- #1
Math100
- 807
- 224
- Homework Statement
- Consider the equation ## \ddot{x}+x+\epsilon(\alpha x^2\operatorname{sgn}(x)+\beta x^3)=0, 0<\epsilon<<1, \alpha>0, \beta>0 ##, where ## \alpha ## and ## \beta ## are constants, and the signum function, ## \operatorname{sgn}(x) ##, is defined by ## \operatorname{sgn}(x)=1 ## for ## x>0 ##, ## \operatorname{sgn}(x)=0 ## for ## x=0 ## and ## \operatorname{sgn}(x)=-1 ## for ## x<0 ##. With initial condition ## x(0)=0 ##, show that solutions with angular frequency ## \omega ## can be expressed as a Fourier series of the form ## x(t)=\sum_{k=1}^{\infty}B_{2k-1}\sin((2k-1)\omega t)=B_{1}\sin\omega t+B_{3}\sin 3\omega t+\cdots ##.
- Relevant Equations
- None.
Proof:
Let ## \epsilon=0 ##.
Then the unperturbed equation is ## \ddot{x}+x=0 ## and the general solution is
## x(t)=A\sin\omega t+B\cos\omega t ## where ## \omega=1 ## is the angular frequency
with the constants ## A ## and ## B ##.
With the initial condition ## x(0)=0 ##, we have that ## B=0 ##.
This gives ## x(t)=A\sin\omega t ## where ## A ## is a constant.
Note that the function of ## x(t)=A\sin\omega t ## is periodic and odd.
From the work/proof shown above, how should I show that solutions with angular frequency ## \omega ##
can be expressed as a Fourier series of the given form on the problem? What needs to be done?
Let ## \epsilon=0 ##.
Then the unperturbed equation is ## \ddot{x}+x=0 ## and the general solution is
## x(t)=A\sin\omega t+B\cos\omega t ## where ## \omega=1 ## is the angular frequency
with the constants ## A ## and ## B ##.
With the initial condition ## x(0)=0 ##, we have that ## B=0 ##.
This gives ## x(t)=A\sin\omega t ## where ## A ## is a constant.
Note that the function of ## x(t)=A\sin\omega t ## is periodic and odd.
From the work/proof shown above, how should I show that solutions with angular frequency ## \omega ##
can be expressed as a Fourier series of the given form on the problem? What needs to be done?