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Applied Mathematics, 2012, 3, 1921-1932 http://dx.doi.org/10.4236/am.2012.312263 Published Online December 2012 (http://www.SciRP.org/journal/am)
Blow-Up and Attractor of Solution for Problems of Nonlinear Schrodinger Equations
Ning Chen*, Jiqian Chen School of Science, Southwest University of Science and Technology, Mianyang, China
Email: *[email protected]
Received September 18, 2012; revised October 18, 2012; accepted October 26, 2012
ABSTRACT
In this paper, the authors study the blow-up of solution for a class of nonlinear Schrodinger equation for some initial boundary problem. On the other hand, the authors give out some analyses and that new conclusion by Eigen-function method. In last section, the authors check the nonlinear parameter for light rule power by using of parameter method to get ground state and excite state correspond case, and discuss the global attractor of some fraction order case, and com- bine numerical test. To illustrate this physics meaning in dimension d = 1, 2 case. So, by numerable solution to give out these wave expression. Keywords: Nonlinear Schrodinger Equation; Eigen-Function Method; Fractional Order; Blow-Up; Glabal Attractor
1. Introduction
The quantum mechanics theory and application in more field in nature science. The non-linear Schrodinger equa- tion is the basic equation in nonlinear science and widely applied in natural science such as the physics, chemistry, biology, communication and nonlinear optics etc. (See [1-9]) We study this equation to extend them are with important meaning (See[10-12]).
As we all know, the nonlinear Schrodinger equation be description quantum state of microcosmic grain by wave, it is variable for dependent time, and that is most essen- tial equation, which position and action similarly Newton equation in position and action classics mechanics, it is apply to field as optics, plasma physics, laser gather, co- hesions etc, particular on that action of power and trap, search analytical solution for Schrodinger equation is also difficult, and more so difficult for complicated power.
Now, we may extend some results in [4] by using Ei- gen-function method in through paper.
As we all know the solution of initial problem for Schrodinger equation bellow
2i , , ,
,0 ,
nt
n
u a u f x t x R t
u x x x R
0, (1.1)
Assume that real part and imaginary part of
, , ,x f x t
are real analytical function for ,nx R then this solu-tion of the problem may expresses in form:
2
0
0
i,
!
, d .
k
k k
k
t k kx
au x t t x
k
t f x
2. Several Theorems
In this section, we consider the blow-up of solutions to the mixed problems for higher-order nonlinear Schrod- inger equation with as bellow.
It is well known the higher order equation:
,i , 0,ktu u f u t x
where
2
22
1
, ,n
i ix
, etc.
that with new results for higher-order case. Now, we consider the blow-up of solutions to the mixed problems for six-order general Schrodinger equation to extend some results [4] that as bellow form:
3
2
i , 0, ,
0, , ,
, ,
tu u f u t x
u x x x
t x u t x u t
0, 0.
(2.1)
Assume that
4 1 60, , ,k
f f c x H x*Corresponding author.
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N. CHEN, J. Q. CHEN 1922
not identical zero. Where f holds complex value function with self-
x. variable for comple x is also complex value
Theorem 2.1. Suppose that nonlinear term
2 2 10, , nt x R R
f of pro- blem (2.1) satisfy , 0C
4 1 60, , ,k
f c x H f
and x) must be
not identical zero then the classical solution of (2.1 for blow-up in finite time in
Proof. Let
1 60, ; .C T H
2d d ,J t
u x uu x
(2.2)
Then
3 3
d
i d
t i
d ,
u u uu x
u u u u x u f u uf u x
(2.3)
By the first Green’s formula, we have
J t
2
1
3 2 2
1
d
d d d
i
i i
xi
n
x xi
u x
u u x u u s u u x
Substituting it into (2.3), then
3 2d di
n
xu u x u u s u
3 3
d
d ,
u uJ t i u
u s
u f u uf u x
We may assume , then we have , 0t x 2u
u u u u u u
2
0.u
Obviously, from 2 ,u f u uf u u f u and 4 1
.k
f u c u Therefore, we have
4d
kd 2J t u x
.
By Schwartz inequality:
u f u uf u x c
22d 1du x x
4
1d , here d 0.u x c x
So,
24 2
1d 1 du x c u x
.
