TITLE: Show $f''+vf' +\alpha^2 f(1-f)=0$ has solutions satisfying $\lim_{x \to - \infty}f=0$ and $\lim_{x \to \infty}f=1$ given $v\leq -2\alpha < 0$
QUESTION [7 upvotes]: I posted this question before but I took a completely different approach here, that's why I reposted as my previous question was already very long and took a different approach from here.
I am given the Fisher equation $$u_t=u_{xx}+\alpha ^2u(1-u)$$ where $\alpha >0$ is a constant. Now assuming that $u(x,t) = f(x-vt)$ I need to show there exists a solution $f$ such that $\lim\limits_{x \to - \infty}f=0$ and $\lim\limits_{x \to  \infty}f=1$ provided that $v \leq -2\alpha$. Now substituting $f$ into the differential equation we get:
$$f''+vf' +\alpha^2 f(1-f)=0$$
Where $f$ is a funtion of a single variable $\eta = x-vt$. 
From here onwards I will go into my attempt to solve using linearization about the stationary points. If someone finds another method to show the asked results I will still award the bounty!
It is easy to see that the two stationary points are $(f_0,f'_0)=(0,0)$ and $(f_1,f'_1)=(1,0)$. Now I will linearize the system in these stationary points, first in $(f_0,f'_0)$ via $f=f_0 +\epsilon w_0$ around the first stationary points. Leaving out terms of order $\epsilon^2$ we obtain the following differential equation via substituting this $f$:
$$w_0''+vw'_0+\alpha^2 w_0=0        $$
The solution to this DE is $w_0=Ae^{r_1 \eta}+Be^{r_2 \eta}$ where $r_1 = \frac{-v + \sqrt{v^2-4\alpha^2}}{2}$ and $r_2 = \frac{-v - \sqrt{v^2-4\alpha^2}}{2}$. Now using $v \leq -2\alpha < 0$ we see that $0 \leq \sqrt{v^2-4\alpha^2}< -v$ such that $-v + \sqrt{v^2-4\alpha^2}$ and $-v-\sqrt{v^2-4\alpha^2 }$ are both larger than zero. Thus around the stationary point $(f,f')=(0,0)$ we can approximate the solution of $f$ by:
$$f = f_0 + w_0 = Ae^{r_1 \eta}+Be^{r_2 \eta}$$
Where both $r_1$ and $r_2$ are positive, thus $f$ is increasing at the stationary point $(f,f')=(0,0)$. This is the first result.
Now for the second stationary point we use the linearization $f=f_1+\epsilon w_1 = 1 + \epsilon w_1$. This turns the DE into:
$$w_0''+vw'_0-\alpha^2 w_0=0        $$
with solution $w_1=Ce^{s_1\eta}+De^{s_2\eta}$
where $s_1 = \frac{-v + \sqrt{v^2+4\alpha^2}}{2} > 0$ and $s_2=\frac{-v - \sqrt{v^2+4\alpha^2}}{2}<0$
So the approximation around $(f,f')=(1,0)$ is given by $$f=f_1+ Ce^{s_1\eta}+De^{s_2\eta}=1+Ce^{s_1\eta}+De^{s_2\eta}$$
which can be either increasing or decreasing, I cant tell unless we determine the constants.
My question is, how do I proceed from here. I think I am going in the right direction as I heard from some of my classmates they solved the problem using linearization... I realize that even the conclusion that the first approximation is increasing is shaky because then I am making the assumption that $A$ and $B$ are not both negative... If anyone could help me out that would be amazing. This question is homework but the deadline is already passes, I handed in what I had and did good on the other questions. I am just frustrated by my inability to get anywhere with this question no matter how much time I spend on it... Thanks in advance!
P.S. if anyone has a better way of solving that would of course be very welcome too! Thanks

REPLY [2 votes]: Added 31 October 2013 11:48 PM PST:  The following, yet another answer, is added as such for the same reasons given in my second:  it is too long for the comments, and after trying to edit it into that answer, and waiting a full two hours while mathjax attempted to render my work, I resorted to breaking up my post(s) into separate answers once again, though they should in fact be considered of a piece, as the saying goes.  And again, I beg my readers' indulgence.
