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Integrate-and-Fire Models I.

Computational Neuroscience. Session 2-2

Dr. Marco A Roque Sol

06/04/2018

Dr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Integrate-and-Fire Models I. Cable theory I .

Cable theory I .

Previously, we have considered only point neuron models.That is, we have assumed that the neuron is electricallycompact, so that it can be represented by a single patch ofmembrane. The gist of this assumption is that themembrane potential is the same everywhere in the neuron.In real neurons, this assumption is not true. Substantialdifferences in potential exist along the length of aneuronâAZs processes and the resulting longitudinalcurrents must be explicitly considered.

Cable theory I .

Previously, we have considered only point neuron models.That is, we have assumed that the neuron is electricallycompact, so that it can be represented by a single patch ofmembrane. The gist of this assumption is that themembrane potential is the same everywhere in the neuron.In real neurons, this assumption is not true. Substantialdifferences in potential exist along the length of aneuronâAZs processes and the resulting longitudinalcurrents must be explicitly considered.

Cable theory I .

When a neuron is excited by a synaptic input, the synapticcurrent induces a membrane polarization which isdistributed nonuniformly over the surface of the cell. Afterthe brief period of synaptic activity, which roughly coincideswith the rising phase of the synaptic potential, the chargeacross the membrane redistributes and decays, producingthe falling phase of the synaptic potential.

Cable theory I .

When a neuron is excited by a synaptic input, the synapticcurrent induces a membrane polarization which isdistributed nonuniformly over the surface of the cell. Afterthe brief period of synaptic activity, which roughly coincideswith the rising phase of the synaptic potential, the chargeacross the membrane redistributes and decays, producingthe falling phase of the synaptic potential.

Cable theory I .

Assuming that the membrane remains passive in itselectrical properties, and assuming that the duration of thesynaptic permeability changes is indeed brief, theredistribution and decay of the membrane potential is welldescribed by the one-dimensional cable equation (Rall,1969 b).

Similarly, the neuron may be polarized by means ofexperimentally imposed voltages and currents. Again, theresulting time course of potential decay and redistributionover the surface of the fiber is well described by theone-dimensional cable equation (Clark and Plonsey, 1966;Pickard, 1969 )

Cable theory I .

The cable equation model of passive nerve fibers canthus be a valuable tool both for analyzing the means bywhich a neuron combines a pattern of synaptic inputs toproduce a specific computation, and for determining thefunctionally significant electrical parameters of the cell bymeasurements of the response to various applied currents.In order for the model to be of practical use, though, it mustbe possible first to set up the equation so that theboundary and initial conditions correspond to theanatomical, physiological, and experimental realities, andsecond, to solve the equation under these conditions.

Cable theory I .

The cable equation model of passive nerve fibers canthus be a valuable tool both for analyzing the means bywhich a neuron combines a pattern of synaptic inputs toproduce a specific computation, and for determining thefunctionally significant electrical parameters of the cell bymeasurements of the response to various applied currents.In order for the model to be of practical use, though, it mustbe possible first to set up the equation so that theboundary and initial conditions correspond to theanatomical, physiological, and experimental realities, andsecond, to solve the equation under these conditions.

Cable theory I .

The solution to the cable equation can be written inclosed form in terms of known functions only for a fewspecial cases, such as the infinite uniform cable (Hodgkinand Rushton, 1946) or the infinite cable terminated in a cellbody (Rall, 1960). more realistic models, such as a finitelength cable, infinite series solutions have been found(Volkov and Platonova, 1970). Using a separation ofvariables approach, Rall (1962, 1969 a) has developed aninfinite series form of solution applicable to the case of afinite length cable with a variety of different types ofterminations.

Cable theory I .

The solution to the cable equation can be written inclosed form in terms of known functions only for a fewspecial cases, such as the infinite uniform cable (Hodgkinand Rushton, 1946) or the infinite cable terminated in a cellbody (Rall, 1960). more realistic models, such as a finitelength cable, infinite series solutions have been found(Volkov and Platonova, 1970). Using a separation ofvariables approach, Rall (1962, 1969 a) has developed aninfinite series form of solution applicable to the case of afinite length cable with a variety of different types ofterminations.

