# Diverse Soliton Structures of the (2+1)-Dimensional Nonlinear Electrical Transmission Line Equation

Authors: Peng-Fei Li; Kang-Jia Wang
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

In this work, the (2+1)-dimensional nonlinear electrical transmission line equation (NETLE) is investigated by applying three recent technologies, namely, the variational approach, Hamiltonian approach, and energy balance approach. Diverse exact soliton solutions such as the bright, bright-like, kinky bright, bright-dark soliton, and periodic soliton solutions are successfully constructed. The outlines of the different solutions are shown in the form of the 3-D plot with the help of the Wolfram Mathematica. It reveals that the used methods are concise and effective and are expected to provide some inspiration for the study of travelling wave solutions of the PDEs in physics.


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## Body

## 1. Introduction

Nonlinear partial differential equations (NLPDEs) appear in mathematics, physics, engineering, and other fields. Many complex phenomena occurring in nature can be described by the NLPDEs. The study of their soliton solutions is of great significance since they can make us more deeply understand the natural phenomena and their internal relations. So far, there are many effective methods available for constructing the soliton solutions such as the exp-function method [1–4], tanh-function method [5–9], (G′/G)-expansion method [10–12], F-expansion method [13, 14], extended rational sine-cosine and sinh-cosh methods [15–18], Sardar-subequation method [19–21], and Sine-Gordon expansion method [22, 23] [24–31]. In the current work, we aim to study the (2+1)-dimensional NETLE, which is expressed by [32]
(1)∂2∂t2u−αu2+βu3−μ02δ12∂2u∂x2+δ1412∂4u∂x4−ϖ02δ22∂2u∂y2+δ2412∂4u∂y4=0.In Eq. (1), α,β, u0, and ϖ0 are nonzero constants, δ1 and δ2 represent the longitudinal and transverse distance between two adjacent sections, respectively. Eq. (1) plays an important role in the field of telecommunication and network engineering. Up to now, some effective approaches have been adopted to solve Eq. (1) such as the modified simple equation method [33], Jacobi elliptic function expansion method [34], sine-Gordon expansion method [35], and the Kudryashov method [36]. In recent years, the variational theory-based methods such as the variational approach (VA), Hamiltonian approach (HA), and energy balance approach (EBA) have caught a wide attention for solving the PDEs since they all probe the problem in view of the energy conservation and obtain the solutions by the stationary conditions. Additionally, these methods can help us insight the problem from a physical perspective. Thus, in this work, we aim to seek for the various soliton solutions by means of the variational theory-based methods, which are the VA, HA, and EBA. The rest content of this paper is arranged as follows. In Section 2, the variational principle and Hamiltonian of the studied problem are presented. In Section 3, the various soliton solutions are derived by applying the three methods. The behaviors of the different solutions are presented through the 3-D plot in Section 4. Finally, we get a conclusion in Section 5.


## 2. Variational Principle and the Hamiltonian

For solving Eq. (1), we introduce the following transformation [36]:
(2)ux,y,t=Uχ,χ=℘x+y−θ0t.In Eq. (2), ℘ and θ0 represent the wave number and wave speed, respectively. Applying Eq. (2) to Eq. (1), integrating the results twice with respect to χ and setting the integrating constants to zero, we get
(3)M−μ02Π1+ω02Π2U+MβU3−αU2−112μ02Π12+ω02Π22d2Udχ2=0,where M=℘θ02, Π1=δ12℘, and Π2=δ22℘.With the aid of the semi-inverse method [37–43], we establish the variational principle of Eq. (3) as
(4)JU=∫12U′2+3Mβμ02Π12+ϖ02Π22U4−4Mαμ02Π12+ϖ02Π22U3+6M−μ02Π1+ϖ02Π2μ02Π12+ϖ02Π22U2dχ.For the convenience of calculation, we reexpress Eq. (4) as
(5)JU=∫12U′2+λ1U4−λ2U3+λ3U2dχ=∫R−Idχ,where λ1=3Mβ/μ02Π12+ϖ02Π22, λ2=4Mα/μ02Π12+ϖ02Π22, and λ3=6M−μ02Π1+ϖ02Π2/μ02Π12+ϖ02Π22. And there are
(6)R=12U′2,I=−λ1U4+λ2U3−λ3U2.Here,R is the kinetic energy, and I indicates the potential energy. Then, we can get the Hamiltonian as [44, 45]
(7)Hm=R+I=12U′2−λ1U4+λ2U3−λ3U2.


## 3. The Solutions

### 3.1. The VA

The main target of this subsection is to apply the VA to search for the abundant exact solutions of Eq. (1).


