Proving existence of solution for ODE $-s\varphi' + f'(\varphi)\varphi' = \varphi''$ 
Let $f : \Bbb{R} \to \Bbb{R}$ be twice differentiable with $f'' > 0$, and let $u_- > u_+$ be real numbers. Show that there exists a solution $\varphi(x)$ to the following differential equation:
$$
-s\varphi' + f'(\varphi)\varphi' = \varphi'' \tag{1}
$$
such that $\lim_{x \to \pm\infty} \varphi(x) = u_\pm$, and where $s = \frac{f(u_+) - f(u_-)}{u_+ - u_-}$.


My initial attempt is to observe that this DE can be nicely integrated to the following:
$$
\varphi' = f(\varphi) - s\varphi + C \tag{2}
$$
Thus, it suffices to show the existence of a solution for this DE instead, where we are free to choose $C$. I attempted to bring RHS over to LHS, which gives:
$$
\int \frac{1}{f(\varphi) - s\varphi + C} \; \mathrm{d}\varphi = x + D
$$
where $D \in \Bbb{R}$. Thus, if we define:
$$
g(x) = \int \frac{1}{f(x) - sx + C} \; \mathrm{d}x
$$
and assuming that $g$ is invertible, then $\varphi(x) = g^{-1}(x)$ would be a solution to $(2)$. However, there are a few issues in this approach that we need to tackle:

*

*The integral will not make sense if $f(\varphi) - s\varphi + C$ vanishes at some point in $\Bbb{R}$. As we are free to choose $C$, if we can show that $f(\varphi) - s\varphi$ is bounded from either above or below, then such a choice of $C$ will exist. I suspect we can use the convexity and the definition of $s$ to prove this, but my attempts are futile so far.

*Should the integral make sense, another problem is if $g$ is invertible. However, this should not be an issue as by FTOC:
$$
g'(x) = \frac{1}{f(x) - sx + C}
$$
so if the denominator does not vanish, $g'$ is continuous and so must be strictly positive or negative, hence $g$ is strictly monotone, thus invertible.

*The biggest issue here is that this definition does not guarantee the requirement of $\lim_{x \to \pm\infty} \varphi(x) = u_\pm$. I tried to manipulate the integral to fit this condition, but to no avail so far.

I also tried other approaches, such as using Picard's iteration, but as this problem is not really an IVP they have not been successful.
Any help is appreciated.
 A: Using the limits at $\pm\infty$, we find
$$
C = su_+ - f(u_+) = su_- - f(u_-) \, ,
$$
$$
\text{and}\qquad
\varphi' = f(\varphi) - f(u_+) - s(\varphi - u_+) = f(\varphi) - f(u_-) - s(\varphi - u_-) \, ,
$$
see this exercise in Evans PDE.
The strict convexity of $\varphi\mapsto \varphi'$ follows from the strict convexity $f''>0$ of $f$. This property yields
$\varphi' < 0$ for $\varphi \in \left]u_+, u_-\right[$. Therefore, $\varphi$ is a smooth decreasing function, that decreases from $u_-$ to $u_+$.
To investigate the stability of the equilibrium $\varphi = u_\pm$, we compute the sign of the derivative $d\varphi'/d\varphi = f'(\varphi) - s$ at equilibrium, which is negative at $\varphi = u_+$ and positive at $\varphi = u_-$ due to strict convexity. Therefore, $u_+$ is an attractive equilibrium and $u_-$ is a repulsive equilibrium. Since the rhs. of the above differential equation is non-singular and does not possess additional roots, any bounded solution will necessarily connect both values $u_\pm$ through a smooth decreasing function $\varphi$.
The integrand in
$$
x+D = \int_{u_+}^{u_-} \frac{\text d \varphi}{f(\varphi) - f(u_+) - s(\varphi - su_+)}
$$
is singular at the bounds $\varphi = u_\pm$.
Convergence of this improper integral follows from its asymptotic behaviour at the bounds.
