ
(
ξ
)
A
+
[
∫
0
ξ
(
∫
0
ξ

y
Φ
(
z
)
B
d
z
)
λ
(
y
)
d
y
]
}
d
ξ

∫
0
x
λ
(
ξ
)
d
ξ
.
Then, integrating again on both sides of (18) over
[
0
,
x
]
with initial value
Φ
(
0
)
=
K
of (8), we conclude
(19)
Φ
(
x
)
=
Δ
Φ
0
(
x
)
+
F
[
Φ
]
(
x
)
,
where
F
[
Φ
]
(
x
)
=
∫
0
x
∫
0
τ
{
Φ
(
ξ
)
A
+
[
∫
0
ξ
(
∫
0
ξ

y
Φ
(
z
)
B
d
z
)
λ
(
y
)
d
y
]
}
d
ξ
d
τ
and
Δ
Φ
0
(
x
)
=
K

∫
0
x
∫
0
τ
λ
(
ξ
)
d
ξ
d
τ
.
Denoting the following iterative relationship:
(20)
Φ
n
+
1
(
x
)
=
Δ
Φ
0
(
x
)
+
F
[
Φ
n
]
(
x
)
,
n
=
0,1
,
2
,
…
,
it suffices to show that if series
{
Φ
n
(
x
)
}
was convergence, then (16) is the unique solution of (19). Considering the difference
(21)
Δ
Φ
n
+
1
(
x
)
=
Φ
n
+
1
(
x
)

Φ
n
(
x
)
=
F
[
Φ
n
]
(
x
)

F
[
Φ
n

1
]
(
x
)
=
F
[
Δ
Φ
n
]
(
x
)
,
now, we will estimate
Δ
Φ
n
(
x
)
by induction. First, for
Δ
Φ
0
(
x
)
, we have
(22)

Δ
Φ
0
(
x
)

≤

K

+
λ

l
2
2
=
M
,
where
λ

=
max
x
∈
[
0
,
l
]

λ
(
x
)

is denoted as above and
l
is the length of the PDE domain. Then, suppose that
(23)

Δ
Φ
n
(
x
)

≤
M
N
n
x
2
n
(
2
n
)
!
with
N
=

A

+
l
2

B

λ

. Then

Ψ
n
+
1
(
x
)

can be estimated as follows:
(24)

Δ
Φ
n
+
1
(
x
)

≤
M
N
n
(
2
n
)
!

A

∫
0
x
∫
0
τ
ξ
2
n
d
ξ
d
τ
+
M

B

λ

N
n
(
2
i
)
!
∫
0
x
∫
0
τ
∫
0
ξ
∫
0
ξ

y
z
2
n
d
z
d
y
d
ξ
d
τ
≤
M
N
n
+
1
x
2
n
+
2
(
2
n
+
2
)
!
.
Noting
x
∈
[
0
,
l
]
, we have
(25)

Φ
(
x
)

≤
∑
n
=
0
∞

Δ
Φ
n
(
x
)

≤
∑
n
=
0
+
∞
M
N
n
l
2
n
(
2
n
)
!
≤
M
e
l
N
.
The series on the righthand side of (25) converges. Hence by Weierstrass’s Discriminance, the series defined by (16) converges absolutely and uniformly on
0
≤
x
≤
l
. Then the existence of the solution to (8) is concluded. To show the uniqueness of the solution (16) to (8), we assume that
Φ

and
Φ
~
are two different solutions of (8). Substituting these two solutions and after some direct calculation, we have
(26)
δ
Φ
(
x
)
=
Φ


Φ
~
=
F
[
δ
Φ
]
(
x
)
.
From (25), we know that

Φ
(
x
)

≤
M
e
l
N
, which means

δ
Φ
(
x
)

≤
2
M
e
l
N
. Next, we will estimate
δ
Φ
by induction. After some direct calculation, we have
(27)
δ
Φ
n
+
1
(
x
)
=
F
[
δ
Φ
n
]
(
x
)
.
Suppose that

δ
Φ
n
(
x
)

≤
2
M
e
l
N
N
n
(
x
2
n
/
(
2
n
)
!
)
, and then
(28)

δ
Φ
n
+
1
(
x
)

≤
2
M
e
l
N
N
n
(
2
n
)
!

A

∫
0
x
∫
0
τ
ξ
2
n
d
ξ
d
τ
+
2
M
e
l
N
N
n
(
2
n
)
!

