# Toronto Math Forum

## APM346-2018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on April 04, 2018, 12:27:11 PM

Title: Week 13 -- BP1
Post by: Victor Ivrii on April 04, 2018, 12:27:11 PM
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.
Title: Re: Week 13 -- BP1
Post by: Andrew Hardy on April 04, 2018, 01:35:49 PM
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

Title: Re: Week 13 -- BP1
Post by: Andrew Hardy on April 04, 2018, 06:33:35 PM
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
$$f' = \overset{\circ}f ' + ((f(a-0) -(f(a+0))\delta(x-a)$$
Title: Re: Week 13 -- BP1
Post by: Victor Ivrii on April 05, 2018, 08:07:12 AM
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
$$f'(x) = \overset{\circ}{f}{}'(x) +\sum_j \bigl(f(a_k+0)-f(a_k-0)\bigr)\delta (x-a_k).$$