I have the following autonomous initial value problem:
$$ \begin{equation*} \left\{ \begin{alignedat}{3} \frac{dy}{dt} & = f(y)=1+k\epsilon-ky -e^{-y} \\ y(0) & = \epsilon \end{alignedat} \right. \end{equation*}$$
where $k,\epsilon>0$. Let $y(t)$ denote the solution of this IVP: I'm told the supremum of $y$ is $y_\infty=\lim\limits_{x \to \infty}{y(t)}$, and its value is found by setting $f(y)=0$. How do I prove it? I tried the following: $f(y)=1+k\epsilon-ky -e^{-y}=0$ has a single solution $\bar{y}>\epsilon$ in $I=[\epsilon,\infty[$ because $f(\epsilon)=1-e^{-\epsilon}>0$ and $\lim\limits_{y \to \infty}{f(y)}=-\infty$. For any $t_0$, $g(t)=\bar{y}$ is the only solution of the IVP
$$ \begin{equation*} \left\{ \begin{alignedat}{3} \frac{dy}{dt} & = f(y) \\ y(t_0) & = \bar{y} \end{alignedat} \right. \end{equation*}$$
because $f(\bar{y})=0$.
Now, $y(t)<\bar{y}\ \forall t\in[0,\infty[$, because $y(0)=\epsilon<\bar{y}$, and if for any $t_0>0$ we had $y(t_0)=\bar{y}$, then we would have two distinct solutions for the same IVP, $g(t)=\bar{y}\ \forall t$, and $y(t)$ which is different from $\bar{y}$ at $t=0$.
Thus I proved that $\bar{y}$ is an upper bound for $y(t)$: how do I also prove that it's the supremum? Since $f\left(y(0)\right)=1-e^{-\epsilon}>0$, $y(t)$ is increasing in a right neighborhood of $t=0$. Now, $f(y(t))=y'(t)$ has to remain positive $\forall t>0$, because in order for it to change sign, it should before become $0$, but it can't, since $y(t)<\bar{y}\ \forall t\in[0,\infty[$. Thus, $y(t)$ is monotone increasing and bounded, so $y_\infty=\lim\limits_{x \to \infty}{y(t)}$ is finite.
- Is my reasoning correct, up to now?
- How do I prove that $y_\infty=\bar{y}$?
Here is my approach to show that $y_{\infty}=\bar{y}$.
For the sake of contradiction assume that $y_{\infty}<\bar{y}$. Because $y(t)$ is a solution to your IVP we must have for any $T>0$ that $$\int_{y(0)}^{y(T)}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=\int_0^T\mathrm{d}t$$ This implies for any $T>0$ $$\int_{\epsilon}^{y(T)}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=T$$ Take $T\mapsto +\infty$ and get $$\int_{\epsilon}^{y_{\infty}}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=+\infty$$ This is a contradiction since $x\mapsto \frac{1}{1+k\epsilon-kx-e^{-x}}$ is continuous on $[\epsilon,y_{\infty}]$.