Show $0 \leq u < 1$ for $u_t - \Delta u = u(u-1).$ Consider the following PDE on the unit sphere with $t > 0$,
$$ u_t - \Delta u = u(u-1),
$$
and $0 \leq u < 1$ on all the boundaries, meaning on $\{(x,t) : |x| = 1, t > 0\} \cup \{(x,t) : |x| < 1, t = 0\}$.  Prove $0 \leq u < 1$ on the entire domain for all finite time.

I'm stumped.  Of course it seems to make intuitive sense, without the RHS we just have the heat equation with Dirichlet boundary conditions.  Knowing that $u$ starts in $[0,1],$ one can be convinced that the nonlinear forcing on the RHS won't push $u$ outside of $[0,1],$ as the forcing gets smaller where the value of $u$ is near 0 or 1.  But this intuition doesn't get me too far when it comes to actually proving this.  Any suggestions? I feel like some sort of perturbation (such as $v = u + \epsilon e^{\lambda t}$) would work, but I haven't been successful with it yet.
All comments are welcome, thank you!
 A: I'll start by showing that $u \geq 0.$  Let $v = u + \epsilon e^{t}.$  Ultimately, if we can show that $v > 0$ on the domain, then as $\epsilon \to 0$ we get $u \geq 0.$  Notice that $v(x,0) > 0.$  Time-differentiating $v,$
$$v_t = \epsilon e^{t} + \Delta v + (v - \epsilon e^{t})(v-\epsilon e^{t}-1).$$
Suppose $v = 0$ at some $(x,t),$ and show a contradiction.  Consider the first time that $v = 0.$ This would be a local minimum, so $\Delta v \geq 0,$ $v_t \leq 0.$  Then, 
$$ 0 = -v_t + \Delta v +\epsilon e^{t}(\epsilon e^{t}+1) + \epsilon e^{t} > 0$$
since $\epsilon  >0.$  Thus, a contradiction, meaning $u \geq 0.$

In a similar vein, we can use this to show $u < 1,$ but we need a different perturbation.  Let $v = u - 1 +\epsilon - \mu e^{kt}.$  Then if we choose $\epsilon > 0$ small enough so that $v(x,0) < 0,$ and we can show that $v < 0$ for any $\mu > 0,$ then $u + \epsilon \leq 1,$ or $u < 1$ strictly.  So we time differentiate $v,$
$$ v_t = \Delta v + (v + 1 -\epsilon + \mu e^{-kt})(v -\epsilon + \mu e^{kt}) - k\mu e^{kt}.
$$ 
Suppose there is a time when $v = 0,$ then at this local maximum, we have,
$$ 0 = -v_t + \Delta v + (1-\epsilon + \mu e^{kt})(\mu e^{kt}-\epsilon) - k\mu e^{kt}
$$
or
$$ 0 = -v_t + \Delta v - (\mu e^{kt}-\epsilon)^2 + \mu e^{kt}-\epsilon  - k\mu e^{kt} < 0 .
$$
for $k$ large enough, thus a contradiction.
