Forced Oscillation of Neutral Impulsive Parabolic Partial Differential Equations with Continuous Distributed Deviating Arguments

(1)

How to cite this paper: Liu, G.J. and Wang, C.Y. (2014) Forced Oscillation of Neutral Impulsive Parabolic Partial Differential Equations with Continuous Distributed Deviating Arguments. Open Access Library Journal, 1: e1168.

http://dx.doi.org/10.4236/oalib.1101168

Forced Oscillation of Neutral Impulsive

Parabolic Partial Differential Equations

with Continuous Distributed

Deviating Arguments

Guangjie Liu1,2_{, Chengyong Wang}1

1_{School of Mathematics and Computer Science, Hubei University of Arts and Science, Xiangyang, China} 2_{School of Mathematics and Physics, China University of Geosciences, Wuhan, China}

Email: [email protected]

Received 26 October 2014; revised 27 November 2014; accepted 18 December 2014

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

Abstract

This paper investigated oscillatory properties of solutions for nonlinear parabolic equations with impulsive effects under two different boundary conditions. By using integral averaging method, variable substitution and functional differential inequalities, we established several sufficient conditions. At last, we provided two examples to illustrate the results.

Keywords

Forced Oscillation, Parabolic Equation, Impulsive, Neutral Type, Continuous Distributed Deviating Arguments

Subject Areas: Integral Equation, Mathematical Analysis, Partial Differential Equation

1. Introduction

In this article, we discuss forced oscillatory properties of solutions for the nonlinear impulsive parabolic equa-tions of neutral type.

( )

(

( )

)

( )

( ) ( )

( )

(

( )

)

(

)

(

( )

)

( )

1

, , , , d

, , , , , , d

, , , , ,

l

i i

i k

u t x g t u t x

t

a t u t x b t u t x q t x f u t x

F t x t t t x G

β α

δ γ

ξ ρ ξ η ξ

τ ζ σ ζ ω ζ

=

+

∂  ₊ 

 

 

∂

= ∆ + ∆ −

+ ≠ ∈ × Ω =

∫

∑

∫



(2)

OALibJ | DOI:10.4236/oalib.1101168 2 December 2014 | Volume 1 | e1168

( ) ( )

_k, _k, _k

(

_k,

)

, 1, 2,

u t+ x −u t− x =b u t x k=  (2) with the boundary conditions

( )

0, , ,

u= t x ∈+× ∂Ω (3)

( )

0, ,

u

cu t x n

+ ∂

+ = ∈ × ∂Ω

∂  (4)

and the initial condition

( )

,

( ) ( )

, , ,

[

, 0

]

.

u t x = Φ t x t x ∈ −λ × Ω

Here Ω ⊂N is a bounded domain with boundary ∂Ω smooth enough and n is the unit exterior normal vector of ∂Ω, max_t_∈_+

{

t−ρ ξ

( )

t, ,t−τi

( )

t t, −σ ζ

( )

t,

}

≤λ a positive constant,

( )

2

(

[

]

)

, , 0 ,

t x C λ

Φ ∈ − × Ω .

We will use the following conditions.

(H1) a t b t

( ) ( )

, i PC

(

,

)

+ +

∈   , τi

( )

t C

(

,

)

+

∈   , g t

( )

,ξ ∈PC

(

+×

[

α β,

]

,+

)

,

( )

t, C

(

[

,

]

,

)

ρ ξ _∈ _+_× α β _

, σ ζ

( )

t, ∈C

(

+×

[ ]

γ δ, ,

)

, q t x

(

, ,ζ

)

∈C

(

+× Ω×

[ ]

γ δ, ,+

)

, such that

( )

, 0

t−ρ ξt > , t> − >t ι σ

( )

t,ζ , limt→∞minξ α β∈_[ _, _]ρ ξ

( )

t, = ∞, limt→∞τi

( )

t = ∞,

[ ],

( )

limt→∞minζ γ δ∈ σ ζt, = ∞, where ι is a constant, PC denote the class of functions which are piecewise continuous in t with discontinuities of first kind only at t=tk, and left continuous at t=tk, k=1, 2,,

(

t_k,

)

t_j

ρ ξ ≠ for j<k, j=1, 2,, g t

( )

_k+,ξ = +

(

1 b_k

)

g t

( )

_k−,ξ .

