The Generalized Projective Riccati Equations Method and its Applications to Nonlinear ... · 2015. 11. 25. · ear PDEs describing nonlinear transmission lines (NLTL). The ﬁrst

Communications on Applied Electronics (CAE) - ISSN : 2394 - 4714Foundation of Computer Science FCS, New York, USAVolume 3 - No. 4, November 2015 - www.caeaccess.org

The Generalized Projective Riccati Equations Methodand its Applications to Nonlinear PDEs Describing

Nonlinear Transmission Lines

E.M.E. ZayedDepartment of Mathematics

Zagazig UniversityP.O.Box 44519, Zagazig, Egypt

K.A.E. AlurrfiDepartment of Mathematics

Zagazig UniversityP.O.Box 44519, Zagazig, Egypt

ABSTRACTIn this article, we apply the generalized projective Riccati equa-tions method with the aid of symbolic computation to constructnew exact traveling wave solutions with parameters for two nonlin-ear PDEs describing nonlinear transmission lines (NLTL). The firstequation describes the model of governing wave propagation in theNLTL as nonlinear low-pass electrical lines. The second equationdescribes pulse narrowing nonlinear transmission lines. The ob-tained solutions include, kink and anti-kink solitons, bell (bright)and anti-bell (dark) solitary wave solutions, hyperbolic solutionsand trigonometric solutions. Based on Kirchhoff’s current law andKirchhoff’s voltage law, the given nonlinear PDEs have been de-rived and can be reduced to nonlinear ordinary differential equa-tions (ODEs) using a simple transformation. The given method inthis article is straightforward and concise, and it can also be appliedto other nonlinear PDEs in mathematical physics.

KeywordsGeneralized projective Riccati equations method, Exact solu-tions, Nonlinear low-pass electrical lines, Pulse narrowing non-linear transmission lines, Kirchhos lawsSS

1. INTRODUCTIONIn the recent years, investigations of exact solutions to nonlinearPDEs play an important role in the study of nonlinear physicalphenomena in such as fluid mechanics, hydrodynamics, optics,plasma physics, solid state physics, biology and so on. Severalmethods for finding the exact solutions to nonlinear equations inmathematical physics have been presented, such as the inversescattering method [1], the Hirota bilinear transform method [2],the truncated Painlevé expansion method [3-6], the Bäcklundtransform method [7,8], the exp-function method [9-11], thetanh-function method [12,13], the Jacobi elliptic function expan-sion method [14-16], the (G

′

G)-expansion method [17-22], the

modified (G′

G)-expansion method [23], the (G

′

G, 1G)-expansion

method [24-27], the modified simple equation method [28-30],the multiple exp-function algorithm method [31,32], the trans-formed rational function method [33], the local fractional seriesexpansion method [34], the first integral method [35,36],the gen-eralized Riccati equation mapping method [37,38], the gener-alized projective Riccati equations method [39-44] and so on.Conte and Musette [39] presented an indirect method to seekmore solitary wave solutions of some NPDEs that can be ex-pressed as polynomials in two elementary functions which sat-isfy a projective Riccati equation [45]. Using this method, manysolitary wave solutions of many NPDEs are found [42,45]. Re-cently, Yan [43] developed further Conte and Musette’s methodby introducing more generalized projective Riccati equations.The objective of this article is to use the generalized projectiveRiccati equations method to construct the exact solutions of thefollowing two nonlinear PDEs:

(1) The nonlineae PDE governing wave propagation in nonlinearlow-pass electrical transmission lines [46]:

∂2V (x, t)

∂t2− α∂

2V 2(x, t)

∂t2+ β

∂2V 3(x, t)

∂t2

−δ2 ∂2V (x, t)

∂x2− δ

4

12

∂4V (x, t)

∂x4= 0, (1.1)

where α, β and δ are constants, while V (x, t) is the voltagein the transmission lines. The variable x is interpreted as thepropagation distance and t is the slow time. The physical detailsof the derivation of Eq. (1.1) using the Kirchhoff’s laws aregiven in [46], which are omitted here for simplicity. Note thatEq. (1.1) has been discussed in [46] using an auxiliary equationmethod [47] and its exact solutions have been found. Also,this equation have been studied in [48] using the new Jacobielliptic function expansion method and its exact traveling wavesolutions have been obtained.

