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

\documentclass{article}
\usepackage{fullpage}
\begin{document}
\begin{enumerate}
\item Macromolecule diffusion into muscle tissue

  \begin{enumerate}
  \item \label{solution} Start with the unsteady spherical diffusion equation
    from the equation sheet:
    $$\frac{\partial C}{\partial t} = \frac{1}{r^2}\frac{\partial}{\partial r}
    \left(r^2D\frac{\partial C}{\partial r}\right) + G.$$
    Steady-state implies $\partial C/\partial t$ is zero, and without other
    information, we can ignore generation, leaving:
    $$\frac{1}{r^2}\frac{\partial}{\partial r}
    \left(r^2\frac{\partial C}{\partial r}\right) = 0.$$
    Cancel the $1/r^2$ up front, and substitute $C=A/r+B$, and simplify:
    $$\frac{\partial}{\partial r}
    \left(r^2\frac{\partial}{\partial r}\left(\frac{A}{r}+B\right)\right) =
    \frac{\partial}{\partial r}\left(r^2\left(-\frac{A}{r^2}+0\right)\right) =
    \frac{\partial}{\partial r}(-A) = 0.$$
    So it works.

  \item Substitute the boundary conditions into $C=A/r+B$; at $r=\infty$,
    $C=0$:
    $$0 = \frac{A}{\infty}+B \Rightarrow B=0,$$
    then at $r=0.5$mm, $C=C_0$:
    $$C_0 = \frac{A}{\rm 0.5mm} \Rightarrow A=C_0\cdot 0.5{\rm mm}.$$
    The solution which fits these boundary conditions is thus:
    $$C = \frac{C_0\cdot 0.5{\rm mm}}{r}.$$

  \item The $r$-component of flux $J_r$ is given by:
    $$J_r = -D\frac{dC}{dr} = -D\frac{d}{dr}\frac{C_0\cdot 0.5{\rm mm}}{r} =
    D\frac{C_0\cdot 0.5{\rm mm}}{r^2}.$$
    At the surface where $r=0.5{\rm mm}$, this is $J_r=DC_0/0.5{\rm mm}$.

    The mass transfer coefficient $h_D$ is then just $D/0.5{\rm mm}$.

  \item \label{halfregion} Set $C\geq C_0/2$, so $C_0/2\leq C$, and solve for
    $r$:
    $$C_0/2 \leq \frac{C_0\cdot 0.5{\rm mm}}{r}$$
    $$r \leq 2\cdot 0.5{\rm mm = 1mm}$$
    So $C\geq C_0/2$ where $r\leq 1$mm, and this region is a sphere with twice
    the diameter of the device.

  \item Initially, the drug will spread rapidly along the direction of the
    cells and slowly across them, due to faster diffusion in the direction of
    the cells, making the region an elongated ellipsoid.  However, as time goes
    on, the drug will diffuse also to the sides, such that at steady state, it
    will still be a sphere with twice the diameter of the device, as if it were
    isotropic.  This is a bit counter-intuitive, and merits explanation.

    One way to think of this is that the $r$-direction diffusivity $D_{rr}$
    depends on the angle $\phi$ away from the direction of muscle cells (where
    the angles $\phi=0$ and $\phi=\pi$ (180$^\circ$) are parallel to the cells,
    and $\phi=\pi/2$ (90$^\circ$) is perpendicular to the cells, but not on
    $\theta$ (longitude-type angle around the cells) or $r$.  So parallel to
    the cells, $D_{rr}(\phi=0)=D_{rr}(\phi=\pi)$ is large; perpendicular to
    them, $D_{rr}(\phi=\pi/2)$ is small; at a 45$^\circ$angle,
    $D_{rr}(\phi=\pi/4)$ is somewhere between the other two.

    Then if we try $C=A/r+B$ in the modified diffusion equation, it still
    works:
    $$\frac{\partial}{\partial r}
    \left(r^2D_{rr}(\phi)\frac{\partial C}{\partial r}\right) =
    \frac{\partial}{\partial r}
    \left(r^2D_{rr}(\phi)\left(-\frac{A}{r^2}+0\right)\right) =
    \frac{\partial}{\partial r} (D_{rr}(\phi)A) = 0.$$

    For those interested in even more details, they go something like this.  In
    an isotropic medium, the diffusivity is a scalar.  In an anisotropic
    medium, it is a tensor, so Fick's first law goes like:
    $$J_i = -D_{ij} C_{,j},$$
    where $D_{ij}$ is the diffusivity tensor and $C_{,j}$ the concentration
    gradient (in indicial notation).  If we point the $z$-axis in the direction
    of the muscle cells, and the $x$ and $y$ axes in orthogonal directions,
    then the diffusivity tensor will look like:
    $$D_{ij} = \left(\begin{array}{ccc} D_\perp&0&0\\ 0&D_\perp&0\\
        0&0&D_\parallel\end{array}\right),$$
    where $D_\perp$ is the low diffusivity across the cells and $D_\parallel$
    is the high diffusivity along the cells.  So $J_z$ is faster for a given
    $\partial C/\partial z$ than $J_x$ would be for the same $\partial
    C/\partial x$.

  \item There are multiple potential complicating factors, any one of which
    receives full credit:
    \begin{itemize}
    \item Blood flow carries the drug around, changing the shape significantly,
      by {\em convection}.
    \item The muscle contracts and extends, changing its shape, also
      effectively a {\em convection} mechanism.
    \item The drug is metabolized or otherwise broken down by the body by
      biochemical reactions, macrophages, etc., resulting in nonzero {\em
        generation}.
    \item Non-uniformities in the muscle such as fluid regions, or else muscle
      damage or scar tissue due to the implantation process, results in {\em
        non-uniform diffusivity} (not only anisotropic).
    \end{itemize}
  \end{enumerate}
\end{enumerate}
\end{document}