Since posting my second answer, which as I explained should really be considered an integral part of my first, I think I have found a way to get substantially closer to finishing this thing off, in the sense of showing that there is indeed a heteroclinic trajectory $\gamma(\eta)$ such that $\lim_{\eta \to -\infty}\gamma(\eta) = (0, 0)$ and $\lim_{\eta \to \infty}\gamma(\eta) = (1, 0)$.  The construction presented in the above discussion, albeit somewhat "hand-wavy", is meant to demonstrate that the necessary existence of a heteroclinic orbit may be mollified by the presence of periodic trajectories; if we can eliminate the possibility that such occur, I think we will be one step closer to a position from which the existence of the sought-for heteroclinic orbit is within reach.
So here we will show that the vector field $F(f, g)$ has no periodic trajectories.  This can be handled by the old method of Bendixson; see this wikipedia entry or this book by Gerald Teschl, the seventh chapter.  (The same book is mentioned in my first answer above.)  The relevant technique is as follows:  suppose $V(f, g)$ is any $C^1$ vector field in the $f$-$g$ plane $\Bbb R^2$, and let $\Omega  \subset \Bbb R^2$ be open and simply connected such that $\nabla \cdot V = \frac{\partial V_f}{\partial f} + \frac{\partial V_g}{\partial g} \ne 0$ on $\Omega$; then $V(f, g)$ can have no periodic orbit in $\Omega$.  To see this, we look at the integral of $\nabla \cdot V$ over a region bounded by such a closed trajectory, assuming one existed, and use the divergence theorem of Gauss in its two-dimensional form, thusly: if $F(f, g)$ has a periodic trajectory $\gamma(\eta)$ in the simply connected region $\Omega$, it bounds some also open simply connected region $\Gamma \subset \Omega$; we then have, since $\nabla \cdot V \ne 0$ on $\Omega$, it is either positive or negative on all of $\Omega$, hence on all of $\Gamma$ as well.  Thus
$\int_\Gamma (\nabla \cdot V) dfdg \ne 0; \tag{1}$
but by Gauss's theorem we have
$\int_\Gamma (\nabla \cdot V) dfdg = \int_{\gamma(\eta)} (V \cdot n)ds; \tag{2}$
but
$\int_{\gamma(\eta)} (V \cdot n)ds = 0, \tag{3}$
since $n$ is the outward pointing normal on $\gamma(\eta)$ and $V$ is tangent to $\gamma(\eta)$; thus $V \cdot n = 0$.  This contradiction proves that $V(f, g)$ has no closed trajectories.  To apply these notions to $F(f, g)$, recall from my first answer that the Jacobean of $F(f, g)$ is
$J_F = \begin{bmatrix} 0 & 1 \\ -\alpha^2(1 - 2 f) & -v \end{bmatrix}, \tag{4}$
whence we have
$\nabla \cdot F = \text{Tr}(J_F) = -v > 0, \, v \, \text{a constant}, \tag{5}$
which in light of the above precludes the possibility of the existence of a periodic trajectory of $F$. Apparently the same argument precludes the existence of a homoclinic orbit joining $(1, 0)$ to itself. i.e, of $W_u(1, 0) = W_s(1, 0)$; the integral over $\gamma(\eta)$ in such as case is well-defined except on a set of measure zero, that being the point $(1, 0)$ itself, where $W_s(1, 0)$ and $W_u(1, 0)$ intersect but have different (non-collinear) tangent vectors.
I seems to me that at this point we are left with the following situation:  the integral curve $\gamma(\eta)$ such that $\gamma(\eta) \to (1, 0)$ as $\eta \to \infty$ cannot, as $\eta \to -\infty$, either loop back to $(1, 0)$ itself (case of $\gamma(\eta)$ homoclinic) or approach a periodic orbit, since there are none.  Furthermore, there is no cycle of equilibria and heteroclinic orbits, since $(0, 0)$ is a purely repelling point; no integral curve can enter it as $\eta \to \infty$.  So by the Poincare-Bendixson theorem, we are left with two prospects:  either $\gamma(\eta) \to (0, 0)$ as $\eta \to -\infty$, or $\gamma(\eta)$ leaves any compact set in $\Bbb R^2$.  Now if I could just rule out this latter possibility . . . 
Ah well, Happy Halloween, my friends!
Hope this helps just a little bit more.  Cheers,
and as always,
Fiat Lux!!!