Properties of the solutions to $x'=t-x^2$

Let $f_c$ be the solution to $$ \left\{ \begin{array}{c} x'=t-x^2 \\ x(0) =c \end{array} \right. $$

I'm trying to prove:

1. If $c \geq 0$ then $f_c(t)$ is defined for all $t>0$
2. There is a unique $c_0$ such that $f_{c_0}(t)$ is defined for all $t>0$ and $\lim_{t\to +\infty} (f_{c_0}(t)+\sqrt t) =0$.

Solutions with $c\ge 0$ drop below $\sqrt{t}$ and stay between $0$ and $\sqrt{t}$

For every $c\ge 0$ there exists $t_0$ such that $f_c(t_0)=\sqrt{t_0}$. Furthermore, $f_c(t)>\sqrt{t_0}$ for $0\le t<t_0$ and $0\le f_c(t)<\sqrt{t_0}$ for $t>t_0$

Proof. By the initial condition, $f_c(0)\ge 0=\sqrt{0}$. Thus, if no $t_0$ exists, then $f_c (t)>\sqrt{t}$ for all $t>0$. According to the ODE, this implies $f_c(t)'<0$, so $f_c$ is decreasing, hence $f_c(t)\le c$ for all $t>0$. But $\sqrt{t}$ is unbounded; a contradiction.

So, let $t_0$ be the first $t$ such that $f_c(t_0)=\sqrt{t_0}$. By construction, $f_c(t)>\sqrt{t_0}$ for $0\le t<t_0$. Suppose that there is $t>t_0$ such that $f_c(t)=\sqrt{t}$; let $t_1$ be the smallest such $t$. Then $f_c'(t_1)=0$, which implies that in some neighborhood of $t_1$, the function $f_c(t)-\sqrt{t}$ is decreasing. But this is impossible, because $f_c(t)-\sqrt{t}<0$ for $t_0<t<t_1$.

Finally, since $f_c'(t)>0$ for $t>t_0$, it follows that $f_c$ never becomes negative. In fact, $t_0$ is its point of minimum, with the minimum value being $\sqrt{t_0}$.

Solutions with $c\ge 0$ exist globally

Global existence of solution follow from the Picard theorem, because

1. the right hand side $F(x,t)=t-x^2$ is Lipschitz on bounded rectangles;
2. on any interval $[a,b]$ the solution stays within the rectangle $[a,b]\times [0,\sqrt{b}]$.

Uniqueness of $c_0$ such that $f_{c_0}(t)=-\sqrt{t}+o(1)$ as $t\to\infty$

By the above, such $c_0$ must be negative. If $b<c_0$, then $f_b(t)-f_{c_0}(t)$ is a decreasing function of $t$, since smaller values of solution yield smaller values of derivative in the negative range. Hence, $f_b(t)-f_c(t)$ is bounded above by $b-c_0<0$, and since $f_{c_0}(t)+\sqrt{t}$ converges to zero, $f_{b}(t)+\sqrt{t}$ cannot converge to zero.

Existence of $c_0$ such that $f_{c_0}(t)=-\sqrt{t}+o(1)$ as $t\to\infty$

The set $U=\{c : \exists t>0 \ f_c(t)>-\sqrt{t}\}$ is open: this follows from the continuity of $f_c$ with respect to $c$. The only possible candidate for $c_0$ is $\inf U$. That $U$ is bounded below can be shown as follows:

1. The IVP $x'=1-x^2$ with $x(0)=-2$ has solution $x(t) = \frac{e^{2t}+3}{e^{2t}-3}$ which escapes to $-\infty$ at $t=\frac12\ln 3<1$.
2. The IVP $x'=t-x^2$ with $x(0)=-2$ has solution that stays below the solution in item 1 on the interval $[0,1]$; therefore, it must also escape to $-\infty$.

So, $\inf U\ge -2$.

I don't see a direct proof of the fact that for $c=\inf U$ the solution is asymptotic to $-\sqrt{t}$. The standard reduction of Riccati equation to 2nd order linear equation (which turns out to be the Airy equation) relates this fact to the existence of an Airy function that is $o(1)$ as $t\to\infty$.

Solutions with $c\ge 0$ satisfy $f_c(t)=\sqrt{t}+o(1)$ as $t\to\infty$

Since we already know that solution drops below $\sqrt{t}$, it suffices to prove: if $c\ge 0$, then $$ f_c(t)>t^{1/2}-t^{-1/2},\quad t\ge 1 \tag{1} $$ Let $g(t)=t^{1/2}-t^{-1/2}$. The key point is that $g$ satisfies the inequality $$g'(t) < t-g(t)^2,\quad t>1\tag{2}$$ because $$g'(t)=\frac12 t^{-1/2}+\frac12 t^{-3/2} < 1$$ and $$ t-g(t)^2 = t - t + 2 - t^{-1} >1 $$ By virtue of (2), $f_c-g$ cannot change sign from positive to negative: at a point where $f_c=g$, the inequality $f_c'>g'$ holds. And since $f_c(1)>0=g(1)$, it follows that $f_c>g$ on $[1,\infty)$.