Investigation of cylindrical shock waves in dusty plasma

Abstract

Electronegative dusty plasma composed of Boltzmann electrons, Boltzmann negative ions, inertial positive ions and charge fluctuating dust has been considered. The fractional modified Burgers’ (FMB) equation, which is derived using Euler–Lagrange variational technique, is analytically obtained and solved for studying the cylindrical geometry effect on the propagation of the dust ion acoustic shock wave. The Laplace homotopy perturbation method, the so-called LHPM is applied to solve the FMB equation. The effect of the fractional parameter, positive ion number density at equilibrium, the number of equilibrium electrons residing on the dust grain surface and shock velocity on the behavior of the shock waves in the dusty plasma has been investigated.

Appendices

Appendix A: Fractional modified Burgers’ equation solution

The general form of one-dimensional nonlinear partial differential equations is considered to illustrate the basic idea of Laplace transform homotopy perturbation method:

$$ L\left( {u\left( {x,t} \right)} \right) + N\left( {u\left( {x,t} \right)} \right) = g\left( x \right). $$

(A1)

where $L$ and $N$ denote a linear and a nonlinear operators respectively. We construct a homotopy as follows:

$$ L\left( {u\left( {x,t} \right)} \right) + pN\left( {u\left( {x,t} \right)} \right) = g\left( x \right). $$

(A2)

Taking the Laplace transform from both sides of Eq. (A2):

$$ {\mathcal{L}}\left\{ {L\left( {u\left( {x,t} \right)} \right)} \right\} + p{\mathcal{L}}\left\{ {N\left( {u\left( {x,t} \right)} \right)} \right\} = {\mathcal{L}}\left\{ {g\left( x \right)} \right\}. $$

(A3)

where ${\mathcal{L}}$ denotes the Laplace transform.

For the linear and nonlinear operators in Eq. (A3), the concept of the homotopy perturbation method with embedding parameter $p$ is used to generate a series expansion for $L\left( {u\left( {x,t} \right)} \right)$ and $N\left( {u\left( {x,t} \right)} \right)$ as follows [41, 61]:

$$ \begin{aligned} L\left( {u\left( {x,t} \right)} \right) & = L\left( {\mathop \sum \limits_{i = 0}^{\infty } p^{i} u_{i} } \right), \\ N\left( {u\left( {x,t} \right)} \right) & = \mathop \sum \limits_{i = 0}^{\infty } p^{i} H_{i} , \\ \end{aligned} $$

(A4)

where, the components $u_{i} , i \ge 0$ are to be determined in a recursive manner, the Adomian polynomials, $H_{i}$, are expressed as:

$$ H_{i} = \frac{1}{i!}\frac{{d^{i} }}{{d\lambda^{i} }}\left[ {N\left( {\mathop \sum \limits_{j = 0}^{n} \lambda^{j} v_{j} } \right)} \right]_{\lambda = 0} , $$

(A5)

By substituting Eq. (A4) into Eq. (A3):

$$ {\mathcal{L}}\left\{ {L\left( {\mathop \sum \limits_{i = 0}^{\infty } p^{i} u_{i} } \right)} \right\} + {\mathcal{L}}\left\{ {\mathop \sum \limits_{i = 0}^{\infty } p^{i + 1} H_{i} } \right\} = {\mathcal{L}}\left( {g\left( x \right)} \right), $$

(A6)

On the other hand, Eq. (A6) can be rewritten in the following manner:

$$ \mathop \sum \limits_{i = 0}^{\infty } p^{i} {\mathcal{L}}\left\{ {L(u_{i} )} \right\} + \mathop \sum \limits_{i = 0}^{\infty } p^{i + 1} {\mathcal{L}}\left\{ { H_{i} } \right\} = {\mathcal{L}}\left( {g\left( x \right)} \right), $$

(A7)

