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2.5 Spatial Structure: The Dendritic Tree

Neurons in the cortex and other areas of the brain often exhibit highly developed dendritic trees that may extend over several hundreds of $ \mu$m. Synaptic input to a neuron is mostly located on its dendritic tree, spikes, however, are generated at the soma near the axon hillock. What are the consequences of the spatial separation of input and output? Up to now we have discussed point neurons only, i.e., neurons without any spatial structure. The electrical properties of point neurons have been described as a capacitor that is charged by synaptic currents and other transversal ion currents across the membrane. A non-uniform distribution of the membrane potential on the dendritic tree and the soma induces additional longitudinal current along the dendrite. We are now going to derive the cable equation that describes the membrane potential along a passive dendrite as a function of time and space. In Section 2.6 we will see how geometric and electrophysiological properties of a certain type of neuron can be integrated in a comprehensive biophysical model.

2.5.1 Derivation of the Cable Equation

Figure 2.16: Part of a dendrite and the corresponding circuit diagram. Longitudinal and transversal resistors are denoted by RL and RT, respectively. The electrical capacity of each small piece of dendrite is symbolized by capacitors C.

Consider a piece of a dendrite decomposed in short cylindric segments of length dx each. The schematic drawing in Fig. 2.16 shows the corresponding circuit diagram. Using Kirchhoff's laws we find equations that relate the voltage u(x) across the membrane at location x with longitudinal and transversal currents. First, a longitudinal current i(x) passing through the dendrite causes a voltage drop across the longitudinal resistor RL according to Ohm's law,

u(t, x + dx) - u(t, x) = RL i(t, x) , (2.23)

where u(t, x + dx) is the membrane potential at the neighboring point x + dx. Second, the transversal current that passes through the RC-circuit is given by C $ \partial$u(t, x)/$ \partial$t + u(t, x)/RT. Kirchhoff's law regarding the conservation of current at each node leads to

i(t, x + dx) - i(t, x) = C $\displaystyle {\frac{{\partial}}{{\partial t}}}$ u(t, x) + $\displaystyle {\frac{{u(t,x)}}{{R_{\text{T}}}}}$ - Iext(t, x) . (2.24)

The values of the longitudinal resistance RL, the transversal conductivity R-1T, the capacity C, and the externally applied current can be expressed in terms of specific quantities per unit length rL, r-1T, c, and iext, respectively, viz.

RL = rL dx ,    R-1T = r-1T dx ,    C = c dx ,    Iext(t, x) = iext(t, x) dx . (2.25)

These scaling relations express the fact that the longitudinal resistance and the capacity increase with the length of the cylinder, whereas the transversal resistance is decreasing, simply because the surface the current can pass through is increasing. Substituting these expressions in Eqs. (2.24) and (2.25), dividing by dx, and taking the limit dx$ \to$ 0 leads to
\begin{subequations}\begin{align}\frac{\partial}{\partial x} \, u(t,x) &= r_{\te...{u(t,x)}{r_{\text{T}}} - i_{\text{ext}}(t,x) \,. \end{align}\end{subequations}

Taking the derivative of these equations with respect to x and crosswise substitution yields
\begin{subequations}\begin{align}\frac{\partial^2}{\partial x^2} \, u(t,x) &= c ...
...c{\partial}{\partial x} \, i_{\text{ext}}(t,x) \,. \end{align}\end{subequations}

We introduce the characteristic length scale $ \lambda^{2}_{}$ = rT/rL (``electrotonic length scale'') and the membrane time constant $ \tau$ = rT c. If we multiply Eq. (2.28) by $ \lambda^{2}_{}$ we get
\begin{subequations}\begin{align}\lambda^2 \, \frac{\partial^2}{\partial x^2} \,...
...c{\partial}{\partial x} \, i_{\text{ext}}(t,x) \,. \end{align}\end{subequations}

