In this Letter, we consider the Jaulent-Miodek equation

$ \begin{equation} \begin{split} &u_{t}+u_{xxx}+\frac{3}{2}vv_{xxx}+\frac{9}{2}v_{x}v_{xx}-6uvv_{x}-\frac{3}{2}u_{x}v^{2}=0, \\ &v_{t}+v_{xxx}-6u_{x}v-6uv_{x}-\frac{15}{2}v_{x}v^{2}=0 \end{split} \end{equation} $

(1.1)

without any initial conditions. There are many methods to solve eq.(1.1) with some initial conditions, such as variational iteration method ^[1], He's homotopy perturbation method ^[2], and homotopy analysis method ^[3]. These methods can only solve a special kind of solutions and obtain solutions which satisfy initial conditions. Recently, some researchers use extended tanh-method ^[4], generalized $(G\prime/G)$-expansion method ^[5], and Riccati equations method ^[6] to investigate the traveling wave solutions of eq. (1.1), and obtained some new types of complex solitary solutions, i.e.various combinations of trigonometric periodic functions and rational function solutions.

The purpose of this work is basing on exp-function method ^[7], use the exp-function method in rational form ^[8] to find the exact solutions of eq.(1.1).

To apply the exp-function method in rational form to eq.(1.1), we make use of the traveling wave transformation $\xi =kx+wt, u=u(\xi), v=v(\xi)$ where and are constants to be determined later. Then eq.(1.1) reduce to ordinary differential equations.

$ \begin{equation} \begin{split} & wu'+k^{3}u^{\prime\prime\prime}+\frac{3}{2}k^{3}vv^{\prime\prime\prime}+\frac{9}{2}k^{3}v'v^{\prime\prime}-6kuvv'-\frac{3}{2}ku'v^{2}=0, \\ & wv'+k^{3}v^{\prime\prime\prime}-6k(uv)'-\frac{15}{2}kv'v^{2}=0, \\ \end{split} \end{equation} $

(2.1)

where the prime denotes the derivative with respect to $\xi$.

The exp-function method in rational form is based on the as

$ \begin{eqnarray} u=\sum\limits_{j=0}^{m}\frac{a_{j}}{(1+e^{\xi})^{j}}, \end{eqnarray} $

(2.2)

$ \begin{eqnarray} v=\sum\limits_{j=0}^{n}\frac{b_{j}}{(1+e^{\xi})^{j}}, \end{eqnarray} $

(2.3)

where $m$ and $n$ are positive integers which are unknown to be further determined, $a_{j}$ and $b_{j}$ are unknown constants. In order to determine the value of $m$ and $n$, we balance the linear term $u^{\prime\prime\prime}$ with the nonlinear term $vv^{\prime\prime\prime}$ in the first equation of (2.1), and the linear term $v^{\prime\prime\prime}$ with the nonlinear term $v'v^{2}$ in the second equation of (2.1), by normal calculation, we have

$ \begin{eqnarray} u^{\prime\prime\prime}&=&\frac{K_{1}}{(1+e^{\xi})^{m+3}}, \end{eqnarray} $

(2.4)

$ \begin{eqnarray} vv^{\prime\prime\prime}&=&\frac{K_{2}}{(1+e^{\xi})^{2n+3}}, \end{eqnarray} $

(2.5)

$ \begin{eqnarray} v^{\prime\prime\prime}&=&\frac{K_{3}}{(1+e^{\xi})^{n+3}}, \end{eqnarray} $

(2.6)

$ \begin{eqnarray} v'v^{2}&=&\frac{K_{4}}{(1+e^{\xi})^{3n+1}}, \end{eqnarray} $

(2.7)

where $K_{1}, K_{2}, K_{3}, $ and $K_{4}$ are determined coefficients only for simplicity. Balancing highest order of exp-function in equations (2.4) and (2.5), we have $m=2n$. Similarly balancing equations (2.6) and (2.7), we obtain $n=1$, so $m=2$. Equations (2.2) and (2.3) become

$ \begin{align} & u=a_{0}+\frac{a_{1}}{1+e^{\xi}}+\frac{a_{2}}{(1+e^{\xi})^{2}}, \end{align} $

(2.8)

$ \begin{align} & v=b_{0}+\frac{b_{1}}{1+e^{\xi}}. \end{align} $

(2.9)

Substituting equations (2.8) and (2.9) into equation (2.1), by help of maple16, we have