B

λ

×
∫
0
x
∫
0
τ
∫
0
ξ
∫
0
ξ

y
z
2
n
d
z
d
y
d
ξ
d
τ
≤
2
M
e
l
N
N
n
+
1
x
2
n
+
2
(
2
n
+
2
)
!
,
which implies the trueness of

δ
Φ
n
(
x
)

≤
2
M
e
l
N
N
n
(
l
2
n
/
(
2
n
)
!
)
. Moreover, since
lim
n
→
+
∞
2
M
e
D
N
N
n
(
l
2
n
/
(
2
n
)
!
)
=
0
,
δ
Φ
(
x
)
≡
0
is easily concluded. Then
Φ

=
Φ
~
, and (16) is the unique solution to (18).
3.2. Design of the StateFeedback Controller with Backstepping Method
Next, we will obtain the backstepping transformation (5). Let
x
=
0
in (5), and we have
(29)
ω
(
0
,
t
)
=
u
(
0
,
t
)

Φ
(
0
)
X
(
t
)
,
and
Φ
(
0
)
=
K
by comparing (4) and (6). The partial derivatives of
ω
(
x
,
t
)
in (5) with respect to
x
are given by
(30)
ω
x
(
x
,
t
)
=
u
x
(
x
,
t
)

k
(
x
,
x
)
u
(
x
,
t
)

∫
0
x
k
x
(
x
,
y
)
u
(
y
,
t
)
d
y

Φ
′
(
x
)
X
(
t
)
,
(31)
ω
x
x
(
x
,
t
)
=
u
x
x
(
x
,
t
)

k
(
x
,
x
)
u
x
(
x
,
t
)

(
d
d
x
k
(
x
,
x
)
)
u
(
x
,
t
)

k
x
(
x
,
x
)
u
(
x
,
t
)

∫
0
x
k
x
x
(
x
,
y
)
u
(
y
,
t
)
d
y

Φ
′′
(
x
)
X
(
t
)
.
The derivative of
ω
(
x
,
t
)
with respect to
t
is
(32)
ω
t
(
x
,
t
)
=
u
x
x
(
x
,
t
)
+
λ
(
x
)
X
(
t
)

k
(
x
,
x
)
u
x
(
x
,
t
)
+
k
(
x
,
0
)
u
x
(
0
,
t
)
+
k
y
(
x
,
x
)
u
(
x
,
t
)

k
y
(
x
,
0
)
u
(
0
,
t
)

∫
0
x
k
y
y
(
x
,
y
)
u
(
y
,
t
)
d
y

∫
0
x
k
(
x
,
y
)
λ
(
y
)
X
(
t
)
d
y

Φ
(
x
)
(
A
X
(
t
)
+
B
u
(
0
,
t
)
)
.
By the target system (6), (31), and (32), we have
(33)
ω
t
(
x
,
t
)

ω
x
x
(
x
,
t
)
=
2
(
d
d
x
k
(
x
,
x
)
)
u
(
x
,
t
)
+
(
λ
(
x
)

∫
0
x
k
(
x
,
y
)
λ
(
y
)
d
y
f
f
i
f
f
i

Φ
(
x
)
A
+
Φ
′′
(
x
)
∫
0
x
)
X
(
t
)
+
∫
0
x
(
k
x
x
(
x
,
y
)

k
y
y
(
x
,
y
)
)
f
f
f
f
i
f
×
u
(
y
,
t
)
d
y
+
(

k
y
(
x
,
0
)

Φ
(
x
)
B
)
u
(
0
,
t
)
=
0
.
This equation should be valid for all
u
and
X
, so we have the following four equations:
(34)
d
d
x
k
(
x
,
x
)
=
0
λ
(
x
)

∫
0
x
k
(
x
,
y
)
λ
(
y
)
d
y

Φ
(
x
)
A
+
Φ
′′
(
x
)
=
0
k
x
x
(
x
,
y
)

k
y
y
(
x
,
y
)
=
0

d
d
y
k
(
x
,
0
)

Φ
(
x
)
B
=
0
.
Let
x
=
0
in (30), which gives
(35)
ω
x
(
0
,
t
)
=

k
(
0,0
)
u
(
0
,
t
)

Φ
′
(
0
)
X
(
t
)
.
Substituting this expression into the boundary condition in (6), we have
(36)

k
(
0,0
)
u
(
0
,
t
)

Φ
′
(
0
)
X
(
t
)
=
0
.
This equation should be valid for all
u
and
X
, so we have two conditions that
k
(
0,0
)
=
0
and
Φ
′
(
0
)
=
0
. In order to satisfy the conditions of the target system (6), the
k
(
x
,
y
)
and
Φ
(
x
)
in (5) should satisfy
(37)
k
x
x
(
x
,
y
)
=
k
y
y
(
x
,
y
)
k
y
(
x
,
0
)
=

Φ
(
x
)
B
d
d
x
k
(
x
,
x
)
=
0
k
(
0,0
)
=
0
,
(38)
λ
(
x
)

