# Differentiating a function with a Heaviside Step Function

• Mar 30th 2008, 05:38 AM
glio
Differentiating a function with a Heaviside Step Function
The function H(x) defined by
$\displaystyle H(x) = 0$ $\displaystyle , x < 0$
$\displaystyle H(x) = 1/2$ $\displaystyle , x = 0$
$\displaystyle H(x) = 1$ $\displaystyle , x > 0$

$\displaystyle F(x) = f(x)H(x - b)$

Find the conditions on $\displaystyle f(x)$ that make $\displaystyle F(x)$
1. Continuous at $\displaystyle x = b$
2. Differentiable at $\displaystyle x = b$
• Mar 30th 2008, 10:10 AM
roy_zhang
Quote:

Originally Posted by glio
The function H(x) defined by
$\displaystyle H(x) = 0$ $\displaystyle , x < 0$
$\displaystyle H(x) = 1/2$ $\displaystyle , x = 0$
$\displaystyle H(x) = 1$ $\displaystyle , x > 0$

$\displaystyle F(x) = f(x)H(x - c)$

Find the conditions on $\displaystyle f(x)$ that make $\displaystyle F(x)$
1. Continuous at $\displaystyle x = c$
2. Differentiable at $\displaystyle x = c$

1. In order to make $\displaystyle F(x)$ continuous at $\displaystyle x = c$, the function $\displaystyle F(x)$ needs to satisfy the following conditions:

(1) $\displaystyle F(c)$ is defined;
(2) $\displaystyle \lim\limits_{x\rightarrow c} F(x)$ exists; and
(3) $\displaystyle \lim\limits_{x\rightarrow c} F(x)=F(c)$

To satisfy (1), consider $\displaystyle F(c)=f(c)H(c-c)=f(c)H(0)=\frac{1}{2}f(c)$, we need $\displaystyle f(c)$ is defined.
To satisfy (2), we need to make sure both $\displaystyle \lim\limits_{x\rightarrow c^-} F(x)$ and $\displaystyle \lim\limits_{x\rightarrow c^+} F(x)$ exist, also $\displaystyle \lim\limits_{x\rightarrow c^-} F(x) = \lim\limits_{x\rightarrow c^+} F(x)$.
Here we have $\displaystyle \lim\limits_{x\rightarrow c^+} F(x)=f(c)\left(\lim\limits_{x\rightarrow c^+}H(x-c)\right)=f(c)(1)=f(c)$ and $\displaystyle \lim\limits_{x\rightarrow c^-} F(x)=f(c)\left(\lim\limits_{x\rightarrow c^-}H(x-c)\right)=f(c)(0)$.
So we need $\displaystyle |f(c)|<\infty$ and $\displaystyle f(c)=f(c)(0)\Rightarrow f(c)=0$.
To satisfy (3), realize that from above, if $\displaystyle \lim\limits_{x\rightarrow c} F(x)$ exists then $\displaystyle \lim\limits_{x\rightarrow c} F(x)=0$. Also we have $\displaystyle F(c)=\frac{1}{2}f(c)$, hence $\displaystyle \frac{1}{2}f(c)=0\Rightarrow f(c)=0$.

In summary, in order to make $\displaystyle F(x)$ continuous at $\displaystyle x = c$, function $\displaystyle f(x)$ has to satisfy $\displaystyle f(c)=0$.

2. Given that $\displaystyle F^\prime(c)=\lim\limits_{x\rightarrow c}\frac{F(x)-F(c)}{x-c}$, so in order to make $\displaystyle F(x)$ differentiable at $\displaystyle x = c$, we need to show the limit $\displaystyle \lim\limits_{x\rightarrow c}\frac{F(x)-F(c)}{x-c}$ exists. More specifically, we need to show the one-sided limits $\displaystyle \lim\limits_{x\rightarrow c^-}\frac{F(x)-F(c)}{x-c}$ and $\displaystyle \lim\limits_{x\rightarrow c^+}\frac{F(x)-F(c)}{x-c}$ exist and are equal. Can you pick up from here?

Roy
• Mar 30th 2008, 10:46 AM
glio
Thanks man.

I finally have ideas how to appreach this (Clapping)

Hmmm... It looks like I'll need more help with this.