For $u(x)=\sin|x|$, the ordinary derivative away from zero is
$$
g(x)=
\begin{cases}
-\cos x,&-\pi<x<0,\\
\cos x,&0<x<\pi.
\end{cases}
$$
This bounded function is locally integrable. For a <test function> $\varphi$, <integration by parts> on $(-\pi,0)$ and $(0,\pi)$ produces boundary terms at zero which cancel because $u$ is continuous there and $u(0)=0$. Hence
$$
\int_{-\pi}^{\pi}u\varphi'=-\int_{-\pi}^{\pi}g\varphi.
$$
Therefore $u$ has the <weak derivative>
$$
\boxed{u'(x)=\operatorname{sgn}(x)\cos|x|\quad\text{for almost every }x.}
$$
The jump in the ordinary derivative creates no <Dirac delta function>; such a term would arise from a jump in the function itself.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1 / b / ii / Solution

 Solution

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