Cable Equation and Its Solutions

Checkout the attachment: 06.1.cable_equation_green_function.pdf

Stationary Equation

The stationary equation is $$\begin{array}{}\text{(1)}& (\frac{d^2}{d{x}^2}-1)u(x)=-i_e(x).\end{array}$$

The form solution using Green’s function method is $$\begin{array}{}\text{(2)}& u(x)=\int G(x,x^{\prime })(-i_e(t,x^{\prime }))d{x}^{\prime },\end{array}$$ where $$G(x,x^{\prime })=\{\begin{array}{lll}\frac{1}{2}{e}^{x-x^{\prime }},& x<x^{\prime };\frac{1}{2}{e}^{x^{\prime }-x},& x>x^{\prime }.\end{array}$$

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, $$(\frac{d^2}{d{x}^2}-1)u(x)=\delta (x^{\prime }-x).$$

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.}

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

Time-dependent Source

To include the time-dependent source we need a two dimensional Dirac delta distribution, $$\begin{array}{}\text{(3)}& \frac{\partial }{\partial t}G(t,t^{\prime };x,x^{\prime })-\frac{{\partial }^2}{\partial x^2}G(t,t^{\prime };x,x^{\prime })+G(t,t^{\prime };x,x^{\prime })=\delta (t^{\prime }-t)\delta (x^{\prime }-x).\end{array}$$

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

Finite Cable

The boundary condition is given by $$\begin{array}{}\text{(5)}& i(t,x=0)=i(t,x=L)=0.\end{array}$$

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.