0:00

[BLANK_AUDIO].

Â Hello.

Â Welcome back.

Â We are now on our sixth, sixth and

Â final lecture on mathematical models of action potentials.

Â In this final lecture on this section in particular, is going to be on.

Â Propagation of action potentials.

Â What we've seen in the first five lectures of this series [COUGH] is

Â a, a series of equations, the

Â Hodgkins-Huxley equations, and the Fitzunit Remote equations.

Â But in all the, the differential equations we've discussed to this point.

Â Voltage or the you know gating variables m, h, and

Â n or capital W in the case of Fitz-Neugumo model.

Â These were always assumed to be spatially-uniform.

Â We had a differential equation for voltage and it was dvdt.

Â We never said anything about location in this case.

Â 0:55

And so then the questions becomes, well, what do you do if your voltage is not

Â just, doesn't just vary with respect to time

Â but it also varies with respect to location?

Â What if voltage at one location is

Â different from a voltage at another location?

Â And what we're talking about in this case is, you know, the voltage on one end

Â of your neuron could be different from the voltage on the other end of your neuron.

Â And when you think about some of the long neurons that

Â are in your body, for instance those that travel from your extremities,

Â then it seems pretty clear that the voltage on, on one

Â end could be much different than the voltage on the other end.

Â 1:26

And in order to discuss how, how we deal with that in

Â terms of mathematical models, we're going to

Â discuss electrical propagation in very conceptual terms.

Â And then, we're going to derive the, the relevant reaction diffusion equation.

Â 1:40

And in terms of the, the material in the course,

Â this is where we transition from ODEs to, to PDEs.

Â With PDE in this case stand-, standing for Partial Differential Equation.

Â So what we've discussed to this point were were

Â systems that were consisted of several ordinary differential equations.

Â Now we're going to talk about systems of Partial Differential Equations.

Â 2:05

When you have a electrical propagation biologically.

Â For instance, when impulses are travelling from the, from

Â one part of your heart to another part of

Â your heart or when electrical impulses are, travelling down

Â your nerves as you are, as you are thinking.

Â 2:19

This electrical propagation results from spatial voltage gradients.

Â In other words, what's key is that there's, there's a gradient.

Â Voltage in one location is different form voltage in, in another location.

Â And we'e going to think about this in just, in conceptual terms.

Â Imagine a very long, one dimen-, dimensional axon.

Â And, this axon is going to start at rest and it's locally stimulated.

Â Now, what we've drawn here is a bunch of positive charges on the outside

Â of the axon, and a bunch of negative charges on the inside of the axon.

Â That's to illustrate the fact that the resting voltage in this axon just like

Â with the voltage in our Hodgkin-Huxley model,

Â is going to be around minus 60 millivolts.

Â 2:59

So the inside of the axon is going to be negative compared

Â with the outside, but you notice it's, it's the same everywhere right?

Â Here you have a negative on the outside and a pos, negative

Â on the inside and a positive on the outside and over at

Â this end you have a negative on the inside and a positive

Â on the outside, but you apply a stimulus just in this region here.

Â That's why we draw the arrow only on the left.

Â 3:19

Well you're going to induce an action potential over on the left hand side.

Â But because you're only stimulating over here, you're not

Â going to immediately induce an action potential over here.

Â So now you're axion is going to depolarized on the

Â left but it's still going to be resting on the right.

Â 3:33

So when we give this electrical stimulus, we're only going to induce action

Â potential on my left hand side here and the way I've represented the

Â action potential is I've flipped these from negative on the inside and positive

Â on the outside to positive on the inside and, and negative on the outside.

Â Remember that at the peak of the action potential

Â the trans-memory potential, the voltage, is positive so we

Â go from 60 at rest to a you know,

Â peak value of you know, somewhere around plus 20.

Â 3:58

It you know, it at the very top of the axis, that's all,

Â but this is only going to occur where we give the electrical stimulus.

Â So it's going to be depolarized on the left

Â and it's going to be resting on the right.

Â The question is, now what happens?

Â 4:11

Well now we're going to have electric current flowing on the inside and on the

Â outside, and when it's at rest there's no reason for electrical current to flow.

Â Left to right, either on the inside or on the

Â outside because the voltage is the same everywhere, at every location.

