Continuous dependence of maximal interval on right hand side for ode

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Let $f,f_n\colon\mathbb{R}\times(0,1)\to\mathbb{R}$ be local Lipschitz continuous functions and assume they form the right-hand-side of some ordinary differential equation's $\dot{x} = f(t,x)$ and $\dot{x_n} = f_n(t,x_n)$ with initial values $(t_n^0,x_n^0)$ and $(t^0,x^0)$ respectively. Moreover, let us assume that $f_n$ converges to $f$ (in some sense) and $(t_n^0,x_n^0)\to(t^0,x^0)$. From standard theory on ordinary differential equations we know that for each ode, we find a maximal time interval $(t_n^\text{min},t_n^\text{max})$ and $(t^\text{min},t^\text{max})$, respectively, where our solutions uniquely exists.

My question is, when do we have $t_n^\text{max}\to t^\text{max}$?

Concrete, I need the result for $f,f_n\colon\mathbb{R}_+\times(0,1)\to\mathbb{R}_+$ with $f_n\to f$ uniformly on all compact subsets of $\mathbb{R}_+\times(0,1)$. Intuitively, I think this should work as the solutions stay close to each other the whole time. Do you maybe have some literature for such questions?

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Theorem 3.2 in P. Hartman's Ordinary Differential Equations, is what you are looking for. The result says $$ \limsup t^{\text{min}}_n\le t^{\text{min}}<t^{\text{max}}\le\liminf t^{\text{max}}_n. $$

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In general, convergence of $t_n^\text{max}\to t^\text{max}$ is not true. The most one can hope for is stated in this answer by Julián Aguirre. Consider the right hand sides (for $n>2$) $$f,f_n\colon\mathbb{R}\times(0,1)\to\mathbb{R}\\ f_n(t,x)=a_n\cos(t),\\f(t,x)=a\cos(t)$$ where $a_n=\frac{1}{2}−\frac{1}{n}$ and $a=\frac{1}{2}$. Clearly, $f,f_n$ are Lipschitz and we have uniform convergence for $n\to\infty$. Let $(t_n^0,x_n^0)=(t^0,x^0)=(0,\frac{1}{2})$. Then these ODEs are solved by $$x_n\colon\mathbb{R}\to(0,1),\quad x_n(t)=a_n\sin(t)+\frac{1}{2},\\ x\colon\Bigl(−\frac{\pi}{2},\frac{\pi}{2}\Bigr)\to(0,1),\quad x(t)=a\sin(t)+\frac{1}{2}.$$ Hence, the times of explosion are $t_n^\text{max}=\infty$ and $t^\text{max}=\frac{\pi}{2}$, which clearly do not converge.