- Numerical Derivative, Normalization, lost here

[FrogPad]

Ok, I'm trying to recreate some results from a paper, and I am really lacking in understanding. I'm getting extremely frustrated.

I'll try to create a similar problem that replicates the original problem. The original problem, is three coupled non-linear differential equations that must be solved numerically. The new problem that I am writing here will just be one differential equation.

If I have the following function:

[tex] \alpha \frac{d \phi}{d\theta} = A_{23} + B_{23} \sin \phi [/tex]

Now they say they have normalized [itex] A_{23}, \,\, B_{23} [/itex], and later they say that [itex] \frac{d \phi}{d \theta} [/itex] is a normlized voltage.

So now should I believe that [itex] \frac{d \phi}{d\theta} [/itex] has no units, and that [itex] A_{23} + B_{23} \sin \phi [/itex] has no units? This would make sense correct?

Now, they define [itex] \alpha [/itex] as:
[tex] \alpha = \frac{h \omega}{2 e I_C R_n} [/tex] and
[tex] \theta = \omega t [/tex]

Now, h is Planck's constant, omega is angular frequency, e is the fundamental quantity of charge (coulombs), and I_C*R_n would have the units of volts. Specifically, [itex] \alpha \sim 10^{-12} seconds [/itex] (with the given I_C*R_n = 0.45mV) -- This is also assuming that I can simply change [itex] d\theta = \omega dt [/itex] and cancel the [itex] \omega [/itex] in teh [itex] \alpha [/itex] term.

Now here is my question (assuming everything else I did was ok). If I use MATLAB's ODE45 solver, with the tspan of [a, b]. What units is this even in? Is this seconds? Nanoseconds... how do I know? Also, why would they put [tex] \alpha [/tex] out there like that?

I hope I have made myself clear. I'm struggling on this one.

 

[dynamicsolo]

I'll try to create a similar problem that replicates the original problem. The original problem, is three coupled non-linear differential equations that must be solved numerically.

I see chaos (theory) in the offing...

If I have the following function:

[tex] \alpha \frac{d \phi}{d\theta} = A_{23} + B_{23} \sin \phi [/tex]

Now they say they have normalized [itex] A_{23}, \,\, B_{23} [/itex], and later they say that is a normlized voltage.

All right, [tex]\phi[/tex] must be dimensionless since it is the argument of a trig function, so A23 and B23 have the same units.

Since [tex] \theta = \omega t [/tex], then [tex]\theta[/tex] is also dimensionless, so [itex] \frac{d \phi}{d \theta} [/itex] is as well.

As for [tex] \alpha = \frac{h \omega}{2 e I_C R_n} [/tex] ,

it is well to keep in mind that Volts = Joules/Coulomb, so [tex]\alpha[/tex] has units of energy/energy, so it too is dimensionless. That means A23 and B23 are as well.

This normalized equation is also now "non-dimensionalized", so your output isn't in any units at all. You will need the values of your constants to interpret your results, as they will set the scale for the actual solution. (This sort of alteration of an equation is pretty commonplace. It is often done to keep the computed values "small", so there are fewer issues of precision or "overflow" in the computer registers -- or just so the model-maker can see the essential behavior of the differential equation without having to think a lot about the specific values. [In the days of analog computers, this was even more important.])

 

[FrogPad]

You are exactly right, this paper deals with chaos theory. Good nose ;)

My professor also mentioned how it is common place to normalize expressions, making them easier for computation. However, my lack of experience has kicked in hard here.

I still have a few questions if you don't mind.

I want to solve the 3 DE's in terms of time. They give [itex] \theta = \omega t [/itex]. So I should be able to rewrite the expression (the example given) as:

[tex] \alpha \frac{d \phi}{\omega dt} = A_{23} + B_{23} \sin \phi [/tex]

This allows me to expand (and cancel) [itex] \alpha [/itex] as:
[tex] \frac{h }{2 e I_C R_n} \frac{d \phi}{dt} = A_{23} + B_{23} \sin \phi [/tex]

Now [itex] \frac{h }{2 e I_C R_n} [/itex] has the units of seconds as I mentioned earlier. So is this a problem? This would not be unitless in this case correct?

Again, thank you very much for the response.

I see chaos (theory) in the offing...



All right, [tex]\phi[/tex] must be dimensionless since it is the argument of a trig function, so A23 and B23 have the same units.

Since [tex] \theta = \omega t [/tex], then [tex]\theta[/tex] is also dimensionless, so [itex] \frac{d \phi}{d \theta} [/itex] is as well.

As for [tex] \alpha = \frac{h \omega}{2 e I_C R_n} [/tex] ,

it is well to keep in mind that Volts = Joules/Coulomb, so [tex]\alpha[/tex] has units of energy/energy, so it too is dimensionless. That means A23 and B23 are as well.

This normalized equation is also now "non-dimensionalized", so your output isn't in any units at all. You will need the values of your constants to interpret your results, as they will set the scale for the actual solution. (This sort of alteration of an equation is pretty commonplace. It is often done to keep the computed values "small", so there are fewer issues of precision or "overflow" in the computer registers -- or just so the model-maker can see the essential behavior of the differential equation without having to think a lot about the specific values. [In the days of analog computers, this was even more important.])

 

[FrogPad]

You are exactly right, this paper deals with chaos theory. Good nose ;)

My professor also mentioned how it is common place to normalize expressions, making them easier for computation. However, my lack of experience has kicked in hard here.

