- #1
EnriqueOrtizMartinez
- 2
- 0
- Homework Statement
- From the equation:
$$\tilde{u}(\xi ,t)=\tilde{f}(\xi)cos(ct\xi)+\tilde{g}(\xi)(c\xi)^{-1}sen(ct\xi)$$
Use the inverse Fourier transform to obtain the D'Alembert equation:
$$u(x,t)=\frac{1}{2}[f(x-ct)+f(x+ct)]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi (y)dy$$
- Relevant Equations
- $$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty }\tilde{u}(\xi ,t)e^{-i\xi x}d\xi$$
Well what I did was first use the inverse Fourier transform:
$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty }\tilde{u}(\xi ,t)e^{-i\xi x}d\xi$$
I substitute the equation that was given to me by obtaining:$$u(x,t)=\frac{1}{2\pi }\left \{ \int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)+\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t) \right \}e^{-i \xi x}d\xi$$$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)e^{-i \xi x}d\xi+\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t)e^{-i \xi x}d\xi$$I solve the integral:$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)e^{-i \xi x}d\xi$$He used the identity:$$cos(c\xi t)=\frac{e^{ict\xi }+e^{-ict\xi }}{2}$$$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{f(\xi)}}{2}\left [ e^{ict\xi }+e^{-ict\xi } \right ]e^{-i \xi x}d\xi$$$$u(x,t)=\frac{1}{2 }\left \{ \frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)e^{-i\xi (x- ct)}d\xi+\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)e^{-i\xi (x+ct)}d\xi \right \}$$
$$\therefore u(x,t)=\frac{1}{2}\left [ f(x-ct)+f(x+ct) \right ]$$Now I solve:
$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t)e^{-i \xi x}d\xi$$But I do not know how to do it, could you help me as I do to solve this integral to find the factor:$$\frac{1}{2c}\int_{x-ct}^{x+ct}\psi (y)dy$$I have been investigating and as the convolution is used, but it does not work out, they could help me, in advance thanks
$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty }\tilde{u}(\xi ,t)e^{-i\xi x}d\xi$$
I substitute the equation that was given to me by obtaining:$$u(x,t)=\frac{1}{2\pi }\left \{ \int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)+\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t) \right \}e^{-i \xi x}d\xi$$$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)e^{-i \xi x}d\xi+\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t)e^{-i \xi x}d\xi$$I solve the integral:$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)cos(c\xi t)e^{-i \xi x}d\xi$$He used the identity:$$cos(c\xi t)=\frac{e^{ict\xi }+e^{-ict\xi }}{2}$$$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{f(\xi)}}{2}\left [ e^{ict\xi }+e^{-ict\xi } \right ]e^{-i \xi x}d\xi$$$$u(x,t)=\frac{1}{2 }\left \{ \frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)e^{-i\xi (x- ct)}d\xi+\frac{1}{2\pi }\int_{-\infty }^{\infty}\tilde{f}(\xi)e^{-i\xi (x+ct)}d\xi \right \}$$
$$\therefore u(x,t)=\frac{1}{2}\left [ f(x-ct)+f(x+ct) \right ]$$Now I solve:
$$u(x,t)=\frac{1}{2\pi }\int_{-\infty }^{\infty}\frac{\tilde{g}(\xi )}{c\xi}sen(c\xi t)e^{-i \xi x}d\xi$$But I do not know how to do it, could you help me as I do to solve this integral to find the factor:$$\frac{1}{2c}\int_{x-ct}^{x+ct}\psi (y)dy$$I have been investigating and as the convolution is used, but it does not work out, they could help me, in advance thanks