We study the instability of the traveling waves of
a sixth-order parabolic equation which arises naturally as a continuum model
for the formation of quantum dots and their faceting. We prove that some
traveling wave solutions are nonlinear unstable under
π»
4
perturbations. These
traveling wave solutions converge to a constant as
π₯
β
β
.
1. Introduction In this paper, we consider the following sixth-order parabolic equation
π
π’
π
π‘
=
π·
6
π’
+
π·
4
ξ·
π’
β
π’
3
ξΈ
+
π
(
π’
)
,
(
π₯
,
π‘
)
β
π
Γ
(
0
,
π
)
,
(
1
.
1
)
where
π
(
π’
)
=
π
(
1
β
π’
2
)
,
π
>
0
.
Equation (1.1 ) arises naturally as a continuum model for the formation of quantum dots and their faceting; see [1 ]. Here
π’
(
π₯
,
π‘
)
denotes the surface slope. The high-order derivatives are a result of the additional regularization energy which is required to form an edge between two-plane surfaces with different orientations.
During the past years, only a few works have been devoted to the sixth-order parabolic equation [2 –7 ]. Barrett et al. [2 ] considered the above equation with
π
=
2
. A finite element method is presented which proves to be well posed and convergent. Numerical experiments illustrate the theory.
Recently, Jüngel and Milišić [5 ] studied the sixth-order nonlinear parabolic equation
π
π’
=
ξ
π’
ξ
1
π
π‘
π’
ξ·
π’
(
l
n
π’
)
π₯
π₯
ξΈ
π₯
π₯
+
1
2
ξ·
(
l
n
π’
)
π₯
π₯
ξΈ
2
ξ
π₯
ξ
π₯
.
(
1
.
2
)
They proved the global-in-time existence of weak nonnegative solutions in one space dimension with periodic boundary conditions.
Evans et al. [3 , 4 ] considered the sixth-order thin film equation containing an unstable (backward parabolic) second-order term
π
π’
ξΊ
π
π‘
=
d
i
v
|
π’
|
π
β
Ξ
2
π’
ξ»
ξ·
β
Ξ
|
π’
|
π
β
1
π’
ξΈ
,
π
>
0
,
π
>
1
.
(
1
.
3
)
By a formal matched expansion technique, they show that, for the first critical exponent
π
=
π
0
=
π
+
1
+
4
/
π
for
π
β
(
0
,
5
/
4
)
, where
π
is the space dimension, the free-boundary problem with zero-height, zero-contact-angle, zero-moment, and zero-flux conditions at the interface admits a countable set of continuous branches of radially symmetric self-similar blow-up solutions
π’
π
(
π₯
,
π‘
)
=
(
π
β
π‘
)
(
β
π
/
(
π
π
+
6
)
)
π
π
(
π¦
)
,
π¦
=
π₯
/
(
(
π
β
π‘
)
(
1
/
(
π
π
+
6
)
)
)
, where
π
>
0
is the blow-up time.
Korzec et al. [8 ] considered the sixth-order equation
π’
π‘
β
π
π’
π’
π₯
β
ξ·
π’
β
π’
3
+
π
2
π’
π₯
π₯
ξΈ
π₯
π₯
π₯
π₯
=
0
.
(
1
.
4
)
New type of stationary solutions is derived by an extension of the method of matched asymptotic expansion.
In this paper, we study instability of the traveling waves of (1.1 ). Our main result is as follows.
Theorem 1.1. All the traveling waves
π
(
π₯
β
π
π‘
)
of (1.1 ) satisfying
π
β
πΏ
β
(
π
)
,
π
(
π
)
β
πΏ
β
(
π
)
β©
πΏ
2
(
π
)
(
π
=
1
,
2
,
β¦
,
6
)
are nonlinearly unstable in the space
π»
4
(
π
)
, where
π
(
π
)
denotes
π
π‘
β
derivative of
π
.
The stability and instability of special solutions for the higher-order parabolic equation are very important in the applied fields. Carlen et al. [9 ] proved the nonlinear stability of fronts for the Cahn-Hilliard, under
πΏ
1
perturbations. Gao and Liu [10 ] prove that it is nonlinearly unstable under
π»
2
perturbations, for some traveling wave solution of the convective Cahn-Hilliard equation. The relevant equations have also been studied in [11 , 12 ]. The main difficulties for treating (1.1 ) are caused by the principal part and the lack of the Lyapunov functional. Our proof is based on the principle of linearization. We invoke a general theorem that asserts that linearized instability implies nonlinear instability.
This paper is organized as follows. In the next section, we find an exact traveling wave solution for (1.1 ). In Section 3 , we give the proof of our main result.
2. Exact Traveling Wave Solutions In this section, we construct an exact traveling wave which satisfies all conditions of Theorem 1.1 .
If
π
(
π₯
β
π
π‘
)
=
π
(
π§
)
is a traveling wave solution of (1.1 ), then
π
satisfies the ordinary differential equation
β
π
π
β²
=
π
(
6
)
+
ξ·
1
β
3
π
2
ξΈ
π
(
4
)
β
3
6
π
ξ
2
π
ξ
ξ
β
1
8
π
π
2
ξ
ξ
β
2
4
π
π
β²
π
ξ
ξ
ξ
ξ·
+
π
1
β
π
2
ξΈ
.
(
2
.
1
)
Let
π
β²
=
π
π
/
π
π§
=
π
(
1
β
π
2
)
. Then
π
ξ
ξ
=
π
ξ·
π
ξ·
π
π§
1
β
π
2
ξΈ
ξΈ
=
β
2
π
2
π
ξ·
1
β
π
2
ξΈ
,
π
ξ
ξ
ξ
=
π
ξ·
π
π§
β
2
π
2
π
ξ·
1
β
π
2
ξΈ
ξΈ
=
2
π
3
ξ·
β
1
+
3
π
2
ξΈ
ξ·
1
β
π
2
ξΈ
,
π
(
4
)
=
π
ξ·
π
π§
2
π
3
ξ·
β
1
+
3
π
2
ξΈ
ξ·
1
β
π
2
ξΈ
ξΈ
=
2
π
4
ξ·
8
π
β
1
2
π
3
ξΈ
ξ·
1
β
π
2
ξΈ
,
π
(
5
)
=
π
ξ·
π
π§
2
π
4
ξ·
8
π
β
1
2
π
3
ξΈ
ξ·
1
β
π
2
ξΈ
ξΈ
=
8
π
5
ξ·
1
β
π
2
ξΈ
ξ·
2
β
1
5
π
2
+
1
5
π
4
ξΈ
,
π
(
6
)
=
π
ξ·
π
π§
8
π
5
ξ·
1
β
π
2
ξΈ
ξ·
2
β
1
5
π
2
+
1
5
π
4
ξΈ
ξΈ
=
1
6
π
6
π
ξ·
1
β
π
2
ξΈ
ξ·
6
0
π
2
β
4
5
π
4
ξΈ
.
