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A Bäcklund transformation (BT) for ae partial differential equation which has ... Substituting (4) into equation (3) and the equating coefficients of the like powers.
Acta Mathematicae Applicatae Sinica, English Series Vol. 23, No. 4 (2007) 563–568

Exact Solutions of Fisher and Generalized Fisher Equations with Variable Coefficients ¨ gu Arzu O˘ ¨ n, Cevat Kart Department of Mathematics, Faculty of Sciences, Ankara University, 06100 Tando˘ gan-Ankara Turkey (E-mail: [email protected], [email protected])

Abstract

In this work, we consider a Fisher and generalized Fisher equations with variable coefficients. Using

truncated Painlev´e expansions of these equations, we obtain exact solutions of these equations with a constraint on the coefficients a(t) and b(t). Keywords

Nonlinear evolution equations, fisher equation, generalized fisher equation, b¨ acklund

transformation, painlev´e property 2000 MR Subject Classification

1

35Q53

Introduction

Many important evolution equations (such as KdV, Burgers and Bousinesq) arise in many physical systems (see [1,9]). These equations can be characterized by having Painlev´e property and B¨acklund transformation [2]. Recently Weiss at al[13] have introduced the Painlev´e PDE test for nonlinear partial differential equations (see 12,13]). There is a connection with integrability. A nonlinear partial differential equation (NPDE) has the Painlev´e property (PP), when the solution of the NPDE is “single-valued” about the movable singularity manifold. If the singularity manifold determined by φ(z1 , z2 , · · · , zn ) = 0

(1)

and u = u(z1 , z2 , · · · , zn ) is a solution of the NPDE, then it is required that u = φα

∞ 

uj φj

(2)

j=0

where u0 = 0 , φ = φ(z1 , z2 , · · · , zn ) and uj = uj (z1 , z2 , · · · , zn ) are analytic functions of (zj ) in a neighborhood of the manifold (1) and α is a negative integer. Substitution of (2) into the NPDE determines the allowed values of α and defines the recursion relation for uj , j = 0, 1. When the ansatz (2) is correct, the NPDE is said to possess the Painlev´e property and is conjectured to be integrable (see [10–12]). A B¨acklund transformation (BT) for ae partial differential equation which has the Painlev´e property can be obtained by truncating the Laurent series (2) at the constant level term. If a BT can be found, then the solution of the NPDE can be used to obtain either a different solution to the same partial differential equation. Kawahara and Tanaka shown [7] that the equation of the form ut = uxx − u(a − u)(1 − u), Manuscript received August 24, 2006.

¨ gu A. O˘ ¨n, C. Kart

564

which is a Fisher type, has the travelling wave solution:  √2ax (a2 − 2a)t  a u= 1 + tanh ± + . 2 4 4 Also, Cariello and Tabor[3] considered the equation ut = uxx + u(1 − 6u) and obtained solutions using truncated Painlev´e expansions of this equation. In this work, we further generalize these equations. In Section 2, we consider the generalized Fisher equation with a variable coefficient ut = a2 (t)uxx + u − u3 and obtain e analytical solutions of this equation with some constraint on the coefficient a(t). Similarly, in Section 3, a variable coefficient Fisher equation ut = b(t)uxx + au(1 − u) is considered, and travelling wave solutions have been found with some constraint on the coefficient b(t). Many authors have obtained certain soliton-typed explicit solutions of some wellknown partial differential equations with some constraints on the coefficients using truncated Painlev´e expansions and symbolic computation method (see [4–6,8,10]).

2

Exact Solutions of a Generalized Fisher Equation with a Variable Coefficient

In this section, we construct analytic solutions of the equation ut = a2 (t)uxx + u − u3 . We use the finite series u=

p 

(3)

uj φj−p ,

j=0

where p is a positive integer. By the leading order analysis of equation (3), we find p = 1. Thus the truncated Painlev´e expansion is u(x, t) =

u0 + u1 , φ

(4)

where u0 = 0. Substituting (4) into equation (3) and the equating coefficients of the like powers of φ to zero give the set of Painlev´e-B¨acklund (PB) equations: φ−3 : − 2a2 u0 φ2x + u30 = 0,

(5a)

φ−2 : − u0 φt + 2a2 u0,x φx + a2 u0 φxx + 3u20 u1 = 0, φ−1 : u0,t − a2 u0,xx − u0 + 3u0 u21 = 0,

(5b) (5c)

φ

(5d)

0

: u1,t − a2 u1,xx − u1 − u31 = 0.

From (5d), u1 is a solution of (3). Equation (5a) gives √ u0 = 2aφx .

(6)

Exact Solutions of Fisher and Generalized Fisher Equations with Variable Coefficients

After substituting (6) into (5b), we obtain √

2 φt u1 = − 6a φx

√ 2a φxx . 2 φx

565

(7)

Substitution of (6) and (7) into (5c) leads to the constraint on the coefficient function as follows 6aat φ2x + 6a2 φx φxt − 6a4 φx φxxx − 6a2 φ2x + φ2t − 6a2 φt φxx + 9a4 φ2xx = 0. We first assume that

φ = 1 + eαx+βt ,

(8) (9)

where α, β are arbitrary real constants. Inserting the expression (9) into equation (8) we obtain the constraint equation on the coefficient a(t) as follow: 6α2 aat + 3α4 a4 − 6α2 a2 + β 2 = 0.

(10)

a2 (t) = b(t).

(11)

Now let Then

3α2 bt + 3α4 b2 − 6α2 b + β 2 = 0.

