The Dirac delta function

Motivation: Pushing a cart, initially at rest.

F dvF m

dt

0 0

0 0

0 0

( ) ( ) (0)t t

dvF t dt m dt m v t v mv t

dt

Applied impulse Acquired momentum

0

0

0

( )t

I F t dt mv t

F

mv

t

t

t

t

Same final momentum, shorter time.

F

mv

t

t

In the limit of short time, we idealize this as an instantaneous, infinitely large force.

F

mv

t

t

F

mv

Dirac’s delta function models for this kind of force.

Dirac delta function

Q: Is this mathematically rigorous, using standard calculus?

A: Not quite. Standard functions do not take infinity as a

value, and that integral would not be valid.

( )t

( ) 0, 0.

(0)

( ) 1

t for t

t dt

Normalization

Representation:

t

• This “unit impulse” function is defined by the conditions:

How to make sense of it then?• What physicists like Dirac (early 1900s) meant is

something like this:

• Mathematical difficulty: in standard calculus, this family of functions would not converge.

• Later on (1950s) mathematicians developed the “theory of distributions” to make this precise. This generalization of calculus is beyond our scope, but we will still learn how to use it for practical problems.

0lim ( ),

1[0, ]

( )0

WW

W

p t where

if t Wp t W

otherwise

0 W

1

W

t

Calculate: ( ) ( )t

u t d

t

t

u1

0 0( )

1 0

if tu t

if t

Remark: value at t = 0is not well defined, we adopt one by convention.

t

u

Unit step function

• In standard calculus, u(t) is only differentiable (and has zero derivative) for nonzero t.

• u(t) is discontinuous at t = 0. Value at point of discontinuity is arbitrarily chosen.

0 0( )

1 0

if tu t

if t

1

( )du

tdt

Step function as a limit.0 0

( ) 0

1

W

for t

tu t for t W

Wfor t W

0

0

lim ( ) ( ), 0

( ),

lim ( ) ( )

WW

WW

WW

u t u t t

dup t

dtp t t

t

1

t

Wdu

dt

Wu

1

W

W

Remark: Physical variables (e.g. the car momentum) are not really discontinuous, they are more like . The step function u(t) is a convenient idealization.

( )Wu t

Example: integrate ( ) ( ) ( )f t t u t

( ) ( ) ( ) ( ) ( ) ( )t t t

g t f d u d u t u d

t

g

1

1

f

0 0( )

0

t if tu d

t if t

Last term:

Step function used to compact formula.

( ) ( )dg

t u tdt

( ) ( ) 1g t u t t

Due tojump

Standard derivativeaway from t = 0.

( )t u t

Scaled steps and deltas

0 0( )

0

if tK u t

K if t

0( ) lim ( )W

W

dKuK t K p t

dt

t

3u

3e.g.:

0 W

K

W

t

Conceptually the same, differentamount of impulse applied

t

(K)Representation:

Negative steps and deltas

0 0( )

1 0

if tu t

if t

( )

d ut

dt

t

-u

-1

t

(-1)Representation:t

(-1)

or

Reverse step and its derivative

1 ( ) ( )u t u t

1 ( )( )

d u tt

dt

t

1-u

1

Strictly speaking, the two expressions differ at t = 0. But recall u(0) was arbitrary.There is no distinction between the two in this calculus.

Differentiating the left-hand side

( )( )

d u tt

dt

Using right-hand side, compositionrule ( )

' ( )dg f t df

g f tdt dt

( ) ( )t t Delta is an even function

Translations and flips of steps and deltas0

( )1

if tu t

if t

t

1

t( ) ( )t t

A pulse function:

( ) ( ) ( ) ( )

1 [ , ]

0

u t a u t b u t a u b t

if t a b

otherwise

ta b

11

( )0

if tu t

if t

Example:

( ) ( ) ( 2)

( ) (2 )

f t u t u t

u t u t

t

f

0 2

t0 2

( ) 2df

t tdt

df

dt

Using the first expression,

Using the second,

( ) (2 ) ( ) 2 ( 1)

( ) (2 ) ( ) 2

dft u t u t t

dtt u t u t t

Different answer?

A basic property of delta

• If f(t) is continuous at 0, ( ) ( ) 0 ( )t f t t f t

• If f(t) continuous at , ( ) ( ) ( )t f t t f t

Let f(t) be a standard function.

t

(1)

f

t

( )f

=

1 1

( ) (2 ) ( ) 2 ( ) (2) (2) 2

( ) 2

dft u t u t t t u u t

dt

t t

Apply to the previous example:

Consistent with previous answer

Consequence of our basic property

( ) ( ) (0) ( ) (0) ( ) (0)f t t dt f t dt f t dt f

( ) ( ) ( )f t t dt f

( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( )

0

b

a

f t t dt u t a u t b f t t dt

f if a bf u a u b

if a or b

Similarly,

t

(1)

f

a b

A general rule on differentiation

Let f(t) have a standard derivative g(t) except at a finite number of points . Then in this calculus,1 2, , , nt t t

1

( ) ( ),n

k kk

dfg t J t t

dt

and we assume the limits exist.

: ( ) ( ) lim ( ) lim ( )k k kk kt t t t

J f t f t f t f t

where the “jumps”

f

t1t 3t2t