Solving a recursive equation of motion

[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 $g=0$

Homework Equations

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.

 

[Ray Vickson]

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 $g=0$

Homework Equations

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.)

 

[rude man]

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

 

[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 Φ_i+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.

 

[rude man]

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! )

 

[Robin04]

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.

 

[Ray Vickson]

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?

 

[timetraveller123]

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]

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?

 

[vela]

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.

 

[rude man]

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.)

 

[Ray Vickson]

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]

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?

 

[Robin04]

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! :)

 

[Robin04]

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

 

[Ray Vickson]

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.