#### 3.1.1. The Bright Soliton Solution

Here, we suppose the solution of Eq. (3) is [46]
(8)Uχ=Ξsechχ.Taking it into Eq. (5) yields
(9)JΞ=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12Ξsechχtanhχ2+λ1Ξsechχ4−λ2Ξsechχ3+λ3Ξsechχ2dχ=Ξ28λ1Ξ2−3Ξλ2π+12λ3+212.According to the Ritz-like method (RLM), we take its stationary condition as(10)dJdΞ=0,which leads to
(11)Ξ32λ1Ξ2−9Ξλ2π+24λ3+412=0.On solving it by the Wolfram Mathematica, we have(12)Ξ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1,Or(13)Ξ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1.So, the solution of Eq. (3) can be obtained as
(14)Uχ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sechχ,or
(15)Uχ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sechχ.In the view of Eq. (2), we have
(16)ux,y,t=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t,or
(17)ux,y,t=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t.


#### 3.1.2. The Bright-Like Soliton Solution

The solution of Eq. (3) is assumed as the following form [47]:
(18)Uχ=Ξ21+coshχ.Putting above equation into Eq. (5), it gives
(19)JΞ2=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12−Ξ2sinhχ1+coshχ22+λ1Ξ21+coshχ4−λ2Ξ21+coshχ3+λ3Ξ21+coshχ2dχ=Ξ224Ξ23λ1Ξ2−7λ2+70λ3+7210.Based on the RLM, there is(20)dJdΞ2=0.It results into(21)Ξ224λ1Ξ22−42λ2Ξ2+70λ3+7150=0.Solving above equation, we have(22)Ξ2=21λ2+2121λ22−80λ1λ3−8λ124λ1,or
(23)Ξ2=21λ2−2121λ22−80λ1λ3−8λ124λ1.So, there are(24)ux,y,t=21λ2+2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t,or
(25)ux,y,t=21λ2−2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t.


#### 3.1.3. The Kinky-Bright Soliton

Inspired by the research results obtained in [47], we can also assume that Eq. (3) has the following solution to construct the kinky-bright soliton:
(26)Uχ=Ξ3sech2χ.By the same way, we have(27)JΞ3=2Ξ3224λ1Ξ32−28λ2Ξ3+35λ3+14105.Computing its stationary condition with respect toΞ3 yields
(28)4Ξ348λ1Ξ32−42λ2Ξ3+35λ3+14105=0,from which, we obtain(29)Ξ3=21λ2+2121λ22−80λ1λ3−32λ148λ1,or
(30)Ξ3=21λ2−2121λ22−80λ1λ3−32λ148λ1.With this, we have(31)ux,y,t=21λ2+2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t,or
(32)ux,y,t=21λ2−2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t.


#### 3.1.4. The Bright-Dark Soliton

Here, we search for a solution of Eq. (3) in the following form [47]:
(33)Uχ=Ξ4sechχtanhχ.Substituting Eq.(33) into Eq. (5), it produces
(34)J=Ξ4221012λ1Ξ42−28λ2Ξ4+70λ3+49.MakingJ stationary with Ξ4 results in
(35)Ξ410524λ1Ξ42−42λ2Ξ4+70λ3+49=0.From Eq. (35), we have
(36)Ξ4=21λ2+2121λ22−80λ1λ3−56λ124λ1,or
(37)Ξ4=21λ2−2121λ22−80λ1λ3−56λ124λ1.So, we have(38)ux,y,t=21λ2+2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t,or
(39)ux,y,t=21λ2−2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t.


### 3.2. The HA

In this section, we aim to use the HA to search for the periodic solution of Eq. (3) as [48–50]
(40)Uχ=ΛcosΩχ,where Λ is the amplitude and Ω is the frequency.Form the obtained Hamiltonian provided by Eq. (7), we construct a new function h⌣U as
(41)h⌣U=∫0T/412U′2−λ1U4+λ2U3−λ3U2dχ.Taking Eq. (40) into above equation, it yields
(42)h⌣U=∫0T/412−ΛΩsinΩχ2−λ1ΛcosΩχ4+λ2ΛcosΩχ3−λ3ΛcosΩχ2dχ=Λ248Ω32λ2Λ−9λ1Λ2π+6Ω2−2λ3π.By the Hamiltonian approach, we set(43)∂∂Λ∂h⌣∂1/Ω=0,which leads to
(44)Ω=8Λλ2π−2λ3−3λ1Λ2.So, the periodic soliton solution of Eq. (1) is found as
(45)ux,y,t=Λcos8Λλ2π−2λ3−3λ1Λ2℘x+y−θ0t.