β
1
7
(
2
.
2
)
Substituting the above equations into (2.1 ), we have
ξ·
β
π
π
β
π
=
3
6
0
π
4
β
7
2
0
π
6
ξΈ
π
5
+
ξ·
9
6
0
π
6
β
4
8
0
π
4
ξΈ
π
3
+
ξ·
1
3
6
π
4
β
2
7
2
π
6
ξΈ
π
.
(
2
.
3
)
Then comparing the order of
π
, we obtain
β
π
π
=
π
,
3
6
0
π
4
β
7
2
0
π
6
=
0
,
9
6
0
π
6
β
4
8
0
π
4
=
0
,
1
3
6
π
4
β
2
7
2
π
6
=
0
.
(
2
.
4
)
A simple calculation shows that
β
π
=
1
/
2
,
β
π
=
β
2
π
. Hence, we get
1
π
β²
=
β
2
ξ·
1
β
π
2
ξΈ
,
(
2
.
5
)
that is,
1
2
l
n
1
+
π
=
1
1
β
π
β
2
π§
,
(
2
.
6
)
that is,
π
π
(
π§
)
=
β
(
1
/
2
)
π§
β
π
β
β
(
1
/
2
)
π§
π
β
(
1
/
2
)
π§
+
π
β
β
(
1
/
2
)
π§
1
=
t
a
n
h
β
2
π§
.
(
2
.
7
)
We easily proved that
l
i
m
π§
β
+
β
π
(
π§
)
=
1
,
l
i
m
π§
β
+
β
π
(
π§
)
=
β
1
(
2
.
8
)
and
π
(
π§
)
satisfies the conditions of the Theorem 1.1 .
3. Proof of The Result To prove the Theorem 1.1 , we first consider an evolution equation
π
π’
π
π‘
=
πΏ
π’
+
πΉ
(
π’
)
,
(
3
.
1
)
where
πΏ
is a linear operator that generates a strongly continuous semigroup
π
π‘
πΏ
on a Banach space
π
, and
πΉ
is a strongly continuous operator such that
πΉ
(
0
)
=
0
. In [13 ], authors considered the whole problem only on space X, that is, the nonlinear operator maps
π
to
π
. However, many equations posses nonlinear terms that include derivatives and therefore,
πΉ
maps into a large Banach space
π
. Hence, they again got the following lemma.
Lemma 3.1 (see [14 ]). Assume the following. (i) X, Z are two Banach spaces with
π
β
π
and
β
π’
β
π
β€
π
1
β
π’
β
π
for
π’
β
π
. (ii) L generates a strongly continuous semigroup
π
π‘
πΏ
on the space Z, and the semigroup
π
π‘
πΏ
maps Z into X for
π‘
>
0
and
β«
1
0
β
π
π‘
πΏ
β
π
β
π
π
π‘
=
πΆ
4
<
β
. (iii) The spectrum of
πΏ
on
π
meets the right half-plane,
{
π
π
π
>
0
}
. (iv)
πΉ
βΆ
π
β
π
is continuous and
β
π
0
>
0
,
πΆ
3
>
0
,
πΌ
>
1
such that
β
πΉ
(
π’
)
β
π
<
πΆ
0
β
π’
β
πΌ
π
, for
β
π’
β
π
<
π
0
.Then the zero solution of (3.1 ) is nonlinearly unstable in the space
π
.
In this paper, we are going to use Lemma 3.1 for the proof of Theorem 1.1 .
Definition 3.2. A traveling wave solution
π
(
π₯
β
π
π‘
)
of (1.1 ) is said to be nonlinearly unstable in the space X , if there exist positive
π
0
and
πΆ
0
, a sequence
{
π’
π
}
of solutions of (1.1 ) and a sequence of time
π‘
π
>
0
such that
β
π’
π
(
0
)
β
π
(
π₯
)
β
π
β
0
but
β
π’
π
(
π‘
π
)
β
π
(
β
β
π
π‘
π
)
β
π
β₯
π
0
.
If
π
(
π₯
β
π
π‘
)
β
π»
4
(
π
)
is a traveling wave solution of (1.1 ), then letting
π
(
π₯
,
π‘
)
=
π’
(
π₯
,
π‘
)
β
π
(
π₯
β
π
π‘
)
, we have
(
π
+
π
)
π‘
=
π
6
π₯
(
π
+
π
)
+
π
4
π₯
ξΊ
(
π
+
π
)
β
(
π
+
π
)
3
ξ»
ξΊ
+
π
1
β
(
π
+
π
)
2
ξ»
=
π
6
π₯
π
+
π
(
6
)
+
π
4
π₯
ξ·
π
+
π
β
π
3
β
π
3
β
3
π
2
π
β
3
π
π
2
ξΈ
ξ·
+
π
1
β
π
2
β
π
2
ξΈ
,
β
2
π
π
(
3
.
2
)
that is,
π
π‘
=
π
6
π₯
π
+
π
4
π₯
ξ·
π
β
π
3
β
3
π
2
π
β
3
π
π
2
ξΈ
ξ·
+
π
β
π
2
ξΈ
,
β
2
π
π
(
3
.
3
)
that is,
π
π‘
β
π
6
π₯
ξ·
π
β
1
β
3
π
2
ξΈ
π
4
π₯
π
+
2
4
π
π
β²
π
3
π₯
ξ
π
+
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
+
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
ξ
π
+
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π
+
2
π
π
=
πΉ
(
π
)
,
(
3
.
4
)
where
πΉ
ξ·
(
π
)
=
3
π
(
4
)
ξΈ
π
β
π
2
+
2
4
π
ξ
ξ
ξ
π
π
π₯
π
+
3
6
π
ξ
ξ
(
π
π₯
π
)
2
+
7
2
π
β²
π
π₯
π
π
2
π₯
π
ξ·
π
+
3
6
π₯
π
ξΈ
2
π
2
π₯
π
+
(
2
4
π
+
2
4
π
)
π
π₯
π
π
3
π₯
π
+
3
6
π
ξ
ξ
π
π
2
π₯
π
ξ·
π
+
(
1
8
π
+
1
8
π
)
2
π₯
π
ξΈ
2
+
2
4
π
β²
π
π
3
π₯
π
+
6
π
π
π
4
π₯
π
+
3
π
2
π
4
π₯
π
,
(
3
.
