to the first term as the order parameter for the ground state is smoothest. Under these approximations, $\int_0^\rho\frac{2}{\rho}\big(\frac{\text{d}m}{\text{d}\rho}\big)^2\text{d}\rho \approx 2\big(\frac{\text{d}m}{\text{d}\rho}\big)^2$. Thus Eq. (\[eqn:43\]) can be rewritten as $-\int_{m(0)}^0\frac{\text{d}m}{\sqrt{-|a|\big(m^2-m^2(0)\big)+\frac{b}{2}\big(m^4-m^4(0)\big)}}\approx\int_0^R\frac{\text{d}\rho}{\sqrt{3}}$, which in turn can be expressed as $$\label{eqn:44}
\frac{\sqrt{3}}{\sqrt{1-\eta}}K\big(\eta/(1-\eta)\big)=R\sqrt{|a|}~,$$ where $K(x)=(\pi/2)[1+x/4+(9/64)x^2+(25/256)x^3+...]$ is the complete elliptic integral of the first kind, and $\eta=bm^2(0)/2|a|\le1/2$. Note that $\eta \le 1/2$, just implies the physical requirement that $m(0)^2 \le m_0^2 = |a|/b$. Based on the approximations above, the mean-field force in Eq. (\[eqn:42\]) can be recast in terms of $\eta$ as $$\begin{aligned}
\label{eqn:45}
F_R^{(\text{mf})}=\frac{2\pi\hbar^2N_0LR}{3M}\frac{\eta|a|^2}{b}(1-\eta),~~\end{aligned}$$ where $\eta$ is to be obtained from the graphical solution of Eq. , which does not appear in closed form. However, an approximate closed form of $\eta$ was obtained by Biswas *et al* in the context of film geometry [@Biswas]. Following their approach, for cylindrical geometry, we get $$\label{eqn:46}
\eta(x)=\frac{1}{2}\text{tanh}^2\bigg(\frac{1}{\sqrt{6}}\sqrt{x^2-\frac{3\pi^2}{4}}\bigg)~,$$ where $x = R \sqrt{|a|}$. We emphasis that Eq.  gives a very close approximation [^3] of the actual roots of Eq. . It is clear from Eq. (\[eqn:46\]) that $\eta\rightarrow1/2$ as $R\rightarrow\infty$, and $\eta=0$ for $R\sqrt{|a|}\le\sqrt{3}\pi/2$. Thus we get the mean-field bulk pressure exerted by the condensate on the cylindrical wall as $F_R^{(\text{mf})}/2\pi RL|_{R\rightarrow\infty}=g\bar{n}^2/6$ in comparison to $g\bar{n}^2/2$ for the film geometry 