Modeling Dendrites and Spatially-Distributed Neuronal Membrane Properties

Abstract

The first step toward understanding the brain is to learn how individual neurons process incoming signals, the vast majority of which arrive in their dendrites. Dendrites were first discovered at the beginning of the twentieth century and were characterized by great anatomical variability, both within and across species. Over the past years, a rich repertoire of active and passive dendritic mechanisms has been unveiled, which greatly influences their integrative power. Yet, our understanding of how dendrites compute remains limited, mainly because technological limitations make it difficult to record from dendrites directly and manipulate them. Computational modeling, on the other hand, is perfectly suited for this task. Biophysical models that account for the morphology as well as passive and active neuronal properties can explain a wide variety of experimental findings, shedding new light on how dendrites contribute to neuronal and circuit computations. This chapter aims to help the interested reader build biophysical models incorporating dendrites by detailing how their electrophysiological properties can be described using simple mathematical frameworks. We start by discussing the passive properties of dendrites and then give an overview of how active conductances can be incorporated, leading to realistic in silico replicas of biological neurons.

Appendix: Mathematical Derivations

1.1 A.1 General Solution to the Linear Cable Equation in Time and Space Given a Current Injection, I(x, t), at Some Point

$$\begin{array}{ll} \displaystyle {\tau}_m\frac{\partial }{\partial t}V\left(x,t\right)&\displaystyle ={\lambda}^2\frac{\partial^2}{\partial {x}^2}V\left(x,t\right)\\&\quad \displaystyle -V\left(x,t\right)+{r}_mI\left(x,t\right)\end{array} $$

(A.1)

The initial voltage at t = 0 is given by V(x, 0) = V₀(t)

To analytically solve Eq. (A.1), we will take advantage of the Fourier transforms. Recall that the Fourier transform is usually used to convert the time domain into a frequency domain. However, here we will use it to solve the cable equation. Thus, we will apply the Fourier transform in the space domain instead of time.

Generally, the Fourier transform of any function is given by the integral

$$\begin{array}{l} \displaystyle f\left(x,t\right)\overset{\mathcal{F}}{\to }F\left(\omega, t\right):\hat{F}\left(\omega, t\right)\\\displaystyle \quad =\underset{-\infty }{\overset{\infty }{\int }}f\left(x,t\right)\exp \left(- i\omega x\right) dx \end{array} $$

The inverse Fourier transform returns us to the space domain

$$\begin{array}{l} \displaystyle \hat{F}\left(\omega, t\right)\overset{{\mathcal{F}}^{-1}}{\to }f\left(x,t\right):f\left(x,t\right)\\\displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}\hat{F}\left(\omega, t\right)\exp \left( i\omega x\right) d\omega \end{array}$$

Applying Fourier transformation to Eq. (A.1) and using the linearity property of the Fourier, i.e.

$$\begin{array}{l} \displaystyle \mathcal{F}\left\{\alpha f\left(x,t\right)+\beta g\left(x,t\right)\right\}\\\displaystyle =\alpha \mathcal{F}\left\{f\left(x,t\right)\right\}+\beta \mathcal{F}\left\{g\left(x,t\right)\right\}\end{array} $$

we obtain the expression in Fourier space

$$\begin{array}{l} \displaystyle\tau \mathcal{F}\left\{\frac{\partial }{\partial t}V\left(x,t\right)\right\}+\mathcal{F}\left\{V\left(x,t\right)\right\}\\\displaystyle -{\lambda}^2\mathcal{F}\left\{\frac{\partial^2}{\partial {x}^2}V\left(x,t\right)\right\}={r}_m\mathcal{F}\left\{I\left(x,t\right)\right\} \end{array}$$

Using the property of derivatives, i.e.:

$$\begin{array}{l} \displaystyle \mathrm{if}\ f\left(x,t\right)\overset{\mathcal{F}}{\to}\hat{F}\left(\omega, t\right),\mathrm{then}\ \mathcal{F}\left\{\frac{\partial^n}{\partial {x}^n}f\left(x,t\right)\right\}\\\displaystyle ={\left( i\omega \right)}^n\hat{F}\left(\omega, t\right)\end{array} $$

Thus, Eq. (A.1) is written as

$$\begin{array}{l} \displaystyle {\tau}_m\frac{\mathrm{d}}{\mathrm{d}t}\hat{V}\left(\omega, t\right)+\hat{V}\left(\omega, t\right)\\\displaystyle -{\lambda}^2{\left( i\omega \right)}^2\hat{V}\left(\omega, t\right)={r}_m\hat{I}\left(\omega, t\right)\Longleftrightarrow \end{array}$$

$$ {\tau}_m\frac{\mathrm{d}}{\mathrm{d}t}\hat{V}\left(\omega, t\right)+\left(1+{\lambda}^2{\omega}^2\right)\hat{V}\left(\omega, t\right)={r}_m\hat{I}\left(\omega, t\right) $$

(A.2)

Equation (A.2) is an inhomogeneous ordinary differential equation, and we will solve it with the method of variation of a constant.

