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Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2009, Article ID 137084, 15 pages
doi:10.1155/2009/137084

Research Article
The Existence of Periodic Solutions for
Non-Autonomous Differential Delay Equations
via Minimax Methods
Rong Cheng1, 2
1

College of Mathematics and Physics, Nanjing University of Information Science and Technology,
Nanjing 210044, China
2
Department of Mathematics, Southeast University, Nanjing 210096, China
Correspondence should be addressed to Rong Cheng,
Received 9 April 2009; Accepted 19 October 2009
Recommended by Ulrich Krause
By using variational methods directly, we establish the existence of periodic solutions for a class of
nonautonomous differential delay equations which are superlinear both at zero and at infinity.
Copyright q 2009 Rong Cheng. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.

1. Introduction and Main Result
Many equations arising in nonlinear population growth models 1 , communication systems
2 , and even in ecology 3 can be written as the following differential delay equation:
x t

−αf x t − 1 ,



1.1

where f ∈ C R, R is odd and α is parameter. Since Jone’s work in 4 , there has been a
great deal of research on problems of existence, multiplicity, stability, bifurcation, uniqueness,
density of periodic solutions to 1.1 by applying various approaches. See 2, 4–23 . But
most of those results concern scalar equations 1.1 and generally slowly oscillating periodic
solutions. A periodic solution x t of 1.1 is called a “slowly oscillating periodic solution” if
there exist numbers p > 1 and q > p 1 such that x t > 0 for 0 < t < p, x t < 0 for p < t < q,
and x t q
x t for all t.
In a recent paper 17 , Guo and Yu applied variational methods directly to study the
following vector equation:
x t

−f x t − r ,

1.2


2

Advances in Difference Equations

where f ∈ C Rn , Rn is odd and r > 0 is a given constant. By using the pseudo index theory
in 24 , they established the existence and multiplicity of periodic solutions of 1.2 with f
satisfying the following asymptotically linear conditions both at zero and at infinity:
f x
f x


B0 x

o |x| ,

as |x| −→ 0,

B∞ x

o |x| ,

as |x| −→ ∞,

1.3

where B0 and B∞ are symmetric n × n constant matrices. Before Guo and Yu’s work, many
authors generally first use the reduction technique introduced by Kaplan and Yorke in 7 to
reduce the search for periodic solutions of 1.2 with n 1 and its similar ones to the problem
of finding periodic solutions for a related system of ordinary differential equations. Then
variational method was applied to study the related systems and the existence of periodic
solutions of the equations is obtained.
The previous papers concern mainly autonomous differential delay equations. In this
paper, we use minimax methods directly to study the following nonautonomous differentialdelay equation:
x t

−f t, x t − r ,

1.4

where f ∈ C R × Rn , Rn is odd with respect to x and satisfies the following superlinear
conditions both at zero and at infinity

lim

|x| → 0

f t, x
|x|

f t, x
lim
|x| → ∞
|x|

0,

uniformly in t,
1.5

∞,

uniformly in t.

When 1.2 satisfies 1.3 , we can apply the twist condition between the zero and at infinity
for f to establish the existence of periodic solutions of 1.2 . Under the superlinear conditions
1.5 , there is no twist condition for f, which brings difficulty to the study of the existence of
periodic solutions of 1.4 . But we can use minimax methods to consider the problem without
twist condition for f.
Throughout this paper, we assume that the following conditions hold.
H1 f t, x ∈ C R × Rn , Rn is odd with respect to x and 2r-periodic with respect to t.
H2 write f


f1 , f2 , . . . , fn . There exist constants μ > 2 and R1 > 0 such that
xi

0<μ

fi t, x1 , . . . , xi−1 , yi , xi 1 , . . . , xn dyi ≤ xi fi t, x

1.6

0

for all x ∈ Rn with |xi | > R1 , for all t ∈ 0, 2r and i

1, 2, . . . , n.

H3 there exist constants c1 > 0, R2 > 0 and 1 < λ < 2 such that
fi t, x

< c1 |xi |λ

for all x ∈ Rn with |xi | > R2 , for all t ∈ 0, 2r and i

1.7
1, 2, . . . , n.


