Stationary solutions for the Cahn - Hilliard equation coupled with Neumann boundary conditions Текст научной статьи по специальности «Математика»

MSC 35K70

DOI: 10.14529/ mm p 160206

STATIONARY SOLUTIONS FOR THE CAHN - HILLIARD EQUATION COUPLED WITH NEUMANN BOUNDARY CONDITIONS

I.B. Krasnyuk, Institute of Applied Mathematics and Mechanics of the NASU, Donetsk, Ukraine, krasnjukigr@rambler.ru,

R.M. Taranets, University of California, Los Angeles, United States of America, taranets_r@yahoo.com,

M. Chugunova, Claremont Graduate University, Claremont, United States of America, marina.chugunova@cgu.edu

The structure of stationary states of the one-dimensional Cahn - Hilliard equation coupled with the Neumann boundary conditions has been studied. Here the free energy is given by a fourth order polynomial. The bifurcation diagram for existence and uniqueness of monotone solutions for this problem has been constructed. Namely, we find the length of the interval on which the solution monotonically increases or decreases and has one zero for some fixed values of physical parameters. Under the non-uniqueness we understand a possibility of existence of more than one monotone solutions for the same values of physical parameters.

Keywords: the Cahn - Hilliard equation; Neumann boundary conditions; steady states.

Introduction

We study the steady state solutions for the Cahn - Hilliard equation (see e.g. [2,4,7,8]) ut = (—au + Yu3 — uxx)xx, 0 < x < l, t > 0, (1)

coupled with the Neumann boundary conditions

UUx — Uxxx — 0, x — 0,/, ^ 2 ^

where a and 7 are some physical parameters. The authors of [2] have considered a structure of stationary solutions of problem (1) - (2) as a function of length l and mass m — J0 u(s)ds. They have counted the total number of monotone solutions, depending on the parameter values of the problem. In [2] phase portraits of the problem have been constructed which describe the monotone solutions for a special stored-energy function f (u) + §2 (u')21 where e is a small parameter. In this case, the function f'(u) has one negative minimum, one positive maximum, and f (±ro) — In this paper, we study the following problem

u" — -au + yu3 - a, u'(0) — u'(l) — 0, (3)

where a reflects influence of the average mass on qualitative behaviour of solutions; a is the main parameter of the problem because it is related to an interaction energy of the phase decomposition in a binary alloy; 7 is a parameter corresponding to thermodynamically stability of the system. In particular, a — fT (1 — T)' wkere Tc is the critical temperature

(¡() Bulletin of the South Ural State University. Ser. Mathematical Modelling, Programming

& Computer Software (Bulletin SUSU MMCS), 2016, vol. 9, no. 2, pp. 60-74

of the phase transition for order-disorder in a disordered binary alloy, which can be measured during the cooling process of the alloy; Eint is the interaction energy between atoms of sorts A and B in the binary alloy. The energy functional corresponding to problem (3) can be written in the form

i

E(u(t)) = J + f («}) dx, E(u(t)) = E(u(0)),

0

where the free energy f (u) is

f (u) = Yui — —u2 — au + C. J \ J 4 2

Note that, in the case a > 0, j > 0, by the re-scaling u = ^J~Yu(y/a,x), problem (3) can be reduced to

W = -u + u3 - a, u'(0) = u'(L) = 0, (4)

where L = y/al and a = This case was studied in [2,7]. To the best of authors'

knowledge all other cases were not considered in a literature. Therefore, in this article we investigate the existence of nontrivial solutions of problem (3) for all possible values of these parameters. In particular, we study an interesting case when y < 0- It should be

mentioned that the case y < 0, a < 0 and |a| < a0 ■= \Ja leads to a non-uniqueness result (i.e. there exist two solutions with the same initial energy), depending on initial energy, but the solution of the problem is unique when |a| ^ a0. Also we describe all possible dynamical scenarios for the parameter values.

