ISSN 2307-7743 http://scienceasia.asia
NEW EXACT SOLUTIONS FOR SOME NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
LUWAI WAZZAN AND M. S. AL-AMRY
Abstract. In this paper we propose a new form of Pade0-II equation, namely, a modified
Pade0-II equation. As the mapping method is a promising method to solve nonlinear evolu-tion equaevolu-tions. We apply it, to solve the Pade0-II, modified Pade0-II equations and the (2+1) dimensional Konopelchenko-Dubrovsky (KD) equation. Exact travelling wave solutions are obtained and expressed in terms of hyperbolic functions, trigonometric functions, rational functions and elliptic functions.
1. Introduction
In recent years, directly searching for exact solutions of nonlinear partial differential e-quations (PDE’s) has become more and more attractive field in different branches of physics and applied mathematics. These equations appear in condensed matter, solid state physics, fluid mechanics, chemical kinetics, plasma physics, nonlinear optics, propagation of fluxions in Josephson junctions, theory of turbulence, ocean dynamics, biophysics and star formation and many others. In order to get exact solutions directly, many powerful methods have been introduced such as the GG0-expansion method [1], inverse scattering method [2,4], Hirota’s bilinear method [5,6], the tanh method [7,8], the sine-cosine method [9,10], Backlund trans-formation method [11,12], the homogeneous balance [13,14], Darboux transtrans-formation [15], the Jacobi elliptic function expansion method [16], A modified tanh-coth method [17,18].
Recently, Yan-Ze Peng [19] introduced a new approach, namely, the mapping method for a reliable treatment of the nonlinear wave equations. The useful mapping method is then widely used by many authors [20,21] and others.
2. Description of the method
Consider the general nonlinear PDE’s, say, in two variables,
(1) P(u, ux, ut, uxx, uxt, ...) = 0.
Let u(x, t) = u(ξ), where ξ = λ(x−ct), then equation (1) reduces to a nonlinear ordinary differential equation (ODE)
(2) Q(u, u0, u00, ...) = 0.
Key words and phrases. Mapping method; (KD) equation; Pade0-II equation; modified Pade0-II equation; (2+1) dimensional Konopelchenko-Dubrovsky equation.
c
2016 Science Asia
Assume the solution of equation (2) takes the form
(3) u(x, t) = u(ξ) = a0+
m
X
i=1
ai(f(ξ)) i
+bi(f(ξ)) i
,
where the coefficients a0, ai, bi(i = 0,1,2, ..., m ), λ, c are constants to be determined, and
f =f(ξ) satisfies a nonlinear ODE
(4) df(ξ)
dξ = r
pf2(ξ) + 1
2qf
4(ξ) +r,
where p, q, r∈R.
Substituting (3) into (2), using (4) repeatedly, the parametermwill be found by balancing the highest-order nonlinear term with the highest-order partial derivative term in the
result-ing equation. Settresult-ing the coefficients of the each order offi(ξ) andfi(ξ)qpf2(ξ) + 1
2qf4(ξ) +r
to zero, we obtain a set of nonlinear algebraic equations fora0, ai, bi(i= 1,2, ...n), λ, c. With the aid of Maple, a symbolic computer program, we can solve the set of nonlinear algebraic equations and obtain all the constants a0, ai, bi(i= 1,2, ...n), λ and c.
The ODE (4) has the following solutions: when p= 1, q=−2, r= 0, then
(5) f(ξ) = sech(ξ)
when p=−2, q= 2, r= 1, then
(6) f(ξ) = tanh(ξ)
when p= 12, q= 12r= 14, then
(7) f(ξ) = tan(ξ)±sec (ξ)
when p=−(1 +k2), q= 2k2, r= 1, then
(8) f(ξ) = sn(ξ)
when p= 2k2−1, q=−2k2, r= 1−k2, then
(9) f(ξ) = cn(ξ)
when p= 2−k2, q=−2, r =−(1−k2), then
(10) f(ξ) = dn(ξ)
when p= 2−k2, q= 2, r = 1−k2, then
(11) f(ξ) = cs(ξ)
when p=−(1 +k2), q= 2, r=k2, then
when p=−1 + 2k2, q= 2, r=−k2(1−k2), then
(13) f(ξ) = ds(ξ)
when p= 0, q= 2, r= 0 andc is an arbitrary constant, then
(14) f(ξ) = c
ξ,
when p= 1, q= 0, r= 0, then
(15) f(ξ) = eξ
3. Applications
3.1. Pade0-II Equation. We consider the Pade0-II equation in the form
(16) ut+ux+uux+auxxx+buxxt= 0,
where u=u(x, t), a, b are real numbers.
