Integral Equation and Fourier Analysis

Question

I am trying to solve the following equation \begin{align*} F(\omega) G(\omega)= 2 \pi \delta(\omega)-2\pi \delta^{(2)}(\omega) \end{align*}

where $F(\omega)$ and $G(\omega)$ are Fourier transforms of $f(t)$ and $g(t)$ and $\delta(\omega)$ is Dirac delta function and $\delta^{(2)}(\omega)$ is the second distributional derivative of $\delta(\omega)$.

I would like to solve for $g(t)$. This is how I proceeded \begin{align*} F(\omega) G(\omega)&= 2 \pi \delta(\omega)-2\pi \delta^{(2)}(\omega)\\ G(\omega)&= 2 \pi \delta(\omega)\frac{1}{F(\omega)}-2\pi \delta^{(2)}(\omega) \frac{1}{F(\omega)}\\ g(t)&=\mathcal{F}^{-1}\left(2 \pi \delta(\omega)\frac{1}{F(\omega)}\right)+ \mathcal{F}^{-1}\left(2\pi \delta^{(2)}(\omega) \frac{1}{F(\omega)}\right) \end{align*}

Now my question is do $\mathcal{F}^{-1}\left(2 \pi \delta(\omega)\frac{1}{F(\omega)}\right)$ and $\mathcal{F}^{-1}\left(2\pi \delta^{(2)}(\omega) \frac{1}{F(\omega)}\right)$ exist? If not under what condition on $f(t)$ can this be solved? This question is somewhat realted to the question I asked before here.

Answer 1

Let $H(\omega)=1/F(\omega)$. Note that

$$H(\omega)\delta(\omega)=H(0)\delta(\omega)\tag{1}$$

and

$$H(\omega)\delta^{(2)}(\omega)=H(0)\delta^{(2)}(\omega)-2H'(0)\delta^{(1)}(\omega)+H''(0)\delta(\omega)\tag{2}$$

where $H'(\omega)$ and $H''(\omega)$ are the first and second derivatives of $H(\omega)$, respectively. So your equation can be solved if $H(0)$, $H'(0)$, and $H''(0)$ exist and are finite. From (1) and (2), the solution is given by

$$G(\omega)=2\pi\delta(\omega)H(\omega)-2\pi\delta^{(2)}(\omega)H(\omega)=\\=2\pi\delta(\omega)(H(0)-H''(0))+4\pi\delta^{(1)}(\omega)H'(0)-2\pi\delta^{(2)}(\omega)H(0)\tag{3}$$

With the Fourier transform relation

$$\delta^{(n)}(\omega)\Longleftrightarrow \frac{1}{2\pi i^n}t^n\tag{4}$$

you obtain

$$g(t)=H(0)-H''(0)-2iH'(0)\cdot t +H(0)\cdot t^2$$

EDIT (explanation of Eq. (2)):

You can apply the product rule also to generalized derivatives, i.e.

$$[H(\omega)\delta(\omega)]^{(1)}=H'(\omega)\delta(\omega)+H(\omega)\delta^{(1)}(\omega)\tag{a}$$

And because $H(\omega)\delta(\omega)=H(0)\delta(\omega)$, (1) also equals $H(0)\delta^{(1)}(\omega)$. So

$$H(0)\delta^{(1)}(\omega)=H'(\omega)\delta(\omega)+H(\omega)\delta^{(1)}(\omega)$$

which gives

$$H(\omega)\delta^{(1)}(\omega)=H(0)\delta^{(1)}(\omega)-H'(\omega)\delta(\omega)\tag{b}$$

You can derive $H(\omega)\delta^{(2)}(\omega)$ in a similar way by considering $[H(\omega)\delta(\omega)]^{(2)}$ and by using the result (b). The general result is

$$H(\omega)\delta^{(n)}(\omega)=\sum_{k=0}^n(-1)^{k}\frac{n!}{k!(n-k)!}f^{(k)}(0)\delta^{(n-k)}(\omega)$$