Moreover, our results can be applied to the Izing model. For example, the Izing model free energy in the vicinity of the phase transition may be written as the following (see, e.g. [6])

f (u) = a + ru2 + su4 + O(u6). (5)

In order for the system to be thermodynamically stable, the parameter on the highest even

s>0

the free energy is bounded. However, for the Neumann boundary value problem we can consider the case when the "open" system is thermodynamically unstable that corresponds to s < 0. For example, in the case s < 0 r > 0 and a e [0, |), a solution is not unique. On the phase plane uOu', this phenomenon is illustrated by two loops with identical lengths

Ou

smooth nontrivial solutions.

1. The Steady State Solutions for the Cahn - Hilliard Equation

L

all possible values of the physical parameters. We proceed by examining all qualitatively different nine cases. Note that since all systems at hands are conservative they cannot have any attracting fixed points. Therefore their phase portraits can have only saddles and/or centers.

,VV'

TTT ,

tun

VVWV

uuhvvvv iiiliUVVVV

MliiiX

IHUJ* 11114

,VV1 Vis

iii

'ITIflfim-K* -<-S V

Fig 1. Phase diagram for the case a > 0, y > 0 with a = 0,1 on the left and a = 1 on the right

1.1. Case a > 0, y > 0

In this case, by the re-scaling u = y^^^/ax), problem (3) can be reduced to

u

-U + U3 - a, U'(0) = U'(L) = 0,

(6)

where L = y/al and a = a\FY- Integrating (6), we find that

U4 U2

(a')2

— - T + y + a U = P —

Note that the equation U3 — U — a = 0 has • three real roots

(U')2 = 1(U4 — 2 U2 — 4aU + 4p).

, 2 /( 2k^, , , 1 3v^3a

Uk = 3 cos(0 + k = 0,1, 2, 0 =3 arccos(——);

provided that a2 < 27 (two minimums and one maximum). Thus, we have two nontrivial

solutions if — + —f^ + aUmax < p < — —4^ + + aUmin and the lengths of intervals, where the corresponding solution has only one zero, are

U2

Lq,I = V2

dt

U4

U1

^/t4 — 2t2 — 4 at + 4p'

L

0,2

V2

dt

U3

^t4 — 2t2 — 4a t + 4p

(7)

where Uk are four real roots of the equation t4 — 2t2 — 4 at + 4p = 0 such that Ui < u2 <

u4 u?

u3 < u4. We have only one nontrivial solution if p < —mfx + + aUmax with the length of interval, where the solution has only one zero, given by

U2

L

0,3

V2

dt

U1

^Jt4 — 2t2 — 4 at + 4p'

(8)

3-

3-

2-

2-

1-

1 -

«' 0-

«' 0-

0

2

3

ttt 11 ttttr ttttr ttttr ttttt ttttt ttttt ttttt ttttt ttttt ttttt ttttt ttttt ttttt

ПШ ttttt ttttt ttttt ttttt

VVHH ^ v V I V 4 4

V V 4 444

V44444 >44444

44444 44444 44444

iiiii 44444 44444

iiiiii , /4 4444 _ W//444 44

-S v•'✓Z 4 4 4 4 4

»■»NSW V 4444 NVVVV444 VV4444 VV4444 44444 44444 44444 44444 44444 44444

ttttt 44444 ,44444

Siittt

/44444 44444 //44444 ««»,<>//44444

Fig 2. Phase diagram for the case a > 0, 7 < 0 with a the right

0,1 от the left and a = 1 on

where uk are two real roots of the equation t4 — 2t2 — 4 at + 4p = 0 such that u1 < u2; • one real root

«min = (2 + \f

4 27/

^ 1 \i/3 (a [a2 1 )1/3

---^ П2 —

4 27

provided that a2 > 27 (one minimum). Thus, we have a nontrivial solution if p < —^ +

u2

+ aumin with the length of interval, where the solution has only one zero, given by

U2

Lo = V2

dt

U1

л/t'4 — 2t2 — 4at + 4p'