Â But here when you have positive charges here

Â on the inside and negative charges here on

Â the inside, you're going to get electrical current flowing

Â from the positive charges to the negative charges.

Â And conversely, on the outside it's going to

Â flow in the other direction and, this is,

Â also true, due to the fact that electrical current always has to flow in a loop.

Â So you're going to get current flowing from left to

Â right on the inside and right to left on the outside.

Â 4:49

Now what's going to happen is these positive charges move towards the right.

Â Well, they're going to take this patch of

Â membrane here and they're going to depolarize that.

Â They're going to make the voltage go up.

Â At these locations here.

Â Next, we'll see what happens once we get this electrical charges unless

Â we get this electrical current flowing on the inside and the outside.

Â 5:10

When an electrical current flows from left to right on

Â the inside, what's going to happen is, more tissue becomes depolarized.

Â The positive charges, as they flow from here.

Â Over here are going to take this patch of membrane and, and bring it up

Â towards threshold and then that area of

Â membrane is going to fire and axion itself.

Â So, what you see in this step number four here is now I have eight positive charges

Â on the inside instead of only three positive charges

Â on the inside which I had during step three.

Â So, now this entire area of membrane here has become depolarized,

Â whereas this area of membrane here is still at the resting level.

Â Then we're going to have more current flowing which

Â is which is going to induce this this

Â section of membrane to fire an action potential

Â etc, etc, so we can continue to draw this.

Â Now that we've drawn the basic step we can continue to draw

Â 6:15

One, one part of membrane requires an action potential.

Â It becomes depolarized, then you have current flowing on

Â the inside and current flowing on the outside and

Â through that mechanism you can depolarize more tissue and

Â get that tissue to also fire an action potential.

Â But what makes this challenging is that voltage

Â depends on both time and location so our voltage

Â over here, across the cell membrane is different

Â from our voltage over here across the cell membrane.

Â 6:39

And so we can't just write down a single differential equation

Â that says, you know, dv, dt, is, is equal to something.

Â Because then that would be ambiguous.

Â Are you talking about this voltage over here?

Â Or are you talking about this voltage over here?

Â So now that voltage depends on both time and location.

Â Now we need to solve a system of partial

Â differential equations, rather than a system of ordinary differential equations.

Â And what we're going to go through in the remainder of this lecture.

Â And then, in the next few lectures that follow, are

Â some of the issues related to solving partial differential equations.

Â When things change as a function of time, and they change as a function of location.

Â 7:18

This shows what we're going to be dealing with the remainder of this lecture.

Â It's a propagated action potential.

Â Well a mean is that if we have an axiom here going from left to

Â right each trace might represent an action potential

Â that you would record from a different location.

Â So the black one would come from here, the red

Â one from here, the green one would come from here.

Â What you see is that if you plot all of these

Â on the same scale, the black one comes first and then

Â the red one and then the green one, so now we

Â have something that is traveling where the action potential occurs here first.

Â And then it occurs here.

Â And then it progressively goes all the way from the left to right.

Â So now our variables in our system v, m, h

Â and n, their functions now are both time and location.

Â 8:03

The relevant equation that we're going to have to

Â solve with respect to voltage is the following.

Â Capacitance times the partial derivative of voltage

Â with respect to time, is equal to

Â a bunch of terms times the second

Â derivative of voltage, with respect to location.

Â Minus the ionic current.

Â You'll notice here that the symbols have changed.

Â Instead of just a dv, dt.

Â It becomes a special, a special delta.

Â And this is to signify.

Â If this is now a partial rather than ordinary differential equation.

Â If you have a, [SOUND] some variable that, that only

Â depends on one other variable, so that if voltage only depends

Â on time then you can compute an ordinary derivative of,

Â it's just a change in voltage with a respect to time.

Â And then you can just a lower vase C.

Â But voltage varies with respect to time and with respect to location.

Â In this case we're representing location by x.

Â Then you could have derivative with respect to time or you could

Â have a derivative with respect to that location with respect to x.

Â And the way that you signify that this is true that you

Â are talking about a partial derivative is with this special symbol over here.

Â So now we have voltage depending on time and voltage that springs on location.

Â That is an x.