### 3.3. The EBA

To use the EBA [50, 51], we first assume the solution of Eq.(3) is
(46)Uχ=ΔcosΘχ.And we have(47)U0=Δ,U′0=0.The energy conservation reveals that the Hamiltonian invariant keep unchanged for the system; so, inserting Eq. (47) into Eq. (7) yields
(48)Hm=R+I=12U′2−λ1U4+λ2U3−λ3U2=−λ1Δ4+λ2Δ3−λ3Δ2.Then, we substitute Eq. (46) into Eq. (48) and set Θχ=π/4, and there is
(49)12−ΔΘsinπ42−λ1Δcosπ44+λ2Δcosπ43−λ3Δcosπ42=−λ1Δ4+λ2Δ3−λ3Δ2.Solving it gives(50)Θ=4−2λ2Δ−2λ3−3λ1Δ2,which has a well agreement with Eq. (44). It strongly proves the correctness of the two different methods. Thus, we get the periodic soliton solution of Eq. (1) as
(51)ux,y,t=Δcos4−2λ2Δ−2λ3−3λ1Δ2℘x+y−θ0t.


## 3.1. The VA

The main target of this subsection is to apply the VA to search for the abundant exact solutions of Eq. (1).


### 3.1.1. The Bright Soliton Solution

Here, we suppose the solution of Eq. (3) is [46]
(8)Uχ=Ξsechχ.Taking it into Eq. (5) yields
(9)JΞ=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12Ξsechχtanhχ2+λ1Ξsechχ4−λ2Ξsechχ3+λ3Ξsechχ2dχ=Ξ28λ1Ξ2−3Ξλ2π+12λ3+212.According to the Ritz-like method (RLM), we take its stationary condition as(10)dJdΞ=0,which leads to
(11)Ξ32λ1Ξ2−9Ξλ2π+24λ3+412=0.On solving it by the Wolfram Mathematica, we have(12)Ξ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1,Or(13)Ξ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1.So, the solution of Eq. (3) can be obtained as
(14)Uχ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sechχ,or
(15)Uχ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sechχ.In the view of Eq. (2), we have
(16)ux,y,t=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t,or
(17)ux,y,t=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t.


### 3.1.2. The Bright-Like Soliton Solution

The solution of Eq. (3) is assumed as the following form [47]:
(18)Uχ=Ξ21+coshχ.Putting above equation into Eq. (5), it gives
(19)JΞ2=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12−Ξ2sinhχ1+coshχ22+λ1Ξ21+coshχ4−λ2Ξ21+coshχ3+λ3Ξ21+coshχ2dχ=Ξ224Ξ23λ1Ξ2−7λ2+70λ3+7210.Based on the RLM, there is(20)dJdΞ2=0.It results into(21)Ξ224λ1Ξ22−42λ2Ξ2+70λ3+7150=0.Solving above equation, we have(22)Ξ2=21λ2+2121λ22−80λ1λ3−8λ124λ1,or
(23)Ξ2=21λ2−2121λ22−80λ1λ3−8λ124λ1.So, there are(24)ux,y,t=21λ2+2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t,or
(25)ux,y,t=21λ2−2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t.


### 3.1.3. The Kinky-Bright Soliton

Inspired by the research results obtained in [47], we can also assume that Eq. (3) has the following solution to construct the kinky-bright soliton:
(26)Uχ=Ξ3sech2χ.By the same way, we have(27)JΞ3=2Ξ3224λ1Ξ32−28λ2Ξ3+35λ3+14105.Computing its stationary condition with respect toΞ3 yields
(28)4Ξ348λ1Ξ32−42λ2Ξ3+35λ3+14105=0,from which, we obtain(29)Ξ3=21λ2+2121λ22−80λ1λ3−32λ148λ1,or
(30)Ξ3=21λ2−2121λ22−80λ1λ3−32λ148λ1.With this, we have(31)ux,y,t=21λ2+2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t,or
(32)ux,y,t=21λ2−2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t.


### 3.1.4. The Bright-Dark Soliton

Here, we search for a solution of Eq. (3) in the following form [47]:
(33)Uχ=Ξ4sechχtanhχ.Substituting Eq.(33) into Eq. (5), it produces
(34)J=Ξ4221012λ1Ξ42−28λ2Ξ4+70λ3+49.MakingJ stationary with Ξ4 results in
(35)Ξ410524λ1Ξ42−42λ2Ξ4+70λ3+49=0.From Eq. (35), we have
(36)Ξ4=21λ2+2121λ22−80λ1λ3−56λ124λ1,or
(37)Ξ4=21λ2−2121λ22−80λ1λ3−56λ124λ1.So, we have(38)ux,y,t=21λ2+2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t,or
(39)ux,y,t=21λ2−2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t.