(9)

where uk are two real roots of the equation t4 — 2t2 — 4 at + 4p = 0 such th at ui < u2;

two real roots

a

u„

2(2>

1/3

ufl

—(-)1/3 v 2

provided that a2 = 247 (one minimum)- Thus, we have a nontriv ial solution if p < — + = 2(|)2/3 with the length of interval, where the solution has only one zero,

mm + a ti 2 \ и iim

given by

Lo = V2

dt

U1

л/t4 — 2t2 — 4at + 4p'

(10)

where uk are the two real roots of the equation t4 — 2t2 + 4 a t — 4p = 0 such th at u < u2.

Thus, if 0 < L < L0 then the corresponding solution of the problem has no zeros located on the interval (0, L). If kL0 < L < (k + 1)L0, where k =1, 2,..., then the solution has exactly k zeros located on the interval (0,L).

1.2. Case a > 0, 7 < 0

In this case, by the re-scaling u = A — -u(y/ax), problem (3) can be reduced to

ua =

—u — u3 — a, u'(0) = u'(L) = 0,

(И)

з

з-

2

2-

1 -

И' 0

И' 0-

-1

-2

-3

0

2

3

0

2

3

where L = y/al and a = £ J — a. Integrating (11), we obtain

+ ^ + ^ + aU = p ^^ (U')2 = U4 — 2U2 — 4aU + 4p). 2 4 2 2

Note that the equation U3 + U + a = 0 has only one real root

. ( a /O2 fx1/3 ( a /"a2 1 \i/3

Umax = {—2^T + Y7> + l—2 ^T+27J

42

for any a G (one maximum). Thus, we have a nontrivial solution if p > + +

aUamax

U2

Lq = V2 i . dt , (12)

J \J—t4 — 2t2 — 4at + 4p V '

U1

where uk are two real roots of the equation t4 + 2t2 + 4 at — 4p = 0 such that u1 <u2. For a = 0 p > 0

(u')2 , (u2 + 1)2 i

A" + B =1' (13)

where A2 = _pг!, B4 = 4p +1, and from (13) it follows that

VB2—! i

L = B2 r dt = ^ I dt

q = a J .jb4 — (t2 +1)2 = J w(1 —12)(1+PP)' (1}

-vB2—! -1

where

B2 B2 1

A =—--, k2 = b~—- G (0,1). (15)

AvBTT B2 + 1 v' ; v ;

Let t = sin p in integral (14). Then

pn/2 i

lQ = 2A P =, lQ g (2.622 A, n A). (16)

Jq v 1 + k2 sin2 p

1.3. Case a < 0, y > 0

In this case, by the re-scaling u = y— ^U(y/—ax), problem (3) can be reduced to

U'' = U + U3 — a, U' (0) = U'(L) = 0, (17)

where L = y/—a l and a = — a\J — a • Integrating (17), we obtain

i^L — !L — !L + aU = p ^^ (U')2 = 1(U4 + 2U2 — 4aU + 4p). 2 4 2 2 '

