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[SOLVED] multi-scale perturbation

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{d^2x}{dt^2} + x + \epsilon\frac{dx}{dt}\left[1 - \left(\frac{dx}{dt}\right)^2 + \beta\left(\frac{dx}{dt}\right)^4\right] = 0,\quad\quad\epsilon\ll 1,
$$

Is there a smart way to do this problem? It will take forever to do directly.
 

pickslides

Member
Feb 1, 2012
57
You can reduce the order of your perturbation expansion to something manageable i.e. order 2 (or just use maple).

Otherwise, looks like you'll have to get your hands dirty.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You can reduce the order of your perturbation expansion to something manageable i.e. order 2 (or just use maple).

Otherwise, looks like you'll have to get your hands dirty.
How can I reduce it?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I have that $\left(\frac{dx}{dt}\right)^4 = \left(\frac{\partial^4}{\partial t^4} +4\epsilon \frac{\partial^4}{\partial t^3\partial T}+6\epsilon^2\frac{\partial^4}{\partial t^2\partial T^2} + 4\epsilon^2\frac{\partial^4}{\partial t\partial T^3} + \epsilon^4\frac{\partial^4}{\partial T^4}\right)(x_0^4 + 4\epsilon x_1x_0^3 + 4\epsilon^2x_0^2x_2+6\epsilon^2 x_1^2x_0^2+\cdots)$

Now when applying the differential, would I have $x_{0tttt}^4$? I am asking is the power affected?
Note $x(t,T)$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I ended up with this:
\begin{alignat*}{3}
R' & = & -\frac{1}{16}(5\beta R^5 + 8R)\\
\theta' & = & 0
\end{alignat*}

When beta is 9/40 it is supposed to be special. I don't see anything different in the plots or limit cycles.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I ended up with this:
\begin{alignat*}{3}
R' & = & -\frac{1}{16}(5\beta R^5 + 8R)\\
\theta' & = & 0
\end{alignat*}

When beta is 9/40 it is supposed to be special. I don't see anything different in the plots or limit cycles.
Here is how I got to this point: Is there a mistake? The professor had the $R_T = \frac{1}{16}(6R^3 - 5\beta r^2 - 8r)$.
However, I was extremely careful and don't see how you get that.

For $\frac{d^4}{dt^4}$, we have
$$
\frac{d^4}{dt^4} = \left(\frac{\partial^4}{\partial t^4} + 4\varepsilon \frac{\partial^4}{\partial t^3\partial T} + 6\varepsilon^2\frac{\partial^4}{\partial t^2\partial T^2} + 4\varepsilon^3\frac{\partial^4}{\partial t\partial T^3} + \varepsilon^4\frac{\partial^4}{\partial T^4}\right).
$$
We now have that
$$
\frac{d^2x}{dt^2} = x_{0tt} + \varepsilon x_{1tt} + 2\varepsilon x_{0tT} + \cdots,
$$
$$
x = x_0 + \varepsilon x_1 + \cdots,
$$
$$
\varepsilon\frac{dx}{dt} = \varepsilon x_{0t} + \cdots,
$$
$$
-\varepsilon\frac{dx}{dt} \left(\frac{dx}{dt}\right)^2 = -\varepsilon x^2_{0ttt} - 2\varepsilon x^2_{0ttT} - \cdots,
$$
and
$$
\varepsilon \beta \frac{dx}{dt} \left(\frac{dx}{dt}\right)^4 = \varepsilon \beta x^5_{0ttttt} + \cdots .
$$
Putting it all together now
$$
x_{0tt} + \varepsilon x_{1tt} + 2\varepsilon x_{0tT} + \cdots + x_0 + \varepsilon x_1 + \cdots + \varepsilon x_{0t} + \cdots -\varepsilon x^2_{0ttt} - 2\varepsilon x^2_{0ttT} - \cdots + \varepsilon\beta x^5_{0ttttt} + \cdots = 0
$$
$$
\begin{array}{lclcl}
\text{Order } 1 & : & x_{0tt} + x_0 & = & 0\\
& & & & \\
\text{Order } \varepsilon & : & x_{1tt} + x_1 & = & x^2_{0ttt} + 2x^2_{0ttT} - x_{0t} - 2x_{0tT} - \beta x^5_{0ttttt}
\end{array}
$$
From the order 1 term, we have that $x_0 = R[T]\cos[t + \theta(T)]$.
Making the substitution into the order $\varepsilon$ term yields:
\begin{alignat*}{3}
x_{1tt} + x_1 & = & R\sin[t + \theta(T)] + 8R^2\cos[t + \theta(T)]\sin[t + \theta(T)] - \beta (1800R^5\cos^2[t + \theta(T)]\sin^3[t + \theta(T)]\\
& & - 1205R^5\cos^4[t + \theta(T)]\sin[t + \theta(T)] - 120R^5\sin^5[t + \theta(T)]) + 2(R_T\sin[t + \theta(T)]\\
& & + R\theta_T\cos[t + \theta(T)]) - 2(4RR_T\cos[t + \theta(T)]^2 - 4R_TR\sin[t + \theta(T)]^2\\
& & - 8R^2\theta_T\cos[t + \theta(T)] \sin[t + \theta(T)])
\end{alignat*}
Let $t + \theta(T) = \theta$.
Then we have
\begin{alignat*}{3}
x_{1tt} + x_1 & = & R\sin\theta + 8R^2\cos\theta\sin\theta - \beta (1800R^5\cos^2\theta\sin^3\theta - 1205R^5\cos^4\theta\sin\theta - 120R^5\sin^5\theta)\\
& & + 2(R_T\sin\theta + R\theta_T\cos\theta) - 2(4RR_T\cos\theta^2 - 4R_TR\sin\theta^2 - 8R^2\theta_T\cos\theta \sin\theta)\\
& = & \left(R + \frac{5}{8}\beta R^5 + 2R_T\right)\sin t + 2R\theta_T\cos t + \text{other terms}
\end{alignat*}
In order to suppress resonance, we must have that
\begin{alignat*}{3}
R_T + \frac{1}{16}(5\beta R^5 + 8R) & = & 0\\
2R\theta_T & = & 0
\end{alignat*}
That is,
\begin{alignat*}{3}
R_T & = & -\frac{1}{16}(5\beta R^5 + 8R)\\
\theta_T & = & 0
\end{alignat*}
The fixed points of $R_T$ are when $R_* = 0, \pm\frac{-(1/5)^{1/4} 2^{3/4}}{\beta^{1/4}}, \pm\frac{(-2)^{3/4}}{5^{1/4}\beta^{1/4}}$.
When $\beta = \frac{9}{40}$, we will either have a limit cycle of radius of approximately 1.625 or the trajectories will go to $(0,0)$ in the phase plane.
The trajectories go to $(0,0)$ when the initial conditions $x_j$ is in the interior of a circle with radius 1.625.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{d^2x}{dt^2} + x + \epsilon\frac{dx}{dt}\left[1 - \left(\frac{dx}{dt}\right)^2 + \beta\left(\frac{dx}{dt}\right)^4\right] = 0,\quad\quad\epsilon\ll 1,
$$

