Model Mathematics

\frac{d}{d time} (R) = kr_plus Ca (1 - R) - (kr_minus R) T = T_bar R

Ca_open = \frac{sigma}{2 Dc r π} sigma = (- 5.182) i_V i_V = \frac{\frac{g_Ca P 2 F V_post}{R_gas_const Temp} Ca_ex}{1 - (ⅇ^{\frac{2 F V_post}{R_gas_const Temp}})} Ca = O Ca_open + 0.1

alpha = 0.45 ⅇ^{\frac{V_post}{22}} alpha_= \frac{alpha}{8} beta = 0.015 ⅇ^{\frac{- V_post}{14}} beta_= beta 8 \frac{d}{d time} (b) = kb_plus T (1 - b) - (kb_minus b) kG_plus = \frac{3 b}{680 + 320 b}

\frac{d}{d time} (C1) = beta C2 + kG_minus C_G1 - (C1 (4 alpha + kG_plus))

\frac{d}{d time} (C2) = 4 alpha C1 + 2 beta C3 + kG2_minus C_G2 - (C2 (beta + 3 alpha + kG_plus))

\frac{d}{d time} (C3) = 3 alpha C2 + 3 beta C4 + kG3_minus C_G3 - (C3 (2 beta + 2 alpha + kG_plus))

\frac{d}{d time} (C4) = 2 alpha C3 + 4 beta O - (C4 (3 beta + alpha))

O = 1 - C1 - C2 - C3 - C4 - C_G1 - C_G2 - C_G3 C_G = C_G1 + C_G2 + C_G3

\frac{d}{d time} (C_G1) = beta_C_G2 + kG_plus C1 - (C_G1 (4 alpha_+ kG_minus))

\frac{d}{d time} (C_G2) = 4 alpha_C_G1 + 2 beta_C_G3 + kG_plus C2 - (C_G2 (beta_+ 3 alpha_+ kG2_minus))

\frac{d}{d time} (C_G3) = 3 alpha_C_G2 + kG_plus C3 - (C_G3 (2 beta_+ kG3_minus))

\frac{d}{d time} (V_post) = \frac{- (i_Na_post + i_K_post + i_leak_post + i_syn)}{Cm}

i_syn = g_syn b (V_post - V_syn)

i_leak_post = 0.3 (V_post + 54)

i_Na_post = 120 {x_infinity}^{3} (1 - n_post) (V_post - 120) x_infinity = \frac{alpha_x}{alpha_x + beta_x} alpha_x = \frac{0.2 (V_post + 40)}{1 - (1 ⅇ^{\frac{- (V_post + 40)}{10}})} beta_x = 8 ⅇ^{\frac{1}{- (V_post + \frac{65}{18})}}

i_K_post = 36 {n_post}^{4} (V_post + 77)

\frac{d}{d time} (n_post) = alpha_n (1 - n_post) - (beta_n n_post) alpha_n = \frac{0.02 (V_post + 55)}{1 - (ⅇ^{\frac{- (V_post + 55)}{10}})} beta_n = 0.25 ⅇ^{\frac{- (V_post + 65)}{80}}