(12) √ This equation is a Riccati equation and can be readily integrated. In fact, if |β| < 3 then b(t) =

 e2γt + c  1  1 + γ , α2 e2γt − c

where β 2 = 3(1 − γ 2 ) , 0 < |γ| ≤ 1 and c is an integrating constant. Thus,  e2γt + c  1 . 1 + γ 2γt a(t) = ± α e −c

(13)

(14)

Finally, combining all terms, we find a family of analytical solutions of the equation (3) as follows: √  e2γt + c  eαx+βt u(x, t) = ± 2 1 + γ 2γt e − c 1 + eαx+βt √ √  e2γt + c  1 2β 2 . (15) − + 1 + γ   6 2 e2γt − c e2γt +c 1 + γ e2γt −c As an example, we choose α = 1 , β = 0 , c = 1. Then a travelling wave solution of the equation (16) ut = (1 + coth t)uxx + u − u3 √ 2√ x u(x, t) = ± 1 + coth t tanh . 2 2

is obtained as

Note that We note that if |β| =

√ 3 then

x lim u(x, t) = tanh . t→∞ 2 1 a(t) = ± α



t+c+1 . t+c

(17) (18)

(19)

¨ gu A. O˘ ¨n, C. Kart

566

Hence the solution of equation (3) is given by √ t + c + 1  αx + √3t  √6 t + c  u(x, t) = ± 2 tanh + , (20) t+c 2 6 t+c+1 √ where c is an integrating constant. If |β| > 3 then equation (12) has the complex solution a(t) = ±

1

1 + δ tan(−δt + c). α

(21)

So the solution of equation (3) is √

 αx + βt  √2β  1

u(x, t) = ± 2 1 + δ tan(−δt + c) tanh + , 2 6 1 + δ tan(−δt + c)

(22)

where β 2 = 3(1 + δ 2 ) and c is an integrating costant.

3 Exact Solutions of a Fisher Equation with a Variable Coefficient Now we consider the variable coefficient Fisher equations of the form ut = b(t)uxx + au(1 − u).

(23)

Truncating the Painlev´e expansion (2) at the constant-level term: u=

p 

uj φj−p

j=0

and balancing the linear term uxx with the nonlinear terms −au2 , we get p = 2. Hence, u(x, t) =

u0 u1 + u2 . + φ2 φ

(24)

Substituting (24) into (23) and equating the coefficients of the like powers of φ to zero give the set of Painlev´e-B¨acklund (PB) equations: − 6b(t)u0 φ2x + u20 = 0, − 2u0 φt + 4b(t)φx u0x + 2b(t)uo φxx − 2b(t)u1 φ2x + 2au0 u1 = 0,

(25a) (25b)

u0t − u1 φt − b(t)u0xx + 2b(t)u1x φx + b(t)u1 φxx − au0 + au21 + 2au0 u2 = 0,

(25c)

u1t − b(t)u1xx − au1 + 2au1 u2 = 0, u2t − b(t)u2xx − au2 + au22 = 0.

(25d) (25e)

From (25a) we find that 6 b(t)φ2x a

(26)

6 6 φt − b(t)φxx . 5a a

(27)

u0 = (26) and (25b) give the result u1 =

Exact Solutions of Fisher and Generalized Fisher Equations with Variable Coefficients

567

By substituting (26) and (27) into (25c) and (25d), we obtain that 12 1 6 bφx φxt + φ2t − bφt φxx 5 25 5 + 3b2 φ2xx − 4b2 φx φxxx + 2abφ2x u2 = 0, 6  6 36 6 φtt − bt − 6b φxx − bφxxt + b2 φxxxx 5a a 5a a   12 6 φt − 12φxx u2 = 0. − φt + 5 5

(bt − ab)φ2x +

(28a)

(28b)

If u2 is eliminated between (28a)-(28b) then the constraint equation on the coefficient a(t) can be given by bt 2 12 1 1 3 7 2 φx φt − φt φx φxt − φ + φ φxx b 5 25 b t 5 t − 9bφt φ2xx + 4bφt φx φxxx + 12bφx φxx φxt + 15b2 φ3xx − 20b2 φx φxx φxxx = 0.

φ2x φtt − 6bφ2x φxxt + 5b2 φ2x φxxxx −

Now choose

φ(x, t) = 1 + eαx+βt ,

(29) (30)

where α, β are arbitrary real constants. If (30) is substituted into the constraint equation (29), then the equation satisfied by b(t) is obtained as α2 bt − α4 b2 +

1 2 β = 0. 25

(31)

This equation (31) is readily integrated and b(t) can be found as follows: b(t) = −

 β β t + c , coth 5α2 5

(32)

where c is an integrating constant. The expression for u2 is obtained from equation (28a) as u2 = −

3β 1 + . 5a 2

(33)

Since of u2 must satisfy equation (25e), we find β=±

5a . 6

(34)

Then from (32) it follows that b(t) = − Thus, for β =

5a 6

 a a t + c . coth 6α2 6

(35)

the solution of equation (23) is given by

u(x, t) =

α a  α 5a  1 5a  1 1 coth t + c sech2 x + t + tanh x+ t + , 4 6 2 12 2 2 12 2

(36)

and for β = − 5a 6 the solution of equation (23) is given as u(x, t) =

α a  α 5a  1 5a  1 1 coth t − c sech2 x − t − tanh x− t + . 4 6 2 12 2 2 12 2

(37)

568

¨ gu A. O˘ ¨n, C. Kart

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