Strange math?

[CycloWizard]

Yes, this is homework, but I already found the solution I'm supposed to find (and supply it below as part of my question). I'm just trying to understand how this all fits together. I'm not sure if anyone else here has seen this sort of thing or not, but here's hoping.

I am to prove that, for u(x)=x^a, 0<x<b, a>1/2 for u(x) to exist in the energy space. For a solution to exist in the energy space, it must be both continuous and finite over the interval considered (0 to b in this case).

The first time I solved this, I attempted to show that a>1/2 for u(x) to be continuous by demonstrating that du/dx is finite on (0,b) only for a>1/2. However, it appears that du/dx is only continuous for a>=1.

Another method proposed is that the integral int((du/dx)^2,x,0,b) must be finite for the solution to exist on the energy space. Solving this, it can easily be shown that a^2*x^(2a-1)/(2a-1) must be finite. Thus, the denominator must be non-zero (and positive) for the solution to exist. So, 2a-1>0, or a>1/2.

So, my question is: since the first approach I attempted indicates that u(x) is discontinuous near 0 for a<1, how can the integral then be finite for 1/2<a<1?

 

[icelazer]

If
u(x)=x^a

And you say
"However, it appears that du/dx is only continuous for a>=1. "

That is FALSE. Check your derivative and evaluation. du/dx is continuous for all a if x != 0

 

[CycloWizard]

I meant to say that u is only continuous for a>=1, since du/dx approaches infinity as x approaches zero. You're right that du/dx is continuous for x>0.

I guess maybe I'm just thinking about it too hard. By taking x to be an arbitrarily small number, du/dx becomes an arbitrarily large number, but maybe not infinite.

 

[DrPizza]

Couldn't you use open intervals instead of closed intervals in your proof? I haven't thought much about it, but something just seems to be jumping out at me along those lines... i.e. the function is continuous on (0,+infinity) but not continuous on [0, +infinity)

 

[CycloWizard]

Originally posted by: DrPizza
Couldn't you use open intervals instead of closed intervals in your proof? I haven't thought much about it, but something just seems to be jumping out at me along those lines... i.e. the function is continuous on (0,+infinity) but not continuous on [0, +infinity)

Yeah, that's sort of what I was trying... Just taking very small values (1E-15) and calculate du/dx. It would come out as some huge number and become increasingly huge as the x value got closer and closer to zero. There was no real difference between a=0.49, 0.50, and 0.51 though, which is what got me wondering. I tried looking at the second derivative as well to see if I could get a sense of how quickly it was diverging, but that didn't do much for me. I guess since I could actually calculate a finite number for any arbitrarily small x, du/dx is finite so the function is continuous. It just strikes me as odd that it balloons so quickly but just stays below infinity.