# Find the largest interval where the Initial Value Problem $y'(t)=t+sin(y(t))$ with $y(2)=1$ has a unique solution. Using Picard-Lindelof Theorem.

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?

• 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? – 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 $${\mathbb R}$$.

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