For simplifying the notation, we let $\hat{V}\equiv \hat{V}\left(\omega, t\right),\hat{I}\equiv \hat{I}\left(\omega, t\right)$

$$ {\tau}_m\frac{\mathrm{d}\hat{V}}{\mathrm{d}t}+\left(1+{\lambda}^2{\omega}^2\right)\hat{V}={r}_m\hat{I} $$

First, we solve the homogenous differential equation, setting the right-hand side of the equation to zero.

$$ {\tau}_m\frac{\mathrm{d}\hat{V}}{\mathrm{d}t}+\left(1+{\lambda}^2{\omega}^2\right)\hat{V} = 0 \Longleftrightarrow $$

$$ \frac{\mathrm{d}\hat{V}}{\mathrm{d}t}=-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}\hat{V}\Longleftrightarrow $$

$$ \frac{\mathrm{d}\hat{V}}{\hat{V}}=-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}\mathrm{d}t $$

Then, we obtain the solution by integrating both sides and using some calculus properties^6,7,8

$$ \int \frac{\mathrm{d}\hat{V}}{\hat{V}}=\int -\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}\mathrm{d}t\Longleftrightarrow \vspace*{-15pt}$$

$$ \ln \left|\hat{V}\right|=-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t+{C}_1,{C}_1\in \mathbb{R} \vspace*{-15pt}$$

$$ \exp \left(\ln \left|\hat{V}\right|\right)=\exp \left(-\frac{\left(1{+}{\lambda}^2{\omega}^2\right)}{\tau_m}t{+}{C}_1\right)\Longleftrightarrow \vspace*{-15pt}$$

$$ \left|\hat{V}\right|=\exp \left({C}_1\right)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\Longleftrightarrow \vspace*{-15pt}$$

$$ \hat{V}=\pm \exp \left({C}_1\right)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\Longleftrightarrow $$

Let C =  ±  exp (C₁)

$$ \hat{V}\left(\omega, t\right)=C\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right),C\in \mathbb{R} $$

(A.3)

Now, to solve the inhomogeneous equation, we plug in the solution (Eq. A.3), assuming that the constant C is some function of time, C(t)

$$ \tau \frac{\mathrm{d}\hat{V}}{\mathrm{d}t}+\left(1+{\lambda}^2{\omega}^2\right)\hat{V}={r}_m\hat{I}\Longleftrightarrow $$

$$\begin{array}{l} \displaystyle \tau \left[\frac{\mathrm{d}}{\mathrm{d}t}C(t)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\right]\\\displaystyle\qquad +\left(1+{\lambda}^2{\omega}^2\right)C(t)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\\\displaystyle\quad ={r}_m\hat{I}\Longleftrightarrow\end{array} $$

where, using the summation and composite rules of derivatives⁹

$$\begin{array}{l} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}C(t)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\\\quad \displaystyle ={C}^{\prime }(t)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\\\qquad \displaystyle -C(t)\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}\\\qquad \displaystyle \times \exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right) \end{array}$$

Canceling out the opposite terms (i.e., same absolute value but different signs),

$$ {C}^{\prime }(t)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)=\frac{r_m}{\tau_m}\hat{I}\Longleftrightarrow \vspace*{-15pt}$$

$$ {C}^{\prime }(t)=\frac{r_m\hat{I}}{\tau_m}\exp \left(\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\Longrightarrow \vspace*{-15pt}$$

$$\begin{array}{ll} C(t)&\displaystyle =\underset{0}{\overset{t }{\int }}\frac{r_m\hat{I}}{\tau_m}\exp \left(\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right) dt\\&\displaystyle \quad +{C}_2,{C}_2\in \mathbb{R}\end{array} \vspace*{-15pt}$$

Plug this into Eq. (A.3)

$$ \begin{array}{ll}\displaystyle \hat{V}\left(\omega, t\right)&\displaystyle =\left(\!\underset{0}{\overset{t}{\int }}\frac{r_m\hat{I}}{\tau_m}\exp \!\left(\!\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}s\!\right)\! ds+{C}_2\!\right)\\&\displaystyle \exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\end{array} $$

