This means that, that, that losing an axial current, the change in axial

current, at each point is actually the second derivative.

This is just the axial current, and I want the change in axial current, so you

have to, once again, take the derivative of the axiom current and this will give

you dV squared to dX squared, because you derived it again, you [UNKNOWN] it again.

And so if you want to eventually think about the cable equation, it will look

something like that. Okay, you will say that the change in

axial current, the change in axial current, which is proportional to the

second derivative of voltage with distance, is equal, basically to the

membrane current, c dV dt plus V divided by r.

Okay, so this is the membrane current, this is the change in axial current and

they are equal. And so the sum of the change in axial

current plus the membrane current must be zero, unless you inject current from the

outside. This is the foundation for Rall's cable

theory. It's a passive cable theory, because all

the parameter, the membrane resistivity, and the capacitance are passive, are

static. Okay, so this is a partial dif, linear

partial differential equation. Because this is linear, and this is

linear. And it's partial differential equation

because you have x and you have t. This means that the voltage changes at

each, each location. The voltage changes with distance and

with time. Okay, and this is the dimensionless cable

equation, and we don't need to go into the details, but this linear partial

differential equation can be solved analytically.

If you have the initial conditions for solving this equation, V0 at some

location, or I0 when you inject I to some location, and if you also have the

boundary condition. Is it an infinite cylinder?

Is it a closed, short cylinder with some boundary at the end?

So you need two additional information, the initial condition and the boundary

condition in order to solve this equation.

And that's exactly what Rall did. So we solve this differential equation,

the cable equation, for different boundary conditions.

I just want to show you one example here just to let you some intuition about what

does it mean to solve the cable equation for particular boundary conditions.

Let's say that you're initital condition is at some location of the cylinder.

It's a cylinder. Is some one voltage at some location is

one. At location zero, the initial condition

is normalized to one. We are now looking at the steady-state

solution, which is easier. Steady-state means that there is no

change in time, so you take this out, and you only look at the change in space, in

distance, from the location, from the initiation of some voltage at this

location. And you can see that the voltage indeed

attenuates with distance. From the site of injection, a way and,

and the, and the slope and the shape of attenuation depends on the property of

the cable. For example, in this case, the membrane

cable is infinite long, infinitely long. In this case, for example, the cable ends

after a certain distance here. And it ends with a sealed end, the

boundary condition. The boundary condition at the end is

sealed. Meaning that there is no axial current at

the boundary. So there is a boundary.

The current cannot cross the boundary. So this is a sealed end boundary

condition. DV DX, the axial, the axial current, is

zero at X equal L, here. So you can see that the slope of

attenuation very strongly depends on the boundary conditions at the end of the

cable. For the infinite case, It attenuates

exponentially with distance. For sealed-end short cylinders, like

dendrites are, it will attenuate less steeply with distance.

As you can see here, or here. So that's the general idea, of solving

the cable equations for dendrites. Eh, taking into account the boundary

conditions. And here is the most fundamental, the

most important solution for the cable equations for, a branched dendritic tree

model here. So this is the model, and you can see

something important. That Rall and Rinzel and Rall, actually,

Rall and Rinzel in '73 already solved the case for this idealized tree, and here is

what they showed. Suppose you have a branched tree like

this. And you inject current here, at the very

distant tip of your dendritic tree, the passive dendritic tree.

So you start with a certain voltage here. So this is voltage.