MHB Can this perturbation problem be solved using a multi-scale approach?

[Dustinsfl]

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

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

 

[pickslides]

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.

 

[Dustinsfl]

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?

 

[Dustinsfl]

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 $x(t,T)$.

 

[Dustinsfl]

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.

 

[Dustinsfl]

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.

 

[Dustinsfl]

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

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}