Â So the pertinent questions that we need to address here

Â are first of all where does this equation come from.

Â I just wrote this down.

Â I show this and I say, this is the equation.

Â Well, I'm not going to leave it at that.

Â I'm going to show you where this equation comes from.

Â That's going to be for the remainder of this lecture here.

Â And then the next thing we want to address after

Â this is, how do you solve this in practice?

Â And this is what the some of the

Â subsequent lectures are, are going to get into.

Â What are the issues involved when you need to

Â solve a partial differential equation rather than an ordinary

Â differential equation, it's more complicated, so we are going to

Â deal with some of the practical aspects of it.

Â 9:48

But now we want to address as well, I said this

Â is the, this is the relevant equation, why did I

Â say that, now this is the, how can we manage

Â to understand Your conceptual terms, why this particular equation makes sense.

Â What we're going to address in the next

Â few slides, is one dimensional electrical propagation.

Â And by that, we mean what I just showed on the

Â previous slide, which is an action

Â potential propagating down a uniform cable.

Â And in order to analyze this in mathematical terms,.

Â What we're going to do is we're going to divide the cable into discrete segments.

Â So we're going to take this long axon and we're going to chop it into pieces.

Â This is piece number one, number two, number three, number four, et cetera.

Â And we're going to analyze the cable as coupled equivalent circuits.

Â So every time we chop this into a

Â piece, we're going to analyze this one piece here.

Â As is there is an equivalent circuit and this should look familiar

Â to you, this is how we represented the membrane when we talked about

Â the Hodgkinâ€“Huxley model where you have a voltage on the outside and then

Â you have a capacitor and the capacitor isn't parallel with the iron current.

Â So this is like what we looked at with the Hodgkinâ€“Huxley model.

Â But remember we had three terms over here when we talked about the ion current.

Â We had a sodium current, we had a

Â potassium current, and we had a wheat current.

Â What I've done here is I've just lumped all three

Â of these together and said something is producing an ionic current.

Â 11:06

And then before we discuss with respect to the Hodgkinâ€“Huxley model.

Â When we talked about it, is ODEs, is that when your

Â ionic current changes then that's charging a capacitor or discharging a capacitor.

Â This current that's flowing is going to either

Â make the membrane voltage go up or down.

Â What we're going to do now is we're going to take these, each patch

Â of membrane here, or each little chunk that we've divided our cable into.

Â We're going to treat as an equivalent circuit which is

Â an ionic current with, in parallel with a capacitor.

Â And now we're going to couple these together.

Â That's what we haven't done yet is we haven't coupled them.

Â And when we couple them together, we're going to see how we can

Â derive, that partial differential equation that we showed on the previous slide.

Â 11:46

And there's another name for the type of equation that were going to derive

Â and the type of analysis that were

Â going to perform and its called cable theory.

Â And in this case in particular were going to keep things simple by

Â analyzing things in only one dimension only along the length of an

Â axon and so therefore its one dimensional cable theory but the same

Â sort of principles apply if you wanted to analyze flow of electrical current.

Â In a tissue where that can occur in

Â multiple dimensions all at once, such as the heart.

Â 12:14

But to keep things simple, just you know, just for the derivation,

Â we're going to, like I say, we're going to do things in one dimension.

Â Then we're going to analyze the cable as a bunch of coupled equivalent circuits.

Â And what I showed you on the last slide was

Â an ionic current in parallel with a capacitor and that represented

Â one patch of membrane or one chunk of our, of our

Â axon when we divided the axon into a bunch of chunks.

Â Now we've actually, actually coupled them together.

Â You can see that there's a resistance here on the inside flowing

Â from the voltage on the inside to the neighboring voltage on the inside.

Â So now when we put these resisters here between the equivalent circuits.

Â Now, we've coupled them together, and this is how we're going to analyze the cable.

Â 12:56

So the terminology we're going to use here is

Â that Ri, the little i stands for intracellular.

Â This represents our intracellular resistance, this says exactly how much

Â resistance to current flow you have from one patch of

Â membrane to the next patch of membrane or from one

Â chunk of the axon to the next chunk of the axon.

Â 13:17

And we discussed before when we talked about you know the difference between

Â a partial differential equation and an

Â ordinary differential equation is that it's not

Â longer correct to just say this is the voltage because if you just

Â say this is the voltage then the question becomes this is the voltage where?