At t = 0, $\hat{V}\left(\omega, 0\right)={\hat{V}}_0\left(\omega \right)$

$$ {C}_2={\hat{V}}_0\left(\omega \right) $$

$$\begin{array}{ll} \displaystyle\hat{V}\left(\omega, t\right)&\displaystyle =\left(\underset{0}{\overset{t}{\int }}\frac{r_m\hat{I}}{\tau_m}\exp \left(\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}s\right)ds\right.\\&\quad \left.\displaystyle \vphantom{\underset{0}{\overset{t}{\int }}}+{\hat{V}}_0\left(\omega \right)\right)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right) \end{array}$$

$$\begin{array}{ll} \displaystyle\hat{V}\left(\omega, t\right)&\displaystyle ={\hat{V}}_0\left(\omega \right)\exp \left(-\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}t\right)\\&\quad \displaystyle +\frac{r_m}{\tau_m}\underset{0}{\overset{t}{\int }}\hat{I}\left(\omega, s\right)\\&\quad \displaystyle \times \exp \left(\frac{\left(1+{\lambda}^2{\omega}^2\right)}{\tau_m}\left(t-s\right)\right) ds\end{array} $$

Thus, we have obtained the solution of the cable equation for any injected current. The next step is to return in space domain applying the inverse Fourier transformation and using the integral summation property¹⁰

$$\begin{array}{l} \displaystyle V\left(x,t\right)\\\quad \displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}{\hat{V}}_0\left(\omega \right)\exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}t\right)\\\qquad \displaystyle\times \exp \left( i\omega t\right) d\omega +\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }} \left(\frac{r_m}{\tau_m}\underset{0}{\overset{t}{\int }}\hat{I}\left(\omega, s\right)\right.\\\qquad \displaystyle \times\left. \exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}\left(t-s\right)\right) d s\right)\\\qquad \displaystyle \times\exp \left( i\omega x\right) d\omega \end{array}$$

Thus, the complete solution of the linear cable equation is

$$\begin{array}{l} \displaystyle V\left(x,t\right)\\\quad \displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}{\hat{V}}_0\left(\omega \right)\exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}t\right)\\\qquad \displaystyle \times \exp \left( i\omega t\right) d\omega +\frac{1}{2\pi}\frac{r_m}{\tau_m}\ \underset{0}{\overset{t}{\int }}\underset{-\infty }{\overset{\infty }{\int }}\hat{I}\left(\omega, s\right)\\\qquad \displaystyle \times \exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}\left(t-s\right)\right)\\\qquad \displaystyle \times \exp \left( i\omega x\right) d\omega ds \end{array}$$

Let

$$\begin{array}{ll} \displaystyle {J}_1&\displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}{\hat{V}}_0\left(\omega \right)\exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}t\right)\\&\quad \displaystyle \times \exp \left( i\omega x\right) d\omega \end{array}$$

and

$$\begin{array}{ll} \displaystyle {J}_2&\displaystyle =\frac{1}{2\pi}\frac{r_m}{\tau_m}\ \underset{0}{\overset{t}{\int }}\underset{-\infty }{\overset{\infty }{\int }}\hat{I}\left(\omega, s\right)\\&\quad \displaystyle \times \exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}\left(t-s\right)\right)\\&\quad \displaystyle \times \exp \left( i\omega x\right) d\omega ds\end{array} $$

First, we solve J₁

$$\begin{array}{ll} \displaystyle {J}_1&=\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}{\hat{V}}_0\left(\omega \right)\exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}t\right)\\&\displaystyle \quad \times \exp \left( i\omega x\right) d\omega \Longleftrightarrow \end{array}$$

$$\begin{array}{ll} \displaystyle {J}_1&\displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}{\hat{V}}_0\left(\omega \right)\\&\quad \displaystyle \times \exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}t+ i\omega x\right) d\omega \Longleftrightarrow \end{array}$$

$$\begin{array}{ll} \displaystyle {J}_1&\displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}\exp \left(-\frac{t}{\tau_m}\right)\\&\displaystyle \quad \times \exp \left(-\frac{t}{\tau_m}{\lambda}^2{\omega}^2\right){\hat{V}}_0\left(\omega \right)\\&\displaystyle \quad \times \exp \left( i\omega x\right) d\omega \end{array}$$

This is the inverse Fourier transform of a multiplication. Here, we will use the convolution property.¹¹

For any arbitrary functions, f(x, t) and h(x, t), according to the convolution property

$$ \mathcal{F}\left\{f\left(x,t\right)\ast h\left(x,t\right)\right\}=F\left(\omega, t\right)H\left(\omega, t\right) $$

where $F\left(\omega, t\right)=\exp \left(-\frac{t}{\tau_m}\right)\exp \left(-\frac{t}{\tau_m}{\lambda}^2{\omega}^2\right)$ and $H\left(\omega, t\right)={\hat{V}}_0\left(\omega \right)$