(2) The nonlinear PDE describing pulse narrowing nonlin-ear transmission lines [49]:

∂2φ(x, t)

∂t2− 1LC0

∂2φ(x, t)

∂x2− b1

2

∂2φ2(x, t)

∂t2

− δ2

12LC0

∂4φ(x, t)

∂x4= 0, (1.2)

where φ(x, t) is the voltage of the pulse and C0, L, δ and b1 areconstants. The physical details of the derivation of Eq. (1.2) iselaborated in [49] using the Kirchhoff’s current law and Kirch-hoff’s voltage law, which are omitted here for simplicity. It iswell-known [49] that Eq. (1.2) has the solution:

φ(x, t) =3(v2−v20)b1v2

sech2[√

3(v2−v20)v0

((x−vt)δ

)], (1.3)

where v is the propagation velocity of the pulse and v0 = 1√LC0provided v > v0 and LC0 > 0. Recently Zayed and Alurrfihave discussed Eq. (1.2) in [50] using the new Jacobi ellipticfunction expansion method and determined its exact travelingwave solutions.This article is organized as follows: In Sec. 2, the description ofthe generalized projective Riccati equations method is given. InSec. 3, we use the given method described in Sec. 2, to find exactsolutions of Eqs. (1.1) and (1.2). In Sec. 4, physical explanationsof some results are presented. In Sec. 5, some conclusions areobtained.

1


2. DESCRIPTION OF THE GENERALIZEDPROJECTIVE RICCATI EQUATIONSMETHOD

Consider a nonlinear PDE in the form

P (u, ux, ut, uxx, utt, ...) = 0, (2.1)

where u = u(x, t) is an unknown function, P is a polynomialin u(x, t) and its partial derivatives in which the highest orderderivatives and nonlinear terms are involved. Let us now givethe main steps of the generalized projective Riccati equationsmethod [39-44]:

Step 1. We use the following transformation:

u(x, t) = u(ξ), ξ = x− vt, (2.2)

where v is velocity of the propagation , to reduce Eq. (2.1) to thefollowing nonlinear (ODE):

H(u, u′, u′′, ...) = 0, (2.3)

where H is a polynomial of u(ξ) and its total derivatives

u′(ξ), u′′(ξ), ... and ′ =d

dξ.

Step 2. We suppose that the solution of Eq. (2.3) has theform:

u(ξ) = A0 +

N∑i=1

σi−1(ξ) [Aiσ(ξ) +Biτ(ξ)] , (2.4)

where A0, Ai and Bi are constants to be determined later. Thefunctions σ(ξ) and τ(ξ) satisfy the ODEs:

σ′(ξ) = εσ(ξ)τ(ξ), (2.5)

τ ′(ξ) = R+ ετ2(ξ)− µσ(ξ), ε = ±1, (2.6)

where

τ2(ξ) = −ε(R− 2µσ(ξ) + µ

2 + r

Rσ2(ξ)

), (2.7)

where r = ±1 and R, µ are nonzero constants.

If R = µ = 0, Eq. (2.3) has the formal solution:

u(ξ) =

N∑i=0

Aiτi(ξ) (2.8)

where τ(ξ) satisfies the ODE:

τ ′(ξ) = τ2(ξ). (2.9)

Step 3. We determine the positive integer N in (2.4) by usingthe homogeneous balance between the highest-order derivativesand the nonlinear terms in Eq. (2.3).

Step 4. Substitute (2.4) along with Eqs. (2.5)-(2.7) intoEq.(2.3) or ((2.8) along with Eq. (2.9) into Eq. (2.3)). Collectingall terms of the same order of σj(ξ)τ i(ξ) (j = 0, 1,...; i = 0, 1)(or τ j(ξ), j = 0, 1, ... ). Setting each coefficient to zero, yields aset of algebraic equations which can be solved to find the valuesof A0, Ai, Bi, v, µ and R.