Ua3 + Ua — a = 0

a a2 1 1/3 a a2 1 1/3

Umin = h + VT + 2TJ + U V T + 27/

"if

4444Й 44444 44444 44444

iiiiit 44444 4444

ttt tt t ttt tttt ttt t tttt tttt ttttt ttttt ttttt ttttt ttttt ttttt ttttt

4444 4444 4444 V

ttttt 444444 444444V 4444444 44444444 44444444

ttttitii

4444444

4444444

шиг

44444 44444 444 44, 4444

Fig 3. Phase diagram for the case a < 0, the right

Y > 0 with a = 0,1 от the left and a = 1 on

5' о

tttrs V V 4 4

ttt V V 4 4 4

itt trUzzzmzz v v ttt

? ? f t tttt 1111 f ¿«^»«»¿¿i-^vY 4 4 4 4 1111 V 4 4 4 4

111 ftl 4 4 4 4

111 tffl I* VV V.**>ftMl4 4 4 4

ttt 4444

1111 >4 «.*»*»/ 44 444

1111Л 4 4 4 4

tttt ^i-*-,'*-*-*-*-*-/! 4444 1111 t/ 4 4 4 4 111X \ 4 444

ttttt 444

tttt \ «.<-«-♦-<-<-<-<-♦-*'/4 444 tttt 4 4 4 4

-3-2-10 1 2 3

5' 0

_ '/У

о

Fig 4. Phase diagram for the case a < 0, the right

Y < 0 with a = 0,1 от the left and a = 1 on

for any a E ^ (one minimum). Thus, we have a nontrivial solution if p > — ^^ — ^^ + aumin with the length of interval, where the solution has only one zero, given by

U2

Lo = v^

dt

U1

лft4 + 2t2 - 4 at + 4p'

(18)

where uk are two real roots of the equation t4 + 2t2 — 4at + 4p = 0 such that ui < u2 1.4. Case a < 0, 7 < 0

In this case, by the re-scaling u = y^HV—ax), problem (3) can be reduced to

и

и - U3 -a, U'(0) = U'(L) = 0,

(19)

з

з

2

2

И' 0

и' 0

-1

-1

-2

-2

-3

-3

3

3

3

2

2

-1

-1

-2

-2

-3

-3

2

3

where L = \J—a ^d a = — Integrating (19), we find that

+ ^ — ^ + aU = p ^^ (U')2 = 1(—U4 + 2U2 — 4aU + 4p). 2 4 2 2

— Ua3 + Ua — a = 0

• three real roots

^ 2 2kn. , ^ ^ . 1 .^v^a.

uk =--= cos(0 +--), k = 0,1, 2, 0 = - arccos(-);

k y/3 yv 3 ' ' ' ^ 3 v 2

provided that a2 < 27 (two maximums and one minimum). Thus, we have two nontrivial

solutions if — + aUmax < p < ^ — Up- + aUmin with the lengths of intervals, where the corresponding solution has only one zero, given by

U2 U4

Lq,i = V2 i dt , Lq,2 = V2 i dt (20)

J yj—t4 + 2t2 — 4at + 4p J yj—t4 + 2t2 — 4at + 4p

U1 U3

Uk t4 — 2t2 + 4 at — 4p = 0 U1 < U2 <

U3 < U4

U4 U2

We have only one nontrivial solution if p > --™ + aUmin with the length of

interval, where the solution has only one zero, given by

U2

Lq,3 = V2 [ dt , (21)

' J ^J—t4 + 2t2 — 4 at + 4p

U1

Uk t4 — 2t2 + 4 at — 4p = 0 U1 < U2

• one real root

a = ( a fa2 fx1/3 ( a fa2 1\!/3 Umax = — 27/ + V 2 V T — 27 J

provided that a2 > 27 (one maximum). Thus, we have a nontrivial solution if p >

2

+ aUmax with the length of interval, where the solution has only one zero, given by

U2

Lq = V2 [ dt , (22)

J ,J—t4 + 2t2 — 4at + 4p

U1

Uk t4 — 2t2 + 4 at — 4p = 0 U1 < U2

= _2(a, 1/, .

max

Umax = —2(2)1/3, f = (2)1/3

provided that a2 = 27 (one maximum). Thus, we have a nontriv ial solution if p >

+ aUmax = —2(|)2/3 with the length of interval, where the solution has only one zero, given by

U2

Lo = V2 I dt , (23)

J ,J—t4 + 2t2 — 4at + 4p

U1

I 1 llkVN-W-H-M-»/ rffltt t irXX fftttt ill л»» f f t

шт

mti

1111Ш V\>-*^Wf/ttttt 11itt 1 t t/t 1111

mi шяяшпп 11111 f 1111 1111 VA t ttt

111 /// N ttt

111 IJk 11

II lyy/ t

-3-2-10 1 2 3

♦ T ♦ » » ■Jl^-t-b-t-^'S.X. <

ttfГ ' ~

a = 0, y > 0 = 0, 1

a = 0, y < 0 with a = 0,1 on the right

3

3

2

2

и 0

и' О

-1

-1

-2

-2

-3

-3

0

2

3

Uk t4 — 2t2 + 4 at — 4p = 0 U1 < U2

a = 0 p > 0

^ + (U2—I = 1, (24)

