Where does the 1/r^(n+1) term come from in the Parabolic Mean Value Formula?

[Kalidor]

A million dollar to anyone who can tell me where that 1/r^(n+1) comes from in the second equality on page 3 of this pdf:

http://www.google.it/url?sa=t&rct=j...dseJTXVXXheev1Dwg&sig2=HjkcJuDcDuUoG-wZFdc4Sw

There are similar passages on Evans` PDE book. Shouldn't it be 1/r^n as shown on the previous page? Thanks to anyone who will dispel my doubts.

 

[Vargo]

Hello.

The exponent n+1 is correct as written. To me the difficulty seems to be that Evans is using the exact same notation (y,s) before and after transforming between E(0,0,r) and E(0,0,1).

So, for this explanation, let (y,s) be the variables in E(0,0,r) and (y',s') the variables in E(0,0,1). (I know primes aren't so great, but Evans is already using (x,t) so I won't use that)

Then the transformation between them is
$$ y=ry' \,\,\, s = r^2 s'.$$
The y variables live in n dimensions and s in 1 D, so the Jacobian will be r^n*r^2.

So let's say you take the final integral over E(0,0,r) and we want to transform that into what is on the line immediately before it. Substitute the change in variable, multiply in the Jacobian and you will end up with the same exponent n+1 for the scale change.

At the bottom of the second page, the exponent is n as you noticed. The difference is in the integrand. Notice the term |y|^2/s^2. When you substitute the change in variables, that produces a factor of (1/r^2).

On the third page, the two terms have factors |y|^2/s and y_i |y|^2/s^2 respectively. When you perform the substitution, there is only the factor (1/r). This accounts for the difference you noticed.

I will be happy to accept your offered payment as cash or check

 

[Kalidor]

Thanks man!