The name is absent

24

denote by £ the set of leaf indices, where a leaf is a branch with no children. We
suppose that each leaf is sealed at its distal end, i.e.,

∂_xv_b(0,t) = 0, be C.                       (2.24)

Initially the neuron is at rest, implying that ∂tv_b{x, 0) = 0. We solve for the rest state
and denote it by υ_b{x), and similarly for the gating variables, which yields the initial
conditions

'(⅛(.τ, 0) = ⅝(χ∙)                                            (2.25)

w_bcf(x, 0) = w_bcf (æ) ≡ w_cftθθ (⅝(x)).                  (2.26)

With the rest state defined, it is easy to modify (2.17) to use current injection instead
of synaptic conductance. If we substitute the rest state t⅛(rr) for the Vb(x) in the
synaptic input term, then this is equivalent to directly injecting current into the cell,
which yields
ɪ s_b

ʃɪɪŋ,b(*) ≡ τr Σ2 ‰(i)<5(≈ - Xb_s)(υ_b(x) - E_bs).             (2.27)

Z7Γ ^z—'
s=l

2.3.2 Linearizing the Cable Equation

The linearization process follows the same technique as that described in §2.2.

Consider (2.17) for branch b at its resting potential t⅛. If g_bs = g_bs + εg_bs(‰ t), where

<<   <   31   32   33   34   35   36   37   >   >>