How many solutions does $x'=\sin\left(x^2+t^2\right)+\frac{t^2}{t^2+1}\left|x\right|, x\left(t_0\right)=x_0$ have and is it extendable on R?

Question

Let $\left(t_0,\:x_0\right)$ from $R^2$ be arbitrary. How many solutions does this Cauchy's problem have:

$$x'=\sin\left(x^2+t^2\right)+\frac{t^2}{t^2+1}\left|x\right|, x\left(t_0\right)=x_0$$

Are they extendable on $\Bbb R$? Justify the answer.

Answer 1

Clearly $$ f(x,t)=\sin\left(x^2+t^2\right)+\frac{t^2}{t^2+1}\left|x\right| $$ is locally Lipschit. So $x'=f(x,t), x(t_0)=x_0$ has a unique solution $x(t)$ near $(x_0,t_0)$. Extend $x(t)$ maximally to $(t_0,t_1)$. Suppose we have $$t_1<\infty, \lim_{t\to t_1^-}|x(t)|=\infty. \tag{1}$$ On the other hand, $$ |x(t)|\le \bigg|x(t_0)+\int_{t_0}^t(\sin\left(x^2(s)+s^2\right)+\frac{s^2}{s^2+1}|x(s)|)ds\bigg|\le |x(t_0)|+(t-t_0)+\int_{t_0}^t|x(s)|ds. $$ By Gronwall's inequality, one has $$ |x(t)|\le(|x(t_0)|+(t-t_0))e^t $$ and hence $$ \lim_{t\to t_1^-}|x(t)|<\infty. \tag{2}$$ Thus (1) and (2) are against each other and hence $t_1=\infty$. So $x(t)$ can be extended to $(t_0,\infty)$. Similarly $x(t)$ can be extended to $(-\infty,t_0)$.