   F = & -\frac{\Gamma w f_0}{a(v/D)} \left[\frac{a}{v} \left(1-\exp  
        \left(-\frac{2vr}{D} \right)\right)\exp\left(-\frac{v}{D}\ell\right)  
        \right. \nonumber \\ &
        \left.-\frac{v}{D}
        \left(1-\exp\left(-\frac{2ar}{v}\right)\right)\right]
        \nonumber\\
        \simeq & -\frac{\Gamma wf_0D}{av}\left(-\frac{v}{D}  
        \right)\left(\frac{2ar}{v} \right)
        \nonumber\\
        = & \frac{2\Gamma wf_0r}{v}.
        \label{eq:F}
    \end{aligned}$$ As $F = K\mu v$ from Eq. (\[eq:motion1\]), $$\begin{aligned}
        K\mu v = \frac{2\Gamma w f_0 r}{v}.
        \label{eq:F1}
    \end{aligned}$$ From Eq. (\[eq:F1\]), we obtain $$\begin{aligned}
        v=\sqrt{\frac{2\Gamma wf_0r}{K\mu}}.
        \label{eq:v}
    \end{aligned}$$

Equation (\[eq:v\]) shows a power law $v\propto\mu^{-1/2}$, if other parameters such as $\Gamma, w$ and $f_0$ are independent of $\mu$. The power law with the index $-1/2$ is an interesting result, since Stokes relation naturally suggests another relation; $v \propto \mu^{-1}$ [@fluid].

Numerical results
=================

In the theoretical analysis, we have assumed the solution depending on $\xi = x - vt$. However, the supposed mathematical model has other symmetries and whether the considered solution depending on $\xi$ is an attractor or not should be checked. Therefore, we performed numerical calculations based on equations in Sec. \[sec:model\]. For numerical calculation, we considered a one-dimensional array with a spatial step of $\Delta x = 0.1$. The spatial size of the considered system was 1000 with periodic boundary condition, and we adopted Euler method with time step $\Delta t = 10^{-3}.$ As for the spatial derivative, we used explicit method. The parameters are set to be $m = 0.1$, $w = 1$, $\Gamma = 1$, $r=1$, $\ell = 1$, $D = 1$, $a = 1$, and $f_0 = 1$. In the discretization process, the first-order interpolation was adopted for Eqs.  and . The parameter $h$ corresponding to the viscosity $\mu$ was changed, and we investigated 