Â So we have to have, have some symbolism to, to

Â show what we mean in terms of where the voltage is.

Â And we're going to use that with the super script here.

Â 13:43

So v super script j means the voltage at the jth element of the cable.

Â So this one we're going to call the jth element and therefore

Â this one to the left is the one preceding the jth element.

Â This is j minus one.

Â And the one on the right is the one after the jth element.

Â So this is j plus one.

Â And one other thing you may have noticed as we,

Â wrote this down as a, as a, you know, electrical circuit.

Â Is that we don't have any resistance here on the outside.

Â We have resistances between locations on the inside.

Â But we don't have resis, resistances on the outside.

Â This is something we've gotten to make our life simple.

Â This is an assumption that extracellular resistance,

Â Re, is equal to zero, that's why

Â I haven't put in an Re, it's because we assume it's equal to zero.

Â So it all extracellular voltages are grounded.

Â They're all equal to, what we would define, as a voltage of zero.

Â 14:35

That makes our life a little bit simpler because normally, remember

Â what's important is voltage on the inside minus voltage on the outside.

Â That's what we were talking about when we were talking about

Â voltage in the Hodgkinâ€“Huxley model when it was series of ODs.

Â So now, we don't have to compute voltage on the outside separate from voltage on

Â the inside; we can just assume voltage on the outside is equal to zero And if

Â voltage on the outside is equal to zero, if v e is equal to zero,

Â then our intracellular potential is equal to our

Â transmembrane potential at every node, at every element.

Â And this is a good assumption when you're talking about an

Â isolated fiber in a bath, which

Â is how Hodgkin-Huxley performed their experiments.

Â Right, they took out, they took a giant axon

Â out of a squid and then they examined it.

Â Through an experimental chamber that had a lot of solution around it.

Â So this is a good assumption when you have an isolated fiber in a bath.

Â In more physiological conditions it might it might

Â not, may be less good of an assumption.

Â But it's, this is a good way for us to derive the, the relevant equation and

Â making it more complicated by considering extra-cellular resistances

Â just makes the equations a little bit more complicated.

Â But it doesn't change things it doesn't change the basic concepts.

Â 15:52

Now that we've set up what our

Â equivalent circuit looks like, and we've discussed some

Â of the assumptions, we want to ask, what

Â equations describe the jth element of the cable?

Â How can we, so we're going to focus on this area right here, voltage.

Â At Jth element vj and these currents that,

Â that can influence the voltage of the Jth element.

Â 16:15

Well we want to think about what's this current

Â going to be and what's this current going to be.

Â And the way that I've the notation I've used, is I've

Â said, I just call this I from J minus one to j.

Â So this is what I mean, it's a current

Â flowing from j minus one element to jth element.

Â 16:33

Depending on how you number it, it can sort of get confusing.

Â If you just call it ij, do the do, that, what, you know, not very clear, you

Â mean the ones that exactly, immediately downstream of

Â j or you mean the one that's immediately upstream?

Â This would be downstream of j, this would be immediately upstream of j.

Â So therefore, I just going to make it absolutely clear.

Â This is the current i.

Â From j minus one traveling to j.

Â Well, how can you compute this current?

Â 17:00

well, it's just the voltage, element j minus one, minus

Â the voltage at element j, divided by this resistance Ri.

Â Similarly, if you wanted to compute this current here, I from j to j plus one,

Â it's this voltage here VJ minus this voltage VJ plus one, again divided by RI.

Â 17:20

What do we call this, how do we

Â derive these particular equations rather than some other equations.

Â Well this is just Ohm's law.

Â Right?

Â Ohm's law just says that the current is equal to the difference in voltage.

Â The voltage gradient divided by the resistance.

Â So, deriving these two equations is, is pretty simple.

Â This is this is Ohm's law.

Â 17:49

Well this is actually pretty simple as well.

Â We can say I, from J minus 100 J this current

Â here is equal to this current here plus this current here.

Â This is another way of saying that current is conserved and in this there

Â is something in the unit that I'm going to get into in a little bit.

Â This IM sub j membrane current at the jth element.

