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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