Quasi hyperbolic function expansion method and tanh-function method for solving vibrating string equation and elastic rod equation

Abstract

This paper applies the quasi hyperbolic function expansion method and the modified extended tanh-function method to exactly solve the nonlinear vibrating string equation and the elastic rod equation. The obtained exact solutions can be expressed by hyperbolic functions, which can outline various vibration properties.

Introduction

Nonlinear vibration equations are widely appeared in many practical application,^1–3 many researches focused on amplitude–frequency relationship, but the wave-like solutions were rarely studied. For example, a bridge vibrating under winds can be expressed by a nonlinear vibration equation; the amplitude and the intrinsic frequency are the main sources of research; however, the bridge vibration behaves sometimes like wave travelling. This paper focuses itself on finding the exact solution of the nonlinear vibration equations.

There are many analytical methods for this purpose, for example the direct method, ⁴ the reduced differential transform method, ⁵ the exp-function method,^6,7 the homotopy perturbation method,^8,9 the variational iteration method,^10,11 and various methods based on variational principles.^12,13 In this paper, we adopt the quasi hyperbolic function expansion method ¹⁴ and modified extended tanh-function (METF) method. ¹⁵

Algorithm of the two proposed analytical methods

Quasi hyperbolic function expansion method

Consider a PDE in two independent variables $x$ and $t$

$$Q(u,u_x,u_t,u_{\mathit{xx}},\cdots )=0$$

(1)

Step 1: Using the transform

$$u(x,t)\text{=}U(\xi ),\begin{array}{cc}& \xi =k(x-\mathit{ct})\end{array}$$

(2)

where $c$ and $k$ are arbitrary constants to be determined; we can rewrite equation (1) in the following ODE

$$R(U,k{U}^{\prime },-\mathit{ck}{U}^{\prime },\cdots )=0$$

(3)

Step 2: Suppose that the wave solution of equation (3) can be expressed in the following form

$$U(\xi )={\displaystyle \sum _{i=0}^m{a}_i}{f}^i+{\displaystyle \sum _{i=1}^m{b}_i}g{f}^{i-1}$$

(4)

where $a_i(0\le i\le m)$ and $b_i(1\le i\le m)$ are constants to be determined, such that $a_m^2+b_m^2\ne 0$ and $f=f(\xi )$ , $g=g(\xi )$ satisfy the following ODE

$$f^{\prime }(\xi )=-f(\xi )g(\xi )$$

(5)

$$g^{\prime }(\xi )=1-g{(\xi )}^2-\mathit{rf}(\xi )$$

(6)

$$g{(\xi )}^2=1-2rf(\xi )+(r^2-1)f{(\xi )}^2$$

(7)

and $f(\xi )=\frac{1}{\cosh (\xi )+r},g(\xi )=\frac{\sinh (\xi )}{\cosh (\xi )+r},(r\ne \pm 1)$ .

Step 3: By balancing the highest order derivative terms with the nonlinear terms of the highest order come out in equation (3), we can evaluate the value of the positive integer $m$ .

Step 4: By substituting equation (4) along with equations (5) to (7) into equation (3), we obtain a system of algebraic equations. The obtained system can be solved to find the value of $c$ , $k$ , $a_i(0\le i\le m),b_i(1\le i\le m)$ ; substituting these terms into equation (4), the determination of solutions of equation (1) will be completed.

METF method

Consider a PDE in two variables given by

$$Q(u,u_x,u_t,u_{\mathit{xx}},\cdots )=0$$

(8)

Step 1: Consider the transform

$$u(x,t)\text{=}U(\xi ),\begin{array}{cc}& \xi =x-\mathit{ct}\end{array}$$

(9)

where $c$ is arbitrary constant to be determined; then equation (8) becomes an ODE

$$R(U,U^{\prime },-c{U}^{\prime },\cdots )=0$$

(10)

Step 2: Suppose that the wave solution of equation (10) can be expressed in the following form

$$U(\xi )=a_0+{\displaystyle \sum _{i=1}^m{a}_i}w{(\xi )}^i+b_{i}w{(\xi )}^{-i}$$