Â We have to multiply this by the surface area, A

Â in order to keep all the currents in the right units.

Â 18:21

But just keeping that in mind for now, we can sort

Â of figure out where this current came from in very simple terms.

Â Whatever current flows in, to this node vj has to equal the amount that flows out.

Â So if you've got this guy flowing out and this guy flowing out, then the sum of

Â those two together has to equal what's flowing in,

Â and this is what we call Kirchoff's current law.

Â You've prolly learned this at some point in in

Â freshman physics or maybe even in high school physics.

Â That anytime you have a node in an electrical circuit, the amount of

Â current that comes into it has to equal the amount that goes out.

Â Current has to be conserved.

Â That's where we get this equation here.

Â So, everything that we're going to show, you

Â know, subsequently, is going to rely on some algebra,

Â 19:03

and then it's going to rely on a little

Â bit of knowledge of calculus, but the basics

Â for how we analyze this circuit here just rely on these two These two laws

Â they should be familiar to you from, from

Â basic physics: Ohm's law and Kirchoff's current law.

Â 19:20

And then we could do the same sort of

Â thing in analyzing our, our membrane current here, right?

Â What, what if we looked at this node right here?

Â Well, we have two choices.

Â The current can either flow under the

Â capacitor or the current can flow through here.

Â And if it flows through here, it's ionic current.

Â So, we can similarly, the same way that we took

Â this current here flowing from j minus one to j inside.

Â It can either go two ways and those two ways have

Â to sum up to equal the amount that is coming in.

Â We need to do the same thing with the memory current.

Â The current is flowing from this node vj to this branch point here.

Â It can either go to the left, or it can go to the right.

Â Therefore, the current is flowing out of

Â the capasiter, which by our definition of capacitance

Â is equal to membrane capacitance is times

Â derivative of voltage, j, with respect to time.

Â Dv, dt times capacitance equals current plus whatever the ionic current is.

Â So, this equation here also just came from Kirchoff's current law.

Â Whatever current is flowing from this node to the branch point is equal

Â to the sum of the two currents that are flowing along either branch.

Â 20:22

And membrane currents, we discussed this a little bit when we talked

Â about the Hodgkin Huxley model, membrane currents are normalized by per unit area.

Â And so these are, these currents are in units of

Â something like amperes per centimeter

Â squared or microamperes per centimeter squared.

Â 20:38

That's why we had to take these,

Â this membrane current and multiply it by surface

Â area here in order to keep it, in the same units that these these currents.

Â So, these currents are just, are just in units

Â of amperes or, or microamperes or something like this.

Â That's why we had to take this membrane current and multiply

Â it by surface area to get it in the same units.

Â Is these two guys here.

Â One of the equation, the equations we had on the current slide was this, right?

Â The current flowing from, node j minus one to node j,

Â is equal to the current flowing from j to j plus one.

Â Plus the membrane surface area times the membrane current.

Â 21:16

This won't remember, we just, we showed, this is [UNKNOWN] current law.

Â So now, what we want to do, is we want to substitute

Â into this equation, some of the other equations we have, right?

Â We calculated this current, j minus one to j with ohm's law.

Â We calculated this current with ohm's law and

Â then this membrane current here, we decided, also

Â based on Kirchoff's current law which is capacitive

Â current, the current that flows across the capacitor.

Â Plus the ionic current.

Â So we're going to take these three equations that we've

Â just discussed and substitute them into this equation here.

Â And if we do that, this is what we got.

Â 21:50

Voltage j minus one, minus voltage j divided by r i, on the left hand side.

Â A similar term j to j plus one on the right hand side

Â [COUGH] and then this term here which are two components of memory current.

Â Also on the right hand side.

Â Now what we do is some rearranging make a couple of assumptions learn a little bit

Â from, from calculus and then we are going to

Â have our reaction diffusion equation that describes the system.

Â 22:20

This is what we had again in the previous side when I put the equations together.

Â Now, we're going to rearrange and this is what we're going to end up with.

Â On the left hand side we're going to

Â have the derivative of voltage with expected time.

Â Times the memory capacitance.

Â Everything else is going to go on to the

Â right hand side and what we're going to see when

Â we have this on the right hand side is

Â that we're going to have a minus I ion over

Â here and then we're going to have voltage the j

Â minus one, minus two times voltage of j, plus

Â voltage of j plus one, divided by the membrane

Â surface area, times the intracellular resistance, time's r i.

