Week 10 Generalised functions, or distributions The Dirac delta function Derivatives of the Dirac delta function Differential equations involving generalised functions 1. The Dirac delta function Consider a family of functions δε(x), where x is the variable and ε is a parameter: (1)

Theorem 1: Consider (2) where f(x) is analytic in a certain interval about the point x = 0. Then (3) Proof: Represent f(x) by its Taylor series: (4)

Substitute (1) and (4) into (2): hence, tends to 0 as ε → 0 which yields (3) as required. █

Consider (2) with f(x) = 1, which yields At the same time, it follows from (1) that Q: what kind of function tends to zero at all points except one, but still has non-zero integral (area under the curve)? A: It’s called the Dirac delta function, or just delta function. It’s not a usual function though, but a generalised function, or distribution.

The delta function [usually denoted by δ(x)] can be defined using infinitely many different families of functions. Introduce, for example, (5) and consider...

where f(x) is analytic in a certain interval about the point x = 0. It can be shown that i.e. the family of functions defined by (5) correspond to the delta function, just like family (1).

Theorem 2: Let a family of functions δε(x) satisfy Then δε(x) corresponds to the delta function, i.e. for any f(x) which is analytic at x = 0.

Example 1: Let For which a does this family of functions correspond to δ(x)?

Comment: Even though generalised functions (GFs) imply an underlying limiting procedure, one often uses a ‘short-hand notation’ treating them as if they were regular functions, e.g. One should keep in mind, however, that the above equality actually means where δε(x) is a suitably defined family of functions. Yet, in many cases, the ‘short-hand notation’ can be used to re-arrange expressions involving GFs, and it yields the correct result!

Example 2: Consider (6) where δ'(x) is the delta function’s derivative (we haven’t defined it yet, but let’s consider it anyway and see what happens). Treating δ(x) as a regular function and δ'(x) as its derivative, we integrate (6) by parts...

Assume that which kind of agrees with the fact that, in the proper definition of δ(x), the function δε(x) vanishes outside the interval (–ε, ε). Thus, Now, recall how δ(x) affects test functions. Recalling also definition (6) of I, we obtain...

(7) Even though this equality was derived without following the proper procedure (families of functions, etc.), we’ll later see that (7) correctly describes how the derivative of the delta function affects a test function.

2. Derivatives of the Dirac delta function Consider the following family of functions: Note that, everywhere except the points x = –ε, 0, +ε, the function δ'ε(x) equals to the derivative of δε(x) defined by (5). At the ‘exceptional’ points, δε(x) doesn’t have a derivative, so the values of δ'ε(x) were chosen, more or less, ad hoc. Now, consider...

where f(x) is analytic in a certain interval about x = 0. Theorem 3: Proof: Represent f(x) by its Taylor series Observe that δ'ε(x) is odd – hence, every other term of the series in [] doesn’t contribute to the integral, and we obtain...

hence, hence, as required. █ tends to 0 as ε → 0

The ‘short-hand’ form of Theorem 3 is Comment: The minus on the r.-h.s. looks ‘unnatural’, but it actually agrees with the result in Example 2 obtained using the short-hand notation. Comment: The family of functions used in Theorem 3 for representing δ'(x) was obtained by differentiating the family of functions defined by (5) and used to represent δ(x). In principle, we could’ve used a different family, but there’s still a general rule: if δε(x) represents δ(x), then the derivative of δε(x) represents δ'(x).

Theorem 4: The n-th derivative of δ(x) [denoted by δ(n)(x)] can be defined through any family of functions such that

We shall also use δ(x – x0) and δ'(x – x0), such that What’s the equivalent of the above equalities for δ"(x – x0)?

Theorem 5: Let a function g(x) be smooth and strictly monotonic, i.e. Let also g(x) have a single zero at x = x0, i.e. Note that, since g(x) is strictly monotonic, Then,

Proof: Let g'(x0) > 0 and consider (8) Let’s change the variable x to y = g(x), so that where x(y) is the inverse function to y = g(x). Observe also that, since y(x0) = 0 and x(y) is the inverse function, it follows that (9)

Now, (8) becomes which can be readily evaluated using the definition of δ(x): Taking into account (9), we obtain

Finally, recalling definition (8) of I, we have which yields the desired result for the case g'(x0) > 0. The case g'(x0) < 0 is similar. █