Cable theory I .

Using a Laplace transform approach, a theoreticalframework can be developed in which both boundary andinitial value conditions are expressed in concise and easilycomputed form. Using this framework, we can develop thegeneral form of solution to the cable equation for arbitraryboundary and initial conditions.

Cable theory I .

Using a Laplace transform approach, a theoreticalframework can be developed in which both boundary andinitial value conditions are expressed in concise and easilycomputed form. Using this framework, we can develop thegeneral form of solution to the cable equation for arbitraryboundary and initial conditions.

Cable theory I.

Equivalent Circuits: The Electrical Analogue.

A very useful way to describe the behavior of themembrane potential is in terms of electrical circuits; this iscommonly called the equivalent circuit model.

The circuit consists of three components: (1) conductors orresistors, representing the ion channels; (2) batteries,representing the concentration gradients of the ions; and(3) capacitors, representing the ability of the membraneto store charge. The equivalent circuit model leads to bothan intuitive and a quantitative understanding of how themovement of ions generates electrical signals in the nervecell.

Cable theory I.

We first consider a membrane that is only permeable topotassium ( the equivalent circuit is shown below ). Thelipid bilayer that constitutes the cell membrane hasdielectric properties and as such behaves in much thesame manner as a capacitor. Recall that capacitors storecharge and then release it in the form of currents. Therelationship between the charge stored and the potential,as we mentioned before is given by

Cable theory I.

Equivalent Circuits: The Electrical Analogue.We first consider a membrane that is only permeable topotassium ( the equivalent circuit is shown below ). Thelipid bilayer that constitutes the cell membrane hasdielectric properties and as such behaves in much thesame manner as a capacitor. Recall that capacitors storecharge and then release it in the form of currents. Therelationship between the charge stored and the potential,as we mentioned before is given by

Cable theory I.

q = CMVM

that is, the total charge q is proportional to the potential V Mwith a proportionality constant CM called the membranecapacitance.

The capacitance per square centimeter is called thespecific membrane capacitance and will be denoted ascM .Hence, the total membrane capacitance CM is thespecific membrane capacitance cM times the total surfacearea of the cell.

Cable theory I.

q = CMVM

Cable theory I.

q = CMVM

Cable theory I.

q = CMVM

Cable theory I.

q = CMVM

Cable theory I.

Since current is the time derivative of charge, we candifferentiate the above equation, divide by the cell’s area,and obtain an expression for the specific capacitancecurrent:

icap = CMdVM

dtThis gives the capacitance current per unit area. We willdenote the total capacitance current as Icap.

Cable theory I.

icap = CMdVM

Cable theory I.

icap = CMdVM

This gives the capacitance current per unit area. We willdenote the total capacitance current as Icap.

Cable theory I.

icap = CMdVM

Cable theory I.

icap = CMdVM

Cable theory I.

In the equivalent circuit,K+ channels are represented as aconductor in series with a battery. If gK is the conductanceof a single K+ channel, then, using Ohm’s law, the ioniccurrent through this channel is

IK = gK (VM − EK )

Here, EK is the potential generated by the battery; this isgiven by the K+ Nernst potential. The driving force is(VM − EK ).

Cable theory I.

Now suppose there are NK K+ channels in a unit area ofmembrane. These can all be combined into the singleequivalent circuit. The conductance per unit area, orspecific membrane conductance (S/cm2 ), is given byNK × gK and the specific membrane resistance (Ωcm2 ) isrK = 1/gK .

Cable theory I.

Since the Nernst potential depends only on theconcentration gradient of K+, and not on the number ofK+ channels, it follows that the K+ current, per unit area,is given by

IK = gK (VM − EK ) =VM − EK

Kirchhoff’s current law states that the total current into thecell must sum to zero. Together with the equivalent circuitrepresentation, this leads to a differential equation for themembrane potential:

IK + icap = CMdVM

VM − EK

rM= 0⇒

dt= gK (EK − VM)⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

IK + icap = CMdVM

VM − EK

rM= 0⇒

Cable theory I.