Â Both of those in the denominator.

Â 23:15

What we want to know now, is how can we relay our intracellular resistance Ri

Â to the cable geometry and this is

Â a formula you probably learned in physics class.

Â But even if you didn't learn it in physics class

Â or even if physics class was a long time ago

Â and you don't remember it, it's it's conceptually, I think

Â we can, we can sort of reason our way through this.

Â 23:37

Your resistance from say the, the left side of

Â this little cylinder to the bot-, or from the

Â top of the cylinder to the bottom, or in

Â other words, from the left here to the right.

Â Depends on a few, a few things.

Â First of all it depends on this resistivity, rho sub I.

Â This intracellular resistivity is just a property

Â of the of the substance that you have.

Â It's a property of the cytoplasm.

Â And resistivity is defined in a way that

Â doesn't, by itself doesn't depend on the geometry.

Â But then there are terms here that depend on the geometry.

Â You can sort of think intuitively.

Â If you had a short, fat cylinder like this, you might have

Â a different resistance compared to if you had a long, thin cylinder.

Â And these aspects related to geometry come in this way.

Â Your delta x comes in the numerator.

Â That's saying, what's our distance from the

Â top of the cylinder to the bottom cylinder?

Â Or in other words, from the left to the right.

Â Go up to the left part of the axons or the right part of the axon.

Â In other words if this were, if this

Â delta X were bigger then you'd have more resistance.

Â 24:41

That makes intuitive sense.

Â The longer your current needs to travel the more resistance that halfway is.

Â And then conversely, what you have in the in

Â the denominator here, is surface area, pi a squared.

Â That's the area of this the, the current flowing from left to right is traveling.

Â Pi times a squared, where a is the little a is the radius here.

Â And we're using a instead of R because we

Â already have a lot of R's in these equations.

Â So, we don't want to confuse R for resistance with R for radius.

Â That's why I'm defining radius as a little a.

Â 25:13

So, pie a squared is the surface area of this, this cylinder here.

Â And if you're imagining a current that's flowing from left

Â to right it can either flow here, or here, or

Â here, or here, and the more pathways you get it,

Â the low the lower the resistance is going to be.

Â That's why the surface areas the denominator here.

Â As the surface area gets bigger and bigger the resistance gets smaller.

Â 25:35

So this is how you can relate this is

Â how you can compute resistance of some arbitrary geometry.

Â It's the the intracellular resistivity, which is inherent property

Â of whatever the substance is, in this case, cytoplasm, times

Â the the distance Delta x divided by the surface area

Â that the current can travel, which Pi times A squared.

Â 26:10

And this surface area of our membrane is pi, is 2 times pi times A.

Â That's the circumference, times delta x.

Â So now if we combine these two.

Â Remember what we had as the denominator on the previous slide, A time R i.

Â 26:26

A is 2 times pie a delta x R i is this R i Delta x over

Â Pie squared, so we can simplify this here.

Â 2 times row i times delta x over a.

Â Now we're going to take this formula here that we

Â just derived and it makes it a symmetrical consideration.

Â Then we are going to substitute them into our previous equation.

Â And if we substitute this, we are going to get an equation that looks like this.

Â Capacitance times a directive of voltage with respect of time.

Â And then on the right hand side we are going to have a couple of

Â constants from here, A over 2 times rho i, and then we're going to have

Â this term here which is voltage at j minus one minus 2 times voltage at

Â j plus voltage at j plus 1 divided by delta x squared, minus I ion.

Â Now the question becomes, what happens if we don't want this to be just a

Â bunch of discreet units that are next to one another with a fine I delta x?

Â What happens if our delta x gets smaller and smaller and smaller?

Â Well, what we're looking at here is voltage to the left.

Â 27:34

Here is voltage to the right and here we

Â have minus 2 times voltage at that particular location.

Â I'll divide it by delta x squared.

Â This term here as delta x goes to

Â 0, becomes the second derivative with respect to voltage.

Â So you know that the, the approximation of the first derivative that

Â we discussed before is going to be, what you do is you go

Â a little bit higher, a little bit lower, and then divide by

Â the, the either the time step or divide by the the delta x.