Inductively, we have
6231 dc u x
etc., 12
d 1u x
24 2
1d 1 d 0. k k
ku x c u x
Then J t increasing function similar in [4 from ]
24 2
1 12 d 2 d
2
12 0,
kk k
kk
J t
c c J t
and then there exists
c c u x c c u x
such that 0
lim ,t T
J t 0 ,T
that is
0
2lim d .x
t Tu
So, we complete the proof of this Theorem 2.1. (As positive integer 1,k we get i 3.1 in
[4
r equ
t is theorem])
onsideRemark. Then we c r that important case is al-ways for the Schrodinge ation may as bellow form
220 : itc u u c u u .
Now, we shall consider also in this similar case:
23
2
i ,tu u c u u
0, , ,
, , 0, 0.
u x x x
t x u t x u t
(2.4)
Therefore, we shall obtain the following theorem. Theorem 2.2. Suppose that non-linear term f of
problem (2.1) satisfy 0c ,
4 10, ,
kf f c
and
6 2 2 1, 0, 0, , ,nx H x t x R R
e classical solution of (2.4) must be for blow-up in finite time in then th
1 60, ;C T H (as positive 1,k then it is theorem 3.2 in [4]).
Proof. Since 20f u ,c u u c satisfies
2 40f c k
c ,
then 4 1.
kf c
Thus, from theorem 2.1, we complete the proof of m 2.2. theore
Now, we shall give out the following theorem form. Here, we shall consider the problem:
23i , 0, ,tu u F u u t x
2
0, , ,
, , 0, 0.
u x x x
t x u t x u t
(2.5)
Theorem 2.3 Suppose that non-linear term of pro- f
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N. CHEN, J. Q. CHEN 1923
blem (2.5) satisfy
0c , 4 3,
kF c x 0,
and
, 6 2 2 1, 0, , nx H t x R R
th bl
(As positive i then it is theorem 3.2 in [4])
oof. Since
en the classical solution of (2.5) must be for ow-up in finite time in 1C 60, ;T H .
nteger 1,k
Pr 2, 0f u F u u c , we have that
2 4 1,
kf F c
and
2 4k0.f c c
eorem 2.1, we complettheorem 2.3. (As it is theorem 3.3 in [4])
Now, we may consider the following problem:
Thus, from th e the proof of 1,k
23i e 1 , 0, ,
0, , ,
p stu u k u t x
u x x x
(2.6)
whTheorem 2.4. that and
then the solution of (2. w-up me in
2, , 0, 0.t x u t x u t
ere constant 0, 0.k p Assume 6 ,x H
6) must be for blo 0,x in finite ti 1 60, ;C T H .
Proof. From
2 22 21e 1
2!p u p u p u ,
then 2
e 1p uf u K u satisfy 0,f and
4kf
1.
It holds the condition of theorem 2.1, then by theorem 2.1 that we know the solution of problem (2. ) must be blow-up in finite time. Therefore, we complete the proof
orem 2.4.
3. Main Results
W
ermore, we will consider eight-order nonlinear Schrodinger equation. In first, stating that lemma 3.1.
Lemma 3.1. This Eigen-value problem (see [4])
kp
6
of the
e consider the initial boundary value of some higher order nonlinear Schrodinger equation. By using of eigen- function method, we can get new results bellow.
Let
,t xDu u D u 1 2, , , ,
nt x x xu u u u
1
1 2
, , , , 1, 2, , ,
, , , .
i i x ii
n
tx x x x xx
x x x x
Du u u u i n
D Du Du Du Du
Furth
0, ,
0.
x
(*)
As we all know the first Eigen valu1e 1 0 of (*), the corresponding Eigen-function 1 0,x assume it with
1 d 1x x
.
Let h co
be bounded closed domaismoot nditions of function
n in nR and by suite f and g that from
G rechrodinge
ight-order case
reen’s second formula, we easy get follow sults. Now, we consider nonlinear S r equation
with e
ing
0 1, 1, 2i i :
2
4 31 2
, ,x xf u D u D u
, 0, ,
4 31 2itu u u u
g u g u (3.1)
g u t x
0 ,,0 ,u x u x x (3.2)
0, , 0.u x t (3.3)
Clearly, 1 2 0, 3.1. Assume
that is theorem 2.1 in [5]. Theorem that problem (3.1)-(3.3) satisfy
(where n out normal direction):
3 2
1 2 1 Im ;u
3 21
0 0,
Re
GI G
n
G g u g u g u
2 Re
2 41 1
32
0
Re , , ,
Im Re ,
Re d , 0.
x xII A f u v D u D u g u
g u u CF u
u x A
F u be continuous, convex and even function, here
1, 0;
1, 0C
0 , and dIII F s s s F s
.