(H2) −η ξ

( )

, ω ζ

( )

are nondecreasing functions on

[

α β,

]

and

[ ]

γ δ, , respectively; f u

( )

∈C

(

 ,

)

,

( )

f u u≥C, g t

( ) ( )

, d h0 1

β

α ξ η ξ

−

_∫

≤ < , where h0 and C are positive constants, u≠0; c>0, bk ≥0, 1, 2,

k= ,

∫

_Ωϕ

( ) ( )

x F t x, dx≤0, limk→∞tk = ∞, 0< < <t1 t2 .

(H3) u t x

( )

, is piecewise continuous in t with discontinuities of first kind only at t=tk, and left continuous at t=tk, u t x

(

k,

)

u t

( )

k,x

−

= , k=1, 2,.

Let us construct the sequence

{ } { }

tk = tk 

{ } { } { }

tjξ  tjτ  tjζ , where

{ }

tjξ =

{

t ρ ξ

( )

t, =tj

}

,

{ }

tjτ =

{

tτi

( )

t =tj

}

,

{ }

tjζ =

{

tσ ζ

( )

t, =tj

}

and tk <tk+1, i=1, 2,,l; k j, =1, 2,.

We introduce the notations:

( )

(

)

{

1

}

0

, : , , , ,

k k k k

k

t x t t t x

∞ +

=

Γ = ∈ ∈ Ω Γ =

_

Γ

( )

(

)

{

1

}

0

, : , , ,

k k k k

k

t x t t t x

∞ +

=

Γ = ∈ ∈ Ω Γ =

_

Γ

( )

( ) ( )

, d ,

( )

, min

(

, ,

)

.

x

v t _Ωϕ x u t x x Q tζ q t xζ

∈Ω

=

_∫

=

The solution u∈C2

( )

Γ C1

( )

Γ of problems (1), (3) ((4)) is called non-oscillatory in the domain G if it is either eventually positive or eventually negative. Otherwise, it is called oscillatory.

(3)

OALibJ | DOI:10.4236/oalib.1101168 3 December 2014 | Volume 1 | e1168 In 1991, the first paper [7] on this class of equations was published. However, we only find a few of papers on oscillation theory of impulsive partial differential equations. Recently, Bainov, Minchev, Fu, and Liu [8]-[12]

investigated the oscillation of solutions of impulsive partial differential equations with or without deviating ar-guments, and Tanaka, Luo, and Shoukaku [13]-[15] discussed the oscillation of solutions of partial differential equations with continuous distributed deviating arguments. However, there is a scarcity in the study of forced oscillation theory of nonlinear impulsive parabolic equations of neutral type with continuous distributed deviat-ing arguments.

2. Oscillation Properties of the Problem (1), (4)

For the main result of this article, we need the following lemma.

Lemma 2.1. Let b0 and ϕ

( )

x be the minimum eigenvalue and the corresponding eigenfunction,

respec-tively, of the problem

( )

x b

( )

x 0, x ,

ϕ ϕ

∆ + = ∈ Ω

( )

x 0, or c

( )

x 0,x ,

x

ϕ

ϕ = ∂ + ϕ = ∈ ∂Ω

∂

where c is a positive constant and n is outer normal vector of ∂Ω at point x. Then b>0 and ϕ

( )

x >0

for x∈Ω.

Lemma 2.2. [16] Let ρ=const>0 , a t0

( ) ( )

,p t ∈

(

[

0,+∞

)

,R

)

be locally summable functions and

( )

0

p t > ; y t

( )

k y t

( )

k −

= , k=1, 2,. If the following condition is satisfied

( )

(

( )

)

(

)

1

0

1

lim inf exp d 1 d

e

k

t s

k

t s

t

s t s

p s a r r d s

ρ ρ _ρ

−

− −

→+∞

∫

_{− < <}

∏

+ > ,

then the following differential inequality has no eventually positive solution.

( )

0

( ) ( )

( ) (

)

0, 0, k

y t′ +a t y t +p t y t−ρ ≤ t≥ t≠t

( ) ( )

_k _k _k

( )

_k , 1, 2, .

y t+ −y t− =d y t k= 

The following theorem is the first main result of this article.