Â If you do that twice, you're going to get a formula that looks like this.

Â So as delta x goes to zero, this thing here

Â approximates the second derivative of voltage with respect to x.

Â That's exactly what this becomes here.

Â And so what I've done in this case is I've dropped

Â the j superscript because this applies for all values of j.

Â And we use J as a, as a generic node in general, but once you like delta x

Â squared is 0, then you don't have to specify

Â exactly which node of J you are talking about here.

Â So now we have this formula that we, we showed up front.

Â Capacitance times the partial derivative of voltage with respect to time.

Â Is these two terms here times the second derivative

Â with respective location with respect to minus i ion.

Â So, what we've done basically just by starting with Ohm's law and Kirchoff's

Â current law and then doing some algebraic manipulation is moving some things around.

Â This is how we've derived the non linear cable equation.

Â Which, as we're going to show in subsequent lectures,

Â is an example of a reaction diffusion equation.

Â 29:16

First of all, this is a reaction diffusion equation.

Â I've already referred to this, in a couple of instances.

Â And we're going to, I'm going to explain in the next couple

Â of lectures in more detail what I really mean by this.

Â But I think it's important to know

Â that these reaction diffusion equations appear in many

Â other contexts and that's how we're going to

Â discuss what we mean by reaction diffuse equations.

Â For instance, if we talk about sub-cellular diffusion of

Â calcium, or other second messengers like such as cyclic ANP.

Â They often obey reaction diffusion equations.

Â And that's why we're going to go, that's why we're, we

Â think it's a category of equations worth explaining and worth describing.

Â 29:54

And so we're going to discuss these other examples.

Â But it, when you look, see an equation like this

Â were you have a first derivative with a respect of time.

Â And then the second derivative with respect

Â to location, that's often a reaction-diffusion equation.

Â 30:10

The other important thing to note about the one

Â dimensional cable equation, which we've already discussed, but, again,

Â worth reiterating this is the partial dfferential equation, which

Â from now on we're going to abbreviate as the PDE.

Â And to obtain a numerical solution of this, you need

Â to convert it to discrete form in both space and time.

Â We already discussed with Oiler's method how the way

Â that Oiler decided to solve differential equations with respect

Â to time was he said the derivative with respect

Â to time is almost equal to something like this, right.

Â Both the next time, the variable and the next time step minus

Â the variable the current time step divided by the time step [UNKNOWN].

Â Well then we can also make an approximation of the second derivative

Â with respect to the location which is something that looks like this.

Â And this is the basis for all solutions of partial differential equations is

Â again you take your derivative and you convert it into a discrete form.

Â So for Oiler's method or ODEs.

Â 31:07

We just took the derivative with the second time and we converted

Â this into a discreet form and that's how we got Oiler's method.

Â When we talk about solutions of

Â partial differential equations the principal's the same.

Â Discrete form of this derivative, the discrete

Â form of this derivative, in this case it's

Â the second derivative and that's why this

Â one's a little more complicated than one here.

Â 31:40

One thing that we've seen is that

Â in neurons, voltage is typically not spacially uniform.

Â So we discuss this with respect to the Hodgkin-Huxley model.

Â When they developed it, the voltage was spatially uniform.

Â They, they made it that way.

Â But in general, you, in the neurons in your body, voltage is traveling.

Â And so, voltage is, instead varies as a function of time

Â and location rather than just saying voltage varies with respect to time.

Â 32:19

In the cable equation that we derived, it is,

Â is an example of a reaction to fusion equation.

Â And the reason I think it's worth going

Â through the derivation of this, and it's worth dwelling

Â on this, is because this is a type

Â of partial differential equation that is frequently encountered biologically.

Â If this were, like an exception, if this were the

Â sort of thing that only came about when you're talking about

Â neurons and didn't come about in other contexts, then it probably

Â wouldn't be worth going through the trouble of, of deriving it.

Â But because it's so frequent encountered in

Â biology that's why I think it's worth going

Â through it and worth explaining what we mean

Â when we talk about a reaction diffusion equation.

Â And that is what we are going to do in some of the

Â next couple of lectures that are coming up on, on specifically on PDEs.

Â