Then up in finite time.
Proof. (I) step, when
the classical solution of (4.1)-(4.3) must be blow-
0, 0A and 0.A In the similar way by [5] from that (4.1) first we take the real part of both sides for (4.1), we get that
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N. CHEN, J. Q. CHEN 1924
4 31 2
4
Re Re ,
2 1
2
4 3
Re Im
Re , ,
Re .
t
x x
4 31 2
3
2 1
Re Rei
Im Im
tu u u u
f g u g u g u
u u
f u D u D u
g u g u g u
.4) u u
(3
Multiplying by x the both sides of (3.4) and in-tegral on fo r x , it is form:
4 31 2
2 4 31 2
Re d Im
Re , , Re d
t
x x
u x u
f u D u D u g u x
Taking then Re d ,a t u x
Re d ,a t u x t
and that
4 31 2
2
4 31 2
2
3 21 2
3 21 2
Im
Re , ,
Re d
d
Re , , d
1 Im d
1 Re d
x x
x x
a t u
f u D u D u
g u x
G G x
f u D u D u x
u x
g u x
(3.5)
By 0G n
in (I) and Green’s semula:
, (3.6)
Substituting (3.6) into (3.5), we get
cond for-
d 0G G x
2
3 21 2
3 21 2
22
3 21 2
Re , ,
Re d
Im
Re d
x xa t f u D u D u
g u x
Imu
2 31Re , ,x xf u D u D u
u
g u x
Hence,
(3.7)
2
4 31 2
4 31 2
Re , ,
Im d
x xa t f u D u D u
g u
u x
From A 0,
2 4 31 2
4 31 2
Re , ,
Im Re
x x
0.
A f u D u D u g u
u CF u
Therefore, we have
2 4 31 2
4 31 2
, ,
Im Re
xRe xf u D D u g u
u CF u
u (3.8)
Combing (3.7)-(3.8), and Jensen’s inequality, we ob- tain
Rea t F u Re dF u x F a t
(3.9)
Here, d d .F a t a t So, d ,
a tt a F a
there exist ,dT a F a
, such that
lim .t T
a t
(3.10)
From Re d ,a t u x
and Holder inequality, we
get 1 1 ,q 1p
Re d Re ,a uq pL Lt u x
that is
1
Re .q pL La t u
Therefore,
1
lim lim Re .q pL Lt T t Ta t u
Hence,
lim Re , 1 .pLt Tu p
(II) step, whe 0, 0,A taking that n
1, ,u x t u x t ,then 1Re Re .u u
, letTherefore 1Re d ,u x
a t we have
0, 1 1 1, ,a t a t a t a t
Combine (4.1)- 0, 1C , (4.8) and we obtain that
A
1 1Re da t F u x
(3.11)
That is also 1 1Re da t F u x
. From Jensen
inequality and F s is even function, we have
1 1 1d d ,F a F a a t
then
1 1dt (3.12) da F a
From (3.12) and similar (I)-step, we can get
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N. CHEN, J. Q. CHEN 1925
lim Re lim Re ,
1 .
p
p p
L
L Lt T t Tu t u t
p
Combine (I)-(II) we comp3.
1lim Re , 1 .t T
u p
lete the proof of theorem 1. Clearly, 0,1 2 that is theorem 2.1 in [5]. Theorem 3.2. that problem (3.1)-(3.3) satisfy: Assume
3 21 2
3 2
1 ReG g u
1 2
0 0,
1 Im ;
GI G
n
u
2 3 21 2Im , ,x x
3 21 2
Im
Re,
Im
f u D u D u g uII B
F u
u
F u
and
0 ,1 0, Im d 0B u
x
where F s is continuous, convex and even function;
0, and d .III F s s s F s
the classical solution for this problem (3.1)-(3.3) in finite time.
Proof. From
Thenis blow-up
1 0,B we discuss two case:
21 0, 0, , i , ,I B u x t u x t
then 2Im Re .u u ing the imagTak inary part for both sides of (3.1),
similar the method of proof for Theorem 3.1, we can easy have
2lim R , 1 .t T
e pLu p
So, we get that
lim Im , 1 .pLt T
u t p
(II) 1 0, 0,B we may let , , ,u x t u x t then
3
So, 3Im Im .u u
3lim Im , 1 .pLt Tu t p
ary part for both sides of (1), by (II) and similar the method of proof for theorem 3.1, we can easy have
Taking the imagin
, 1pL
We
3lim Re .t T
u p
get that
lim Im , 1 . pL
u t p
3.3. Clearly
t T
Combine (I)-(II), we complete the proof of theorem 3.2.
