Symmetry analysis for three-dimensional dissipation Rossby waves

Abstract

Rossby waves, belonging to the most important waves in the atmosphere and ocean, can affect the energy transfer of the atmosphere and ocean and have significant theoretical meaning and research value. In previous research performed with the theory and calculation method limit, the dissipation effect was commonly ignored. However, under the conditions of the weak linear approximation, the magnitude difference between nonlinear and dissipation is small, and the dissipation effect must be considered. In this paper, based on the classic Lie group approach, the $(3+1)$-dimensional quasi-geodetiophic vorticity equation with dissipation effect is solved. With the help of the solutions, we can better comprehend the influence of the dissipation effect on the propagation of Rossby waves.

1 Introduction

Rossby waves, which are caused by the rotation of the earth and the influence of the sphere effect, are long and large-scale permanent waves in the ocean and atmosphere, such as the huge red spots in Jupiter’s atmosphere and eddy currents in the gulf of Mexico. Rossby waves determine the ocean’s response to the climate and atmospheric change and have significant theoretical meaning and research value. However, in recent years, many researchers have focused on the traveling-wave solutions for handling nonlinear problems [1–3]. Few researchers have paid attention to the solution of the Rossby wave. Thus, with the development of theory, the study of Rossby waves is an important research direction [4–7].

Nonlinear partial differential equations [8–10] play an important role in the field of Rossby waves. Many models have been derived [11–14], and many methods have been used to solve the nonlinear partial differential equations, such as the algebro-geometric method [15, 16], Hirota method [17], Painlevé analysis method [18], Darboux transformations [19–21], Lie symmetry method [22–24] and so on [25–29]. Based on the classic Lie group method, the $(2+1)$-dimensional nonlinear inviscid barotropic nondivergent vorticity equation was studied by Huang and Lou [30], and the $(3+1)$-dimensional nonlinear Charney–Obukhov equation was studied by Kudryavtsev and Myagkov [31]. However, the dissipation effect was ignored in these studies of Rossby waves. Friction dissipation, one of the external forces in the atmosphere and ocean, plays an increasingly vital role in atmospheric circulation. Under the conditions of weak linear approximation, the magnitude difference between nonlinear and dissipation is very small, i.e., the dissipation effect should be considered in the research of Rossby waves.

In this paper, we consider the $(3+1)$-dimensional quasi-geodetiophic vorticity equation with dissipation effect

$$ \frac{\partial }{\partial t}\Delta u+J(u,\Delta u)+\alpha \Delta u+ \beta \frac{\partial }{\partial x}u=0, $$

(1)

where the three-dimensional Laplacian can be expressed by $\Delta =\frac{ \partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{ \partial^{2}}{\partial z^{2}}$, the dimensionless stream function can be described by u, the Jacobian operator can be introduced by $J(a,b)=\frac{\partial a}{\partial x}\frac{\partial b}{\partial y}-\frac{ \partial a}{\partial y}\frac{\partial b}{\partial x}$, $\beta =\beta _{0}(L^{2}/U)$ and $\beta_{0}=(\omega_{0}/R_{0})\cos \phi_{0}$, in which $\omega_{0}$ is the angular frequency of the Earth’s rotation, L depicts the characteristic horizontal length, U expresses the velocity scales, $\phi_{0}$ and $R_{0}$ are the latitude and the Earth’s radius, respectively, and $\alpha \Delta u$ denotes the dissipation effect, in which α is the dissipation coefficient.

The structure of the full paper is as follows. We apply the classic Lie group method to acquire the solution of a $(3+1)$-dimensional dissipation Rossby wave in Sect. 2. In Sect. 3, we discuss the approximate analytical solution of a $(2+1)$-dimensional dissipation Rossby wave. Finally, the dissipation effect is researched, and some conclusions are reported in Sect. 4.