Theorem 2.1. Suppose that the conditions (H1)-(H3) and the following condition (5) hold

( )

(

)

(

)

( )

(

) (

) ( )

1

0 0

,

1

lim inf exp d 1 d 1 , d .

e

k

t s

k

t s

t

t t t

b a r r b s δC h Q s

ι ι _{σ ζ} γ ζ ω ζ

−

− −

→+∞

∫

∏

_{≤ <} +

∫

− > (5)

Then each solution of the problem (1)-(3) oscillates in G.

Proof. Suppose that the assertion is not true and u t x

( )

, is a non-oscillatory solution of problem (1), (3) in G. Without loss of generality, we may assume that there exists a t0 >0 such that u t x

( )

, >0, u

(

ρ ξ

( )

t, ,x

)

>0,

( )

(

i ,

)

0

u τ t x > , i=1, 2,,l, u

(

σ ζ

( )

t, ,x

)

>0, for any

( )

t x, ∈

[

t0,+∞ × Ω

)

.

For t≥t0, t≠tk, k=1, 2,, multiplying (1) by ϕ

( )

x and then integrating it with respect to x over Ω yields

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

( ) (

)

(

( )

)

( ) ( )

1

d

, d , d , , d

d

d , d

d , , , , d , d .

l

i i

i

x u t x x g t x u t x x

t

a t x u x b t x u t x x

x q t x f u t x x x F t x x

β α

δ γ

ϕ ξ η ξ ϕ ρ ξ

ϕ ϕ τ

ω ζ ϕ ζ σ ζ ϕ

Ω Ω

=

Ω Ω

 ₊ 

 

 

= ∆ + ∆

− +

∫

∑

∫

By Green’s formula and the boundary condition, we have

( )

x u xd u

( )

x dx u sd u ds 0.

n n

ϕ

ϕ ϕ ϕ

Ω Ω ∂Ω ∂Ω

∂ ∂

∆ − ∆ = − =

∂ ∂

∫

It follows that

( )

x u xd u

( )

x dx b0 u

( )

x d ,x

ϕ ϕ ϕ

Ω ∆ = Ω ∆ = − Ω

(4)

( )

x u

(

i

( )

t ,x

)

dx u

(

i

( )

t ,x

)

( )

x dx b0 u

(

i

( )

t ,x

)

( )

x d .x

ϕ τ τ ϕ τ ϕ

Ω ∆ = Ω ∆ = − Ω

∫

From condition (H2), we can easily obtain

( ) (

x q t x, ,

)

f u

(

( )

t, ,x

)

dx CQ t

( )

,

( )

x u

(

( )

t, ,x

)

d .x

ϕ ζ σ ζ ζ ϕ σ ζ

Ω ≥ Ω

∫

( ) ( )

x F t x, dx 0. ϕ

Ω ≤

∫

From the above it follows that

( )

(

( )

)

( )

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

0 0 1 d

, , d

d

, , d 0.

l

i i

i

v t g t v t b a t v t

t

b b t v t C Q t v t

β α

δ γ

ξ ρ ξ η ξ

τ ζ σ ζ ω ζ

=  ₊ ₊     + + ≤

∫

∑

∫

(6)

In inequality (6), set

( )

(

) ( )

0

1

k k

t t t

w t =

∏

_{≤ <} +b − v t , we can obtain the following results: 1) w t

( )

is continuous on

[

t0,+∞

)

; 2) Inequality (6) has no eventually positive solution if the following inequality (7) has no

eventually positive solution.

( )

(

( )

)

( )

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

0

0 0 0

1

d

, , d

d

, , d 0, , ,

l

i i k

i

w t G t w t b a t w t

t

b B t w t C Q t w t t t t t

β α

δ γ

ξ ρ ξ η ξ

τ ζ σ ζ ω ζ

=  ₊ ₊     + + ≤ ≥ ≠

∫

∑

∫

(7) where

( )

(

) ( )

( )

(

)

( )

(

)

( )

( ) 1 1 , 1 0 ,

, 1 , , 1 ,

, 1 , .

k i k

k

k i k i

t t t t t t

k t t t

G t b g t B t b b t

Q t b Q t

ρ ξ τ

σ ζ

ξ ξ

ζ ζ

− −

≤ < ≤ <

− ≤ < = + = + = +

∏

In fact, v t

( )

is continuous on each interval

(

t tk, k+1

]

, and in view of v t

( )

k

(

1 b v tk

) ( )

k + _{= +}

, it follows that for all t≥t0,

( )

(

)

( )

(

)

( )

0 0

1 1

1 1 ,

j k j k

k j k j k k

t t t t t t

w t+ b − v t+ b − v t w t

≤ ≤ ≤ <

=

∏

+ =

∏

+ =

( )

(

)

( )

(

)

( )

0 1 0

1 1

j k j k

k j k j k k

t t t t t t

w t b v t b v t w t

−

− −

≤ ≤ ≤ <

=

∏

+ =

∏

+ =

which implies that w t

( )

is continuous on

[

t0,+∞

)

.

