Problem 2: Homogenized Bidomain Model

Define transmembrane potential as

$$\phi = (\phi_i - \phi_e)\mid_{\partial\Omega_m}.\nonumber$$

and transmembrane current by

$$I_m = C_m{d\phi\over dt} + {1\over R_m}F_m(\phi)\nonumber$$

From above, we know that

$$\phi_i = \psi_i(x) + \epsilon W_i(z)\cdot T^{-1}\nabla \psi_i(x) + O(\epsilon^2 \psi_i)\nonumber$$

and

$$\phi_e = \psi_e(x) +\epsilon W_e(z)\cdot T^{-1}\nabla \psi_e(x) + O(\epsilon^2\psi_e)\nonumber$$

and that $\psi_i(x)$ and $\psi_e(x)$ satisfy

$$\nabla \cdot(\sigma_i \nabla\psi_i)= - \nabla\cdot (\sigma_e... ...i_e) = {1\over V}\int_{\partial\Omega_m} I_m(x,z)dS_z\nonumber$$

It follows that

$$\nabla \cdot(\sigma_i \nabla\psi_i)= - \nabla\cdot (\sigma_e... ...}F_m\left(\psi + \epsilon H(z,x)\right)dS_z \right). \nonumber$$

where

$$H(z,x) = W_i(z)\cdot T^{-1}\nabla \psi_i(x) - W_e(z)\cdot T^{-1}\nabla \psi_e(x) \nonumber$$

$$\psi = \psi_i - \psi_e.\nonumber$$