ODE solution question

  • I
  • Thread starter Mentz114
  • Start date
  • #1
5,428
291

Main Question or Discussion Point

I've been solving these two ODEs

##\frac{d}{d\,r}\,A=F(A,r) + \epsilon f(r)## and ##\frac{d}{d\,r}\,A=F(A,r)##.

If the solutions are respectively ##A_1(r,\epsilon)## and ##A_2(r)## then will ##A_1(r,0) = A_2(r)## ?
I realize the answer could depend on the actual functions but with the ones I'm using it appears that setting ##\epsilon=0## does not recover ##A_2##.

I'd be grateful for any advice on this.
 

Answers and Replies

  • #2
Math_QED
Science Advisor
Homework Helper
2019 Award
1,658
678
Weird stuff. In this answer, I assume that your differential equation has a unique solution (which is not always the case)

So for every ##\epsilon##, you have the ODE
$$\frac{d}{dr} A = F(A,r) + \epsilon f(r)$$

with solution ##A_1(r,\epsilon)##. In particular, for ##\epsilon = 0## you get the solution ##A_1(r,0)## which is a solution for ##\frac{d}{dr} A = F(A,r)##, so you should get the same solutions (if they agree at a point and the differential equation behaves nicely enough to get a unique solution).

Maybe post the exact functions/problem so we can see what goes wrong.
 
  • #3
7
0
Yes, I see it now. I'm not sure how I missed it. At least it's a constant as I assumed. :cool:






Lucky Patcher Kodi Nox
 
  • #4
1
0
I'd be grateful for any advice on this.
 

Related Threads on ODE solution question

Replies
1
Views
1K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
900
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
10
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
0
Views
2K
Top