## 3.1.1. The Bright Soliton Solution

Here, we suppose the solution of Eq. (3) is [46]
(8)Uχ=Ξsechχ.Taking it into Eq. (5) yields
(9)JΞ=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12Ξsechχtanhχ2+λ1Ξsechχ4−λ2Ξsechχ3+λ3Ξsechχ2dχ=Ξ28λ1Ξ2−3Ξλ2π+12λ3+212.According to the Ritz-like method (RLM), we take its stationary condition as(10)dJdΞ=0,which leads to
(11)Ξ32λ1Ξ2−9Ξλ2π+24λ3+412=0.On solving it by the Wolfram Mathematica, we have(12)Ξ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1,Or(13)Ξ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1.So, the solution of Eq. (3) can be obtained as
(14)Uχ=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sechχ,or
(15)Uχ=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sechχ.In the view of Eq. (2), we have
(16)ux,y,t=9λ2π+81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t,or
(17)ux,y,t=9λ2π−81λ22π2−3072λ1λ3−512λ164λ1sech℘x+y−θ0t.


## 3.1.2. The Bright-Like Soliton Solution

The solution of Eq. (3) is assumed as the following form [47]:
(18)Uχ=Ξ21+coshχ.Putting above equation into Eq. (5), it gives
(19)JΞ2=∫0∞12U′2+λ1U4−λ2U3+λ3U2dχ=∫0∞12−Ξ2sinhχ1+coshχ22+λ1Ξ21+coshχ4−λ2Ξ21+coshχ3+λ3Ξ21+coshχ2dχ=Ξ224Ξ23λ1Ξ2−7λ2+70λ3+7210.Based on the RLM, there is(20)dJdΞ2=0.It results into(21)Ξ224λ1Ξ22−42λ2Ξ2+70λ3+7150=0.Solving above equation, we have(22)Ξ2=21λ2+2121λ22−80λ1λ3−8λ124λ1,or
(23)Ξ2=21λ2−2121λ22−80λ1λ3−8λ124λ1.So, there are(24)ux,y,t=21λ2+2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t,or
(25)ux,y,t=21λ2−2121λ22−80λ1λ3−8λ124λ111+cosh℘x+y−θ0t.


## 3.1.3. The Kinky-Bright Soliton

Inspired by the research results obtained in [47], we can also assume that Eq. (3) has the following solution to construct the kinky-bright soliton:
(26)Uχ=Ξ3sech2χ.By the same way, we have(27)JΞ3=2Ξ3224λ1Ξ32−28λ2Ξ3+35λ3+14105.Computing its stationary condition with respect toΞ3 yields
(28)4Ξ348λ1Ξ32−42λ2Ξ3+35λ3+14105=0,from which, we obtain(29)Ξ3=21λ2+2121λ22−80λ1λ3−32λ148λ1,or
(30)Ξ3=21λ2−2121λ22−80λ1λ3−32λ148λ1.With this, we have(31)ux,y,t=21λ2+2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t,or
(32)ux,y,t=21λ2−2121λ22−80λ1λ3−32λ148λ1sech2℘x+y−θ0t.


## 3.1.4. The Bright-Dark Soliton

Here, we search for a solution of Eq. (3) in the following form [47]:
(33)Uχ=Ξ4sechχtanhχ.Substituting Eq.(33) into Eq. (5), it produces
(34)J=Ξ4221012λ1Ξ42−28λ2Ξ4+70λ3+49.MakingJ stationary with Ξ4 results in
(35)Ξ410524λ1Ξ42−42λ2Ξ4+70λ3+49=0.From Eq. (35), we have
(36)Ξ4=21λ2+2121λ22−80λ1λ3−56λ124λ1,or
(37)Ξ4=21λ2−2121λ22−80λ1λ3−56λ124λ1.So, we have(38)ux,y,t=21λ2+2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t,or
(39)ux,y,t=21λ2−2121λ22−80λ1λ3−56λ124λ1sechkx+y−v0ttanh℘x+y−θ0t.


