- #1

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## Main Question or Discussion Point

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.

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.