HA7-P2

[Victor Ivrii]

Problem 2

http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter5/S5.2.P.html#problem-5.2.P.2

[Xi Yue Wang]

Noted that the Fourier transform of $f(x)$ is given as
correct formula
$$\hat{f}(\omega) = \frac{k}{2\pi}\int_{-\infty}^{\infty} f(x)e^{-i\omega x}dx$$
 where
$$k = 1, \omega = \frac{\pi n}{l}$$ WTH? There is no $l$ or $n$, $\omega\in \mathbb{R}$ is continuous
For part (a), given $\alpha > 0$
$$f(x) = e^{-\alpha |x|}\\
 \hat{f}(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-\alpha |x|}e^{-i\omega x} dx \\
= \frac{1}{2\pi}[\int_{-\infty}^{0} e^{\alpha x - i\omega x} dx + \int_{0}^{\infty} e^{-\alpha x - i\omega x} dx]\\
 = \frac{1}{2\pi}[\int_{0}^{\infty} e^{-\alpha x + i\omega x} dx + \int_{0}^{\infty} e^{-\alpha x - i\omega x} dx]\\
 = \frac{1}{2\pi}[\frac{1}{\alpha - i\omega}+\frac{1}{\alpha + i\omega}]\\ = \frac{\alpha}{\pi(\alpha^2 + \omega^2)}$$

[Xi Yue Wang]

For part (b), let $f(x)=e^{-\alpha|x|}$, for $g(x) = e^{-\alpha |x|}\cos(\beta x)$, where $\beta >0$.
$$\cos(\beta x) = \frac{1}{2}[e^{i\beta x} + e^{-i\beta x}]\\ g(x) = \frac{1}{2}[f(x)e^{i\beta x }+f(x)e^{-i\beta x}]\\\hat{g}(\omega) = \frac{1}{2}[\hat{f}(\omega - \beta)+\hat{f}(\omega +\beta)]$$
Combine the solution that we found in part (a), we have $$\\\hat{g}(\omega) = \frac{1}{2}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}+\frac{\alpha}{\pi(\alpha^2 + (\omega +\beta)^2)}]$$
Similarly, for $g(x) = e^{-\alpha |x|}\sin(\beta x)$, where $\beta >0$.$$\sin(\beta x) = \frac{1}{2i}[e^{i\beta x} - e^{-i\beta x}]\\\hat{g}(\omega) = \frac{1}{2i}[\hat{f}(\omega - \beta)-\hat{f}(\omega +\beta)]\\\hat{g}(\omega) = \frac{1}{2i}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}-\frac{\alpha}{\pi(\alpha^2 + (\omega +\beta)^2)}]$$

[Xi Yue Wang]

For part (c),
      Given $g(x) = xe^{-\alpha |x|}$ where $f(x) = e^{-\alpha |x|}$. Hence, $g(x) = xf(x)\ then, \hat{g}(\omega) = i\hat{f'}(\omega)$
We have $$\hat{f}(\omega) = \frac{\alpha}{\pi(\alpha^2 +\omega^2)}\\ \hat{f'}(\omega) = -\frac{2\alpha\omega}{\pi(\alpha^2+\omega^2)^2}$$
Then we get $$\hat{g}(\omega) = -\frac{2i\alpha\omega}{\pi(\alpha^2+\omega^2)^2}$$

[Xi Yue Wang]

For part (d), given $g(x) = xe^{-\alpha |x|}\cos(\beta x)$, where $f(x) =  e^{-\alpha |x|}\cos(\beta x)$
By theorem $\hat{g}(\omega) = i \hat{f'}(\omega)$
By part (b), we have $\hat{f}(\omega) = \frac{1}{2}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)}+\frac{\alpha}{\pi(\alpha^2+(\omega + \beta)^2)}]$
Then, $$ \hat{f'}(\omega) = -\frac{\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} + \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Hence, $$ \hat{g}(\omega) = -\frac{i\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} + \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Similarly, for $g(x) =  xe^{-\alpha |x|}\sin(\beta x)$, where $f(x) =  e^{-\alpha |x|}\sin(\beta x)$
By part (b), we have $\hat{f}(\omega) = \frac{1}{2i}[\frac{\alpha}{\pi(\alpha^2+(\omega - \beta)^2)} - \frac{\alpha}{\pi(\alpha^2+(\omega + \beta)^2)}]$
Then, $$ \hat{f'}(\omega) = -\frac{\alpha}{i\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} - \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$
Hence, $$ \hat{g}(\omega) = -\frac{\alpha}{\pi}[\frac{(\omega - \beta)}{(\alpha^2 + (\omega - \beta)^2)^2} - \frac{(\omega + \beta)}{(\alpha^2 + (\omega + \beta)^2)^2}]$$

[Catch Cheng]

For the f^(ω) formula in (a), I think there should be a negative sign before iωx, however the final answer is right.

[Xi Yue Wang]

For the f^(ω) formula in (a), I think there should be a negative sign before iωx, however the final answer is right.

Yes, I miss the negative sign in my code.