After a transformation to unit-free coordinates,

x$\displaystyle \to$$\displaystyle \hat{{x}}$ = x/$\displaystyle \lambda$ ,    t$\displaystyle \to$$\displaystyle \hat{{t}}$ = t/$\displaystyle \tau$ , (2.29)

and rescaling the current variables,

i$\displaystyle \to$$\displaystyle \hat{{i}}$ = $\displaystyle \sqrt{{r_{\text{T}} \, r_{\text{L}}}}$  i ,    iext$\displaystyle \to$$\displaystyle \hat{{i}}_{{\text{ext}}}^{}$ = rT iext , (2.30)

we obtain the cable equations (where we have dropped the hats)
\begin{subequations}\begin{align}\frac{\partial}{\partial t} \, u(t,x) &= \frac{...{\partial}{\partial x} \, i_{\text{ext}}(t,x) \,,\end{align}\end{subequations}

in a symmetric, unit-free form. Note that it suffices to solve one of these equations due to the simple relation between u and i given in Eq. (2.27a).

The cable equations can be easily interpreted. These equations describe the change in time of voltage and longitudinal current. Both equations contain three different contributions. The first term on the right-hand side of Eq. (2.32) is a diffusion term that is positive if the voltage (or current) is a convex function of x. The voltage at x thus tends to decrease, if the values of u are lower in a neighborhood of x than at x itself. The second term on the right-hand side of Eq. (2.32) is a simple decay term that causes the voltage to decay exponentially towards zero. The third term, finally, is a source term that acts as an inhomogeneity in the otherwise autonomous differential equation. This source can be due to an externally applied current, to synaptic input, or to other (non-linear) ion channels; cf.Section 2.5.3. Example: Stationary solutions of the cable equation

In order to get an intuitive understanding of the behavior of the cable equation we look for stationary solutions of Eq. (2.32a), i.e., for solutions with $ \partial$u(t, x)/$ \partial$t = 0. In that case, the partial differential equation reduces to an ordinary differential equation in x, viz.

$\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ u(t, x) - u(t, x) = - iext(t, x) . (2.32)

The general solution to the homogenous equation with iext(t, x) $ \equiv$ 0 is

u(t, x) = c1 sinh(x) + c2 cosh(x) , (2.33)

as can easily be checked by taking the second derivative with respect to x. Here, c1 and c2 are constants that are determined by the boundary conditions.

Solutions for non-vanishing input current can be found by standard techniques. For a stationary input current iext(t, x) = $ \delta$(x) localized at x = 0 and boundary conditions u$ \infty$) = 0 we find

u(t, x) = $\displaystyle {\frac{{1}}{{2}}}$ e-$\scriptstyle \left\vert\vphantom{ x }\right.$x$\scriptstyle \left.\vphantom{ x }\right\vert$ , (2.34)

cf. Fig. 2.17. This solution is given in units of the intrinsic length scale $ \lambda$ = (rT/rL)1/2. If we re-substitute the physical units we see that $ \lambda$ is the length over which the stationary membrane potential drops by a factor 1/e. In the literature $ \lambda$ is refered to as the electrotonic length scale (Rall, 1989). Typical values for the specific resistance of intracellular medium and the cell membrane are 100 $ \Omega$ cm and 30 k$ \Omega$ cm2, respectively. In a dendrite with radius $ \rho$ = 1 $ \mu$m this amounts to a transversal and a longitudinal resistance of rL = 100 $ \Omega$ cm/($ \pi$$ \rho^{2}_{}$) = 3 . 105 $ \Omega$ $ \mu$m-1 and rT = 30 k$ \Omega$ cm2/(2$ \pi$$ \rho$) = 5 . 1011 $ \Omega$ $ \mu$m. The corresponding electrotonic length scale is $ \lambda$ = 1.2 mm. Note that the electrotonic length can be significantly smaller if the transversal conductivity is increased, e.g., due to open ion channels.

Figure 2.17: Stationary solution of the cable equation with a constant current of unit strength being injected at x = 0, i.e., iext(t, x) = $ \delta$(x). The electrotonic length scale $ \lambda$ is the distance over which the membrane potential drops to 1/e of its initial value.