Corollary 1 2 0, g it for some
that is theorem 2.2 in [5]. By ([13] )lookin applications.
4. Some Higher-Order Case
In the same way, we can consider the higher-order case (integer 0k ):
4 4 1
, 2 , 0g u g u t
1 2
4 1
i
,
k kt
k
k
u u u u
x
2 41, ,x xf u D u D u g u
(4.1)
0,0 ,u x u x x (4.2)
0, , 0.u x t (4.3)
Clearly, 1,k that is problem of eight order case. Theorem 4.1. ssume that problem (4.1)A -(4.3) satisfy
Im ;u
4 4 12 1 2
4 4
0 0, ,
and Re
k kk
k k
GI G P
n
G P g u P
2Im , Re4 4,
Im
x x k k,
II B
f u D u P g u P u D u
and
F u
01 0, Im d 0,B u
x
F s is continuous, convex and even function; where
0,III F s s and d .s F s
Then the classical solution for this problem (4.1)-(4.3) w-up in finite time.(omit this similar proof )
Remark 4.2. Assume that (here ) is blo
4 3k
4 4 1 4 2 4k k k kkP 3
then we will obtain similar results of theorem 3.2 with more case.
s in an al
4 1 2 3 4 ,
Remark 4.3. (See [6,14]) According to the direction of [6], we may consider that coupled nonlinear Schrod- inger equation as in the following iterative formula
gorithmic form by VIM:
2 2
0i i d ,
t xx
t
n n n n nA A A B A
1 , ,n nA x t A x t
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N. CHEN, J. Q. CHEN 1926
1
2 2
0i i d .
t xx
n n
t
n n n n nB B A B B
, ,x t B x tB
The solution procedure with initial approximations (omit the details ):
0
0 0
,0 1 cos ,
,0 1 cos .
A A x f x a x
B B x g x b x
The other components can be obtained directly:
1 1, , , , , etc.A x t B x t
Furthermore, the conserved quantities:
2 2
2, d
s
sE A A x t x
and
2 2
2, d
sE B B x t x
,
s
where 2π .s This numerical results is with higher accuracy.
5. The Global Attractor of the Fractional NSE
ntly, they also showed that dynamic behavior of large time action to investigate for [15,16], they are deep- going study global attractor and dimen on estimate of
rder nonor sea
linear Schrodinger equation in [17]. The au- of
[17] an sed on [16-18], and combine [19] obtained the
Rece
siinteger o -linear Schrodinger equation in [16].
The auth rch the Cauchy problem for fractional order non-thor search the global attractor problem for a classfractional order non-linear Schrodinger equation in
d we bacondition of existence of solution for following fractional order non-linear Schrodinger equation:
i ip
tu u F u u u
ma, express the field of electricity [20], where
is with standard perpendicular base, i is imthe function
0
, , 0,
,0 , ,
, , , , 0.i
f x x t
u x u x x
u x Le t u x t x t
(5.1)
Physics background of (1) is arise the main part of nonlinear interaction for laser and plas u
0, , 0, ,0,1,0, ,0 , 1, 2, ,n
il e i n
aginary unit, F is with one order derivative
2 , 0,n p 0, where 0 with some con- sume effect, and as 0 express the integral system with soliton solution.
As 3, , 1c for (3.1), and (4. ) thirdly
section case, we will obtain global attracto
1
r of initial value problem (5.1) that first give out Lemma as llows.
