1550078481-Ordinary_Differential_Equations__Roberts_

12 Ordinary Differential Equations

For example, the piecewise defined function

{

x^3 + 2,

y( x)= x2-1,

x<O

0:::; x

is defined on the interval (-oo, oo ). It is continuous on the intervals (-oo, 0)

and (0, oo), but it is not continuous at x = 0, since

lim y(x ) = 2 =f. -1 = lim y( x ) = y(O).

x->O- x->O+

Because y(x) is not continuous at x = 0, the function y( x ) is not differentiable

at x = O; however , y(x) is differentiable on (-oo,O) and (O,oo). In fact,

'( ) { 3x

2

y x = ,

2 x,

The absolute value function

y(x)=lxl= { '

• x
  x,

x<O

0 < x.

x<O

0:::; x

is a piecewise defined function which is continuous on (-oo, oo ). Computing
the left-hand derivative of y( x ) = lx l at x = 0, we find

y' (0) = lim IO+ hi - IOI = lim 0:1 = lim -h = -1.

• h->O- h h->O- h h->O- h

And computing the right-hand derivative of y( x ) = lx l at x = 0, we find

. 10 +hi - 101. lhl. h

y~ (0) = hm = hm - = hm - = 1.

h ->O+ h h ->O+ h h->O- h

Since y'__(O) = -1=f.1 = y~(O), the a bsolute value function, y(x) = lx l, is not

differentiable at x = 0. However, the absolute value function is different iable

on ( -oo, 0) and (0, oo), and its derivative is

y'(x) = dlxl = {-1,

dx 1,

x < 0} = -lx l = sgn(x )

0 < x x

where sgn(x) is an abbreviation for the signum function.

The previous two examples illustrat e that points where piecewise defined

functions may not b e continuous or m ay not b e differentiable are points at

which the definition of the function ch a nges from one mathematical expression

to another.