Pade0-II equation is a nonlinear wave equation modeling unidirectional propagation of long wave in dispersive media. It is originally derived by using a Pade0 (2,2) approximation of the phase velocity that arises in linear wave theory [22].
We will apply the mapping method to solve the nonlinear general Pade0-II equation. Sub-stituting u(x, t) = u(ξ), where ξ =λ(x−ct), into Eq. (16) and integrating once yields
(17) (1−c)u+ u
2
2 +λ
2(a−bc)u00= 0.
Balancing the order of the nonlinear termu2with the highest derivativeu00 gives 2m=m+ 2
that gives m= 2. Thus, the solution of Eq. (17) has the form
(18) u(ξ) =a0+a1f(ξ) +a2(f(ξ)) 2
+b1(f(ξ))
−1
+b2(f(ξ))
−2
.
Substituting (18) in (17) and using (4), collecting the coefficients of each power of fi,0 ≤
i≤ 8, setting each coefficient to zero, and solving the resulting system, obtain three sets of solutions.
(A) a0 =−2 + 2c , a1 =a2 = 0 , λ =λ
(B) a0 = (1−c) (4αp−1), a1 = 0 , a2 =−6α(c−1)q , λ=±
q
c−1
a−bcβ
(C) a0 = (c−1) (4αp+ 1), a1 = 0 , a2 = 6α(c−1)q , λ=±
q
1−c a−bcβ
where α=q16p2−124rq and β = q
2
2p2−3rq.Note that b1 =b2 = 0 for all sets of solutions. • Using (5), a solution of the ODE (4), and (A), (B) and (C), Eq. (18) gives
forc > 1
u2(x, t) = 3 (c−1) sech2
"
1 2
r c−1
a−bc(x−ct) #
,
u3(x, t) = −2 + 2c+ 3 (1−c) sec2
"
1 2
r c−1
a−bc(x−ct) #
.
respectively, for c <1
u4(x, t) = 3 (c−1) sec2
"
1 2
r
1−c
a−bc(x−ct) #
,
u5(x, t) =−2 + 2c+ 3 (1−c) sech2
"
1 2
r
1−c
a−bc(x−ct) #
.
• Using (6) and (B) and (C), Eq. (18) gives u2 and u3 for c > 1 and u4 and u5 for
c <1, respectively.
• Using (8) and (B) and (C), Eq. (18) gives
u6,7(x, t) = a0+a2sn2(λξ),
respectively, wherea0, a2 are defined in (B) and (C).
When k → 0, u6,7 become u1, when k →1, we obtain u2 and u3 for c > 1 and u4
and u5 for c <1, respectively.
• Using (9) and (B) and (C), Eq. (18) gives
u8,9(x, t) =a0+a2cn2(λξ),
respectively, wherea0, a2 are defined in (B) and (C). Ask →0,u8,9 becomeu1 and
ask →1 we obtain u2 and u3.
• Using (10), and (B) and (C), Eq. (18) gives
u10,11(x, t) =a0+a2dn2(λξ),
respectively, wherea0, a2 are defined in (B) and (C). As k → 0, we obtain constant
solution, ask →1, we obtainu1.
• Using (11), and (B) and (C), Eq. (18) gives
u12,13(x, t) = a0+a2cs2(λξ),
respectively, where a0, a2 are defined in (B) and (C). As k → 0 and c > 1, u12,13
become
u14(x, t) = 1−c+ 3 (1−c) cot2
"
1 2
r c−1
a−bc(x−ct) #
,
u15(x, t) = 3c−3−3 (c−1) coth2
"
1 2
r c−1
a−bc(x−ct) #
and for c <1, we obtain
u16(x, t) = 1−c−3 (1−c) coth2
"
1 2
r
1−c
a−bc(x−ct) #
,
u17(x, t) = 3c−3 + 3 (c−1) cot2
"
1 2
r
1−c
a−bc(x−ct) #
.