2 $(3+1)$-Dimensional dissipation Rossby wave

To discuss the dissipation effect of three-dimensional dissipation Rossby waves, we first study the solution of Eq. (1). In the following, we introduce the vector field

$$ \begin{aligned} V&=\xi (x,y,z,t,u)\frac{\partial }{\partial x}+\eta (x,y,z,t,u) \frac{ \partial }{\partial y}+\lambda (x,y,z,t,u) \frac{\partial }{\partial z} \\ &\quad{} +\tau (x,y,z,t,u)\frac{\partial }{\partial t}+\phi (x,y,z,t,u)\frac{\partial }{\partial u}. \end{aligned} $$

The first-order propagator is defined as

$$ \mathit {Pr}^{(1)}V=V+\phi^{x}\frac{\partial }{\partial u_{x}}+\phi^{y} \frac{ \partial }{\partial u_{y}}+\phi^{z}\frac{\partial }{\partial u_{z}}+ \phi^{t} \frac{\partial }{\partial u_{t}}, $$

and the second-order propagator is defined as

$$ \begin{aligned} \mathit {Pr}^{(2)}V &=\mathit {Pr}^{(1)}V+ \phi^{xx}\frac{\partial }{\partial u_{xx}}+\phi ^{xy}\frac{\partial }{\partial u_{xy}}+ \phi^{yy}\frac{\partial }{ \partial u_{xx}}+\phi^{xz}\frac{\partial }{\partial u_{pz}}+ \phi^{zz}\frac{ \partial }{\partial u_{zz}}+\phi^{y z} \frac{\partial }{\partial u_{y z}} \\ & \quad {} +\phi^{xt}\frac{\partial }{ \partial u_{xt}}+\phi^{yt} \frac{\partial }{\partial u_{\theta t}}+ \phi^{zt}\frac{\partial }{\partial u_{zt}}+\phi^{tt} \frac{\partial }{ \partial u_{tt}}. \end{aligned} $$

Similarly, the third-order propagator has the form

$$ \begin{aligned}[b] \mathit {Pr}^{(3)}V &=\mathit {Pr}^{(2)}V+ \phi^{xxx}\frac{\partial }{\partial u_{xxx}}+ \phi^{xxy}\frac{\partial }{\partial u_{xxy}}+ \phi^{xyy}\frac{\partial }{\partial u_{xyy}}+\phi^{yyy}\frac{\partial }{\partial u_{yyy}}+ \phi^{\rho \rho z}\frac{\partial }{\partial u_{xxz}} \\ & \quad {} +\phi^{xzz}\frac{ \partial }{\partial u_{xzz}}++\phi^{zzz} \frac{\partial }{\partial u _{zzz}}+\cdots +\phi^{yyz}\frac{\partial }{\partial u_{yyz}}+\cdots . \end{aligned} $$

(2)

According to the Lie group method, by substituting (2) into (1), we obtain

$$ \begin{aligned}[b] & \phi^{xxt}+\phi^{yyt}+ \phi^{xxy}+\phi^{yyy}u_{x}+\phi^{yzz}u_{x}+ \phi^{x}(u_{xxy}+u_{yyy}+u_{yzz})-\bigl( \phi^{xxx}+\phi^{xyy}+\phi^{xzz}\bigr)u _{y} \\ & \quad {} +\alpha \bigl(\phi^{xx}+\phi^{yy}+ \phi^{zz}\bigr)+\beta \bigl(\phi^{x}+\phi ^{y}+ \phi^{z}\bigr)=0, \end{aligned} $$

(3)

where

$$ \begin{aligned} & \phi^{x}=D_{x}(\phi -\xi u_{x}- \eta u_{y}-\lambda u_{z}-\tau u_{t})+ \xi u_{xx}+\eta ux_{xy}+\lambda u_{xz}+\tau u_{xt}, \\ & \phi^{xx}=D ^{2}_{x}(\phi -\xi u_{x}- \eta u_{y}-\lambda u_{z}-\tau u_{t})+\xi u _{xxx}+\eta ux_{xxy}+\lambda u_{xxz}+\tau u_{xxt}, \\ & \phi^{xxx}=D ^{3}_{x}(\phi -\xi u_{x}- \eta u_{y}-\lambda u_{z}-\tau u_{t})+\xi u _{xxxx}+\eta ux_{xxxy}+\lambda u_{xxxz}+\tau u_{xxxt}. \end{aligned} $$

