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 $v\leq -2\alpha < 0$ 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
 A: To make things slightly easier I consider the equation
$$u_t=u_{xx}+u(1-u),
$$
which can be obtained from the original one by scaling the spatial and temporal variables.
Now consider 
$$
u(x,t)=U(z)=U(x\pm ct),
$$
where I choose "plus", but "minus" is treated similarly. We have
$$
cU'=U(1-U)+U'',
$$
which can be rewritten as the system
$$
U'=V\\
V'=cV-U(1-U)
$$
with two equilibria: $O=(0,0)$ and $P=(1,0)$.
At $O$ the Jacobi matrix has two eigenvalues
$$
\lambda_{1,2}=\frac 12 (c\pm\sqrt{c^2-4}),
$$
which are not complex if $c\geq 2$ (we cannot have complex eigenvalues, hence this would mean that $U$ changes sign). Point $P$ is a saddle for any parameter values. This point has stable manifold with the direction $(-c,1)$ --- this is an eigenvector corresponding to the negative eigenvalue, hence pointing "north-west" from the point. This stable manifold has to cross the unstable manifold of $O$, since the latter is two dimensional, hence there is a unique homoclinic trajectory, connecting $O$ and $P$ and positive. This trajectory corresponds to the traveling wave solution of the original problem.
A: 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!!!
A: Well, I guess I hadn't typed quite enough!
In light of the extended discussion of notions surrounding stable manifold theory in both answers and in the comments, I wish to add a few remarks which further explicate, and hopefully clarify, the situation regarding its application to the problems presented in this question.  I am appending these "comments" as a separate answer, for two reasons:  i.)  they are probably too long to fit comfortably in even several comment boxes; and ii.)  inserting them via the editing process into my already-posted answer will take a long, long while, due to the time it takes the mathjax system to render long posts with substantial amounts of Latex.  So rather than strain my patience I beg the indulgence of my readers and continue the discussion in this, a separate answer to the OP's question.  This answer can and should, for all practical purposes, be considered of a piece with the one I previously posted.
These things being said . . . 
As mentioned by Slugger in his comment on my previous answer, it is indeed the stable manifold theorem which affirms that the limiting behaviors of $F(f, g)$ and $DF$ are the same in the neighborhood of a critical point of $F$.  Without some result to this effect, there is no guarantee that the linear system will, in the limit, exhibit the same structure as the original, non-linearized equation(s).  A key observation here is that as long as the independent variable remains finite, i.e. we are only interested in $\eta$ in some bounded interval $(\eta_{\text{min}}, \eta_{\text{max}})$ with $-\infty < \eta_{\text{min}} < \eta_{\text{max}} < \infty$, then it is a relatively simple matter to affirm that the solutions of the linearized system stay arbitrarily close to those of the original one, provided they are initialized so and, again, the dependent variable is bounded.  This is if you will a standard, somewhat elementary result in the theory ODEs; a thorough discussion can by found, for example, in Chapter 17 of Differential Equations, Dynamical Systems and and Introduction to Chaos, by Morris Hirsch, Stephen Smale, and Robert L. Devaney; for a more advanced treatment see
Jack K. Hale's Ordinary Differential Equations, the citings in these books, or any of a number of other standard texts and references.  But it is the limiting cases as $\eta \to \pm \infty$ which require a more delicate approach, and this is what the theory of stable manifolds provides.