Ask →1, we obtain u14, ..., u17, for c >0 and c <0, respectively. • Using (12), and (B) and (C), Eq. (18) gives
u18,19(x, t) = a0+a2ns2(λξ),
respectively, where a0, a2 are defined in (B) and (C). As k → 1, u18,19 become u14
and u15 and as k→0, we obtainu14, ..., u17, forc >0 and c <0, respectively. • Using (7), and (B) and (C), respectively, Eq. (18) gives
forc > 1
u20(x, t) = 1−c+ 3 (1−c) tan
r c−1
a−bc(x−ct)±sec r
c−1
a−bc(x−ct) !2
,
u21(x, t) = 3 (c−1)
1 + itanh r
c−1
a−bc(x−ct)±sech r
c−1
a−bc(x−ct) !2
.
forc < 1
u22(x, t) = 1−c+ 3 (1−c) itanh
r
1−c
a−bc(x−ct)±sech r
1−c
a−bc(x−ct) !2
,
u23(x, t) = 3 (c−1)
1 + tan r
1−c
a−bc(x−ct)±sec r
1−c
a−bc(x−ct) !2
.
3.2. Modified Pade0-II Equation. In this section, we browse our proposed equation, namely, a modified Pade0-II equation as the form
(19) ut+ux+u2ux+auxxx+buxxt = 0,
where u=u(x, t), a, b are real numbers.
Now, we apply the mapping method to solve our equation, as a consequence, we get the original solutions for our new equation, as the follows
Substituting u(x, t) =u(ξ) , ξ=λ(x−ct),into Eq. (19) and integrating once yields
(20) (1−c)u+ u
3
3 +λ
2(a−bc)u00
Balancing the order of the nonlinear termu3with the highest derivativeu00 gives 3m=m+ 2
that gives m= 1. Thus, the solution of (20) has the form
(21) u(ξ) = a0+a1f(ξ) +b1(f(ξ))
−1
Substituting (21) in (20) and using (4), collecting the coefficients of each power of fi,0 ≤
i≤6,setting each coefficient to zero, and solving the resulting system, obtain the following sets of solutions.
(1) a0 = √
−3 + 3c , a1 =b1 = 0 , λ=λ
(2) a0 = 0 , a1 =
q
3pq(1−c), b1 = 0 , λ=±
q
c−1
p(a−bc)
(3) a0 = 0 , a1 =−
q
3pq(1−c), b1 = 0 , λ=±
q
c−1
p(a−bc)
(4) a0 = 0 , a1 = 0 , b1 =
q
6r
p (1−c), λ=±
q
c−1
p(a−bc)
(5) a0 = 0 , a1 = 0 , b1 =−
q
6r
p (1−c), λ=±
q
c−1
p(a−bc)
(6) a0 = 0, a1 =±
(c−1)[p(a−bc)(3√2qr−p)−1]
(a−bc)(18qr−p2)
r
6r(c−1)3
√
2qr−p p2−18qr
,
b1 =∓
q
6r(c−1)3
√
2qr−p p2−18qr, λ=
q
18qr(c−1)(a−bc)(p2−1)(p−3√2qr)
(a−bc)(18qr−p2)
(7) a0 = 0, a1 =±
(c−1)[p(a−bc)(3√2qr−p)−1]
(a−bc)(18qr−p2)
r
6r(c−1)3
√
2qr−p p2−18qr
,
b1 =∓
q
6r(c−1)3
√
2qr−p
p2−18qr, λ=−
q
18qr(c−1)(a−bc)(p2−1)(p−3√2qr)
(a−bc)(18qr−p2)
Using (21), the solution of Eq. (4) when p = 1, q = −2, r = 0, and the sets of solutions 2-7, we get
u1(x, t) =± √
−3 + 3c.
for c >1 and a > bc u2(x, t) =±
√
−6 + 6csechqac−−bc1 (x−ct).
for c <1 and a < bc u3(x, t) =±i
√
6−6csechqbc1−−ca(x−ct).
for c <1 and a > bc u4(x, t) =±i
p
6 (1−c) secqa1−−bcc (x−ct).
for c >1 and a < bc u5(x, t) =±
p
6 (c−1) sec
q
c−1
bc−a(x−ct).
