1

I'm currently studying Rall cable theory which is a concept from neuroscience and I've come across this differential equation that governs change of axial current depending of time and distance:

$$\partial^2 V/\partial x^2 = \partial V/\partial t + V(x, t)$$ I have some basic knowledge of single variable calculus and can solve basic first order differential equations but this is a second order equation and V is a multi-variable function so... how do I go about solving this? What initial conditions should I choose? Would not it be too much to ask someone to solve this with comments?

src091
  • 53
  • 1
    Your application will dictate the selection of initial ($t=0$) and boundary conditions (is the space dimension $x$ a finite interval, or the entire real line?). This is similar to the heat equation. – hardmath Jan 03 '14 at 22:38
  • @hardmath Thanks, dendritic tree in neuron is surely finite. I'll read into the provided article. – src091 Jan 03 '14 at 23:00
  • From the mention of "dendritic tree" I begin to suspect a one-dimensional but branching geometry is involved. Google search led me to a couple of Wikipedia articles, on cable theory and on Wilfrid Rall. – hardmath Jan 05 '14 at 14:41

1 Answers1

1

The first solution technique that occurred to me is separation of variables, a topic one covers in undergraduate differential equations classes, applicable to many constant coefficient parabolic DEs as here.

There's a lengthy chapter (59 pages) by Rall in the Handbook of Physiology -- The Nervous System (Supp. 1), and the "dimensionless" form of the cable equation of neurophysiology appears about one-third of the way through:

$$ \partial^2 V/ \partial X^2 = \partial V/ \partial T + V $$

The solution by separation of variables appears a couple of pages later:

$$ V(X,T) = [A \sin(\alpha X) + B \cos(\alpha X)] \exp(-(1+\alpha^2) T) $$

which involves three "arbitrary constants" $A,B,\alpha$.

Because the cable equation is linear and homogeneous, it obeys the superposition principle. Thus more general solutions can be obtained by summing ones shown above, suitably chosen to fit initial and boundary conditions.

Here the coefficients of a Fourier expansion of the initial condition ($T=0$) may be expected to appear. Notice that the individual components are all decaying exponentially with time $T \gt 0$.

A more sophisticated solution approach involves Green's functions, which model a response to an instantaneous impulse. The usefulness of such an approach is illustrated by a recent (2010) PubMed paper.


Connecting with solutions of the heat equation can help to visualize solutions of the cable equation. The heat equation says:

$$ \partial^2 U/ \partial X^2 = \partial U/ \partial T $$

and solutions of the cable equation above are related by an integrating factor:

$$ V(X,T) = e^{-T} U(X,T) $$

The Reader is invited to verify this by directly substituting into the cable equation and simplifying, cancelling the nonzero factor of $e^{-T}$ on both sides. Thus a solution $V$ of the cable equation is exactly a solution of the heat equation times the extra $e^{-T}$, an exponentially decaying-in-time factor.

The solutions of the heat equation are known for smoothing of uneven "temperature" distribution in the initial condition, as "heat" flows from warmer areas to cooler ones. Given constant Dirichlet boundary conditions, say $U(0,T) = U_0, U(1,T) = U_1$, the solution evolves toward the "steady state" (flat) linear interpolation between these endpoints.

Solutions of the cable equation will likewise exhibit this "smoothing out" of peaks in the initial condition, combined with the extra factor of simple exponential decay.

hardmath
  • 37,015
  • Wow, all this seems far ahead of my current math capabilities, lots of learning has to be done. I have one last question: in paper I read it's said that voltage in cable decreases depending on both time and distance. And it's clearly seen from solution that voltage attenuates as time goes by but if time stays the same while only distance is increased it does not but swings back and forth due to sin/cos properties? This conflicts with my understanding of what should have happened. – src091 Jan 06 '14 at 12:25
  • 1
    If the initial condition were a sine/cosine wave with frequency parameter $\alpha$, then of course that feature at $T=0$ is frozen in time and doesn't "attenuate" with distance. However the larger the parameter $\alpha$, the more quickly the peaks of the wave dissipate going forward in time. Now apply this, not to a pure sine or cosine wave, but to the sort of "impulse" modelled by the Green's function, and what you'd see is the peak broadening out with distance (as time progresses). I'll add some insight from connecting to solutions of the heat equation. – hardmath Jan 06 '14 at 13:05