How should I use the averaging approximation to find this?

[Math100]

Homework Statement

Consider the differential equation $\ddot{x}+\epsilon h(x, \dot{x})+x=0, 0<\epsilon<<1$. Let $x(t)$ and $y(t)=\dot{x}(t)$ be expressed in terms of the polar coordinates $a(t)$ and $\theta(t)$ through $x(t)=a(t)\cos\theta(t), y(t)=a(t)\sin\theta(t)$. Note that $a(t)$ and $\theta(t)$ satisfy the following differential equations:
$\dot{a}=-\epsilon\sin\theta h(a\cos\theta, a\sin\theta)$,
$\dot{\theta}=-1-\frac{\epsilon}{a}\cos\theta h(a\cos\theta, a\sin\theta)$.

a) By letting $\theta(t)=\psi(t)-t$, transform Equations $\dot{a}=-\epsilon\sin\theta h(a\cos\theta, a\sin\theta)$ and $\dot{\theta}=-1-\frac{\epsilon}{a}\cos\theta h(a\cos\theta, a\sin\theta)$ into non-autonomous differential equations for $a(t)$ and $\psi(t)$. Hence use an averaging theorem to show that provided that $\epsilon$ is small enough, Equations $\dot{a}=-\epsilon\sin\theta h(a\cos\theta, a\sin\theta)$ and $\dot{\theta}=-1-\frac{\epsilon}{a}\cos\theta h(a\cos\theta, a\sin\theta)$ can be well approximated by $\dot{a}=-\epsilon p_{0}(a), \dot{\theta}=-1-\frac{\epsilon r_{0}(a)}{a}$, where $p_{0}(a)=\frac{1}{2\pi}\int_{0}^{2\pi}\sin u h(a\cos u, a\sin u)du, r_{0}(a)=\frac{1}{2\pi}\int_{0}^{2\pi}\cos u h(a\cos u, a\sin u)du$.

b) Now consider the equation $\ddot{x}+\epsilon(x^4-b^4)\dot{x}+x=\epsilon\gamma x\dot{x}^4, 0<\epsilon<<1, b>0$, where $b$ and $\gamma$ are constants. Use the averaging approximation of part (a) to show that the system has a stable limit cycle with amplitude $A$ and angular frequency $\omega$ given approximately by $A=8^{1/4}b, \omega=1-\frac{1}{2}\epsilon\gamma b^4$. (Hint: Note the following identity: $cos^{4}\theta\sin^{2}\theta=\frac{1}{16}+\frac{1}{32}\cos 2\theta-\frac{1}{16}\cos 4\theta-\frac{1}{32}\cos 6\theta.$)

Relevant Equations

Not given.

a) Proof:

Consider the equations $\dot{a}=-\epsilon\sin\theta h(a\cos\theta, a\sin\theta)$ and $\dot{\theta}=-1-\frac{\epsilon}{a}\cos\theta h(a\cos\theta, a\sin\theta)$.
Let $\theta(t)=\psi(t)-t$.
Then $\dot{\theta}(t)=\dot{\psi}(t)-1$.
By direct substitution of $\dot{\theta}(t)=\dot{\psi}(t)-1$, we have $\dot{\psi}(t)-1=-1-\frac{\epsilon}{a}\cos\theta h(a\cos\theta, a\sin\theta)\implies \dot{\psi}(t)=-\frac{\epsilon}{a}\cos(\psi(t)-t)\cdot h(a(t)\cos(\psi(t)-t), a(t)\sin(\psi(t)-t))$.
Hence, the non-autonomous differential equations for $a(t)$ and $\psi(t)$ are $\dot{a}(t)=-\epsilon\sin(\psi(t)-t)\cdot h(a(t)\cos(\psi(t)-t), a(t)\sin(\psi(t)-t))$ with $\dot{\psi}(t)=-\frac{\epsilon}{a}\cos(\psi(t)-t)\cdot h(a(t)\cos(\psi(t)-t), a(t)\sin(\psi(t)-t))$.

b) Proof:

Consider the equation $\ddot{x}+\epsilon(x^4-b^4)\dot{x}+x=\epsilon\gamma x\dot{x}^4, 0<\epsilon<<1, b>0$, where $b$ and $\gamma$ are constants.
By definition, the system $\ddot{x}+\epsilon h(x, \dot{x})+x=0, 0<\epsilon<<1$ has an approximately circular limit cycle given by the equations $x=a_{0}\cos\omega t, y=\dot{x}=-a_{0}\sin\omega t, \omega=1$, where $a_{0}$ satisfies the equation $g(a)=0$ such that $g(a)=\epsilon a\int_{0}^{2\pi} h(a\cos t, -a\sin t)\sin t dt$. The limit cycle is stable if $g'(a_{0})<0$ and unstable if $g'(a_{0})>0$.
Note that the first-order approximation to the frequency $\omega$ of the limit cycle equation is given by $\omega=1+\frac{\epsilon}{2\pi a_{0}}\int_{0}^{2\pi} h(a_{0}\cos\theta, a_{0}\sin\theta)\cos\theta d\theta$.
Then we have $h(x, y)=h(x, \dot{x})=(x^4-b^4)\dot{x}$ where $x=a\cos t$ and $y=\dot{x}=-a\sin t$.
This gives $g(a)=\epsilon a\int_{0}^{2\pi} h(a\cos t, -a\sin t)\sin t dt=\epsilon a\int_{0}^{2\pi} (a^{4}\cos^{4} t-b^{4})(-a\sin t) dt=-\epsilon a^{2}\int_{0}^{2\pi}(a^{4}\cos^{4} t-b^{4})\sin t dt$.
Observe that $g(a)=-\epsilon a^{2}\int_{0}^{2\pi}(a^{4}\cos^{4}t-b^{4})\sin t dt\implies g(a)=-\epsilon a^{2}(\int_{0}^{2\pi} a^{4}\cos^{4} t\cdot\sin t dt)+\epsilon a^{2}\int_{0}^{2\pi} b^{4}\sin t dt\implies g(a)=-\epsilon a^{6}\int_{0}^{2\pi}\cos^{4} t\cdot\sin t dt+\epsilon a^{2}b^{4}\int_{0}^{2\pi}\sint dt\implies g(a)=-\epsilon a^{6}(-\frac{1}{5}+\frac{1}{5})+\epsilon a^{2}b^{4}(-1+1)$.
Hence, $g(a)=0$.