А2

B 4

where A2 = _pг!, B4 = 4p +1, and from (24) it follows that

Ln = ^

/B2+T

B2 r dt

A J л/В4 - (t2 - 1)2

-VB2+T

A

dt

V(1 -12)(1 + k2t2)

(25)

where

A

B2

^B2 - 1

, k

2 = B2 + 1

B2 - 1

> 1.

Let t = sin p in integral (25). Then

(26)

Ln = 2A

rn/2

dp

л/1 + k2

sin2 p

<nA.

(27)

1.5. Case a = 0, y = 0

If a = 0, y > 0 by the re-scaling u = U(y/jx), then problem (3) can be reduced to the

one

u" = u3 -a, u'(0) = u'(L) = 0,

(28)

where L = V/Yl and a = — -. Integrating (28), we find that

(u')

2 u4

2 - -j + au = p

(u')2 = 2(u4 - 4 au + 4p).

i

n

Thus, problem (28) has a nontrivial solution provided that p > |a4/3, i.e. p > 3(Y)4/3-The length of interval, where the solution has only one zero, is

u2

L0 = V2 i . d\ , (29)

J y/14 — 4 at + 4p

u1

where uk are two real roots of the equation t4 — 4at + 4p = 0.

If a = 0, 7 < 0, by the re-scaling u = u(y/—7x), then problem (3) can be reduced to

W = —u3 — a, u'(0) = u'(L) = 0, (30)

where L = ^J—7I and a = Integrating (30), we find that

+ "4 + au = p ^^ (u')2 = 2(—u4 — 4au + 4p).

Thus, problem (30) has a nontrivial solution provided that p > — |a4/3, i. e. p > — 3(Y)4/3-The length of interval, where the solution has only one zero, is

u2

L0 = V2 I' dt-, (31)

J ^—t4 — 4at + 4p

u1

where uk are two real ^^^s of the equation t4 + 4at — 4p = 0. For example, if a = 0 and p > 0

(u')2 u4 1

A- + B4 = 1, (32)

A2 = 2p B4 = 4p

b 1

B2 f dt 2B f dt B (11) B 2,622

L0 = — , =— , =— B -^)« 2, 62^— = ^^, 33)

0 A nT4 U A K U O A \A ' oj 'A ..i '

A j VB4 — t4 a j 2A M' 2' ' A

-B

p

4

1

where B(a,6) ■= f xa 1(1 — x)b 1dx is the beta function.

0

1.6. Case 7 = 0, a e R1

If a = 0, 7 = 0 then problem (3) can be reduced to

u" = — a, u' (0) = u' (l) = 0. (34)

This problem has the following solution

u(x) = C VC e R1 and a = 0, and it has no solutions if a = 0.

///4V4W

V У

t Г r tfr Л**^-**»v\VV V V V V 11 r i Yfc iiv v v v

4/4 444

\ \ \ Vfc /4 4 4

N X \ \ / / / V

WW S /////

< i

\ * ч / / *

vv vvv WW W wv WW w V V wv uum

ШШГ

44444 44444 4 4 44 4.

л *

гггг trrtt rttttt .tttttt tttttttt tttttttt \tttttt

wwt

> S*

Fig 6. Phase diagram for the case a > 0, 7 = 0 with a = 0,1 on the left and for the case a < 0, 7 = 0 with a = 0,1 on the right

If a > 0, 7 = 0, by the re-scaling u = a(y/ax), problem (3) can be reduced to

u'' = —u — a, u'(0) = u'(L) = 0,

where L = ^/al and a = a- This problem has the following solution

nk

u(x) = C cos(vax) — a, l = , k e Z+, C e R1.

a

(35)

If a < 0, 7 = 0, by the re-scaling u = u(y/—ax), problem (3) can be reduced to

u'' = u — a, u'(0) = u'(L) = 0, where L = \J—a ^d a = — This problem has the following solution

u(x) = a.