5
)
with the initial value
π
(
π₯
,
0
)
=
π
0
(
π₯
)
β‘
π’
0
(
π₯
)
β
π
(
π₯
)
.
(
3
.
6
)
So the stability of traveling wave solutions of (1.1 ) is translated into the stability of the zero solution of (3.4 ). In order to prove Theorem 1.1 , taking
π
=
πΏ
2
(
π
)
,
π
=
π»
4
(
π
)
, we need to prove that the four conditions of Lemma 3.1 are satisfied by the associated equation (3.4 ). The condition (i) is satisfied by our choice of
π
and
π
.
Denote the linear partial differential operator in (3.4 ) by
πΏ
=
(
π
6
π₯
+
π
4
π₯
)
β
[
3
π
2
π
4
π₯
+
2
4
π
π
β²
π
3
π₯
+
(
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
)
π
2
π₯
+
(
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
)
π
π₯
+
(
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
+
2
π
π
)
]
=
πΏ
0
β
[
3
π
2
π
4
π₯
+
2
4
π
π
β²
π
3
π₯
+
(
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
)
π
2
π₯
+
(
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
)
π
π₯
+
(
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
+
2
π
π
)
]
with
πΏ
0
=
π
6
π₯
+
π
4
π₯
. Then (3.4 ) may be rewritten in the form of (3.1 )
π
π‘
=
πΏ
π
+
πΉ
(
π
)
.
(
3
.
7
)
Note the F maps
π»
4
(
π
)
into
πΏ
2
(
π
)
, using the Sobolev embedding theorem, we have
β
β
πΉ
(
π
)
πΏ
2
β€
πΆ
β
π
β
2
π»
4
,
πΆ
>
0
,
f
o
r
β
π
β
π»
4
<
1
.
(
3
.
8
)
So, the condition (iv) is satisfied.
To prove condition (ii) in Lemma 3.1 , we need the following two lemmas.
Lemma 3.3. Let
πΏ
0
=
π
6
π₯
+
π
4
π₯
. Then
β
β
π
π‘
πΏ
0
β
β
π»
π
β
π»
π
β€
π
(
4
/
2
7
)
π‘
,
f
o
r
π
β
π
+
β
β
π
,
0
β€
π‘
<
β
,
(
3
.
9
)
π‘
πΏ
0
β
β
πΏ
2
β
π»
4
β€
π
(
π‘
)
β‘
5
π‘
β
2
/
3
,
f
o
r
0
<
π‘
β€
1
.
(
3
.
1
0
)
Proof. We write
π’
(
π₯
,
π‘
)
=
π
π‘
πΏ
0
π’
0
(
π₯
)
. By Fourier transformation
Μ
π’
(
π
,
π‘
)
=
π
β
π‘
(
π
6
β
π
4
)
Μ
π’
0
(
π
)
,
β
π’
β
2
π»
π
β‘
ξ
β
β
β
ξ·
1
+
π
2
ξΈ
π
|
|
|
|
Μ
π’
(
π
,
π‘
)
2
=
ξ
π
π
β
β
β
ξ·
1
+
π
2
ξΈ
π
π
β
2
π‘
(
π
6
β
π
4
)
|
|
Μ
π’
0
|
|
(
π
)
2
π
π
β€
s
u
p
π
β
π
π
β
2
π‘
(
π
6
β
π
4
)
ξ
β
β
β
ξ·
1
+
π
2
ξΈ
π
|
|
Μ
π’
0
|
|
(
π
)
2
π
π
=
π
(
8
/
2
7
)
π‘
β
β
π’
0
β
β
2
π»
π
.
(
3
.
1
1
)
Hence,
β
β
π
π‘
πΏ
0
β
β
π»
π
β
π»
π
β€
π
(
4
/
2
7
)
π‘
.
(
3
.
1
2
)
On the other hand, letting
π
=
π
2
, we have
β
π’
β
2
π»
4
β€
s
u
p
π
β
π
β«
π
(
π
)
β
β
β
|
|
Μ
π’
0
|
|
(
π
)
2
π
π
,
(
3
.
1
3
)
with
π
(
π
)
=
(
1
+
π
)
4
π
β
2
π‘
(
π
3
β
π
2
)
,
π‘
>
0
. Elementary computation shows that
s
u
p
π
>
0
ξ
4
π
(
π
)
β€
3
+
1
6
π‘
β
4
/
3
ξ
π
(
8
/
2
7
)
π‘
.
(
3
.
1
4
)
Thus,
β
π’
(
π₯
,
π‘
)
β
π»
4
β€
ξ
4
3
+
1
6
π‘
β
4
/
3
ξ
1
/
2
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
πΏ
2
,
β
β
π
π‘
πΏ
0
β
β
πΏ
2
β
π»
4
β€
ξ
4
3
+
1
6
π‘
β
4
/
3
ξ
1
/
2
π
(
4
/
2
7
)
π‘
β€
5
π‘
β
2
/
3
,
f
o
r
0
<
π‘
β€
1
,
(
3
.
1
5
)
since
π
(
4
/
2
7
)
π‘
β€
π
4
/
2
7
<
2
. Thus, Lemma 3.3 has been proved.
Lemma 3.4. Let
πΏ
=
(
π
6
π₯
+
π
4
π₯
)
β
[
3
π
2
π
4
π₯
+
2
4
π
π
ξ
π
3
π₯
+
(
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
)
π
2
π₯
+
(
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
β²
β²
)
π
π₯
+
(
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
+
2
π
π
)
]
=
πΏ
0
β
[
3
π
2
π
4
π₯
+
2
4
π
π
β²
π
3
π₯
+
(
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
)
π
2
π₯
+
(
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
)
π
π₯
+
(
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
+
2
π
π
)
]
with
πΏ
0
=
π
6
π₯
+
π
4
π₯
,
π
(
π
)
β
πΏ
β
(
π
)
,
π
=
0
,
1
,
2
,
3
,
4
. Then
β
β
e
t
L
β
β
L
2
β
H
4
β€
C
1
t
β
2
/
3
β
β
e
,
f
o
r
0
<
t
β€
1
,
(
3
.