It is important to emphasize that

$$ \begin{aligned} \frac{\partial^{3} u}{\partial x^{2} \partial t}+\frac{\partial^{3} u}{ \partial y^{2} \partial t}+\frac{\partial^{3} u}{\partial z^{2} \partial t}&= - \frac{\partial u}{\partial x}\frac{\partial^{2} u}{ \partial x^{3} \partial y}-\frac{\partial u}{\partial x}\frac{\partial ^{3} u}{\partial y^{3}}- \frac{\partial u}{\partial x}\frac{\partial ^{3} u}{\partial y \partial z^{2}}+\frac{\partial u}{\partial y}\frac{ \partial^{3} u}{\partial x^{3}}+ \frac{\partial u}{\partial y}\frac{ \partial^{3} u}{\partial x \partial z^{2}} \\ &\quad {} -\alpha \biggl(\frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+ \frac{ \partial^{2} u}{\partial z^{2}}\biggr)-\beta \biggl(\frac{\partial u}{\partial x}+\frac{ \partial u}{\partial y}+ \frac{\partial u}{\partial z}\biggr) \end{aligned} $$

and

$$ \begin{aligned} \frac{\partial^{3} u}{\partial x^{2} \partial t}&= -\frac{\partial^{3} u}{\partial y^{2} \partial t}-\frac{\partial^{3} u}{\partial z^{2} \partial t}- \frac{\partial u}{\partial x}\frac{\partial^{2} u}{\partial x^{3} \partial y}-\frac{\partial u}{\partial x}\frac{\partial^{3} u}{ \partial y^{3}}- \frac{\partial u}{\partial x}\frac{\partial^{3} u}{ \partial y \partial z^{2}} \\ & \quad {} +\frac{\partial u}{\partial y}\frac{ \partial^{3} u}{\partial x^{3}}+\frac{\partial u}{\partial y} \frac{ \partial^{3} u}{\partial x \partial z^{2}}-\alpha \biggl(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+ \frac{ \partial^{2} u}{\partial z^{2}}\biggr)-\beta \biggl(\frac{\partial u}{\partial x}+\frac{ \partial u}{\partial y}+ \frac{\partial u}{\partial z}\biggr), \end{aligned} $$

and that $u_{z},u_{y},u_{x},u_{t},u_{xt},u_{xz},u_{xy},u_{xyz},\ldots$ are not independent.

According to a complicated calculation and the above transformation, we obtain

$$ \textstyle\begin{cases} \frac{\partial \xi }{\partial u}=0,\quad\quad \frac{\partial \eta }{\partial u}=0,\quad\quad \frac{\partial \lambda }{\partial u}=0,\quad\quad \frac{\partial \tau }{ \partial u}=0,\\ \frac{\partial \xi }{\partial y}=0,\quad\quad \frac{\partial \xi }{\partial z}=0,\quad\quad \frac{\partial \eta }{\partial x}=0,\quad\quad \frac{\partial \eta }{\partial z}=0,\\ \frac{\partial \lambda }{\partial x}=0,\quad\quad \frac{ \partial \lambda }{\partial y}=0,\quad\quad \frac{\partial \lambda }{\partial t}=0, \\ \frac{\partial \tau }{\partial x}=0,\quad\quad \frac{\partial \tau }{ \partial y}=0,\quad\quad \frac{\partial \tau }{\partial z}=0,\\ \frac{\partial ^{2} \phi }{\partial u^{2}}=0,\quad\quad \frac{\partial^{2}\phi }{\partial u \partial x}=0,\quad\quad \frac{\partial^{2}\phi }{\partial uy}=0,\quad\quad \frac{\partial ^{2}\phi }{\partial u\partial z}=0,\quad\quad \frac{\partial^{2} \phi }{\partial u\partial t}=0,\\ \frac{\partial^{2} \xi }{\partial x^{2}}=0,\quad\quad \frac{ \partial^{2}\eta }{\partial y^{2}}=0,\quad\quad \frac{\partial^{2}\lambda }{ \partial z^{2}}=0. \end{cases} $$