For convenience, we summarize our existence results in the form of Table.

(36)

1.7. Example

As shown in [3], the possible dimensionless function f (u,T) at fixed T can be written

as

f (u, T ) = T (1 - u2)2 + aT-T- (и + 1)2 + fO T ln (,

(37)

4T0 T0 f0 T0

where T0 is a critical temperature, c is the heat capacity, a = f- is dimensionless, 1 is the f0

f(u)

Y

T T - 2a(T - To) 2a(To - T)

t, a =-t-' a = —t—

To To To

C = n( П + T + 4a(T - To)

C = fo T I To) + 4To ■

»

3

3

2

2

и 0

и' О

-1

-1

-2

-2

-3

-3

0

2

3

0

2

3

Table

Conditions on the problem's parameters for existence and non-existence of nontrivial solutions

a (J Y p Lo

> 0 a2 < 4a3 < 277 < 4a3 < 27y < 4a3 < 27y \ 4a3 — 27y \ 4a3 — 27y > 0 p [pmin)pmax) Yes (2 sol's)

> 0 a2 > 0 p < pmin Yes

> 0 a2 > 0 p — pmax No

> 0 a2 > 0 p < pmin Yes

> 0 a2 > 0 P — Pmin No

> 0 R1 < 0 p > pmin Yes

> 0 R1 < 0 P < Pmin No

> 0 R1 = 0 R1 Yes (family sol's)

< 0 R1 > 0 P > Pmin Yes

< 0 R1 > 0 P < Pmin No

< 0 a2 < 4a3 < 27y < 4a3 < 27y < 4a3 < 27y > 4a3 — 27y > 4a3 — 27y < 0 P ^ (Pmin, Pmax] Yes (2 sol's)

< 0 a2 < 0 P > Pmax Yes

< 0 a2 < 0 P < Pmin No

< 0 a2 < 0 P > Pmin Yes

< 0 a2 < 0 P < Pmin No

< 0 R1 = 0 R1 No

= 0 R1 > 0 p > 4 (Y)4/3 Yes

= 0 R1 > 0 P < 3 ( Y )4/3 No

= 0 R1 < 0 p> - 3 (Y)4/3 Yes

= 0 R1 < 0 p < I ( Y )4/3 No

= 0 R1 = 0 R1 No

For example, if 0 < T < To then a > 0 Y > 0, and a > 0. In this case, according to the paragraph 2.1, problem (3) has nontrivial solutions. On the other hand, if T > To and a < 2(t-t0) n a > 0 Y > 0 a < 0 and we again have the existence of nontrivial solutions.

2. The Steady States for General Free Energy

In the previous section, we have studied the case when the free energy contains the linear term but it has no cubic term. Note that the presence of a linear term breaks down the symmetry of the free energy. Next, we will examine how the length of the existence interval can be effected by the cubic term. It is possible to introduce asymmetry in a phase diagram by adding odd powers to the free energy expansion so that

2 3 4

?/ ?/ u

f (u,T) = ao(T) + ai(T )u + a2(T)- + a3(T)- + a4(T)- + O(u5). (38)

Note that the special case a1(T) = a3(T) = 0 was introduced in [6, p. 19, (2.41)]. The authors assumed that a4(T) > 0 and a2(T) < 0. Then f (u,T) has maximum at u = 0 and minimum at u± = iy—^2. By definition, it is easy to verify that f (0,T) = ao(T) > 0

and f (u±,T) < 0 as a2 > 4aoa4 > 0. It means that there is a point uo e (0, ^J—j^) such that f (±uo,T) = 0. Thus, we can calculate the minimal length as

Lo = W-

uo

dt

-uo

V(t2 - u0)(t2 - u?)

