There are two degenerate cases of interest. The first is the \emph{cylinder}, for which the radii at each end are euqal, i.e. $a_\ell= a_r = a$. In this case the lateral area of the surface is
\begin{equation}
\sigma_{\text{lateral}} = 2\pi a \Delta x.
\end{equation}
The second is a cone, for which $a_\ell=0$ and $a_r=a$:
\begin{equation}
\sigma_{\text{lateral}} = \pi a \sqrt{\Delta x^2 + a^2}.
Note that the standard convention is followed, whereby membrane and synapse currents ($i_m$) are positive when outward, and electrod currents ($i_e$) are positive inward.
The PDE in (\ref{eq:cable}) is derived from the following mass balance expression for a cable segment $i$:
where $\int_\Omega\cdot\deriv{v}$ is shorthand for the volume integral over the segment $\Omega$, and $\int_\Gamma\cdot\deriv{s}$ is shorthand for the surface integral over the surface $\Gamma$.
\end{equation}
where
\begin{itemize}
\item$\int_\Omega\cdot\deriv{v}$ is shorthand for the volume integral over the segment $\Omega_i$
\item$\int_\Gamma\cdot\deriv{s}$ is shorthand for the surface integral over the surface $\Gamma$
\item$q_{i,j}=-\frac{1}{r_L}\pder{V}{x} n_{i,j}$ is the flux per unit area of current \emph{from segment $i$ to segment $j$} over the interface $\Gamma_{i,j}$ between the two segments.
\item$q_i=i_m - i_e$ is the flux per unit area over the cell membrane $\Gamma_i$ due to ion channels, synapses and electrodes.
\item the set $\mathcal{N}_i$ is the set of segments that are neighbours of $\Omega_i$
\end{itemize}
The surface of the cable segment is sub-divided into the internal and external surfaces.
The external surface $\Gamma_{ext}$ is the cell membrane at the interface between the extra-cellular and intra-cellular regions.
The current, which is the conserved quantity in our conservation law, over the surface is composed of the synapse and ion channel contributions.
This is derived from a thin film approximation to the cell membrane, whereby the membrane is treated as an infinitesimally thin interface between the intra and extra cellular regions.
The internal surfaces are the interface between the cable segment and its neighbour segments which are denoted by the set $\mathcal{N}_i$.
Equation~\eq{eq:cable_balance} handles the general case where a cable might lie at a branch, and can be simplified for a one dimensional segment:
Note that some information is lost when going from a three-dimensional description of a neuron to a system of branching one-dimensional cable segments.
If the cell is represented by cylinders or frustrums\footnote{a frustrum is a truncated cone, where the truncation plane is parallel to the base of the cone.}, the three-dimensional values for volume and surface area at branch points can't be retrieved from the one-dimensional description.
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@@ -104,7 +107,7 @@ The finite volume method is a natural choice for the solution of the conservatio
where $\Omega_i$ is the control volume with the point $x_i$ at its centroid illustrated in \fig{fig:segment}, and $v_i$ is the volume of the control volume $\Omega_i$.
The integral one the left hand side of~\eq{eq:cable_balance} can be expressed in terms of the volume average of $V$
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@@ -149,7 +152,7 @@ where $\sigma_{i,j}=\pi a_{i,j}^2$ is the area of the surface $\Gamma_{i,j}$, wh
Some symmetries
\begin{itemize}
\item$\sigma_{i,j}=\sigma_{j,i}$ : surface area of $\Gamma_{i,j}$
\item$\Delta_{i,j}=\Delta_{j,i}$ : distance between $x_i$ and $x_j$
\item$\Delta x_{i,j}=\Delta x_{j,i}$ : distance between $x_i$ and $x_j$
where $a_{i,\ell}$ and $a_{i,r}$ are the radii of at the left and right end of the segment respectively (see~\eq{eq:frustrum_area} for derivation of this formula).
By substituting the volume averaging of the temporal derivative in~\eq{eq:dvdt} approximations for the flux over the surfaces in~\eq{eq:q_ij} and~\eq{eq:frustrum_volume} respectively into the conservation equation~\eq{eq:cable_balance} we get the following ODE defined for each node in the cell
By substituting the volume averaging of the temporal derivative in~\eq{eq:dvdt} approximations for the flux over the surfaces in~\eq{eq:q_ij} and~\eq{eq:cv_volume} respectively into the conservation equation~\eq{eq:cable_balance} we get the following ODE defined for each node in the cell
\item$\sigma_{i,j}=\pi a_{i,j}^2$ is the area of the surface between two adjacent segments $i$ and $j$. Care must be taken if the neighbour is a child or parent of the other at a branch point. In this case, the radius is that of the child segment (which ever has the largest index in the minimum degree ordering used for numbering segments). \todo{what about if we balance the tree so that the parent-child relationship is swapped?, maybe the relationship ``farther from soma'' has to used to choose.}
\item$\sigma_{i}=(a_{i,\ell}+ a_{i,r})+\sqrt{\Delta x_i^2+(a_{i,\ell}- a_{i,r})^2}.$ is the lateral area of the conical frustrum describing segment $i$. For this area, we take the value of the segment radius at the left and right of the frustrum iteself, regardless of whether either end lies at a branch point.
\end{itemize}
%\begin{itemize}
% \item $\sigma_{i,j}=\pi a_{i,j}^2$ is the area of the surface between two adjacent segments $i$ and $j$.
% \item $\sigma_{i}=\pi(a_{i,\ell} + a_{i,r})+\sqrt{\Delta x_i^2 + (a_{i,\ell} - a_{i,r})^2}.$ is the lateral area of the conical frustrum describing segment $i$.
% \item $\Delta_{i}=\frac{\pi\Delta x_i}{2} \left( a_{i,l}^2 + a_{i,r}^2 \right)$ is the volume of the segment $\Omega_i$.
%\end{itemize}
\begin{equation}
\sigma_{i,j} = \pi a_{i,j}^2
\label{eq:sigma_ij}
\end{equation}
is the area of the surface between two adjacent segments $i$ and $j$, and
The value of the lateral area and volume, $\sigma_i$ and $\Delta_i$ in~\eq{eq:sigma_i} and~\eq{eq:delta_i} respetively, must include contributions from each branch at branch points.
\todo{a picture of a branching point to illustrate}
\todo{a picture of a soma to illustrate the ball and stick model with a sphere for the soma and sticks for the dendrites branching off the soma.}