1
6
)
t
L
β
β
H
4
β
H
4
β€
C
2
<
β
,
f
o
r
0
<
t
β€
1
.
(
3
.
1
7
)
Proof. Consider the initial value problem
π’
π‘
=
πΏ
π’
=
πΏ
0
π’
β
3
π
2
π
4
π₯
π’
β
2
4
π
π
β²
π
3
π₯
ξ
π’
β
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π’
β
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
ξ
π’
β
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
+
2
π
π
π’
,
π’
(
π₯
,
0
)
=
π’
0
(
π₯
)
.
(
3
.
1
8
)
Then
π’
(
π₯
,
π‘
)
=
π
π‘
πΏ
π’
0
(
π₯
)
,
π‘
β₯
0
,
π₯
β
π
, thus
π’
(
π₯
,
π‘
)
=
π
π‘
πΏ
0
π’
0
β
ξ
π‘
0
π
(
π‘
β
π
)
πΏ
0
ξ
3
π
2
π
4
π₯
π’
+
2
4
π
π
β²
π
3
π₯
ξ
π’
+
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π’
+
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π’
+
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π’
ξ
+
2
π
π
π
π
.
(
3
.
1
9
)
Denote
π΄
=
β
π
β
πΏ
β
,
π΅
=
β
π
β²
β
πΏ
β
,
πΆ
=
β
π
ξ
ξ
β
πΏ
β
,
π·
=
β
π
ξ
ξ
ξ
β
πΏ
β
,
πΈ
=
β
π
(
4
)
β
πΏ
β
and
π
=
3
π΄
2
+
2
4
π΄
π΅
+
3
6
π΄
πΆ
+
3
6
π΅
2
+
2
4
π΄
π·
+
7
2
π΅
πΆ
+
1
8
πΆ
2
+
2
4
π΅
π·
+
6
π΄
πΈ
+
2
π
π΄
.
(
3
.
2
0
)
Then, we have
(
β
π’
π‘
)
β
π»
4
β€
β
β
π
π‘
πΏ
0
β
β
π»
4
β
π»
4
β
β
π’
0
β
β
π»
4
+
ξ
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
π»
4
β
π»
4
3
β
π
β
2
πΏ
β
β
β
π
4
π₯
π’
β
β
πΏ
2
ξ
π
π
+
2
4
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
π»
4
β
π»
4
β
π
β
πΏ
β
β
π
β²
β
πΏ
β
β
β
π
3
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
π»
4
β
π»
4
ξ
3
6
β
π
β
πΏ
β
β
β
π
ξ
ξ
β
β
πΏ
β
β
β
π
+
3
6
ξ
2
β
β
πΏ
β
ξ
β
β
π
2
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
π»
4
β
π»
4
ξ·
2
4
β
π
β
πΏ
β
β
β
π
ξ
ξ
ξ
β
β
πΏ
β
+
7
2
β
π
β²
β
πΏ
β
β
β
π
ξ
ξ
β
β
πΏ
β
ξΈ
β
β
π
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
π»
4
β
π»
4
ξ
β
β
π
1
8
ξ
ξ
β
β
2
πΏ
β
+
2
4
β
π
β²
β
πΏ
β
β
β
π
ξ
ξ
ξ
β
β
πΏ
β
+
6
β
π
β
πΏ
β
β
β
π
(
4
)
β
β
πΏ
β
+
2
π
β
π
β
πΏ
β
ξ
β
π’
β
πΏ
2
π
π
β€
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
ξ
+
π
π‘
0
π
(
4
/
2
7
)
(
π‘
β
π
)
β
π’
(
π
)
β
π»
4
π
π
,
(
3
.
2
1
)
where we use
π’
(
π‘
)
to denote
π’
(
β
,
π‘
)
. By iteration,
(
β
π’
π‘
)
β
π»
4
β€
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
ξ
+
π
π‘
0
π
(
4
/
2
7
)
(
π‘
β
π
)
ξΈ
π
(
4
/
2
7
)
π
β
β
π’
0
β
β
π»
4
ξ
+
π
π
0
π
(
4
/
2
7
)
(
π
β
π
)
(
β
π’
π
)
β
π»
4
ξΉ
π
π
π
π
=
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
ξ
+
π
π‘
0
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
π
π
+
π
2
ξ
π‘
0
ξ
π
0
π
(
4
/
2
7
)
(
π‘
β
π
)
(
β
π’
π
)
β
π»
4
π
π
π
π
β€
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
+
π
π‘
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
+
π
2
ξ
π‘
0
ξΈ
ξ
π‘
π
π
(
4
/
2
7
)
(
π‘
β
π
)
β
π’
(
π
)
β
π»
4
ξΉ
π
π
π
π
β€
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
+
π
π‘
π
(
4
/
2
7
)
π‘
β
β
π’
0
β
β
π»
4
+
π
2
π‘
π
(
4
/
2
7
)
π‘
ξ
π‘
0
β
π’
(
π
)
β
π»
4
π
π
β€
π
4
/
2
7
β
β
π’
0
β
β
π»
4
+
π
π
4
/
2
7
β
β
π’
0
β
β
π»
4
+
π
4
/
2
7
π
2
ξ
π‘
0
β
β
π’
(
π
)
π»
4
π
π
,
f
o
r
0
<
π
β€
π
β€
π‘
β€
1
.
(
3
.
2
2
)
Let
β«
π£
(
π‘
)
=
π‘
0
β
π’
(
π
)
β
π»
4
π
π
. Then
π
π£
(
π‘
)
β€
ξ·
π
π
π‘
4
/
2
7
+
π
4
/
2
7
π
ξΈ
β
β
π’
0
β
β
π»
4
+
π
4
/
2
7
π
2
π£
(
π‘
)
,
f
o
r
0
<
π‘
β€
1
.
(
3
.
2
3
)
Multiplying both sides of the above inequality by
π
β
π
(
4
/
2
7
)
π
2
π‘
, we have
π
ξ
π
β
π
4
/
2
7
π
2
π‘
ξ
π£
(
π‘
)
π
π‘
β€
π
β
π
4
/
2
7
π
2
π‘
π
4
/
2
7
β
β
π’
(
1
+
π
)
0
β
β
π»
4
,
w
h
e
r
e
0
<
π‘
β€
1
.
(
3
.
