I have to calculate the derivatives of order $\le 2$ of for example $f(x) = |x|$, is it the same as the second derivate, what does this "of order $\le 2$" mean? calculating distributionell derivatives is easy, for my example it would be $$ \langle f', \phi \rangle = - \int_{-\infty}^0 -1 \phi' \mathrm{d}x - \int_{0}^{\infty} 1 \phi' \mathrm{d}x = 2\int_{0}^{\infty} \phi' \mathrm{d}x = -2\phi(0) $$ is guess.
whats the order of a distributional derivate?
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terminology
distribution-theory
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0for $f(x)=[x]$ from (-infinity,0) $f(x)=-x$ and $f'(x)=-1$,from [0,infnity) $f(x)=x$ so $f'(x)=1$ – 2012-06-26
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0By the way, your computation of the first distributional derivative is incorrect. $$ \langle f',\phi\rangle = \int_{-\infty}^0 x\phi' \mathrm{d}x - \int_{0}^\infty x\phi' \mathrm{d}x $$ Perhaps you intended to compute the second distributional derivative of $f$ in your question statement? – 2012-06-26
1 Answers
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A distributional derivative of order 1 is the first derivative.
A distributional derivative of order 2 is the second derivative.
A distributional derivative of order 3 is the third ...
We write "of order $N$" instead of the $N$th derivative because, among other reasons, (1) that in the multivariable context there isn't just the second derivative. When your domain is $\mathbb{R}^2$ there are three independent second derivatives; (2) it makes possible for expression like "of order $\leq N$" which is a short hand for "the collection of first, second, ... , all the way up to $N$th derivatives".