Differentiating a function with a Heaviside Step Function

**glio** · March 30th 2008, 05:38 AM

The function H(x) defined by
$H(x) = 0$ $, x < 0$
$H(x) = 1/2$ $, x = 0$
$H(x) = 1$ $, x > 0$

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

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

**roy_zhang** · March 30th 2008, 10:10 AM

Originally Posted by glio

The function H(x) defined by
$H(x) = 0$ $, x < 0$
$H(x) = 1/2$ $, x = 0$
$H(x) = 1$ $, x > 0$

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

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

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

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

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

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

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

Roy

**glio** · March 30th 2008, 10:46 AM

Thanks man.

I finally have ideas how to appreach this

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

Thread: Differentiating a function with a Heaviside Step Function

LinkBack

Thread Tools

Display

Differentiating a function with a Heaviside Step Function

Similar Math Help Forum Discussions

Laplace transform with heaviside step function

D.E involving the heaviside step function

Heaviside Step Function?

Heaviside step function graph

Unit step function/Heaviside function.

Search Tags