Advances in Difference Equations

3


Then our main result can be read as follows.
Theorem 1.1. Suppose that f t, x ∈ C R × Rn , Rn satisfies 1.5 and the conditions H1 – H3
hold. Then 1.4 possesses a nontrivial 4r-periodic solution.
Remark 1.2. We shall use a minimax theorem in critical point theory in 25 to prove our main
result. The ideas come from 25–27 . Theorem 1.1 will be proved in Section 2.

2. Proof of the Main Result
First of all in this section, we introduce a minimax theorem which will be used in our
discussion. Let E be a Hilbert space with E E1 ⊕ E2 . Let P1 , P2 be the projections of E onto
E1 and E2 , respectively.
Write
Λ

ϕ ∈ C 0, 2r × E | ϕ 0, u

u and P2 ϕ t, u

P2 u − Φ t, u ,

2.1

where Φ : 0, 2r × E → E2 is compact.
Definition 2.1. Let S, Q ⊂ E, and Q be boundary. One calls S and ∂Q link if whenever ϕ ∈ Λ
and ϕ t, ∂Q ∩ S ∅ for all t, then ϕ t, Q ∩ S / ∅.
Definition 2.2. A functional φ ∈ C1 E, R satisfies P S condition, if every sequence that
{xm } ⊂ E, φ xm → 0 and φ xm being bounded, possesses a convergent subsequence.
Then 25, Theorem 5.29 can be stated as follows.
Theorem A. Let E be a real Hilbert space with E
φ ∈ C1 E, R satisfies P S condition,


E1 ⊕E2 , E2


E1 and inner product ·, · . Suppose

A1 P1 x A2 P2 x and Ai : Ei → Ei is bounded

1/2 Ax, x ψ x , where A z
I1 φ x
and selfadjoint, i 1, 2,
I2 ψ is compact, and

I3 there exists a subspace E ⊂ E and sets S ⊂ E, Q ⊂ E and constants α > ω such that
i S ⊂ E1 and φ|S ≥ α,
ii Q is bounded and φ|∂Q ≤ ω,
iii S and ∂Q link.
Then φ possesses a critical value c ≥ α.
Let
x1

F t, x

f1 t, y1 , x2 , . . . , xn dy1
0

···

xn

fn t, x1 , . . . , xn−1 , yn dyn .


2.2

0

Then F t, 0
0 and F t, x
f1 , f2 , . . . , fn , where F denotes the gradient of F with respect
to x. We have the following lemma.


4

Advances in Difference Equations

Lemma 2.3. Under the conditions of Theorem 1.1, the function F satisfies the following.
i F t, x ∈ C1 0, 2r × Rn , R is 2r-periodic with respect to t and F t, x ≥ 0 for all t, x ∈
0, 2r × Rn ,
ii
F t, x

lim

|x| → 0

lim

F t, x

|x| → ∞


uniformly in t,

0,

|x|2

∞,

|x|2

2.3

uniformly in t.

2.4

x1 , . . . , xn ∈ Rn with

iii There exist constants c2 , L > 0, and R > 0 such that for all x
|x| > L and |xi | ≥ R, i 1, 2, . . . , n, and t ∈ 0, 2r
0 < μF t, x ≤ x, F t, x ,
F t, x

2.5

λ

≤ c2 |x| ,


2.6

where ·, · denotes the inner product in Rn .
Proof. The definition of F implies i directly. We prove case ii and case iii .
Case ii . Let
x1
x2
x3

r sin θ1 ,

r sin θ1 conθ2 ,

r sin θ1 sin θ2 conθ3 ,

2.7

···
xn−1

r sin θ1 sin θ2 sin θ3 · · · sin θn−2 cos θn−1 ,
r sin θ1 sin θ2 sin θ3 · · · sin θn−2 sin θn−1 .

xn

Then |x|2 r 2 and |x| → 0 or |x| → ∞ is equivalent to r → 0 or r → ∞, respectively.
From 1.5 and L’Hospital rules, we have 2.3 by a direct computation.

nR1 such that 0 < μF t, x ≤ x, F t, x
Case iii . By H2 , we have a constant L1

for |x| > L1 with |xi | ≥ R1 .

nR2 with |xi | ≥ R2 , that is,
Now we prove |F t, x | ≤ c2 |x|λ for |x| > L2
2
f1 t, x1

···

2
2
2
fn t, xn ≤ c2 x1

···

2
xn

λ

.