(11)

where $a_i(0\le i\le m)$ and $b_i(1\le i\le m)$ are constants to be determined, such that $w(\xi )$ satisfies the following ODE

$$w^{\prime }(\xi )=b+w{(\xi )}^2$$

(12)

where $b$ is arbitrary constant. Equation (12) has the general solutions

• If $b<0$

  $$w=-\sqrt{-b}\tanh [\sqrt{-b}\xi ]$$
  $$w=-\sqrt{-b}\coth [\sqrt{-b}\xi ]$$

• If $b\text{=}0$

  $$w=-1/\xi$$

• If $b>0$

  $$w=\sqrt{b}\tan [\sqrt{b}\xi ]$$
  $$w=-\sqrt{b}\cot [\sqrt{b}\xi ]$$

Step 3: By balancing the highest order derivative terms with the nonlinear terms of the highest order come out in equation (10), we can evaluate the value of the positive integer $m$ .

Step 4: By substituting equation (11) along with equation (12) into equation (10) will yield a system of algebraic equations with respect to $c$ , $b$ , $a_i(0\le i\le m),b_i(1\le i\le m)$ , and the determination of solutions of equation (8) will be completed.

Exact travelling wave solutions to the nonlinear vibrating string equation

Quasi hyperbolic function expansion method

Consider the following nonlinear vibrating string equation (2)

$$u_{\mathit{tt}}-2{\alpha }^2{u}_x{u}_{\mathit{xx}}-2\mu u_{\mathit{xxxx}}=0,\begin{array}{cc}& (\alpha \mu \ne 0)\end{array}$$

(13)

Step 1: Using the wave transformation

$$u(x,t)\text{=}U(\xi ),\begin{array}{cc}& \xi =k(x-\mathit{ct})\end{array}$$

(14)

then the nonlinear partial differential equation (13) is reduced to a nonlinear ordinary differential equation

$$c^2{k}^2{U}^{\prime \prime }(\xi )-2k^3{\alpha }^2{U}^{\prime }(\xi )U^{\prime \prime }(\xi )-2k^4\mu U^{(4)}(\xi )=0$$

(15)

integrating equation (15) with respect to variable $\xi$ , we have

$$c^2{k}^2{U}^{\prime }(\xi )-k^3{\alpha }^2{U}^{\prime }{(\xi )}^2-2k^4\mu U^{(3)}(\xi )=0$$

(16)

now, setting

$$\phi (\xi )=U^{\prime }(\xi )$$

(17)

then equation (16) reduces to

$$c^2{k}^2\phi (\xi )-k^3{\alpha }^2\phi {(\xi )}^2-2k^4\mu {\phi }^{(2)}(\xi )=0$$

(18)

Step 2: Balancing the term $\phi {(\xi )}^2$ with term ${\phi }^{(2)}(\xi )$ in equation (18) gives $m=2$ .

Step 3: Taking into account the quasi hyperbolic function expansion method, we seek solutions to equation (18) by using the expansion

$$\phi (\xi )=a_0+a_{1}f(\xi )+a_{2}f{(\xi )}^2+b_{1}g(\xi )+b_{2}g(\xi )f(\xi )$$

(19)

substituting equation (19) along with equations (5), (6), and (7), a set of algebraic equations for $c$ , $k$ , $a_i(0\le i\le m)$ , $b_i(1\le i\le m)$ is obtained as follows

$$a_0{c}^2{k}^2-a_0{\hspace{0.17em}}^2{k}^3{\alpha }^2-b_1^2{k}^3{\alpha }^2=0,$$

$$a_1{c}^2{k}^2-2a_0{a}_1{k}^3{\alpha }^2-2b_1{b}_2{k}^3{\alpha }^2\text{+}2b_1^2{k}^{3}r{\alpha }^2-2a_1{k}^4\mu =0,$$

$$a_2{c}^2{k}^2-a_1^2{k}^3{\alpha }^2-2a_0{a}_2{k}^3{\alpha }^2\text{+}{b}_1^2{k}^3{\alpha }^2-b_2^2{k}^3{\alpha }^2+4b_1{b}_2{k}^{3}r{\alpha }^2-b_1^2{k}^3{r}^2{\alpha }^2-8a_2{k}^4\mu +6a_1{k}^{4}r\mu =0,$$