( )

(

( )

)

( )

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

(

) ( )

(

) ( )

( ) ( )

(

)

(

( )

)

( )

(

) ( )

(

)

( )

(

)

(

( )

)

0 0 0 0

0 0 0

1

1 1 1

, ,

1 1 1

0 0

d

, , d , , d

d d

1 1 , 1 , d

d

1 1 1

k k k

k i k k

l

i i

i

k k k

t t t t t t t t t

k k i k i

t t t t t t t t

w t G t w t b a t w t b B t w t C Q t w t

t

b v t b g t b v t

t

b a t b v t b b b t b v t

β δ

α γ

β

α_{ρ ξ} _{ρ ξ}

τ

ξ ρ ξ η ξ τ ζ σ ζ ω ζ

ξ ρ ξ η ξ

τ =

− − −

≤ < ≤ < ≤ <

− − −

≤ < ≤ < ≤

 ₊ ₊ ₊ ₊       =  + + + +     + + + + +

∑

∫

∏

∫

∏

( )

(

)

( )

( ) ( )

(

)

(

( )

)

( )

(

)

( )

(

( )

)

( )

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

0 0 1 1 1 , , 1 0 0 1

1 , 1 , d

d

1 , , d

d

, , d 0,

i k k k l i t k k

t t t t t t

k t t t

l

i i

i

C b Q t b v t

b v t g t v t b a t v t

t

b b t v t C Q t v t

τ δ

γ

σ ζ σ ζ

β α

δ γ

ζ σ ζ ω ζ

ξ ρ ξ η ξ

τ ζ σ ζ ω ζ

= <

− −

≤ < ≤ <

− ≤ < = + + +  _ _ = + _ _ + _+     + + _≤ 

∑

∏

∫

∏

∫

∑

∫

(5)

OALibJ | DOI:10.4236/oalib.1101168 5 December 2014 | Volume 1 | e1168 Now in inequality (7), set

( )

,

(

( )

,

)

d

( )

.

y t =w t +

∫

_αβG tξ w ρ ξt η ξ (8) Hence we have

( )

0

( ) ( )

0

( )

(

( )

)

0

( )

(

( )

)

( )

0

1

, , d 0, , .

l

i i k

i

y t b a t w t b B t w t C δQ t w t t t t t

γ

τ ζ σ ζ ω ζ

=

′ + +

∑

+

_∫

≤ ≥ ≠ (9)

For t≥t t0, =tk,k=1, 2,, since w t

( )

is continuous on

[

t0,+∞

)

and G t

(

k ,ξ

)

G t

(

k,ξ

)

+ ₌

, it is easy to verify that

( )

k

( )

k .

y t+ =y t (10) From inequality (9) and (10), it is easy to see that y t

( )

is nonincreasing on

[

t0,+∞

)

, so we have that

( )

lim_t_→∞ y t =L. Now we discuss L.

1) If we suppose that L= −∞, then lim_t_→∞ y t

( )

= −∞. From inequality (8), we can get that w t

( )

is un- bounded; consequently, there exists

{

sk:k→ ∞,sk → ∞

}

such that y s

( )

k <0, w s

( )

k =maxr∈[t₀,sk]

{

w r

( )

}

. Therefore y s

( )

k w s

( )

k G s

(

k,

)

w

(

sk,

)

d

( )

1 G s

(

k,

)

d

( )

w s

( )

k 0

β β

α ξ ρ ξ η ξ  α ξ η ξ 

= +

_∫

≥ +__

_∫

__ ≥ . This contradicts

( )

_k 0

y s < .