Step 5. It is well known [41,44] that Eqs. (2.5) and (2.6)admits the following solutions:

Case 1. When ε = −1, r = −1, R > 0,

σ1(ξ) =Rsech(

√Rξ)

µsech(√Rξ)+1

, τ1(ξ) =√R tanh(

√Rξ)

µsech(√Rξ)+1

, (2.10)

Case 2. When ε = −1, r = 1, R > 0,

σ2(ξ) =Rcsch(

√Rξ)

µcsch(√Rξ)+1

, τ2(ξ) =√R coth(

√Rξ)

µcsch(√Rξ)+1

. (2.11)

Case 3. When ε = 1, r = −1, R > 0,

σ3(ξ) =R sec(

√Rξ)

µ sec(√Rξ)+1

, τ3(ξ) =√R tan(

√Rξ)

µ sec(√Rξ)+1

, (2.12)

σ4(ξ) =R csc(

√Rξ)

µ csc(√Rξ)+1

, τ4(ξ) = −√R cot(

√Rξ)

µ csc(√Rξ)+1

. (2.13)

Case 4. when R = µ = 0,

σ5(ξ) =C

ξ, τ5(ξ) =

1

εξ, (2.14)

where C is a nonzero constant.

Step 6. Substituting the values of A0, Ai, Bi, v, µ and Ras well as the solutions (2.10)-(2.14) into (2.4) we obtain theexact solutions of Eq. (2.1).

3. EXACT SOLUTIONS OF EQUATIONS (1.1)AND (1.2) USING THE GIVEN METHOD OFSEC. 2

In this section, we apply the generalized projective Riccati equa-tions method of Sec. 2 to find families of new exact solutions ofEqs. (1.1) and (1.2).

3.1 Exact solutions of the nonlinear PDE (1.1)In this subsection, we find the exact wave solutions of Eq. (1.1).To this end, we use the transformation

V (x, t) = V (ξ), ξ =√k (x− vt) , (3.1)

to reduce Eq. (1.1) to the following nonlinear ODE:

d2

dξ2

{k2δ4

12d2Vdξ2

+ (kδ2 − kv2)V + αkv2V 2 − βkv2V 3}= 0.

(3.2)

Integrating Eq. (3.2) twice and vanishing the constants ofintegration, we find the following ODE:

K2

12d2Vdξ2

+ (K − U)V + αUV 2 − βUV 3 = 0. )(3.3)

where K = kδ2 and U = kv2.Balancing d

2Vdξ2

with V 3 gives N = 1. Therefore, (2.4) reducesto

V (ξ) = A0 +A1σ(ξ) +B1τ(ξ), (3.4)

where A0, A1 and B1 are constants to be determined such thatA1 6= 0 or B1 6= 0.Substituting (3.4) and using (2.5)-(2.7) into Eq. (3.3), the

2


left-hand side of Eq. (3.3) becomes a polynomial in σ(ξ) andτ(ξ). Setting the coefficients of this polynomial to be zero,yields the following system of algebraic equations:

σ3 : −UβA31 − 1Rε(16K2ε2A1 − 3UβA1B21

)(µ2 + r) = 0,

σ2 : − εR(UαB21 − 3UβA0B21) (µ2 + r)− 112K

2µεA1− 1

12K2µεA1 − 3UβA0A21 = 0,

+UαA21 = 0,

σ2τ : 1Rε(UβB31 − 16K

2ε2B1)(µ2 + r)− 3UβA21B1 = 0,

σ : A1 (K − U)− εR(16K2ε2A1 − 3UβA1B21

)+2εµ (UαB21 − 3UβA0B21)− 3UβA20A1+2UαA0A1 +

112K2RεA1 = 0,

στ : −2µε(UβB31 − 16K

2ε2B1)− 1

4K2µεB1

+2UαA1B1 − 6UβA0A1B1 = 0,

τ : B1 (K − U) +Rε(UβB31 − 16K

2ε2B1)

−3UβA20B1 + 2UαA0B1 + 16K2RεB1 = 0,

σ0 : A0 (K − U)−Rε (UαB21 − 3UβA0B21)+UαA20 − UβA30 = 0. (3.5)

Case 1. If we substitute ε = −1 into the algebraic equations(3.5) and solve them by Maple 14, we have the following results:

Result 1. We have

K = − 24α2(µ2+r)

(−9βµ2+2rα2+2µ2α2)R , U =216α2βµ2(µ2+r)

(−9βµ2+2rα2+2µ2α2)2R ,

A0 = 0, A1 =2α(µ2+r)