For arbitrary stationary input current iext(x) the solution of Eq. (2.32a) can be found by a superposition of translated fundamental solutions (2.35), viz.,

u(t, x) = $\displaystyle \int$dx'  $\displaystyle {\frac{{1}}{{2}}}$ e-$\scriptstyle \left\vert\vphantom{ x-x' }\right.$x - x'$\scriptstyle \left.\vphantom{ x-x' }\right\vert$ iext(x') . (2.35)

This is an example of the Green's function approach applied here to the stationary case. The general time-dependent case will be treated in the next section.

2.5.2 Green's Function (*)

In the following we will concentrate on the equation for the voltage and start our analysis by deriving the Green's function for a cable extending to infinity in both directions. The Green's function is defined as the solution of a linear equation such as Eq. (2.32) with a Dirac $ \delta$-pulse as its input. It can be seen as an elementary solution of the differential equation because - due to linearity - the solution for any given input can be constructed as a superposition of these Green's functions.

In order to find the Green's function for the cable equation we thus have to solve Eq. (2.32a) with iext(t, x) replaced by a $ \delta$ impulse at x = 0 and t = 0,

$\displaystyle {\frac{{\partial}}{{\partial t}}}$ u(t, x) - $\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ u(t, x) + u(t, x) = $\displaystyle \delta$(t$\displaystyle \delta$(x) . (2.36)

Fourier transformation with respect to the spatial variable yields

$\displaystyle {\frac{{\partial}}{{\partial t}}}$ u(t, k) + k2 u(t, k) + u(t, k) = $\displaystyle \delta$(t)/$\displaystyle \sqrt{{2 \pi}}$ . (2.37)

This is an ordinary differential equation in t and has a solution of the form

u(t, k) = exp$\displaystyle \left[\vphantom{ - \left (1 + k^2 \right ) \, t }\right.$ - $\displaystyle \left(\vphantom{1 + k^2 }\right.$1 + k2$\displaystyle \left.\vphantom{1 + k^2 }\right)$ t$\displaystyle \left.\vphantom{ - \left (1 + k^2 \right ) \, t }\right]$/$\displaystyle \sqrt{{2 \pi}}$ $\displaystyle \Theta$(t) (2.38)

with $ \Theta$(t) denoting the Heaviside function. After an inverse Fourier transform we obtain the desired Green's function G$\scriptstyle \infty$(t, x),

u(t, x) = $\displaystyle {\frac{{\Theta (t)}}{{\sqrt{4 \pi \, t}}}}$ exp$\displaystyle \left[\vphantom{ -t - \frac{x^2}{4 \, t} }\right.$ - t - $\displaystyle {\frac{{x^2}}{{4 \, t}}}$$\displaystyle \left.\vphantom{ -t - \frac{x^2}{4 \, t} }\right]$ $\displaystyle \equiv$ G$\scriptstyle \infty$(t, x) . (2.39)

The general solution for an infinitely long cable is therewith given through

u(t, x) = $\displaystyle \int_{{-\infty}}^{t}$dt'$\displaystyle \int_{{-\infty}}^{\infty}$dx'  G$\scriptstyle \infty$(t - t', x - x'iext(t', x') . (2.40) Example: Checking the Green's property

We can check the validity of Eq. (2.40) by substituting G$\scriptstyle \infty$(t, x) into the left-hand side of Eq. (2.37). After a short calculation we find

$\displaystyle \left[\vphantom{ \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1 }\right.$$\displaystyle {\frac{{\partial}}{{\partial t}}}$ - $\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ + 1$\displaystyle \left.\vphantom{ \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1 }\right]$ G$\scriptstyle \infty$(t, x) = $\displaystyle {\frac{{1}}{{\sqrt{4\pi\,t}}}}$ exp$\displaystyle \left(\vphantom{ -t -\frac{x^2}{4 t} }\right.$ - t - $\displaystyle {\frac{{x^2}}{{4 t}}}$$\displaystyle \left.\vphantom{ -t -\frac{x^2}{4 t} }\right)$ $\displaystyle \delta$(t) , (2.41)

where we have used $ \partial$$ \Theta$(t)/$ \partial$t = $ \delta$(t). As long as t$ \ne$ 0 the right-hand side of Eq. (2.42) vanishes, as required by Eq. (2.37). For t$ \to$ 0 we find

