ODE Solution Question: Comparing A1(r,ε) and A2(r) with ε=0

[Mentz114]

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.

 

[member 587159]

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.

 

[hilbert2]

The question is quite ambiguous, too, because it would seem to suggest that ##F## is an integral function of ##f## but the two don't depend on same variables.

 

[jim mcnamara]

The question is old and the OP has not been on PF in a long time. If you want to discuss this please create a new thread. This one is locked.