# Green's Function Method

## Green’s Function Method

In this section, we demonstrate Green’s function method of solving differential equations.

### Cable Equation

The cable equation is written as ^{1}

\begin{equation} \frac{\partial}{\partial t} u(t,x) = \frac{\partial^2}{\partial x^2} u(t,x) - u(t,x) + i_{e}(t,x), \label{eqn-cable-equation-potential} \end{equation}

or

\begin{equation} \frac{\partial}{\partial t} i(t,x) = \frac{\partial^2}{\partial x^2} i(t,x) - i(t,x) + \frac{\partial}{\partial x} i_e (t,x), \label{eqn-cable-equation-current} \end{equation}

where $t$, $x$, $i$, $i_e$ are all renormalized unit less quantities. For the meaning and definition of them, ref. page 55 of Gerster 2002 ^{1}.

### Solutions to Cable Equation

#### Stationary Solution

Equation \ref{eqn-cable-equation-potential} can be solved for stationary case, which is

\begin{equation} \frac{\partial^2}{\partial x^2} u(t,x) - u(t,x) =- i_{e}(t,x) . \label{eqn-cable-equation-stationary-equation-potential} \end{equation}

While many methods can be used to solve second order nonhonogenerous differential equations, Green’s function is the most general and useful one.

The stationary equation \ref{eqn-cable-equation-stationary-equation-potential} can be written as

\begin{equation} \hat L_x u(x) = - i_e(t,x), \end{equation}

where $\hat L_x = \frac{d^2}{dx^2} -1$. The boundary condition is the vanishing wave at infinity $u(\pm\infty)=0$. As we are talking about stationary equation, the source should be time-independent, thus we take only a one dimension Dirac distribution $\delta(x)$ to solve for GF.

The general Green’s function is ^{2}

\begin{equation} G(x,x’) = \begin{cases} C_1 e^{-x} + D_1 e^{x}, & \qquad x\leq z,\\ C_2 e^{-x} + D_2 e^{x}, & \qquad x\geq z. \end{cases} \end{equation}

In this simple case, BC can be applied to Green’s function first ^{3}, which means

These conditions can significantly simplify the GF,

\begin{equation} G(x,x’) = \begin{cases} D_1 e^{x}, & \qquad x<x’,\\ C_2 e^{-x}, & \qquad x>x’. \end{cases} \end{equation}

Then we use the continuity condition and discontinuity condition,

$$ \begin{align} G(x'_-,x') - G(x'_+,x') &= 0 \\ \left.\frac{d}{dx}G(x,x')\right\vert_{x=x'_+} - \left.\frac{d}{dx}G(x,x')\right\vert_{x=x'_-} &= 1, \end{align} $$which is

$$ \begin{align} D_1 e^{x'} - C_2 e^{-x'} &= 0,\\ - C_2 e^{-x'} - D_1 e^{x'} & =1. \end{align} $$Solving out the coefficients, we get

$$ \begin{align} D_1 & = \frac{1}{2}e^{-x'},\\ C_2 &= \frac{1}{2}e^{x'}. \end{align} $$Then we reached the complete and final GF,

$$ G(x,x’) = \begin{cases} \frac{1}{2}e^{x-x’}, & \qquad x<x’\\ \frac{1}{2}e^{x’ - x}. & \qquad x>x' \end{cases} $$

Given any general source $-i_e(t,x)$, we can write down the solution

$$ u(x) = \int G(x,x') (-i_e(t,x') ) dx'. $$As a verification, we integrate out for $i_e(t,x) = 1$,

$$ u(x) = -\int_{-\infty}^{x} \frac{1}{2}e^{x'-x} dx' - \int_{x}^{\infty} \frac{1}{2}e^{x - x'} dx' = 1, $$which is exactly the solution given by Mathematica and makes sense.

#### Physical Meaning

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,

$$ \left(\frac{d^2}{dx^2} -1\right) u(x) = \delta(x'-x). $$On the RHS, source term is a delta function, which is just a stimulation to the system at point $x’$. The textbook shows a graph ^{4} for the case $x’=0$, where we see the stimulation is given for point $x’=0$ and the potential drops as we deviate from the stimulated point.

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 ^{5}

#### Non-stationary Solution

To solve the most general non-homogeneous cable equation even for non-stationary case, we have to introduce a two-dimensional Dirac distribution $\delta^2(t,x) = \delta(t)\delta(x)$.

Green’s function for the most general case should satisfy ^{1}

Again to save ink we define

$$ \hat L_{t,x} = \hat L_t - \hat L_x, $$where $\hat L_t = \frac{\partial}{\partial t}$ and $\hat L_x = \frac{\partial^2}{\partial x^2} - 1$.

The trick is to solve for time dependence first by Fourier transforming the equation to frequency space. To achieve that, we define

$$ \begin{equation} G(t,t';x,x') = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty G(t,t';k,x')e^{ikx}dk. \label{eqn-green-function-fourier-transform} \end{equation} $$On the other hand, Dirac delta is Fourier transformed to

$$ \begin{equation} \delta(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \delta(\bar x) e^{- ik \bar x} d\bar x = \frac{1}{\sqrt{2\pi}} , \label{eqn-dirac-delta-fourier-transform} \end{equation} $$which infact gives one of the representations of Dirac delta distribution

$$ \delta(\bar x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{ik \bar x} dk . $$Applying the transforms of \ref{eqn-green-function-fourier-transform} and \ref{eqn-dirac-delta-fourier-transform} to the equation we have

$$ \hat L_{t,x}\left( \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty G(t,t';k,x') e^{i kx} dk \right) = \delta(t'-t) \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{i k(x'-x)} dk, $$which becomes

$$ \begin{align} &\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{\partial}{\partial t} G(t,t';k,x') e^{ikx}dk - \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty G(t,t';k,x') \frac{\partial^2}{\partial x^2} e^{ikx}dk \\ &+ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty G(t,t';k,x')e^{ikx}dk = \delta(t'-t) \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{ik(x'-x)} dk, \end{align} $$which is then simplified by removing the integral and common parts

$$ \frac{\partial}{\partial t} G(t,t';k,x') + k^2 G(t,t';k,x') + G(t,t';k,x') = \delta(t'-t) \frac{1}{\sqrt{2\pi}} . $$Then we solve this first-order differential equation.

Wulfram Gerstner and Werner M. Kistler, Spiking Neuron Models, (2002). ↩︎ ↩︎ ↩︎

In fact this is can be obtained by using the Fourier transform method. ↩︎

Because the only possibility to make the integral $u(\pm\infty)=\int G(\pm\infty,x’) dx’=0$ satisfy $u(\pm\infty)=0$ is to make sure GF vanish on the boundaries. ↩︎

Wulfram Gerstner and Werner M. Kistler, Spiking Neuron Models, (2002), Fig. 2.17. ↩︎

Since opening of ion channesl can significantly change the transverse conductivity, this estimation can change significantly in different situations. ↩︎

Lei Ma (2020). 'Green's Function Method', Intelligence, 10 April. Available at: https://intelligence.leima.is/toolbox/equation-solving-in-neuroscience/greens-function/.