Using Eq. (21), the solution of (4) when p = −2, q = 2, r = 1, and the sets of solutions 2-7, we get for c <1 and a < bc
u6(x, t) =± √
3−3ctan
1
√
2
q
1−c
bc−a(x−ct)
,
u7(x, t) =± √
3−3ccot√1
2
q
1−c
bc−a(x−ct)
,
u8(x, t) =±
q
3
2(1−c)
h
tanh12qbc1−−ca(x−ct)−coth21qbc1−−ca(x−ct)i,
u9(x, t) =±12
p
3 (1−c)htan12
q
1 2
1−c bc−a
(x−ct)−cot12
q
1 2
1−c bc−a
for c >1 and a > bc u10(x, t) = ±i
√
−3 + 3ctan
1
√
2
q
c−1
a−bc(x−ct)
,
u11(x, t) = ±i √
−3 + 3ccot√1
2
q
c−1
a−bc(x−ct)
,
u12(x, t) = ±i
q
3
2 (c−1)
h
tanh12qac−−bc1 (x−ct)−coth21qac−−bc1 (x−ct)i,
u13(x, t) = ±12i
p
3 (c−1)htan12q21 ac−−bc1(x−ct)−cot12q12 ac−−bc1(x−ct)i.
for c >1 and a < bc u14(x, t) = ±
√
−3 + 3ctanh
1
√
2
q
c−1
bc−a(x−ct)
,
u15(x, t) = ± √
−3 + 3ccoth√1
2
q
c−1
bc−a(x−ct)
,
u16(x, t) = ±
q
3
2(c−1)
h
tan12qbcc−−1a(x−ct)+ cot21qbcc−−1a(x−ct)i,
u17(x, t) = ±12
p
3 (c−1)htanh12q21 bcc−−1a(x−ct)+ coth12q12 bcc−−1a(x−ct)i.
for c <1 and a > bc u18(x, t) = ±i
√
3−3ctanh√1
2
q
1−c
a−bc(x−ct)
,
u19(x, t) = ±i √
3−3ccoth√1
2
q
1−c
a−bc(x−ct)
,
u20(x, t) = ±i
q
3
2 (1−c)
h
tan12qa1−−bcc (x−ct)+ cot21qa1−−bcc (x−ct)i,
u21(x, t) = ±12i
p
3 (1−c)htanh 1
2√2
q
1−c
a−bc(x−ct)
+ coth 1
2√2
q
1−c
a−bc(x−ct)
i .
Using Eq.. (21), the solution of (4) when p =−(k2+ 1), q = 2k2, r = 1, and the sets of
solutions 2-7, we get
u22,23,...,27(x, t) = a0 +a1snλξ+b1nsλξ, where a0, a1 and b1 are defined in the sets of
solutions 2-7.
Note that, when k →0 we obtain constant solution, when k →1 we get, [u6, u7, ..., u21].
Using Eq. (21), the solution of (4) when p = 2k2 −1, q =−2k2, r = 1−k2, and the sets of solutions 2-7, we get
u28,29...,33(x, t) = a0 +a1cnλξ +b1ncλξ, where a0, a1 and b1 are defined in the sets of
solutions 2-7.
When k →0 we obtain constant solution, when k →1 we get, [u2, u3, ..., u5].
Using Eq. (21), the solution of (4) when p= 2−k2, q =−2, r=−(1−k2), and the sets
of solutions 2-7, we get
u34,35,...,39(x, t) = a0 +a1dnλξ+b1ndλξ, where a0, a1 and b1 are defined in the sets of
solutions 2-7.
As k →0 we obtain constant solutions, ask →1 we get, [u2, u3, ..., u5].
Using Eq. (21), the solution of (4) when p = 2−k2, q = 2, r = 1−k2, and the sets of solutions 2-7, we get
u40,41,...,45(x, t) = a0 +a1csλξ +b1scλξ, where a0, a1 and b1 are defined in the sets of
solutions 2-7.
Using Eq. (21), the solution of (4) when p = −(1 +k2), q = 2, r = k2, and the sets of
solutions 2-7, we get
u46,47,...,51(x, t) = a0 +a1nsλξ+b1snλξ, where a0, a1 and b1 are defined in the sets of
solutions 2-7.
As k →0 we obtain constant solution, ask →1 we get, [u6, u7, ..., u21].
Using Eq. (21), the solution of (4) whenp= 12, q= 12, r = 14, and the sets of solutions 2-7, we get
forc <1 and a < bc
u52(x, t) = √
3−3c tan √2
r
1−c
bc−a(x−ct) !
±sec √2
r
1−c
bc−a(x−ct) !!
,
u53(x, t) =− √
3−3c tan √2
r
1−c
bc−a(x−ct) !
