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August 9th, 2010, 11:16 AM   #1
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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 and , , be the solution to the initial value problem




where is a bounded function (which doesn't depend on ). The conjecture is that



as , where a, b are constants. I actually think is real analytic in .

The existence of the constant follows easily since the function is decreasing and bounded below, as decreases to zero. I don't know how to show that



exists. Any help or idea is appreciated.
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August 14th, 2010, 12:05 PM   #2
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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.
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August 15th, 2010, 04:44 PM   #3
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Re: A 1st order non-linear ODE

Quote:
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 . For , the solution is defined and satisfies 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 . The differential equation implies . This contradicts the choice of s.
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August 15th, 2010, 04:48 PM   #4
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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



In other words, the question is whether there exist constants a, b such that



Any help is appreciated.
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August 17th, 2010, 09:16 AM   #5
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Re: A 1st order non-linear ODE

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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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