Check whether there is a global existence of solution of function $y'(x) = \frac{-y(x)}{1+e^{y(x)}}$?

Question

I know that the derivative needs to be bounded in order to have a global solution but I got stuck here and don't know how to proceed further.

$$\left|\frac{df}{dy}\right|= \left|\frac{y.e^y-(e^y+1)}{(1+e^y)^2}\right|$$

I tried applying reverse triangular inequality but don't know whether am I going right or not. Is it possible to proceed this way?

$$\left|(1+e^y)^2\right| \geq\left||1|-|e^y|\right|^2$$

So,$$\left|\frac{y.e^y-(e^y+1)}{(1+e^y)^2}\right| \leq \frac {|y.e^y-(e^y+1)|}{\left||1|-|e^y|\right|^2}$$

May I know how to simply further from here? Moreover, is there any other way to solve this problem? I request anyone who can take a look and help me understand this.

Answer 1

You can also directly use $e^y>0$ to get $$ |y'(x)|\le |y(x)|\implies |y(x)|\le e^{|x|}|y(0)|. $$ This bound holds wherever a solution $y$ exists. At the same time the bound prevents divergence at finite times, so that the domain of the maximal solution has to be $\Bbb R$.

This is a special case of the similar claim for linearly bounded ODE functions, $|f(x,y)|\le C+M|y|$.

Answer 2

Though this won't give you explicit bounds, you can apply the following fact: If the function $g:\mathbb{R}\to\mathbb{R}$ is continuous and both limits $\lim_{x\to\pm\infty} g(x)$ exist and are finite, then $g$ is bounded (see here for a proof).

Using $\lim_{y\to-\infty}ye^y=\lim_{y\to-\infty}e^y=0$, we see that $\lim_{y\to-\infty}f'(y)=-1$. Writing the derivative as $$f'(y)=\frac{y/e^y-1/e^y-1/e^{2y}}{(1+1/e^y)^2} $$ and using $\lim_{y\to\infty}y/e^y=0$, we obtain that $\lim_{y\to\infty}f'(y)=0$. Taking into the account that $f'$ is continuous on $\mathbb{R}$, it follows that it's bounded.