By definition,

$$ h\left(x,t\right)={V}_0(x) $$

Thus, we have to find the inverse Fourier transform of f(x, t)

$$\begin{array}{ll} \displaystyle f\left(x,t\right)&\displaystyle =\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}\exp \left(-\frac{t}{\tau_m}\right)\exp \left(-\frac{t}{\tau_m}{\lambda}^2{\omega}^2\right)\\&\quad \displaystyle \times \exp \left( i\omega x\right) d\omega \end{array}$$

The first exponential does not depend on ω, thus, it can go outside of the integral, and completing the square over ω inside the second exponential:¹²

$$\begin{array}{l} \displaystyle -\frac{t}{\tau_m}{\lambda}^2{\omega}^2+ ix\omega =-\frac{t}{\tau_m}{\lambda}^2\!{\left(\!\omega +\frac{(ix)^2}{2\!\left(\!-\frac{t}{\tau_m}{\lambda}^2\!\right)}\!\right)}^2\\\displaystyle -\frac{(ix)^2}{4\left(-\frac{t}{\tau_m}{\lambda}^2\right)}==-\frac{t}{\tau_m}{\lambda}^2{\left(\omega -\frac{x^2}{2\left(-\frac{t}{\tau_m}{\lambda}^2\right)}\right)}^2\\\displaystyle +\frac{x^2}{4\left(-\frac{t}{\tau_m}{\lambda}^2\right)}\end{array} $$

Thus,

$$\begin{array}{l} \displaystyle f\left(x,t\right)\displaystyle =\exp \left(-\frac{t}{\tau_m}\right)\exp \left(-\frac{x^2}{4\left(\frac{t}{\tau_m}{\lambda}^2\right)}\right)\frac{1}{2\pi}\\\quad \displaystyle \times \underset{-\infty }{\overset{\infty }{\int }}\exp \left(-\frac{t}{\tau_m}{\lambda}^2{\left(\omega +\frac{x^2}{2\left(\frac{t}{\tau_m}{\lambda}^2\right)}\right)}^2\right) d\omega\end{array} $$

Let

$$ {J}_3=\frac{1}{2\pi}\!\underset{-\infty }{\overset{\infty }{\int }}\!\exp\! \left(\!-\frac{t}{\tau_m}{\lambda}^2{\left(\!\omega +\frac{x^2}{2\left(\frac{t}{\tau_m}{\lambda}^2\right)}\!\right)}^2\right)\! d\omega $$

Let $\psi =\omega +\frac{x^2}{2\left(\frac{t}{\tau_m}{\lambda}^2\right)}\Longrightarrow d\psi = d\omega$

$$ {J}_3=\frac{1}{2\pi}\underset{-\infty }{\overset{\infty }{\int }}\exp \left(-\frac{t}{\tau_m}{\lambda}^2{\psi}^2\right) d\omega $$

Let $\xi =\sqrt{\frac{t{\lambda}^2}{\tau_m}}\psi \Longrightarrow d\xi =\sqrt{\frac{t{\lambda}^2}{\tau_m}} d\psi \Longrightarrow d\psi =\sqrt{\frac{\tau_m}{t{\lambda}^2}}\ d\xi$

$$ {J}_3=\frac{1}{2\pi}\sqrt{\frac{\tau_m}{t{\lambda}^2}}\underset{-\infty }{\overset{\infty }{\int }}\exp \left(-{\xi}^2\right) d\xi $$

This integral is the famous Gaussian integral and is equal to $\sqrt{\pi }$, i.e.

$$ {J}_3=\frac{1}{2\pi}\sqrt{\frac{\tau_m}{t{\lambda}^2}}\sqrt{\pi}\Longleftrightarrow {J}_3=\sqrt{\frac{\tau_m}{4\pi {\lambda}^2t}} $$

Therefore,

$$\begin{array}{ll} \displaystyle f\left(x,t\right)&\displaystyle =\exp \left(-\frac{t}{\tau_m}\right)\exp \left(-\frac{x^2}{4\left(\frac{t}{\tau_m}{\lambda}^2\right)}\right)\\&\quad \displaystyle \times \sqrt{\frac{\tau_m}{4\pi {\lambda}^2t}}\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle f\left(x,t\right)&\displaystyle =\sqrt{\frac{1}{4\pi {\lambda}^2\frac{t}{\tau_m}\ }}\ \exp \left(-\frac{t}{\tau_m}\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2\frac{t}{\tau_m}}\right)\end{array} $$

