z
I have a bit of difficulty regarding how you start.
Let P=f(e)--this is fine. Then (you claim):
P_0+P_e=f(p_0+p_e)
However, this is a bit of a mystery since, in general, f(x+y)≠f(x)+f(y) unless f is a homomorphism. In the interest of non-mathematical pursuit, I think you can safely get away with this kind of sleight of hand.
However! It is absolutely NOT true that
f(p_0+p_e)-f(p_e)
------------------ = f'(p_0)
p_e
Indeed, this is only a true statement if you include the limit, i.e.
f(p_0+p_e)-f(p_e)
lim ---------------------- = f'(p_0)
p_e-->0 p_e
I don't mean to be a spoil-sport; it's just a nontrivial amount of hand-waving to achieve your results. A less...suspicious way to do things might be something like the following:
Let u(x,t) represent a displacement of a particle in some direction of the point x at time t>0. Let U be an open set and V an smooth subregion of U. The acceleration within V is then
/ /
d^2 | |
---- | u dx = | u_{tt} dx
dt^2 / /
V V
where u_{tt} represents the second partial derivative of u with respect to t, and the net contact force is
/
|
- | F µ dS ,
/
∂V
where F denotes the force acting on V through the boundary ∂V and the mass density is taken to be unity. Newton's law asserts the mass times the acceleration equals the net force, i.e.:
/ /
| |
| u_{tt} dx = - | F µ dS .
/ /
V ∂V
This identity holds for each subregion V and so
u_{tt} = - div(F).
Since we can safely assume air is an "elastic" body, F is a function of the displacement gradient Du; whence
u_{tt}+div(F(Du))=0
and for small Du, the linearization F(Du) ~ -aDu is often appropriate; and so
u_{tt} - a∆u = 0 .
This is the wave equation for propagation of sound when a, some arbitary constant, is given by a=c^2 and c is the speed of sound.
Gender:
Points: 3891
Reviews: 3821