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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