New Complex Hyperbolic Structures to the Lonngren-Wave Equation by Using Sine-Gordon Expansion Method

Over some decades, the field of nonlinear evolution equations (NEEs) has attracted the attention of many researchers. NEEs are broadly used to describe problems in science, engineering and mathematical physics such as fluid dynamics, plasma physics, hydro magnetic waves, optic fibers, solid state physics and many others. NEEs can also be used to describe the propagation of a nonlinear dispersive waves in inhomogeneous media [1, 2]. It has become an important bottom-line to find the analytical solutions to these types of equations. Several methods for finding the solutions of various NEEs have been proposed and/or improved by many scholars [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71].

The aim of this paper was to apply the sine-Gordon expansion method (SGEM) to find a new solutions to the Lonngren-wave equation [14].

(uxx−αu+βu2)tt+uxx=0, $$\begin{array}{} \displaystyle (u_{xx}-\alpha u+\beta u^{2})_{tt}+u_{xx}=0, \end{array}$$(1)

where α and β are real constants. The equation describes the electric signals in telegraph lines on the basis of the tunnel diode [15, 16]. The Lonngren-wave equation was used as an example by Akcagil and Aydemir [14] to show the existence of strong connection between the (G′G) $\begin{array}{} (\frac{G^{'}}{G}) \end{array}$-expansion method and the modified extended tanh method.

SGEM is a method for solving different nonlinear partial differential equations that is developed based on wave transformation and te sine-Gordon expansion method [17]. A new hyperbolic function solutions to the Davey-Stewartson equation with power-law nonlinearity was obtained in [18] by using SGEM. With the aid of symbolic computation, a new transformation was developed using the general sine-Gordon travelling wave reduction equation and a generalized transformation to obtain the solutions of various types of nonlinear differential equations [19]. A considerable investigation has been implemented by Yan [20] to sine-Gordon-type equations where the equations are systematically solved by using the Jacobi elliptic function expansion method.

The remaining parts of this paper are organized as follows: In Section 2, we discuss the general facts of the SGEM. In Section 3, we apply the SGEM to the Lonngren-wave equation given in Eq. (1). Section 4 is about results, discussion and some remarks. We finally, give the conclusion of this paper in Section 5.

In this section we discuss the general facts of SGEM.

Consider the following sine-Gordon equation [17, 21, 22]:

uxx−utt=m2sin(u), $$\begin{array}{} \displaystyle u_{xx}-u_{tt}=m^{2}sin(u), \end{array}$$(2)

where u = u(x, t) and m is a real constant.

Applying the wave transformation u = u(x, t) = U(ξ), ξ = μ (x − ct) to Eq. (2), yields the following nonlinear ordinary differential equation (NODE):

U″=m2μ2(1−c2)sin(U), $$\begin{array}{} \displaystyle U^{''}=\frac{m^{2}}{\mu^{2}(1-c^{2})}sin(U), \end{array}$$(3)

where U = U(ξ), ξ is the amplitude of the travelling wave and c is the velocity of the travelling wave. Reconsidering Eq. (3), we can write its full simplification as:

[(U2)′]2=m2μ2(1−c2)sin2(U2)+K, $$\begin{array}{} \displaystyle \bigg[\Big(\frac{U}{2}\Big)^{'} \bigg]^{2}=\frac{m^{2}}{\mu^{2}(1-c^{2})}sin^{2}\Big(\frac{U}{2}\Big)+K, \end{array}$$(4)

where K is the integration constant.