Using Eq. (A7), we introduce the recursive relations as:

$$ \begin{aligned} & p^{0} : {\mathcal{L}}\left\{ {L\left( {u_{0} } \right)} \right\} = {\mathcal{L}}\left\{ g \right\}, \\ & p^{1} : {\mathcal{L}}\left\{ {L\left( {u_{1} } \right)} \right\} + {\mathcal{L}}\left\{ {H_{0} } \right\} = 0, \\ & p^{2} : {\mathcal{L}}\left\{ {L\left( {u_{2} } \right)} \right\} + {\mathcal{L}}\left\{ {H_{1} } \right\} = 0, \\ & p^{3} : {\mathcal{L}}\left\{ {L\left( {u_{3} } \right)} \right\} + {\mathcal{L}}\left\{ {H_{2} } \right\} = 0, \\ & \vdots \\ & p^{k} : {\mathcal{L}}\left\{ {L\left( {u_{k} } \right)} \right\} + {\mathcal{L}}\left\{ {H_{k - 1} } \right\} = 0, \\ \end{aligned} $$

(A8)

Using the Maple symbolic code, the first part of Eq. (A8), $p^{0}$, gives the value of ${\mathcal{L}}\left\{ {u_{0} } \right\}$. First, applying the inverse Laplace transform to ${\mathcal{L}}\left\{ {u_{0} } \right\}$ gives the value of $u_{0} ,$ that will define the Adomian polynomial, $H_{0}$ using the first part of Eq. (A5). In the second part Eq. (A8), the Adomian polynomial $H_{0}$ will enable us to evaluate ${\mathcal{L}}\left\{ {u_{1} } \right\}$. Second, applying the inverse Laplace transform to ${\mathcal{L}}\left\{ {u_{1} } \right\}$ gives the value of $u_{1} ,$ that will define the Adomian polynomial $H_{1}$ using the second part of Eq. (A5) and so on. This in turn will lead to the complete evaluation of the components of $u_{k} , k \ge 0$ upon using different corresponding parts of Eqs. (A8) and (A5). Therefore, the series solution follows immediately after using the first part of Eq. (A4) with embedding parameter $p = 1$.

Appendix B: Fractional modified Burgers’ equation

Using the potential function, $\Upsilon \left( {\xi ,\tau } \right)$, where the $\varphi_{1} =\Upsilon _{\xi }$, gives the potential equation of the modified Burgers’ equation, Eq. (10), in the form:

$$ \Upsilon _{\xi \tau } + A\Upsilon _{\xi }\Upsilon _{\xi \xi } - B\varphi_{1\xi \xi } + \varphi_{1} /2\tau = 0, $$

(B1)

The functional of the Eq. (B1) can be represented [62, 63]:

$$ J\left(\Upsilon \right) = \int d\xi \int d\tau\Upsilon [a_{1}\Upsilon _{\xi \tau } + a_{2} A\Upsilon _{\xi }\Upsilon _{\xi \xi } - a_{3} B\varphi_{1\xi \xi } + a_{4} \varphi_{1} /2\tau ] $$

(B2)

Where $a_{1} , a_{2} , a_{3}$ and $a_{3}$ are unknown constants to be determined. Integrating this equation by parts where $\left. {\Upsilon_{\tau } } \right|_{\text{R}} = \left. {\Upsilon_{\xi } } \right|_{\text{T}} = 0$ gives:

$$ J\left( \Upsilon \right) = \int {d\xi \int {d\tau \left[ { - a_{1} \Upsilon_{\xi } \Upsilon_{\tau } - \frac{1}{2}a_{2} A\Upsilon_{\xi }^{3} - a_{3} B\Upsilon \varphi_{1\xi \xi } + a_{4} \Upsilon \varphi_{1} /2\tau } \right]} } $$

(B3)

Taking the first variations for this equation with respect to the dependent variable,$\Upsilon$, and assuming that $\varphi_{1}$ is a fixed function lead to,

$$ a_{1} = 1/2,\quad a_{2} = 1/3,\quad a_{3} = 1,\quad a_{4} = 1, $$

(B4)

So, according to the Eq. (B2), the Lagrangian of FMB equation, Eq. (B3), is given by

$$ F\left( {\Upsilon_{\tau } ,\Upsilon_{\xi } ,\Upsilon_{\xi \xi } } \right) = - \frac{1}{2}\Upsilon_{\xi } \Upsilon_{\tau } - \frac{1}{6}A\Upsilon_{\xi }^{3} - B\Upsilon \varphi_{1\xi \xi } + \Upsilon \varphi_{1} /2\tau , $$

(B5)