3βµR, B1 = 0. (3.6)

From (2.10), (2.11), (3.4) and (3.6), we deduce that if r = −1,then we have the exact wave solution

V (ξ) = 2α(µ2−1)

3βµR

[Rsech(

√Rξ)

µsech(√Rξ)+1

], (3.7)

whereξ =

√− 24α

2(µ2−1)δ2(−9βµ2−2α2+2µ2α2)Rx−

√216α2βµ2(µ2−1)

(−9βµ2−2α2+2µ2α2)2R t,provided that β(µ2 − 1) > 0 and 9βµ2 > 2α2(µ2 − 1).

while if r = 1, then we have the exact wave solution

V (ξ) = 2α(µ2+1)

3βµR

[Rcsch(

√Rξ)

µcsch(√Rξ)+1

], (3.8)

whereξ =

√− 24α2(µ2+1)δ2(−9βµ2+2α2+2µ2α2)Rx−

√216α2βµ2(µ2+1)

(−9βµ2+2α2+2µ2α2)2R t,provided that β > 0 and 9βµ2 > 2α2(µ2 + 1).

Result 2. We have

K = − 24α2(2α2−9β)R , U =

216α2β(2α2−9β)2R , A0 =

α

3β,

A1 = ±α√µ2+r

βR, B1 = ± α3β√R . (3.9)

In this case, we deduce that if r = −1, then we have the exactwave solution

V (ξ) = α3β

[1±√µ2−1sech(

√Rξ)+tanh(

√Rξ)

µsech(√Rξ)+1

], (3.10)


V (ξ) = α3β

[1±√µ2+1csch(

√Rξ)+coth(

√Rξ)

µcsch(√Rξ)+1

], (3.11)

whereξ =

√− 24α2δ2(2α2−9β)Rx −

√216α2β

(2α2−9β)2R t, provided that β > 0

and 9β > 2α2.

Case 2. If we substitute ε = 1 and r = −1 into the alge-braic equations (3.5) and solve them by Maple 14, we have thefollowing results:

Result 1. We have

K = 24α2(µ2−1)

(−9βµ2−2α2+2µ2α2)R , U = −216α2βµ2(µ2−1)

(−9βµ2−2α2+2µ2α2)2R ,

A0 = 0, A1 =2α(µ2−1)

3βµR, B1 = 0. (3.12)

From (2.12), (2.13), (3.4) and (3.12), we deduce the followingexact wave solutions

V (ξ) = 2α(µ2−1)

3βµR

[R sec(

√Rξ)

µ sec(√Rξ)+1

], (3.13)

or

V (ξ) = 2α(µ2−1)

3βµR

[R csc(

√Rξ)

µ csc(√Rξ)+1

], (3.14)

whereξ =

√24α2(µ2−1)

δ2(−9βµ2−2α2+2µ2α2)Rx−√− 216α

2βµ2(µ2−1)(−9βµ2−2α2+2µ2α2)2R t,

provided that β(µ2 − 1) < 0 and 9βµ2 > 2α2(µ2 − 1).

3.2 Exact solutions of the nonlinear PDE (1.2)In this subsection, we find the exact solutions of Eq. (1.2). Tothis end, we use the transformation (2.2) to reduce Eq. (1.2) tothe following nonlinear ODE:

φ′′(ξ) + k1φ(ξ) + k2φ2(ξ) = 0, (3.15)

where

k1 = −12(v2−v20)δ2v20

, k2 = 6b1v2

δ2v20. (3.16)

Balancing φ′′ with φ2 gives N = 2. Therefore, (2.4) re-duces to

φ(ξ) = A0 +A1σ(ξ) +A2σ2(ξ)

+B1τ(ξ) +B2σ(ξ)τ(ξ), (3.17)

where A0, A1 , A2, B1and B2 are constants to be determinedsuch that A2 6= 0 or B2 6= 0.Substituting (3.17) and using (2.5)-(2.7) into Eq. (3.15), theleft-hand side of Eq. (3.15) becomes a polynomial in σ(ξ) andτ(ξ). Setting the coefficients of this polynomial to be zero,

3


yields the following system of algebraic equations:

