Distributional derivative of $\sin { (\pi|x|/2))} \chi_{\{|x|<2\}}$

Question

I have to find the first and the second distributional derivative of the function : $$f(x) = \begin{cases} \sin { (\pi|x|/2))} \quad & |x| <2 \\ 0 \quad & |x| \geq 2 \end{cases}$$ but I am not sure if I am doing it in the right way.

The first derivative should be
$$ g'(x) = \cos { (\pi|x|/2))}\operatorname{sgn}(x) \pi/2 $$
because $f(x)$ is a continuous function with no jumps in any point so the delta does not appear.

The second derivative should be :

$$ g''(x) = - \sin { (\pi|x|/2))}[\operatorname{ sgn}(x)]^2 (\pi/2)^2 + \delta _2 -\delta _{(-2)} $$

I have never seen a distribution with jumps in different points before, am I correct if I apply the delta in each of this points? Also I am not sure if the jump is $+1$ or $-1$.

Answer 1

The distributional derivative of a piecewise smooth function consists of two terms:

• pointwise derivative, found as in calculus
• singular part: if $f$ is discontinuous at $a$, add $(f(a+)-f(a-))\delta_a$; this is done for each discontinuity.

You correctly found that $f'$ has no singular part. However, note that the formula for $f'$ should also say that it's zero when $|x|>2$.

The function $f'$ is discontinuous at three points: $-2,0,2$.

• at $-2$ the left limit is $0$, the right limit is $\pi/2$
• at $0$ the left limit is $-\pi/2$, the right limit is $\pi/2$
• at $ 2$ the left limit is $-\pi/2$, the right limit is $0$

So, the singular part of $f''$ is $$\frac{\pi}{2} (\delta_{-2} +2\delta_0 +\delta_2)$$ You found the continuous part of $f''$ correctly, but it can be simplified to $$- \sin { (\pi|x|/2))} (\pi/2)^2$$ And again, $f''(x)$ is zero when $|x|>2$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\fermi\pars{x} \equiv \Theta\pars{2 - \verts{x}}\sin\pars{{\pi \over 2}\,\verts{x}}}$

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\begin{align} \color{#66f}{\large\fermi'\pars{x}}& =\delta\pars{2 - \verts{x}}\bracks{-\sgn\pars{x}} \sin\pars{{\pi \over 2}\,\verts{x}} +\Theta\pars{2 - \verts{x}}\cos\pars{{\pi \over 2}\,\verts{x}} \,{\pi \over 2}\,\sgn\pars{x} \\[5mm]&=\color{#66f}{\large{\pi \over 2}\,\Theta\pars{2 - \verts{x}} \cos\pars{{\pi \over 2}\,\verts{x}}\sgn\pars{x}} \\[1cm]\color{#66f}{\large\fermi''\pars{x}}& ={\pi \over 2}\,\delta\pars{2 - \verts{x}}\bracks{-\sgn\pars{x}} \cos\pars{{\pi \over 2}\,\verts{x}}\sgn\pars{x} \\[5mm]&+{\pi \over 2}\,\Theta\pars{2 - \verts{x}} \bracks{-\sin\pars{{\pi \over 2}\,\verts{x}} \,{\pi \over 2}\,\sgn\pars{x}}\sgn\pars{x} \\[5mm]&+{\pi \over 2}\,\Theta\pars{2 - \verts{x}} \cos\pars{{\pi \over 2}\,\verts{x}}\bracks{2\,\delta\pars{x}} \\[5mm]&={\pi \over 2}\,\delta\pars{\verts{x} - 2} -{\pi^{2} \over 4}\,\Theta\pars{2 - \verts{x}}\sin\pars{{\pi \over 2}\,\verts{x}} +\pi\,\delta\pars{x} \\[5mm]&=\color{#66f}{\large\pi\,\delta\pars{x} + {\pi \over 2}\,\delta\pars{x + 2} + {\pi \over 2}\,\delta\pars{x - 2}} \\[5mm]& \color{#66f}{\large\phantom{=}-{\pi^{2} \over 4}\,\Theta\pars{2 - \verts{x}}\sin\pars{{\pi \over 2}\,\verts{x}}} \end{align}