In addition, we can obtain the following coefficients:

$$ \textstyle\begin{cases} 1: \quad \frac{\partial^{3}\phi }{\partial x^{2}\partial t}+\frac{ \partial^{3}\phi }{\partial y^{2}\partial t}+\frac{\partial^{3}\phi }{ \partial z^{2}\partial t}+\alpha (\frac{\partial^{2}\phi }{\partial x ^{2}}+\frac{\partial^{2}\phi }{\partial y^{2}}+\frac{\partial^{2} \phi }{\partial z^{2}})+\beta \frac{\partial \phi }{\partial x}=0,\\ u_{x}: \quad \beta (\frac{\partial \xi }{\partial x}+\frac{\partial \tau }{\partial t})=0,\\ u_{yyt}: \quad \frac{\partial \xi }{\partial x}-\frac{ \partial \eta }{\partial y}=0,\\ u_{zzt}: \quad \frac{\partial \xi }{ \partial x}-\frac{\partial \lambda }{\partial z}=0,\\ u_{x}u_{yyy}:\quad \frac{\partial \phi }{\partial u}+\frac{\partial \xi }{\partial x}-3\frac{ \partial \eta }{\partial y}+\frac{\partial \tau }{\partial t}=0,\\ u_{x}u_{xxy}:\quad \frac{\partial \phi }{\partial u}-\frac{\partial \xi }{\partial x}-\frac{\partial \eta }{\partial y}+\frac{\partial \tau }{\partial t}=0,\\ u_{y}u_{xxx}: \quad -\frac{\partial \phi }{ \partial u}+\frac{\partial \xi }{\partial x}+\frac{\partial \eta }{ \partial y}-\frac{\partial \tau }{\partial t}=0,\\ u_{y}u_{xyy}:\quad -\frac{ \partial \phi }{\partial u}-\frac{\partial \xi }{\partial x}+3\frac{ \partial \eta }{\partial y}-\frac{\partial \tau }{\partial t}=0. \end{cases} $$

By comparing the coefficients of $u_{z},u_{y},u_{x},u_{t},u_{xt},u _{xz},u_{xy},u_{xyz},\ldots$ , the general solutions can be written as

$$ \textstyle\begin{cases} \xi =C_{1}+C_{5}x+C_{7}\int h_{2}(t)\,dt,\\ \eta =C_{2}+C_{5}y+C_{10} \int h_{5}(t)\,dt,\\ \lambda =C_{3}+C_{5}z,\\ \tau =C_{4}-C_{5}t,\\ \phi =3C_{5}u+C_{6}h_{1}(t)-C_{7}h_{2}(t)y+C_{8}h_{3}(z)+C_{9}h_{4}(t)z \\ \hphantom{\phi =}{}+C _{10}[h_{5}(t)x-\frac{\beta }{2}h_{5}(t)z^{2}+\frac{\alpha }{2}e^{-t}z ^{2}], \end{cases} $$

(4)

where $C_{1},C_{2}, \ldots, C_{10}$ are arbitrary constants, and $h_{1}(t),h_{2}(t), \ldots, h_{5}(t)$ are arbitrary functions of t.

Thus, we obtain the Lie algebra basis of the classic symmetry group for Eq. (1):

$$ \textstyle\begin{cases} V_{1}=\frac{\partial }{\partial x},\quad\quad V_{2}=\frac{\partial }{\partial y},\quad\quad V_{3}=\frac{\partial }{\partial z},\quad\quad V_{4}=\frac{\partial }{\partial t}, \\ V_{5}=x\frac{\partial }{\partial x}+y\frac{\partial }{ \partial y}+z\frac{\partial }{\partial z}-t\frac{\partial }{\partial t}+3u\frac{\partial }{\partial u},\quad\quad V_{6}=h_{1}(t)\frac{\partial }{ \partial u}, \\ V_{7}=(\int h_{2}(t)\,dt) \frac{\partial }{\partial x}-h_{2}(t)y\frac{\partial }{\partial u},\\ V_{8}=h_{3}(z)\frac{\partial }{\partial u},\quad\quad V_{9}=h_{4}(t)z\frac{ \partial }{\partial u},\\ V_{10}=(\int h_{5}(t)\,dt)\frac{\partial }{ \partial y}+[h_{5}(t)x-\frac{\beta }{2}h_{5}(t)z^{2}+ \frac{\alpha }{2}e^{-t}z^{2}]\frac{\partial }{\partial u}, \end{cases} $$

(5)

where $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$, $h_{5}$ are continuous functions.

If z is eliminated in $V_{5}$, operators $V_{1}$, $V_{2}$, $V_{3}$, $V_{4}$, $V _{5}$, $V_{6}$, $V_{7}$ are the Lie algebra basis of the classic symmetry group for the $(2 + 1)$-dimensional case. Specifically, $V_{8}$, $V_{9}$, $V _{10}$ can be extended to the classic symmetry group for the $(3 + 1)$-dimensional case.

