# Solving a recursive equation of motion

• Robin04
In summary: Thanks!)In summary, the equation of motion for a series of pendulums linked by a torsion spring is given by:-The equation of motion for the pendulum at the initial time is given by:-The equation of motion for the pendulum after a time step is given by:-The equation of motion for the pendulum after a second time step is given by:-The equation of motion for the pendulum after a third time step is given by:-The equation of motion for the pendulum after a fourth time step is given by:-The equation of motion for the pendulum after a fifth time step is given by:-The
Robin04

## Homework Statement

##\ddot{\Phi_i}-\frac{k}{ml^2}\Phi_i+\frac{k}{ml^2}\Phi_{i+1}=0##
This is the equation of motion of a series of pendulums linked by a torsion spring in the case where ##g=0##

## The Attempt at a Solution

I haven't really met this type of equation before. I should look for wave solutions but have no idea how to give this explicitly.

Robin04 said:

## Homework Statement

##\ddot{\Phi_i}-\frac{k}{ml^2}\Phi_i+\frac{k}{ml^2}\Phi_{i+1}=0##
This is the equation of motion of a series of pendulums linked by a torsion spring in the case where ##g=0##

## The Attempt at a Solution

I haven't really met this type of equation before. I should look for wave solutions but have no idea how to give this explicitly.

If you know ##\Phi_0## you get a simple, non-differential equation for ##\Phi_1##. Then you can take its appropriate derivatives to get a non-differential equation for ##\Phi_2##, etc. It may get impossibly messy after a few iterations (or it may not, depending on the exact form of ##\Phi_0.##)

Sometimes people attack such problem using "generating functions": let
$$G(z,t) = \sum_{i=0}^\infty z^i \Phi_i(t).$$ Then, with ##c = k/ml^2## and ##L = (d/dt)^2 -c## we have ##L\Phi_0 + (1/z) z c\Phi_1 = 0,## ##L \Phi_1 + (1/z) z^2 c \Phi_2 = 0,## etc. Summing, we get
$$L G(z,t) + \frac1 z c [G(z,t) - \Phi_0] = 0,$$
so
$$\frac{\partial^2 G}{{\partial t}^2} - c G + \frac{c}{z} G = \frac{c}{z} \Phi_0.$$
The left-hand-side is an ordinary SHM DE with ##\omega^2 = c(1-z)/z##. Maybe you can do something with that. (The point is that if you can manage to get an explicit solution for ##G(z,t)## and can then find it series expansion in the variable ##z##, the coefficients of that expansion will by your desired ##\Phi_(t)## functions.)

Robin04
I've never seen a finite-difference equation with a continuous time derivative in the same discrete variable. I would have expected a term Φi+2 instead, accounting for the second-order analogous ODE.

rude man said:
I've never seen a finite-difference equation with a continuous time derivative in the same discrete variable. I would have expected a term Φi+2 instead, accounting for the second-order analogous ODE.
http://www.physics.sk/aps/pubs/2011/aps-11-02/aps-11-02.pdf
Page 61 (in PDF).
It's just the Lagragian there, but this is the equation of motion you get from that in the case where g = 0. Oh and I made sign errors: ##\Phi_i## and ##\Phi_{i+1}## have the opposite signes.

Thanks.
I'm afraid this is beyond my ken.
It belongs in the Advanced Physics forum and I am reporting it as such. (Don't worry, you won't be reprimanded! )

rude man said:
Thanks.
I'm afraid this is beyond my ken.
It belongs in the Advanced Physics forum and I am reporting it as such. (Don't worry, you won't be reprimanded! )
Ah sorry. I'm used to posting in the introductory forum as I'm a first year undergraduate. Just doing this in an extracurricular project.

Robin04 said:
http://www.physics.sk/aps/pubs/2011/aps-11-02/aps-11-02.pdf
Page 61 (in PDF).
It's just the Lagragian there, but this is the equation of motion you get from that in the case where g = 0. Oh and I made sign errors: ##\Phi_i## and ##\Phi_{i+1}## have the opposite signes.
So, did you find my answer useless?

i am just making a guess experts correct me please thanks
by looking for solutions of the form of
##
\Phi_i = A_i e^{i \omega t}
##
we can get
##
-A_i \omega^2 = \frac{k}{m l ^2} (A_{i+1} - A_i)
##
this gives
##
A_{i+1} = c A_i
##
once again i am completely doubtful so don't trust this why is this wrong or correct?

