Studies on the Dust Acoustic Shock, Solitary, and Periodic Waves in an Unmagnetized Viscous Dusty Plasma with Two-Temperature Ions

Abstract

The non-linear propagation of dust acoustic wave (DAW) in collisionless, unmagnetized, viscous dusty plasma containing electrons with Maxwell-Boltzmann distribution, two-temperature ions, and highly negatively charged dust grains, are investigated through the fabric of the Kadomtsev-Petviashvili-Burgers (KPb) equation. Following the approach of the reductive perturbation technique (RPT), the KPb equation is constructed from the governing equation, and further, utilizing traveling wave transformation, the KPb equation is converted into an ordinary differential equation (ODE). Analyzing phase portraits for the KPb system varying different plasma parameters, it is found that the KPb system includes shock, solitary as well as periodic solutions. But, in case of lack of Burgers term, the system only provides the solitary and periodic solutions which are directly generated from the KP equation, and further, a shock solution of the KPb equation is derived by applying the $(G'/G)$-expansion method. Introducing the indirect F-function method and incorporating a Jacobi elliptic function, a finite-amplitude periodic solution for the KPb equation is also constructed. Finally, the impacts of the physical parameters on wave propagation in the present system are illustrated from a numerical standpoint.

Appendix

In order to acquire the periodic wave solution of KP Eq. (35), utilizing the transformation (36), we finally get,

$$\begin{aligned} (-Vl+Dm^2)U'' + Al^2(UU')' + Bl^4U^{(4)} = 0. \end{aligned}$$

(99)

Twice integration yields,

$$\begin{aligned} (-Vl+Dm^2)U +\frac{Al^2}{2}U^2 + Bl^4U'' = 0, \end{aligned}$$

After, multiplication by $2U'$ we again integrate the above equation and get

$$\begin{aligned} (U')^2 =-\frac{A}{3Bl^2}U^3+\frac{(Vl-Dm^2)}{Bl^4}U^2+ \frac{\Gamma }{Bl^4}, \end{aligned}$$

(100)

To approve a periodic solution, the constant $\Gamma$ is chosen as $\Gamma =\frac{2(Dm^2-Vl)^3}{3A^2l^4}$. Due to a particular initial condition, $\Gamma$ possibly takes the aforesaid form. Then for $A>0$ and $A<0$, we can express (48) respectively as

$$\begin{aligned} (U')^2 =\frac{A}{3Bl^2}[(e_1-U)(U-e_2)(U-e_3)], \end{aligned}$$

(101)

and

$$\begin{aligned} (U')^2 =(-\frac{A}{3Bl^2})[(e_3-U)(e_2-U)(U-e_1)], \end{aligned}$$

(102)

where $e_1=(1+\sqrt{3})\frac{Vl-Dm^2}{Al^2},e_2=\frac{Vl-Dm^2}{Al^2}, e_3=(1-\sqrt{3})\frac{Vl-Dm^2}{Al^2}$. Here, B is always positive, and so in accordance with the nonlinear coefficient A, the term $\frac{A}{3Bl^2}$ becomes positive or negative. The real numbers $e_1,e_2,e_3$ are ordered as $e_1>e_2>e_3$ and $e_1>U \ge e_2$ when $\frac{A}{3Bl^2}>0$. On the other hand, if $\frac{A}{3Bl^2}<0$, then $e_1,e_2,e_3$ are arranged as $e_1<e_2<e_3$ and $e_1<U \le e_2$. Now, we write Eq. (101) as

$$\begin{aligned} \sqrt{\frac{A}{3Bl^2}}\theta =\int \frac{dU}{\sqrt{(e_1-U)(U-e_2)(U-e_3)}}. \end{aligned}$$

(103)

Using the transformation

$$\begin{aligned} U=e_1-(e_1-e_2)sin^2\Theta \end{aligned}$$

(104)

we have

$$\begin{aligned} \begin{aligned} \sqrt{\frac{A}{3Bl^2}}\theta&=\int \frac{dU}{\sqrt{(e_1-U)(U-e_2)(U-e_3)}} \\&=\frac{2}{\sqrt{e_1-e_3}}\int \frac{d\Theta }{\sqrt{1-\kappa ^2sin^2\Theta }}, \end{aligned} \end{aligned}$$

(105)

where $\kappa ^2=\frac{e_1-e_2}{e_1-e_3}$, and from the definition of Jacobi elliptic sine function [69] we get

$$\begin{aligned} sin\Theta =sn\left( \frac{\sqrt{\frac{A}{3Bl^2}(e_1-e_3)}}{2}\theta ,\kappa \right) \end{aligned}$$

(106)

Hence, the solution of Eq. (100) becomes,

$$\begin{aligned} U(\theta )=e_1-(e_1-e_2)sn^2\left( \frac{\sqrt{\frac{A}{3Bl^2}(e_1-e_3)}}{2}\theta ,\kappa \right) . \end{aligned}$$

(107)

Therefore, the periodic wave solution of KP Eq. (35) is

$$\begin{aligned} \phi (\xi ,\eta ,\tau )=e_1-(e_1-e_2)sn^2\left( \frac{\sqrt{\frac{A}{3Bl^2}(e_1-e_3)}}{2}(l\xi +m\eta -V\tau ),\kappa \right) . \end{aligned}$$

(108)

Similarly for $A<0$, $e_1<e_2<e_3$ and $e_1<U \le e_2$ we can write (102) using (104) as

$$\begin{aligned} \begin{aligned} \sqrt{-\frac{A}{3Bl^2}}\theta&=\int \frac{dU}{\sqrt{(e_3-U)(e_2-U)(U-e_1)}} \\&=\frac{2}{\sqrt{e_3-e_1}}\int \frac{d\Theta }{\sqrt{1-\kappa ^2sin^2\Theta }}, \end{aligned} \end{aligned}$$

(109)

and from the definition of Jacobi elliptic sine function [69] we get

$$\begin{aligned} sin\Theta =sn\left( \frac{\sqrt{-\frac{A}{3Bl^2}(e_3-e_1)}}{2}\theta ,\kappa \right) . \end{aligned}$$

(110)

Hence, the solution of (100) will be

$$\begin{aligned} U(\theta )=e_1-(e_1-e_2)sn^2\left( \frac{\sqrt{-\frac{A}{3Bl^2}(e_3-e_1)}}{2}\theta ,\kappa \right) . \end{aligned}$$

(111)

Finally, the periodic wave solution of KP Eq. (35) can be expressed as,

$$\begin{aligned} \phi (\xi ,\eta ,\tau )=e_1-(e_1-e_2)sn^2\left( \frac{\sqrt{-\frac{A}{3Bl^2}(e_3-e_1)}}{2}(l\xi +m\eta -V\tau ),\kappa \right) . \end{aligned}$$

(112)