The figure below shows an equivalent circuit with threeparallel conductances and a current source, I(t). Here thecapacitance current must be equal to the sum of the ioniccurrents and the current source. As before, thecapacitance current, per unit area, is given by the previousequation and the ionic current, per unit area, is given by

Cable theory I.

Equivalent Circuits: The Electrical Analogue.The figure below shows an equivalent circuit with threeparallel conductances and a current source, I(t). Here thecapacitance current must be equal to the sum of the ioniccurrents and the current source. As before, thecapacitance current, per unit area, is given by the previousequation and the ionic current, per unit area, is given by

Cable theory I.

iion = −gCl(VM − ECl)− gK (VM − EK )− gNa(VM − ENa)

The current source is not typically expressed as currentper unit area, so we must divide I(t) by the total surfacearea of the neuron, A. It then follows that

dt= −gCl(VM −ECl)−gK (VM −EK )−gNa(VM −ENa)+

I(t)/A ⇒

Cable theory I.

I(t)/A ⇒

Cable theory I.

I(t)/A ⇒

Cable theory I.

I(t)/A ⇒

Cable theory I.

I(t)/A ⇒

Cable theory I.

I(t)/A ⇒

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

ER = (gClECl + gK EK + gNaENa)rM ; rM =1

gCl + gK + gNa

For a passive membrane in which the conductances andcurrents are all constant, VM will reach a steady state(dVM

dt = 0):

Vss =gClECl + gK EK + gNaENa + I/A

gCl + gK + gNa

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

gCl + gK + gNa

dt = 0):

gCl + gK + gNa

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

gCl + gK + gNa

dt = 0):

gCl + gK + gNa

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

gCl + gK + gNa

dt = 0):

gCl + gK + gNa

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

gCl + gK + gNa

dt = 0):

gCl + gK + gNa

Cable theory I.

dt= −VM − ER

rM+ I(t)/A

gCl + gK + gNa

dt = 0):

gCl + gK + gNa

Cable theory I.

In the absence of the applied current, the steady-statepotential is a weighted sum of the equilibrium potentials ofthe three currents. This is similar to the GHK − equation ,in which the contribution to the resting potential by eachion is weighted in proportion to the permeability of themembrane to that particular ion.

Cable theory I.

In the absence of the applied current, the steady-statepotential is a weighted sum of the equilibrium potentials ofthe three currents. This is similar to the GHK − equation ,in which the contribution to the resting potential by eachion is weighted in proportion to the permeability of themembrane to that particular ion.

Cable theory I.

The Membrane Time Constant.

Now, we consider how a passive, isopotential cell respondsto an applied current. This will help explain how eachcomponent of the electrical circuit con- tributes to changesin the membrane potential. The cell is said to be passive ifits electrical properties do not change during signaling.

The cell is said to be isopotential if the membrane potentialis uniform at all points of the cell; that is, the membranepotential depends only on time. To simplify the analysis,we will consider a spherical cell with radius ρ.

Cable theory I.

Suppose this cell is injected with an applied current, I(t)that is turned on at t = 0 to some constant value, I0 , andturned off at t = T . Here, we assume (without loss ofgenerality) I0 > 0.Note that for an isopotential cell, theinjected current distributes uniformly across the surface.

Cable theory I.

Suppose this cell is injected with an applied current, I(t)that is turned on at t = 0 to some constant value, I0 , andturned off at t = T . Here, we assume (without loss ofgenerality) I0 > 0.Note that for an isopotential cell, theinjected current distributes uniformly across the surface.

Cable theory I.

It follows that for a spherical cell, the current flowing acrossa unit area of the membrane is

IM(t) =I(t)

4πρ2 =

4πρ2 ; 10 < t < T0 otherwise

Cable theory I.

IM(t) =

I(t)4πρ2 =

Cable theory I.

IM(t) =I(t)

4πρ2 =

Cable theory I.

IM(t) =I(t)

4πρ2 =

Cable theory I.

IM(t) =I(t)

4πρ2 =

Cable theory I.

As before, suppose cM is the specific membranecapacitance, rM is the specific membrane resistance, andER is the cell’s resting potential. To simplify things, we takeER = 0 so that VM measures the deviation of themembrane potential from rest.