2
2
Firstly, it follows from |f1 t, x1 | ≤ c1 |x1 |λ that f1 t, x1 ≤ c1 |x1 |2λ .
2
2
2



Now we show f1 t, x1 f2 t, x2 ≤ c1 |x1 |
|x2 | . Let |x1 |λ τ cos θ, |x2 |λ
λ
By 1 < λ < 2, 1 − sin2 θ ≥ 1 − sin2/λ , that is, cos2/λ sin2/λ λ ≥ 1. Then

2
x1

2
x2

λ

τ 2 cos2/λ θ

sin2/λ θ ≥ τ 2

|x1 |2λ

|x2 |2λ .

2.8

τ sin θ.

2.9


Advances in Difference Equations


5

By reducing method, we have
···


x1


2
xn ≤ x1

···

2
xn

λ

|x|2λ .

2.10

Thus, the inequality |F t, x | ≤ c2 |x|λ for |xi | ≥ R2 holds.
Take L max{L1 , L2 } and R max{R1 , R2 }. Then 2.5 and 2.6 hold with |x| > L and
|xi | > R.
Below we will construct a variational functional of 1.4 defined on a suitable Hilbert
space such that finding 4r-periodic solutions of 1.4 is equivalent to seeking critical points
of the functional.

Firstly, we make the change of variable
π
t
2r

t −→

ν−1 t.

2.11

Then 1.4 can be changed to
−νf t, x t −

x t

π
2

,

2.12

where f is π-periodic with respect to t. Therefore we only seek 2π-periodic solution of 2.12
which corresponds to the 4r-periodic solution of 1.4 .
We work in the Sobolev space H W 1/2,2 S1 , R2N . The simplest way to introduce this
space seems as follows. Every function x ∈ L2 S1 , Rn has a Fourier expansion:


x t


a0

am cos mt

bm sin mt ,

2.13

m 1

where am , bm are n-vectors. H is the set of such functions that
x

2

|a0 |2



m |am |2

|bm |2 < ∞.

2.14

m 1

With this norm · , H is a Hilbert space induced by the inner product ·, · defined by



x, y

2π a0 , a0

m am , am

π

bm , bm

,

2.15

F t, x t dt.

2.16

m 1

bm sin mt .
where y a0
m 1 am cos mt
We define a functional φ : H → R by


φ x
0


1
x t
2

π
, x t dt
2



ν
0


6

Advances in Difference Equations

By Riesz representation theorem, H identifies with its dual space H∗. Then we define
an operator A:H → H∗ H by extending the bilinear form:


π
, y t dt,
2

x t

Ax, y
0


∀x, y ∈ H.

It is not difficult to see that A is a bounded linear operator on H and kerA
Define a mapping ψ : H → R as

2.17
Rn .



ψ x

ν

2.18

F t, x t dt,
0

Then the functional φ can be rewritten as
φ x

1
Ax, x
2

∀x ∈ H.

ψ x ,


2.19

According to a standard argument in 24 , one has for any x, y ∈ H,


φ x ,y
0

1
x t
2

π
2

−x t−

π
, y t dt
2



ν

f t, x t , y t dt.

2.20


0

Moreover according to 28 , ψ : H → H is a compact operator defined by


ψ x ,y

ν

f t, x t , y t dt.

2.21

0

Our aim is to reduce the existence of periodic solutions of 2.12 to the existence of
critical points of φ. For this we introduce a shift operator Γ : H → H defined by
Γx t

x t

π
.
2

2.22

It is easy to compute that Γ is bounded and linear. Moreover Γ is isometric, that is, Γx
and Γ4 id, where id denotes the identity mapping on H.
Write

E

x ∈ H : Γ2 x t

−x t

.

x

2.23

Lemma 2.4. Critical points of φ|E over E are critical points of φ on H, where φ|E is the restriction of
φ over E.
Proof. Note that any x ∈ E is 2π-periodic and f is odd with respect to x. It is enough for us to
0 for any y ∈ H and x being a critical point of φ in E.
prove φ x , y


Advances in Difference Equations

7

For any y ∈ H, we have
Γ2 φ x , y



π
,y t − π

2

x t
0


π
2

x t
0


−x t





dt

ν
ν

0

f t, x t , y t − π dt
f t

π, x t


π , y t dt

0


2.24

f t, −x t , y t dt

ν

π
, y t dt − ν
2

x t

ψ x , Γ−2 y

0


π , y t dt

π
, y t dt
2

0




Ax, Γ−2 y

Γ2 ψ x , y

Γ2 Ax, y

0


f t, x t , y t dt
0

−φ x , y .
−φ x , that is, φ x ∈ E.
This yields Γ2 φ x
Suppose that x is a critical point of φ in E. We only need to show that φ x , y
any y ∈ H. Writing y y1 ⊕ y2 with y1 ∈ E, y2 ∈ E⊥ and noting φ x ∈ E, one has
φ x ,y

φ x , y1

φ x , y2

0.