Up to here, I don't think the result I've got from part b) is correct, since $g(a)=0$. But I want to know what exactly is the averaging approximation from part (a) that I should use/apply in part b) to obtain the given amplitude and the angular frequency $\omega$. Also for part (a), after I've got non-autonomous differential equations, how should I use/apply the averaging theorem? I have limited resource and couldn't find the exact definition for the averaging theorem from my book.

 

[pasmith]

What "averaging theorems" do you know?

 

[Math100]

What "averaging theorems" do you know?

I think I've made mistakes in part b). Here's what I've revised so far:
Part b) Proof:

By definition, the system $\ddot{x}+\epsilon h(x, \dot{x})+x=0, \lvert\epsilon\rvert<<1$ has an approximately circular limit cycle given by the equations $x=a_{0}\cos\omega t, y=\dot{x}=-a_{0}\sin\omega t, \omega=1$, where $a_{0}$ satisfies the equation $g(a)=0$ such that $g(a)=\epsilon a\int_{0}^{2\pi} h(a\cos t, -a\sin t)\sin t dt$. The limit cycle is stable if $g'(a_{0})<0$ and unstable if $g'(a_{0})>0$.
Consider the equation $\ddot{x}+\epsilon(x^4-b^4)\dot{x}+x=\epsilon\gamma x\dot{x}^{4}, 0<\epsilon<<1, b>0$, where $b$ and $\gamma$ are constants.
Then we have $h(x, y)=h(x, \dot{x})=(x^4-b^4)\dot{x}-\gamma x\dot{x}^{4}$, because $\ddot{x}+\epsilon(x^4-b^4)\dot{x}+x=\epsilon\gamma x\dot{x}^{4}\implies \ddot{x}+\epsilon(x^4-b^4)\dot{x}+x-\epsilon\gamma x\dot{x}^{4}=0\implies \ddot{x}+\epsilon[(x^4-b^4)\dot{x}-\gamma x\dot{x}^{4}]+x=0$.
Let $x=a\cos t$ and $y=\dot{x}=-a\sin t$.
Note that $h(x, y)=h(x, \dot{x})=(x^4-b^4)\dot{x}-\gamma x\dot{x}^{4}\implies h(x, y)=h(x, \dot{x})=(a^4\cos^{4} t-b^4)(-a\sin t)-\gamma(a\cos t)(-a\sin t)^{4}$.
This gives $h(x, y)=h(x, \dot{x})=ab^{4}\sin t-a^{5}\cos^{4} t\cdot\sin t-\gamma a^{5}\cos t\cdot\sin^{4} t$.
Observe that $g(a)=\epsilon a\int_{0}^{2\pi} h(a\cos t, -a\sin t)\sin t dt\implies g(a)=\epsilon a\int_{0}^{2\pi}(ab^{4}\sin t-a^{5}\cos^{4} t\cdot\sin t-\gamma a^{5}\cos t\cdot\sin^{4} t)\sin t dt\implies g(a)=\epsilon a^{2}\int_{0}^{2\pi}(b^{4}\sin^{2} t-a^{4}\cos^{4} t\cdot\sin^{2} t-\gamma a^{4}\cos t\cdot\sin^{5} t)dt\implies g(a)=\epsilon a^{2}[b^{4}\int_{0}^{2\pi}\sin^{2} t dt-a^{4}\int_{0}^{2\pi}\cos^{4} t\cdot\sin^{2} t dt-\gamma a^{4}\int_{0}^{2\pi}\cos t\cdot\sin^{5} t dt]\implies g(a)=\epsilon a^{2}[b^{4}(\pi-0)-a^{4}(\frac{24\pi}{192}-0)-\gamma a^{4}(0)]$.
Hence, $g(a)=\epsilon a^{2}(b^{4}\pi-\frac{24a^{4}\pi}{192})=0\implies g(a)=\epsilon a^{2}(b^{4}\pi-\frac{a^{4}\pi}{8})=0\implies g(a)=\epsilon a^{2}\pi(b^{4}-\frac{a^{4}}{8})=0$, where $b^{4}-\frac{a^{4}}{8}=0$ for $a\neq 0$, so $a^{4}=8b^{4}\implies a=\pm 2^{3/4}\cdot b$.
Since there are limit cycles at $a=\pm 2^{3/4}\cdot b$, it follows that $g'(a)=\epsilon\pi(2ab^{4}-\frac{3a^{5}}{4})$.

Up to here, what should I do for part b) in order to find the given amplitude $A$ and the angular frequency $\omega$? As for part a) and the averaging theorem, I've done research and found out that the averaging theorem talks about the system with the following form: $\dot{x}=\epsilon f(x, t, \epsilon), 0\leq\epsilon<<1$ of a phase space variable $x$. The fast oscillation is given by $f$ versus a slow drift of $\dot{x}$. The averaging method yields an autonomous dynamical system $\dot{y}=\epsilon\frac{1}{T}\int_{0}^{T} f(y, s, 0)ds=: \epsilon\bar{f}(y)$. But how should I apply this on part a)?

 

[pasmith]

How does the unperturbed system in part (a) behave?