±sec √2
r
1−c
bc−a(x−ct) !!
,
u54(x, t) =
√
3−3c
tan
√
2qbc1−−ca(x−ct)
±sec
√
2qbc1−−ca(x−ct)
,
u55(x, t) =
−√3−3c
tan√2qbc1−−ca(x−ct)±sec√2qbc1−−ca(x−ct)
,
u56(x, t) = −
r
3 2
√
1−c tanh
r
1−c
bc−a(x−ct) !
∓isech
r
1−c
bc−a(x−ct) !!
+
q
3 2
√
1−c
tanhqbc1−−ca(x−ct)∓isechqbc1−−ca(x−ct)
,
u57(x, t) =
r
3 2
√
1−c tanh
r
1−c
bc−a(x−ct) !
∓isech
r
1−c
bc−a(x−ct) !!
−
q
3 2
√
1−c
tanhqbc1−−ca(x−ct)∓isechqbc1−−ca(x−ct)
,
u58(x, t) = √
3 2
√
1−c tan √1
2
r
1−c
bc−a(x−ct) !
±sec √1
2
r
1−c
bc−a(x−ct) !!
−
√
3 2
√
1−c
tan√1
2
q
1−c
bc−a(x−ct)
±sec√1
2
q
1−c
bc−a(x−ct)
,
u59(x, t) =− √
3 2
√
1−c tan √1
2
r
1−c
bc−a(x−ct) !
±sec √1
2
r
1−c
bc−a(x−ct) !!
√
3 2
√
1−c
tan√1
2
q
1−c
bc−a(x−ct)
±sec√1
2
q
1−c
bc−a(x−ct)
.
for c >1 and a > bc
u60(x, t) =i √
−3 + 3c tan √2
r c−1
a−bc(x−ct) !
±sec √2
r c−1
a−bc(x−ct) !!
,
u61(x, t) =−i √
−3 + 3c tan √2
r c−1
a−bc(x−ct) !
±sec √2
r c−1
a−bc(x−ct) !!
,
u62(x, t) =
i√−3 + 3c
tan√2qac−−bc1 (x−ct)±sec√2qac−−bc1 (x−ct)
,
u63(x, t) =
−i√−3 + 3c
tan√2qac−−bc1 (x−ct)±sec√2qac−−bc1 (x−ct)
,
u64(x, t) =
r
3 2
√
−1 +c itanh
r c−1
a−bc(x−ct) !
±sech
r c−1
a−bc(x−ct) !! + q 3 2 √
−1 +c
itanhqac−−bc1 (x−ct)±sechqac−−bc1 (x−ct)
,
u65(x, t) = −
r
3 2
√
−1 +c itanh
r c−1
a−bc(x−ct) !
±sech
r c−1
a−bc(x−ct) !! − q 3 2 √
−1 +c
itanhqbc1−−ca(x−ct)±sechqbc1−−ca(x−ct)
,
u66(x, t) = i √
3 2
√
c−1 tan √1
2
r c−1
a−bc(x−ct) !
±sec √1
2
r c−1
a−bc(x−ct) !! − i √ 3 2 √
1−c
tan√1 2
q
c−1
a−bc(x−ct)
±sec√1 2
q
c−1
a−bc(x−ct)
,
u67(x, t) =−i √
3 2
√
c−1 tan √1
2
r c−1
a−bc(x−ct) !
±sec √1
2
r c−1
a−bc(x−ct) !! + i √ 3 2 √
c−1
tan√1
2
q
c−1
a−bc(x−ct)
±sec√1
2
q
c−1
a−bc(x−ct)
for c >1 and a < bc
u68(x, t) = √
−3 + 3c tanh √2
r c−1
bc−a(x−ct) !
∓isech √2
r c−1
bc−a(x−ct) !!
,
u69(x, t) =− √
−3 + 3c tanh √2
r c−1
bc−a(x−ct) !
∓isech √2
r c−1
bc−a(x−ct) !!
,
u70(x, t) =
√
−3 + 3c
tanh√2qbcc−−1a(x−ct)±isech√2qbcc−−1a(x−ct)
,
u71(x, t) =
−√−3 + 3c
tanh√2qbcc−−1a(x−ct)±isech√2qbcc−−1a(x−ct)
,
u72(x, t) =
r
3 2
√
−1 +c tan
r c−1
bc−a(x−ct) !
