Delta Function Rules
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Short note by Phil dated 4.28.11, in his Stakgold folder, reviewing distribution theory. It starts from Stakgold's translation, scaling, derivative and function rules and applies them to delta functions. It derives f(x)δ^(n)(x-b) as a sum of lower delta derivatives, gets identities such as xδ'(x) = -δ(x) and x^nδ^(n)(x) = (-1)^n n! δ(x), and ends with a summary of the rules.
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Delta Function Rules PhL 4.28.11
See summary of rules at the end.
Review of Distribution Theory
Back on page 35 Stak derives four rules for distributions:
Translation Rule: <t(x-a),φ(x)> = <t(x),φ(x+a)>
For example, applied to t = δ(x) this would say
<δa,φ(x)> = <δ0,φ(x+a)> = φ(0+a) = φ(a)
which is consistent with what we know. Then we have
Scaling Rule:
<t(ax),φ(x)> = 1/|a| <t(x),φ(x/a)>
and we can apply this to a delta function
< δ(ax),φ(x)> = 1/|a| <δ(x), φ(x/a)> = 1/|a| φ(0) = 1/|a| <δ, φ(x)>
Comparing both sides we would say in terms of symbolic functions
δ(ax) = δ(x)/|a|
which we also know to be true. Finally, we have the
Derivative Rule:
<∂xt, φ> = - <t,∂xφ>
which we can apply to t = δb to get
<δ'b, φ> = - <δb,φ'> = - φ'(b)
and there is not much more we can say here. The idea is just parts integration with a φ test function having compact support so the parts go away. This does not give us a result δ'(x-b) = something in terms of symbolic functions. More generally
<δ(n)b, φ> = (-1)n <δb,φ(n)> = (-1)n φ(n)(b)
Function Rule:
<f t, φ> = <t,fφ>
This is a sort of nothing rule, but applied to t = δb it says
<f δb, φ> = <δb,fφ> = f(b)φ(b) = f(b) <δb,φ> = < f(b)δb,φ>
from which we conclude in terms of symbolic functions that f(x)δ(x-b) = f(b)δ(x-b) which is not too earth shattering.
Combination of the Function Rule with a First Derivative Rule:
<f ∂xt, φ> = < ∂xt, f φ> = - <t,∂x(fφ)>
which we can apply to t = δb to get
<f δ'b, φ> = < δ'b, f φ> = - <δb,∂x(fφ)> = - ∂x(fφ)|x=b = - [ f(b)φ'(b) + f'(b)φ(b)]
= - f(b)φ'(b) - f'(b)φ(b) = - f(b) <δb, φ'> - f'(b) <δb, φ> = + f(b) <δ'b, φ> - f'(b) <δb, φ>
= < f(b) δ'b - f'(b) δb, φ>
and the symbolic function result is then
f(x) δ'(x-b) = f(b) δ'(x-b) - f'(b) δ(x-b)
a result which Maple verifies
In the case that b = 0 we get
f(x) δ'(x) = f(0) δ'(x) - f'(0) δ(x)
and if we set f(x) = xα we get,
xα δ'(x) = - α(xα-1)x=0 δ(x)
Therefore we only get well-defined results for two ranges of α:
xα δ'(x) = 0 Re(α) > 1
xα δ'(x) = -δ(x) α = 1
For example
x δ'(x) = -δ(x)
x2δ'(x) = x3δ'(x) = ... = 0
Combination of the Function Rule with a Second Derivative Rule:
<f ∂x2t, φ> = < ∂x2t, f φ> = + <t,∂x2(fφ)>
Applied to t = δb we get
<f δ"b, φ> = <δb,∂x2(fφ)> = [∂x2(f(x)φ(x)]x=b
= [ f"(b)φ(b) + 2 f'(b)φ'(b) + f(b)φ"(b)
= f"(b)<δb,φ> - 2 f'(b)<δ'b,φ> + f(b) <δ''b,φ>
and we conclude that
f(x) δ"(x-b) = f"(b)δ(x-b) - 2f'(b)δ'(x-b) + f(b) δ"(x-b)
And if b = 0 and f(x) = xn this says
xn δ"(x) = n(n-1)0n-2δ(x) - 2n0n-1δ'(x) + 0n δ"(x)
and now we need n ≥ 2 to get a reasonable result, such as
x2 δ"(x) = 2 δ(x)
x3 δ"(x) = 0 and same for higher powers
Here are some general rules on this:
xnδ(n)(x) = (-1)n n! δ(x)
xn+1δ(n)(x) = 0 and of course for higher powers of x as well
Combination of the Function Rule with an nth Derivative Rule:
<f ∂xnt, φ> = < ∂xnt, f φ> = + <t,∂xn(fφ)>
We use the result
∂xn(fφ) = Σk=0n ∂xk f ∂xn-kφ
to find in the end that
f(x) δ(n)(x-b) = Σk=0n (-1)k f(n-k)(b) δ(k)(x-b)
Summary of Delta Function Rules
δ(ax) = δ(x)/|a|
f(x)δ(x-b) = f(b)
f(x) δ'(x-b) = f(b) δ'(x-b) - f'(b) δ(x-b)
f(x) δ"(x-b) = f"(b)δ(x-b) - 2f'(b)δ'(x-b) + f(b) δ"(x-b)
f(x) δ(n)(x-b) = Σk=0n(-1)k f(n-k)(b) δ(k)(x-b)
xnδ(n)(x) = (-1)n n! δ(x)
x δ'(x) = -δ(x) n=1
x2 δ"(x) = 2δ(x) n=2
etc
xn+1δ(n)(x) = 0 eg
x δ(x) = 0 n=1
x2 δ'(x) = 0 n=2
etc