σ4 : A22k2 − 1Rε (µ2 + r) (6A2ε

2 + k2B22) = 0,

σ3 : 2εµ (6A2ε2 + k2B

22)− 2µεA2 + 2A1A2k2

− εR(µ2 + r) (2A1ε

2 + 2B1B2k2) = 0,

σ3τ : 2A2B2k2 − 6Rε3B2 (µ

2 + r) = 0,

σ2 : k2 (A21 + 2A0A2)− εR (6A2ε2 + k2B22)

+2εµ (2A1ε2 + 2B1B2k2)− εRB

21k2 (µ

2 + r)+A2k1 + 2RεA2 − µεA1 = 0,

σ2τ : ε(12µε2B2 − 2Rε

2B1 (µ2 + r)

)+2k2 (A1B2 +A2B1)− 6µεB2 = 0,

σ : −ε (R (2A1ε2 + 2B1B2k2)− 2µB21k2)+A1k1 +RεA1 + 2A0A1k2 = 0,

στ : B2k1 + ε (4µε2B1 − 6Rε2B2)− 3µεB1

+2k2 (A0B2 +A1B1) + 5RεB2 = 0,

τ : 2RB1ε− 2RB1ε3 +B1k1 + 2A0B1k2 = 0,

σ0 : A20k2+A0k1−RεB21k2 = 0. (3.18)

Case 1. If we substitute ε = −1 into the algebraic equations(3.18) and solve them by Maple 14, we have the followingresults:

Result 1. We have

R = −k1, A0 = 0, A1 =3µ

k2, A2 =

3(µ2 + r)

k1k2,

B1 = 0, B2 = ±3

k2

√− (µ

2 + r)

k1, (3.19)

where k1 < 0, µ2 + r > 0.From (2.10), (2.11), (3.17) and (3.19), we deduce that if r = −1,then we have the exact wave solution

φ(ξ) = −3k1sech(

√−k1ξ)

(µ+sech(

√−k1ξ)±

√µ2−1 tanh(

√−k1ξ)

)k2(µsech(

√−k1ξ)+1)

2 ,

(3.20)


φ(ξ) = −3k1csch(

√−k1ξ)

(µ−csch(

√−k1ξ)±

√µ2+1coth(

√−k1ξ)

)k2(µcsch(

√−k1ξ)+1)

2 ,

(3.21)

Result 2.

A0 = −k1k2, A1 =

3µ

k2, A2 = − 3(µ

2+r)k1k2

,

B1 = 0, B2 = ±3

k2

√(µ2 + r)

k1, R = k1. (3.22)

where k1 > 0, µ2 + r > 0In this case, we deduce that if r = −1, then we have the exactwave solution

φ(ξ) = − k1k2

(1 +

3sech(√k1ξ)

(−µ−sech(

√k1ξ)±

√µ2−1 tanh(

√k1ξ)

)(µsech(

√k1ξ)+1)

2

),

(3.23)


φ(ξ) = − k1k2

(1 +

3csch(√k1ξ)

(−µ+csch(

√k1ξ)±

√µ2+1coth(

√k1ξ)

)(µcsch(

√k1ξ)+1)

2

),

(3.24)

Case 2. If we substitute ε = 1 and r = −1 into the algebraicequations (3.18) and solve them by Maple 14, we have thefollowing results:

Result 1. We have

A0 = 0, A1 = −3µ

k2, A2 =

3(µ2 − 1)k1k2

,

B1 = 0, B2 = ±3

k2

√− (µ

2 − 1)k1

, R = k1, (3.25)

where k1 > 0, µ2 − 1 < 0.From (2.10), (2.11), (3.17) and (3.25), we deduce the followingexact wave solutions

φ(ξ) = −3k1 sec(

√k1ξ)

(µ+sec(

√k1ξ)±

√−(µ2−1) tan(

√k1ξ)

)k2(µ sec(

√k1ξ)+1)

2 ,

(3.26)

or

φ(ξ) = −3k1 csc(

√k1ξ)

(µ+csc(

√k1ξ)±

√−(µ2−1) cot(

√k1ξ)

)k2(µ csc(

√k1ξ)+1)

2 ,

(3.27)

Result 2.