Thus,the membrane potential satisfies the ordinarydifferential equation

dt= −VM

rM+ IM(t)

Cable theory I.

dt= −VM

rM+ IM(t)

Cable theory I.

dt= −VM

rM+ IM(t)

Cable theory I.

dt= −VM

rM+ IM(t)

Cable theory I.

If the cell starts at rest, then the solution of this linearequation ( it is also separable ... why ???) satisfies

VM =rM I04πρ2

(1− e−

)0 < t < T

where τM = cM rM is the membrane time constant and

VM = VM(T )e−(t−T )

τM T ≤ t

Cable theory I.

The Membrane Time Constant.If the cell starts at rest, then the solution of this linearequation ( it is also separable ... why ???) satisfies

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

VM =rM I04πρ2

(1− e−

)0 < t < T

τM T ≤ t

Cable theory I.

Once the current is turned on, the membrane potentialasymptotically approaches the steady-state value rM I0

4πρ2

The membrane time constant also determines the rate atwhich the membrane potential decays back to rest after thecurrent is turned off.

Cable theory I.

4πρ2

Cable theory I.

4πρ2

Cable theory I.

Finally, The steady-state membrane potential satisfies

rM I04πρ2 = I0RINP

where RINP is the input resistance of the cell.

Cable theory I.

The Cable Equation.

We consider a cell that is shaped as a long cylinder, orcable, of radius a. We assume the current flow is along asingle spatial dimension, x , the distance along the cable.In particular, the membrane potential depends only on thex variable, not on the radial or angular components.

The cable equation is a partial differential equation thatdescribes how the membrane potential VM(x , t) dependson currents entering, leaving, and flowing within the neuron

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

The equivalent circuit is shown below. In what follows, wewill assume Re = 0, so that the extracellular space isisopotential. This assumption is justified if the cable is in abath with large cross-sectional area.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

We first consider the axial current flowing along the neurondue to voltage gradients. Note that the total resistance ofthe cytoplasm grows in proportion to the length of the cableand is inversely proportional to the cross-sectional area ofthe cable.

The specific intracellular resistivity, which we denote as rL ,is the constant of proportionality. Hence, a cable of radiusa and length ∆x has a total resistance ofRL = rL∆x/(πa2). It follows from Ohm’s law that at anypoint x , the decrease in VM with distance is equal to thecurrent times the resistance. That is,

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

VM(x + ∆x , t)− VM(x , t) = −Ilong(x , t)RL = −Ilong(x , t)∆xπa2 rL

There is a minus sign because of the convention thatpositive current is a flow of positive charges from left toright. If voltage decreases with increasing x , then thecurrent is positive. In the limit ∆x → 0 ,

Ilong(x , t) = −πa2

(Now, we are using partial derivatives because VM is afunction of two variables. )

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Cable theory I.

The Cable Equation.

Let iion be the current per unit area due to ions flowing intoand out of the cell. Then the total ionic current that flowsacross a membrane of radius a and length ∆x is given byIion = (2πa∆x)/iion .

Recall that the rate of change of the membrane potential isdetermined by the capacitance .

Hence, for a cable of radius a and length ∆x , the totalcapacitance is given by CM = (2πa∆x)cM and the amountof current needed to change the membrane potential at arate ∂VM /∂x is

Icap(x , t) = (2πa∆x)cM∂VM

Cable theory I.

The Cable Equation.

Let iion be the current per unit area due to ions flowing intoand out of the cell. Then the total ionic current that flowsacross a membrane of radius a and length ∆x is given byIion = (2πa∆x)/iion .

Cable theory I.

The Cable Equation.Let iion be the current per unit area due to ions flowing intoand out of the cell. Then the total ionic current that flowsacross a membrane of radius a and length ∆x is given byIion = (2πa∆x)/iion .

Cable theory I.

∂tDr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

∂tDr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

The Cable Equation.