2) If we suppose that L≠0 is limited, then integrating inequality (9) from t0 to t, Noting that −η ξ

( )

is

nondecreasing, then we can obtain

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

( ) ( )

0 0 0 ₁ 0 0 0 0

d d d , , d .

l

t t t

i i

t t t

i

b a s w s s b B s w s s C s δQ s w s y t y t

γ

τ ζ σ ζ ω ζ

=

+

∑

+ ≤ −

∫

∫ ∫

This implies that w t

( )

→0; hence, we have y t

( )

→0. This contradicts L≠0. It follows that L=0. Since y t

( )

is nonincreasing, then y t

( )

>0.

Noting that −η ξ

( )

is nondecreasing, then we can obtain

( )

(

( )

)

( )

(

( )

)

(

( )

)

(

( )

)

( )

(

( )

)

( )

(

( ) ( )

)

( ) (

0

) ( )

, , d

, , , , , , d d

, , d 1 , d 1 ,

w t y t G t w t

y t G t y t G t w t

y t G t y t G t y t h y t

β α

β β

α α

β β

α α

ξ ρ ξ η ξ

ξ ρ ξ ρ ξ ξ ρ ρ ξ ξ η ξ η ξ

ξ ρ ξ η ξ ξ η ξ

= −

 

= − __ − __

≥ − ≥ − ≥ −

∫

( )

(

,

)

(

1 0

)

(

( )

,

)

(

1 0

) ( )

.

w σ ζt ≥ −h y σ ζt ≥ −h y t−ι

From inequality (8), we get y t

( )

<w t

( )

. Then from (9), we obtain

( )

0

( ) ( )

(

1 0

) (

)

0

( ) ( )

, d 0, 0, k.

y t′ +b a t y t +C −h y t−ι

∫

_γδQ tζ ω ζ ≤ t≥t t≠t (11) Hence, we can obtain that y t

( )

≥0 is an eventually positive solution of differential inequality (10), (11). But according to Lemma 2.2 (where ₀

( )

_{( )}

(

)

1

( )

,

, 1 ,

k k

t t t

Q tζ =

∏

_{σ ζ} _{≤ <} +b − Q tζ , dk =0) and condition (5), the differential inequality (10), (11) has no eventually positive solution. This is a contradiction. This ends the proof of the theorem.

3. Oscillation Properties of the Problem (1), (4)

The following theorem is the second main result of this article.

Theorem 3.1. Suppose that conditions (H1)-(H3) and the following conditions (12) hold

( )

(

)

(

)

( )

(

) (

) ( )

1

0 0

,

1

lim inf exp d 1 d 1 , d .

e

k

t s

k

t s

t

t t t

b a r r b s δC h Q s

ι ι γ

σ ζ

ζ ω ζ

−

− −

(6)

OALibJ | DOI:10.4236/oalib.1101168 6 December 2014 | Volume 1 | e1168 Then every solution of the problem (1), (4) oscillates in G.

Proof. Suppose that the assertion is not true and u t x

( )

, is a non-oscillatory solution of problem (1), (4) in G. Without loss of generality, we may assume that there exists a t0 >0 such that u t x

( )

, >0, u

(

ρ ξ

( )

t, ,x

)

>0,

( )

(

_i ,

)

0

u τ t x > , i=1, 2,,l, u

(

σ ζ

( )

t, ,x

)

>0, for any

( )

t x, ∈

[

t0,+∞ × Ω

)

.

For t≥t0, t≠tk, k=1, 2,, multiplying (1) by ϕ

( )

x and then integrating it with respect to x over Ω yields

( ) ( )

( )

(

( )

)

( )

(

( )

)

( )

( ) (

)

(

( )

)

( ) ( )

1

d

, d , d , , d

d

d , d

d , , , , d , d .

l

i i

i

x u t x x g t x u t x x

t

a t x u x b t x u t x x

x q t x f u t x x x F t x x

β α

δ γ

ϕ ξ η ξ ϕ ρ ξ

ϕ ϕ τ

ω ζ ϕ ζ σ ζ ϕ

Ω Ω

Ω ₌ Ω

Ω Ω

 ₊ 

 

 

= ∆ + ∆

− +

∫

∑

∫

By Green’s formula and the boundary condition, we have

( )

x u xd u

( )

x dx u sd u ds c u sd cu ds 0.