$$\text{-}2a_1{a}_2{k}^3{\alpha }^2\text{+}2b_1{b}_2{k}^3{\alpha }^2+2b_2^2{k}^{3}r{\alpha }^2\text{-2}{b}_1{b}_2{k}^3{r}^2{\alpha }^2+4a_1{k}^4\mu +20a_2{k}^{4}r\mu -4a_1{k}^4{r}^2\mu =0$$

(20)

$$\text{-}{a}_2^2{k}^3{\alpha }^2\text{+}{b}_2^2{k}^3{\alpha }^2-b_2^2{k}^3{r}^2{\alpha }^2+12a_2{k}^4\mu -12a_2{k}^4{r}^2\mu =0,$$

$$b_1{c}^2{k}^2-2a_0{b}_1{k}^3{\alpha }^2=0,$$

$$b_2{c}^2{k}^2-2a_1{b}_1{k}^3{\alpha }^2-2a_0{b}_2{k}^3{\alpha }^2-2b_2{k}^4\mu +2b_1{k}^{4}r\mu =0,$$

$$\text{-}2a_2{b}_1{k}^3{\alpha }^2-2a_1{b}_2{k}^3{\alpha }^2+4b_1{k}^4\mu +12b_2{k}^{4}r\mu -4b_1{k}^4{r}^2\mu =0,$$

$$\text{-}2a_2{b}_2{k}^3{\alpha }^2\text{+}12b_2{k}^4\mu -12b_2{k}^4{r}^2\mu =0$$

solving the above algebraic equation (20), the following sets of coefficients for the solutions generated from equation (19) are obtained as

Case 1

$$a_0=-\frac{8k\mu }{{\alpha }^2},a_1=0,\hspace{1em}{a}_2=\frac{12k\mu }{{\alpha }^2},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=0,\hspace{1em}18\mu (c^2+8k^2\mu )=0$$

now, the exact solution of equation (18) can be written as

$$\phi (x,t)=-\frac{8k\mu }{{\alpha }^2}+\frac{12k\mu }{{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}$$

and $18\mu (c^2+8k^2\mu )=0.$ For the values $k=c=1,r=1/5,\alpha =1,\mu =-1/8$ , fixed a local point x = 4, we have depicted the graphs of attained solution to visualize the internal structure of physical phenomena within the range $0\le t\le 7$ , which is shown in Figure 1.

Figure 1.

The wave profile of $\phi (4,t)$ .

Case 2

$$a_0=0,\hspace{1em}{a}_1=0,\hspace{1em}{a}_2=\frac{12k\mu }{{\alpha }^2},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=0,\hspace{1em}18\mu (c^2-8k^2\mu )=0$$

now, the exact solution of equation (18) can be identified as

$$\phi (x,t)=\frac{12k\mu }{{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}$$

and $18\mu (c^2-8k^2\mu )=0$ .

Case 3

$$a_0=0,\hspace{1em}{a}_1=\frac{6kr\mu }{{\alpha }^2},\hspace{1em}{a}_2=-\frac{6(-k\mu +k{r}^2\mu )}{{\alpha }^2},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=\pm \frac{6(-k\mu +k{r}^2\mu )}{\sqrt{-1+r^2}{\alpha }^2},18(-1+r^2)\mu (c^2-2k^2\mu )=0$$

now, the exact solution of equation (18) can be written as

$$\phi (x,t)=-\frac{6(-k\mu +k{r}^2\mu )}{{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}+\frac{6kr\mu }{{\alpha }^2(r+\cosh (k(x-\mathit{ct})))}\pm \frac{6(-k\mu +k{r}^2\mu )\sinh (k(x-\mathit{ct}))}{\sqrt{-1+r^2}{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}$$

and $18(-1+r^2)\mu (c^2-2k^2\mu )=0$ .