Substituting K = 0, w(ξ) = U2 $\begin{array}{} \displaystyle \frac{U}{2} \end{array}$ and a² = m2μ2(1−c2) $\begin{array}{} \displaystyle \frac{m^{2}}{\mu^{2}(1-c^{2})} \end{array}$ in Eq. (4), gives:

w′=asin(w), $$\begin{array}{} \displaystyle w^{'}=a sin(w), \end{array}$$(5)

Putting a = 1 in Eq. (5), we have:

w′=sin(w), $$\begin{array}{} \displaystyle w^{'}= sin(w), \end{array}$$(6)

Equation (6) is variables separable equation, we obtain the following two significant equations from solving it:

sin(w)=sin(w(ξ))=2peξp2e2ξ+1|p=1=sech(ξ), $$\begin{array}{} \displaystyle sin(w)=sin(w(\xi))=\frac{2pe^{\xi}}{p^{2}e^{2\xi}+1}\Bigg |_{p=1}=sech(\xi), \end{array}$$(7)

cos(w)=cos(w(ξ))=p2e2ξ−1p2e2ξ+1|p=1=tanh(ξ), $$\begin{array}{} \displaystyle cos(w)=cos(w(\xi))=\frac{p^{2}e^{2\xi}-1}{p^{2}e^{2\xi}+1}\Bigg |_{p=1}=tanh(\xi), \end{array}$$(8)

where p is the integral constant.

For the solution of the following nonlinear partial differential equation;

P(u,ux,ut,uxx,utt,uxt…), $$\begin{array}{} \displaystyle P(u,\;u_{x},\;u_{t},\;u_{xx},\;u_{tt},\;u_{xt} \ldots), \end{array}$$(9)

we consider,

U(ξ)=∑i=1ntanhi−1(ξ)[Bisech(ξ)+Aitanh(ξ)]+A0, $$\begin{array}{} \displaystyle U(\xi)=\sum_{i=1}^{n}tanh^{i-1}(\xi)\big[B_{i}sech(\xi)+A_{i}tanh(\xi)\big]+A_{0}, \end{array}$$(10)

Equation (10) can be rewritten according to Eqs. (7) and (8) as follows:

U(w)=∑i=1ncosi−1(w)[Bisin(w)+Aicos(w)]+A0. $$\begin{array}{} \displaystyle U(w)=\sum_{i=1}^{n}cos^{i-1}(w)\big[B_{i}sin(w)+A_{i}cos(w)\big]+A_{0}. \end{array}$$(11)

We determine the value n under the terms of NODE by the balance principle. Letting the coefficients of sinⁱ(w)cos^j(w) to be all zero, yields a system of equations. Solving this system by using Wolfram Mathematica 9 gives the values of A_i, B_i, μ and c. Finally, substituting the values of A_i, B_i, μ and c in Eq. (10), we obtain the new travelling wave solutions to Eq. (9)

Consider the Lonngren-wave equation given in Eq. (1);

Applying the transformation u = u(x, t) = U(ξ), ξ = μ (x − ct) to Eq. (1), we have:

c2μ2U″+(1−c2α)U+c2βU2=0 $$\begin{array}{} \displaystyle c^{2}\mu^{2}U^{''}+(1-c^{2}\alpha)U+c^{2}\beta U^{2}=0 \end{array}$$(12)

We obtain n = 2, by applying the balance principle on Eq. (12). With n = 2 and using Eq. (11), yields:

U(w)=B1sin(w)+A1cos(w)+B2cos(w)sin(w)+A2cos2(w)+A0. $$\begin{array}{} \displaystyle U(w)=B_{1}sin(w)+A_{1}cos(w)+B_{2}cos(w)sin(w)+A_{2}cos^{2}(w)+A_{0}. \end{array}$$(13)

Differentiating Eq. (13) twice, gives:

U″(w)=B1cos2(w)sin(w)−B1sin3(w)−2A1sin2(w)cos(w)+B2cos3(w)sin(w)−5B2sin3(w)cos(w)−4A2cos2(w)sinw(w)+2A2sin4(w), $$\begin{array}{} \displaystyle U^{''}(w)=B_{1}cos^{2}(w)sin(w)-B_{1}sin^{3}(w)-2A_{1}sin^{2}(w)cos(w)+B_{2}cos^{3}(w)sin(w)\\\displaystyle\qquad\qquad\qquad\qquad\qquad -5B_{2}sin^{3}(w)cos(w)-4A_{2}cos^{2}(w)sin^{w}(w)+2A_{2}sin^{4}(w), \end{array}$$(14)

Putting Eqs. (13) and (14) in Eq. (12), yields;

