Consider the radial diffusion of material in a sphere of radius \(a\), with a concentration distribution \(C(r, t)\) and constant diffusivity \(D\). It is assumed that: (i) the concentration at the center of the sphere is finite for \(t>0\), (ii) the surface of the sphere is maintained at a zero concentration for \(t>0\), and (iii) initially there is a uniform concentration \(C_{o}\) throughout the sphere \(0
Defining the variables \(x=\frac{r}{a}, \tau=\frac{D t}{a^{2}}\), and \(u=\frac{C}{C_{o}} \frac{r}{a}=\frac{C}{C_{o}} x\), the transformed problem is \(\frac{\partial u}{\partial \tau}=\frac{\partial^{2} u}{\partial x^{2}}\) with conditions \(u(x, \tau)=x, 00\), and \(u(x, \tau)=0, x=1, \tau>0\).

Using the Laplace transform method with respect to the variable \(\tau\), such that \(U(x, s)=\mathscr{L}\{u(x, \tau)\}\), show that the transformed solution is \(U(x, s)=\) \(\frac{1}{s}\left[x-\frac{\sinh x \sqrt{s}}{\sinh \sqrt{s}}ight]\). What are the corresponding solutions for \(u(x, \tau)\) and \(C(r, t)\) ?

(a) The release fraction \(F(t)\) can be defined as the total amount of material which has diffused through the surface of the sphere at time \(t\) divided by the initial amount of material in the sphere, \(F(t)=\frac{4 \pi a^{2} \int_{0}^{t} J(t) d t}{\frac{4}{3} \pi a^{3} C_{o}}=\frac{3}{a C_{o}} \int_{0}^{t} J(t) d t\). The flux of material \(J(t)\) is evaluated from Fick's law of diffusion: \(J(t)=-\left.D \frac{\partial C(r, t)}{\partial r}ight|_{r=a}\).

Hence, show that \(F(\tau)=\frac{3 a}{D C_{o}} \int_{0}^{\tau} J(\tau) d \tau=-3 \int_{0}^{\tau}\left(\frac{d u}{d x}ight)_{x=1} d \tau\), and defining the Laplace transform \(F(s)=\mathscr{L}\{F(\tau)\}\), show that \(F(s)=-\left.\frac{3}{s} \frac{d U(x, s)}{d x}ight|_{x=1}\).

(b) Using the solution for \(U(x, s)\) and the latter expression for \(F(s)\) from part (a), show that \(F(s)=3\left[\frac{\operatorname{coth} \sqrt{s}}{s^{3 / 2}}-\frac{1}{s^{2}}ight]\).

(c) An infinite series will eventually result for the release fraction \(F(t)\) which is more difficult to evaluate. However, an analytical form for \(F(t)\) is possible by considering a "short-time" approximation for the condition that \(\tau<<1\). Show that this condition is equivalent to \(s>>1\) in Laplace transform space. Determine an analytic 
expression for \(F(t)\) by applying the condition \(s>>1\) and taking the inverse transform of this resultant expression.