This is a homework question. Given an initial value problem, $\frac{dx}{dt} = f(t,x(t)), \text{ } x(0) = 0$ with the function defined as:
$$ f(t,x(t)) = \begin{cases} \hfill 0 \hfill & t \leq 0 \\ \hfill 2t \hfill & t>0, \text{ } x \leq 0 \\ \hfill 2t-4\frac{x}{t} \hfill & t>0, \text{ } 0<x<t^2\\ \hfill -2t \hfill & t>0, \text{ } x \geq t^2, \end{cases} $$ I have to show that the $x_n$ sequence(acquired by Picard-iteration) is not convergent, nor does it have a convergent subsequence which converges to a solution. I'd like to be provided the least help possible.

What I've so far tried is:

  1. Showing f is not locally Lipschitz-continuous. This is part of the problem ("Notice that f is continuous, but not locally Lipschitz-continuous!"). As I remember, this is done by evaluating the partial derivatives. The derivative with respect to t is 0 in the first, and 2 in the second interval (with x being x(t), I'm not sure this is the correct method, though), so I did no further evaluation.
  2. Calculating the $x_n$ members. This goes as follows ($x_0 := x(0) = 0) $: $$ x_{n+1} = x(0) + \int_\limits{0}^t f(s, x_n (s))ds. $$ What I got is $$ x_n(t) = \begin{cases} \hfill 0 \hfill & t \leq 0 \\ \hfill \frac{2t^{n+1}}{(n+1)!} \hfill & t>0, \text{ } x<t^2 \\ \hfill -\frac{2t^{n+1}}{(n+1)!} \hfill & t>0, \text{ } x \geq t^2. \end{cases} $$ At this point, I ask the question: are these two steps correct (in showing f is not Lipschitz-continuous locally, and acquiring $x_n(t)$)? My next step would be first to substitute $x_n$ into the initial value problem, getting: $$ \begin{cases} \hfill 0 = 0 \hfill & t \leq 0 \\ \hfill \frac{2t^{n}}{n!}=2t \hfill & t>0, \text{ } x_n \leq 0 \\ \hfill \frac{2t^{n}}{n!}=2t-4\frac{\frac{2t^{n+1}}{(n+1)!}}{t} \hfill & t>0, \text{ } 0<x_n<t^2\\ \hfill -\frac{2t^{n+1}}{(n+1)!}=-2t \hfill & t>0, \text{ } x_n \geq t^2, \end{cases} $$ $$ \begin{cases} \hfill 0 = 0 \hfill & t \leq 0 \\ \hfill \frac{t^{n-1}}{n!}=1 \hfill & t>0, \text{ } x_n \leq 0 \\ \hfill \frac{(n+5)t^{n-1}}{(n+1)!}=1 \hfill & t>0, \text{ } 0<x_n<t^2\\ \hfill \frac{t^{n}}{(n+1)!}=1 \hfill & t>0, \text{ } x_n \geq t^2. \end{cases} $$ This leads to quite strange inequalities. Substituting into $x_n$ I could get contradictions on all except the first line, I believe, and thus I could show that for any n, it cannot satisfy the initial value problem. But how do I show that the limit does not satisfy it either (with t remaining as a variable)?

Edit: This seems to be an exercise in the book Solving Ordinary Differential Equations I: Nonstiff Problems (I.8 / 5.).

  • $\begingroup$ Are you given what is $x_0(t)$? $\endgroup$ – user99914 Oct 4 '15 at 10:04
  • $\begingroup$ According to this, $x_0(t) := x_0$ for every t (this means $x_0(t) = 0$). $\endgroup$ – Mitlasóczki Bence Oct 4 '15 at 11:18

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