$\displaystyle \lim_{{t \to 0}}^{}$$\displaystyle {\frac{{1}}{{\sqrt{4\pi\,t}}}}$ exp$\displaystyle \left(\vphantom{ -t -\frac{x^2}{4 t} }\right.$ - t - $\displaystyle {\frac{{x^2}}{{4 t}}}$$\displaystyle \left.\vphantom{ -t -\frac{x^2}{4 t} }\right)$ = $\displaystyle \delta$(x) , (2.42)

which proves that the right-hand side of Eq. (2.42) is indeed equivalent to the right-hand side of Eq. (2.37).

Having established that

$\displaystyle \left[\vphantom{ \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1 }\right.$$\displaystyle {\frac{{\partial}}{{\partial t}}}$ - $\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ + 1$\displaystyle \left.\vphantom{ \frac{\partial}{\partial t} - \frac{\partial^2}{\partial x^2} + 1 }\right]$ G$\scriptstyle \infty$(t, x) = $\displaystyle \delta$(x$\displaystyle \delta$(t) , (2.43)

we can readily show that Eq. (2.41) is the general solution of the cable equation for arbitrary input currents iext(t0, x0). We substitute Eq. (2.41) into the cable equation, exchange the order of integration and differentiation, and find

\left [
\frac{\partial}{\partial t} -
...elta(t-t') \, i_{\text{ext}} (t',x')
= i_{\text{ext}} (t,x)
\end{multline} Example: Finite cable

Real cables do not extend from - $ \infty$ to + $ \infty$ and we have to take extra care to correctly include boundary conditions at the ends. We consider a finite cable extending from x = 0 to x = L with sealed ends, i.e., i(t, x = 0) = i(t, x = L) = 0 or, equivalently, $ {\frac{{\partial}}{{\partial x}}}$u(t, x = 0) = $ {\frac{{\partial}}{{\partial x}}}$u(t, x = L) = 0.

The Green's function G0, L for a cable with sealed ends can be constructed from G$\scriptstyle \infty$ by applying a trick from electro-statics called ``mirror charges'' (Jackson, 1962). Similar techniques can also be applied to treat branching points in a dendritic tree (Abbott, 1991). The cable equation is linear and, therefore, a superposition of two solutions is also a solution. Consider a $ \delta$ current pulse at time t0 and position x0 somewhere along the cable. The boundary condition $ {\frac{{\partial}}{{\partial x}}}$u(t, x = 0) = 0 can be satisfied if we add a second, virtual current pulse at a position x = - x0 outside the interval [0, L]. Adding a current pulse outside the interval [0, L] comes for free since the result is still a solution of the cable equation on that interval. Similarly, we can fulfill the boundary condition at x = L by adding a mirror pulse at x = 2 L - x0. In order to account for both boundary conditions simultaneously, we have to compensate for the mirror pulse at - x0 by adding another mirror pulse at 2 L + x0 and for the mirror pulse at x = 2 L - x0 by adding a fourth pulse at -2 L + x0 and so forth. Altogether we have

G_{0,L}(t_0,x_0; t,x)= \\
...2 \, n \, L - x_0) +
G_\infty(t-t_0,x - 2 \, n \, L + x_0)

We emphasize that in the above Green's function we have to specify both (t0, x0) and (t, x) because the setup is no longer translation invariant. The general solution on the interval [0, L] is given by

u(t, x) = $\displaystyle \int_{{-\infty}}^{t}$dt0  $\displaystyle \int_{0}^{L}$dx0  G0, L(t0, x0;t, xiext(t0, x0) . (2.44)

An example for the spatial distribution of the membrane potential along the cable is shown in Fig. 2.18A, where a current pulse has been injected at location x = 1. In addition to Fig. 2.18A, subfigure B exhibits the time course of the membrane potential measured in various distances from the point of injection. It is clearly visible that the peak of the membrane potential measured at, e.g., x = 3 is more delayed than at, e.g., x = 2. Also the amplitude of the membrane potential decreases significantly with the distance from the injection point. This is a well-known phenomenon that is also present in neurons. In the absence of active amplification mechanisms, synaptic input at distal dendrites produces broader and weaker response at the soma as compared to synaptic input at proximal dendrites.