From Kirchhoff’s law, the change in intracellular axialcurrent is equal to the amount of current that flows acrossthe membrane. Hence,

Icap(x , t) + Iion(x , t) = −[Ilong(x − ∆x , t)− Ilong(x , t)

(2πa∆x)cM∂VM

∂t+(2πa∆x)iion =

∂x(x +∆x , t)−−πa2

∂x(x , t)

We divide both sides of this equation by (2πa∆x) and let∆x → 0 to obtain the cable equation:

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

The Cable Equation.From Kirchhoff’s law, the change in intracellular axialcurrent is equal to the amount of current that flows acrossthe membrane. Hence,

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

(2πa∆x)cM∂VM

∂x(x +∆x , t)−−πa2

∂x(x , t)

cM∂VM

∂2VM

∂x2 − iion

Cable theory I.

The Cable Equation.

For a passive cable, in which the resting potential isassumed to be zero, iion = VM(x , t)/rM where rM is thespecific membrane resistance. Then:

cM∂VM

∂2VM

∂x2 −VM(x , t)

We can rewrite this equation( The cable equation ) as

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

where λ2 = arM /2rL and τ = cM rM are the space or lengthconstant and the membrane time constant, respectively.

Cable theory I.

The Cable Equation.

cM∂VM

∂2VM

∂x2 −VM(x , t)

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

Cable theory I.

The Cable Equation.

cM∂VM

∂2VM

∂x2 −VM(x , t)

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

Cable theory I.

The Cable Equation.

cM∂VM

∂2VM

∂x2 −VM(x , t)

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

Cable theory I.

The Cable Equation.

cM∂VM

∂2VM

∂x2 −VM(x , t)

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

Cable theory I.

The Cable Equation.

cM∂VM

∂2VM

∂x2 −VM(x , t)

τM∂VM

∂t= λ2 ∂2VM

∂x2 − VM

Cable theory I.

Partial Differential Equations.

A few words are necessary to say about PartialDifferential Equations (PDE’s), at this point of thediscussion.

A partial differential equation (PDE), is an equation thatcontains partial derivatives of an unknown multi-valuedfunction u. Thus, for a function of the form u = u(x , y),denoting the partial derivative by ∂u/∂x = ux , ∂u/∂y = uy .We can write the general first order PDE for u(x , y) as

F (x , y ,u(x , y),ux (x , y),uy (x , y)) = F (x , y ,u,ux ,uy ) = 0

or the general second order PDE for u(x , y) as

F (x , y ,u,ux ,uy ,uxx ,uyy ,uxy ,uyx ) = 0

Cable theory I.

Partial Differential Equations.A few words are necessary to say about PartialDifferential Equations (PDE’s), at this point of thediscussion.

Cable theory I.

F (x , y ,u,ux ,uy ,uxx ,uyy ,uxy ,uyx ) = 0Dr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

F (x , y ,u,ux ,uy ,uxx ,uyy ,uxy ,uyx ) = 0Dr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

A solution to the PDE written above is a functionu(x , y)(u(x , t)) which satisfies for all values of thevariables x and y such a equation. Some examples ofPDEs (of physical significance) are:

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equationut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equation

ut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equationut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equation

uxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equationut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equation

utt − uxx = 0 Wave equationut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equation

ut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equationut − uxx = 0 Heat equation

ut + uux + uxxx = 0 KdV equationiut − uxx = 0 Scrodinger equation

Cable theory I.

ux + ux = 0 Transport equationut + uux = 0 Burguer’s equationuxx + uyy = 0 Laplace equationutt − uxx = 0 Wave equationut − uxx = 0 Heat equationut + uux + uxxx = 0 KdV equation

iut − uxx = 0 Scrodinger equation

Cable theory I.

Order. The order of a partial differential equation is theorder of the highest derivative entering the equation.Linearity. Linearity means that all instances of theunknown and its derivatives enter the equation linearly (raised to the power 1 ).

One of the classical partial differential equation ofmathematical physics is the equation describing theconduction of heat in a solid body (Originated in the 18thcentury) [ Heat Equation ]. A modern one is the spacevehicle reentry problem: Analysis of transfer anddissipation of heat generated by the friction with earth’satmosphere.

Cable theory I.

For example: Consider a straight bar with uniformcross-section and homogeneous material. We wish todevelop a model for heat flow through the bar.