A₀ − c²α A₀ + c²β A₀² + A₁ cos(w) − c²α A₁ cos(w) - 2c²μ²A₁sin²(w)cos(w) + 2c²β A₀A₁ cos(w) + c²β A₁²cos²(w) + A₂cos²(w) − c²α A₂ cos²(w)

−4c²μ²A₂cos²(w)sin²(w) + 2c²μ²A₂sin⁴(w) + 2c²β A₀A₂cos²(w)+

2c²β A₁A₂cos³(w) + c²β A₂²cos⁴(w) + B₁sin(w) − c²α B₁sin(w) +

c²μ²B₁cos²(w)sin(w) − c²μ²B₁sin³(w) + 2c²β A₀B₁sin(w)+

2c²β A₁B₁ cos(w)sin(w)+ 2c²β A₂B₁cos²(w)sin(w) + c²β B₁²sin²(w)

+ B₂cos(w)sin(w) − c²α B₂cos(w)sin(w) + c²μ²B₂cos³(w)sin(w)−

5c²μ²B₂sin³(w)cos(w) + 2c²β A₀B₂cos(w)sin(w) + 2c²β A₁B₂cos²sin(w)

+ 2c²β A₂B₂cos³sin(w) + 2c²β B₁B₂sin²(w)cos(w) + c²β B₂²cos²(w)sin²(w) = 0.

We collect a set of algebraic equations by equating each summation of the coefficients of the trigonometric terms of the same power to zero in the abovementioned equation. We solve the set of generated to obtained the values of the coefficients. To get the new solitary solutions, u(x, t) to Eq. (1), we substitute in each case the obtained results of the coefficients into Eq. (10) along with n = 2.

Case-1

When we consider following coefficients:

A0=2μ2β,A1=0,B1=0,A2=−6μ2β,B2=0,α=1c2−4μ2, $$\begin{array}{} \displaystyle A_{0}=\frac{2\mu^{2}}{\beta}, A_{1}=0, B_{1}=0, A_{2}=-\frac{6\mu^{2}}{\beta}, B_{2}=0, \alpha=\frac{1}{c^{2}}-4\mu^{2}, \end{array}$$

these produce new dark solution as:

u1(x,t)=2μ2β(1−3tanh[μ(x−ct)]2). $$\begin{array}{} \displaystyle u_{1}(x,\;t)=\frac{2\mu^{2}}{\beta}\big(1-3tanh[\mu(x-ct)]^{2}\big). \end{array}$$(15)

Fig. 1

Case-3

When we take

A0=2μ2β,A1=0,B1=0,A2=−3μ2β,B2=3iμ2β,α=1c2−μ2, $$\begin{array}{} \displaystyle A_{0}=\frac{2\mu^{2}}{\beta}, A_{1}=0, B_{1}=0, A_{2}=-\frac{3\mu^{2}}{\beta}, B_{2}=\frac{3i\mu^{2}}{\beta}, \alpha=\frac{1}{c^{2}}-\mu^{2}, \end{array}$$

they give mixed complex singular solution as:

u3(x,t)=μ2β(−1+3ii+sinh(μ(x−ct))). $$\begin{array}{} \displaystyle u_{3}(x,\;t)=\frac{\mu^{2}}{\beta}\Big(-1+\frac{3i}{i+sinh(\mu(x-ct))} \Big). \end{array}$$(17)

Fig. 2

Fig. 3

Case-4

A0=3(c2α−1)c2β,A1=0,B1=0,A2=3−3c2αc2β,B2=3i(c2α−1)c2β,μ=−c2α−1c, $$\begin{array}{} \displaystyle A_{0}=\frac{3(c^{2}\alpha -1)}{c^{2}\beta}, A_{1}=0, B_{1}=0, A_{2}=\frac{3-3c^{2}\alpha}{c^{2}\beta}, B_{2}=\frac{3i(c^{2}\alpha -1)}{c^{2}\beta}, \mu=-\frac{\sqrt{c^{2}\alpha -1}}{c}, \end{array}$$

give mixed complex rational solution as:

u4(x,t)=3i(c2α−1)c2β(i−sinh[(c2α−1c)(x−ct)]). $$\begin{array}{} \displaystyle u_{4}(x,\;t)=\frac{3i(c^{2}\alpha -1)}{c^{2}\beta \Big(i-sinh[(\frac{\sqrt{c^{2}\alpha -1}}{c})(x-ct)]\Big)}. \end{array}$$(18)