When we know a particular solution, a new solution of the differential equation can be acquired by the classic Lie symmetry group method [32–34]. Suppose that a solution of Eq. (1) is expressed by $u_{s}(t,x,y,z)$. It is easy to infer that operators $V_{6}$, $V_{7}$, $V_{8}$, $V_{9}$, $V_{10}$ have the following formulas for the new solution $u_{\mathrm {new}}(t,x,y,z)$:

$$ \textstyle\begin{cases} u_{\mathrm {new}}(t,x,y,z)=h_{1}(t)+u_{s}(t,x,y,z), & \text{(a)} \\ u_{\mathrm {new}}(t,x,y,z)=-h_{2}(t)y+u_{s}(t,x+\int h_{2}(t)\,dt,y,z), & \text{(b)} \\ u_{\mathrm {new}}(t,x,y,z)=h_{3}(z)+u_{s}(t,x,y,z), & \text{(c)} \\ u_{\mathrm {new}}(t,x,y,z)=h_{4}(t)z+u_{s}(t,x,y,z), & \text{(d)} \\ u_{\mathrm {new}}(t,x,y,z)=h_{5}(t)x-\frac{\beta }{2}h_{5}(t)z^{2}+\frac{\alpha }{2}e^{-t}z^{2}+u_{s}(t,x,y+\int h_{5}(t)\,dt,z). & \text{(e)} \end{cases} $$

(6)

According to the nontrivial transformations (6b), (6d), and (6e), we obtain

$$ \begin{aligned}[b] u_{\mathrm {new}}(t,\rho ,\theta ,z)&= h_{5}(t)x-h_{2}(t)y+h_{4}(t)z- \frac{ \beta }{2}h_{5}(t)z^{2}+\frac{\alpha }{2}e^{-t}z^{2} \\ & \quad {} +u_{s}\biggl(t,x+ \int h_{2}(t)\,dt,y+ \int h_{5}(t)\,dt,z\biggr). \end{aligned} $$

(7)

Clearly, Eq. (7) has the following transformations for the component velocities:

$$ \textstyle\begin{cases} l_{\mathrm {new}}(t,x,y,z)=-h_{2}(t)+l_{s}(t,x+\int h_{2}(t)\,dt,y+\int h_{5}(t)\,dt,z), & \text{(a)} \\ m_{\mathrm {new}}(t,x,y,z)=-h_{5}(t)+m_{s}(t,x+\int h_{2}(t)\,dt,y+\int h_{5}(t)\,dt,z), & \text{(b)} \\ n_{\mathrm {new}}(t,x,y,z)=-[h_{4}(t)-\frac{\beta }{2}h_{5}(t)z+ \frac{\alpha }{2}e^{-t}z] \\ \hphantom{n_{\mathrm {new}}(t,x,y,z)=}{}+n_{s}(t,x+\int h_{2}(t)\,dt,y+\int h_{5}(t)\,dt,z), & \text{(c)} \end{cases} $$

(8)

where

$$ \textstyle\begin{cases} l_{s}(t,x,y,z)=-\frac{\partial u_{s}(t,x,y,z)}{\partial y}, \\ m _{s}(t,x,y,z)=\frac{\partial u_{s}(t,x,y,z)}{\partial x},\\ n_{s}(t,x,y,z)=-(\frac{ \partial }{\partial t}+l_{s}(x,y,z)\frac{\partial }{\partial x}+m_{s}(x,y,z)\frac{ \partial }{\partial y})\frac{\partial u_{s}(t,x,y,z)}{\partial z}. \end{cases} $$

Based on Eqs. (7)–(8), we can note that in order to get the new form of a $(3+1)$-dimensional dissipation Rossby wave, we need to find the solution of a $(2+1)$-dimensional dissipation Rossby wave.

