Consider the DE $y'=1+y^2 , y(0) = 0$

Show that the associated integram map $T$ is not a contraction on $C[-r,r]$ for any $r\gt 0$ Then, find an $r \gt 0$ sich that $T$ maps the unit ball of $C[-r,r]$ into itself is a contraction mapping on this ball.

My logic, T wont be a contraction if it has multiple fixed points, so since y = tanx if we take $r = \pi$, tan (0) = tan($\pi$) = 0 so the fixed point is not unique. More over, if we choose $r \lt \pi$ clearly y is one to one so all fixed points are unique.


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    $\begingroup$ What is an 'integram map'? Integral map? $\endgroup$ – copper.hat Mar 8 '13 at 18:03

Choose $x \in C[-r,r]$, and define $Tx(t) = \int_0^t (1+x^2(\tau)) d\tau$ (I believe this is the associated map you asked about?).

Then $(Tx_1-Tx_2 )(t) = \int_0^t (x_1^2(\tau)-x_2^2(\tau)) d\tau = \int_0^t (x_1(\tau)+x_2(\tau)) (x_1(\tau)-x_2(\tau)) dt$.

Choose $K >0$ and let $x_1(t) = \frac{K+1}{2}$, and $x_2(t) = \frac{K-1}{2}$. Then $(Tx_1-Tx_2 )(t) = Kt$, and so $\|Tx_1-Tx_2\| = Kr$, while $\|x_1-x_2\| = 1$. Clearly for all $r>0$, we can choose $K$ such that $T$ is not a contraction.

However, if $x_1,x_2 \in B(0,L)$ for some $L>0$, then, from above, we have the estimate $|(Tx_1-Tx_2 )(t) | \le \int_0^t |x_1(\tau)+x_2(\tau)| |x_1(\tau)-x_2(\tau)| dt \le2Lt \|x_1-x_2\| $, which gives $\|T x_1 - Tx_2\| \le 2Lr \|x_1-x_2\| $. Hence if we choose $r < \frac{1}{2L}$, $T$ will be a contraction on $B(0,L)$.


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