Case 4

$$a_0=\frac{c^2}{k{\alpha }^2},\hspace{1em}{a}_1=\frac{6kr\mu }{{\alpha }^2},\hspace{1em}{a}_2=-\frac{6(-k\mu +k{r}^2\mu )}{{\alpha }^2},\hspace{1em}{b}_1=0,b_2=\pm \frac{3\sqrt{2}\sqrt{c^2\mu (1-r^2)}}{{\alpha }^2},-\frac{18{(-1+r^2)}^2\mu (c^2+2k^2\mu )}{{\alpha }^4}=0$$

now, the exact solution of equation (18) can be identified as

$$\phi (x,t)=\frac{c^2}{k{\alpha }^2}-\frac{6(-k\mu +k{r}^2\mu )}{{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}+\frac{6kr\mu }{{\alpha }^2(r+\cosh (k(x-\mathit{ct})))}\pm \frac{3\sqrt{2}\sqrt{c^2\mu (1-r^2)}\sinh (k(x-\mathit{ct}))}{{\alpha }^2{(r+\cosh (k(x-\mathit{ct})))}^2}$$

and $-\frac{18{(-1+r^2)}^2\mu (c^2+2k^2\mu )}{{\alpha }^4}=0$ .

Exp-function method ¹⁶

Suppose that the wave solution of equation (18) can be expressed in the following form

$$\phi (\xi )\text{=}\frac{a_1\hspace{0.17em}\exp (\xi )+a_0+a_2\hspace{0.17em}\exp (-\xi )}{b_1\hspace{0.17em}\exp (\xi )+b_0+b_2\hspace{0.17em}\exp (-\xi )}$$

(21)

for simplicity, we let $b_1\text{=}1$ . Substituting equation (21) into equation (18), we have

$$\frac{1}{A}(C_0\text{+}{C}_1\hspace{0.17em}\exp (\xi )+C_2\hspace{0.17em}\exp (2\xi )+\cdots +C_6\hspace{0.17em}\exp (6\xi ))\text{=}0$$

(22)

where $A={(b_2+e^{\xi }(b_0+e^{\xi }))}^3$ ; collecting $C_i(i=0,1,\cdots ,6)$ and simultaneously equating to zero, a set of algebraic equations is obtained; solving the algebraic equations, the following results are obtained:

Case 1

$$a_0=-\frac{2b_0{c}^2}{k{\alpha }^2},a_1=-\frac{a_0}{2b_0},a_2=-\frac{a_0{b}_0}{8},b_2=\frac{b_0^2}{4},a_0^2{\alpha }^4+8b_0^2{c}^2\mu =0$$

Case 2

$$a_0=\frac{3b_0{c}^2}{k{\alpha }^2},a_1=0,a_2=0,b_2=\frac{b_0^2}{4},\text{-}{a}_0^2{\alpha }^4+18b_0^2{c}^2\mu =0$$

Case 3

$$a_0=\text{-}\frac{2b_{0}k\mu }{{\alpha }^2},a_1=\text{-}\frac{2k\mu }{{\alpha }^2},a_2=0,b_2=0,a_0^2{\alpha }^4+2b_0^2{c}^2\mu =0$$

Case 4

$$a_0=\frac{b_0{c}^2}{k{\alpha }^2},a_1=\frac{c^2}{k{\alpha }^2},a_2=0,b_2=0,a_0^2{\alpha }^4+2b_0^2{c}^2\mu =0$$

Case 5

$$a_0=\frac{a_2{b}_0}{b_2},a_1=\frac{a_2}{b_2},-k{a}_0{\alpha }^2+b_0{c}^2=0$$

Case 6

$$a_1=\frac{a_0}{b_0},a_2=0,b_2=0,-k{a}_0{\alpha }^2+b_0{c}^2=0$$

Case 7

$$a_0=0,a_1=\frac{c^2}{k{\alpha }^2},b_0=0,-k{a}_2{\alpha }^2+b_2{c}^2=0$$

For the sake of simplicity, we consider only the solution with respect to Case 7, the other solutions can be obtained in a similar way

$$\phi (\xi )\text{=}\frac{a_2{e}^{-\xi }+\frac{c^2{e}^{\xi }}{k{\alpha }^2}}{b_2{e}^{-\xi }+e^{\xi }}$$