0 for


2.25

The proof is complete.
Remark 2.5. By Lemma 2.4, we only need to find critical points of φ|E over E. Therefore in the
following φ will be assumed on E.
For x ∈ E, x t π
−x t yields that a0
x. Thus kerA|E {0}. Moreover for any x, y ∈ E,


x t

Ax, y
0




0

0, where a0 is in the Fourier expansion of

π
, y t dt
2

π
x t ,y t −
2






x t
0


dt

π
, y t dt
2

x t ,y t
0

π
2

dt

2.26

x, Ay .
Hence A is self-adjoint on E.
Let E and E− denote the positive definite and negative definite subspace of A in E,
respectively. Then E E ⊕ E− . Letting E1 E , E2 E− , we see that I1 of Theorem A holds.
Since ψ is compact, I2 of Theorem A holds. Now we establish I3 of Theorem A by the
following three lemmas.

Lemma 2.6. Under the assumptions of Theorem 1.1, i of I3 holds for φ.
Proof. From the assumptions of Theorem 1.1 and Lemma 2.3, one has
F t, x ≤ c3

c4 |x|λ 1 ,

∀ t, x ∈ 0, π × Rn .

2.27


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Advances in Difference Equations

By 2.3 , for any ε > 0, there is a δ > 0 such that
F t, x ≤ ε|x|2 ,
Therefore, there is an M

∀t ∈ 0, π , |x| ≤ δ.

2.28

M ε > 0 such that

F t, x ≤ ε|x|2

M|x|λ 1 ,

∀ t, x ∈ 0, π × Rn .


2.29

Since E is compactly embedded in Ls S1 , Rn for all s ≥ 1 and by 2.29 , we have


F t, x dt ≤ ε x

0

Consequently, for x ∈ E1

2
L2

3νc5

−1

λ 1
Lλ 1

≤ εc5

Mc6 x

λ−1

x 2.


2.30

E ,
φ x ≥ x

Choose ε

M x

2

− ν εc5

Mc6 x

x 2.

2.31

1. Then for any x ∈ ∂Bρ ∩ E1 ,

and ρ so that 3νMc6 ρλ−1
φ x ≥

1 2
ρ .
3

∂Bρ ∩ E1 and α


Thus φ satisfies i of I3 with S

λ−1

2.32
1/3 ρ2 .

Lemma 2.7. Under the assumptions of Theorem 1.1, φ satisfies ii of I3 .
Proof. Set e ∈ S

∂Bρ ∩ E1 and let
Q

{se : 0 ≤ s ≤ 2s1 } ⊕ B2s1 ∩ E2 ,

2.33

where s1 is free for the moment.
Let E E− ⊕ span{e}. Write
K

x∈E: x

1 ,

λ−

inf

x∈E− , x


1

Ax− , x− ,

λ

sup
x∈E , x

| Ax , x |.
1

2.34
Case 1 . If x− > γ x
φ sx

λ /λ− , one has

with γ

1
Asx , sx
2
1
≤ − λ− s2 x−
2

2


1
Asx− , sx− − ν
2
1
2
λ s2 x
≤ 0.
2



F t, sx dt
0

2.35


Advances in Difference Equations

9

Case 2 . If x− ≤ γ x , we have
1

x

2

x


x−

2

2

≤ 1

γ2

x

2

.

2.36

That is
x

Denote K

2



1
1


γ2

> 0.

2.37

{x ∈ K : x− ≤ γ x }. By appendix, there exists ε1 > 0 such that ∀u ∈ K,
meas{t ∈ 0, π : |u t | ≥ ε1 } ≥ ε1 .

2.38

x− ∈ K, set Ωx {t ∈ 0, π : |x t | ≥ ε1 }. By 2.4 , for a constant M0
Now for x x
3
A /νε1 > 0, there is an L3 > 0 such that
F t, z ≥ M0 |x|2 ,

∀|x| ≥ L3 uniformly in t.