Taking all together,

$$ \begin{array}{l}\displaystyle \mathcal{F}\left\{f\left(x,t\right)\ast h\left(x,t\right)\right\}=F\left(\omega, t\right)H\left(\omega, t\right)\\\displaystyle \quad \Longleftrightarrow f\left(x,t\right)\ast h\left(x,t\right)\\\displaystyle \quad ={\mathcal{F}}^{-1}\left\{F\left(\omega, t\right)H\left(\omega, t\right)\right\}\end{array}\vspace*{-15pt} $$

$$ f\left(x,t\right)\ast h\left(x,t\right)\stackrel{\scriptscriptstyle\mathrm{def}}{=}\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t\right)h\left(\zeta, t\right) d\zeta \vspace*{-15pt}$$

$$ {J}_1=\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t\right){V}_0(x) d\zeta $$

For the second term, we have the integral

$$\begin{array}{ll} \displaystyle {J}_2&\displaystyle =\underset{-\infty }{\overset{\infty }{\int }}\exp \left(-\frac{1+{\lambda}^2{\omega}^2}{\tau_m}\left(t-s\right)\right)\hat{I}\left(\omega, s\right)\\&\displaystyle \quad \times \exp \left( i\omega x\right) d\omega =\underset{-\infty }{\overset{\infty }{\int }}F\left(\omega, t-s\right)\hat{I}\left(\omega, s\right)\\&\displaystyle \quad \times \exp \left( i\omega x\right) d\omega \end{array}$$

where, as before,

$$ F\left(\omega, t\right)=\exp \left(-\frac{t}{\tau_m}\right)\exp \left(-\frac{t}{\tau_m}{\lambda}^2{\omega}^2\right) $$

Thus,

$$ {J}_2=\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t-s\right)I\left(\zeta, t-s\right) d\zeta $$

The complete solution of the linear cable equation is to any current injection, I_inj(x, t):

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t\right){V}_0(x) d\zeta\\&\quad \displaystyle +\frac{r_m}{\tau_m}\underset{0}{\overset{t}{\int }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t-s\right){I}_{inj}\\&\quad \displaystyle \times \left(\zeta, t-s\right) d\zeta ds \end{array}$$

where

$$\begin{array}{ll} \displaystyle f\left(x,t\right)&\displaystyle =\sqrt{\frac{1}{4\pi {\lambda}^2\frac{t}{\tau_m}\ }}\ \exp \left(-\frac{t}{\tau_m}\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2\frac{t}{\tau_m}}\right)\end{array} $$

1.2 A.2 Solution to a Constant, Localized Current

The solution of the linear cable equation, when a constant current is applied at x = 0, i.e., I(x, t) = I₀δ(x)/πd, and the initial voltage is zero, V₀(x) = 0

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m}{\tau_m\pi d}\underset{0}{\overset{t}{\int }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t-s\right)\\&\quad \displaystyle \times {I}_0\left(\zeta, t-s\right) d\zeta ds\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m}{\tau_m\pi d}\underset{0}{\overset{t}{\int }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x-\zeta, t-s\right)\\&\quad \displaystyle \times {I}_0\delta \left(\zeta \right) d\zeta ds \end{array}$$

Convolution of any function with the delta function returns the function itself, thus:

$$ V\left(x,t\right)=\frac{r_m{I}_0}{\tau_m\pi d}\underset{0}{\overset{t}{\int }}f\left(x,t-s\right) ds\Longleftrightarrow \vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{\tau_m\pi d}\underset{0}{\overset{t}{\int }}\sqrt{\frac{1}{4\pi {\lambda}^2\frac{t-s}{\tau_m}\ }}\ \exp \left(-\frac{t-s}{\tau_m}\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2\frac{t-s}{\tau_m}}\right) ds \end{array}$$

Let ϕ = t − s ⟹ dϕ =  − ds, and absorb the minus by inverting the integral limits

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{\tau_m\pi d}\underset{t}{\overset{0}{\int }}-\sqrt{\frac{1}{4\pi {\lambda}^2\frac{\phi }{\tau_m}\ }}\ \exp \left(-\frac{\phi }{\tau_m}\right)\\&\displaystyle \quad \times \exp \left(-\frac{x^2}{4{\lambda}^2\frac{\phi }{\tau_m}}\right) d\phi \Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{\tau_m\pi d}\underset{0}{\overset{t}{\int }}\frac{1}{\sqrt{4\pi {\lambda}^2\frac{\phi }{\tau_m}\ }}\ \exp \left(-\frac{\phi }{\tau_m}\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2\frac{\phi }{\tau_m}}\right) d\phi\end{array} $$

Let $T=\frac{\phi }{\tau_m}\Longrightarrow dT=\frac{1}{\tau_m} d\phi$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{\tau_m\pi d\sqrt{4\pi {\lambda}^2}}\underset{0}{\overset{t/{\tau}_m}{\int }}\frac{1}{\sqrt{T}}\ \exp \left(-T\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2T}\right){\tau}_m dT\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{\pi d\sqrt{4\pi {\lambda}^2}}\underset{0}{\overset{t/{\tau}_m}{\int }}\frac{1}{\sqrt{T}}\ \exp \left(-T\right)\\&\quad \displaystyle \times \exp \left(-\frac{x^2}{4{\lambda}^2T}\right) dT\end{array} $$

To solve this integral, we will use the Laplace transform because its properties involve integrals of this form.