and $b_2{c}^2-a_{2}k{\alpha }^2=0.$

Exact travelling wave solutions of nonlinear elastic rod equation

Consider the nonlinear wave equation of longitudinal oscillation of a nonlinear elastic rod with lateral inertia as ³

$$\frac{{\partial }^{2}u}{\partial t^2}-c_0^2[1+n{a}_n{\left(\frac{\partial u}{\partial x}\right)}^{n-1}]\frac{{\partial }^{2}u}{\partial x^2}-\frac{v^2{J}_p}{s}\frac{{\partial }^{4}u}{\partial t^2\partial x^2}=0$$

(23)

where $s$ , $J_p$ and $c_0^2=E/\rho$ , $v$ , $E$ , and $\rho$ are the cross-section area of the rod, the polar moment of inertia, the square of the linear elastic longitudinal wave velocity, Poisson’s ratio, the Young’s modulus, and the density of the rod, respectively, $a_n$ is the material constant, $n$ is an integer. For the soft nonlinear materials $a_n<0$ , for example, majority of the metals. For the hard nonlinear materials such as rubbers and polymers, $a_n>0$ when $n=2,3$ .

METF method

Step 1: Consider n = 2, using the transform

$$u(x,t)\text{=}U(\xi ),\begin{array}{cc}& \xi =x-\mathit{ct}\end{array}$$

(24)

then equation (23) becomes an ODE

$$c^2{U}^{\prime \prime }(\xi )-c_0^2(1+2a_2{U}^{\prime }(\xi ))U^{\prime \prime }(\xi )-\frac{c^2{J}_p{v}^2{U}^{(4)}(\xi )}{s}=0$$

(25)

integrating equation (25) with respect to variable $\xi$ , we have

$$c^2{U}^{\prime }(\xi )-c_0^2{U}^{\prime }(\xi )-a_2{c}_0^2{U}^{\prime }{(\xi )}^2-\frac{c^2{J}_p{v}^2{U}^{(3)}(\xi )}{s}=0$$

(26)

now, setting $\phi (\xi )=U^{\prime }(\xi ),$ then equation (26) reduces to

$$c^2\phi (\xi )-c_0^2\phi (\xi )-a_2{c}_0^2\phi {(\xi )}^2-\frac{c^2{J}_p{v}^2{\phi }^{(2)}(\xi )}{s}=0$$

(27)

Step 2: Balancing the term $\phi {(\xi )}^2$ with term ${\phi }^{(2)}(\xi )$ in equation (27) gives $m=2$ .

Step 3: Taking into account the METF method, the solution of equation (27) reads

$$\phi (\xi )=d_0+d_{1}w(\xi )+d_{2}w{(\xi )}^2+b_1\frac{1}{w(\xi )}+b_2\frac{1}{w{(\xi )}^2}$$

(28)

substituting equation (28) along with equation (12) into equation (27), a set of algebraic equations for $c$ , $b$ , $d_i(0\le i\le 2)$ , $b_i(1\le i\le 2)$ is obtained as follows

$$c^2{d}_0-c_0^2{d}_0-a_2{c}_0^2{d}_0^2-2a_2{b}_1{c}_0^2{d}_1-2a_2{b}_2{c}_0^2{d}_2-\frac{2b_2{c}^2{J}_p{v}^2}{s}-\frac{2b^2{c}^2{d}_2{J}_p{v}^2}{s}=0,$$

$$c^2{d}_1-c_0^2{d}_1-2a_2{c}_0^2{d}_0{d}_1-2a_2{b}_1{c}_0^2{d}_2-\frac{2b{c}^2{d}_1{J}_p{v}^2}{s}=0,-a_2{c}_0^2{d}_1^2+c^2{d}_2-c_0^2{d}_2-2a_2{c}_0^2{d}_0{d}_2-\frac{8b{c}^2{d}_2{J}_p{v}^2}{s}=0,$$

$$-2a_2{c}_0^2{d}_1{d}_2-\frac{2c^2{d}_1{J}_p{v}^2}{s}=0,-a_2{c}_0^2{d}_2^2-\frac{6c^2{d}_2{J}_p{v}^2}{s}=0,$$