2.39

Choosing s1 ≥ L3 /ε1 , for s ≥ s1 ,
F t, sx t

2
≥ M0 |sx t |2 ≥ M0 s2 ε1 ,

∀t ∈ Ωx .

2.40


For s ≥ s1 , we have
φ sx

1 2
1 2
s Ax , x
s Ax− , x− − ν
2
2
1
≤ A s2 − ν
F t, sx dt
2
Ωx
1
2
≤ A s2 − M0 s2 ε1 meas Ωx
2
1
1
3
≤ A s2 − M0 s2 ε1 − A s2 < 0.
2
2



F t, sx dt
0


2.41

Henceforth, φ sx ≤ 0 for any x ∈ K and s ≥ s1 , that is, φ|∂Q ≤ 0. Then ii of I3 holds.
Lemma 2.8. S and ∂Q link.
Proof. Suppose ϕ ∈ Λ and ϕ ∂Q ∩S ∅ for all t ∈ 0, π . Then we claim that for each t ∈ 0, π ,
there is a w t ∈ Q such that φ t, w t ∈ S, that is,
Pϕ w t

0,

w t

ρ,

2.42

where P : E → E− is a projection. Define
G : 0, π × Q −→ E− × Re

2.43


10

Advances in Difference Equations

as follows:
G t, u


se

1−t u

Pϕ u

1−t s

se

t I −P ϕ u

se

− ρ e.

2.44

It is easy to see that
G t, u

se

s − ρ e / 0,

u

as u

se ∈ ∂Q.


2.45

However,
G 1, u

se

Pϕ u
G 0, u

se

I −P ϕ u

se
u

se

−ρ e

2.46

s − ρ e.

According to topological degree theory in 29 , we have
deg G 1, · ; Q, 0

deg G 0, · ; Q, 0

deg idE− ; E− ∩ B2s1 , 0 deg s − ρ, 0, 2s1 , 0

1

2.47

since ρ ∈ 0, 2s1 . Therefore S and ∂Q link.
Now it remains to verify that φ satisfies P S -condition.
Lemma 2.9. Under the assumptions of Theorem 1.1, φ satisfies P S -condition.
Proof. Suppose that
φ xm

≤M,

φ xm −→ 0,

as m −→ ∞.

2.48

We first show that {xm } is bounded. If {xm } is not bounded, then by passing to a subsequence
if necessary, let xm → ∞ as m → ∞.
By 2.4 , there exists a constant M > 0 such that F t, x > c7 |x|2 as |x| > M . By 2.5 ,
one has


2φ xm − φ xm , xm

xm , νF t, xm



0


− 2νF t, xm dt

ν μ − 2 F t, xm dt

2.49

0

≥ c7 ν μ − 2



|xm |2 dt.

0

This yields

2
dt
0 |xm |

xm

−→ 0 as m −→ ∞.


2.50


Advances in Difference Equations
Write κ

11

1/2 λ − 1 . By 2.6 , there is a constant c9 > 0 such that
F t, x

κ

κ
≤ c2 |x|λκ

∀ t, x ∈ 0, π × Rn .

c8 ,

2.51

Therefore,


F t, xm

k

dt ≤


0


0

κ
c2 |xm |λκ


≤ c9

c8 dt
1/2

xm

0


≤ c11

1/2



xm 2 dt

2 κλ−1


dt

c10

2.52

−→ 0

2.53

0
1/2

xm 2 dt

xm

κλ−1

c12 .

0

This inequality and 2.50 imply that





|F

0

t, xm |κ dt
xm

1/κ ⎞1/κ

κ






c11


0

xm 2 dt

1/2

xm
xm

1/2

xm


κλ−1
κ−1/2

c12
xm

κ

as m → ∞, since κ > 1.

