Green's function in acoustics,method of descent

[davon806]

Homework Statement

Hi there,
I was reading a book discussing on the topic of compact Green's function in 2D. However,I have been stuck for a while on some mathematical manipulations depicted below.

Homework Equations

The Attempt at a Solution

In 2nd box,I guess the author was trying to pull out the differentiation sign and make use of the fact that the delta function is an even function.However,the delta function in the first box is differentiating with respect to the single term within the bracket rather than t?

 

[Delta2]

if $f(x)$ is a function and $x=g(t)$ is any linear function of t (such that $\frac{\partial g}{\partial{t}}=c_1$) then from the chain rule it follows that $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}c_1$.

To state it more clearly, the operators $\frac{\partial}{\partial t}, \frac{\partial}{\partial Y}$ have the same effect (up to a constant) because the argument $x=g(t,Y)$ of the delta function is linear both in t and Y.

 

[davon806]

if $f(x)$ is a function and $x=g(t)$ is any linear function of t (such that $\frac{\partial g}{\partial{t}}=c_1$) then from the chain rule it follows that $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial x}c_1$.

To state it more clearly, the operators $\frac{\partial}{\partial t}, \frac{\partial}{\partial Y}$ have the same effect (up to a constant) because the argument $x=g(t,Y)$ of the delta function is linear both in t and Y.

For the primed delta in 2nd box, what is delta differentiating with respect to?

 

[Delta2]

I don't see any primed delta in the 2nd box, if you mean at first box, there it is like saying $\delta'(x)$ that is differentiate w.r.t x and then replace $x=t-\tau-...$

 

[Delta2]

I see now Y is a vector $(y_1,y_2)$ if we want to do this in detail we ll have to use taylor's theorem with two variables but again the key point is that the term $\hat x \cdot Y=x_1y_1+x_2y_2$ is linear in $y_1 , y_2$