Pramana method. PACS Nos 05.45.Yv; 02.30.Jr; 03.65.Vf

March 9, 2019 Physics

Pramana – J. Phys. (2018) 91:9 © Indian Academy of Sciences
https://doi.org/10.1007/s12043-018-1583-4
Exact solutions to (2+1)-dimensional Chaffee–Infante equation
YUANYUAN MAO
Department of Mathematics, Northeast Petroleum University, Daqing 163318, China
E-mail: [email protected]
MS received 28 November 2017; revised 4 January 2018; accepted 8 January 2018
Abstract. In this paper, the canonical-like transformation method and trial equation method are applied to(2+1)-
dimensional Chaffee–Infante equation, and some exact solutions are obtained. In particular, a new solution in terms
of elliptic functions is given.
Keywords. Chaffee–Infante equation; travelling wave solutions; canonical-like transformation method; trial
equation method.
PACS Nos 05.45.Yv; 02.30.Jr; 03.65.Vf
1. Introduction
The construction of exact solutions to nonlinear
differential equations is an important and difcult task.
It also plays a signicant role in mathematics and
physics. Various powerful methods such as inverse
scattering method 1, direct method 2, symmetri-
cal method 3, the complete discrimination system for
polynomial method 4–14 and so on have been pro-
posed for obtaining approximate and exact solutions for
various nonlinear equations.
In this paper, we consider (2+1)-dimensional
Chaffee–Infante equation 15
uxt+ (?uxx+au3?au)x+?uyy=0,(1)
where aand?are arbitrary constants. Assume that a
substance spreads in a region with the concentration of
the diffusion as u(x,y,z).LetD(x,y,z,t)be the diffu-
sion coefcient, then according to the law of diffusion,
we have
d m=?D ? u
?n d sdt,(2)
where mrepresents the amount of diffusion material and
D >0.Accordingtoeq.(2) and the law of conservation
of mass yields
m =
t2
t1

D ? u
?n d s

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d t
=

u(x,y,z,t2)?u(x,y,z,t1)dxdydz,(3)
that is,
?
? x

D ? u
?x

+ ?
? x

D ? u
?y

+ ?
? z

D ? u
?x

= ? u
?t . (4)
Assume that D=1 and the inuential factor isf(u)=
u3? u.Weget
? u
?t = ?2u
? x2+ ?2u
? y2+ ?2u
? z2+?(u?u3),(5)
where parameter ?adjusts the relative balance of the
diffusion term and the nonlinear term. So the (1+1)-
dimensional Chaffee–Infante equation is
ut? uxx=?(u?u3),(6)
and the derivation process of (2+1)-dimensional
Chaffee–Infante equation is similar to it.
The (2+1)-dimensional Chaffee–Infante equation is
a well-known reaction diffusion equation. It describes
the physical process of mass transport and particle dif-
fusion, and has been widely used in environmental
science, uid dynamics, high-energy physics, electronic
science, and so on. Therefore, building exact solutions
to this equation has great scientic signicance and
a broad application background (see 16 and refer-
ences therein). In 15, the authors use exp-function
method to obtain solutions to eq. ( 1), and by this method,
we cannot get solutions of other types such as ellip-
tic function solutions. It is also difcult to get elliptic
function solutions to eq. ( 1) by some other methods,
but canonical-like transformation method can be used

9 Page 2 of 4Pramana – J. Phys. (2018) 91:9
to get elliptic function solutions. The canonical-like
transformation method was proposed by Liu 17to
obtain exact travelling wave solutions to some diffusion-
reaction equations such as Fisher equation, Burgers–
KdV equation and Newell–Whitehead–Kwahara
equation and so on. In this paper, we use Liu’s canonical-
like transformation method to nd solutions to eq. (1)
and get a new solution in terms of elliptic functions. In
addition, we use trial equation method 18–23toget
exact solutions of other forms. Further studies for exact
solutions and integrability of differential equations in
mathematical physics can be found in many papers (see,
for example, 24–30).
2. Elliptic functions solutions by canonical-like
transformation method
According to ref. 17, the canonical-like transforma-
tion method is as follows. We consider the following
ordinary differential equation:
u
(? )?Au(? )=Bu(? )+Du(? ).(7)
If we take?=u
,eq.(7) becomes
d?
du=Bu
+Du+A?
?.(8)
In order to solve the rst-order nonlinear ODE (8), we
re-parametriseuand?by a new parameters,thatis,we
take the canonical-like transformations
u=a
11(s)v(s)+a12(s)v(s),(9)
?=a
21(s)v(s)+a22(s)v(s),(10)
wherea
ij(s)(i,j=1,2)andv(s)are functions to be
determined. In order to obtain the parameters, we set
a
12=0. After a lot of calculations, we get
a
11=
+3(?1)A
1/2
(s?s0)2/(?1),(11)
a
22=
(?1)A+3
1/2
(s?s0)2/(?1),(12)
a
21=A?h2
+3(?1)A
1/2
(s?s0)2/(?1).(13)
Then eq. (7) becomes
v
(s)=Mv(s),(14)
andsis determined by
d?
ds=1?(s),(15)
whereM=Bh
?(+3)/2,h=±?4D+A2and?(s)=
h(s?s
0).The general solution of eq. (14)isgivenby
±(s?s
0)=
dv 2M+1v+1+c,(16)
wherecis an integral constant. Ifc=0, we have
v(s)=
±1?
2