The Laplace transform is defined as

$$ f(t)\overset{\mathcal{L}}{\to }F(s),F(s)=\underset{0}{\overset{\infty }{\int }}f(t)\exp \left(- st\right) dt \vspace*{-15pt}$$

$$ F(s)\overset{{\mathcal{L}}^{-1}}{\to }f(t),f(t)=\frac{1}{2\pi i}\underset{-\infty }{\overset{\infty }{\int }}F(s)\exp (st) ds $$

Using the time-domain integration Laplace transform property, i.e.

$$ \mathrm{if}\ f(t)\overset{\mathcal{L}}{\to }F(s),\mathrm{then}\ \mathcal{L}\left\{\underset{0}{\overset{t}{\int }}f(z) dz\right\}=\frac{1}{s}F(s) $$

Let $f(T)=\frac{1}{\sqrt{T}}\exp \left(-T-{\left(\frac{x}{2\lambda}\right)}^2\frac{1}{T}\ \right)$

$$\begin{array}{ll} \displaystyle \mathcal{L}\left\{f(T)\right\}&\displaystyle =F(s)=\underset{0}{\overset{+\infty }{\int }}\frac{1}{\sqrt{T}}\exp \left(-T-b/T\right)\\&\quad \displaystyle \times \exp \left(- sT\right) dT,b\\& \displaystyle ={\left(\frac{x}{2\lambda}\right)}^2\mathrm{thus}\ b\ge 0 \vspace*{-15pt}\end{array}$$

$$ F(s)=\underset{0}{\overset{+\infty }{\int }}\frac{1}{\sqrt{T}}\exp \left(-T-b/T- sT\right) dT\Longleftrightarrow \vspace*{-15pt}$$

$$ F(s)=\underset{0}{\overset{+\infty }{\int }}\frac{1}{\sqrt{T}}\exp \left(-\left(s+1\right)T-b/T\right) dT $$

Let $u=\sqrt{T}\Longrightarrow du=\frac{1}{2}\frac{dT}{\sqrt{T}},{u}_1=0,{u}_2=\infty$

$$ F(s)=\underset{0}{\overset{+\infty }{\int }}2\exp \left(-\left(s+1\right){u}^2-b/{u}^2\right) du $$

We complete the square inside the exponential, i.e.

$$\begin{array}{ll} \displaystyle -\left(s+1\right){u}^2-\frac{b}{u^2}&\displaystyle =-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\\&\quad \displaystyle -2\sqrt{b\left(s+1\right)}\end{array} $$

Thus,

$$\begin{array}{ll} \displaystyle F(s)&\displaystyle =2\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right.\\&\displaystyle \quad \left.-2\sqrt{b\left(s+1\right)}\right) du\Longleftrightarrow\end{array} $$

$$\begin{array}{l} \displaystyle F(s)\displaystyle =2\underset{0}{\overset{+\infty }{\int }}\exp \left(-2\sqrt{b\left(s+1\right)}\right)\\\displaystyle\quad \times \exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right) du\Longleftrightarrow \end{array}$$

$$\begin{array}{ll} \displaystyle F(s)&\displaystyle =2\exp \left(-2\sqrt{b\left(s+1\right)}\right)\underset{0}{\overset{+\infty }{\int }}\\&\displaystyle\quad \times \exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right) du \end{array}$$

Let

$$ {J}_1=\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right) du $$

(A.4)

Also, let $k=\sqrt{\frac{b}{s+1}}\frac{1}{u}\Longrightarrow u=\sqrt{\frac{b}{s+1}}\frac{1}{k}\Longrightarrow du=-\sqrt{\frac{b}{s+1}}\frac{1}{k^2}dk, {k}_1=\infty, {k}_2=0$

$$ \begin{array}{ll}\displaystyle {J}_1&\displaystyle =-\underset{\infty }{\overset{0}{\int }}\exp \left(-{\left(\frac{\sqrt{b}}{k}-\sqrt{s+1}k\right)}^2\right)\\&\displaystyle \quad \times \sqrt{\frac{b}{s+1}}\frac{1}{k^2} dk\Longleftrightarrow \vspace*{-15pt}\end{array}$$

$$\begin{array}{ll} \displaystyle {J}_1&\displaystyle =\sqrt{\frac{b}{s+1}}\underset{0}{\overset{\infty }{\int }}\\&\quad \displaystyle\times \exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right)\frac{1}{u^2} du \end{array}$$

