Activating function

In a compartment model of an axon, the activating function of compartment n, ${\displaystyle f_{n}}$, is derived from the driving term of the external potential, or the equivalent injected current

${\displaystyle f_{n}=1/c\left({\frac {V_{n-1}^{e}-V_{n}^{e}}{R_{n-1}/2+R_{n}/2}}+{\frac {V_{n+1}^{e}-V_{n}^{e}}{R_{n+1}/2+R_{n}/2}}+...\right)}$,

where ${\displaystyle c}$ is the membrane capacity, ${\displaystyle V_{n}^{e}}$ the extracellular voltage outside compartment ${\displaystyle n}$ relative to the ground and ${\displaystyle R_{n}}$ the axonal resistance of compartment ${\displaystyle n}$.

The activating function represents the rate of membrane potential change if the neuron is in resting state before the stimulation. Its physical dimensions are V/s or mV/ms. In other words, it represents the slope of the membrane voltage at the beginning of the stimulation.^[8]

Following McNeal's^[9] simplifications for long fibers of an ideal internode membrane, with both membrane capacity and conductance assumed to be 0 the differential equation determining the membrane potential ${\displaystyle V^{m}}$ for each node is:

${\displaystyle {\frac {dV_{n}^{m}}{dt}}=\left[-i_{ion,n}+{\frac {d\Delta x}{4\rho _{i}L}}\cdot \left({\frac {V_{n-1}^{m}-2V_{n}^{m}+V_{n+1}^{m}}{\Delta x^{2}}}+{\frac {V_{n-1}^{e}-2V_{n}^{e}+V_{n+1}^{e}}{\Delta x^{2}}}\right)\right]/c}$,

where ${\displaystyle d}$ is the constant fiber diameter, ${\displaystyle \Delta x}$ the node-to-node distance, ${\displaystyle L}$ the node length ${\displaystyle \rho _{i}}$ the axomplasmatic resistivity, ${\displaystyle c}$ the capacity and ${\displaystyle i_{ion}}$ the ionic currents. From this the activating function follows as:

${\displaystyle f_{n}={\frac {d\Delta x}{4\rho _{i}Lc}}{\frac {V_{n-1}^{e}-2V_{n}^{e}+V_{n+1}^{e}}{\Delta x^{2}}}}$.

In this case the activating function is proportional to the second order spatial difference of the extracellular potential along the fibers. If ${\displaystyle L=\Delta x}$ and ${\displaystyle \Delta x\to 0}$ then:

${\displaystyle f={\frac {d}{4\rho _{i}c}}\cdot {\frac {\delta ^{2}V^{e}}{\delta x^{2}}}}$.

Thus ${\displaystyle f}$ is proportional to the second order spatial differential along the fiber.

Positive values of ${\displaystyle f}$ suggest a depolarization of the membrane potential and negative values a hyperpolarization of the membrane potential.