±sec
r c−1
bc−a(x−ct) !! + q 3 2 √
−1 +c
tan
q
c−1
bc−a(x−ct)
±sec
q
c−1
bc−a(x−ct)
,
u73(x, t) = −
r
3 2
√
−1 +c tan
r c−1
bc−a(x−ct) !
±sec
r c−1
bc−a(x−ct) !! − q 3 2 √
−1 +c
tanqbcc−−1a(x−ct)±secqbcc−−1a(x−ct)
,
u74(x, t) = − √
3 2
√
c−1 tanh √1
2
r c−1
bc−a(x−ct) !
∓isech √1
2
r c−1
bc−a(x−ct) !! − √ 3 2 √
c−1
tanh 1 √ 2 q
c−1
bc−a(x−ct)
∓isech
1
√
2
q
c−1
bc−a(x−ct)
,
u75(x, t) = √
3 2
√
c−1 tanh √1
2
r c−1
bc−a(x−ct) !
∓isech √1
2
r c−1
bc−a(x−ct) !! + √ 3 2 √
c−1
tanh√1
2
q
c−1
bc−a(x−ct)
∓isech√1
2
q
c−1
bc−a(x−ct)
.
for c <1 and a > bc
u76(x, t) = √
3−3c itanh √2
r
1−c
a−bc(x−ct) !
±sech √2
r
1−c
a−bc(x−ct) !!
,
u77(x, t) =− √
3−3c itanh √2
r
1−c
bc−a(x−ct) !
±sech √2
r
1−c
a−bc(x−ct) !!
u78(x, t) =
√
3−3c
itanh
√
2
q
1−c
a−bc(x−ct)
±sech
√
2
q
1−c
a−bc(x−ct)
,
u79(x, t) =
−√3−3c
itanh√2qa1−−bcc (x−ct)±sech√2qa1−−bcc (x−ct)
,
u80(x, t) = i
r
3 2
√
1−c tan
r
1−c
a−bc(x−ct) !
±sec
r
1−c
a−bc(x−ct) !! + i q 3 2 √
1−c
tanqa1−−bcc (x−ct)±secqa1−−bcc (x−ct)
,
u81(x, t) = −i
r
3 2
√
1−c tan
r
1−c
a−bc(x−ct) !
±sec
r
1−c
a−bc(x−ct) !! − i q 3 2 √
1−c
tanqa1−−bcc (x−ct)±secqa1−−bcc (x−ct)
,
u82(x, t) = √
3 2
√
1−c itanh √1
2
r
1−c
a−bc(x−ct) !
±sech √1
2
r
1−c
a−bc(x−ct) !! − √ 3 2 √
1−c
itanh√1
2
q
1−c
a−bc(x−ct)
±sech√1
2
q
1−c
a−bc(x−ct)
,
u83(x, t) = − √
3 2
√
1−c itanh √1
2
r
1−c
a−bc(x−ct) !
±sech √1
2
r
1−c
a−bc(x−ct) !! + √ 3 2 √
1−c
itanh√1
2
q
1−c
a−bc(x−ct)
±sech√1
2
q
1−c
a−bc(x−ct)
.
3.3. The (2+1) Dimensional KD Equation. In this section, we will solve the nonlinear
(2+1) dimensional Konopelchenko-Dubrovsky (KD) equation of the form
uy = vx,
ut−uxxx−6buux+ 3 2a
2u2u
x−3vy+ 3auxv = 0, (22)
where u=u(x, y, t), v =v(x, y, t) and a, bare real numbers.
We apply the mapping method, to solve the nonlinear (2+1) dimensional Konopelchenko-Dubrovsky (KD) equation. Substitutingu(x, y, t) =u(ξ), v(x, y, t) =v(ξ), ξ =λ(x+y−βt),
into Eq. (16) and integrating once yields
u = v,
−βu−λ2u00−3bu2+ a
2
6u
3−3u+3
2au
2 = 0.
(23)
Balancing the order of the nonlinear termu3with the highest derivativeu00 gives 3m=m+ 2 that gives m= 1. Thus, the solution of (23) has the form
(24) u(ξ) = v(ξ) = a0+a1f(ξ) +b1(f(ξ))
−1
.
Substituting (24) in (23) and using (4), collecting the coefficients of each power of fi,0 ≤
i≤6,setting each coefficient to zero, and solving the resulting system, obtain the following sets of solutions.
