Computational Neuroscience. Session 1-2roquesol/Computational_Neuroscience_Summer...Deﬁnitions...

Computational Neuroscience. Session 1-2

Dr. Marco A Roque Sol

05/29/2018

Dr. Marco A Roque Sol Computational Neuroscience. Session 1-2

Definitions

Differential Equations

A differential equation is any equation which containsderivatives, either ordinary or partial derivatives of an unknownfunction.

Ordinary and Partial Differential EquationsA differential equation is called an ordinary differential equation,abbreviated by ODE, if it has only ordinary derivatives in it,having the form:

F (y (n), y (n−1), ..., y ′, y(t), t) = 0

Likewise, a differential equation is called a partial differentialequation, abbreviated by PDE, if it has partial derivatives in it.


Definitions

There many situations where we can find such an object. Thus,for instance in the case of the study of Classical Mechanics inPhysics, if an object of mass m is moving with acceleration aand being acted on with force F then NewtonâAZs Second Lawtells us.

F = ma

To see that this is in fact a differential equation. First, rememberthat we can rewrite the acceleration, a, in one of two ways.

a =dvdt

or a =d2udt2

Where v is the velocity of the object and u is the positionfunction of the object at any time t . We should also rememberat this point that the force, F may also be a function of time,velocity, and/or position.


Definitions

mdvdt

= F (t , v) or md2udt2 = F (t , u, v)

More examples of differential equations

2y ′′ + 3y ′ − 5y = 0

cos(y)d2ydx2 − (1 + y)

dydx

+ y3e−y = 0

y (4) + 5y ′′′ − 4y ′′ + y = sin(x)

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2

∂u3

∂2x∂t= 1 +

∂u∂y


Classification

OrderThe order of a differential equation is the largest derivativepresent in the differential equation. The equation

mdvdt

= F (t , v)

is a first order differential equation, the equations

md2udt2 = F (t , u, v)

2y ′′ + 3y ′ − 5y = 0


dydx

+ y3e−y = 0


Classification

a2 ∂2u∂x2 =

∂u∂t

a2 ∂2u∂x2 =

∂2u∂t2

are second order differential equations, the equation∂u3

∂2x∂t= 1 +

∂u∂y

is a third order differential equation and finally, the equation

y (4) + 5y ′′′ − 4y ′′ + y = sin(x)

is a fourth order differential equation.

Note that the order does not depend on whether or not you’vegot ordinary or partial derivatives in the differential equation.


Definitions

Solution

A solution to an ordinary differential equation (ODE) on aninterval α < t < β is any function y(t) which satisfies thedifferential equation in question on the interval. It is important tonote that the solutions are often accompanied by intervals andthese intervals can impart some important information aboutthe solution.


Classification

Linear Differential Equations

A linear differential equation is any differential equation thatcan be written in the following form.

an(t)y (n) + an−1(t)y (n−1) + ... + a1(t)y ′ + a0(t)y = g(t)

The important thing to note about linear differential equations isthat there are no products of the function and its derivatives,and neither the function or its derivatives occur to any powerother than the first power.


Classification

The coefficients an(t), ..., a0(t), g(t) can be any function. Onlythe function y(t) and its derivatives are used in determining if adifferential equation is linear. If a differential equation cannot bewritten in the above form, then it is called a non-lineardifferential equation. In the examples above only


dydx

+ y3e−y = 0

is a non-linear equation. Note that we can’t classify Newton’sSecond Law equations since we do not know what form thefunction F has. These could be either linear or non-lineardepending on F .


Basic Examples

Differential Equations in Physics

Schrodinger’s Equation.

− h2

2m∇2Ψ + V (r)Ψ = i h ∂Ψ∂t

Is the starting point for non-relativistic Quantum Mechanics,here the solution Ψ(r , t) is the wave function.

In the one-dimensional case, If we substitute into the aboveequation, the proposed solution

Ψ(x , t) = ψ(x)φ(t)


Basic Examples

we get

− h2

2mψ′′(x)φ(t) + V (x)ψ(x)φ(t) = i hψ(x)φ′(t)

Dividing the whole equation by ψ(x)φ(t) we get

− h2

2mψ′′(x)ψ(x)

+ V (x) = i hφ′(t)φ(t)

= constant = E

So we obtain two ODE’s.


Basic Examples

First, The Time Independent Schrodinger Equation

− h2

2md2ψ(x)

dt2 + V (x)ψ(x) = Eψ(x)

and second, The Energy Eigenvalue Equation

i hdφ(t)

dt= Eφ(t)


Basic Examples

Maxwell’s equations:

∇ · E =ρ

ε0, (1a)

∇ · B = 0, (1b)

∇× E = −∂B∂t

, (1c)

∇× B = µ0ε0∂E∂t

+ µ0J, (1d)

This is the set of fundamental equations for ClassicalElectromagnetism, here the solutions E and B represent theelectrical and magnetic fields.