Let u(x , t) be the temperature on a cross section locatedat x and at time t . We shall follow some basic principles ofphysics:

A) The amount of heat per unit time flowing through a unitof cross-sectional area is proportional to ∂u/∂x withconstant of proportionality k(x) called the thermalconductivity of the material.

Cable theory I.

B) Heat flow is always from points of higher temperatureto points of lower temperature.

C) The amount of heat necessary to raise the temperatureof an object of mass m by an amount c(x)m∆u, wherec(x) is known as the specific heat capacity of the material.

Thus to study the amount of heat H(x) flowing from left toright through a surface A of a cross section during the timeinterval ∆t can then be given by the formula:

H(x) = −k(x)(area of A)∆t∂u∂x

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

(x , t)

Cable theory I.

Heat flowing from left to right across the plane during antime interval ∆t is:

H(x + ∆x) = −k(x + ∆x)(area of B)∆t∂u∂x

(x + ∆x , t)

If on the interval [x , x + ∆x ], during time ∆t , additional heatsources were generated by, say, chemical reactions,heater, or electric currents, with energy density Q(x , t),then the total change in the heat ∆E is given by the formula∆E = Heat entering A - Heat leaving B + Heatgenerated .

Cable theory I.

(x + ∆x , t)

Cable theory I.

(x + ∆x , t)

Cable theory I.

(x + ∆x , t)

Cable theory I.

(x + ∆x , t)

If on the interval [x , x + ∆x ], during time ∆t , additional heatsources were generated by, say, chemical reactions,heater, or electric currents, with energy density Q(x , t),then the total change in the heat ∆E is given by the formula

∆E = Heat entering A - Heat leaving B + Heatgenerated .

Cable theory I.

(x + ∆x , t)

Cable theory I.

(x + ∆x , t)

Cable theory I.

And taking into simplification the principle C above,∆E = c(x)m∆u, where m = ρ(x)∆V . After dividing by(∆x)(∆t), and taking the limits as ∆x → 0 and ∆t → 0 weget:

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

If we assume k , c, ρ are constants, then the equationbecomes:

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

∂u∂x

)+ Q(x , t) = c(x)ρ(x)

∂u∂t

= β∂2u∂x2 +p(x , t) = c(x)ρ(x); β =

; p(x , t) =Q(x , t)

Cable theory I.

Boundary and Initial conditions. Now, we need to giveinitial or boundary conditions to specify a particularproblem. We thus obtain the mathematical model for theheat flow in a uniform rod without internal sources (p = 0)with homogeneous boundary conditions and initialtemperature distribution f (x), the following Initial BoundaryValue Problem:∂u∂t = β ∂2u

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

Boundary and Initial conditions. Now, we need to giveinitial or boundary conditions to specify a particularproblem. We thus obtain the mathematical model for theheat flow in a uniform rod without internal sources (p = 0)with homogeneous boundary conditions and initialtemperature distribution f (x), the following Initial BoundaryValue Problem:

∂u∂t = β ∂2u

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

∂x2 + p(x , t) = c(x)ρ(x); 0 < x < L; t > 0

u(0, t) = u(L, t) = 0, t > 0

u(x ,0) = f (x), 0 < x < L

Cable theory I.

The method of separation of variables. We are going topropose a solution of the form

u(x , t) = X (x)T (t)

Substituting into the Boundary problem, we obtain:

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

and applying boundary conditions, we obtain:

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

u(x , t) = X (x)T (t)

X ′′(x)X (x)

=T ′(t)T (t)

= β; 0 < x < L; t > 0⇒

T ′(t)− βT (t) = 0 X ′′(x)− kX (x) = 0⇒

X ′′(x)− kX (x) = 0; X (0) = X (L) = 0

Cable theory I.

We shall consider 3 cases:k = 0, k > 0, k < 0

k = 0. In this case X (x) = 0

k > 0. In this case k = λ2 and the ODE that needs to besolved is X ′′ − λ2X = 0, whose general solution isX (x) = c1eλx + c2e−λx , and applying boundary conditionsis ... X (x) = 0 ... !!!!!

k < 0. In this case k = −λ2 and the ODE that needs to besolved is X ′′ + λ2X = 0, whose general solution isX (x) = c1cos(λx) + c2sin(λx), and applying boundaryconditions: c1 = 0 and c2sin(λL) = 0, for this to happen,we need λL = nπ, i.e.,λ = nπ/L or k = −(nπ/L)2

We set Xn(x) = ansin([nπ/λ]x) : n = 1,2,3, ...