(1) a0 =−a−a22b, a1 = 0, b1 =±
q
−2r p (a−2b)
a2 , λ=
q
−1 2p(a−2b)
a , β =−4
(a2−ab+b2)
a2
(2) a0 =−a−a22b, a1 = 0, b1 =±
q
−2r p (a−2b)
a2 , λ=−
q
−1 2p(a−2b)
a , β =−4
(a2−ab+b2)
a2
(3) a0 =−a−a22b, a1 =±
q−
q p(a−2b)
a2 , b1 = 0, λ=
q−
1 2p(a−2b)
a , β =−4
(a2−ab+b2)
a2
(4) a0 =−a−a22b, a1 =±
q−
q p(a−2b)
a2 , b1 = 0, λ=−
q
−1 2p(a−2b)
a , β =−4
(a2−ab+b2)
a2
(5) a0 =−a−a22b, a1 =±
r
pq+3q√2qr
18qr−p2 (a−2b)
a2 , b1 =±13
p(p+3√2qr)(a−2b) 18qr−p2 +(a−2b)
a2
r
pq+3q√2qr
18qr−p2
,
λ= √1
2a q
(18qr−p2)(p+3√2qr)(a−2b)
18qr−p2 , β =−4
(a2−ab+b2)
a2
(6) a0 =−a−a22b, a1 =±
r
pq+3q√2qr
18qr−p2 (a−2b)
a2 , b1 =±13
p(p+3√2qr)(a−2b) 18qr−p2 +(a−2b)
a2
r
pq+3q√2qr
18qr−p2 ,
λ=−√1 2a
q
(18qr−p2)(p+3√2qr)(a−2b)
18qr−p2 , β =−4
(a2−ab+b2)
a2
Using Eq.. (24), the solution of (4) when p = 1, q = −2, r = 0, and the above sets of solutions, we get
u1(x, y, t) =a0, (constant solution),
u2(x, y, t) =−
a−2b a2 ±
√
2 (a−2b)
a2 sec
1
√
2
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
.
Note thatu(x, y, t) =v(x, y, t), for all cases.
Using Eq. (24), the solution of (4) when p = −2, q = 2, r = 1, and the sets of solutions 1-6, we get
u3(x, y, t) = −
a−2b a2 ±
a−2b a2 tanh
1 2
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
,
u4(x, y, t) = −
a−2b a2 ±
a−2b a2 coth
1 2
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
u5(x, y, t) = −
a−2b a2 ±
a−2b
2a2 tanh
1 4
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
±a−2b
2a2 coth
1 4
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
,
u6(x, y, t) = −
a−2b a2 ±
√
2 (a−2b)
2a2 tan
√
2 4
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
!
± √
2 (a−2b)
2a2 cot
√
2 4
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
! .
Using Eq.. (24), the solution of (4) when p =−(1 +k2), q = 2k2, r = 1, and the sets of solutions 1-6, we get
u7,...,12(x, y, t) =a0 +a1sn (λξ) +b1ns (λξ),
where a0, a1 and b1 are defined in the sets of solutions 1-6. Note that, as k→0, we obtain
u13(x, y, t) = −
a−2b a2 ±
√
2 (a−2b)
a2 csc
1
√
2
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
.
As k →1, we get u3, u4..., u6.
Using Eq. (24), the solution of (4) when p= 2k2−1, q=−2k2, r = 1−k2 and the sets of solutions 1-6, we obtain
u14,...,119(x, y, t) = a0+a1cn (λξ) +b1nc (λξ),
where a0, a1 and b1 are defined in the sets of solutions 1-6. Whenk →0,we obtain u2, also
we get u2 when k →1.
Using Eq. (24), the solution of (4) when p = 2−k2, q = −2, r = −(1−k2), and above
sets of solutions 1-6, we get
u20,...,25(x, y, t) = a0+a1dn (λξ) +b1nd (λξ),
where a0, a1 and b1 are defined in the sets of solutions 1-6. As k →0, we get u1 and v1, as
k →1, we obtain u2 .
Using Eq. (24), the solution of (4) when p= 2−k2, q = 2, r = (1−k2), and the sets of
solutions 1-6, we get
u26,...,31(x, y, t) =a0+a1cs (λξ) +b1sc (λξ),
wherea0, a1andb1 are defined in the sets of solutions 1-6. Ask →0,we obtain, [u3, u4..., u6],
as k→1, we get u13.
