Mathematical Tools for Physics - Department of Physics - University

17—Densities and Distributions 418

This turns out to have all the properties that you need, though again, you don’t have to invoke its
explicit form.

What isδ(ax)? Integrateδn(ax)with a test function.

lim

n

∫∞

−∞

dxδn(ax)φ(x) = lim

n

∫∞

−∞

dy

a

δn(y)φ(y/a)

wherey=ax. Actually, this isn’t quite right. Ifa > 0 it is fine, but ifais negative, then when

x→ −∞you havey→+∞. You have to change the limits to put it in the standard form. You can

carry out that case for yourself and verify that the expression covering both cases is

limn

∫∞

−∞

dxδn(ax)φ(x) = limn

∫∞

−∞

dy

|a|

δn(y)φ(y/a) =

1

|a|

φ(0)

Translate this into the language of delta functions and it is

δ(ax) =

1

|a|

δ(x) (17.27)

You can prove other relations in the same way. For example

δ(x^2 −a^2 ) =

1

2 |a|

[

δ(x−a) +δ(x+a)

]

or δ

(

f(x)

)

=

∑

k

1

|f′(xk)|

δ(x−xk) (17.28)

In the latter equation,xkis a root off, and you sum over all roots. Notice that it doesn’t make any

sense if you have a double root. Just try to see whatδ(x^2 )would mean. The last of these identities

contains the others as special cases. Eq. (17.27) implies thatδis even.

17.6 Differential Equations
Where do you use these delta functions? Start with differential equations. I’ll pick one that has the
smallest number of technical questions associated with it. I want to solve the equation

d^2 f

dx^2

−k^2 f=F(x) (17.29)

subject to conditions thatf(x)should approach zero for large magnitudex. I’ll assume that the given

functionFhas this property too.

But first: Letkandybe constants, then solve

d^2 g

dx^2

−k^2 g=δ(x−y) (17.30)

I want a solution that is well-behaved at infinity, staying finite there. This equality is “in the sense of
distributions” recall, and I’ll derive it a couple of different ways, here and in the next section.

First way: Treat theδas simply a spike atx=y. Everywhere else on thex-axis it is zero. Solve

the equation for two cases then,x < yandx > y. In both cases the form is the same.

g′′−k^2 g= 0, so g(x) =Aekx+Be−kx

Forx < y, I wantg(x)to go to zero far away, so that requires the coefficient ofe−kxto be zero. For

x > 0 , the reverse is true and only thee−kxcan be present.

g(x) =

{

Aekx (x < y)

Be−kx (x > y)