Д2

V aU J

' -l

dt

- t2)(1 - k2t2):

(39)

where

uo

J- — - —л/a2 - 4aoa4, ui = a - — + —л[Ж—4a0a4, k2 = U0 e (0,1). (40)

V a4 a4 v V a4 a4 v u2

Substituting the new variable t = sin p into (39), we have

Lo = 2

\j a4u?

en/2

dp

Next, consider the equation

then

a4u2 Jo y/1 - k2 si

1(u' )2 = R(u)

Sin2 p

(41)

Lo = —

u2

V2 f dz

2 J vRz)'

where u^ are real roots of R(u) = 0 such that u1 <u2. Here

R(z) = az4 + bz3 + cz2 + dz + e = a(z2 + pz + q)(z2 + p'z + q') where a, b, c, d, e,p, q,p', q' e R, a = 0, i. e.

(42)

(43)

be + (ad - bc)q , be - adq , e 11 c c2 - 4ae \

P a(2e - cq) ' P a(2e - cq)' ^ aq ^ 2 I a V a2 J

Note that (43) has four real roots

c2 4ae.

Ul2

2 - ± VP2 - 4q) , U3,4 = 2 ± V(p')2 - 4q')

provided that

p2 - 4q > ^d (p')2 - 4q' > 0

Assume that p = p'. By change of variables

at + v и - v z =-, dz = --— dt

t + 1

(t +1)2

(44)

in (42), we deduce that

U2

Lo = —

y/2 t dz

2 j VW)

/J.U2+V

«2+1

A

dt

/J.U 1 + v

U1+1

V(i + fcit2)(i + k2t2)

(45)

where

A

ki

V2

i — V

2 yja(v2 + pv + q)(v2 + p'v + q')

12 + pi + q _ 12 + p'l + q'

: ~~ô-, k2 — -;-7 •

v2 + pv + q v2 + p'v + q'

¡V

p(i + v ) + 2(q + ¡1v ) — 0, p' (1 + v ) + 2(q' + ¡v ) — 0,

hence

— ±\ l(—) p — p p — p

pq — p q p —p

¡

——t\l( q-- ) p — p p — p

pq — p q p —p

(46)

(47)

If (43) has four real roots then ki < 0. By change of variables t — (—k1) 1/2 sin p in (45) we deduce that

¥2

dp

Lo — A

¥1

where

A

A

V-h

k2

k3 — ~T > 0, Pi k1

\Jl — k3 sin2 p

yJ—ki(iUi + v )

(48)

arcsin

Ui + 1

k1 < 0 k2 > 0 k1 > 0 k2 < 0

a > 0

case a < 0 only. Assume that k1 < 0 Then we obtain (48) with k3 < 0.

In the case p — p', by change of variables z — t — p in (42), we deduce that

L0 — —

_ U2

V2 f dz

A

dt

U1

V^M u-p \]a(1 — 4it2)(1 — 42t2)

(49)

where

A

2^2

4i —

> 0, 42 —

vV — 4q)(p2 — 4q' )' p2 — 4q ' p2 — 4q'

By change of variables t — (41)-1/2 sin p in (49), we find that

> 0-

¥>2

Lo — A

dp

¥ 1

a(1 — h3 sin2 p)

(50)

2

2

p

2

4

4

where

A A f f2 A = —=, f3 = — > 0, ф kf i kf i

arcsin

in ^]/k~i(ui - p/2)^

Next, we consider the special cases p2 = 4q or (p')2 = 4q'. For example, if (p')2 = 4q' then

R(u) = a u -

u

-p - \Jp2 - 4q

u-

-p + у/p2 - 4q W p -2- u + 2

') (u + p)2.

If a < 0 then two cases are possible:

(0-2e(

p

(ii) - 2

К

-p - \Jp2 - 4q -p + \Jp2 - 4q 2 ' 2

-p - \Jp2 - 4q -p + \Jp2 - 4q

In the first case (i), we find that Lo = to. In the second case (ii), we have (48). On the

a > 0 we have Lo = to.