Is there a smart way to do this problem? It will take forever to do directly.
$x(t,\varepsilon) = x_0(t,T)+\varepsilon x_1(t,T)+\cdots$ where the slow time is $T=\varepsilon t$. Let $f(x,x') = \frac{dx}{dt}\left[1 - \left(\frac{dx}{dt}\right)^2 + \beta\left(\frac{dx}{dt}\right)^4\right]$.
$$
\frac{d}{dt}=\frac{\partial }{\partial t} +\varepsilon\frac{\partial }{\partial T}
$$
Then
$$
x_{0tt} + \varepsilon x_{1tt} + 2\varepsilon x_{0tT} + \cdots + x_0 + \varepsilon x_1 +\cdots +\varepsilon f(x_0,x_{0t})+\cdots = 0
$$
\begin{alignat}{3}
x_{0tt} + x_0 & = & 0\\
x_{1tt} + x_1 & = & - 2x_{0tT} - f(x_0,x_{0t})
\end{alignat}
Then $x_0(t,\varepsilon)=R(T)\cos(t+\phi(T))$.
$$
x_{1tt} + x_1 = 2R'\sin(t+\phi) +2R\phi'\cos(t+\phi) - f(R\cos(t+\phi),-R\sin(t+\phi))
$$
Let $\theta = t+\phi$.
Then
$$
f(R\cos\theta,-R\sin\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty}[a_n\cos n\theta + b_n\sin n\theta]
$$
where resonance is at the $n = 1$ terms.
$$
\sin\theta(2R' - b_1) = 0\quad \cos\theta(2R\phi' -a_1) = 0
$$
So
\begin{alignat}{3}
R_T & = & \frac{1}{2\pi}\int_{-\pi}^{\pi}f(R\cos\theta,-R\sin\theta)\sin\theta d\theta\\
R\phi_T & = & \frac{1}{2\pi}\int_{-\pi}^{\pi}f(R\cos\theta,-R\sin\theta)\cos\theta d\theta
\end{alignat}
Solving for $R$ first,
\begin{alignat}{3}
R(T) & = & \frac{1}{2\pi}\int_{-\pi}^{\pi}(-R\sin\theta)(1-R^2\sin^2\theta+\beta R^4\sin^4\theta)\sin\theta d\theta\\
& = & \frac{1}{2\pi}\int_{-\pi}^{\pi}[-R\sin^2\theta + R^3\sin^4\theta - \beta R^5\sin^6\theta]d\theta
\end{alignat}
Using the orthonormal basis $\left\{\frac{1}{\sqrt{2}},\cos\theta,\cos 2\theta,\ldots,\sin\theta,\ldots\right\}$, we can obtain the integral by its inner products.
\begin{alignat}{3}
R_T & = & -R\langle\sin^2\theta\rangle + R^3\langle\sin^4\theta\rangle -\beta R^5\langle\sin^6\theta\rangle\\
& = & \frac{-R}{2} + \frac{3R^3}{8} - \frac{5\beta R^5}{16}\\
& = & \frac{R}{16}(6R^2 - 8 - 5\beta R^4)
\end{alignat}
Let $\omega = R^2$.
Then
$$
6\omega - 8 - 5\beta \omega^2 = 0\Rightarrow R^2 = \omega = \frac{-6\pm\sqrt{36 - 160\beta}}{10\beta} = \frac{-3\pm\sqrt{9 - 40\beta}}{5\beta}.
$$
So
$$
R_T = \pm\sqrt{\frac{-3\pm\sqrt{9 - 40\beta}}{5\beta}}.
$$
If $9 - 40\beta < 0$, we will only have one fixed point at $(0,0)$. So the critical $\beta$ is $\beta = \frac{9}{40}$. Therefore, $0\leq\beta\leq\frac{9}{40}$. When $\beta$ is in this range, we will have 5 fixed points.
\begin{alignat}{3}
\phi(T) & = & \frac{1}{2\pi}\int_{-\pi}^{\pi}[-\sin\theta\cos\theta + R^2\sin^3\theta\cos\theta - \beta R^4\sin^5\theta\cos\theta]d\theta\\
& = & -\langle\sin\theta\cos\theta\rangle + R^2\langle\sin^3\cos\theta\rangle - \beta R^4\langle\sin^5\theta\cos\theta\rangle\\
& = & \phi_0
\end{alignat}
 
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