3 $(2+1)$-Dimensional dissipation Rossby wave

In this section, we consider the $(2+1)$-dimensional approximate analytical solution of Eq. (1)

$$ \frac{\partial }{\partial t}\nabla^{2} u+J\bigl(u, \nabla^{2} u\bigr)+\alpha \nabla^{2}u+\beta \frac{\partial }{\partial x}u=0, $$

(9)

where the Jacobian operator can be introduced by $J(a,b)=\frac{\partial a}{\partial x}\frac{\partial b}{\partial y}-\frac{\partial a}{\partial y}\frac{\partial b}{\partial x}$, and $\nabla^{2}=\frac{\partial^{2}}{ \partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}$ depicts the two-dimensional Laplacian.

First, when dissipation does not exist, we introduce the following transform:

$$ \xi =x-y-ct, $$

(10)

where c expresses the phase speed of the wave. Substituting (10) into (9), we obtain

$$ -c\frac{\partial }{\partial \xi }\biggl[2\frac{\partial^{2} u}{\partial \xi ^{2}}- \frac{\beta }{c}u\biggr]=0. $$

(11)

It is easy to write the general solution of (11)

$$ u=C_{1}e^{\frac{\sqrt{2\beta }\xi }{\sqrt{c}}}+C_{2}e^{-\frac{\sqrt{2 \beta }\xi }{\sqrt{c}}}. $$

(12)

Equation (12) can be rewritten as

$$ \begin{aligned}[b] u&= C_{1}\biggl(\cosh \biggl( \frac{\sqrt{2\beta }(x-y-ct)}{\sqrt{c}}\biggr)+ \operatorname {sech}\biggl(\frac{\sqrt{2\beta }(x-y-ct)}{\sqrt{c}}\biggr)\biggr) \\ & \quad {} +C_{2}\biggl( \cosh \biggl(\frac{\sqrt{2\beta }(x-y-ct)}{\sqrt{c}}\biggr)-\operatorname {sech}\biggl(\frac{\sqrt{2 \beta }(x-y-ct)}{\sqrt{c}}\biggr)\biggr), \end{aligned} $$

(13)

where $C_{1}$, $C_{2}$ are constants.

Then, we consider the impact of dissipation on Eq. (9). Assume $\alpha \ll 1$ and $\alpha \ll \beta$, and take a new space coordinate:

$$ \rho =x+y- \int_{0}^{t}\frac{au_{0}}{2}\,dt. $$

(14)

Suppose that $u_{0}=u_{0}(\alpha t)$ varies slowly with time; we then obtain

$$ 2\frac{\partial }{\partial t}\biggl( \frac{\partial^{2} u}{\partial \rho^{2}}\biggr)-au_{0} \frac{\partial }{ \partial \rho }\biggl(\frac{\partial^{2} u}{\partial \rho^{2}}\biggr)+2\alpha \frac{ \partial^{2} u}{\partial \rho^{2}}+ \beta \frac{\partial u}{\partial \rho }=0, $$

(15)

by substituting (14) into (9). Let

$$ \tau =t,\quad\quad \eta =\alpha t, $$

(16)

and the solution has the following form:

$$ u(\rho ,t)=u_{1}(\rho ,\tau ,\eta )+\alpha u_{2}(\rho ,\tau )+\cdots . $$

(17)

Substituting (17) into (15), we have

$$\begin{aligned}& \alpha^{0}: \quad 2\frac{\partial }{\partial \tau }\biggl( \frac{\partial^{2} u _{1}}{\partial \rho^{2}}\biggr)-au_{0}\frac{\partial }{\partial \rho }\biggl( \frac{ \partial^{2} u_{1}}{\partial \rho^{2}}\biggr)+\beta \frac{\partial u_{1}}{ \partial \rho }=0, \end{aligned}$$

(18)

$$\begin{aligned}& \alpha^{1}: \quad 2\frac{\partial }{\partial \tau }\biggl( \frac{\partial^{2} u _{2}}{\partial \rho^{2}}\biggr)-au_{0}\frac{\partial }{\partial \rho }\biggl( \frac{ \partial^{2} u_{2}}{\partial \rho^{2}}\biggr)+2\frac{\partial^{2} u_{1}}{ \partial \rho^{2}}+\beta \frac{\partial u_{2}}{\partial \rho }=-2 \frac{ \partial }{\partial \eta }\biggl(\frac{\partial^{2} u_{1}}{\partial \rho ^{2}}\biggr). \end{aligned}$$

(19)

Then, let

$$ \zeta =\rho +\frac{au_{0}}{2}\tau . $$

(20)