2
4
)
Integrating the above inequality with respect to
π‘
over
(
0
,
π‘
)
, we obtain
π
β
π
4
/
2
7
π
2
π‘
ξ
π£
(
π‘
)
β€
π‘
0
π
β
π
4
/
2
7
π
2
π
π
4
/
2
7
(
β
β
π’
1
+
π
)
0
β
β
π»
4
π
π
,
(
3
.
2
5
)
that is,
π£
(
π‘
)
β€
π
π
4
/
2
7
π
2
π‘
ξ
π‘
0
π
β
π
4
/
2
7
π
2
π
π
4
/
2
7
(
β
β
π’
1
+
π
)
0
β
β
π»
4
π
π
.
(
3
.
2
6
)
Observing that
β«
π£
(
π‘
)
=
π‘
0
β
π’
(
π
)
β
π»
4
π
π
is bounded and substituting the above inequality into (3.22 ), we get
(
β
π’
π‘
)
β
π»
4
β€
π
4
/
2
7
β
β
π’
0
β
β
π»
4
+
π
π
4
/
2
7
β
β
π’
0
β
β
π»
4
+
π
4
/
2
7
π
2
ξ
π‘
0
(
β
π’
π
)
β
π»
4
π
π
β€
π
2
<
β
,
f
o
r
0
<
π‘
β€
1
,
π
2
>
0
,
(
3
.
2
7
)
thus (3.17 ) has been proven. Next, we prove the inequality (3.16 ). Clearly, we have
(
β
π’
π‘
)
β
π»
4
β€
β
β
π
π‘
πΏ
0
β
β
πΏ
2
β
π»
4
β
β
π’
0
β
β
πΏ
2
+
ξ
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
πΏ
2
β
π»
4
3
β
π
β
2
πΏ
β
β
β
π
4
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
πΏ
2
β
π»
4
2
4
β
π
β
πΏ
β
β
π
β²
β
πΏ
β
β
β
π
3
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
πΏ
2
β
π»
4
ξ
3
6
β
π
β
πΏ
β
β
β
π
ξ
ξ
β
β
πΏ
β
β
β
π
+
3
6
ξ
2
β
β
πΏ
β
ξ
β
β
π
2
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
πΏ
2
β
π»
4
ξ·
2
4
β
π
β
πΏ
β
β
β
π
ξ
ξ
ξ
β
β
πΏ
β
+
7
2
β
π
β²
β
πΏ
β
β
β
π
ξ
ξ
β
β
πΏ
β
ξΈ
β
β
π
π₯
π’
β
β
πΏ
2
+
ξ
π
π
π‘
0
β
β
π
(
π‘
β
π
)
πΏ
0
β
β
πΏ
2
β
π»
4
ξ
β
β
π
1
8
ξ
ξ
β
β
2
πΏ
β
+
2
4
β
π
β²
β
πΏ
β
β
β
π
ξ
ξ
ξ
β
β
πΏ
β
+
6
β
π
β
πΏ
β
β
β
π
(
4
)
β
β
πΏ
β
+
2
π
β
π
β
πΏ
β
ξ
β
π’
β
πΏ
2
β
β
π’
π
π
β€
π
(
π‘
)
0
β
β
πΏ
2
ξ
+
π
π‘
0
π
β
(
π‘
β
π
)
β
π’
(
π
)
π»
4
π
π
,
(
3
.
2
8
)
where
π
(
π‘
)
is defined in Lemma 3.3 , and we use
π’
(
π‘
)
to denote
π’
(
β
,
π‘
)
. By iteration,
(
β
π’
π‘
)
β
π»
4
β
β
π’
β€
π
(
π‘
)
0
β
β
πΏ
2
ξ
+
π
π‘
0
ξΈ
β
β
π’
π
(
π‘
β
π
)
π
(
π
)
π
β
β
πΏ
2
ξ
+
π
π
0
π
(
π
β
π
)
β
π’
(
π
)
β
π»
4
ξΉ
β
β
π’
π
π
π
π
=
π
(
π‘
)
0
β
β
πΏ
2
ξ
+
π
π‘
0
β
β
π’
π
(
π‘
β
π
)
π
(
π
)
0
β
β
πΏ
2
π
π
+
π
2
ξ
π‘
0
ξ
π
0
π
(
π‘
β
π
)
π
(
π
β
π
)
β
π’
(
π
)
β
π»
4
π
π
π
π
.
(
3
.
2
9
)
The second term on the right of (3.29 ) is
π
ξ
π‘
0
β
β
π’
π
(
π‘
β
π
)
π
(
π
)
0
β
β
πΏ
2
β
β
π’
π
π
=
π
0
β
β
πΏ
2
ξ
π‘
0
5
(
π‘
β
π
)
β
2
/
3
5
π
β
2
/
3
β
β
π’
π
π
=
2
5
π
0
β
β
πΏ
2
ξ
π‘
0
π‘
β
4
/
3
ξ
π
1
β
π‘
ξ
β
2
/
3
ξ
π
π‘
ξ
β
2
/
3
π
π
=
2
5
π
πΆ
3
π‘
β
1
/
3
β
β
π’
0
β
β
πΏ
2
,
0
<
π‘
<
1
,
(
3
.
3
0
)
where
πΆ
3
=
β«
1
0
(
1
β
π
)
β
1
/
4
π
β
1
/
4
π
π
. By exchanging the order of integration, we get from the third term on the right side of (3.29 ),
ξ
π‘
0
ξ
π
0
π
(
π‘
β
π
)
π
(
π
β
π
)
β
π’
(
π
)
β
π»
4
ξ
π
π
π
π
=
π‘
0
ξΈ
ξ
π‘
π
ξΉ
(
π
(
π‘
β
π
)
π
(
π
β
π
)
π
π
β
π’
π
)
β
π»
4
π
π
,
(
3
.
3
1
)
then
ξ
π‘
π
ξ
π
(
π‘
β
π
)
π
(
π
β
π
)
π
π
=
2
5
π‘
π
(
π‘
β
π
)
β
2
/
3
(
π
β
π
)
β
2
/
3
π
π
=
2
5
πΆ
3
(
π‘
β
π
)
β
1
/
3
,
0
<
π
β€
π‘
β€
1
.
(
3
.
