Absolute value function is the Heaviside step function

[Ted123]

Homework Statement

[PLAIN]http://img845.imageshack.us/img845/1706/delta.png

The Attempt at a Solution

This question has come after a chapter on Dirac's delta-function although the question doesn't say use it. Could I use it? If not how do I do it directly?

 

[Wingeer]

Yes you could. The derivative of the absolute value function is the Heaviside step function, which is the integral of Dirac's delta function.

 

[Ted123]

How?

 

[HallsofIvy]

"The derivative of the absolute value is the Heaviside function"- so use the basic sum property and chain rule to differentiate this sum of absolute values.

" the Heaviside step function is the integral of Dirac's delta function" so the derivative of the Heaviside function, and so the second derivative of the absolute value is the Dirac Delta function. You have a sum of three absolute values so the first derivative will be a sum of three Heaviside functions and the second derivative a sum of three delta functions.

As a check, you could also do this is a very basic way: If x< -1, then all of x+ 1, x, and x- 1 are negative: for x< -1, f(x)= -(x+1)- 3(-x)- (x- 1)= x. What is the derivative of that? What is the second derivative? If -1< x< 1, then x+ 1 is positive but x and x- 1 are still negative. f(x)= x+ 1- 3(-x)- (x- 1)= 3x+ 2. What is the derivative of that? What is the second derivative? If 0< x< 1, then x+1 and x are positive but x- 1 is still negative. f(x)= x+ 1- 3x- (x- 1)= -3x+ 2. What is the derivative of that? What is its second derivative? If x> 1, all three of x+1, x, and x-1 are positive so f(x)= x+ 1- 3x+ x- 1= -x. What is the derivative of that? What is its second derivative? Be sure to calculate the derivative and second derivative at x= -1, x= 0, and x= 1 separately.

 

[Ted123]

$\begin{displaymath} f(x) = \left\{ \begin{array}{lr} 0, & \;x\leq -1\\ 2x+2, & \;-1<x\leq0\\ -2x +2, & \;0<x\leq1\\ 0, & \;x > 1 \end{array} \right. \end{displaymath}$

The function $f(x)$ being continuous and piecewise smooth, its first derivative can be obtained by piecewise differentiation:

$\begin{displaymath} f'(x) = \left\{ \begin{array}{lr} 0, & \;x\leq -1\\ 2, & \;-1<x\leq 0\\ -2, & \;0<x\leq1\\ 0, & \;x > 1\\ \end{array} \right. \end{displaymath}$

i.e. $f'(x) = 2H(x+1) -4H(x) + 2H(x-1)$ where H is the Heaviside function.

$f'(x)$ is piecewise constant, therefore $f''(x)=0$ for all $x$ except $x=0,\pm 1$.

The 3 points contribute to $f''$ the sum of delta-functions:

$f''(x) = 2\delta(x+1) -4\delta(x) + 2\delta(x-1)$

 

[Ted123]

Does this look right?