Cable theory I.

Finally for T ′(t) + βT = 0. Whose solutions areT (t) = be−β(nπ/λ)t2

Thus the functions un = Xn(x)Tn(t), satisfies theBoundary Problem. To satisfy the initial condition we tryu(x , t) = ∑ un(x , t) ( by linearity, a superposition ofsolutions ... !!!!) over all n. More precisely

u(x , t) =∞

∑n=1

cne−β(nπ/λ)t2sin([nπ/λ]x)

Cable theory I.

u(x , t) =∞

∑n=1

Cable theory I.

u(x , t) =∞

∑n=1

Cable theory I.

u(x , t) =∞

∑n=1

Cable theory I.

u(x , t) =∞

∑n=1

Cable theory I.

u(x , t) =∞

∑n=1

Cable theory I.

we must have

u(x ,0) =∞

∑n=1

cnsin([nπ/λ]x) = f (x)

This leads to the question of when it is possible to find cn

from a given f (x) or when it is possible to represent f (x)by such a series ? This problem was studied and solved byJean-Baptiste Joseph Fourier.

Cable theory I.

we must have

u(x ,0) =∞

∑n=1

Cable theory I.

we must have

u(x ,0) =∞

∑n=1

Cable theory I.

we must have

u(x ,0) =∞

∑n=1

Cable theory I.

we must have

u(x ,0) =∞

∑n=1

Cable theory I.

Jean Baptiste Joseph Fourier (1768 - 1830). Developedthe equation for heat transmission and obtained solutionunder various boundary conditions (1800− 1811).

Under Napoleon he went to Egypt as a soldier and workedwith G. Monge as a cultural attache for the French army.

Cable theory I.

Integrate-and-fire neuron model.

This section builds on the simple model neuron that wedeveloped in the past by adding in action potentials (APs).Rather than modeling the biophysical basis of the AP, inthis model we manually cause the neuron to spike when itsmembrane voltage reaches a threshold value.

Our working model is that neurons are like capacitorsconnected to resistors - the cell membrane stores charge,which can leak out through ion channels. As developed inthe last lecture, the equation weâAZll be using for all modelneurons is:

Cable theory I.

τmdVdt

= E − V + RmIe

where V (t) is given as follows:

V (t) = V∞ + [V (0)− V∞]e−t

Running a simulation- the “integrate” part of the model.Running a simulation entails computing changes in thecell’s membrane voltage for each iteration of the simulation(dt ms). We can use the same equation as above, butreplace V (t) with V (t + dt) (i.e. the voltage in the nexttime step of the simulation) and V (0) (the starting voltage)with V (t):

In practice, these computations work best for small valuesof dt (in most cases we’ll use dt ≤ 0.1ms).

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

τmdVdt

= E − V + RmIe

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

In practice, these computations work best for small values of dt(in most cases we’ll use dt ≤ 0.1ms).

Running a simulation - the “fire” part of the model. Nowthe integrate-and-fire model is almost entirely in place. Theonly thing we need to add is the rule that whenV (t + dt) ≥ V thresh , simulate an action potential bysetting V (t) = Vpeak and V (t + dt) = Vreset . Vthresh isgenerally somewhere around −55mV , Vpeak is around40mV , and Vreset is around −80mV . In the future we’lldiscuss the biophysical basis of why the neurondepolarizes (increases its membrane voltage) suddenlyduring the start of the action potential and why themembrane voltage becomes hyperpolarized (decreased)after the action potential is fired.

Cable theory I.

Computing firing rate. This is straightforward. We cansimply count up the number of spikes that were fired duringthe simulation (i.e. times when V ≥ V thresh and divide bythe length of time we were simulating.