(29)

$$b_1{c}^2-b_1{c}_0^2-2a_2{b}_1{c}_0^2{d}_0-2a_2{b}_2{c}_0^2{d}_1-\frac{2b{b}_1{c}^2{J}_p{v}^2}{s}=0,$$

$$b_2{c}^2-a_2{b}_1^2{c}_0^2-b_2{c}_0^2-2a_2{b}_2{c}_0^2{d}_0-\frac{8b{b}_2{c}^2{J}_p{v}^2}{s}=0,$$

$$-2a_2{b}_1{b}_2{c}_0^2-\frac{2b^2{b}_1{c}^2{J}_p{v}^2}{s}=0,-a_2{b}_2^2{c}_0^2-\frac{6b^2{b}_2{c}^2{J}_p{v}^2}{s}=0$$

solving the above algebraic equation (29), the following sets of coefficients for the solutions generated from equation (28) are obtained as

Case 1

$$d_0=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=-\frac{6J_p{v}^2}{a_2(s-4b{J}_p{v}^2)},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=0,\hspace{1em}{c}^2=\frac{c_0^{2}s}{s-4b{J}_p{v}^2}$$

Case 2

$$d_0=-\frac{6b{J}_p{v}^2}{a_2(s\text{+}4b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=-\frac{6J_p{v}^2}{a_2(s\text{+}4b{J}_p{v}^2)},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=0,\hspace{1em}{c}^2=\frac{c_0^{2}s}{s\text{+}4b{J}_p{v}^2}$$

Case 3

$$d_0=-\frac{6b{J}_p{v}^2}{a_2(s\text{+}4b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=0,\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=-\frac{6b^2{J}_p{v}^2}{a_2(s\text{+}4b{J}_p{v}^2)},\hspace{1em}{c}^2=\frac{c_0^{2}s}{s\text{+}4b{J}_p{v}^2}$$

Case 4

$$d_0=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=0,\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=-\frac{6b^2{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)},\hspace{1em}{c}^2=\frac{c_0^{2}s}{s-4b{J}_p{v}^2}$$

Case 5

$$d_0=-\frac{12b{J}_p{v}^2}{a_2(s\text{+16}b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=-\frac{6J_p{v}^2}{a_2(s\text{+16}b{J}_p{v}^2)},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=-\frac{6b^2{J}_p{v}^2}{a_2(s\text{+}16b{J}_p{v}^2)},\hspace{1em}{c}^2=\frac{c_0^{2}s}{s\text{+}16b{J}_p{v}^2}$$

Case 6

$$d_0=\frac{4b{J}_p{v}^2}{a_2(s-\text{16}b{J}_p{v}^2)},\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=-\frac{6J_p{v}^2}{a_2(s-\text{16}b{J}_p{v}^2)},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=-\frac{6b^2{J}_p{v}^2}{a_2(s-16b{J}_p{v}^2)},\hspace{1em}{c}^2=\frac{c_0^{2}s}{s-16b{J}_p{v}^2}$$

Case 7

$$d_0=0,\hspace{1em}{d}_1=0,\hspace{1em}{d}_2=-\frac{6J_p{v}^2}{a_{2}s},\hspace{1em}{b}_1=0,\hspace{1em}{b}_2=0,\hspace{1em}c=\pm c_0,\hspace{1em}b=0$$

For the sake of simplicity, we consider only the solution with respect to Case 1, the other solutions can be obtained in a similar way