## 3.2. The HA

In this section, we aim to use the HA to search for the periodic solution of Eq. (3) as [48–50]
(40)Uχ=ΛcosΩχ,where Λ is the amplitude and Ω is the frequency.Form the obtained Hamiltonian provided by Eq. (7), we construct a new function h⌣U as
(41)h⌣U=∫0T/412U′2−λ1U4+λ2U3−λ3U2dχ.Taking Eq. (40) into above equation, it yields
(42)h⌣U=∫0T/412−ΛΩsinΩχ2−λ1ΛcosΩχ4+λ2ΛcosΩχ3−λ3ΛcosΩχ2dχ=Λ248Ω32λ2Λ−9λ1Λ2π+6Ω2−2λ3π.By the Hamiltonian approach, we set(43)∂∂Λ∂h⌣∂1/Ω=0,which leads to
(44)Ω=8Λλ2π−2λ3−3λ1Λ2.So, the periodic soliton solution of Eq. (1) is found as
(45)ux,y,t=Λcos8Λλ2π−2λ3−3λ1Λ2℘x+y−θ0t.


## 3.3. The EBA

To use the EBA [50, 51], we first assume the solution of Eq.(3) is
(46)Uχ=ΔcosΘχ.And we have(47)U0=Δ,U′0=0.The energy conservation reveals that the Hamiltonian invariant keep unchanged for the system; so, inserting Eq. (47) into Eq. (7) yields
(48)Hm=R+I=12U′2−λ1U4+λ2U3−λ3U2=−λ1Δ4+λ2Δ3−λ3Δ2.Then, we substitute Eq. (46) into Eq. (48) and set Θχ=π/4, and there is
(49)12−ΔΘsinπ42−λ1Δcosπ44+λ2Δcosπ43−λ3Δcosπ42=−λ1Δ4+λ2Δ3−λ3Δ2.Solving it gives(50)Θ=4−2λ2Δ−2λ3−3λ1Δ2,which has a well agreement with Eq. (44). It strongly proves the correctness of the two different methods. Thus, we get the periodic soliton solution of Eq. (1) as
(51)ux,y,t=Δcos4−2λ2Δ−2λ3−3λ1Δ2℘x+y−θ0t.


## 4. Results and Discussion

This section will give the graphical representations and physical interpretation of the obtained solutions in Section3 by using proper parameters.For℘=1, θ0=1, δ12=δ22=1, ϖ0=1,, μ0=1, α=1, and β=1, Figure 1 plots the performance of the bright solitons obtained by Eqs. (16) and (17) within the interval −10<x<10 and −10<y<10. Obviously, they have the bright soliton characteristics. M=1.Figure 1
The outline of the bright soliton. (a) for Eq. (16) and (b) for Eq. (17).
(a)(b)We plot the behaviors of the bright-like soliton solutions given by Eqs. (24) and (25) in Figure 2 by choosing ℘=1, θ0=1, δ12=δ22=1, ϖ0=1, M=1, μ0=1,α=1, and β=1, where the performance are like bright soliton.Figure 2
The behaviors of the bright-like soliton. (a) for Eq. (24) and (b) for Eq. (25).
(a)(b)Solutions of Eqs. (31) and (32) are the kinky-bright solitons. We plot their behaviors in Figure 3 within the interval on −10<x<10 and −10<y<10 with for ℘=1, θ0=1, δ12=δ22=1, ϖ0=1, M=1, μ0=1, α=1, and β=1. As expected, they all have the bright soliton characteristics.Figure 3
The outline of the kinky-bright soliton. (a) for Eq. (31) and (b) for Eq. (32).
(a)(b)Selecting℘=1, θ0=1, δ12=δ22=1, ϖ0=1, M=1, μ0=1, α=1, and β=1, the 3-D plots of the bright-dark solitons given by Eqs. (38) and (39) are presented in Figure 4 within the interval −10<x<10 and −10<y<10. It can be observed that the performances are bright-dark soliton.Figure 4
The outline of the bright-dark soliton. (a) for Eq.(38) and (b) for Eq.(39).
(a)(b)Figure5 plots the profile of Eqs. (45) and (51) within the interval −5<x<5 and −5<y<5 by using the parameters as ℘=1, θ0=1, δ12=δ22=1, ϖ0=1, M=1, μ0=1, Λ=1 (Δ=1), α=1, and β=1. It can be seen that the two profiles are all perfect periodic waves. In addition, the two contours are basically the same.Figure 5
The outline of the periodic soliton. (a) for Eq. (45) and (b) for Eq. (51).
(a)(b)


## 5. Conclusion

In this work, diverse soliton solutions of (2+1)-dimensional nonlinear electrical transmission line equation like the bright soliton, bright-like soliton, kinky-bright soliton, bright-dark soliton, and periodic soliton solutions were constructed by using the variational approach, Hamiltonian approach, and energy balance approach. The profiles of the solutions were presented through the 3-D plots via selecting the appropriate parameters by means of the Wolfram Mathematica. The results revealed that the proposed methods were straightforward, simple, and effective, which can be adopted to study the traveling wave theory of the PDEs arising in physics.



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