$ \frac{1}{A}[C_{1}e^{4\xi}+C_{2}e^{3\xi}+C_{3}e^{2\xi}+C_{4}e^{\xi}]=0 $

and

$ \frac{1}{B}[D_{1}e^{3\xi}+D_{2}e^{2\xi}+D_{3}e^{\xi}]=0, $

where

$ \begin{eqnarray*} A&=&(1+e^{\xi})^{5}, B=(1+e^{\xi})^{4}, \\ C_{1}&=&-wa_{1}-\frac{3}{2}k^{3}b_{1}b_{0}+6ka_{0}a_{1}-k^{3}a_{1}+6kb_{1}a_{0}b_{0}+\frac{3}{2}ka_{1}b_{0}^{2}, \\ C_{2}&=& 18kb_{1}a_{0}b_{0}+18ka_{0}a_{1}+6ka_{1}^{2}-8k^{3}a_{2}+12ka_{0}a_{2}+9kb_{1}a_{1}b_{0}+\\ &&3k^{3}a_{1} +\frac{9}{2}ka_{1}b_{0}^{2}+6kb_{1}^{2}a_{0}\\ && + 3ka_{2}b_{0}^{2}+\frac{9}{2}k^{3}b_{1}b_{0}-6k^{3}b_{1}^{2}-3wa_{1}-2wa_{2}, \\ C_{3}&=&-4wa_{2}+18kb_{1}a_{0}b_{0}+6ka_{2}b_{0}^{2}+\frac{15}{2}kb_{1}^{2}a_{1}+\\&&3k^{3}a_{1}+12kb_{1}a_{2}b_{0} +\frac{9}{2}k^{3}b_{1}b_{0}+24ka_{0}a_{2}\\ && + 18kb_{1}a_{1}b_{0}+\frac{9}{2}ka_{1}b_{0}^{2}+12ka_{1}^{2} +12kb_{1}^{2}a_{0}+\frac{21}{2}k^{3}b_{1}^{2}+\\&&18ka_{0}a_{1}+18ka_{1}a_{2}+14k^{3}a_{2}-3wa_{1}, \\ C_{4}&=&\frac{15}{2}kb_{1}^{2}a_{1}+9kb_{1}^{2}a_{2}+18ka_{1}a_{2}+\\&&12kb_{1}a_{2}b_{0}+6kb_{1}^{2}a_{0} +\frac{3}{2}ka_{1}b_{0}^{2}-\frac{3}{2}k^{3}b_{1}b_{0}+6ka_{0}a_{1}+3ka_{2}b_{0}^{2}\\ &&+ 12ka_{2}^{2}+6kb_{1}a_{0}b_{0} +12ka_{0}a_{2}-wa_{1}-2wa_{2}-k^{3}a_{1}+\\&&9kb_{1}a_{1}b_{0}-\frac{3}{2}k^{3}b_{1}^{2} +6ka_{1}^{2}-2k^{3}a_{2}, \\ D_{1}&=&-k^{3}b_{1}+6kb_{1}a_{0}-wb_{1}+\frac{15}{2}kb_{1}b_{0}^{2}+6ka_{1}b_{0}, \\ D_{2}&=& 12kb_{1}a_{0}+12kb_{1}a_{1}+12ka_{1}b_{0}+12ka_{2}b_{0}-2wb_{1}+4k^{3}b_{1}+15kb_{1}b_{0}^{2} +15kb_{1}^{2}b_{0}, \\ D_{3}&=&-wb_{1}+6kb_{1}a_{0}+6ka_{1}b_{0}+12ka_{2}b_{0}+\frac{15}{2}kb_{1}b_{0}^{2}-k^{3}b_{1}+15kb_{1}^{2}b_{0} +\\&&12kb_{1}a_{1}+\frac{15}{2}kb_{1}^{3}+18kb_{1}a_{2}. \end{eqnarray*} $

Equating the coefficients of all powers of $e^{n\xi}$ to be zero, we obtain

$ \begin{equation} [C_{1}=0, C_{2}=0, C_{3}=0, C_{4}=0, D_{1}=0, D_{2}=0, D_{3}=0]. \end{equation} $

(2.10)

Solving the system, equations (2.10), simultaneously, we get the following solution

$ \begin{equation} a_{0}=-\frac{1}{4}b_{0}^{2}, a_{1}=-\frac{1}{2}k(k\pm b_{0}i), a_{2}=\frac{3}{4}k^{2}, b_{1}=\pm ik, w=\pm 3k^{2}b_{0}i-k^{3}+3kb_{0}^{2}. \end{equation} $

(2.11)