(A.5)

Summing Eqs. (A.4) and (A.5)

$$\begin{array}{ll} \displaystyle 2{J}_1&\displaystyle =\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right) du\\&\quad \displaystyle +\sqrt{\frac{b}{s+1}}\underset{0}{\overset{\infty }{\int }}\!\exp\! \left(\!-{\left(\!\sqrt{s+1}u-\frac{\sqrt{b}}{u}\!\right)}^{\!2}\right)\\&\quad \displaystyle \times \frac{1}{u^2} du\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle 2{J}_1&\displaystyle =\underset{0}{\overset{+\infty }{\int }}\left(\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right)\right.\\&\quad \displaystyle +\sqrt{\frac{b}{s+1}}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right)\\&\quad \displaystyle \left.\times \frac{1}{u^2}\right) du\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle 2{J}_1&\displaystyle =\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right)\\&\displaystyle \quad \times \left(1+\sqrt{\frac{b}{s+1}}\frac{1}{u^2}\right) du\end{array} $$

Multiplying both sides with $\sqrt{s+1}$

$$ \begin{array}{ll}\displaystyle 2\sqrt{s+1}{J}_1&\displaystyle =\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s+1}u-\frac{\sqrt{b}}{u}\right)}^2\right)\\&\displaystyle \quad \times \left(\sqrt{s+1}+\sqrt{b}\frac{1}{u^2}\right) du\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle {J}_1&\displaystyle {\,=\,}\frac{1}{2\sqrt{s+1}}\underset{0}{\overset{+\infty }{\int }}\exp \left(-{\left(\sqrt{s{\,+\,}1}u{\,-\,}\frac{\sqrt{b}}{u}\right)}^2\right)\\&\quad \displaystyle \times \left(\sqrt{s+1}+\sqrt{b}\frac{1}{u^2}\right) du\end{array} $$

Let $w=\sqrt{s+1}u-\frac{\sqrt{b}}{u}\Longrightarrow dw=\sqrt{s+1}+\frac{\sqrt{b}}{u^2},{w}_1=-\infty, {w}_2=\infty$

$$ {J}_1=\frac{1}{2\sqrt{s+1}}\underset{-\infty }{\overset{+\infty }{\int }}\exp \left(-{w}^2\right) dw \vspace*{-15pt}$$

$$ {J}_1=\frac{\sqrt{\pi }}{2\sqrt{s+1}} $$

Therefore,

$$ F(s)=2\exp \left(-2\sqrt{b\left(s+1\right)}\right)\frac{\sqrt{\pi }}{2\sqrt{s+1}}\Longleftrightarrow $$

$$ F(s)=\exp \left(-2\sqrt{b\left(s+1\right)}\right)\frac{\sqrt{\pi }}{\sqrt{s+1}} $$

Thus,

$$ V\left(x,T\right)=\frac{r_m{I}_0}{\sqrt{4\pi {\lambda}^2}}{\mathcal{L}}^{-1}\left\{\frac{1}{s}F(s)\right\} $$

Let $G(s)=\frac{1}{s}F(s)$

$$ G(s)=\frac{1}{2}\sqrt{\pi}\frac{2s}{\sqrt{s+1}}\exp \left(-2\sqrt{b\left(s+1\right)}\right) $$

Using the identity

$$ \begin{array}{ll}\displaystyle \frac{2}{s\sqrt{s+1}}&\displaystyle =\frac{1}{\left(s+1\right)-\sqrt{s+1}}\\&\quad \displaystyle -\frac{1}{\left(s+1\right)+\sqrt{s+1}} \end{array}$$

Another important property of the Laplace transform is the shift in frequency domain

$$\begin{array}{ll} \displaystyle if\ f(T)&\displaystyle \overset{\mathcal{L}}{\to }F(s), then\exp \left(\epsilon T\right)\\\displaystyle f(T)&\displaystyle \overset{\mathcal{L}}{\to }F\left(s-\epsilon \right) \end{array}$$

Here, ϵ =  − 1

In addition, from Bateman (1954), page 261, Table 16, we have the inverse Laplace of a function

$$\begin{array}{l} \displaystyle {\mathcal{L}}^{-1}\left\{\frac{1}{s+\gamma \sqrt{s}}\exp \left(-\eta \sqrt{s}\right)\right\}\\\displaystyle\quad =\exp \left(\eta \gamma +{\gamma}^2T\right) \operatorname {erfc}\left(\frac{1}{2}\eta {T}^{-\frac{1}{2}}+\gamma {T}^{\frac{1}{2}}\right) \end{array}$$