We have

$$\begin{aligned}& \alpha^{0}: \quad 2\frac{\partial }{\partial \tau }\biggl( \frac{\partial^{2} u _{1}}{\partial \rho^{2}}\biggr)+\beta \frac{\partial u_{1}}{\partial \rho }=0, \end{aligned}$$

(21)

$$\begin{aligned}& \alpha^{1}: \quad 2\frac{\partial }{\partial \tau }\biggl( \frac{\partial^{2} u _{2}}{\partial \rho^{2}}\biggr)+\beta \frac{\partial u_{2}}{\partial \rho }=-2\frac{ \partial }{\partial \eta }\biggl( \frac{\partial^{2} u_{1}}{\partial \rho ^{2}}\biggr)-2\frac{\partial^{2} u_{1}}{\partial \rho^{2}}. \end{aligned}$$

(22)

The solution of (21) is

$$ \begin{aligned}[b] u&= \frac{u_{0}}{C_{2}}\biggl(\cosh \biggl( \frac{\sqrt{2\beta }(\zeta -\frac{au _{0}}{2}\tau )}{\sqrt{c}}\biggr)+\operatorname {sech}\biggl(\frac{\sqrt{2\beta }( \zeta -\frac{au_{0}}{2}\tau )}{\sqrt{c}}\biggr)\biggr) \\ & \quad {} +\frac{u_{0}}{C_{1}}\biggl( \cosh \biggl(\frac{\sqrt{2\beta }(\zeta -\frac{au_{0}}{2}\tau )}{ \sqrt{c}}\biggr)- \operatorname {sech}\biggl(\frac{\sqrt{2\beta }(\zeta -\frac{au_{0}}{2} \tau )}{\sqrt{c}}\biggr)\biggr), \end{aligned} $$

(23)

where $u_{0}=C_{1}C_{2}$. By using (14) and (20), we obtain

$$ \begin{aligned}[b] u&= \frac{u_{0}}{C_{2}}\biggl(\cosh \biggl( \frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{au _{0}}{2}\,dt)}{\sqrt{c}}\biggr)+\operatorname {sech}\biggl(\frac{\sqrt{2\beta }(x+y-\int _{0}^{t}\frac{au_{0}}{2}\,dt)}{\sqrt{c}}\biggr)\biggr) \\ &\quad {} +\frac{u_{0}}{C_{1}}\biggl( \cosh \biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{au_{0}}{2}\,dt)}{ \sqrt{c}}\biggr)- \operatorname {sech}\biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{au _{0}}{2}\,dt)}{\sqrt{c}}\biggr)\biggr). \end{aligned} $$

(24)

Next, setting

$$ u_{2}=M(N),\quad\quad N=\zeta -v\tau . $$

(25)

By substituting (25) into (22), we obtain

$$ -2v\frac{\partial^{3} M}{\partial N^{3}}+\beta \frac{\partial M}{ \partial N}=W(u_{1}), $$

(26)

where

$$ W(u_{1})=-2\frac{\partial }{\partial \eta }\biggl(\frac{\partial^{2} u_{1}}{ \partial N^{2}}\biggr)-2 \frac{\partial^{2} u_{1}}{\partial N^{2}}. $$

The solvability condition of Eq. (26) is

$$ \int_{-\infty }^{+\infty }F(N)W(u_{1})\,dN=0, $$

(27)

where

$$ -2v\frac{\partial^{3} F }{\partial N^{3}}+\beta \frac{\partial F}{ \partial N}=0. $$

(28)

The solution of Eq. (28) is easy to obtain:

$$ \begin{aligned}[b] F&=\frac{u_{0}}{C_{2}}\biggl(\cosh \biggl(\frac{\sqrt{2\beta }N}{\sqrt{c}} \biggr)+ \operatorname {sech}\biggl(\frac{\sqrt{2\beta }N}{\sqrt{c}}\biggr)\biggr) \\ &\quad{} + \frac{u_{0}}{C_{1}}\biggl( \cosh \biggl(\frac{\sqrt{2\beta }N}{\sqrt{c}}\biggr)- \operatorname {sech}\biggl(\frac{\sqrt{2\beta }N}{\sqrt{c}}\biggr) \biggr),\end{aligned} $$