A0 = −k1k2, A1 = −

3µ

k2, A2 = −

3(µ2 − 1)k1k2

,

B1 = 0, B2 = ±3

k2

√(µ2 − 1)k1

, R = −k1. (3.28)

where k1 < 0, µ2 − 1 < 0In this case, we deduce the following exact wave solutions

φ(ξ) = k1k2

(−1 +

3 sec(√−k1ξ)

(µ+sec(

√−k1ξ)±

√−(µ2−1) tan(

√−k1ξ)

)(µ sec(

√−k1ξ)+1)

2

),

(3.29)

or

φ(ξ) = − k1k2

(1 +

3csc(√−k1ξ)

(−µ−csc(

√−k1ξ)±

√−(µ2−1) cot(

√−k1ξ)

)(µ csc(

√−k1ξ)+1)

2

).

(3.30)

4. PHYSICAL EXPLANATIONS OF SOMERESULTS

Solitary waves can be obtained from each traveling wave solu-tion by setting particular values to its unknown parameters. Inthis section, we have presented some graphs of solitary wavesconstructed by taking suitable values of involved unknown pa-rameters to visualize the underlying mechanism of the originalequation. Using mathematical software Maple 14, three dimen-sional plots of some obtained exact traveling wave solutions havebeen shown in Figure 1- Figure. 6.

4


Fig. 1. The plot of the solution (3.7) when µ = 2, α = 1, δ = 1,β = 2, R = 1.

4.1 The nonlinear PDE (1.1) governing wavepropagation in nonlinear low-pass electricaltransmission lines

The obtained solutions for the nonlinear PDE (1.1) incorporatethree types of explicit solutions namely, hyperbolic and trigono-metric. From these explicit results it is easy to say that the so-lution (3.7) is a bell-shaped soliton solution; the solution (3.8)is a singular bell-shaped soliton solution; the solution (3.10) isa bell-kink shaped soliton solution; the solution (3.11) is a sin-gular bell-kink shaped soliton solution and the solutions (3.13),(3.14) are periodic solutions. The graphical representation of thesolutions (3.7), (3.10) and (3.14) can be plotted as follows:

4.2 The nonlinear PDE (1.2) describing pulsenarrowing nonlinear transmission lines

The obtained solutions for the nonlinear PDE (1.2) are hyper-bolic and trigonometric. From the obtained solutions for thisequation we observe that the solutions (3.20), (3.23) are bell-kinkshaped soliton solutions; the solutions (3.21), (3.24) are singularbell-kink shaped soliton solutions and the solutions (3.26)-(3.30)are periodic solutions. The graphical representation of the solu-tions (3.21), (3.23) and (3.29) can be plotted as follows:

5. CONCLUSIONSThe generalized projective Riccati equations method describedin Section 2 of this article has been applied to construct manynew exact solutions of the nonlinear PDEs (1.1) and (1.2) whichdescribe the nonlinear low-pass electrical transmission lines andpulse narrowing nonlinear transmission lines respectively, withthe aid of Maple 14. On comparing our results obtained in thisarticle with the well-known results obtained in [46,48,49,50] we

Fig. 2. The plot of the solution (3.10) when µ = 2, α = 1, δ = 2,β = 1, R = 2.

deduce that our results are new and not published elsewhere. Theproposed method of this paper is effective and can be appliedto many other nonlinear PDEs. Finally, all solutions obtained inthis article have been checked with the Maple 14 by putting themback into the original equations.

AcknowledgementThe authors wish to thank the referees for their comments on thepaper.

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Fig. 3. The plot of the solution (3.14) when µ = 2, α = 1, δ = 2,β = −1, R = 2.

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IntroductionDescription of the generalized projective Riccati equations methodExact solutions of equations (1.1) and (1.2) using the given method of Sec. 2Exact solutions of the nonlinear PDE (1.1)Exact solutions of the nonlinear PDE (1.2)

Physical explanations of some resultsThe nonlinear PDE (1.1) governing wave propagation in nonlinear low-pass electrical transmission linesThe nonlinear PDE (1.2) describing pulse narrowing nonlinear transmission lines

ConclusionsReferences

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The Generalized Projective Riccati Equations Method and its Applications to Nonlinear ... · 2015. 11. 25. · ear PDEs describing nonlinear transmission lines (NLTL). The ﬁrst