# Cannot make sense of a derivation step involving dirac delta

1. Jun 26, 2014

### D_Tr

I am self studying the 17th Chapter of "Mathematical Methods for Physics and Engineering", Riley, Hobson, Bence, 3rd Edition. It is about eigenfunction methods for the solution of linear ODEs.

1. The problem statement, all variables and given/known data

On page 563, it states:

"As noted earlier, the eigenfunctions of a Hermitian operator may be shown to form a complete set over the relevant interval. ... Working in terms of the normalized eigenfunctions $\hat{y}_n$, we may thus write:"

$$f(x)=\sum_{n}\hat{y}_n(x)\int_{a}^{b}\hat{y}_n^*(z)f(z)w(z)dz = \int_{a}^{b}f(z)w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z)dz$$
from the definition of the dirac delta and the above relation, we get:
$$w(z)\sum_{n}\hat{y}_n(x)\hat{y}_n^*(z) = \delta(x-z)~~~~~(17.28)$$
now we are at the stage where I am getting confused:
"We also note that the LHS of (17.28) is a delta function and so it is only non-zero when z=x; thus w(z) on the LHS can be replaced by w(x) if required"

The problem is, I cannot why this replacement can be made, since w(x) can be zero in isolated points. w(x) is the weight function for calculating inner products of functions.

2. Relevant equations

The relevant relations are those in the previous section.

3. The attempt at a solution

When x is not equal to z, 17.28 is zero on both sides. If we suppose that w(x) is nowhere zero, then it follows that the sum is zero, since the RHS is zero, and we can make the replacement. BUT if we allow w(x) to be zero in a set of points, which is allowed for a weight function as stated earlier in the chapter and in various online sources, I am having problem justifying the validity of this replacement. This is because, if w(z) is zero for a particular z, then the sum can be non zero for this z and a particular x, and 17.28 stlil holds. But if we replace w(z) with w(x), the LHS may become non-zero. What am I missing?

Last edited: Jun 26, 2014
2. Jun 27, 2014

### haruspex

Consider two cases.
When z is not x, the summation on the left is zero. In that case, it does not matter what you change w(z) to, the equation is still 0 = 0.
When z = x, w(z) = w(x), so the substitution is still valid.

3. Jun 27, 2014

### D_Tr

Thank you for the reply. What you wrote is clear to me for the case where w(z) is non-zero for all z. What confuses me is that the weight function w(z) is allowed to be zero in some points. First of all, if w(z) = 0 for some z = z0, 17.28 does not hold because the LHS is zero for all pairs (z0, x). Secondly, if w(z0) = 0, then the sum could be non zero for x and z0, so in this case changing w(z0) to w(x) could make the LHS non zero for x not equal to z0. Is what I'm writing making any sense?

What I am thinking now is that since (as I have read online) w(x) can be zero only in isolated points, we could just change w(x) so that it has a non zero value at those points, and then everything seems ok to me, since the integrals containing w(x) would not change by changing w(x) in a set of isolated points. But still, 17.28 is derived in the text with the assumption that w(x) can indeed take the value zero... I would be glad if you could enlighten me further :)

Last edited: Jun 27, 2014
4. Jun 27, 2014

### xaos

In terms of distributions and Lebesque measures, up to any countable null set, w(z) can be whatever it wants to be, so this function is defined as the equivalence class of all such functions.

[deleted]

[The] Dirac distribution should only be seen within the definition of an integral. Do not treat it like a function, it is a distribution.

Last edited: Jun 27, 2014
5. Jun 28, 2014

### haruspex

I'm sorry, I misread the OP. I thought the text was saying that the summation was in itself a delta function (based on some prior work). I understand what you are saying now: if w(z) = 0 for some z then the sum may be nonzero for that z and some x not equal to z.
It seems to me your own explanation is good.

6. Jun 28, 2014

### D_Tr

Thank you both very much, you really helped me.