06.Cable Equation and Its Solutions

Cable Equation and Its Solutions

Checkout the attachment: 06.1.cable_equation_green_function.pdf

Stationary Equation

The stationary equation is \begin{equation} \left(\frac{d^2}{dx^2} -1\right) u(x) = - i_e(x). \end{equation}

The form solution using Green’s function method is \begin{equation} u(x) = \int G(x,x’) (-i_e(t,x’) ) dx’, \end{equation} where \begin{equation} G(x,x’) = \begin{cases} \frac{1}{2}e^{x-x’}, & \qquad x<x’;
\frac{1}{2}e^{x’ - x}, & \qquad x>x’. \end{cases} \end{equation

So far we have been dealing with math. What is the actual meaning of GF? To dive into this question we need to review the equation for GF, in this case, \begin{equation} \left(\frac{d^2}{dx^2} -1\right) u(x) = \delta(x’-x). \end{equation}

In a stimulation-response system, one of the most important properties is the resonance width, or reaction width, which means the deviation required for the amplitude to drop to $1/e$ of the peak value. In this stationary solution, the distance is $1$ in renormalized unit. To transform back to to SI unit, recall that the characteristic length is this problem is $\lambda = \sqrt{\frac{r_T}{r_L}}$.

Just to build a picture, this length is around\footnote{Since opening of ion channesl can significantly change the transverse conductivity, this estimation can change significantly in different situations.}

\begin{equation} \lambda = \sqrt{ \frac{r_T}{r_L}} = \sqrt{ \frac{30\mathrm{k\Omega\cdot cm^2}/(2\pi \rho)}{ 100 \mathrm{k\Omega\cdot cm}/(2\pi \rho) } } = \sqrt{ \frac{5\times 10^{11} \mathrm{\Omega \cdot \mu m} }{ 3\times 10^{5} \mathrm{\Omega \cdot \mu m^{-1}} } } = 1.2\mathrm{mm} \end{equation}

Time-dependent Source

To include the time-dependent source we need a two dimensional Dirac delta distribution, \begin{equation} \frac{\partial}{\partial t} G(t,t’;x,x’) - \frac{\partial^2}{\partial x^2} G(t,t’;x,x’) + G(t,t’;x,x’) = \delta(t’-t)\delta(x’-x). \end{equation}

The Green’s function is solved out \begin{equation} G(t,0;x,0) = \frac{\Theta(t)}{\sqrt{4\pi t}} \exp\left( -t - \frac{x^2}{4t} \right). \end{equation}

Finite Cable

The boundary condition is given by \begin{equation} i(t,x=0) = i(t,x=L) = 0. \end{equation}

Non-linear Cable Equation

The current for the ion channels is not simply proportional to the membrane potential which we used for the previous cable equations. Introducing such a time dependent conductivity for the transverse current will significantly increase the complexity of the equations.

comments powered by Disqus