0:00

[BLANK_AUDIO].

Hello.

Welcome back.

We are now on our sixth, sixth and

final lecture on mathematical models of action potentials.

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

Propagation of action potentials.

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

a, a series of equations, the

Hodgkins-Huxley equations, and the Fitzunit Remote equations.

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

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

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

These were always assumed to be spatially-uniform.

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

We never said anything about location in this case.

0:55

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

just, doesn't just vary with respect to time

but it also varies with respect to location?

What if voltage at one location is

different from a voltage at another location?

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

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

And when you think about some of the long neurons that

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

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

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

1:26

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

terms of mathematical models, we're going to

discuss electrical propagation in very conceptual terms.

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

1:40

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

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

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

So what we've discussed to this point were were

systems that were consisted of several ordinary differential equations.

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

2:05

When you have a electrical propagation biologically.

For instance, when impulses are travelling from the, from

one part of your heart to another part of

your heart or when electrical impulses are, travelling down

your nerves as you are, as you are thinking.

2:19

This electrical propagation results from spatial voltage gradients.

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

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

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

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

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

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

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

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

with the voltage in our Hodgkin-Huxley model,

is going to be around minus 60 millivolts.

2:59

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

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

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

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

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

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

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

3:19

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

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

going to immediately induce an action potential over here.

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

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

3:33

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

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

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

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

Remember that at the peak of the action potential

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

go from 60 at rest to a you know,

peak value of you know, somewhere around plus 20.

3:58

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

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

So it's going to be depolarized on the left

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

The question is, now what happens?

4:11

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

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

Left to right, either on the inside or on the

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

But here when you have positive charges here

on the inside and negative charges here on

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

from the positive charges to the negative charges.

And conversely, on the outside it's going to

flow in the other direction and, this is,

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

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

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

4:49

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

Well, they're going to take this patch of

membrane here and they're going to depolarize that.

They're going to make the voltage go up.

At these locations here.

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

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

5:10

When an electrical current flows from left to right on

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

The positive charges, as they flow from here.

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

towards threshold and then that area of

membrane is going to fire and axion itself.

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

on the inside instead of only three positive charges

on the inside which I had during step three.

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

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

Then we're going to have more current flowing which

is which is going to induce this this

section of membrane to fire an action potential

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

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

6:15

One, one part of membrane requires an action potential.

It becomes depolarized, then you have current flowing on

the inside and current flowing on the outside and

through that mechanism you can depolarize more tissue and

get that tissue to also fire an action potential.

But what makes this challenging is that voltage

depends on both time and location so our voltage

over here, across the cell membrane is different

from our voltage over here across the cell membrane.

6:39

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

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

Because then that would be ambiguous.

Are you talking about this voltage over here?

Or are you talking about this voltage over here?

So now that voltage depends on both time and location.

Now we need to solve a system of partial

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

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

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

some of the issues related to solving partial differential equations.

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

7:18

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

It's a propagated action potential.

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

right each trace might represent an action potential

that you would record from a different location.

So the black one would come from here, the red

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

What you see is that if you plot all of these

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

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

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

And then it occurs here.

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

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

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

8:03

The relevant equation that we're going to have to

solve with respect to voltage is the following.

Capacitance times the partial derivative of voltage

with respect to time, is equal to

a bunch of terms times the second

derivative of voltage, with respect to location.

Minus the ionic current.

You'll notice here that the symbols have changed.

Instead of just a dv, dt.

It becomes a special, a special delta.

And this is to signify.

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

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

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

on time then you can compute an ordinary derivative of,

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

And then you can just a lower vase C.

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

In this case we're representing location by x.

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

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

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

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

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

That is an x.

So the pertinent questions that we need to address here

are first of all where does this equation come from.

I just wrote this down.

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

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

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

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

And then the next thing we want to address after

this is, how do you solve this in practice?

And this is what the some of the

subsequent lectures are, are going to get into.

What are the issues involved when you need to

solve a partial differential equation rather than an ordinary

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

deal with some of the practical aspects of it.

