Find the second distributional derivative of $\left|\sin(x)\right|$

Question

I am trying to find the second distributional derivative of $f(x)=\left|\sin(x)\right|$, defined for $x$ in $\mathbb R$.

I have not got far but I started like this:

Let $\varphi$ be defined in $S(\mathbb R)$, then $$ \frac{d^2}{dx^2} f(\varphi) = \int_{\mathbb R} f(x) \,\frac{d^2}{dx^2} \varphi \,\mathrm dx $$

what to do next?

By the way, where can I find the syntax for doing all the symboles?

Answer 1

My strategy for computing distributional derivatives is to compute symbolically/naively first, then confirm my answer using integration by parts. So, symbolically, the first derivative of $f(x)$ should be (by the chain rule):

$$ f^\prime(x)=\text{sgn}(\sin(x))\cos(x) $$ Where $\text{sgn}(x)$ is the signum function, defined as: $$ \text{sgn}(x)=\left\{\begin{array}{cc} -1 & x<0\\ 0 & x=0\\ 1 & x>0 \end{array}\right. $$

Then, using the product rule, we get

$$ f^{\prime\prime}(x)=\text{sgn}^\prime(\sin(x))\cos^2(x)-\sin(x)\text{sgn}(\sin(x)) $$ Recall that the derivative of the signum function is $\text{sgn}^\prime(x)=2\delta(x)$, so $\text{sgn}^\prime(\sin(x))\cos^2(x)=2\sum_{k\in\Bbb{Z}}2\delta(x-k\pi)$. This is because when $\sin(x)=0$, $\cos^2(x)=1$. Also note that $\sin(x)\text{sgn}(\sin(x))$ will be exactly $\vert \sin(x)\vert$, since $\text{sgn}(\sin(x))=-1$ when $\sin(x)<0$ and $\text{sgn}(\sin(x))=1$ when $\sin(x)>0$. Hence we have

$$ f^{\prime\prime}(x)=2\sum_{k\in\Bbb{Z}}\delta(x-k\pi)-\vert\sin(x)\vert $$

From here, you should be able to verify this using integration by parts. The goal is to show that for every $\varphi\in\mathcal{S}$, we have

$$ \langle f^{\prime\prime},\varphi\rangle=\langle f,\varphi^{\prime\prime}\rangle $$ where $\langle\cdot,\cdot\rangle$ is the inner product.

Answer 2

Let's apply the jump formula for derivation of distributions. $|\sin x|$ is continuous and piecewise $C^1$, we write $$ \left|\sin x\right|'=\begin{cases}\cos x,&x\in [2\pi k,\pi+2\pi k),\\-\cos x,&\text{otherwise}.\end{cases} $$ Clearly, this function is no longer continuous, but still piecewise $C^1$, so we will get $\delta$-functions in the second derivative. The jumps are all equal to $2$ and they occur in points of the form $\pi\mathbb Z$.

We can write it as $$ \left|\sin x\right|''=g+2\sum_{k\in\pi\mathbb Z}\delta_k, $$ where $$ g(x)=\begin{cases}-\sin x,&x\in [2\pi k,\pi+2\pi k),\\\sin x,&\text{otherwise},\end{cases}=-\left|\sin x\right|, $$ That is, $$ \left|\sin x\right|''=-\left|\sin x\right|+2\sum_{k\in\pi\mathbb Z}\delta_k. $$