I still have a few questions if you don't mind.

I want to solve the 3 DE's in terms of time. They give [itex] \theta = \omega t [/itex]. So I should be able to rewrite the expression (the example given) as:

[tex] \alpha \frac{d \phi}{\omega dt} = A_{23} + B_{23} \sin \phi [/tex]

This allows me to expand (and cancel) [itex] \alpha [/itex] as:
[tex] \frac{h }{2 e I_C R_n} \frac{d \phi}{dt} = A_{23} + B_{23} \sin \phi [/tex]

Now [itex] \frac{h }{2 e I_C R_n} [/itex] has the units of seconds as I mentioned earlier. So is this a problem? This would not be unitless in this case correct?

Again, thank you very much for the response.

I see chaos (theory) in the offing...
All right, [tex]\phi[/tex] must be dimensionless since it is the argument of a trig function, so A23 and B23 have the same units.

Since [tex] \theta = \omega t [/tex], then [tex]\theta[/tex] is also dimensionless, so [itex] \frac{d \phi}{d \theta} [/itex] is as well.

As for [tex] \alpha = \frac{h \omega}{2 e I_C R_n} [/tex] ,

it is well to keep in mind that Volts = Joules/Coulomb, so [tex]\alpha[/tex] has units of energy/energy, so it too is dimensionless. That means A23 and B23 are as well.

This normalized equation is also now "non-dimensionalized", so your output isn't in any units at all. You will need the values of your constants to interpret your results, as they will set the scale for the actual solution. (This sort of alteration of an equation is pretty commonplace. It is often done to keep the computed values "small", so there are fewer issues of precision or "overflow" in the computer registers -- or just so the model-maker can see the essential behavior of the differential equation without having to think a lot about the specific values. [In the days of analog computers, this was even more important.])

 

[dynamicsolo]

This allows me to expand (and cancel) [itex] \alpha [/itex] as:
[tex] \frac{h }{2 e I_C R_n} \frac{d \phi}{dt} = A_{23} + B_{23} \sin \phi [/tex]

Now [itex] \frac{h }{2 e I_C R_n} [/itex] has the units of seconds as I mentioned earlier. So is this a problem? This would not be unitless in this case correct?

Depending upon what you're wanting to do with this equation, I guess it's all right. There is no problem on the left-hand side because while [itex] \frac{h }{2 e I_C R_n} [/itex] is in seconds,

[tex]\frac{d \phi}{dt}[/tex] has units of (sec)^(-1), so the left-hand side is still dimensionless. (This factor now will set a scale for the solution.)

It looks like this leaves a separable differential equation (not so pretty, but separable...).

 

[FrogPad]

[tex]\frac{d \phi}{dt}[/tex] has units of (sec)^(-1), so the left-hand side is still dimensionless. (This factor now will set a scale for the solution.)

Could you elaborate on this?
So let's say that after performing the steps I talked about, we can reduce [itex] \alpha [/itex], and therefore get an expression to something like this:

[tex] (10^{-12}sec) \frac{d \phi}{dt} = A_{23} + B_{23} \sin \phi
[/tex]

It looks like this leaves a separable differential equation (not so pretty, but separable...).

Actually, the equations in the paper are much much longer, and add many new layers of complexity. So they must be solved numerically. And this is actually where a lot of my confusion comes in.

So let's say I want to solve the example equation given. It would be solved numerically over some time range. So if I choose a time range of [itex] [0,10] [/itex], and used a method (say Runge-Kutta), I will get a solution that is now an approximation (now it is discrete), that is valid for a time range between [0,10]. But, of what units? How can I say that these are seconds. Why not micro seconds, or 10^-10 seconds. To me it seems saying that the time range is in seconds is just as arbitrary as picking any other unit of time.

Thanks again.

 

[dynamicsolo]

Could you elaborate on this?

So let's say that after performing the steps I talked about, we can reduce [itex] \alpha [/itex], and therefore get an expression to something like this:

[tex] (10^{-12}sec) \frac{d \phi}{dt} = A_{23} + B_{23} \sin \phi
[/tex]

So let's say I want to solve the example equation given. It would be solved numerically over some time range. So if I choose a time range of [itex] [0,10] [/itex], and used a method (say Runge-Kutta), I will get a solution that is now an approximation (now it is discrete), that is valid for a time range between [0,10]. But, of what units? How can I say that these are seconds. Why not micro seconds, or 10^-10 seconds. To me it seems saying that the time range is in seconds is just as arbitrary as picking any other unit of time.

Sorry, I had to absent myself for a while: I had a grading deadline for early today.

What I'm saying is that, provided you know that value of omega you extracted, so that you know that constant multiplying [tex\frac{d \phi}{dt}[/tex], then you know the left-hand side still equals [tex]A_{23} + B_{23} \sin \phi [/tex] and that both sides are still the dimensionless value they were before. You could divide through by that constant, so that both sides now are in sec^(-1) and the values now have a defined scale.

But, for the purpose of computing, you could just leave that constant out altogether and calculate [tex]A_{23} + B_{23} \sin \phi [/tex], the original dimensionless expression. This may be easier for the computer to deal with by keeping the numbers small, letting you keep higher precision perhaps. For the purpose of interpretation, though, since you know what the omitted constant is, you can translate the results back into a physically sensible form. This is what I mean by the constant "setting the scale" of the problem.

I don't know whether that will be useful or helpful to you for your purposes, but it is one way of working with such equations.