Figure 2.18: Spatial distribution (A) and temporal evolution (B) of the membrane potential along a dendrite (L = 5) with sealed ends ( $ \left.\vphantom{ \frac{\partial} {\partial x} u}\right.$$ {\frac{{\partial}}{{\partial x}}}$u$ \left.\vphantom{ \frac{\partial} {\partial x} u}\right\vert _{{x\in\{0,L\}}}^{}$ = 0) after injection of a unit current pulse at x = 1 and t = 0. The various traces in A show snapshots for time t = 0.1, 0.2,..., 1.0, respectively (top to bottom). The traces in B give the membrane potential as a function of time for different locations x = 1.5, 2.0, 2.5..., 5.0 (top to bottom) along the cable.
{\bf A}
...rline{\includegraphics[width=\textwidth]{}} \end{minipage}

2.5.3 Non-linear Extensions to the Cable Equation

In the context of a realistic modeling of `biological' neurons two non-linear extensions of the cable equation have to be discussed. The obvious one is the inclusion of non-linear elements in the circuit diagram of Fig. 2.16 that account for specialized ion channels. As we have seen in the Hodgkin-Huxley model, ion channels can exhibit a complex dynamics that is in itself governed by a system of (ordinary) differential equations. The current through one of these channels is thus not simply a (non-linear) function of the actual value of the membrane potential but may also depend on the time course of the membrane potential in the past. Using the symbolic notation iion[u](t, x) for this functional dependence the extended cable equation takes the form

$\displaystyle {\frac{{\partial}}{{\partial t}}}$ u(t, x) = $\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ u(t, x) - u(t, x) - iion[u](t, x) + iext(t, x) . (2.45)

A more subtle complication arises from the fact that a synapse can not be treated as an ideal current source. The effect of an incoming action potential is the opening of ion channels. The resulting current is proportional to the difference of the membrane potential and the corresponding ionic reversal potential. Hence, a time-dependent conductivity as in Eq. (2.19) provides a more realistic description of synaptic input than an ideal current source with a fixed time course.

If we replace in Eq. (2.32a) the external input current iext(t, x) by an appropriate synaptic input current - isyn(t, x) = - gsyn(t, x)[u(t, x) - Esyn] with gsyn being the synaptic conductivity and Esyn the corresponding reversal potential, we obtain2.2

$\displaystyle {\frac{{\partial}}{{\partial t}}}$ u(t, x) = $\displaystyle {\frac{{\partial^2}}{{\partial x^2}}}$ u(t, x) - u(t, x) - gsyn(t, x)[u(t, x) - Esyn] . (2.46)

This is still a linear differential equation but its coefficients are now time-dependent. If the time course of the synaptic conductivity can be written as a solution of a differential equation then the cable equation can be reformulated so that synaptic input reappears as an inhomogeneity to an autonomous equation. For example, if the synaptic conductivity is simply given by an exponential decay with time constant $ \tau_{{\text{syn}}}^{}$ we have
\begin{subequations}\begin{align}\frac{\partial}{\partial t} \, u(t,x) - \frac{\...
...xt{syn}}^{-1} \, g_{\text{syn}}(t,x) &= S(t,x) \,. \end{align}\end{subequations}

Here, S(t, x) is a sum of Dirac $ \delta$ functions which describe the presynaptic spike train that arrives at a synapse located at position x. Note that this equation is non-linear because it contains a product of gsyn and u which are both unknown functions of the differential equation. Consequently, the formalism based on Green's functions can not be applied.

next up previous contents index
Next: 2.6 Compartmental Models Up: 2. Detailed Neuron Models Previous: 2.4 Synapses
Gerstner and Kistler
Spiking Neuron Models. Single Neurons, Populations, Plasticity
Cambridge University Press, 2002

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