Some phase plane analysis for $u_t=u_{xx}+f(u)$ with assumptions on $f$ Consider the semilinear diffusion equation
$$
u_t=u_{xx}+f(u)
$$
with the following assumptions on $f$ for some $\alpha\in (0,1)$:


*

*$f\in C^1[0,1]$

*$f(0)=f(1)=0$

*$f'(0) < 0$

*$f(u)<0$ in $(0,\alpha)$

*$f(u)>0$ in $(\alpha,1)$

*$\int_0^1 f(u)\, du>0$
Introduce the moving coordinates $\xi=x-ct$. Plugging this into the equation gives 
$$
-cU_\xi=U_{\xi\xi}+f(U)
$$
which as a system is
$$
\begin{align*}
U_\xi&=V\\
V_\xi&=-cV-f(U)
\end{align*}
$$
For $c=0$, we have the "Energy"
$$
E(U,U_\xi)=\frac{1}{2}(U_\xi)^2+F(U),\qquad F(U):=\int_0^U f(y)\, dy
$$
and, in particular, each trajectory satisfies an equation
$$
E(U,U_\xi)=\textrm{const}.
$$
Now, there is the following part in a paper which I do not understand:


Under our hypotheses on $f(u)$ [which I listed above], there is an $\eta\in (0,1)$ such that $F(\eta)>0$. For any $0 < \nu < (2F(\eta))^{1/2}$ the trajectory through $(0,-\nu)$ lies in the strip $U\in [0,1)$ and contains a point of the positive $V$-axis.


1.) Why is there an $\eta\in (0,1)$ with $F(\eta)>0$? 
Am I right that $F(0)=0$ and $F(1)>0$; if yes then this implies the existence of such an $\eta$ (Intermediate value theorem?)
2.) Where does the condition $0<\nu<(2 F(\eta))^{1/2}$ come from?
From the energy picture, it is clear to me that by assumption $\int_0^1 f(u)\, du=F(1)-F(0)>0$, we have that there are to hill tops positioned at $U=0, F(0)$ and $U=1, F(1)$ and that the second hill top is higher than the first one. Hence, from the phase plane it is clear to me that there are trajectories through $(u,0)$ and $(0,-v)$ for $u\approx 1$ and some $v>0$. I think the condition $0<\nu <(2 F(\eta))^{1/2}$ has something to do with this energy picture.
Edit.
My idea would be to use
$$
E(U,V)=C \Leftrightarrow V=\pm\sqrt{2C-2F(U)}
$$
which makes sense iff $F(U)\leqslant C$.
If we choose $C=F(\eta)$ then for $0<F(U)<F(\eta)$ we have 
$$
0<\lvert V\rvert<\sqrt{2F(\eta)}
$$
and the trajectories containing $(0,-V)$ are of the described type?
 A: For the first question your argument with the IVT is correct.  For the second note that along any trajectory (writing $' = d/d\xi$)
$$
 \left( \frac{1}{2}|U'(\xi)|^2 + F(U(\xi)) \right)' + c|U'(\xi)|^2 = 0.
$$
This follows by multiplying $-c U' = U'' + F'(U)$ by $U'$ and noticing the total derivative structure.   Now, if $c \ge 0$, then this is a dissipation relation and we can integrate to conclude that 
$$
 \frac{1}{2}|U'(\xi)|^2 + F(U(\xi)) \le \frac{1}{2}|U'(0)|^2 + F(U(0)).
$$
Now, if the trajectory starts with $U(0) =0$ and $U'(0) = -\nu$, then the above inequality tells us that
$$
 \frac{1}{2}|U'(\xi)|^2 + F(U(\xi)) \le \frac{1}{2}\nu^2 .
$$
Assume for the moment that along the trajectory there exists $\xi_0$ such that $U(\xi_0) = \eta $.   Then
$$
F(\eta) = F(U(\xi_0)) \le \frac{1}{2} \nu^2 < F(\eta)
$$
by hypothesis, which is a contradiction.  Thus by the IVT $U(\xi) < \eta$ for all $\xi$.  
This is only part of what you want, but I don't think it's actually true that $U(\xi) \ge 0$ for all $\xi$ given what is stated.  In fact, if $U(0) = 0$ and $U'(0) = -\nu$, then by continuity there exists $\epsilon >0$ such that $U'(\xi) \le -\nu/2$ for $\xi \in [0,\epsilon)$ and hence 
$$
U(\xi) = U(0) + \int_0^\xi U'(s) ds = \int_0^\xi U'(s) ds \le -\frac{\nu \xi}{2} < 0 \text{ for } \xi \in (0,\epsilon).
$$
This seems to be at odds with the fact that $f$ is only defined on $[0,1]$.  Maybe there's a typo and $U'(0) = \nu$ is supposed to be the initial condition?
