# Lipschitz continuous

• Dec 2nd 2012, 02:26 PM
lahuxixi
Lipschitz continuous
http://i822.photobucket.com/albums/z...whatisthis.png
This question is about lipschitz continuous, i think the way to check if the solutions can be found as fixed points is just differentiating f(t), but i'm not sure about this. Can anyone give me some hints please? I will really appreciate if you can give me some small hints.
• Dec 2nd 2012, 02:58 PM
emakarov
Re: Lipschitz continuous
You need to check one of the assumptions of Picard–Lindelöf theorem. Namely, denote the Picard operator above by $P(f)$ and the right-hand side of the differential equation by $F(t,x(t))$. If for some constant $q$ we have $|P(f)(t)-P(g)(t)|\le q\|f-g\|_\infty$ for all $t\in[0,T]$, then $\|P(f)-P(g)\|_\infty\le q\|f-g\|_\infty$. We want to find $T$ such that $0\le q<1$. For any $t\in[0,T]$,

\begin{align*}|P(f)(t)-P(g)(t)|&\le\left|\int_0^t F(s,f(s))\,ds-\int_0^t F(s,g(s))\,ds\right|\\&=\left|\int_0^t (F(s,f(s))-F(s,g(s)))\,ds\right|\\&\le\int_0^t |F(s,f(s))-F(s,g(s))|\,ds\\&\le\int_0^t L|f(s)-g(s)|\,ds\\&\le\int_0^t L\|f(s)-g(s)\|_\infty\,ds\\&\le TL\|f(s)-g(s)\|_\infty\end{align*}

where $L$ is the Lipschitz constant of $F$ with respect to the second argument.
• Dec 2nd 2012, 03:05 PM
lahuxixi
Re: Lipschitz continuous
How can we find the Lipschitz constant? and i'm not sure about the first part, the checking part either, can you explain it more clearly?
• Dec 2nd 2012, 03:07 PM
lahuxixi
Re: Lipschitz continuous
"check that the solutions can be found as fixed points of the map"
• Dec 2nd 2012, 04:04 PM
emakarov
Re: Lipschitz continuous
Quote:

Originally Posted by lahuxixi
How can we find the Lipschitz constant? and i'm not sure about the first part, the checking part either, can you explain it more clearly?

You said that you wanted only small hints. (Smile) You don't have to follow the link to the theorem; I wrote the relevant part of the proof. The restriction on T that guarantees that $P$ is a contraction constitutes one of the assumptions of the theorem.

So, we have $F(t,x)=2\cos(tx^2)$. To prove that for every $t\in[0,T]$, $F$ is Lipschitz continuous w.r.t. $x$, we need to find a constant $L$ such that for every $t$, $x_1$ and $x_2$ it is the case that $|F(t,x_1)-F(t,x_2)|\le L|x_1-x_2|$. Using trigonometry,

\begin{align*}|F(t,x_1)-F(t,x_2)|&=4|\sin(t(x_1^2+x_2^2)/2)\sin(t(x_1^2-x_2^2)/2)|\\&\le4|\sin(t(x_1^2-x_2^2)/2)|\\&\le2|t(x_1^2-x_2^2)|\\&=2t|x_1+x_2|\cdot|x_1-x_2|\\&\le2T|x_1+x_2|\cdot|x_1-x_2|\end{align*}

So, we need to find $L$ such that $2T|x_1+x_2|\le L$ for all $t\in[0,T]$. In particular, we need to bound $|x_1+x_2|\le|x_1|+|x_2|$ from above. It is clear that $F(t,x)$ is not Lipschitz on the whole $[0,T]\times\mathbb{R}$. Indeed, given any $t$, if we take larger and larger $x$, the cosine starts to oscillate faster and faster. For this reason, I am starting to doubt that $P$ is a contraction on all continuous functions on $[0, T]$. However, it is possible to find a positive constant $U$ such that $P$ is a contraction in $\{f:C[0,T]:\|f\|\le U\}$. This is sufficient for Picard theorem if we show that $P$ maps $\{f:C[0,T]:\|f\|\le U\}$ into itself. In this respect, note that $|P(f)|\le 1+\int_0^t 2|\cos(sf^2(s))|\,ds\le1+2T$.
• Dec 2nd 2012, 05:45 PM
lahuxixi
Re: Lipschitz continuous
thank you for your help, its because i didnt think that i wouldnt be able to understand, guess i will have to revise now.