Lemma 5.1. Let fo
2 20 , , ,u x L f x L u x t
(5.1), an
is the solution of problem d
2 2 22
0, e .tu x t u f x (5.2) x
Proof. Multiply u for the both sides of (**) act as inner product, we have
,
i ,tu u ,
, i ,p
u u
F u u u u u
(**)
and take real part,
f x u
2 212 d du t u
22 11
Im ,
2 2 ,
f x u f x u
u f x
(5.3)
From (5.3) and by use of Gronwall inequality, we ob-tain
2,u x t
2 22
0 e ( ) .tu x f x
Lemma 5.2. Let
,u x t 20 , ,pu x H L f x H
2
2, pL
u u
is the solution of problem (1), then
woof. To establish inner product for both sides of eq-
uation (5.1) with for
ith uniform bounded. Pr
tu , and take real part, we have that
2212 u
d Im ,
Re , ,
p
t tt
t
F u u u x u u
f x u
easy get that by (5.1),
22
Re , ,t
Im ,
d
t
p
u u u
F u u u x
f x u
where
,
Re , ,
pRe , Imtf x u f x u F u u
f x u
by use of Jensen’s inequality, we have
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N. CHEN, J. Q. CHEN 1927
12
1
2211
22
1
2 2
2 21 1
,
21 2 1
11 2
1
2 d
d
,
.2
d
2 d 2 d
2
2
P
P
p p
tt
p p
p
P
t tLL L t
t t
p
t
P
L
t
u F u u u u u x
u F u u u u u x
f x u F u u
f x u
f x F u u u
u x
F u u x u x
F u
u
Im
Im ,p p
F u u u F u u
1
2 2 2 12, 2 .P
P
t t Ltu u F u
So,
221
221
2 d
2 2 d
p
tt
p
u F u u u x
u F u u u x
C
by use of Gronwall inequality, we obtain
2, d
p
tu F u u u
x
uniform boundary. Lemma 5.3. Let
,
is the solution of problem (5.1), then
2 20 , ,u x H f x L u x t
u with
uniform bounded. Proof. To derivative both sides of Equation (5.1) for and take inner product fort tu , and taking also imagi-
nary part, we have
2 212 Im ,p
t tt
u F u u u 0t u
Then
2 21
21
2 t tt
P
t L
u u
By Lemma 5.2 and Young inequality, the (5.4) with form
2 2 12
,
2
tL
P
tt
u u
u u F u
1 2
Im ,
2
p
tF u u u
F
2 2 22t t t
tu u u
C
by use of Gronwall inequality, we obtain
22 10 ,0 e ,t
tu u x C
Because hold these inequality bellow
2
0 0 0 0
0
0
1 2
,0
,
,
p
t
H
t
P
t L
u x C u F u u u
C u
u F u u u
pu u F u u u
Hence ,tu u are uniform boundary. Similar
method of [19,20], we give out that condition of yield global attractor of problem (5.1).
Theorem 5.4. Assume hat
2 20 , ,
2 , 0, 0,
u x H f x L
n p
then the periodic global attractor of initial value problem (4.1-4.3):
00
,s t s
A S t B
here t S or semi-group with needing de-e in prov with the bounded attractor set in
w for operatfin e an for
using of similar proof m
Remark 5.5. Furthermore, we shall study global at-tractor of fraction order non-linear Schrodinger type equation, and the estimate for its dimensions, and that
6. Some Notes for Shake Power and Light Power
omic transition, is one of the
d 0Bfollowing in prove processes.
Proof. We omit the proof (byethod in [18,19]).
blowing-up of solution for some fraction order non-linear Schrodinger type equation.
We consider some meaning of physic and Energy for nonlinear Schrodinger equation.The numerical test for solution of nonlinear Schrodinger equation with ground state and excite state.
Atoms absorb energy from the ground state transition to the excited state, learned through experiments in ex- treme case, the ground state solution is not controlled so- lution-Blow-up solution.
Thus, strictly control the number and perturbation for impulsive velocity of the at
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ict co e atomic transition to the first, second and
212 V main methods to produce new material structure. Strntrol of th
ingly.
(6.3)
third excited state is more practical significance, espe- cially the transition to the first excited state. As we all now, the ultra-low temperatures, the atomic gas in the magnetic potential well Boer-Einstein condensation ex- pe
So, by check parameter method in [23] we check nonlinear parameter for light rule power, then we get ground state and excite state correspond
riments [21], promotion of scholars study the macro- scopic quantum behavior of atoms and kinetic character- istics.
By using of above stating method we consider calcu- lat
6.1. One Dimension Case (d = 1)
( )a We consider two class powers (shake power and light power) in (6.3), Setting shake power
e to the ground state solution and excite state of d- dimension BECS (Bose-Einstein condensate) with mix harmonic potential and crystal lattice potential.
