Maximal interval of existence for $x'=\frac{xt}{\sqrt{x^2+1}}$

Question

I want to find the maximal interval of existence for $$x'=f(x,t)=\frac{xt}{\sqrt{x^2+1}},\ x(0)=1$$ and I want to use a theorem I found in one of my books. It is a certain statement of boundary behaviour of maximal solutions which I couldn't find in a wikipedia article to link. So just to make sure we are on the same page I will provide the theorem:

Let $D\subseteq \mathbb R\times \mathbb R^n$ be a domain and $f:D\to\mathbb R^n,\ (t,x)\mapsto f(t,x)$ be continuous and locally Lipschitz with respect to $x$. For $(\tau,\zeta)\in D$ let $\lambda:]a,b[\to\mathbb R^n$ be a solution of the IVP $$x'=f(t,x),\ x(\tau)=\zeta$$ Then $\lambda$ is a maximal solution to the IVP iff one of the conditions hold:

1. $a=-\infty$
2. $a>-\infty$ and $\limsup_{t\to a}||\lambda(t)||=\infty$
3. $a>-\infty,\partial D=\emptyset$ and $\lim_{t\to a}\text{distance}(\partial D,(t,\lambda(t)))=0$

Replacing the $-\infty$ with $\infty$ yields the conditions for $b$ so I will not list them here.

Now back to the problem: Since $f$ is continuously differentiable with respect to $x$ it is clear that $f$ is locally Lipschitz and therefore we have a unique maximal solution to the IVP with I will call $\lambda:\ ]a,b[\to\mathbb R$. The argument for $a$ should be the same as for $b$ so I will only care about the first one. Obviously we have $\partial D=\emptyset$ so (3.) is not an option and we only need to take care of (2.). So let's assume $a>-\infty$, then I am using $$f(t,x)\leq\text{sgn}(x)t$$ to obtain $$\limsup_{t\to a}|\lambda(t)|=\limsup_{t\to a}|\lambda(0)+\int_0^t\lambda'(s)ds|=\limsup_{t\to a}|1+\int_0^t f(s,x)ds|\leq$$ $$\leq\limsup_{t\to a}|1+\text{sgn}(x)\int_0^t sds|=|1+\text{sgn}(x)\cdot\frac{a^2}{2}|<\infty$$ so only option (1.) is left and the same goes for $b$, so in total we have that the maximal interval of existence is $\mathbb R$. Is my reasoning correct here? And if not, what is wrong and how do I fix it?

Answer 1

Using $\frac{|x|}{\sqrt{1+x^2}}\le 1$ gives $$ |f(t,x)|\le |t|\implies |x(t)-x(0)|\le\int_0^t|f(t,x)|\,dt\le\frac{|t|^2}2 $$ so that the solution to the IVP, where it is defined, satisfies $$ 1-\frac12t^2\le x(t)\le 1+\frac12t^2. $$ Per the cited possibilities for the behavior at the end points of the maximal domain, this excludes any finite end points.