n n

ϕ

ϕ ϕ ϕ ϕ ϕ

Ω Ω ∂Ω ∂Ω ∂Ω ∂Ω

∂ ∂

∆ − ∆ = − = − ⋅ + ⋅ =

∂ ∂

∫

It follows that

( )

x u xd u

( )

x dx b0 u

( )

x d ,x

ϕ ϕ ϕ

Ω ∆ = Ω ∆ = − Ω

∫

( )

x u

(

i

( )

t ,x

)

dx u

(

i

( )

t ,x

)

( )

x dx b0 u

(

i

( )

t ,x

)

( )

x d .x

ϕ τ τ ϕ τ ϕ

Ω ∆ = Ω ∆ = − Ω

∫

The rest of the proof is similar to the one in Theorem 2.1, so we omit it.

4. Remarks and Examples

Remarks. From the theoretical viewpoint, the results of this paper, uncovered the essential difference between partial differential equations with impulses, functional arguments and partial differential equations without im-pulses, functional arguments; from a practical standpoint, they are very convenient because these criteria only depend on the coefficients of the equations, impulsive term and the time-delays. The results of this article im-prove the results in the papers [17]-[19]. For example, paper [19] discussed the case with distributed deviating arguments; however, we consider a more complex case with continuous distributed deviating arguments.

The following are examples to illustrate the applicability of the conditions.

Example 4.1. Consider the equation

( )

(

)

(

)

( )

2

π 0

3π

,

2 2 2 2

π 2

e 3π

, , d

4 2

π π

e , , 1 e , e d

2 2

cos sin , 1, 2 , , 0,π ,

t

u t x

t t

k

u t x u t x

t

u u u t x u t x x u t x

x t t t t x G

ξ

ζ ζ

ξ ξ

ζ ζ

− −

+

 

∂ ₊  _{+ −}  ₋

   

∂    

   

= ∆ + _ − _∆ _ − _− + −

   

+ > ≠ ∈ × =

∫



( )

(

)

(

( )

)

1

( )

2 , 2 , 2 , , 1, 2, ,

2

k k k

k

u + x −u − x = u x k= 

and the boundary condition

( )

0,

( )

π, 0, .

u t =u t = t∈+

Here N=1 , Ω =

( )

0,π ,

( )

, e 4 t g t

ξ

ξ = − − ,

( )

, 3π

2

t t

ρ ξ = + −ξ , η ξ

( )

= −ξ , a t

( )

=1 , 1

( )

e

t b t = ,

( )

2

1

h u =u , 2k k

t = , 1

( )

π 2

t t

τ = − ,

(

, ,

)

(

2 1 e

)

t

q t xζ = x + −ζ ,

( )

eu2

f u =u , σ

( )

t,ζ = −t ζ ,

( )

, cos sin

(7)

Example 4.2. Consider the equation

( )

(

) ( )

(

)

(

)

( )

2

5π 2 3π 2

2π ,

2 2 2

π

e

, , d

4

π π

e , , 1 e , e d

2 2

cos sin , 1, 2 , , 0,π ,

t

u t x

t t

k

u t x u t x

t

u u u t x u t x x u t x

x t t t t x G

ξ

ζ ζ

ξ ξ

ζ ζ

− −

+

 

∂

+ − −

 

∂  

   

= ∆ + _ − _∆ _ − _− + −

   

+ > ≠ ∈ × =

∫



( )

(

)

(

( )

)

1

( )

2 , 2 , 2 , , 1, 2, ,

2

k k k

k

u + x −u − x = u x k= 

with the boundary condition

( ) ( )

0, 0, 0,

( ) ( )

π, π, 0, .

u u

t u t t u t t

n n

+

∂ ∂

+ = + = ∈

∂ ∂ 

Here N=1, Ω =

( )

0,π ,

( )

, e 4 t g t

ξ

ξ = − − , ρ ξ

( )

t, = −t ξ , η ξ

( )

= −ξ , a t

( )

=1, 1

( )

e

t

b t = , h u1

( )

=u2,

2k k

t = , 1

( )

2

t t

τ = −π, q t x

(

, ,ζ

)

=

(

x2+1 e

)

t−ζ , f u

( )

=ueu2, σ

( )

t,ζ = −t ζ , F t x

( )

, =cos sinx t, c=1. It is easy to verify that the conditions (H1)-(H3) and the condition of Theorem 3.1 are satisfied. Hence all solu-tions of the above problem oscillate.

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