3
2
)
Therefore (3.28 )–(3.32 ) imply
β
π’
(
π‘
)
β
π»
4
β€
ξΊ
π
(
π‘
)
+
2
5
πΆ
3
π
π‘
β
1
/
3
ξ»
β
β
π’
0
β
β
πΏ
2
+
2
5
πΆ
3
π
2
ξ
π‘
0
(
π‘
β
π
)
β
1
/
3
β
π’
(
π
)
β
π»
4
π
π
,
0
<
π‘
β€
1
.
(
3
.
3
3
)
From (3.17 ), we know
β
π’
(
π‘
)
β
π»
4
β€
πΆ
2
β
π’
0
β
π»
4
,
0
<
π‘
β€
1
. Then
β
π’
(
π‘
)
β
π»
4
β€
ξΊ
π
(
π‘
)
+
2
5
πΆ
3
π
π‘
β
1
/
3
ξ»
β
β
π’
0
β
β
πΏ
2
+
7
5
2
πΆ
3
πΆ
2
π
2
β
β
π’
0
β
β
π»
4
π‘
2
/
3
,
0
<
π‘
β€
1
.
(
3
.
3
4
)
Therefore, there exists a
π‘
β
, such that
β
β
π’
(
π‘
)
π»
4
β€
πΆ
1
π‘
β
2
/
3
β
β
π’
0
β
β
πΏ
2
,
0
<
π‘
β€
π‘
β
β€
1
,
πΆ
1
>
0
.
(
3
.
3
5
)
So, we proved the inequality (3.16 ). Hence (3.17 ) is proven and proof of Lemma 3.4 is finished.
By Lemma 3.4 , the condition (ii) is proved.
We now proceed to verify condition (iii) of Lemma 3.1 . Observing that if
π’
(
π₯
,
π‘
)
satisfies
π
π’
(
π₯
,
π‘
)
=
π
π
π‘
6
π’
π
π₯
6
+
π
4
π’
π
π₯
4
β
3
π
2
π
4
π’
π
π₯
4
π
β
2
4
π
π
β²
3
π’
π
π₯
3
β
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π’
π
π₯
2
β
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π’
β
ξ
π
π₯
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
+
2
π
π
π’
,
(
3
.
3
6
)
then
π’
(
π₯
,
π
+
π‘
)
also satisfies the above equation. By uniqueness of solution, we know that
πΏ
generates a strongly continuous semigroup on the Banach space
π»
4
(
π
)
(see [15 ] p.344). By Fourier transformation, the essential spectrum of
πΏ
0
on
π»
4
(
π
)
is
π
ξ·
πΏ
0
ξΈ
β
ξ½
β
π
6
+
π
4
ξΎ
.
β£
π
β
π
(
3
.
3
7
)
The curve
π
=
β
π
6
+
π
4
meets the vertical lines
π
π
π
=
πΌ
for
β
β
<
πΌ
β€
4
/
2
7
because
β
β
<
β
π
6
+
π
4
β€
4
/
2
7
.
We now prove that the same curve belongs to the essential spectrum of
πΏ
.
Lemma 3.5. The essential spectrum of
πΏ
on
π»
4
(
π
)
contains that of
πΏ
0
.
Proof. Let
π
β
π
and let
π
=
π
(
π
)
=
β
π
6
+
π
4
. Following Schechter [16 ],
π
β
π
(
πΏ
)
if there exists a sequence
{
π
π
}
β
π»
4
(
π
)
with
β
β
π
π
β
β
π»
4
β
β
=
1
,
(
πΏ
β
π
)
π
π
β
β
π»
4
βΆ
0
,
(
3
.
3
8
)
and
{
π
π
}
does not have a strongly convergent subsequence in
π»
4
(
π
)
. Here we use the definition
π
β
π
(
πΏ
)
if and only if
πΏ
β
π
is Fredholm with index zero. Now let
π
0
β’
0
be a
πΆ
β
function with compact support in
(
0
,
β
)
. Define
π
π
π
(
π₯
)
=
π
π
π
π
π₯
π
0
(
π₯
/
π
)
β
π
,
π
=
1
,
2
,
β¦
,
(
3
.
3
9
)
where
π
π
is chosen so that
β
π
π
β
π»
4
=
1
. In fact,
β
β
π
π
β
β
πΏ
2
=
π
π
β
β
π
0
β
β
πΏ
2
β
β
π
,
1
=
π
β
β
π»
4
β€
π
π
π
,
(
3
.
4
0
)
for some positive constant
π
. Hence
π
π
β₯
1
/
π
>
0
. Since
β
π
π
β
πΏ
β
β
0
but
β
π
π
β
πΏ
2
is bounded away from zero,
{
π
π
}
can have no convergent subsequence in
πΏ
2
(
π
)
. It remains to show that
β
(
πΏ
β
π
)
π
π
β
π»
4
β
0
. We write
πΏ
β
π
=
πΏ
0
ξ
β
π
β
3
π
2
π
4
π₯
+
2
4
π
π
β²
π
3
π₯
+
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
+
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
+
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
.
+
2
π
π
ξ
ξ
(
3
.
4
1
)
A simple calculation shows that
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
=
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
)
(
π₯
/
π
)
π
!
π
(
1
/
2
)
+
π
,
π
ξ·
πΏ
0
ξΈ
π
β
π
π
ξ·
πΏ
(
π₯
)
=
π
π
0
ξΈ
π
β
π
π
(
π₯
)
+
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
1
)
(
π₯
/
π
)
π
!
π
(
3
/
2
)
+
π
,
π
2
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
=
β
π
2
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
+
2
π
π
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
1
)
(
π₯
/
π
)
π
!
π
(
3
/
2
)
+
π
+
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
2
)
(
π₯
/
π
)
π
!
π
(
5
/
2
)
+
π
,
π
3
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
=
β
π
π
3
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
β
3
π
2
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
1
)
(
π₯
/
π
)
π
!
π
(
3
/
2
)
+
π
+
3
π
π
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
2
)
(
π₯
/
π
)
π
!
π
(
5
/
2
)
+
π
+
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
3
)
(
π₯
/
π
)
π
!
π
(
7
/
2
)
+
π
,
π
4
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
=
π
4
ξ·
πΏ
0
ξΈ
π
β
π
π
(
π₯
)
β
4
π
π
3
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
1
)
(
π₯
/
π
)
π
!
π
(
3
/
2
)
+
π
β
6
π
2
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
2
)
(
π₯
/
π
)
π
!
π
(
5
/
2
)
+
π
+
4
π
π
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
3
)
(
π₯
/
π
)
π
!