Denote xm xm xm ∈ E ⊕ E− . We have

φ xm , xm



Axm , xm −


≥ Axm , xm −


≥ Axm , xm −




xm , F t, xm dt

0




xm F t, xm dt

0

2.54
1/κ



F t, xm

κ

dt

0


Cκ xm ,

where Cκ > 0 is a constant independent of m.
By the above inequality, one has



φ xm
xm

Axm , xm



xm xm
xm xm


0 |F

t, xm |κ dt
xm

1/κ


Cκ xm
−→ 0

xm

2.55

as m → ∞. This yields

xm
−→ 0 as m −→ ∞.
xm

2.56



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Advances in Difference Equations

Similarly, we have
xm
−→ 0 as m −→ ∞.
xm

2.57

Thus it follows from 2.56 and 2.57 that
1


x
xm
xm
≤ m
−→ 0 as m −→ ∞,
xm
xm

which is a contradiction. Hence {xm } is bounded.
Below we show that {xm } has a convergent subsequence. Notice that kerA|E
ψ : H → H is compact. Since {xm } is bounded, we may suppose that
ψ xm −→ y


as m −→ ∞.

2.58

{0} and

2.59

Since A has continuous inverse A−1 in E, it follows from
Axm

φ xm

ψ xm

2.60

that
xm

A−1 φ xm

ψ xm

−→ A−1 y

as m −→ ∞.

2.61


Henceforth {xm } has a convergent subsequence.
Now we are ready to prove Theorem 1.1.
Proof of Theorem 1.1. It is obviously that Theorem 1.1 holds from Lemmas 2.3, 2.4, 2.6, 2.7, 2.8,
and 2.9 and Theorem A.

Appendix
The purpose of this appendix is to prove the following lemma. The main idea of the proof
comes from 26 .
Lemma A.1. There exists ε1 > 0 such that, for all u ∈ K,
meas{t ∈ 0, π : |u t | ≥ ε1 } ≥ ε1 .

A.1

Proof. If A.1 is not true, ∀k > 0, there exists uk ∈ K such that
meas t ∈ 0, π : |u t | ≥

1
k



1
.
k

A.2


Advances in Difference Equations
Write uk

we have

13

u− uk ∈ E. Notice that dimspan{e} < ∞ and uk ≤ 1. In the sense of subsequence,
k

uk −→ u0 ∈ span{e} as k −→ ∞.

A.3

Then 2.37 implies that
2

u0



1
γ2

1

> 0.

Note that u− ≤ 1, in the sense of subsequence u−
k
k
sense of subsequence,
uk


u0

u−
0

A.4

u− ∈ E− as k →
0

∞. Thus in the

as k −→ ∞.

u0

A.5

This means that uk → u0 in L2 , that is,
π

|uk − u0 |2 dt −→ 0

as k −→ ∞.

A.6

0


By A.4 we know that u0 > 0. Therefore,
that

π
|u0 |2 dt
0

> 0. Then there exist δ1 > 0, δ2 > 0 such

meas{t ∈ 0, π : |u0 t | ≥ δ1 } ≥ δ2 .

A.7

Otherwise, for all n > 0, we must have
meas t ∈ 0, π : |u0 t | ≥

1
n

0,

A.8

π

that is, meas{t ∈ 0, π : |u0 t | ≤ 1/n} 1, 0 < 0 |u0 |2 dt < 1/n2 → 0 as n → ∞. We get a
contradiction. Thus A.7 holds. Let Ω0 {t ∈ 0, π : |u0 t | ≥ δ1 }, Ωk {t ∈ 0, π : |u0 t | ≤
0, π \ Ωk . By A.2 , we have
1/k}, and Ω⊥
k

meas Ωk ∩ Ω0

meas Ω0 − Ω0 ∩ Ω⊥ ≥ meas Ω0 − meas Ω0 ∩ Ω⊥ ≥ δ2 −
k
k

1
.
k

A.9

Let k be large enough such that δ2 − 1/k ≥ 1/2 δ2 and δ1 − 1/k ≥ 1/2 δ1 . Then we have
|uk − u0 |2 ≥

δ1 −

1
k

2



1
δ1
2

2


,

∀t ∈ Ωk ∩ Ω0 .

A.10


14

Advances in Difference Equations

This implies that
π

|uk − u0 |2 dt ≥

0



Ωk ∩Ω0

1
δ1
2

1
δ1
2


|uk − u0 |2 dt ≥
2

1
· δ2 −
k



1
δ1
2

2

· meas Ωk ∩ Ω0
2

1
δ2
2

A.11
> 0.

This is a contradiction to A.6 . Therefore the lemma is true and A.1 holds.

Acknowledgments
This work is supported by the Specialized Research Fund for the Doctoral Program of Higher
Education for New Teachers and the Science Research Foundation of Nanjing University of

Information Science and Technology 20070049 .

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