(29)

in the case of $G(\pm \infty )=0$. Equation (29) can be rewritten as

$$ \begin{aligned}[b] F&= \frac{u_{0}}{C_{2}}\biggl(\cosh \biggl( \frac{\sqrt{2\beta }(\zeta -v\tau )}{ \sqrt{c}}\biggr)+\operatorname {sech}\biggl(\frac{\sqrt{2\beta }(\zeta -v\tau )}{ \sqrt{c}}\biggr)\biggr) \\ &\quad {} +\frac{u_{0}}{C_{1}}\biggl(\cosh \biggl(\frac{\sqrt{2\beta }( \zeta -v\tau )}{\sqrt{c}}\biggr)- \operatorname {sech}\biggl(\frac{\sqrt{2\beta }( \zeta -v\tau )}{\sqrt{c}}\biggr)\biggr). \end{aligned} $$

(30)

Substituting (30) into (27), we obtain

$$ u_{0}=\bar{u}_{0}e^{-\alpha t}, $$

(31)

where $\bar{u}_{0}=u_{0}$. Therefore, we get

$$\begin{aligned} \begin{aligned}[b] u&= \frac{\bar{u}_{0}e^{-\alpha t}}{C_{2}}\biggl(\cosh \biggl( \frac{\sqrt{2 \beta }(x+y-\int_{0}^{t}\frac{a\bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{ \sqrt{c}}\biggr)+\operatorname {sech}\biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{a \bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{\sqrt{c}}\biggr)\biggr) \\ &\quad {} +\frac{\bar{u}_{0}e ^{-\alpha t}}{C_{1}}\biggl(\cosh \biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{a \bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{\sqrt{c}}\biggr) \\ &\quad{} - \operatorname {sech}\biggl(\frac{\sqrt{2 \beta }(x+y-\int_{0}^{t}\frac{a\bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{ \sqrt{c}}\biggr)\biggr). \end{aligned} \end{aligned}$$

(32)

It is easy to cheek that (32) is the approximate analytical solution of Eq. (9). According to (7) and (32), we obtain the new solution of the $(3+1)$-dimensional quasi-geodetiophic vorticity equation with dissipation

$$ \begin{aligned}[b] u_{\mathrm {new}}(t,x,y,z) &=h_{5}(t)x-h_{2}(t)y+h_{4}(t)z- \frac{\beta }{2}h_{5}(t)z ^{2}+\frac{\alpha }{2}e^{-t}z^{2} \\ &\quad {} +\frac{\bar{u}_{0}e^{-\alpha t}}{C _{2}}\biggl(\cosh \biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{a\bar{u} _{0}e^{-\alpha t}}{2}\,dt)}{\sqrt{c}}\biggr)\\ &\quad {}+\operatorname {sech}\biggl(\frac{\sqrt{2 \beta }(x+y-\int_{0}^{t}\frac{a\bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{ \sqrt{c}}\biggr)\biggr) \\ &\quad {} +\frac{\bar{u}_{0}e^{-\alpha t}}{C_{1}}\biggl(\cosh \biggl(\frac{\sqrt{2 \beta }(x+y-\int_{0}^{t}\frac{a\bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{ \sqrt{c}}\biggr)\\ &\quad {}-\operatorname {sech}\biggl(\frac{\sqrt{2\beta }(x+y-\int_{0}^{t}\frac{a \bar{u}_{0}e^{-\alpha t}}{2}\,dt)}{\sqrt{c}}\biggr)\biggr). \end{aligned} $$

(33)

4 Results and discussion

Based on the classic Lie group method, the solution of the $(3+1)$-dimensional quasi-geodetiophic vorticity equation with dissipation is derived. On the one hand, the small dissipation effect can result in a decrease in amplitude $e^{-\alpha t}$, where α is the dissipation coefficient from (33). On the other hand, the small dissipation effect can result in a decrease in velocity of a Rossby wave in the process of propagation. Hence, with the help of the new solution (33) and Fig. 1, we can better comprehend the influence of the dissipation effect on the propagation of Rossby waves.

Figure 1

The solution of $(3+1)$-dimensional quasi-geodetiophic vorticity equation with dissipation represented by (33), where $h_{5}(t)=\cos (t)$, $h _{2}(t)=\sin (t)$, $h_{4}(t)=\cos (t)$