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.,

xvb(0,t) = 0, be C.                       (2.24)

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

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

wbcf(x, 0) = wbcf (æ) ≡ wcftθθ (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
sb

ʃɪɪŋ,b(*) ≡ τr Σ2 ‰(i)<5(≈ - Xbs)(υb(x) - Ebs).             (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 gbs = gbs + εgbs(‰ t), where



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