Robin04
Ray Vickson said:
So, did you find my answer useless?
No, absolutely not. Sorry for my late reaction.
I tried it with a specific ##\Phi_1 = Asin{\omega t}## and I got ##\Phi_i=sin{\omega t}[A(1-\frac{ml^2\omega^2}{k})^{i-1}]## and it surprises me as I got no phase difference. Do you think this is correct in the g=0 case?

A traditional problem in classical mechanics is that of a series of masses connected by springs, and it's covered in upper-division textbooks. Your problem sounds similar, so it might be worth seeing how the mass-spring system is analyzed.

Robin04 said:
Ah sorry. I'm used to posting in the introductory forum as I'm a first year undergraduate. Just doing this in an extracurricular project.
If you're doing this as a 1st-yr undergraduate I predict a very bright career for you! Hope you go into physics or related & not to Wall Street!

(BTW I think I see now that the subscripts on Φ are pendulum identifiers so this makes sense now.. But still I'll leave it in others' hands.)

Robin04
Robin04 said:
No, absolutely not. Sorry for my late reaction.
I tried it with a specific ##\Phi_1 = Asin{\omega t}## and I got ##\Phi_i=sin{\omega t}[A(1-\frac{ml^2\omega^2}{k})^{i-1}]## and it surprises me as I got no phase difference. Do you think this is correct in the g=0 case?

You can do something more general: let ##\omega^2 = \frac{k}{ml^2}##. Then, setting ##D = d/dt##, your DE reads as
$$(D^2 - \omega^2) \Phi_i + \omega^2 \Phi_{i+1} = 0.$$ If you assume
$$\Phi_0 = A \cos(\mu t) + B \sin (\mu t)\hspace{4ex} (1)$$ we have
$$-(\mu^2+\omega^2) \Phi_0 + \omega^2 \Phi_1 = 0,$$ hence
$$\Phi_1 = \beta \Phi_0, \; \text{where} \; \beta = 1 + \frac{\mu^2}{\omega^2}.$$Similarly, ##\Phi_2 = \beta \Phi_1 = \beta^2 \Phi_0,## etc.
In general, ##\Phi_n = \beta^n \Phi_0,## for the specific form of ##\Phi_0## in eq. (1).

More generally, if you have a Fourier series expansion for ##\Phi_0## of the form
$$\Phi_0 = \frac{1}{2} A_0 + \sum_{i=1}^\infty \left( A_i \cos(i\, \mu t) + B_i \sin(i \, \mu t) \right),$$ then you have
$$\Phi_n = \frac{1}{2} A_0 + \sum_{i=1}^\infty \beta_i^n \left( A_i \cos(i\, \mu t) + B_i \sin(i\, \mu t) \right),$$
where
$$\beta_i = 1 + \frac{i^2 \,\mu^2}{\omega^2}.$$

Robin04
Ray Vickson said:
You can do something more general: let ##\omega^2 = \frac{k}{ml^2}##. Then, setting ##D = d/dt##, your DE reads as
$$(D^2 - \omega^2) \Phi_i + \omega^2 \Phi_{i+1} = 0.$$ If you assume
$$\Phi_0 = A \cos(\mu t) + B \sin (\mu t)\hspace{4ex} (1)$$ we have
$$-(\mu^2+\omega^2) \Phi_0 + \omega^2 \Phi_1 = 0,$$ hence
$$\Phi_1 = \beta \Phi_0, \; \text{where} \; \beta = 1 + \frac{\mu^2}{\omega^2}.$$Similarly, ##\Phi_2 = \beta \Phi_1 = \beta^2 \Phi_0,## etc.
In general, ##\Phi_n = \beta^n \Phi_0,## for the specific form of ##\Phi_0## in eq. (1).
Thank you, this looks more elegant. :)
More generally, if you have a Fourier series expansion for ##\Phi_0## of the form
$$\Phi_0 = \frac{1}{2} A_0 + \sum_{i=1}^\infty \left( A_i \cos(i\, \mu t) + B_i \sin(i \, \mu t) \right),$$ then you have
$$\Phi_n = \frac{1}{2} A_0 + \sum_{i=1}^\infty \beta_i^n \left( A_i \cos(i\, \mu t) + B_i \sin(i\, \mu t) \right),$$
where
$$\beta_i = 1 + \frac{i^2 \,\mu^2}{\omega^2}.$$
Why ##\beta_i## instead of ##\beta## ,and also why is the ##i^2## in the second term? Why didn't you multiply ##\frac{1}{2}A_0## with ##\beta##, or ##\beta_i^n## in your case?

rude man said:
If you're doing this as a 1st-yr undergraduate I predict a very bright career for you! Hope you go into physics or related & not to Wall Street!