2M
+1(s?s0)
2/(1?)
.(17)
And in the special case ofc=1and=3, we get
v(s)=
2
M
1/4asn(?(s?s0),m)+bcn(?(s?s0),m)
csn(?(s?s0),m)+dcn(?(s?s0),m,
(18)
where
a=3?2?
2,b=2?2?1,
c=2?
2?4,d=?2,
m=16+12?
2
17+12?2,?=
19+12?2
52?32?2.
So, we can get solutions of eq. (7) from the formulas
(17)and(18)
u(? )=6
B
(?1)A+3
(+2)/2e2A+3(???0)
f
e(?1)A+3(???0)?s0

2?1
(19)
and
u(? )=
3
A
1/2
(eh(???0)?s0)
2M
1/4
×asn(?(e
h(???0)?s0),m)+bcn(?(eh(???0)?s0),m)
csn(?(eh(???0)?s0),m)+dcn(?(eh(???0)?s0),m).
(20)
Now, we use the canonical-like transformation
method to obtain the solutions of eq. (1). Letu=u(? )
and?=kx+ly+wt, then eq. (1) becomes
(kw+?l
2)u?k3u+3aku2u?aku=0,(21)
and integrating it yields
(kw+?l
2)u?k3u+aku3?aku=c0.(22)
Ifc
0=0, by (19)and(20), the solutions of eq. (1)
can be written as
u(x,y,t)=
6k2
a
fkw+?l2
3k3

5/2ekw+?l23k3(kx+ly+?t??0)
ekw+?l23k3(kx+ly+?t??0)?s0
(23)

Pramana – J. Phys. (2018) 91:9 Page 3 of 4 9
and
u(x,y,t)=
3
A
1/2
(eh(kx+ly+?t??0)?s0)
2M
1/4
×a1sn(?(eh(kx+ly+?t??0)?s0),m)+b1cn(?(eh(kx+ly+?t??0)?s0),m)
c1sn(?(eh(kx+ly+?t??0)?s0),m)+d1cn(?(eh(kx+ly+?t??0)?s0),m),(24)
where
A=kw+?l
2
k3,h=±(kw+?l2)2+4ak4
k3,
M=a
k2h?3,a1=3?2?2,b1=2?2?1,
c
1=2?2?4,d1=?2,
m=16+12?
2
17+12?2,?=
19+12?2
26?16?2.
Furthermore, according to ref. 17, we have
2(kw+?
2)2=?9ak4,(25)
from which we have
h=kw+?l
2
3k3.
Formula (24) is a new solution represented by elliptic
functions with double period.
3. Other solutions by trial equation method
Ifc
0=0, by taking
u=a
0+a1?+b1
?(26)
and
?=?
,(27)
where?is a function to be determined, eq. (22) becomes

a
1?b1
?2

??
?
Aa1?Ab1
?2

?+2b1
?3?2
=B

a0+a1?+b1
?
3
?
a0+a1?+b1
?
+c
0,(28)
where
A=k?+?l
2
k3,B=ak2,
and the prime indicates differential with respect to?,
e.g.,?
=d?/d?. Without loss of generality, we take
the solution of eq. (28) as follows:
?=?
2+b0,(29)whereb
0is a constant to be determined. Substituting
(29)into(28) yields a polynomial of?. Setting all coef-
cients of this polynomial to zero, we get a system of
algebraic equations
2b
1b2
0?Bb3
1=0,(30)
Ab
1b0?Ba0b2
1=0,(31)
4b
1b0?2b1b0?Ba1b2
1?Ba2
0b1+Bb1=0,(32)
?Aa
1b0+Ab1?Ba3
0?Ba0a1b1+Ba0?c=0,
(33)
2a
1b0?2b1+2b1?Ba2
0a1?Ba2
1b1+Ba1=0,
(34)
?Aa
1?Ba0a2
1=0,(35)
2a
1?Ba3
1=0.(36)
Solving this algebraic equation system, we obtain a
family of values of parameters
a
0=?A?2B,a1=
2
B,b0=?2B?A
2
8,
b
1=?A
2?2B
4?2B.(37)
So we have
?=?
2B?A2
8tanh?
?
2B?A2
8???0
?
?
(38)
and
?=?
2B?A2
8coth?
?
2B?A2
8???0
?
?
,(39)
where?
0is an arbitrary constant. Therefore, the solu-
tions of eq. (1)aregivenby
u(x,y,t)=?kw+?l
2
k2?2a
?m
2k2?atanh
m2?2k3(kx+ly+wt)??0

+m
2k2?acoth
m2?2k3(kx+ly+wt)??0

(40)

9 Page 4 of 4Pramana – J. Phys. (2018) 91:9
and
u(x,y,t)
=?kw+?l
2
k2?2a
?m
2k2?acoth
m2?2k3(kx+ly+wt)??0

+m
2k2?atanh
m2?2k3(kx+ly+wt)??0

,
(41)
wherem=?
2ak4?k2?2?2k??l2??2l4. Expres-
sions (40)and(41) are solutions represented by hyper-
bolic tangent function and hyperbolic cotangent
function.
4. Conclusion
In this paper, the canonical-like transformation method
and trial equation method are applied to obtain exact
solutions to(2+1)-dimensional Chaffee–Infante equa-
tion. Among those, some new solutions are given. The
results show that canonical-like transformation method
and trial equation method are powerful for solving non-
linear problems arising in mathematical physics.
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