Basic Examples

In the vacuum ρ = J = 0. Maxwell Equations can be written as

∇2E = µ0ε0∂2E∂t2 ∇2B = µ0ε0

∂2B∂t2

Each one of these is 3 PDE’s. Consider the one-dimensionalcase for the Electrical Field

∂2E∂x2 = µ0ε0

∂2E∂t2

Now, suppose that E(x , t) = X(x)T (t) making this substitutionwe get


Basic Examples

X′′(x)T (t) = µ0ε0X(x)T ′′(t)

dividing by X(x)T (t)

X′′(x)X(x)

= µ0ε0T ′′(t)T (t)

= constant = a

Thus, we have to solve a couple of ODE’s

d2X(x)dx2 = aX(x); µ0ε0T ′′(t) = aT (t)


Basic Examples

Newton’s Second Law.

md2rdt2 = F (r , t)

Is the fundamental equation for Classical Mechanics, here thesolution r (t) is the position as a function of time. In this casewe have one independent variable, the time t . Solving anyparticular case of a Force bring us immediately to solving anODE.


Basic Examples

Consider the Gravity near earth’s surface

md2rdt2 = −mgk

That vector equation is equivalent to the next three ODE’s

md2xdt2 = 0

md2ydt2 = 0

md2zdt2 = −mg


Basic Examples

Example 1.1Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 for x > 0 .

SolutionWe will need the first and second derivatives :

y ′(x) = −32x−5/2 , y ′′(x) = 15

4 x−7/2

Plug these as well as the function into the differential equation:


Basic Examples

4x2(154 x−7/2) + 12x(−3

2x−5/2) + 3x−3/2 = 0

15x−3/2 − 18x−3/2 + 3x−3/2 = 0

0 = 0

So, y(x) = x−3/2 does satisfy the differential equation andhence is a solution. Why then didn’t I include the condition thatx > 0 ? I did not use this condition anywhere in the workshowing that the function would satisfy the differential equation.


Basic Examples

To see why recall that

y(x) = x−3/2 = 1x3/2

In this form it is clear that we will need to avoid x = 0 at theleast as this would give division by zero.

So, we saw in the last example that even though a function maysymbolically satisfy a differential equation, because of certainrestrictions brought about by the solution we cannot use allvalues of the independent variable and hence, must make arestriction on the independent variable.


Basic Examples

In the last example, note that there are in fact many morepossible solutions to the differential equation. For instance all ofthe following are also solutions

y(x) = x−1/2

y(x) = 5x−3/2

y(x) = 3x−1/2

y(x) = 2x−1/2 + 3x−3/2


Basic Examples

I’ll leave the details to you to check that these are in factsolutions. Given these examples can you come up with anyother solutions to the differential equation?

y(x) = c1x−1/2 + c2x−3/2

There are in fact an infinite number of solutions to thisdifferential equation.

So, given that there are an infinite number of solutions to thedifferential equation in the last example, we can ask a naturalquestion. Which is the solution that we want or does it matterwhich solution we use? This question leads us to the nextmaterial in this section.


Basic Examples

Example 1.2Show that y(x) = x−3/2 is a solution to4x2y ′′ + 12xy ′ + 3y = 0 with initial conditions y(4) = 1

8 andy ′(4) = − 3

64 .

SolutionAs we saw in previous example the function y(x) = x−3/2 is asolution and we can then note that

y(4) = 4−3/2 =1

43/2 =18

y ′(4) = −32

4−5/2 = −32

145/2 = − 3

64

and so this solution also meets the initial conditions.


Basic Examples

In fact, y(x) = x−3/2 is the only solution to this differentialequation that satisfies these two initial conditions.

Initial Value ProblemAn Initial Value Problem (or IVP) is a differential equation alongwith an appropriate number of initial conditions.

Example 1.3The following is an example of IVP

4x2y ′′ + 12xy ′ + 3y = 0, y(4) = 18 , and y ′(4) = − 3

64 .


Basic Examples

Example 1.4This is another example of an IVP

2ty ′ + 4y = 3, y(1) = −4 .

As you can notice, the number of initial conditions required willdepend on the order of the differential equation.

Interval of ValidityThe interval of validity for an IVP with initial condition(s)

y(t0) = y0, y ′(t0) = y1, ..., yk (t0) = yk

.


Basic Examples

is the largest possible interval on which the solution is valid andcontains t0 . These are easy to define, but can be very difficultto find in the practice.

General SolutionThe general solution to a differential equation is the mostgeneral form that the solution can take and doesn’t take anyinitial conditions into account.


