Dirac delta solutions

Question

I am going through some lecture notes on Fourier transforms (here) and it is stated without proof (example 2.16 on page 29) that the general solution to the equation
$$x f(x) = a$$ is given by $$f(x) = a/x + b\, \delta(x)$$ and the general solution to
$$x^2 f(x) = a$$ is given by $$f(x) = a/x^2 + b \delta(x)/x + c \delta(x) + d\delta'(x).$$ I don't really understand how $\delta$ appears here. Could anyone please give a hint/proof?

I know that $\delta$ is a distribution so I tried integrating out both sides of these equations w.r.t $x$, but I failed to show that they give similar results. Also, I think an expression like $\delta(x)/x$ is ambiguous except if integrated against another function that can possibly cancel out the denominator.

Answer 1

First, if $f_1$ and $f_2$ are solutions to $Tf=g,$ where $T$ is some linear operator and $g$ is given, then $f_1-f_2$ is a solution to $Tf=0.$ Therefore we will study $x f(x)=0.$ It can easily be shown in distribution theory that $x\delta(x)=0,$ $x^2\delta(x)=0,$ and $x^2\delta'(x)=0,$ but since you're studying Fourier transforms I will give an explanation using Fourier transforms:

Take the equation $x f(x) = 0$ and apply the Fourier transform to both sides. You get $i\hat{f}'(\xi) = 0.$ This is a differential equation with solutions $\hat{f}(\xi) = C,$ where $C$ is a constant. Taking the inverse Fourier transform gives us $f(x) = C\delta(x).$

Likewise, $x^2 f(x) = 0$ transforms to $-\hat{f}''(\xi)=0$ with solutions $\hat{f}(\xi) = A\xi+B,$ i.e. $f(x) = -iA\delta'(x)+B\delta(x).$

Answer 2

So, informally, the Dirac $\delta$ is zero everywhere except at $0$ and has integral $1$. So, informally, $\delta$ is infinite at $0$, therefore $\delta$ is not admitted by traditional analysis. In regular analysis, given $x.f(x) = a$, we divide both sides by $x$ to obtain $f(x) = a/x$ but we can add any number (say $b$) times $\delta(x)$ on to $a/x$ as when $x$ is not zero any number times $\delta(x)$ is just $0$ and when $x$ is $0$ then $x.f(x)$ is still zero and so adding $b. \delta (x)$ on to $a/x$ does not change the truth of the fact that $x.f(x)=a$.

Now perhaps the other solution will make sense but it might help to know that if $\delta^{'}(x)$ is the derivative of the Dirac function then $\delta^{'}(x)=-\delta(x)/x$ so $\delta^{'}$ is 'even more infinite' than $\delta(x)$ :).

Answer 3

To complete the good answer given by md2perpe, you just need to get one particular solution of the equations. In dimension $1$ however, $1/x$ and $1/x^2$ are not locally integrable functions, and so one should define them as principal values (and one sometimes writes $\mathrm{P}(\tfrac{1}{x}) = \mathrm{pv.}(\tfrac{1}{x})$ and $\mathrm{fp.}(\tfrac{1}{x^2})$ for principal value and finite part). For any smooth and compactly supported function $\varphi$, they are defined by $$ \langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)}{x}\,\mathrm{d} x $$ which can also be written $\langle\mathrm{P}(\tfrac{1}{x}),\varphi\rangle = \lim_{\varepsilon\to 0}\int_{|x|>\varepsilon} \frac{\varphi(x)}{x}\,\mathrm{d} x$. One can easily verify that $$ x\, \mathrm{P}(\tfrac{1}{x}) = 1 $$

So the general solution for the first equation is $$ f(x) = a \, \mathrm{P}(\tfrac{1}{x}) + b \, \delta_0 $$

In the same way, one can define $$ \langle\mathrm{pf.}(\tfrac{1}{x^2}),\varphi\rangle = \int_{\mathbb{R}} \frac{\varphi(x)-\varphi(0)- x \varphi'(0)}{x^2}\,\mathrm{d} x $$ and then the general solution of the second equation is $$ f(x) = a \, \mathrm{pf.}(\tfrac{1}{x^2}) + b \, \delta_0 + c \, \delta_0' $$

Edit: $\delta_0(x)/x$ has no clear meaning in distribution theory. However, as indicated by Simon Terrington, one could define $\delta_0(x)/x = -\delta_0'(x)$ since it is one of the solution of the equation $$ x\,g(x) = -\delta_0(x). $$ The general solution being $g(x) = -\delta_0' + c\, \delta_0$. It is better to use just $\delta_0'$.

Answer 4

$\def\a{\alpha}$Here is another approach.

We begin by taking the Fourier transform of each side of the original equation: \begin{align*} x f(x) &= a \\ \int_{-\infty}^\infty x f(x) e^{ikx}dx &= \int_{-\infty}^\infty a e^{ikx}dx \\ -i\frac{\partial}{\partial k} \int_{-\infty}^\infty f(x) e^{ikx}dx &= 2\pi a\delta(k) \\ \hat f'(k) &= 2\pi i a \delta(k) \end{align*} This differential equation can be solved by standard methods, with the result \begin{align*} \hat f(k) &= 2\pi i a\Theta(k) + c, \end{align*} where $\Theta$ is the Heaviside step function.

All that remains is to perform the inverse transform: \begin{align*} f(x) &= \frac{1}{2\pi}\int_{-\infty}^\infty (2\pi i a \Theta(k)+c)e^{-ikx}dk \\ &= ia\int_{-\infty}^\infty \Theta(k)e^{-ikx}dk + c\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}dk \\ &= i a\left(-\frac{i}{x}+\pi\delta(x) \right)+c\delta(x) \\ &= \frac{a}{x} + b\delta(x), \end{align*} where $b=ia\pi+c$.

Aside: Fourier transform of the Heaviside step function

\begin{align*} \int_{-\infty}^\infty \Theta(k)e^{-ikx}dk &= \lim_{\a\rightarrow 0^+} \int_{-\infty}^\infty \Theta(k)e^{-\a k}e^{-ikx}dk \\ &= \lim_{\a\rightarrow 0^+} \int_0^\infty e^{-k(\a+i x)}dk \\ &= \lim_{\a\rightarrow 0^+} \left.\frac{-e^{-k(\a+i x)}}{\a+i x}\right|_0^\infty \\ &= \lim_{\a\rightarrow 0^+} \frac{1}{\a+i x}\\ &= \lim_{\a\rightarrow 0^+} \left(\frac{\a}{\a^2+x^2}-i\frac{x}{\a^2+x^2}\right) \\ &= \pi \delta(x) - \frac{i}{x}. \end{align*} In the last step we use that $\a/(\pi(\a^2+x^2))$ is a standard nascent delta function.