Diff Eq Region of Unique Solution

  • Thread starter Thread starter DieCommie
  • Start date Start date
  • Tags Tags
    Diff eq
DieCommie
Messages
156
Reaction score
0
I have just started my first diff eq class, and I am struggling with the concepts. In the section 'Preliminary Theory' we are given Picard's Theorem and a few odd examples. Then this is the question from the homework:

"Determine a region of the xy-plane for which the given differential equation would have a unique solution through a point (x_0, y_0) in the region. dy/dx = y^(2/3) "

The wording has got me confused, and I am not really sure what they want. The solution is a family of curves, so they want the particular curve at the point (x_0, y_0)? Or they want a region of something?

Thx for any help you have.
 
Physics news on Phys.org
DieCommie said:
I have just started my first diff eq class, and I am struggling with the concepts. In the section 'Preliminary Theory' we are given Picard's Theorem and a few odd examples. Then this is the question from the homework:

"Determine a region of the xy-plane for which the given differential equation would have a unique solution through a point (x_0, y_0) in the region. dy/dx = y^(2/3) "

The wording has got me confused, and I am not really sure what they want. The solution is a family of curves, so they want the particular curve at the point (x_0, y_0)? Or they want a region of something?

Thx for any help you have.
The "wording" is "Determine a region" and doesn't say anything about solving the equation. I don't see anything confusing about that. Picard's theorem says that the d.e. dy/dx= f(x,y), with boundary condition y(x0)= y0 has a unique solution in a region containing (x0, y0) if f(x,y) is continuous in that region and f(x,y) is "Lipschitz" in y in that region (beginning texts use the simpler but less general condition that fy(x,y) be continuous in the region). So: where are those not true? As long as your region avoids places where that is not true, you are okay.

In this particular case, it is easy to find the general solution and to see if it is possible to find another solution for some points (x0,y0) but that is not required.
 
Ok, so the only place place these conditions are not true is at y=0, so my region is everywhere but y=0. And this region is the region where the family of solutions occupy?

THx for the help, I think I can do the problems though I don't fully grasp what I am doing.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top