∫
0
x
k
(
x
,
y
)
λ
(
y
)
d
y

Φ
(
x
)
A
+
Φ
′′
(
x
)
=
0
Φ
(
0
)
=
K
Φ
′
(
0
)
=
0
.
Note that (37) is a secondorder hyperbolic PDE about
k
(
x
,
y
)
and the boundary condition is related to
Φ
(
x
)
, and (38) is a secondorder integraldifferential equation about
Φ
(
x
)
associated with
λ
(
x
)
and
k
(
x
,
y
)
. Next, we will obtain
k
(
x
,
y
)
from (37) and
Φ
(
x
)
from (38).
Suppose
k
(
x
,
y
)
=
Θ
(
x

y
)
+
Υ
(
x
+
y
)
; it can be easily obtained by (37) that
(39)
k
(
x
,
y
)
=
∫
0
x

y
Φ
(
z
)
B
d
z
.
Substituting this expression into (38), we get
(40)
λ
(
x
)

∫
0
x
∫
0
x

y
Φ
(
z
)
B
d
z
λ
(
y
)
d
y

Φ
(
x
)
A
+
Φ
′′
(
x
)
=
0
Φ
(
0
)
=
K
Φ
′
(
0
)
=
0
.
For
λ
(
x
)
∈
P
[
x
]
n
, by Lemma 1, (40) has a unique solution as follows:
(41)
Φ
(
x
)
=
Γ
(
0
)
e
D
x
E
,
and thus the kernel function
k
(
x
,
y
)
can be expressed as follows by (39):
(42)
k
(
x
,
y
)
=
∫
0
x

y
Γ
(
0
)
e
D
z
E
B
d
z
.
For
λ
(
x
)
∈
C
[
0
,
D
]
∞
, by Lemma 2, (40) has a unique series solution as follows:
(43)
Φ
(
x
)
=
∑
n
=
0
+
∞
Δ
Φ
n
(
x
)
,
and thus the kernel function
k
(
x
,
y
)
can be expressed as follows by (39):
(44)
k
(
x
,
y
)
=
∫
0
x

y
∑
n
=
0
+
∞
Δ
Φ
n
(
x
)
B
d
z
.
Next, we will obtain the inverse transformation of (5) by using a process similar to the one we used above in obtaining the kernels
k
(
x
,
y
)
and
Φ
(
x
)
. Actually, the inverse of the transformation
(
X
,
ω
)
↦
(
X
,
u
)
can be found as follows:
(45)
u
(
x
,
t
)
=
ω
(
x
,
t
)
+
∫
0
x
ι
(
x
,
y
)
ω
(
y
,
t
)
d
y
+
Ψ
(
x
)
X
(
t
)
.
The kernel functions
ι
(
x
,
y
)
and
Ψ
(
x
)
can be easily obtained by a method similar to that above
(46)
ι
(
x
,
y
)
=
∫
0
x

y
Ψ
(
z
)
B
d
z
,
Ψ
(
x
)
=
(
Ξ
(
0
)
e
D
x
+
∫
0
x
C
e
F
(
x

τ
)
d
τ
)
G
,
where
Ξ
(
0
)
=
(
K
0
n
×
n
)
n
×
(
2
n
)
,
F
=
(
0
n
×
n
A
+
B
K
I
n
×
n
0
n
×
n
)
(
2
n
)
×
(
2
n
)
,
C
=
(
0
n
×
n
λ
(
x
)
)
n
×
(
2
n
)
, and
G
=
(
I
n
×
n
0
n
×
n
)
(
2
n
)
×
n
.
Evaluating (5) at
x
=
l
, and by the boundary condition of (4) and (6), a controller is obtained as follows:
(47)
U
(
t
)
=
∫
0
l
k
(
l
,
y
)
u
(
y
,
t
)
d
y
+
Φ
(
l
)
X
(
t
)
.
Furthermore, the explicit solution to the closedloop system (6) under the controller (47) can also be obtained if the initial state
(
X
(
0
)
,
u
(
x
,
0
)
)
is known. The solution in (6) is
(48)
X
(
t
)
=
X
(
0
)
e
(
A
+
B
K
)
t
+
∫
0
t
e
(
A
+
B
K
)
(
t

τ
)
B
ω
(
0
,
τ
)
d
τ
ω
(
x
,
t
)
=
2
l
∑
m
=
1
∞
∫
0
l
ω
0
(
ξ
)
cos
(
(
m
+
1
/
2
)
π
l
ξ
)
d
ξ
f
f
f
f
f
f
f
f
f
×
e

(
(
m
+
1
/
2
)
2
π
2
/
l
2
)
t
f
f
f
f
f
f
f
f
f
×
cos
(
(
m
+
1
/
2
)
π
l
x
)
and the initial condition
ω
0
(
x
)
is calculated by the initial state
(
X
(
0
)
,
u
(
x
,
0
)
)
through (5).