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For the first order system $$x_1' = x_2, \hspace{1cm} x_2' = -\sin(x_1) - x_2|x_2| + \cos(t),$$ show that the right-hand side satisifies a Lipschitz condition in the domain $|x| \leq \beta$ for $|t| \leq \alpha$, where $\alpha$ and $\beta$ are arbitrary but finite numbers. Use Picard Iteration to deduce that the initial-value problem $$x_1(0)=0, \hspace{1cm} x_2(0)=0$$ has a unique solution for $|t|\leq\delta$, and obtain an estimate for $\delta$. By allowing $\alpha$ and $\beta$ to be as large as possible, attempt to improve your estimate of $\delta$.

I am confused on how to estimate $\delta$, and consequently how to improve my choice of $\delta$. Anyway, my work up until that point is:

First we will show that the right hand side satisifies a Lipschitz condition on the specified domain. We have that $x_1$ satisifies the Lipschitz condition since $$x_1' = x_2 \leq |x| \leq \beta,$$ so that its derivative is bounded. Moreover, $x_2$ satisifies the Lipschitz condition since $$x_2' = -\sin(x_1) - x_2|x_2| + \cos(t) \leq 2 + x_2|x_2| \leq 2 + |x|^2 \leq 2 + \beta^2$$ so we see that $x_2'$ is bounded as well, implying that it is Lipschitz. Now by Picard's Iterations since $D:=|x-0| \leq \beta$ and $I:=|t-0| \leq \alpha$ for some $\alpha,\beta>0$ satisfy the Lipschitz condition with some $L>0$, we have that the IVP: $$\begin{cases} x'(t) &= f(x,t) \\ x_1(0) &= 0 \\ x_2(0) &=0 \end{cases}$$ has a unqiue solution on $|t| \leq \delta$, where $$\delta := \min\left\{\alpha, \dfrac{\beta}{M}\right\} \hspace{1cm} \text{and} \hspace{1cm} M:=\max_{D\times I} \|\ f(x,t) \|\ .$$ An estimate for $\delta$ would clearly be either $\alpha$ or $\beta/M$ by how $\delta$ is defined, but even if that is the estimate we're looking for, how would I improve that $\delta$ with $\alpha$ and $\beta$ getting larger.

Dragonite
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  • It is not $x_1$ and $x_2$ which need to satisfy a Lipschitz-condition, but the right-handed side $f$ (but your proof for $x_2'$ is indeed enough for the first component of $f$). – Viertel Jan 26 '18 at 19:45
  • Oh, so for the second component of $f$, would I differentiate the $-\sin(x_1) - x_2|x_2| + \cos(t)$ with respect to $x_2$? If so I get, $\dfrac{-2x_2^2}{|x_2|} \leq 2x_2 \leq 2|x| \leq 2\beta$ to get Lipschitz condition in the second component. Or am I still misunderstanding? – Dragonite Jan 26 '18 at 20:22
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    No, that's not correct (my first comment was probably also misleading). $f$ satisfying a Lipschitz condition means that there exists a constant $L(\alpha, \beta)$ (depending on $\alpha$ and $\beta$) such that for $|t| \leq \alpha$ and $||x||, ||y|| \leq \beta$: $||f(t,x) - f(t,y)|| \leq L(\alpha, \beta) ||x - y||$. It suffices to consider $f_1$ and $f_2$ and since $f_1(t,x) = x_2$ you clearly already have $|f_1(t, x) - f_2(t, y)| = |x_2-y_2| \leq ||x - y||$. For $f_2$ you need to find something similar but differentiating with respect to one variable won't do it. – Viertel Jan 26 '18 at 20:55
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    Once you've got that, this link might give you an idea on how to optimize $\delta$ – Viertel Jan 27 '18 at 07:38
  • So, let $Q:=\max(|y_2|,|x_2|)$ and then we have: $|\ f(t,x) - f(t,y) || = |\ (-\sin(x_1) - x_2|x_2| + \cos(t)) - (-\sin(y_1) - y_2|y_2| + \cos(t)) |\ = |\ (-\sin(x_1) - x_2|x_2| + \sin(y_1) + y_2|y_2| |\ \leq |\ 2 - x_2|x_2| + y_2|y_2| |\ \leq |\ 2 + Q(y_2 - x_2) |\ \leq 2 + Q |\ y_2 - x_2 |\ \leq 2 + Q |\ x - y |\ $. Probably more long-winded than it needed to be, but will that suffice? – Dragonite Jan 28 '18 at 14:50
  • Replace $f$ with $f_2$, typo. Won't let me edit for some reason. :/ – Dragonite Jan 28 '18 at 15:13
  • No, it doesn't suffice as 2 + ... $Q ||x-y||$ is not a Lipschitz condition. So you need to do this differently. Also on a smaller note, you're probably aware of this, you should not write 2 inside the norm. – Viertel Jan 28 '18 at 18:39

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