- #1
Math100
- 780
- 220
- 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.
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.