π
(
7
/
2
)
+
π
+
π
π
π
π
π
π₯
ξ
1
β€
π
β€
6
(
β
π
)
π
π
(
π
)
(
π
)
π
0
(
π
+
4
)
(
π₯
/
π
)
π
!
π
(
9
/
2
)
+
π
.
(
3
.
4
2
)
Thus,
β
β
ξ·
πΏ
0
ξΈ
π
β
π
π
β
β
(
π₯
)
π»
4
β€
ξ
|
|
π
|
|
+
|
|
π
|
|
1
+
2
+
|
|
π
|
|
3
+
|
|
π
|
|
4
ξ
ξ
1
β€
π
β€
6
|
|
π
(
π
)
|
|
π
(
π
)
π
β
β
π
0
(
π
)
β
β
(
π₯
/
π
)
πΏ
2
π
!
π
(
1
/
2
)
+
π
+
ξ
|
|
π
|
|
|
|
π
|
|
1
+
2
+
3
2
|
|
π
|
|
+
4
3
ξ
ξ
1
β€
π
β€
6
|
|
π
(
π
)
(
|
|
π
π
)
π
β
β
π
0
(
π
+
1
)
(
β
β
π₯
/
π
)
πΏ
2
π
!
π
(
3
/
2
)
+
π
+
ξ
|
|
π
|
|
|
|
π
|
|
1
+
3
+
6
2
ξ
ξ
1
β€
π
β€
6
|
|
π
(
π
)
|
|
π
(
π
)
π
β
β
π
0
(
π
+
2
)
β
β
(
π₯
/
π
)
πΏ
2
π
!
π
(
5
/
2
)
+
π
+
ξ·
|
|
π
|
|
ξΈ
ξ
1
+
4
1
β€
π
β€
6
|
|
π
(
π
)
|
|
π
(
π
)
π
β
β
π
0
(
π
+
3
)
β
β
(
π₯
/
π
)
πΏ
2
π
!
π
(
7
/
2
)
+
π
+
ξ
1
β€
π
β€
6
|
|
π
(
π
)
|
|
π
(
π
)
π
β
β
π
0
(
π
+
4
)
β
β
(
π₯
/
π
)
πΏ
2
π
!
π
(
9
/
2
)
+
π
βΆ
0
,
a
s
π
βΆ
β
.
(
3
.
4
3
)
Moreover, for any positive integer
π
,
β
π
π
π₯
π
π
β
πΏ
β
β
0
, as
π
β
β
, we have
β
β
3
π
2
π
4
π₯
π
π
β
β
2
πΏ
2
β€
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
3
π
2
β
β
2
πΏ
2
β
β
π
βΆ
0
,
π₯
ξΊ
3
π
2
π
4
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
5
π₯
π
π
β
β
2
πΏ
β
β
β
3
π
2
β
β
2
πΏ
2
+
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
6
π
π
β²
β
2
πΏ
2
β
β
π
βΆ
0
,
2
π₯
ξΊ
3
π
2
π
4
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
6
π₯
π
π
β
β
2
πΏ
β
β
β
3
π
2
β
β
2
πΏ
2
β
β
π
+
2
5
π₯
π
π
β
β
2
πΏ
β
β
6
π
π
β²
β
2
πΏ
2
+
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
6
π
β²
2
+
6
π
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
3
π₯
ξΊ
3
π
2
π
4
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
7
π₯
π
π
β
β
2
πΏ
β
β
β
3
π
2
β
β
2
πΏ
2
β
β
π
+
3
6
π₯
π
π
β
β
2
πΏ
β
β
6
π
π
β²
β
2
πΏ
2
β
β
π
+
3
5
π₯
π
π
β
β
2
πΏ
β
β
β
6
π
β²
2
+
6
π
π
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
β²
π
ξ
ξ
+
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
4
π₯
ξΊ
3
π
2
π
4
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
8
π₯
π
π
β
β
2
πΏ
β
β
β
3
π
2
β
β
2
πΏ
2
β
β
π
+
4
5
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
β²
π
ξ
ξ
+
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
4
7
π₯
π
π
β
β
2
πΏ
β
β
6
π
π
β²
β
2
πΏ
2
β
β
π
+
6
6
π₯
π
π
β
β
2
πΏ
β
β
β
6
π
β²
2
+
6
π
π
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
ξ
ξ
2
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
β
β
2
πΏ
2
βΆ
0
.
(
3
.
4
4
)
From the assumptions on
π
, we obtain
β
β
2
4
π
π
β²
π
3
π₯
π
π
β
β
2
πΏ
2
β€
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
2
4
π
π
β²
β
2
πΏ
2
β
β
π
βΆ
0
,
π₯
ξΊ
2
4
π
π
ξ
π
3
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
2
4
π
π
β²
β
2
πΏ
2
+
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
β²
2
+
2
4
π
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
2
π₯
ξΊ
2
4
π
π
ξ
π
3
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
5
π₯
π
π
β
β
2
πΏ
β
β
2
4
π
π
β²
β
2
πΏ
2
β
β
π
+
2
4
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
β²
2
+
2
4
π
π
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
7
2
π
β²
π
ξ
ξ
+
2
4
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
3
π₯
ξΊ
2
4
π
π
ξ
π
3
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
6
π₯
π
π
β
β
2
πΏ
β
β
2
4
π
π
β²
β
2
πΏ
2
β
β
π
+
3
5
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
β²
2
+
2
4
π
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
3
4
π₯
π
π
β
β
2
πΏ
β
β
β
7
2
π
β²
π
ξ
ξ
+
2
4
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
7
2
π
2
ξ
ξ
+
9
6
π
β²
π
ξ
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
4
π₯
ξΊ
2
4
π
π
ξ
π
3
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
7
π₯
π
π
β
β
2
πΏ
β
β
2
4
π
π
β²
β
2
πΏ
2
β
β
π
+
4
6
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
β²
2
+
2
4
π
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
6
5
π₯
π
π
β
β
2
πΏ
β
β
β
7
2
π
β²
π
ξ
ξ
+
2
4
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
4
4
π₯
π
π
β
β
2
πΏ
β
β
β
7
2
π
2
ξ
ξ
+
9
6
π
β²
π
ξ
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
+
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
2
0
π
β²
π
(
4
)
+
2
4
π
π
(
5
)
β
β
2
πΏ
2
βΆ
0
.
(
3
.