Let

$$\begin{array}{ll} \displaystyle H(s)&\displaystyle =\frac{1}{2}\sqrt{\pi}\exp \left(-\left(2\sqrt{b}\right)\sqrt{s}\right)\\&\quad \displaystyle \times \left(\frac{1}{s-\sqrt{s}}-\frac{1}{s+\sqrt{s}}\right),H\left(s+1\right)\\&\displaystyle =G(s), thus\ g(T)=\exp \left(-T\right)h(T) \end{array}$$

$$ \begin{array}{ll}\displaystyle H(s)&\displaystyle =\frac{1}{2}\sqrt{\pi}\left(-\left(2\sqrt{b}\right)\sqrt{s}\right)\left(\frac{1}{s-\sqrt{s}}\right)\\&\displaystyle \quad -\frac{1}{2}\sqrt{\pi}\exp \left(-2\sqrt{b}\right)\sqrt{s}\frac{1}{s+\sqrt{s}} \end{array}$$

In our case, $\gamma =-1,\eta =2\sqrt{b}$, for the first term, and $\gamma =1,\eta =2\sqrt{b}$ for the second. Due to the linearity property of the Laplace transform,¹³

$$\begin{array}{ll} \displaystyle h(T)&\displaystyle =\frac{1}{2}\sqrt{\pi }\ \left[\exp \left(-2\sqrt{b}+T\right)\right.\\&\displaystyle\quad \times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}-{T}^{\frac{1}{2}}\right)\\&\displaystyle\quad -\exp \left(2\sqrt{b}+T\right)\\&\displaystyle\quad \left.\times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}+{T}^{\frac{1}{2}}\right)\right] \end{array}$$

Thus,

$$ \begin{array}{ll}\displaystyle g(T)&\displaystyle =\exp \left(-T\right)\frac{1}{2}\sqrt{\pi }\ \left[\exp \left(-2\sqrt{b}+T\right)\right.\\&\quad \displaystyle\times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}-{T}^{\frac{1}{2}}\right)\\&\quad \displaystyle -\exp \left(2\sqrt{b}+T\right)\\&\quad \displaystyle \left.\times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}+{T}^{\frac{1}{2}}\right)\right]\Longleftrightarrow \end{array}\vspace*{-15pt}$$

$$\begin{array}{ll} \displaystyle g(T)&\displaystyle =\frac{1}{2}\sqrt{\pi }\ \left[\exp \left(-2\sqrt{b}\right)\right.\\&\displaystyle \quad \times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}-{T}^{\frac{1}{2}}\right)-\exp \left(2\sqrt{b}\right)\\&\displaystyle \quad \left.\times \operatorname{erfc}\left(\sqrt{b}{T}^{-\frac{1}{2}}+{T}^{\frac{1}{2}}\right)\right] \end{array}$$

Thus, changing back the expressions for b and T = t/τ_m

$$ \begin{array}{ll}\displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{2\pi d\lambda \sqrt{\pi }}\frac{\sqrt{\pi }}{2}\left\{\exp \left(-\frac{\left|x\right|}{\lambda}\right)\right.\\&\quad \displaystyle \times \operatorname{erfc}\left(\frac{\left|x\right|}{2\lambda}\sqrt{\frac{\tau_m}{t}}-\sqrt{\frac{t}{\tau_m}}\ \right)\\&\quad \displaystyle -\exp \left(\frac{\left|x\right|}{\lambda}\right) \\&\quad \displaystyle \left.\times \operatorname{erfc}\left(\frac{\left|x\right|}{2\lambda}\sqrt{\frac{\tau_m}{t}}+\sqrt{\frac{t}{\tau_m}}\ \right)\ \right\} \end{array}$$

$$\begin{array}{ll} \displaystyle V\left(x,t\right)&\displaystyle =\frac{r_m{I}_0}{4\pi d\lambda}\left\{\exp \left(-\frac{\left|x\right|}{\lambda}\right)\right.\\&\quad \displaystyle \times \operatorname{erfc}\left(\frac{\left|x\right|}{2\lambda}\sqrt{\frac{\tau_m}{t}}-\sqrt{\frac{t}{\tau_m}}\ \right)\\&\quad \displaystyle -\exp \left(\frac{\left|x\right|}{\lambda}\right)\\&\quad \displaystyle \left.\times \operatorname{erfc}\left(\frac{\left|x\right|}{2\lambda}\sqrt{\frac{\tau_m}{t}}+\sqrt{\frac{t}{\tau_m}}\ \right)\ \right\} \end{array}$$

Setting t ⟶ ∞, we obtain the steady-state solution and using the erfc properties, i.e. erfc(−∞) = 2, erfc (∞) = 0,

$$ V\left(x,\infty \right)\equiv {V}_{\infty }(x)=\frac{r_m{I}_0}{2\pi d\lambda}\exp \left(-\frac{\left|x\right|}{\lambda}\right) $$