9:48

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

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

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

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

What we're going to address in the next

few slides, is one dimensional electrical propagation.

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

previous slide, which is an action

potential propagating down a uniform cable.

And in order to analyze this in mathematical terms,.

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

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

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

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

So every time we chop this into a

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

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

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

the Hodgkin–Huxley model where you have a voltage on the outside and then

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

So this is like what we looked at with the Hodgkin–Huxley model.

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

We had a sodium current, we had a

potassium current, and we had a wheat current.

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

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

11:06

And then before we discuss with respect to the Hodgkin–Huxley model.

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

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

This current that's flowing is going to either

make the membrane voltage go up or down.

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

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

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

an ionic current with, in parallel with a capacitor.

And now we're going to couple these together.

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

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

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

11:46

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

and the type of analysis that were

going to perform and its called cable theory.

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

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

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

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

In a tissue where that can occur in

multiple dimensions all at once, such as the heart.

12:14

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

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

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

And what I showed you on the last slide was

an ionic current in parallel with a capacitor and that represented

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

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

Now we've actually, actually coupled them together.

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

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

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

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

12:56

So the terminology we're going to use here is

that Ri, the little i stands for intracellular.

This represents our intracellular resistance, this says exactly how much

resistance to current flow you have from one patch of

membrane to the next patch of membrane or from one

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

13:17

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

a partial differential equation and an

ordinary differential equation is that it's not

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

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

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

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

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

13:43

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

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

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

This is j minus one.

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

So this is j plus one.

And one other thing you may have noticed as we,

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

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

We have resistances between locations on the inside.

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

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

This is an assumption that extracellular resistance,

Re, is equal to zero, that's why

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

So it all extracellular voltages are grounded.

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

14:35

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

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

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

voltage in the Hodgkin–Huxley model when it was series of ODs.

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

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

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

then our intracellular potential is equal to our

transmembrane potential at every node, at every element.

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

isolated fiber in a bath, which

is how Hodgkin-Huxley performed their experiments.

Right, they took out, they took a giant axon

out of a squid and then they examined it.

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

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

In more physiological conditions it might it might

not, may be less good of an assumption.

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

making it more complicated by considering extra-cellular resistances

just makes the equations a little bit more complicated.

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

15:52

Now that we've set up what our

equivalent circuit looks like, and we've discussed some

of the assumptions, we want to ask, what

equations describe the jth element of the cable?

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

At Jth element vj and these currents that,

that can influence the voltage of the Jth element.

16:15

Well we want to think about what's this current

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

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

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

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

flowing from j minus one element to jth element.

16:33

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

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

mean the ones that exactly, immediately downstream of

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

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

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

This is the current i.

From j minus one traveling to j.

Well, how can you compute this current?

17:00

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

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

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

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

17:20

What do we call this, how do we

derive these particular equations rather than some other equations.

Well this is just Ohm's law.

Right?

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

The voltage gradient divided by the resistance.

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

This is this is Ohm's law.

17:49

Well this is actually pretty simple as well.

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

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

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

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

This IM sub j membrane current at the jth element.

We have to multiply this by the surface area, A

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

18:21

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

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

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

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

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

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

You've prolly learned this at some point in in

freshman physics or maybe even in high school physics.

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

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

Current has to be conserved.

That's where we get this equation here.

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

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

19:03

and then it's going to rely on a little

bit of knowledge of calculus, but the basics

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

they should be familiar to you from, from

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

19:20

And then we could do the same sort of

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

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

Well, we have two choices.

The current can either flow under the

capacitor or the current can flow through here.

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

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

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

It can either go two ways and those two ways have

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

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

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

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

Therefore, the current is flowing out of

the capasiter, which by our definition of capacitance

is equal to membrane capacitance is times

derivative of voltage, j, with respect to time.

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

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

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

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

20:22

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

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

And so these are, these currents are in units of

something like amperes per centimeter

squared or microamperes per centimeter squared.

20:38

That's why we had to take these,

this membrane current and multiply it by surface

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

So, these currents are just, are just in units

of amperes or, or microamperes or something like this.

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

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

Is these two guys here.

