Modeling the release of a reagent from an inwardly tapered disk with a central hole

Abstract

A theoretical analysis on the controlled release of an active reagent from an inwardly tapered polymeric disk with a central releasing hole proposed originally by Bechard and McMullen (J Pharm Sci 77:222–228, 1988) is conducted. Analytic expressions for the temporal variations in the amount of reagent released and the size of the unreleased portion of the device are derived. Relaxing their assumptions of pseudo-steady and constant reagent concentration in the hole, the problem is reanalyzed. The resultant model is suitable to the whole release period, and successfully describes their experimental data. To make the analytic result derived more readily applicable, the results for some special cases are also presented.

Appendix

Suppose that the solution to (5) takes the following form:

$$\begin{aligned} {w}({x,t})={X}({x}){T}({t}). \end{aligned}$$

(23)

The boundary conditions (5a) and (5b), suggest that X(x) has the form $\sin [{n}\pi {x/}({R}-{A})]$, $n=1,2,{\ldots }$ Therefore, by applying the principle of superposition,

$$\begin{aligned} {w}({x,t})=\sum _{{n}=1}^\infty {\sin \left( {\frac{{n\pi x}}{{R-A}}} \right) } {T}_{n} ({t}),\quad 0<x<(R-A). \end{aligned}$$

(24)

Substituting this expression into (5) yields

$$\begin{aligned} \sum _{n=1}^\infty {\sin \left( {\frac{n\pi x}{R-A}} \right) \frac{\mathrm{d}T_n }{\mathrm{d}t}} =\left[ {\sum _{n=1}^\infty {-D\left( {\frac{n\pi }{R-A}} \right) ^{2}\sin \left( {\frac{n\pi x}{R-A}} \right) T_n (t)} } \right] +S(x,t), \end{aligned}$$

(25)

where $S(x,t)=-[(R-x-A)A(\mathrm{d}C_\mathrm{b}/\mathrm{d}t)]/(R-A)$. Expand S(x, t) as

$$\begin{aligned} {S}({x},{t})=\sum _{{n}=1}^\infty {{S}_{n} ({t})\sin \left( {\frac{{n\pi x}}{{R-A}}} \right) } ,\quad 0<x<R-A. \end{aligned}$$

(26)

That is, $S_{n}(t)$ is the coefficient of the Fourier sine series of S(x,t) on the interval $[0,R-A$]. It can be shown that

(27)

Substituting (26) in (25) yields

$$\begin{aligned} \frac{{\mathrm{d}T}_{n} ({t})}{{\mathrm{d}t}}=-{D}\left( {\frac{{n\pi }}{{R-A}}} \right) ^{2}{T}_{n} ({t})+{S}_{n} ({t}),\quad {n}=1,2,\ldots \end{aligned}$$

(28)

Equations (24) and (5c) suggest that

$$\begin{aligned} {w}({x},0)= & {} -\frac{{xRC}_\mathrm{s}}{{R}-{A}}=\sum _{{n}=1}^\infty {\sin \left( {\frac{{n\pi x}}{{R-A}}} \right) } {T}_{n} (0) ,\quad 0<x<R-A. \end{aligned}$$

(29)

It can be shown that

$$\begin{aligned} T_n (0)= & {} \frac{2}{R-A}\int \limits _0^{R-A} {-\frac{xRC_\mathrm{s} }{R-A}} \sin \left( {\frac{n\pi x}{R-A}} \right) \mathrm{d}x =(-1)^{n}\frac{2RC_\mathrm{s} }{n\pi } , \quad n=1,2,{\ldots } \end{aligned}$$

(30)

Solving (28) subject to (30) and noting that $S_{n}(0)=0$, (8a) in the text can be recovered.