Basic Examples

Example 1.5

y(t) = 34 + c

t2 is the general solution to the equation

2ty ′ + 4y = 3

Check it out !!!!!! In fact, all solutions to this differential equationwill be in this form. This is one of the first differential equationsthat we will learn how to solve and we will be able to verify thisshortly.


Basic Examples

Particular SolutionThe particular solution to a differential equation is the specificsolution that not only satisfies the differential equation, but alsosatisfies the given initial condition(s).

Example 1.6What is the particular solution to the following IVP?

2ty ′ + 4y = 3 y(1) = −4


Basic Examples

SolutionThis is actually easier to do than it might, the general solution isof the form:

y(t) =34+

ct2

All what we need is to determine the value of c that will give usthe solution that we are after. To find this, all we need to do isuse our initial condition as follows:

−4 =34+

c12


Basic Examples

c = −4− 34= −19

4

So, the particular solution to the IVP is:

y(t) =34− 19

4t2


Basic Examples

Implicit/Explicit SolutionAn explicit solution is any solution that is given in the formy = y(t).

An implicit solution is any solution that is not in explicit form,that is, y can not be expressed explicitly as a function of t .Note that it is possible to have either general implicit/explicitsolutions and particular implicit/explicit solutions.

Example 1.7Show that y2 = t2 − 3 is an implicit solution of the differentialequation yy ′ = t .


Basic Examples

SolutionUsing implicit derivation, the solution follows:

2yy ′ = 2t + 0

yy ′ = t

Example 1.8Find a particular explicit solution to the IVP

yy ′ = t y(2) = −1


Basic Examples

SolutionWe already know from the previous example that an implicitsolution to this IVP is y2 = t2 − 3. To find the explicit solution allwe need to do is solve for y(t) from

y(t) = ±√

t2 − 3

Now, we’ve got a problem here. There are two functions hereand we only want one and in fact only one will be the correct !!!We can determine the correct function by reapplying the initialcondition.


Basic Examples

Only one of them will satisfy the initial condition.

In this case we can see that the negative solution, will be thecorrect one. The explicit solution is then

y(t) = −√

t2 − 3

In this case we were able to find an explicit solution to thedifferential equation. It should be noted however that it will notalways be possible to find an explicit solution.


Basic Examples

Example 1.9

A spherical raindrop evaporates at a rate proportional to itssurface area. Write a differential equation for the volume of theraindrop as a function of time.

Solution

If the volume at time t is denoted by V (t) and the surface attime t is denoted by S(t), then the fact that raindrop evaporatesat a rate proportional to its surface area can be write as

dV (t)dt

proportional to S(t)

dV (t)dt

= −αS(t)


Basic Examples

were α is a constant of proportionality. But, we have thefollowing relationship

V =43

πr3 =⇒ r = [3V4π

]1/3

this implies that

dV (t)dt

= −αS(t) = 4πr2 = −4απ[3V4π

]2/3 = −4απ[3

4π]2/3V 2/3

dV (t)dt

= −cV 2/3

where c is a constant.


Basic Examples

Example 1.10

A certain drug is being administered intravenously to a hospitalpatient. Fluid containing 5mg/cm3 of the drug enters thepatientâAZs bloodstream at a rate of 100cm3/h. The drug isabsorbed by body tissues or otherwise leaves the bloodstreamat a rate proportional to the amount present, with a rateconstant of 0.4(h)−1.

(a) Assuming that the drug is always uniformly distributedthroughout the bloodstream, write a differential equation for theamount of the drug that is present in the bloodstream at anytime.

(b) How much of the drug is present in the bloodstream after along time?


Basic Examples

Solution

(a) If the drug is always uniformly distributed throughout thebloodstream, and the amount of the drug in mg at time t isdenoted by Q(t), then the equation that we will use is abalance equation

Rate of change of Q(t) =

Rate at which Q(t) enters the bloodstream -

Rate at which Q(t) exits the bloodstream

dQdt

= drug entering − drug exiting

dQdt

= (concentration)(rate of entering)− (concentration)(rate of exiting)


Basic Examples

dQdt

= (5)(100)−Q(t)(0.4)

and the final equation is

dQdt

= 500−Q(t)(0.4)

(b)

Q′(t) = 500−Q(t)(0.4)

Q′(t)500−Q(t)(0.4)

= 1


Basic Examples

− ddt

ln|500−Q(t)(0.4)|/0.40 = 1

ln|500−Q(t)(0.4)| = −0.4t + c′

500−Q(t)(0.4) = e−0.4t+c′ = ce−0.4t

Q(t) =500− ce−t

0.4Thus, in the long run the amount of drug present is ... 500/0.4 =20 mg !!!


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Computational Neuroscience. Session 1-2roquesol/Computational_Neuroscience_Summer...Deﬁnitions...

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