As for the existence of a heteroclinic trajectory $\gamma(\eta)$  "connecting" the equilibrium points of $F(f, g)$ in the sense that $\lim_{\eta \to -\infty} \gamma(\eta) = (0, 0)$ and $\lim_{\eta \to \infty} \gamma(\eta) = (1, 0)$, stable manifold theorems provide only a piece of the puzzle, for they are essentially local in nature; that is, they only give a detailed picture of $W_s$, $W_u$ in some neighborhood $U$ of the critical point to which they correspond.  Of course, such stable and unstable manifolds can be globalized by the action of $\phi_\eta$; for example, if $W_s$ is defined in a small neighborhood $U$ of a critical point, then the set 
$\bigcup_{\eta \in \Bbb R} \phi_\eta(W_s) = \{(f, g) \in \Bbb R^2 \mid \phi_\eta(f, g) \in W_s \, \, \text{for some} \, \,\eta \in \Bbb R \}, \tag{1}$
consists of all points in $\Bbb R^2$ which $\phi_\eta$ takes into $W_s$, and hence eventually into the critical point associated with $W_s$ as $\eta \to \infty$.  But despite the fact that we can in principle construct global stable and unstable manifolds, in the sense of establishing their existence, we can't say much more about their properties without resorting to a more detailed, refined analysis of the vector field $F(f, g)$ and its flow $\phi_\eta$.  Indeed, it is possible, with a little effort and imagination, to describe a vector field $F_1(f, g)$ on $\Bbb R^2$ which has exactly the same critical point structure as $F(f, g)$, and even the same Jacobean matrices at these critical points, but for which there is no heteroclinic trajectory running between one critical point and the other.  To wit:
Let $F_1(f, g)$ have critical points at $(0, 0)$ and $(0, 1)$, as does $F(f, g)$, and suppose that in fact we arrange things so that $J_{F_1} = J_F$ at each of these points.  Consider the two circles $C_1$ and $C_2$ of radius $r_1$ and $r_2$ respectively, with $0 < r_1 < r_2 < 1$, both centered at $(0, 0)$, and further arrange things so that $F_1(f, g)$ is of constant magnitude and tangent to each of these circles, with the direction  of rotation the same on each.  Inside $C_1$, choose $F_1(f, g)$ differentiable so that, in addition to agreeing together with its Jacobean with $F(f, g)$ and its Jacobean at $(0, 0)$, the trajectories of $F_1(f, g)$ form spirals, as we approach $C_1$ from the inside, which converge to $C_1$ as $\eta \to \infty$.  In the annulur region between $C_1$ and $C_2$, choose $F_1(f, g)$ differentiable so that its trajectories spiral inward from $C_2$ and converge to $C_1$ as $\eta \to \infty$, and to $C_2$ as $\eta \to -\infty$; outside of $C_2$ choose $F_1(f, g)$ so that, near $C_2$, the trajectories spiral outward from $C_2$ and approach $C_2$ as $\eta \to -\infty$; in addition, outside of $C_2$ we arrange things so that exactly one of the orbits $\gamma(\eta)$ of $F_1(f, g)$ which "unwind from", that is to say, spiral outward from $C_2$, is in fact the stable manifold of $(1, 0)$;  that is, $\gamma(\eta) \to (1, 0)$ as $\eta \to \infty$; the other trajectories spiraling outward from $C_2$ wend off and become unbounded as they may.  What I have attempted to give here is a verbal description, albeit possibly a crude one, of the phase portrait of a system $(f', g')^T = F_1(f, g)$ which completely agrees with $(f', g')^T = F(f, g)$ insofar as critical point structure is concerned, but whose global trajectories exhibit radical differences from those of $(f', g')^T = F(f, g)$.  In particular, one should observe that (i.)  the unstable manifold of the point $(0, 0)$ is precisely the interior of the disk of radius $r_1$ centered at $(0, 0)$, the boundary of which is the circle $C_1$; and (ii.)  there is no heteroclinic trajectory of $F_1(f, g)$ connecting $(0, 0)$ and $(1, 0)$.  This illustrates the point that, even with the aid of stable manifold theory, the assessment of global orbit structure may be an arduous task, requireing much more work than the present answers present.  Finally, though my informal description of $F_1(f, g)$ and its flow ultimatley lacks rigor, I believe it is possible to analytically define $F_1(f, g)$ if sufficient care is taken, perhaps using partitions of unity in $\Bbb R^2$, but unfortunately I have neither the time nor space to present such a development here.  It is also unfortunate that I haven't given any pictures, undoubtedly worth several thousands of words in this context, but at present I lack the necessary graphics software to do so.
When I read our OP Slugger's question, I took the phrase "solutions satisfying $\lim_{x \to -\infty} f = 0$ and $\lim_{x \to \infty} f = 1$" to mean separate solutions of each type, rather than solutions satisfying both properties together.  And such we have show to exist; but the demonstration of the existence of a true heteroclinic integral curve of $(f', g')^T = F(f, g)$ is evidently a far more arduous task, one which makes me wonder just what level of experience and knowledge is assumed in the class for which this "little exercise" is homework.
Well, now I will attempt, once again, to stop typing, at least for the present.
And I really do hope this helps.  Once again I say, "Cheers!"
and once again, as always, I say
Fiat Lux!!!