When $b<0$

$$\phi (\xi )=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)}\text{+}\frac{6b{J}_p{v}^2\mathrm{Tanh}{(\sqrt{-b}\xi )}^2}{a_2(s-4b{J}_p{v}^2)},$$

$$\phi (\xi )=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)}\text{+}\frac{6b{J}_p{v}^2\text{Coth}{(\sqrt{-b}\xi )}^2}{a_2(s-4b{J}_p{v}^2)}$$

and $c^2=\frac{c_0^{2}s}{s-4b{J}_p{v}^2}.$ When $b>0$

$$\phi (\xi )=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)}-\frac{6b{J}_p{v}^2\mathrm{Tan}{(\sqrt{b}\xi )}^2}{a_2(s-4b{J}_p{v}^2)},$$

$$\phi (\xi )=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)}-\frac{6b{J}_p{v}^2\text{Cot}{(\sqrt{b}\xi )}^2}{a_2(s-4b{J}_p{v}^2)}$$

and $c^2=\frac{c_0^{2}s}{s-4b{J}_p{v}^2}.$ When $b\text{=}0$

$$\phi (\xi )=-\frac{2b{J}_p{v}^2}{a_2(s-4b{J}_p{v}^2)}-\frac{6J_p{v}^2}{a_2(s-4b{J}_p{v}^2){\xi }^2}$$

and $c^2=c_0^2.$

Exp-function method

Suppose that the wave solution of equation (27) can be expressed in the following form

$$\phi (\xi )\text{=}\frac{d_1\hspace{0.17em}\exp (\xi )+d_0+d_2\hspace{0.17em}\exp (-\xi )}{b_1\hspace{0.17em}\exp (\xi )+b_0+b_2\hspace{0.17em}\exp (-\xi )}$$

(30)

for simplicity, we let $b_1\text{=}1$ . Substituting equation (30) into equation (27), we have

$$\frac{1}{A}(C_0\text{+}{C}_1\hspace{0.17em}\exp (\xi )+C_2\hspace{0.17em}\exp (2\xi )+\cdots +C_6\hspace{0.17em}\exp (6\xi ))\text{=}0$$

(31)

where $A={(b_2+e^{\xi }(b_0+e^{\xi }))}^{3}s$ ; collecting $C_i(i=0,1,\cdots ,6)$ and simultaneously equating to zero, a set of algebraic equations is obtained; solving the algebraic equations, the following results are obtained

Case 1

$$d_0=-\frac{b_0{J}_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_1=-\frac{J_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_2=-\frac{b_2{J}_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{c}^2{J}_p{v}^2+c^{2}s-c_0^{2}s=0$$

Case 2

$$d_0=\frac{2b_0{J}_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_1=-\frac{J_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_2=-\frac{b_0^2{J}_p{v}^2}{4a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{b}_2=\frac{b_0^2}{4},\hspace{1em}{c}^2{J}_p{v}^2+c^{2}s-c_0^{2}s=0$$

Case 3

$$d_0=\text{-}\frac{b_0{J}_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_1=-\frac{J_p{v}^2}{a_2(s\text{+}{J}_p{v}^2)},\hspace{1em}{d}_2=0,\hspace{1em}{b}_2=0,\hspace{1em}{c}^2{J}_p{v}^2+c^{2}s-c_0^{2}s=0$$

Case 4

$$d_0=\frac{3b_0{J}_p{v}^2}{a_2(s-J_p{v}^2)},d_1=0,d_2=0,b_2=\frac{b_0^2}{4},-c^2{J}_p{v}^2+c^{2}s-c_0^{2}s=0$$

For the sake of simplicity, we consider only the solution with respect to Case 4, the other solutions can be obtained in a similar way

$$\phi (\xi )\text{=}\frac{3b_0{J}_p{v}^2}{a_2(b_0+\frac{1}{4}{b}_0^2{e}^{-\xi }+e^{\xi })(s-J_p{v}^2)}$$

and $-c^2{J}_p{v}^2+c^{2}s-c_0^{2}s=0$ .

Conclusions

In this article, we investigated the exact travelling wave solutions to some nonlinear partial differential equations, namely the nonlinear vibrating string equation and the elastic rod equation by quasi hyperbolic function expansion method and METF method; the obtained solutions can be expressed by hyperbolic functions, which can outline various vibration properties. The paper gave some exact solutions which cannot be obtained by other analytical methods, for example the homotopy perturbation method^8,9 and the variational iteration method.^10,11 To the best of our knowledge, the results achieved in this article have not been reported in open literature, and we will further study the physical meaning of the obtained exact solutions.