Week 13 -- BP1

[Victor Ivrii]

a. Check that if $\theta(x)$ is a Heaviside function: $\theta(x)=1$ as $x> 0$ and $\theta(x)=0$ as $x\le 0$ then $\theta'(x)=\delta(x)$.

b. Check that if $f(x)$ is a smooth function as $x<  a$ and as $x> a$ but has a jump at $a$ then
\begin{equation*}
f'=\overset{\circ}{f}{}'+ \bigl(f(a+0)-f(a-0)\bigr)\delta (x-a),
\end{equation*}
where $f'$ is understood in the sense of distributions and $\overset{\circ}{f}' (x)$ is an ordinary function equal to derivative of $f$ as $x\ne a$. Generalize to piecewise differentiable functions.

[Andrew Hardy]

a) we know $$ \int_{-\infty}^{\infty} f(x)\delta(x) = f(0) $$
then derivatives of distributions $$ \int_{-\infty}^{\infty} f(x)\theta'(x) = - \int_{-\infty}^{\infty} f(x)'\theta(x)  $$ 

Now $$ - \int_{-\infty}^{\infty} f'(x)\theta(x) = - \int_{0}^{\infty} f'(x)\theta(x) = -f(\infty)+ f(0) = f(0)=  \int_{-\infty}^{\infty} f(x)\delta(x)  $$
therefore $$ \theta'(x) = \delta(x) $$

if I figure out B I will update, otherwise best of luck to someone faster than me

[Andrew Hardy]

b)
$$ f'(\theta) = -f(\theta') = - \int_{-\infty}^{\infty} f(x)\theta'(x)$$
$$ -\int_{-\infty}^a f\theta'(x) + \int_a^{\infty}  f\theta'(x) $$
if I understand your notation correctly
$$=  \overset{\circ}f '(\theta) + f (\theta) \Big|_{-\infty}^a - f (\theta) \Big|_a^\infty $$
$$ = \overset{\circ}f ' -  \lim{x\to a-}f(x)\theta(a) + \lim{x\to a+}f(x)\theta(a) $$
$$ = \overset{\circ}f ' + ((f(a-0) -(f(a+0))(\theta(a)) $$
where this is defined in the sense of distributions
$$ f' =  \overset{\circ}f '  + ((f(a-0) -(f(a+0))\delta(x-a) $$

[Victor Ivrii]

The easier way: $f(x)= f_c(x) + \theta (x-a)$, where $f_c(x)= f(x)$ for $x<a$ and $f_c(x)= f(x)-J$ for $x>a$ is a continuous function, $J=(f(a+0)-f(a-0))$. Then $f_c (x)'=\overset{\circ}{f}{}'(x)$ and for the second term we apply Part a.


If we have jumps at points $a_k$, then
$$f'(x) = \overset{\circ}{f}{}'(x) +\sum_j \bigl(f(a_k+0)-f(a_k-0)\bigr)\delta (x-a_k).$$