Fig. 4

Case-5

A0=2−2c2αc2β,A1=0,B1=0,A2=3(c2α−1)c2β,B2=−3i(c2α−1)c2β,μ=−ic2α−1c, $$\begin{array}{} \displaystyle A_{0}=\frac{2-2c^{2}\alpha}{c^{2}\beta}, A_{1}=0, B_{1}=0, A_{2}=\frac{3(c^{2}\alpha -1)}{c^{2}\beta}, B_{2}=-\frac{3i(c^{2}\alpha -1)}{c^{2}\beta}, \mu=-\frac{i \sqrt{c^{2}\alpha -1}}{c}, \end{array}$$

which produces the following trigonometric travelling wave solution as:

u5(x,t)=1c2β(c2α−1+3(c2α−1)−1+sin[(x−ct)c2α−1c]). $$\begin{array}{} \displaystyle u_{5}(x,\;t)=\frac{1}{c^{2}\beta}\Big(c^{2}\alpha -1+ \frac{3(c^{2}\alpha -1)}{-1+sin[(x-ct)\frac{\sqrt{c^{2}\alpha -1}}{c}]}\Big). \end{array}$$(19)

Fig. 5

Case-6

A0=3μ2β,A1=0,B1=0,A2=−3μ2β,B2=3iμ2β,c=−1α−μ2, $$\begin{array}{} \displaystyle A_{0}=\frac{3\mu^{2}}{\beta}, A_{1}=0, B_{1}=0, A_{2}=-\frac{3\mu^{2}}{\beta}, B_{2}=\frac{3i\mu^{2}}{\beta}, c=-\frac{1}{\sqrt{\alpha-\mu^{2}}}, \end{array}$$

which introduces the following complex mixed solution as:

u7(x,t)=3μ2βsech[μ(x+tα−μ2)](sech[μ(x+tα−μ2)]+itanh[μ(x+tα−μ2)]). $$\begin{array}{} \begin{split} \displaystyle u_{7}(x,\;t)=\frac{3\mu^{2}}{\beta}sech\bigg[\mu\Big(x+\frac{t}{\sqrt{\alpha -\mu^{2}}}\Big)\bigg]\bigg(sech\Big[\mu \Big(x+\frac{t}{\sqrt{\alpha -\mu^{2}}}\Big)\Big]\\+itanh\Big[\mu\Big(x+\frac{t}{\sqrt{\alpha-\mu^{2}}}\Big)\Big]\bigg). \end{split} \end{array}$$(20)

Fig. 6

The powerful SGEM as one of the prominent methods for obtaining the some new travelling wave solutions to the nonlinear partial differential equations has been used in this paper. This method is based on both important properties of the sine-Gordon equation such as Eqs. (7) and Eq.(8). The SGEM includes trigonometric functions, which will be used later for obtaining novel solutions in Eq.(11). Many new solutions can be obtained by using the properties of these trigonometric functions. This is one of the main properties of SGEM. Therefore, it gives many coefficients to the considered model such as complex, exponential and trigonometric.

In this manuscript, by selecting of some of them, we have obtained the same solution, Eq.(15); moreover, we have found some entirely new complex, exponential, dark and hyperbolic solutions to the model considered when we compared the solutions obtained with the help of exp the (G′/G)-expansion method and the modified extended tanh method used in [14]. These solutions are new physical properties of model equation Eq.(1.1). The effectiveness and the simplicity of the method show that its powerful and reliable mathematical tool that can be applied in solving various NEEs. For computational calculations, we have used the packet programs for drawing graphical surfaces in this paper. To the best of our knowledge, the application of SGEM to the Lonngren-wave equation has not been submitted to the literature beforehand.