the pressure of the shocked region $P_{\rm sh}$ becomes equal to that of the ISM $p$. If the outer shock decaying time is shorter than the lifetime of MSP, the adiabatic cooling is negligible. In order to estimate the outer shock decaying timescale, we solve the equation of motion and the energy conservation law at the outer shock front. The equation of motion can be described as $$\label{eos}
\frac{d}{dt} \left\{ \frac{4 \pi}{3} R_{\rm out}^3 \rho \frac{dR_{\rm out}}{dt} \right\} = 4\pi R_{\rm out}^2 P_{\rm sh},$$ where $R_{\rm out}$ is the radius of the outer shock front, $P_{\rm sh}$ is the pressure of the shocked region and $\rho$ is the density of ISM. The energy equation is $$\label{ece}
\frac{d}{dt} \left\{ \frac{4 \pi}{3} R_{\rm out}^3 \frac{3}{2} P_{\rm sh} \right\} = L_{\rm sd} - P_{\rm sh} \frac{d}{dt} \left\{ \frac{4 \pi}{3} R_{\rm out}^3 \right\} ,$$ where $L_{\rm sd}$ is the spin-down luminosity of MSP, $$L_{\rm sd} =\frac{B^2_0\Omega^4R^6}{2c^3} = 2\times 10^{33}\left(\frac{B_0}{10^{8.5}{\rm G}}\right)^2\left(\frac{\Omega}{10^3{\rm s^{-1}}}\right)^4\left(\frac{R}{10^{6}{\rm cm}}\right)^6 {\rm erg}\ {\rm s}^{-1}.$$ In the derivation of eq.(\[ece\]), we assume that in the shocked region the internal energy of particles is $3P_{\rm sh}/2$ because the energy of particles is the relativistic regime. Using the typical value for the density of ISM 