In particular, if p2 = 4q and (p')2 = 4q' then

R(u) = a(u + p) и + p^ .

Obviously, if a < 0 ^^^n ^te problem has no nontrivial solutions but if a > 0 then L0 = то.

References / Литература

1. Cahn J.W., Hilliard J.E. Free Energy of a Nonuniform System, I. Interfacial Free Energy. The Journal of Chemical Physics, 1958, vol. 28, pp. 258-267. DOI: 10.1063/1.1744102

2. Carr J., Gurtin M.E., Slemrod M. Structured Phase Transition on a Finite Interval. Archive for Rational Mechanics and Analysis, 1984, vol. 86, pp. 317-351. DOI: 10.1007/BF00280031

3. Fife P.C., Penrose O. Interfaced Dynamics for Thermodinamically Consistent Phase-Field Models with Nonconcerved Order Parameter. Electronic Journal of Differential Equations, 1995, vol. 16, pp. 1-49.

4. Grinfeld M., Novick-Cohen A. Counting Stationary Solutions of the Cahn - Hilliard Equation by Transversality Arguments. Proceedings of the Royal Society of Edinburgh: Section A Mathematics, 1995, vol. 125, no. 2, pp. 351-370. DOI: 10.1017/S0308210500028079

5. Grinfeld M., Novick-Cohen A. The Viscous Cahn-Hilliard Equation: Morse Decomposition and Structure of the Global Attractor. Transactions of the American Mathematical Society 6, 1999, vol. 351, no. 6, pp. 2375-2406.

6. Provatas N., Elder K. Phase-Field Methods in Materials Science and Engineering. Weinheim, Wiley-VCH, 2010. 312 p.

7. Novick-Cohen A., Peletier L.A. Steady States of the One-Dimensional Cahn - Hilliard Equation. Proceeding of the Royal Society of Royal of Edinburg, 1993, vol. 123A, pp. 1071-1098. DOI: 10.1017/s0308210500029747

8. Smoller J., Wasserman A. Global Bifurcation of Steady-State Solutions. Journal of

Differential Equations, 1981, vol. 39, pp. 269-290. DOI: 10.1016/0022-0396(81)90077-2

Received February 28, 2016

УДК 517.912 DOI: 10.14529/mmpl60206

СТАЦИОНАРНЫЕ РЕШЕНИЯ УРАВНЕНИЯ КАНА -ХИЛАРДА С ГРАНИЧНЫМ УСЛОВИЕМ НЕЙМАНА

И.Б. Краснюк, P.M. Таранец, М. Чугунова

Исследована структура стационарного состояния одномерного уравнения Кана -Хилларда в сочетании с граничными условиями Неймана. Здесь свободная энергия задается полиномом четвертого порядка. Была построена диаграмма бифуркации существования и единственности монотонных решений этой задачи. А именно, найдена длина интервала, на котором решение монотонно возрастает или убывает и имеет один нуль для некоторых фиксированных значений физических параметров. Под НбОД-нозначностью понимается возможность существования более чем одного монотонного решения для некоторых значений физических параметров.

Ключевые слова: уравнение Кана - Хиларда; граничное условие Неймана; устойчивые состояния.

Игорь Борисович Краснюк, КсШДИДсХТ физико-математических наук, старший научный сотрудник, Институт прикладной математики и механики НАН Украины (г. Донецк, Украина), krasnjukigr@rambler.ru.

Роман Михайлович Таранец, КсШДИДсХТ физико-математических наук, научный сотрудник, Калифорнийский университет (г. Лос-Анджелес, Соединенные Штаты Америки), taranets_r@yahoo.com.

Марина Васильевна Чугунова, профессор, Клермонтский университет (г. Клер-монт, Соединенные Штаты Америки), marina.chugunova@cgu.edu.

Поступила в редакцию 28 февраля 2016 г.