not be any Bogoliubov excitation. So, fluctuation part of the Casimir force would be zero. But, the mean field part of the Casimir force is not necessarily zero, as because, the profile of the order parameter is still given by $m_{p_z,0,1}(z,\rho,\theta)=\sqrt{\frac{2}{R^2L}}\text{e}^{i\frac{p_zz}{\hbar}}\frac{J_0(\alpha_{0,1}\rho/R)}{J_{1}(\alpha_{0,1})}$[^4] whose derivative at the cylindrical surface ($\rho=R$) is nonzero. Our present calculation can be extended for finite temperature by further consideration of the confinement of thermal fluctuations of the thermal cloud over the condensate [@Biswas2007].

In a realistic experimental scenario, liquid $^4$He can be placed inside a cylindrical shell. Well below the $\lambda$-point ($T_\lambda=2.18$ K), the liquid can become a Bose-Einstein condensate, and the condensate would exert Casimir force on the cylindrical surface. For 1 $\mu$m radius of the shallow of the ultra-cold $^{4}$He liquid ($a_s=2.5\text{\AA}$ [@Leggett]), where there is no classical critical fluctuations, the Casimir pressure is predicted from Eq.  to be of the order of $-20\times10^{-12}$ N/m$^2$ for the unit dimensionless density ($\tilde{\rho}\rightarrow1$).

While McMahon asymptotic expansion of roots of the Bessel functions and their first derivative fails to reproduce the fluctuation part of the Casimir force as one can obtain from the prediction of Milton, DeRaad Jr., Gosdzinsky and Romeo [@Milton1981; @Gosdzinsky], analytic continuation of $\sum_{n=1}^\infty\alpha^{-2}_{\sigma,n}$ 