The Gross-Pitaevskii equation:
212, 250, 2V x x b b
taking initial wave
21
20
2e
π
b
,x
x (6.4)
2
220
,i
, ,2
r th
t
hV r NU r t r t
m
(6.1)
where , 1,2,3. 0,dr R d t m expresses mass of at
to calculate ground state .g For (6.4) we calculate first arouse state 1, space field for the time step for
10 10,x 0.2.t
( )b Similar above way, taking
2 212 25sin π 4 , 500, 2,V x x x b b oms, h be planck constant, N be number of atoms
in cohesion system, V r be outer power,
20 4πU h a m describe interaction between the at-
oms cohesion ( 0,sa means repel; 0,sa shows at appropriate im
and (6.3) for
21
20
2: e
π
b
,x
g x x tract each other). Thus, by pass
e (6.1) may be measur-
able process, then th written:
22, 1i
2
r tV r r t r t
t
, ,
(6.2)
The parameter
and 10 10,x and 0.2.t On the other hand, by the MATLAB search the solu-
tion of Equation (6.3) in case (1) and (2) as follow with
0 1 2 3, , , (See Figures 1 and 2). tract corres
for positive, or negative, describe that repel or at ponding, out power V r be
by phys stem for us to sa c
defined tudy things. By using of the im ginary time method to alculate it in [22] that let it
ic sy 6.2. Two-Dimension Case (d = 2)
Consider shake power in [14,24] substitutin nto (6.2), we have g it i
-15 -10 -5 0 5 10 15-150
-100
-50
0
50
100
150
200
250
300
Phi0
Phi1Phi2
Phi3
V=x2/2
V(x
),P
hi
X
Figure 1. Ground state phi0. First excited state phi1. V = x2/2; b = 500, bi = 2.
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N. CHEN, J. Q. CHEN 1929
0,0 ,0 ,
,, other.
x a yV x y
b
The grain energy:
1 2
2 22 21 1
1 22 2
π, , 1,2,3, ; 5, 2.
2n n
n nhE n n a
m a b
b
We take initial wave function for
1 2
1 20 1
π π4, sin sin ,
n x n yx y n n
ab a b
2 1.
To calculate ground state g ; For
1 2
1 210
π π4, sin sin
n x n yx y x
ab a b
1 2
1 201
π π4, sin sin
n x n yx y y
ab a b
,
and
1 2
1 211 ,
π π4, sin sin
n x n yx y xy
ab a b
1 2where 2n n .
10 ,x y By calculating along the direction of axe ,x and 10 ,x y
cited of in direction of axe y, and calculating
first ex 11 ,x y alld for
, and space fie
ong direction for axe x and y 2, time step: axe 0 5,0x y
0.05, 0.02.x yt t
,
Combine these cases as Fig: (See Figures 3(a) and (b), Figures 4-6)
-15 -10 -5 0 5 10 15-150
-100
-50
0
300
50
100
150
200
250
X
Phi0
Phi1
Phi2Phi3
Phi4
V(x
),P
hi
2Figure 2. Ground state phi0. First excited state phi1. V = x /2 + 25*(sin(pi*x/4))2; b = 500, bi = 2.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Phi0
Phi1Phi2
Phi3
Phi
X (a)
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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Y
Phi
Phi0Phi1Phi2Phi3
(b)
Figure 3. (a) Ground state phi0. First excited state phi1. a = 5, b = 2. (b) Ground state phi0. First excited state phi1. a = 5, b = 2.
01
23
45
0
0.5
1
1.5
20
0.5
1
X
Y
Phi
0
Figure 4. Ground state phi0 a = 5, b = 2.
01
23
45
00.5
1
1.52
-3
-2
-1
0
1
2
3
X
Y
Phi
Figure 5. First excited state phi1-x a = 5, b = 2.
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01
23
45
0
0.5
1
1.5
2-2
-1
0
1
2
X
Y
Phi
Figure 6. First excited state phi2-y a = 5, b = 2.
01
23
45
00.5
1
1.52
-1.5
-1
-0.5
0
0.5
1
1.5
X
Y
Phi
Figure 7. First excited state phi3-xy a = 5, b = 2.
We consider three-dimension case, Figure 4 for ground
state ( )0 ,x yϕ corresponding case, the ( )10 ,x yϕ as with express along direction of axe x (wave surface) in Figure 5, the ( )01 ,x yϕ as with express along direction of axe y (wave surface) in Figure 6, the ( )11 ,x yϕ as for express along direction of axe x and axe y (wave sur- face) in Figure 7.
7. Concluding Remarks Recently, the higher-order Schrodinger differential equa- tions is also a very interesting topic, and that application of some physics and mechanics of for some more fields as nonlinear Schrodinger equations and some compute methods etc. In our future work, we may obtain some better results.
The application of some physics and mechanics of for some more fields with some combine equations (look [7,
13]).
8. Acknowledgements This work is supported by the Nature Science Foundation (No.11ZB192) of Sichuan Education Bureau (No.11zd 1007 of Southwest University of Science and Technol- ogy).
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