Analytic solution for firing rate. While the fullintegrate-and-fire simulation is often useful (and isnecessary if you want to model things like spike timing), itturns out that there is an analytic method for computing theexpected firing rate of the model, given a constant externalcurrent Ie. We start with the equation for finding themembrane voltage at time t :

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

V (t) = V∞ + [V (0)− V∞]e−t

Cable theory I.

We can then solve for t as follows:V (t)− V∞ = [V (0)− V∞]e−

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

We can then solve for t as follows:V (t)− V∞ = [V (0)− V∞]e−

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

Integrate-and-fire neuron model.We can then solve for t as follows:

V (t)− V∞ = [V (0)− V∞]e−t

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

Integrate-and-fire neuron model.We can then solve for t as follows:V (t)− V∞ = [V (0)− V∞]e−

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

V (t)−V∞V (0)−V∞

= e−t

V (t)−V∞V (0)−V∞

]= − t

τmln[

V (t)−V∞V (0)−V∞

]= −t

t = −τmln[

V (t)−V∞V (0)−V∞

Cable theory I.

Now let’s suppose our model neuron has just fired a spikein the previous timestep of our simulation. We start bysetting V (0) = Vreset . We next need to know how long it isuntil the neuron next fires a spike (the inter-spike interval,tisi - or, in other words, the time t at which V (t) = Vthreshafter starting at V (0) = Vreset . Plugging in the appropriatevalues, we can compute tisi as follows:

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

tisi = −τmln[

Vtresh − V∞

Vreset − V∞

Cable theory I.

Recall that V∞ = E + RmIe . Thus

tisi = −τmln[

Vtresh − (E + RmIe)Vreset − (E + RmIe)

]The firing rate (risi ) is the reciprocal of the inter-spike interval:

risi =

(−τmln

[Vtresh − (E + RmIe)Vreset − (E + RmIe)

])−1

Cable theory I.

tisi = −τmln[

risi =

(−τmln

])−1

Cable theory I.

tisi = −τmln[

The firing rate (risi ) is the reciprocal of the inter-spike interval:

risi =

(−τmln

])−1

Cable theory I.

tisi = −τmln[

risi =

(−τmln

])−1

Cable theory I.

tisi = −τmln[

risi =

(−τmln

])−1

Cable theory I.

tisi = −τmln[

risi =

(−τmln

])−1

Cable theory I.

Putting it all together. To run the simulation, start with thebasic model neuron equation:

τmdVdt

= E − V + RmIe

Now solving for dV :

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

τmdVdt

= E − V + RmIe

=E − V + RmIe

dV =E − V + RmIe

Cable theory I.

Start the simulation by setting V (0) = E . With eachtimestep set V (t + dt) = V (t) + dV . You’ll need tore-compute dV for each time-step given V (t) and I(t) forthe appropriate time t . Remember to include the rule forfiring a spike (and resetting) when V > Vthresh - otherwisethe neuron won’t fire spikes.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mVDr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

Matlab Code

V_reset = -0.080; % -80mV

V_e = -0.075; % -75mV

V_th = -0.040; % -40mVDr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

Rm = 10e6; % membrane resistance

tau_m = 10e-3; % membrane time constant

dt = 0.02;

T = 0:dt:1; % 1 second simulation

Vm(1) = V_reset;

Im = 5e-9;

Cable theory I.

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

dt = 0.02;

Vm(1) = V_reset;

Im = 5e-9;

Cable theory I.

V_e = -0.075; % -75mV

V_th = -0.040; % -40mV

dt = 0.02;

Vm(1) = V_reset;

Im = 5e-9;

Cable theory I.

for t=1:length(T)-1,

if Vm(t) > V_th,

Vm(t+1) = V_reset;

Vm(t+1) = Vm(t) + dt * ( -(Vm(t) - V_e) + Im * Rm) / tau_m;

plot(T,Vm,’b-’);

xlabel(’Time(s)’);

ylabel(’Voltage (V)’);Dr. Marco A Roque Sol Computational Neuroscience. Session 2-2

Cable theory I.

if Vm(t) > V_th,

Vm(t+1) = V_reset;

Cable theory I.

if Vm(t) > V_th,

Vm(t+1) = V_reset;

Cable theory I.

Integrate-and-fire neuron model. For different values of dt weobtain:

Cable theory I.

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