My Math Forum A 1st order non-linear ODE

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 August 9th, 2010, 11:16 AM #1 Newbie   Joined: Aug 2010 Posts: 8 Thanks: 0 A 1st order non-linear ODE I am proving a theorem and I need to understand the asymptotic behavior of the solution to a differential equation, which probably can't be solved explicitly. Let $\epsilon > 0$ and $u_{\epsilon}(t)$, $t \geq 0$, be the solution to the initial value problem $u'(t) = \frac{1}{u(t)} + \psi(t)$ $u(0)= \epsilon > 0=$ where $\psi(t)$ is a bounded $C^{\infty}$ function (which doesn't depend on $\epsilon$). The conjecture is that $u_{\epsilon}(1)= a + b \epsilon^2 + O(\epsilon^4)$ as $\epsilon \to 0$, where a, b are constants. I actually think $u_{\epsilon}(1)$ is real analytic in $\epsilon^2$. The existence of the constant $a= \lim_{\epsilon \to 0} u_{\epsilon}(1)$ follows easily since the function is decreasing and bounded below, as $\epsilon$ decreases to zero. I don't know how to show that $\lim_{\epsilon \to 0} \frac{u_{\epsilon}(1) - a}{\epsilon^2}$ exists. Any help or idea is appreciated.
 August 14th, 2010, 12:05 PM #2 Member     Joined: Aug 2007 Posts: 42 Thanks: 4 Re: A 1st order non-linear ODE I am not sure your solutions must be defined up to x=1. Also, for smooth but not analytic psi, the expansion for u(1) is not valid.
August 15th, 2010, 04:44 PM   #3
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Re: A 1st order non-linear ODE

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 Originally Posted by Rebesques I am not sure your solutions must be defined up to x=1. Also, for smooth but not analytic psi, the expansion for u(1) is not valid.
Let $M= \sup_{t} \left| \psi(t) \right|$. For $0 < \epsilon < \frac{1}{M}$, the solution is defined and satisfies $u(t) \geq \epsilon$ for all t >= 0. This can be proved by contradiction. Suppose the solution is defined only up to a finite time T > 0. Necessarily, u(T) = 0. (Otherwise, the local existence and uniqueness theorem says we can extend the solution for a little longer time.) Consider the last time s < T for which $u(s)= \epsilon$. The differential equation implies $u'(s) > 0$. This contradicts the choice of s.

 August 15th, 2010, 04:48 PM #4 Newbie   Joined: Aug 2010 Posts: 8 Thanks: 0 Re: A 1st order non-linear ODE Actually I am not sure about the error term. I will be equally happy if anyone can prove or disprove $u_{\epsilon}(1)= a + b \epsilon^2 + o(\epsilon^2)$ In other words, the question is whether there exist constants a, b such that $\lim_{\epsilon \downarrow 0} \frac{u_{\epsilon}(1) - a - b \epsilon^2}{\epsilon^2}= 0$ Any help is appreciated.
August 17th, 2010, 09:16 AM   #5
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Re: A 1st order non-linear ODE

Quote:
 Originally Posted by Rebesques Also, for smooth but not analytic psi, the expansion for u(1) is not valid.
For smooth but not anlytic psi, we can see from the differential equation that u_{epsilon}(t) is not analytic in t, but how do you know it is not analytic in epsilon?

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