Consider the initial value problem (IVP) \begin{cases} y'(t)= t + \sin(y(t)), \\ y(2) = 1. \\ \end{cases} Find the largest interval $\mathcal{I}\subset \mathbb{R}$ containing $t_0=2$ so that the problem has a unique solution $y$ in $\mathcal{I}$.

My proof attempt:

Let $L>\frac{\pi}{2}>0$. Define $$ R := \{(t,y)\in \mathbb{R}^2:|t-2|\leq L, |y-1|\leq L \} $$ Then $$ 2-L\leq t \leq 2+L \text{ and } 1-L\leq y \leq 1+L $$ Since $L>\frac{\pi}{2} \Rightarrow \exists y_0\in (1-L, 1+L)$ so that $\sin(y_0)=1$. Hence, $$ M = \underset{R}{\sup}|F(t,y)|=3+L $$ Then $$ |\partial_y F(t,y)|=|\cos(y)|\leq 1 $$ Put $c = 1$, then by the Mean Value theorem, for $(t,y),(t,u)\in R \Rightarrow$ $$ |F(t,y)-F(t,u)|\leq c|y-u| $$ Let $a_{*}=\min \left(L, \frac{L}{M}\right)=\min \left(L, \frac{L}{3+L}\right)=\frac{L}{3+L}$. Using Picard-Lindelof's theorem, there should exist a unique solution on the interval $\mathcal{I}=\left[2-\frac{L}{3+L}, 2+\frac{L}{3+L}\right]$.

Since $\underset{L\rightarrow \infty}{\lim}\frac{L}{3+L}=1$, our largest interval should be $\mathcal{I}=\left[1+\varepsilon,3-\varepsilon\right]$ where $\varepsilon \in (0,1)$.

Am I on the right track here?

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    $\begingroup$ I have also read somewhere that there is a global solution (i.e. $\mathcal{I}=\mathbb{R}$) if F(t,y) is globally lipschitz. Which I believe it is since $\nabla F=(1, \cos(y))$. Which should be bounded under the operator norm. So how do I reconcile this fact with what I did up above? Is what I did wrong? $\endgroup$ – Joe Man Analysis Nov 19 '18 at 4:49

From the Picard theorem one can infer the following: Given an ODE $y'=f(t,y)$ with $f$ defined in an open set $\Omega\subset{\mathbb R}^2$ and fulfilling the assumptions of the theorem in the neighborhood of each point $(t,y)\in\Omega$, any solution of an IVP $y(t_0)=y_0$ can be extended to a maximal solution $$\tilde y:\quad J\mapsto{\mathbb R},\quad x\mapsto \tilde y(x)$$ in an unique way. Here $J$ is an open interval which may depend on the given initial point $(t_0,y_0)$. Furthermore the graph ${\cal G}(\tilde y)\subset\Omega$ of this maximal solution will "ultimately" leave any compact set $K\subset\Omega$ given in advance. (For example, the solution cannot develop a $x\mapsto\sin{1\over x}$ singularity in the interior of $\Omega$.)

In the case at hand we have $f(t,y)=t+\sin y$ and $\Omega={\mathbb R}^2$. It follows that for any solution $t\mapsto y(t)$ one has $|y'(t)|\leq |t|+1$. This allows to conclude that $|\tilde y(t)|\leq C(1+t^2)$ for a suitable $C>0$; hence $\tilde y$ cannot drift away to $\pm \infty$ in finite time. Since ${\cal G}(\tilde y)$ will ultimately leave any compact rectangle $K:[-M,M]\times[-C(2+M^2),C(2+M^2]$, and cannot do so across the bottom and top edges, it follows that $\tilde y$ is defined on all of ${\mathbb R}$.

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    $\begingroup$ Could you please explain how the bound on the maximal solution was obtained? $\endgroup$ – Joe Man Analysis Nov 19 '18 at 10:49

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