necessary that the boundary condition on the condensate has to be the same as that assigned to the elementary excitation in the condensate as they are independent. In a realistic situation of a superfluid/condensate in contact with a metallic plate the boundary condition is considered to be of Dirichlet type [@Chan; @Krech; @Biswas]. For Dirichlet boundary condition on a cylindrical surface and the ground sate of the system, [*i.e. *]{} for $ m(R)=0, \frac{dm}{d\rho}|_{\rho=0}=0$[^2], we obtain, $$\begin{aligned}
\label{eqn:42}
F_R^{(\text{mf})}=\frac{\pi\hbar^2N_0}{M} LR\bigg(\frac{\text{d}m}{\text{d}\rho}\bigg)^2\bigg|_{\rho=R}.\end{aligned}$$ Eq.  is physically appealing as it relates the energy density at the boundary ($\propto (d_\rho m|_R)^2$) to the mean-field part of the Casimir pressure, and a similar form is also expected in other confined geometries [@Biswas2010]. The value of $\big(\frac{\text{d}m}{\text{d}\rho}\big)\big|_{\rho=R}$ is to be obtained from the non-linear Eq.  which does not have a closed form analytical solution to the best of our knowledge. However, we can obtain an approximate form for the same. Multiplying Eq.  by $2\frac{\text{d}m}{\text{d}\rho}$ and integrating over $\rho$, we get $$\begin{aligned}
\label{eqn:43}
\left(\frac{\text{d}m}{\text{d}\rho}\right)^2+\int_0^\rho\frac{2}{\rho}\left(\frac{\text{d}m}{\text{d}\rho}\right)^2\text{d}\rho&=&-|a|\left[m^2-m^2(0)\right]\nonumber\\&+&\frac{b}{2}\left[m^4-m^4(0)\right].\end{aligned}$$ The second term in the LHS in Eq. , can be integrated by parts: $\int_0^\rho\frac{2}{\rho}\big(\frac{\text{d}m}{\text{d}\rho}\big)^2\text{d}\rho=2\big(\frac{\text{d}m}{\text{d}\rho}\big)^2|_{0}^\rho-\{\frac{\text{d}}{\text{d}\rho}\big(\frac{1}{\sqrt{\rho}}\frac{\text{d}m}{\text{d}\rho}\big)^2\}\rho^2|_{0}^\rho+\int_0^\rho\{\frac{\text{d}^2}{\text{d}\rho^2}\big(\frac{1}{\sqrt{\rho}}\frac{\text{d}m}{\text{d}\rho}\big)^2\}\rho^2\text{d}\rho$. Now as $\rho\rightarrow0$, $\frac{1}{\sqrt{\rho}}\frac{\text{d}m}{\text{d}\rho}\rightarrow0$ for the ground state. Additionally, the derivatives of $\frac{1}{\sqrt{\rho}}\frac{\text{d}m}{\text{d}\rho}$ can also be neglected in comparison 