My Math Forum Limit of solutions of ODE solves ODE?

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 March 13th, 2011, 01:17 PM #1 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 Limit of solutions of ODE solves ODE? Here's the thing: you've got an decreasing and (uniformly) bounded sequence of functions $\varphi_n$ with $\varphi_n'(t)=f(t,\varphi_n(t))$ and $\varphi_n(A_n)=B_n$. Take $\varphi$ to be the pointwise limit of $\varphi_n$; what sort of conditions on f can be given to ensure that $\varphi(t)=f(t,\varphi(t))$? (Conditions on An and Bn are also negotiable, but I'd prefer not to go down that road.)
 March 14th, 2011, 12:03 AM #2 Member   Joined: Nov 2007 Posts: 73 Thanks: 0 Re: Limit of solutions of ODE solves ODE? There are different sort of conditions for f. If you think in the integral form of the equation, you obtain: $\varphi(t)=\varphi(0)+\lim_{n\rightarrow \infty} \int_0^t f(s,\varphi_n(s))$ Then your problem is reduced to apply a certain convergence theorem. A couple of options: -If you want to apply the Lebesgue's dominated convergence theorem, you must require $f(s,\varphi_n(s))$ bounded by a function $g(s)$, not depending on $n$. -If you want to apply monotone convergence theorem, you need nonnegativity of f, and $f(s,\varphi_n(s))$ increasing in n (or viceversa). Probably you can look for more precise conditions on f, using better convergence theorems. Be happy!
 June 16th, 2011, 02:18 PM #3 Member   Joined: Nov 2009 Posts: 72 Thanks: 0 Re: Limit of solutions of ODE solves ODE? Thanks, that was very helpful. Where could I find some better convergence theorems? (essentially all I'm familiar with are MCT, DCT, and Fatou's lemma).
 June 24th, 2011, 01:17 AM #4 Member   Joined: Nov 2007 Posts: 73 Thanks: 0 Re: Limit of solutions of ODE solves ODE? Probably, the "best" one is Vitali's convergence theorem. Roughly speaking, Vitali's theorem is the equivalence between L^p convergence, and the fact that the integral goes to zero when the measure of the region of integration goes to zero. The point with this theorem is the equivalence. You can use it in both ways.

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