Existence of Solution for Initial value Problem $y'(t)=\cos(ty), t>0$ and $y(0)=\frac{1}{n}, n\in\mathbb{N}$ using Ascoli Arzela

Question

Consider the cauchy problem

(1)\begin{cases} y'(t)=\frac{1}{1+ty}, & t>0 \\ y(0)=1+\frac{1}{n} & n\in\mathbb{N} \end{cases}

(2)\begin{cases} y'(t)=\cos(ty), & t>0 \\ y(0)=\frac{1}{n} & n\in\mathbb{N} \end{cases} (a) Prove that for every $$ there exists a solution $y_n(.)$ defined on the whole $\mathbb{R}^+$.

(b) Prove that the sequence $(y_n)$ admits a subsequence uniformly converging on any compact subinterval of $\mathbb{R}^+$ to the solution of the Cauchy problem $y'(t)=\cos(ty), t>0$ and $y(0)=0$.

My attempt

(1a) Since $t>0, y>0$, $y'(t)>0$. This and definition of $y'$ implies $0<y_n'(t)<1, \forall n\in\mathbb{N}$. Thus \begin{equation} 0<y_n(t)=y_n(0)+\int_0^ty_n'(s)ds\leq 2+t, t>0 \end{equation} which is the required solution.

(1b) Let $I=[0,1]$. Since $0\leq y_n'(t)\leq 1$ and $y_n'$ is Lipschitz.Thus, $(y_n)$ is Equicontinous and by Arzela Ascoli the result follows.

(2) My problem is here. Since $-1\leq y'(t)\leq 1, t>0$, graph of $y$ alternate and I do not know how to get boundedness for the family $(y_n)$

Answer 1

(a) Since $-1\leq y'(t)\leq 1, t>0$, we have that $-1\leq y_n'(t)\leq 1, t>0, n\in\mathbb{N}$. Thus \begin{equation} y_n(t)=\frac{1}{n}+\int_0^t y_n'(s)ds\end{equation} which implies \begin{equation}-t\leq y_n(t)\leq \frac{1}{n}+t\leq 1+t.\end{equation} Thus $(y_n)_n$ is the required solution.

(b) Let $I=[0,1]$. This implies $-1\leq y_n(t)\leq 2$ equivalently $|y_n(t)|\leq 2$ (uniformly bounded).

Since $y_n'$ is continuous for each $n\in\mathbb{N}$, $(y_n)_n$ is a Lipschitz family. So they are Equicontinuous. By Arzela Ascoli, there exist $(y_{n_k})_k$ converging to $y$ where $y(0)=\lim \frac{1}{n_k}=0$. Hence, the result.

Answer 2

The task is easily solvable using the Grönwall theorem, as the right sides have easy Lipschitz constants $L(t)=t$.

For two solutions $y,z$ you get for the difference the upper bounds $$ |\dot z(t)-\dot y(t)|\le t|z(t)-y(t)| \implies |z(t)-y(t)|\le e^{t^2/2}\,|z(0)-y(0)|. $$ This implies uniform convergence on any finite intervals, as for $t\in[0,T]$ one has $$ |y_n(t)-y_\infty(t)|\le \frac1n\, e^{T^2/2} $$