I have a function $q(t)$ that starts at $q(0)=q_0$ and needs to get to $q(1)=q_1>q_0$. I have a free parameter $z$ that I can wiggle around to control $q'(t)$. Namely, I have a function $$q'(t)=Q(t,q(t),z)>0.$$ I know that $\forall t$, $\forall q(t)$, $\lim_{z \rightarrow -\infty}Q(t,q(t),z)=0$, and $\lim_{z \rightarrow +\infty}Q(t,q(t),z)=+\infty$. I know that $Q(t,q(t),z)$ is increasing in $z$.
1) Do I understand correctly that, if $Q(t,q(t),z)$ is continuous in $z$, I can always find $z$ such that I will hit exactly $q(1)=q_1$? Is it obvious, or some sort of a proof is in order?
2) My $Q(t,q(t),z)$ is not continuous in $z$, but the jumps, if they happen, only happen upwards, and they are not "frequent" in the sense that if there is a jump for a given $z$ at a given $t$ and $q(t)$, there is no jump for all $z$ in a small neighborhood of my $z$ for the same $t$ and $q(t)$. Can I still argue that I can always hit $q(1)=q_1$? I mean, my $Q(t,q(t),z)$ is still continuous where there are no jumps...
2.1) Would it help if I could impose that, generically, if $Q(t,q(t),z)$ has a jump at $z$ for a given $t$ and $q(t)$, $Q(t,q(t)+\varepsilon,z)$ is locally continuous at the same $z$?
3) My $Q()$ turns out to be continuously differentiable, except for the points where it jumps. That probably implies Lipschitz continuity almost everywhere, which guarantees the existence of the solution to the initial value problem if I start from $q(0)=q_0$? upd I can probably only care about $q\in[q_0+\infty)$, in which case my $Q()$ is bounded, and therefore continuous differentiability in $q$, which I have, is sufficient for being Lipschitz?
4) This guy (http://www.math.washington.edu/~burke/crs/555/555_notes/continuity.pdf, p23) says that for continuity in $z$, my $Q$ should have a Lipschitz constant independent of $z$. Would boundedness and continuous differentiability suffice to get a Lipschitz constant independent of z?
Well, for your question 2.0, here is a counterexample:
Counterexample for 2.0: Define $q_0=1, q_1=2$. Define $f(z)$ as follows: $$ f(z) = \left\{ \begin{array}{ll} 0 &\mbox{ if $z < \ln(4)$} \\ z& \mbox{ if $z\geq \ln(4)$} \end{array} \right.$$ Define $Q(t,q,z) = f(z)q$. Then $\lim_{z\rightarrow\infty} Q(t,q,z)=\infty$ and $\lim_{z\rightarrow-\infty} Q(t,q,z) = 0$, and we have only one jump discontinuity at $z=\ln(4)$. But the solution to the ODE is $q(t) = q(0)e^{f(z)t}$. So $q(1) = q(0)=1$ if $z < \ln(4)$ and $q(1) = e^{f(z)}\geq 4 > 2$ if $z\geq \ln(4)$.