Suppose we are given a PDE $$ u_t=u_{xx}+f(u), $$ where $f$ is a non-linearity with zeros, say, $0$ and $1$. That is, $0$ and $1$ are equilibria. Moreoever, suppose that both $0$ and $1$ are saddles, i.e. the linearization in $0$ and $1$ both has a negative and a positive eigenvalue. Denote the negative eigenvalue of $0$ by $\lambda$ and the positive one by $\mu$.
Now, let $\varphi(z:=x-ct), c>0$, be some decaying ($\varphi'<0$) travelling wave solution with $\lim_{z\to-\infty}\varphi(z)=1$ and $\lim_{z\to\infty}\varphi(z)=0$. That is, $\varphi$ is a heteroclinic solution which means that $\varphi\in W^s(0)\cap W^u(1)$, i.e. lies in the intersection of the stable and unstable manifolds of $0$ and $1$, respectively.
Thus, we surely have by the stable/ unstable manifold theorem that heteroclinic decay exponentially. That is, we surely have $$ \lvert\varphi(x)\rvert=-\varphi'(x)\leqslant Ce^{\lambda x}, x>0,\tag{1} $$ $$ \lvert\varphi'(x)\rvert=-\varphi'(x)\leqslant D e^{\mu x}, x<0.\tag{2} $$ for some positive constants $C$ and $D$.
My question is if we can also write $$ -\varphi'(x)=C(x)e^{\lambda x}, x>0 $$ $$ -\varphi'(x)=D(x)e^{\mu x}, x<0 $$ where $C$ and $D$ are now not constants but positive, non-linear, bounded, convergent functions? With convergent I mean that $\lim_{x\to\infty} C(x)=:C(\infty)$ and $\lim_{x\to-\infty}D(x)=:D(-\infty)$ exist.
Edit
My feeling is that this is no big thing: The function $C(x)$ can just be defined to be $$ C(x):=\lvert\varphi'(x)\rvert e^{-\lambda x}, x>0. $$ Then, of course, $C(x)$ is non-linear and, moreover, bounded for $x>0$ due to (1), namely $$ C(x)\leqslant C, x>0 $$ and, consequently, $C(\infty)\in\mathbb{R}$ has to exists.
Maybe that's all... maybe it is that simple.