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

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

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

Plus the membrane surface area times the membrane current.

21:16

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

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

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

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

We calculated this current with ohm's law and

then this membrane current here, we decided, also

based on Kirchoff's current law which is capacitive

current, the current that flows across the capacitor.

Plus the ionic current.

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

just discussed and substitute them into this equation here.

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

21:50

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

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

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

Also on the right hand side.

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

from, from calculus and then we are going to

have our reaction diffusion equation that describes the system.

22:20

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

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

On the left hand side we're going to

have the derivative of voltage with expected time.

Times the memory capacitance.

Everything else is going to go on to the

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

we have this on the right hand side is

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

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

minus one, minus two times voltage of j, plus

voltage of j plus one, divided by the membrane

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

Both of those in the denominator.

23:15

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

to the cable geometry and this is

a formula you probably learned in physics class.

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

or even if physics class was a long time ago

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

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

23:37

Your resistance from say the, the left side of

this little cylinder to the bot-, or from the

top of the cylinder to the bottom, or in

other words, from the left here to the right.

Depends on a few, a few things.

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

This intracellular resistivity is just a property

of the of the substance that you have.

It's a property of the cytoplasm.

And resistivity is defined in a way that

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

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

You can sort of think intuitively.

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

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

And these aspects related to geometry come in this way.

Your delta x comes in the numerator.

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

top of the cylinder to the bottom cylinder?

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

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

In other words if this were, if this

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

24:41

That makes intuitive sense.

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

And then conversely, what you have in the in

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

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

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

And we're using a instead of R because we

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

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

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

25:13

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

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

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

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

the low the lower the resistance is going to be.

That's why the surface areas the denominator here.

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

25:35

So this is how you can relate this is

how you can compute resistance of some arbitrary geometry.

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

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

the the distance Delta x divided by the surface area

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

26:10

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

That's the circumference, times delta x.

So now if we combine these two.

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

26:26

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

Pie squared, so we can simplify this here.

2 times row i times delta x over a.

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

just derived and it makes it a symmetrical consideration.

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

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

Capacitance times a directive of voltage with respect of time.

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

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

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

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

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

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

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

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

27:34

Here is voltage to the right and here we

have minus 2 times voltage at that particular location.

I'll divide it by delta x squared.

This term here as delta x goes to

0, becomes the second derivative with respect to voltage.

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

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

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

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

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

So as delta x goes to zero, this thing here

approximates the second derivative of voltage with respect to x.

That's exactly what this becomes here.

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

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

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

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

exactly which node of J you are talking about here.

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

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

Is these two terms here times the second derivative

with respective location with respect to minus i ion.

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

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

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

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

is an example of a reaction diffusion equation.

29:16

First of all, this is a reaction diffusion equation.

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

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

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

But I think it's important to know

that these reaction diffusion equations appear in many

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

discuss what we mean by reaction diffuse equations.

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

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

They often obey reaction diffusion equations.

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

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

29:54

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

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

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

And then the second derivative with respect

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

30:10

The other important thing to note about the one

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

worth reiterating this is the partial dfferential equation, which

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

And to obtain a numerical solution of this, you need

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

We already discussed with Oiler's method how the way

that Oiler decided to solve differential equations with respect

to time was he said the derivative with respect

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

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

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

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

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

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

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

So for Oiler's method or ODEs.

31:07

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

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

When we talk about solutions of

partial differential equations the principal's the same.

Discrete form of this derivative, the discrete

form of this derivative, in this case it's

the second derivative and that's why this

one's a little more complicated than one here.

31:40

One thing that we've seen is that

in neurons, voltage is typically not spacially uniform.

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

When they developed it, the voltage was spatially uniform.

They, they made it that way.

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

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

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

32:19

In the cable equation that we derived, it is,

is an example of a reaction to fusion equation.

And the reason I think it's worth going

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

on this, is because this is a type

of partial differential equation that is frequently encountered biologically.

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

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

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

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

But because it's so frequent encountered in

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

through it and worth explaining what we mean

when we talk about a reaction diffusion equation.

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

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