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Multiple Scales and Non-Linear Oscillator.

  1. May 21, 2008 #1
    Hello everyone. My question is quite long, so please bear with me; my professor is very busy and cannot help me at the moment, and I can't contact the course tutor.

    We have the DE

    [tex]\ddot \theta + \alpha \dot \theta + \sin{\theta } = \epsilon \cos{\omega t} [/tex]

    where theta is the angle the pendulum makes with the vertical, and the RHS represents the driving force. [tex]\omega [/tex] is not near any resonant frequencies. [tex]\epsilon [/tex] is a small perturbation, and [tex]\alpha [/tex] is small.

    We should solve this DE for small angles theta. The assignment says
    [tex]\theta (t) = \epsilon \phi (t) [/tex] and [tex]\phi [/tex] is of order 1.

    So, the Taylor seres expansion of sine gives

    [tex]\sin{\theta } = \theta - \theta ^3/6 + ...[/tex]

    I am doing a first order expansion (only keeping terms up to the order of epsilon^1), therefore we can keep only the first term in the expansion, as [tex]\sin{\theta}[/tex] is of the order [tex]\epsilon[/tex]. But then the governing DE is linear! Is this correct, because this is meant to be a non-linear oscillator assignment? Anyway, if I then keep the first two terms in the expansion of sine we get

    [tex] \ddot \theta + \alpha \dot \theta + \theta - \frac{\theta ^3}{6} = \epsilon \cos{\omega t} [/tex]

    If I try to solve this directly by the multiple scale method, I do
    [tex]\theta = \theta _0 + \epsilon \theta _1[/tex]
    where [tex]\theta _n =\theta _n(T_0,T_1,...)[/tex] and [tex]T_n=\epsilon ^nt[/tex].
    If I define
    [tex]D_n = \frac{\partial }{\partial T_n} [/tex]
    then multiple scales gives:
    collecting terms of coefficient [tex]\epsilon ^0[/tex]:

    [tex]{D_0}^2\theta _0 + \alpha D_0\theta _0 + \theta_0 - \frac{\theta_0 ^3}{6} = 0[/tex]

    This is just the homogeneous form of the governing DE. Therefore, if you try solving this by multiple scales, and collecting terms [tex]\epsilon ^0[/tex] again, you will find the method requires you to solve the exact same DE again, an infinite regression of the same DE! So how do we go about this? I could use a straight expansion [tex]\theta = \theta + \epsilon \theta [/tex] but I encounter the same problem. I cannot use other techniques to solve this.

    Now, if instead I make the substitution [tex]\theta = \epsilon \phi [/tex], the governing DE becomes

    [tex]\ddot \phi + \alpha \dot \phi + \phi - \frac{\epsilon ^2\phi ^3}{6} = \cos{\omega t} [/tex]

    Since I only do a first order expansion, this reduces to

    [tex]\ddot \phi + \alpha \dot \phi + \phi = \cos{\omega t}[/tex]

    But this is simply the linear, damped, driven oscillator!! So solving this gives an amplitude and phase which are constants, and that is obviously not correct for a non-linear oscillator, surely? Help, Im confused.
     
  2. jcsd
  3. May 21, 2008 #2

    Mute

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    Homework Helper

    Does the assignment itself give you any indication that you want a solution beyond the linear approximation?

    If you want more than a linear approximation, you could suppose that [itex]\theta = \epsilon \phi_0 + \epsilon^2 \phi_1 + \dots[/itex], and then collect terms in each order of epsilon, and suppose that they approximately vanish separately (i.e., collect all the [itex]\mathcal{O}(\epsilon)[/itex] terms, that gives you one DE to solve, then collect all the [itex]\mathcal{O}(\epsilon^2)[/itex] terms, giving you another DE to solve, which will require your solution to the first DE). For instance, for [itex]\phi_0[/itex] you would get the same linear equation you just had, but you would also generate an equation for [itex]\phi_1[/itex]:

    [tex]\ddot{\phi}_1 + \alpha \dot{\phi_1} + \phi_1 = \frac{\phi_0^3}{6}[/tex]

    From your first equation you have the solution for [itex]\phi_0[/itex], so you plug that into this expression and solve the equation for [itex]\phi_1[/itex]. You can keep generating equations in this manner to keep adding corrections to your linear solution. You have to be careful not to lose any terms, though. This means as you go to higher order corrections, you'll have to keep more terms in your expansion of sine, etc. (So, check to make sure I didn't miss any terms in the [itex]\mathcal{O}(\epsilon^2)[/itex] expression above - I did it quickly). There might be a caveat about the expansion - something to do with making sure you don't generate any resonant terms, but I can't remember if this is already taken care of the way I wrote things.

    But anyways, that's the general idea you can use.
     
    Last edited: May 21, 2008
  4. May 22, 2008 #3
    Ok, so it is established that the zeroth order gives a linear DE to solve. For the first order term in epsilon, I do not get the same expression as you. Is that what you have when you eliminate secular terms? Because I get something very complicated.
     
  5. May 22, 2008 #4

    Mute

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    Like I said, I did that quickly. Looking at it again, I might have made a mistake with the orders of terms, namely that DE I wrote for [itex]\phi_1[/itex] shouldn't have the [itex]\phi_0[/itex] term in it - that's a next order term, I think. I think the following should be mostly correct, but I am a bit rusty at this so I may have missed a few details:

    Let [itex]\theta = \epsilon \phi_0 + \epsilon^2 \phi_1 + \hat{\phi}[/itex], where [itex]\phi_0, \phi_1 = \mathcal{O}(1)[/itex] and [itex]\hat{\phi} = \mathcal{O}(\epsilon^3)[/itex]. Your DE is

    [tex]\ddot{\theta} + \alpha \dot{\theta} + \left[\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots\right] = \epsilon \cos \omega t[/tex]

    Suppose we keep only order 1 terms in [itex]\theta/\epsilon[/itex]. That gives your original linear DE [itex]\ddot{\phi_0} + \alpha \dot{\phi_0} + \phi_0 = \cos \omega t[/itex]. If you keep order epsilon terms, then you get [itex]\ddot{\phi_1} + \alpha \dot{\phi_1} + \phi_1 = 0[/itex] - there is no contribution from higher order terms in the sine since [itex]\theta^3/\epsilon[/itex] is order epsilon^2 at best. If you let [itex]\hat{\phi} = \epsilon^3 \phi_2 + \hat{\phi}'[/itex], where [itex]\hat{\phi}' = \mathcal{O}(\epsilon^4)[/itex], then the next order in epsilon equation is

    [tex]\ddot{\phi}_2 + \alpha \dot{\phi_2} + \phi_2 = \frac{\phi_0^3}{6}[/tex]

    and so on. (No other terms from [itex]\theta^3[/itex] appear since they are all next order in epsilon. If you expand to fourth order in epsilon you will have to consider other terms that go as [itex]\phi_0\phi_1[/itex], for instance. There are no contributions from the [itex]\theta^5[/itex] term until you expand to fifth order in epsilon).

    Essentially what I'm suggesting is a power series solution in the small parameter [itex]\epsilon[/itex]:

    [tex]\theta = \epsilon \sum_{n=0}^\infty \phi_n \epsilon^{n}[/tex]

    The right hand side of your equation is also treated as a power series sum in [itex]\epsilon[/itex], of which only one term is nonzero. You then suppose that you can get your solution by equating coefficients of each power series that are the same order in [itex]\epsilon[/itex].

    One thing to keep in mind here is that this is all still effectively a small angle approximation; you're just calculating small nonlinear corrections to the solution.
     
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