According to Eq. (B5), the Lagrangian of FSE can be written as:

$$ G\left( {{}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon ,\Upsilon_{\xi } ,\Upsilon_{\xi \xi } } \right) = - \frac{1}{2}\left[ {{}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon } \right]\Upsilon_{\xi } - \frac{1}{6}A\Upsilon_{\xi }^{3} - B\Upsilon \varphi_{1\xi \xi } + \Upsilon \varphi_{1} /2\tau ,\quad 0 < \alpha \le 1, $$

(B6)

where the left Riemann–Liouville fractional derivative ${}_{0}^{{}} D_{\tau }^{\alpha }$ is reperesented by [36, 37]

$$ {}_{a}^{{}} D_{t}^{\alpha } f\left( t \right) = \frac{1}{{\Gamma \left( {k - \alpha } \right)}} \frac{{d^{k} }}{{dt^{k} }} \left[ {\mathop \int \limits_{a}^{t} d\tau \left( {t - \tau } \right)^{k - \alpha - 1} f\left( \tau \right)} \right]\quad k - 1 < \alpha \le k,\quad t\in\left[ {a,b} \right] $$

(B7)

The Laplace transform of the fractional derivative, ${}_{a}^{{}} D_{t}^{\alpha } f\left( t \right)$, is given by:

$$ {\mathcal{L}}\left( {{}_{a}^{{}} D_{t}^{\alpha } f\left( t \right);s} \right) = s^{\alpha } {\mathcal{L}}\left( {f\left( t \right)} \right) - \mathop \sum \limits_{k = 0}^{n - 1} s^{k} \left[ {{}_{a}^{{}} D_{t}^{\alpha - k - 1} f\left( t \right)} \right]_{t = a} ,\quad k - 1 < \alpha \le k, $$

(B8)

The functional of the time fractional potential equation can be represented in the form,

$$ J\left( \Upsilon \right) = \mathop \int \limits_{R}^{{}} d\xi \mathop \int \limits_{T}^{{}} d\tau G\left( {{}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon ,\Upsilon_{\xi } ,\Upsilon_{\xi \xi } } \right), $$

(B9)

where the time-fractional Lagrangian is defined by (B6).

Taking the first variations of Eq. (B9) with respect to the $\Upsilon$ leads to:

$$ \delta J\left( \Upsilon \right) = \mathop \int \limits_{R}^{{}} d\xi \mathop \int \limits_{T}^{{}} d\tau \left[ {\left( {\frac{\partial G}{{\partial {}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon }}} \right)\delta {}_{0}^{{}} D_{t}^{\alpha } \Upsilon + \left( {\frac{\partial G}{{\partial \Upsilon_{\xi } }}} \right)\delta \Upsilon_{\xi } + \left( {\frac{\partial G}{\partial \Upsilon }} \right)\delta \Upsilon } \right] $$

(B10)

Integrating by parts lead to:

$$ \delta J\left( \Upsilon \right) = \mathop \int \limits_{R}^{{}} d\xi \mathop \int \limits_{T}^{{}} d\tau \left[ {{}_{\tau }^{{}} D_{{T_{0} }}^{\alpha } \left( {\frac{\partial G}{{\partial {}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon }}} \right) - \frac{\partial }{\partial \xi }\left( {\frac{\partial G}{{\partial \Upsilon_{\xi } }}} \right) + \left( {\frac{\partial G}{\partial \Upsilon }} \right)} \right]\delta \Upsilon , $$

(B11)

Here we assume that $\updelta\Upsilon {\mid }_{\text{T}} =\updelta\Upsilon |_{\text{R}} =\updelta\Upsilon_{\upxi} {\mid }_{\text{R}} = 0.$

Optimizing Eq. (B11), $\updelta{\text{J}}\left( \Upsilon \right) = 0$, gives the following Euler–Lagrange equation:

$$ {}_{\tau }^{{}} D_{{T_{0} }}^{\alpha } \left( {\frac{\partial G}{{\partial {}_{0}^{{}} D_{\tau }^{\alpha } \Upsilon }}} \right) - \frac{\partial }{\partial \xi }\left( {\frac{\partial G}{{\partial \Upsilon_{\xi } }}} \right) + \left( {\frac{\partial G}{\partial \Upsilon }} \right) = 0. $$

(B12)

Substituting from the Lagrangian equation, Eq. (B6), into Euler–Lagrange formula, Eq. (B12), and making use of Eq. (14) and $\varphi_{1} = \Upsilon_{\xi }$ lead to the FMB equation, Eq. (13).