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.