Using Eq. (24), the solution of (4) when p = −(1 +k2), q = 2, r = k2, and the sets of
u32,...,37(x, y, t) = a0+a1dc (λξ) +b1cd (λξ),
where a0, a1 and b1 are defined in the sets of solutions 1-6. As k → 0, we obtain u2, as
k →1, we get u1.
Using Eq. (24), the solution of (4) when p = −1 + 2k2, q = 2, r = −k2(1−k2) and the
sets of solutions 1-6, we get
u38,...,43(x, y, t) =a0+a1ds (λξ) +b1sd (λξ),
where a0, a1 and b1 are defined in the sets of solutions 1-6. As k → 0, we obtain u13, as
k →1, we get also u13 and.
Using Eq. (24), the solution of (4) when p= 0, q = 2, r= 0,and the sets of solutions 1-6, we get u1.
Using Eq. (24), the solution of (4) when p= 1, q = 0, r= 0,and the sets of solutions 1-6, we obtainu1.
Using Eq. (24), the solution of (4) whenp= 12, q= 12, r = 14, and the sets of solutions 1-6, we get
u44(x, y, t) = −
a−2b a2 +
(a−2b)
√
2a2
±tan
a−2b
√
2a
x+y+ 4(a
2−ab+b2)
a2 t
+(a√−2b)
2a2 sec
a−2b
√
2a
x+y+ 4(a
2−ab+b2)
a2 t
+
(a−2b)
√
2a2±tana√−2b
2a
x+y+ 4(a2−aab2+b2)t
+ seca√−2b
2a
x+y+ 4(a2−aab2+b2)t ,
u45(x, y, t) = −
a−2b a2 −
(a−2b)
√
2a2
±tan
a−2b
√
2a
x+y+ 4(a
2−ab+b2)
a2 t
−(a√−2b)
2a2 sec
a−2b
√
2a
x+y+ 4(a
2−ab+b2)
a2 t
−√ (a−2b)
2a2±tana√−2b
2a
x+y+ 4(a2−aab2+b2)t
+ seca√−2b
2a
x+y+ 4(a2−aab2+b2)t ,
u46(x, y, t) = −
a−2b a2 +
(a−2b)
a2 tanh
a−2b
a x+y+ 4
(a2 −ab+b2)
a2 t
±i(a−2b)
a2 sech
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
,
u47(x, y, t) = −
a−2b a2 −
(a−2b)
a2 tanh
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
±i(a−2b)
a2 sech
a−2b
a x+y+ 4
(a2−ab+b2)
a2 t
u48(x, y, t) =−
a−2b a2 +
(a−2b)
a2±tanh a−2b a
x+y+ 4(a2−aab2+b2)t
+isech
a−2b a
x+y+ 4(a2−aab2+b2)t ,
u49(x, y, t) =−
a−2b a2 −
(a−2b)
a2±tanh a−2b a
x+y+ 4(a2−aab2+b2)t
+isech
a−2b a
x+y+ 4(a2−aab2+b2)t ,
u50(x, y, t) = −
a−2b a2 −
(a−2b) 2a2
±tanh
a−2b
2a
x+y+ 4(a
2−ab+b2)
a2 t
−(a−2b)
2a2 isech
a−2b
2a
x+y+ 4(a
2 −ab+b2)
a2 t
− (a−2b)
2a2±tanha−2b
2a
x+y+ 4(a2−aab2+b2)t
+isecha−2a2bx+y+ 4(a2−aab2+b2)t ,
u51(x, y, t) = −
a−2b a2 +
(a−2b) 2a2
±tanh
a−2b
2a
x+y+ 4(a
2−ab+b2)
a2 t
+(a−2b) 2a2 isech
a−2b
2a
x+y+ 4(a
2−ab+b2)
a2 t
+ (a−2b)
2a2±tanha−2b
2a
x+y+ 4(a2−aab2+b2)t
+isecha−2a2bx+y+ 4(a2−aab2+b2)t .
Note that, if we replace sec by −sec and sech by −sech in u44, ..., u51,we obtain also true
solutions.
By comparing our results with the results in [17], it can be seen that some of the obtained results are new, and the rest solutions are the same.
4. Conclusion
Acknowledgements
This project was founded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant no (306/130/1431). The authors, therefore, acknowledge with thanks DSR technical and financial support.
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