root/trunk/matml/transport/problems/liposomes/liposomes-solution.tex

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\begin{enumerate}
\item Encapsulated liposomes for long-term drug delivery

  \begin{enumerate}
  \item The ``thickness'' of the encapsulant is $R_2-R_1=0.01$ cm, so
    $$\frac{\rm thickness^2}{D}=
    \frac{\rm (0.01 cm)^2}{\rm 10^{-7}\frac{cm^2}{sec}}={\rm 1000 seconds}$$
    This is short compared to the service life of the device (for any
    definition of ``long-term''), so we can safely assume quasi-steady-state
    diffusion in the encapsulant.

  \item It's worth pointing out that ``lipid bilayer membrane'' means two
    molecular layers, so the membrane is {\em really} thin ({\em i.e.}
    submicron, ask a biomat person for molecular details).

    \hspace{1.1in}Membrane-controlled \hspace{1.4in}Diffusion-controlled

    \hspace{.2in}
    \PSbox{encapmem.ps}{190pt}{135pt}
    \PSbox{encapdiff.ps}{190pt}{135pt}

  \item \label{eldiffeq} The universal conservation equation:
    $$\rm accumulation=in-out+generation$$
    Steady state indicates accumulation=0, there should be no generation term.
    We'll use a spherical ``shell'' of thickness $\Delta r$:
    $$0 = \left[A\cdot J_r\right]_r - \left[A\cdot J_r\right]_{r+\Delta r}$$
    $$0 = \left[4\pi r^2\cdot J_r\right]_r -
    \left[4\pi r^2\cdot J_r\right]_{r+\Delta r}$$
    Divide by $4\pi\Delta r$, let $\Delta r$ go to zero:
    $$0 = -\frac{d}{dr}(r^2J_r)$$
    Substitute $J_r=-D\frac{dC}{dr}$:
    $$0 = \frac{d}{dr}\left(Dr^2\frac{dC}{dr}\right)$$
    For constant $D$, we divide by $D$ and are left with:
    $$0 = \frac{d}{dr}\left(r^2\frac{dC}{dr}\right).$$

  \item \label{solut} Start from the equation in part \ref{eldiffeq} and
    integrate:
    $$-A = r^2\frac{dC}{dr}$$
    ($A$ is an arbitrary constant, negative here because we'll make it positive
    below.)  Divide by $r^2$ and integrate again to yield the general solution:
    $$C = \frac{A}{r} + B$$
    Now apply the boundary conditions from above:
    $$C_2 = \frac{A}{R_1} + B$$
    $$0   = \frac{A}{R_2} + B$$
    There are a couple of ways to go from here.  I like to subtract the second
    from the first to eliminate $B$, this gives
    $$C_2 = \frac{A}{R_1} - \frac{A}{R_2}
    = A \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
    $$A = \frac{C_2}{\frac{1}{R_1} - \frac{1}{R_2}}$$
    Then from the second BC:
    $$0 = \frac{C_2}{\frac{1}{R_1} - \frac{1}{R_2}} \frac{1}{R_2} + B$$
    $$B = -\frac{C_2/R_2}{\frac{1}{R_1}-\frac{1}{R_2}}$$
    Plugging this $A$ and $B$ into the general solution and dividing by $C_2$
    gives:
    $$\frac{C}{C_2} =
    \frac{\frac{1}{r} - \frac{1}{R_2}} {\frac{1}{R_1} - \frac{1}{R_2}}$$

  \item\label{diffcont} We begin with the flux:
    $$J_r = -D \frac{dC}{dr} = -D \frac{d}{dr}
    \left[\frac{C_2}{\frac{1}{R_1} - \frac{1}{R_2}}
      \left(\frac{1}{r}-\frac{1}{R_2}\right)\right]$$
    $$J_r=\frac{DC_2}{\frac{1}{R_1} - \frac{1}{R_2}}\cdot\frac{1}{r^2}$$
    The area $A$ is simply $4\pi r^2$, so when we multiply them the $r^2$s
    cancel:
    $$J_rA = \frac{4\pi DC_2}{\frac{1}{R_1} - \frac{1}{R_2}}$$
    This is independent of $r$.

    If this is diffusion-limited, then $C_2\simeq C_1$, and:
    $$J_rA = \frac{4\pi DC_1}{\frac{1}{R_1} - \frac{1}{R_2}}$$
    Plugging in the problem parameters gives us:
    $$J_rA = \frac{\rm4\pi \cdot 10^{-7}\frac{cm^2}{sec}}
    {\rm \frac{1}{0.01cm} - \frac{1}{0.02cm}} \cdot C_1=
    2.5\times10^{-8}\frac{\rm cm^3}{\rm sec}C_1$$
    Since $C_1$ is in $\rm\frac{moles}{cm^3}$ or $\rm\frac{mg}{cm^3}$ etc.,
    this gives the overall rate of drug delivery from the device---if it's
    diffusion-limited.

  \item\label{memcont} The flux equation was given, and here $C_2=0$ modifies
    it slightly:
    $$J_r=h_D(C_1-C_2)\simeq h_DC_1$$
    Because the membrane is at $R_1$, the relevant area $A$ is $4\pi R_1^2$,
    which gives us:
    $$J_r\cdot A = h_DC_1 \cdot 4\pi R_1^2 =
    {\rm 4\pi\cdot(0.01cm)^2\cdot1.4\times10^{-6}\frac{cm}{s}}C_1 =
    {\rm 1.8\times10^{-9}\frac{cm^3}{s}}C_1$$
    
  \item Because the result of part \ref{memcont} is so much lower than that of
    part \ref{diffcont}, the membrane definitely controls the rate of drug
    delivery.  The actual rate of delivery will thus be close to what's
    predicted in part \ref{memcont}, or about
    $2\times10^{-9}{\rm\frac{cm^3}{s}}C_1$.  That's the end of the story as far
    as diffusion is concerned, and the answer this question was looking for.

    However, if you read the reference given in the problem, you'll notice that
    the drug delivery rate for the encapsulated liposomes is actually {\em
      higher} than for the bare liposomes.  This is because the encapsulation
    process weakens the membrane so that it is not as effective a barrier as
    without the encapsulant.
  \end{enumerate}
\end{enumerate}
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