I'm studying a problem from an old ODE exam and I have some general ideas on how to solve it but I feel as though these arguments are kind of heuristically clear but wanting in formality. We're given the initial value problem
$$ y' = e^{-t} - \arctan(y),$$$$ y(0) = 0.$$
And we're asked to
- Prove there is a unique global solution to this problem.
- Study the monotonicity of the solution.
- Calculate the limits $\lim_{t \to \pm\infty} y(t)$ of the solution.
- Sketch a graph of the solutions.
For the first item, clearly $f(t,y) = e^{-t} - \arctan(y)$ is continous on $t$, and
$$ \frac{\partial f(t,y)}{\partial y} = - \frac{1}{1+y^2}, $$$$ \left| \frac{\partial f(t,y)}{\partial y} \right| \leq 1. $$
Since this partial derivative is bounded, we know the function must be Lipschitz continous everywhere and thus by the Picard-Lindelöf theorem we have the existence of global solutions.
For the second item, note that monotonicity in some interval means that the derivative is strictly positive or strictly negative in this interval. So either we have a decreasing function if $e^{-t} < \arctan{y}$ or an increasing function if $e^{-t} > \arctan{y}$. Since $y(0)=0$, by continuity, we ought to have an interval around the origin such that $y'>0$ if $t$ stays within this interval. So we have at least locally that the solutions are strictly increasing around $0$. Now, the general idea that I have is that whatever $y$ ends up being, it's still instide $\arctan$, which can only go from $-\pi/2$ to $\pi/2$. So it has to happen that at some (negative) $t_1$, $e^{-t} > \arctan(y(t_1))$ for every $t<t_1$. Similarly, we know heuristically that $e^{-t}$ ought to go to zero faster than $\arctan(y)$, so for some positive $t_2$ we should have $e^{-t_2} < \arctan(y(t))$ for every $t_2<t$. So $y$ should be strictly decreasing beyond this point. I checked this out on Mathematica and indeed we have the following picture:
The point at which this changes is not entirely clear to me however.
Now, for the third item, we know that
$$ \lim_{t \to \infty} y'+ \arctan(y) = \lim_{t \to \infty} e^{-t} = 0,$$
$$ \lim_{t \to -\infty} y'+ \arctan(y) = \lim_{t \to \infty} e^{t} = \infty.$$
We know that $\arctan(y)$ is bounded and that $y'$ is bounded and decreasing for large $t$ so we can guarantee that at least $\lim_{t \to \infty} y'$ exists, so the other limit must exist and go to zero as well. The second limit I found tricky to use, since we can't really get information about $y$ from this. Best I could do here was argue that we can deduce from this limit that $\lim_{t \to -\infty} y' = \infty$. So, since we have a global solution, $y$ should go to $-\infty$ when $t\to -\infty$.
That's more or less my approach for this problem. Any help on how to fill in the gaps and improve my writing will be appreciated.