4
5
)
Similarly, we have
β
β
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
π
β
β
2
πΏ
2
β€
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
β
β
2
πΏ
2
β
β
π
βΆ
0
,
π₯
ξ
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
β
β
2
πΏ
2
+
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
β²
π
ξ
ξ
+
3
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
2
π₯
ξ
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
β
β
2
πΏ
2
β
β
π
+
2
3
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
β²
π
ξ
ξ
+
3
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
2
ξ
ξ
+
1
4
4
π
β²
π
ξ
ξ
ξ
+
3
6
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
3
π₯
ξ
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
5
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
β
β
2
πΏ
2
β
β
π
+
3
4
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
β²
π
ξ
ξ
+
3
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
3
3
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
2
ξ
ξ
+
1
4
4
π
β²
π
ξ
ξ
ξ
+
3
6
π
π
(
4
)
β
β
2
πΏ
2
+
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
8
0
π
β²
π
(
4
)
+
3
6
π
π
(
5
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
4
π₯
ξ
ξ
3
6
π
π
ξ
ξ
+
3
6
π
ξ
2
ξ
π
2
π₯
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
6
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
π
π
ξ
ξ
+
3
6
π
β²
2
β
β
2
πΏ
2
β
β
π
+
4
5
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
β²
π
ξ
ξ
+
3
6
π
π
ξ
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
6
4
π₯
π
π
β
β
2
πΏ
β
β
β
1
0
8
π
2
ξ
ξ
+
1
4
4
π
β²
π
ξ
ξ
ξ
+
3
6
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
+
4
3
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
8
0
π
β²
π
(
4
)
+
3
6
π
π
(
5
)
β
β
2
πΏ
2
+
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
3
6
0
π
2
ξ
ξ
ξ
+
5
4
0
π
ξ
ξ
π
(
4
)
+
2
1
6
π
β²
π
(
5
)
+
3
6
π
π
(
6
)
β
β
2
L
2
β
β
ξ·
βΆ
0
,
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π
π
β
β
2
πΏ
2
β€
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
π₯
ξΊ
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
β
β
2
πΏ
2
+
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
9
6
π
β²
π
ξ
ξ
ξ
+
7
2
π
2
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
2
π₯
ξΊ
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
2
2
π₯
π
π
β
β
2
πΏ
β
β
β
9
6
π
β²
π
ξ
ξ
ξ
+
7
2
π
2
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
+
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
2
0
π
β²
π
(
4
)
+
2
4
π
π
(
5
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
3
π₯
ξΊ
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
4
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
3
3
π₯
π
π
β
β
2
πΏ
β
β
β
9
6
π
β²
π
ξ
ξ
ξ
+
7
2
π
2
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
+
3
2
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
2
0
π
β²
π
(
4
)
+
2
4
π
π
(
5
)
β
β
2
πΏ
2
+
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
2
ξ
ξ
ξ
+
3
6
0
π
ξ
ξ
π
(
4
)
+
1
4
4
π
β²
π
(
5
)
+
2
4
π
π
(
6
)
β
β
2
πΏ
2
β
β
π
βΆ
0
,
4
π₯
ξΊ
ξ·
2
4
π
π
ξ
ξ
ξ
+
7
2
π
ξ
π
ξ
ξ
ξΈ
π
π₯
π
π
ξ»
β
β
2
πΏ
2
β€
β
β
π
5
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
π
π
ξ
ξ
ξ
+
7
2
π
β²
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
+
4
4
π₯
π
π
β
β
2
πΏ
β
β
β
9
6
π
β²
π
ξ
ξ
ξ
+
7
2
π
2
ξ
ξ
+
2
4
π
π
(
4
)
β
β
2
πΏ
2
β
β
π
+
6
3
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
ξ
ξ
π
ξ
ξ
ξ
+
1
2
0
π
β²
π
(
4
)
+
2
4
π
π
(
5
)
β
β
2
πΏ
2
β
β
π
+
4
2
π₯
π
π
β
β
2
πΏ
β
β
β
2
4
0
π
2
ξ
ξ
ξ
+
3
6
0
π
ξ
ξ
π
(
4
)
+
1
4
4
π
β²
π
(
5
)
+
2
4
π
π
(
6
)
β
β
2
πΏ
2
+
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
8
4
0
π
ξ
ξ
ξ
π
(
4
)
+
5
0
4
π
ξ
ξ
π
(
5
)
+
1
6
8
π
β²
π
(
6
)
+
2
4
π
π
(
7
)
β
β
2
πΏ
2
βΆ
0
.
(
3
.
4
6
)
In addition,
β
β
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π
+
2
π
π
π
β
β
2
πΏ
2
β€
β
β
π
π
β
β
2
πΏ
β
β
β
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
β
β
+
2
π
π
2
πΏ
2
β
β
π
βΆ
0
,
π₯
ξ
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π
+
2
π
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
β
β
+
2
π
π
2
πΏ
2
+
β
β
π
π
β
β
2
πΏ
β
β
β
6
0
π
ξ
ξ
π
ξ
ξ
ξ
+
3
0
π
β²
π
(
4
)
+
6
π
π
(
5
)
β
β
+
2
π
π
β²
2
πΏ
2
β
β
π
βΆ
0
,
2
π₯
ξ
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π
+
2
π
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
2
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
β
β
+
2
π
π
2
πΏ
2
β
β
π
+
2
π₯
π
π
β
β
2
πΏ
β
β
β
6
0
π
ξ
ξ
π
ξ
ξ
ξ
+
3
0
π
β²
π
(
4
)
+
6
π
π
(
5
)
β
β
+
2
π
π
β²
2
πΏ
2
+
β
β
π
π
β
β
2
πΏ
β
β
β
6
0
π
2
ξ
ξ
ξ
+
9
0
π
ξ
ξ
π
(
4
)
+
3
6
π
β²
π
(
5
)
+
6
π
π
(
6
)
+
2
π
π
ξ
ξ
β
β
2
πΏ
2
β
β
π
βΆ
0
,
3
π₯
ξ
ξ
1
8
π
2
ξ
ξ
+
2
4
π
ξ
π
ξ
ξ
ξ
+
6
π
π
(
4
)
ξ
π
+
2
π
π
π
ξ
β
β
2
πΏ
2
β€
β
β
π
3
π₯
π
π
β
β
2
πΏ
β
β
β
1
8
π
2
ξ
ξ
+
2
4
π
β²
π
ξ
ξ
ξ
+
6
π
π
(
4
)
β
β
+