(BTW I think I see now that the subscripts on Φ are pendulum identifiers so this makes sense now.. But still I'll leave it in others' hands.)
Aah, thank you, but I barely understand the big picture of what I'm doing, haha :D I just got into this topic a few weeks ago, but my mentor is very helpful and I really enjoy it. And I'll stay in physics for sure! :)

I learned a method for solving the forced harmonic oscillator with Green's function. Do you think it can be applied in this case?

Robin04 said:
Thank you, this looks more elegant. :)

Why ##\beta_i## instead of ##\beta## ,and also why is the ##i^2## in the second term? Why didn't you multiply ##\frac{1}{2}A_0## with ##\beta##, or ##\beta_i^n## in your case?

(1) If ##\Phi_0 = C,## a constant, what do you get for ##\Phi_1?##
(2) If you had ##\Phi_0 = \cos(2 \mu t)## --- instead of ##\cos(\mu t)## --- what would you get for ##\Phi_1?##
(3) As to the use of Green's function: in this case you really need an "anti-Green's function", because you do not have a differential equation for ##\Phi_1## with ##\Phi_0## on the right-hand-side. You have it the other way round. (Of course, you could use the DE to determine ##\Phi_0(t)## in terms of a given ##\Phi_1(t)##, and for doing that a Green's function might be helpful. However, that seems to have the problem backwards.)
(4) If you know about Green's functions as a first-year undergraduate, I am very impressed; I did not get that material until a 4th year "Mathematical Methods for Physics" course, back in the Stone Age.

Robin04
Sorry for my late reaction again. I realized that my equations of motion were wrong. I left out the term ##\Phi_{i-1}## which resulted in the counterintuitive solutions I got (having no phase difference between the pendulums). So I corrected my mistake and started to work on it, and I seem to be okay now. If you're interested the correct equation is: ##\ddot{\Phi_i} =-\frac{k}{ml^2}(2\Phi_i-\Phi_{i+1} - \Phi_{i-1})## I managed to solve it with some basic linear algebra.

## 1. What is a recursive equation of motion?

A recursive equation of motion is a mathematical representation of a system's motion that includes terms that depend on the system's previous position, velocity, and acceleration. It is used to describe the motion of objects in a dynamic system, such as a pendulum or a bouncing ball.

## 2. How do you solve a recursive equation of motion?

To solve a recursive equation of motion, you must first determine the initial conditions of the system, such as the initial position and velocity. Then, you can use a recursive formula to calculate the position, velocity, and acceleration at each time step. This process is repeated until you reach the desired time or position.

## 3. What are the applications of solving a recursive equation of motion?

Solving a recursive equation of motion is useful in various fields such as physics, engineering, and computer science. It can be used to model and predict the behavior of complex systems, design control systems, and simulate real-world scenarios.

## 4. What are the challenges of solving a recursive equation of motion?

One of the main challenges of solving a recursive equation of motion is ensuring that the initial conditions and the recursive formula accurately represent the system's behavior. Small errors in these values can lead to significant deviations in the predicted motion. Additionally, solving complex recursive equations can be computationally intensive and may require advanced mathematical techniques.

## 5. Are there any limitations to solving a recursive equation of motion?

One limitation of solving a recursive equation of motion is that it assumes a constant acceleration and does not account for external forces or factors that may affect the system's motion. Additionally, it may not accurately predict the behavior of systems with nonlinear or chaotic dynamics. It is essential to consider these limitations when using recursive equations of motion in practical applications.

Replies
2
Views
1K
Replies
16
Views
1K
Replies
10
Views
2K
Replies
4
Views
3K
Replies
2
Views
2K
Replies
7
Views
1K
Replies
1
Views
2K
Replies
6
Views
2K