Using the transformations

$ \left\{\begin{array}{rl} \xi=i\zeta=Kx+Wt, \\ e^{\xi}=\cos(\zeta)+i\sin(\zeta), \end{array}\right. $

and $k=iK, w=iW$, where $K$ and $W$ are real number, eq.(2.11) becomes

$ \begin{equation} a_{0}=-\frac{1}{4}b_{0}^{2}, a_{1}=\frac{1}{2}K^{2}\pm \frac{1}{2}Kb_{0}, a_{2}=-\frac{3}{4}K^{2}, b_{1}=\mp K, w=\mp 3K^{2}b_{0}-K^{3}+3Kb_{0}^{2}. \end{equation} $

Inserting equation (2.12) into equations (2.8) and (2.9) yields the following exact solution

$ \begin{array}{l} u = - \frac{1}{4}b_0^2 + \frac{{\frac{1}{2}{K^2} \pm \frac{1}{2}K{b_0}}}{{1 + {e^{i\zeta }}}} - \frac{{ - \frac{3}{4}{K^2}}}{{{{(1 + {e^{i\zeta }})}^2}}}\\ \;\;\;\; = \frac{{ - \frac{3}{2}b_0^2 + \frac{3}{4}{K^2} \pm \frac{3}{2}{b_0}K + ( \pm 2{b_0}K - 2b_0^2 + \frac{1}{2}{K^2})\cos (\zeta ) + ( \pm \frac{1}{2}{b_0}K - \frac{1}{2}b_0^2 - \frac{1}{4}{K^2})\cos (2\zeta )}}{{6 + 8\cos (\zeta ) + 2\cos (2\zeta )}}\\ \;\;\;\; + i\frac{{({b_0}K \mp \frac{1}{2}{K^2})\sin (\zeta ) + (\frac{1}{2}{b_0}K \mp \frac{1}{4}{K^2})\sin (2\zeta )}}{{6 + 8\cos (\zeta ) + 2\cos (2\zeta )}}. \end{array} $

(2.13)

If we search for a periodic solution or compaction-like solution, the imaginary part of eq.(2.13) must be zero, that requires

$ \begin{equation} (b_{0}K\mp\frac{1}{2}K^{2})\sin(\zeta)+(\frac{1}{2}b_{0}K\mp\frac{1}{4}K^{2})\sin(2\zeta)=0. \end{equation} $

(2.14)

Solving eq.(2.14), we obtain

$ \begin{equation} b_{0}=\pm\frac{1}{2}K. \end{equation} $

(2.15)

Substituting eq.(2.15) into eq.(2.12), then eq.(2.12) becomes

$ \begin{equation} a_{0}=-\frac{1}{16}K^{2}, a_{1}=\frac{3}{4}K^{2}, a_{2}=-\frac{3}{4}K^{2}, b_{1}=\mp K, W=-\frac{7}{4}K^{3}. \end{equation} $

(2.16)

Inserting equation (2.16) into the real part of equations (2.13) and (2.9) yields the following exact solution

$ \begin{equation} \begin{split} & u_{1}=\frac{-\frac{3}{8}K^{2}-\frac{1}{2}K^{2}\cos(i(Kx-\frac{7}{4}K^{3}t)) -\frac{5}{8}K^{2}\cos(2i(Kx-\frac{7}{4}K^{3}t))}{6+8\cos(i(Kx-\frac{7}{4}K^{3}t))+2\cos(2i(Kx-\frac{7}{4}K^{3}t))}, \\ & v_{1}=\frac{-iK\sin(i(Kx-\frac{7}{4}K^{3}t)}{2+2\cos(i(Kx-\frac{7}{4}K^{3}t)}, \\ & u_{2}=\frac{\frac{9}{8}K^{2}+K^{2}\cos(i(Kx-\frac{7}{4}K^{3}t)) -\frac{1}{8}K^{2}\cos(2i(Kx-\frac{7}{4}K^{3}t))}{6+8\cos(i(Kx-\frac{7}{4}K^{3}t))+2\cos(2i(Kx-\frac{7}{4}K^{3}t))}, \\ & v_{2}=\frac{iK\sin(i(Kx-\frac{7}{4}K^{3}t)}{2+2\cos(i(Kx-\frac{7}{4}K^{3}t)}, \end{split} \end{equation} $

(2.17)

where $K$ is free parameter.

The graph of the solution (2.17) is given in the following figures with $K=2$.

In this study, exp-function method in rational form with a computerized symbolic computation has been successfully applied to find generalized complex solitary solutions of Jaulent-Miodek equation without any initial conditions, so the solutions are more general. The results are simpler with the fewest free parameters and can be shown graphically.