Solution of a differential equation hitting a constant - possible? In my ODE lecture, we came across the DE
$$\frac{dy}{dx} = y(1-y)$$
I know that the solution to this is simple to derive but there are also 2 constant solutions $y \equiv 0$ and $y \equiv 1$.
I'm wondering why there can't be a solution that is above $1$ and eventually reaches the constant $1$. I asked my lecturer and he said you can think of the solution backwards (in the negative $x$ direction) and it won't make sense that a solution would suddenly jump up from $1$.
I'm not very satisfied with the answer.
My best explanation is that if a solution hits 1 at say $x = k$, then at $x \leq k - c$, the solution does not touch $1$ at all, meaning that the method of integration should supply this solution but since none of the solutions here touch 1 at all, we have a contradiction.
Is there a more elegant way to show this?
 A: Reaching $1$ and staying constant, or touching $1$ at any point at all cannot happen unless it is $1$ throughout, this ODE is locally lipschitz continuous so that a "solutions can't cross" lemma applies: formally if $x:I \rightarrow \mathbb{R}$ and $y: J \rightarrow \mathbb{R}$ are two solutions that agree at some point in $I \cap J$ then they agree on all of $I \cap J$. In this case we know $x = 1$ is a constant solution defined for all forward and backward times, so if any other solution $y:J \rightarrow 1$ is equal to $1$ at some $t \in J$, $y = 1$ on $J$.
Since your RHS is $C^1$ it is locally lipschitz.
Note that $I$ and $J$ in the statement of that lemma are intervals.
EDIT: I'll include a proof of the "solutions don't cross" lemma for these types of ODE's
You'll have to accept the following:
Let $x' = f(t,x)$ be an ODE, with $f$ locally lipschitz continuous with respect to $x$, $f:E \rightarrow \mathbb{R}^d$, $E \subset \mathbb{R}^{1+d}$ , $E$ open. Then for any $(t_0,x_0) \in E$ the initial value problem given by $x(t_0) = x_0$ , $x' = f(t,x)$ has a unique solution in some interval $[t_0 - h,t_0 + h]$, for some $h>0$.
If you want to see a proof of this , google "Picard - Lindelof"

Now let $x$, $y$ be two solutions on the intervals $I$, $J$ to the ODE. Further suppose that $t_0 \in I \cap J$ is such that $x(t_0) = y(t_0)$. Define $A = \{t \in I \cap J | x(t) = y(t), t \in [t_0, t]\}$, we already know that $x$ and $y$ solve the same IVP in a neigbourhood of $t_0$, so this set is certainly nonempty by the above result, we want to show that $\sup A = \sup I \cap J$.
Suppose not, then $\sup A < \sup I \cap J$, put $a = \sup A$, $b = \sup I\cap J$, (note we can have $b = + \infty$), then $x$ and $y$ agree up to $a$, and agree at $a$ by continuity, solving the IVP $(a,x(a))$ gives us a unique local solution in a neighbourhood of $a$ that agrees with both $x$ and $y$ in that neighbourhood, contradicting the definition of $a$. Thus $a$ = $b$ and we are done.
The same thing can be done for infimums.
As an interesting last point, this is precisely why the orbits of an autonomous ODE partition the phase space.
EDIT2:
It is also interesting to observe that solutions to 1-d autonomous ODEs that are locally lipchitz (like this one) have solutions that are strictly monotone or constant - which would also be enough to demonstrate why a solution like you are proposing cannot happen. You could try to prove this via the tools I've just given you as it doesn't require anything more.
A: If I differentiate both sides of the equation I get 
$$\frac{d^2y}{{dx}^{2}}= \frac{dy}{dx}(1-y) -\frac{dy}{dx}y$$
$$\frac{d^2y}{{dx}^{2}}= y(1-y)(1-y) -y(1-y)y$$
So the second derivative must also be $0$ at $y=1$
You can continue this process to get that the $n$th derivative is $0$ at $y=1$ and if you assume that the solution is analytic, this implies that it is constant.
A: Of the ODE is locally Lipschitz with a Lipschitz constant $L$, then the Grönwall lemma tells us that for two solutions (and in a time interval that has them in the set where $L$ is valid)
$$
|y_1(t)-y_2(t)|\le e^{L|t-s|}|y_1(s)-y_2(s)|.
$$
The usual conclusion is that the solution depends continuously on the initial point. And also that two solutions that share an initial point remain identical for their whole domain.
Interesting here is that this inequality is valid in both directions, for $t>$ as well as for $t<s$. Thus it can be reversed to 
$$
|y_1(t)-y_2(t)|\ge e^{-L|t-s|}|y_1(s)-y_2(s)|.
$$
This means that the distance between two different solutions can not fall to zero in finite time, that is, that the scenario you envision can not happen for sufficiently smooth right sides.
