I) Here, we would like to give a Hamiltonian formulation of a point particle in a constant gravitational field with quadratic air resistance
$$\tag{1}\label{eq:1} \dot{u}~=~ -\alpha u \sqrt{u^2+v^2}, \qquad \dot{v}~=~ -\alpha v \sqrt{u^2+v^2} -g. $$
The $u$ and $v$ are the horizontal and vertical velocities, respectively. A dot on top denotes differentiation with respect to time $t$. The two positive constants $\alpha>0$ and $g>0$ can be put to one by scaling the three variables
$$\tag{2} t'~=~\sqrt{\alpha g}t, \qquad u'~=~\sqrt{\frac{\alpha}{g}}u, \qquad v'~=~\sqrt{\frac{\alpha}{g}}v. $$
See, e.g. Ref. [1] for a general introduction to Hamiltonian and Lagrangian formulations.
II) Define two canonical variables (generalized position and momentum) as
$$\tag{3} q~:=~ -\frac{v}{|u|}, \qquad p~:=~ \frac{1}{|u|}~>~0.$$
(The position $q$ is (up to signs) Shouryya Ray's $\psi$ variable, and the momentum $p$ is (up to a multiplicative factor) Shouryya Ray's $\dot{\Psi}$ variable. We assume$^\dagger$ for simplicity that $u\neq 0$.) Then the equations of motion $\eqref{eq:1}$ become
$$\tag{4a}\label{eq:4a} \dot{q}~=~ gp, $$
$$\tag{4b}\label{eq:4b} \dot{p}~=~ \alpha \sqrt{1+q^2}. $$
III) Equation $\eqref{eq:4a}$ suggests that we should identify $\frac{1}{g}$ with a mass
$$\tag{5} m~:=~ \frac{1}{g}, $$
so that we have the standard expression
$$\tag{6} p~=~m\dot{q}$$
for the momentum of a non-relativistic point particle. Let us furthermore define kinetic energy
$$\tag{7} T~:=~\frac{p^2}{2m}~=~ \frac{gp^2}{2}. $$
IV) Equation $\eqref{eq:4b}$ and Newton's second law suggest that we should define a modified Hooke's force
$$\tag{8} F(q)~:=~ \alpha \sqrt{1+q^2}~=~-V^{\prime}(q), $$
with potential given by (minus) the antiderivative
$$ V(q)~:=~ - \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$ $$\tag{9} ~=~ - \frac{\alpha}{2} \left(q \sqrt{1+q^2} + \ln(q+\sqrt{1+q^2})\right). $$
Note that this corresponds to an unstable situation because the force $F(-q)~=~F(q)$ is an even function, while the potential $V(-q) = - V(q)$ is a monotonic odd function of the position $q$.
It is tempting to define an angle variable $\theta$ as
$$\tag{10} q~=~\tan\theta, $$
so that the corresponding force and potential read
$$\tag{11} F~=~\frac{\alpha}{\cos\theta} , \qquad
V~=~- \frac{\alpha}{2} \left(\frac{\sin\theta}{\cos^2\theta}
+ \ln\frac{1+\sin\theta}{\cos\theta}\right). $$
V) The Hamiltonian is the total mechanical energy
$$ H(q,p)~:=~T+V(q)~=~\frac{gp^2}{2}- \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$
$$\tag{12}\label{eq:12}~=~\frac{g}{2u^2} +\frac{\alpha}{2} \left( \frac{v\sqrt{u^2+v^2}}{u^2} + {\rm arsinh}\frac{v}{|u|}\right). $$
Since the Hamiltonian $H$ contains no explicit time dependence, the mechanical energy $\eqref{eq:12}$ is conserved in time, which is Shouryya Ray's first integral of motion.
$$\tag{13} \frac{\mathrm{d}H}{\mathrm{d}t}~=~ \frac{\partial H}{\partial t}~=~0. $$
VI) The Hamiltonian equations of motion are eqs. $\eqref{eq:4a}$-$\eqref{eq:4b}$. Suppose that we know $q(t_i)$ and $p(t_i)$ at some initial instant $t_i$, and we would like to find $q(t_f)$ and $p(t_f)$ at some final instant $t_f$.
The Hamiltonian $H$ is the generator of time evolution.
If we introduce the canonical equal-time Poisson bracket
$$\tag{14} \{q(t_i),p(t_i)\}~=~1,$$
then (minus) the Hamiltonian vector field reads
$$\tag{15}\label{eq:15} -X_H~:=~-\{H(q(t_i),p(t_i)), \cdot\} ~=~ gp(t_i)\frac{\partial}{\partial q(t_i)} + F(q(t_i))\frac{\partial}{\partial p(t_i)}. $$
For completeness, let us mention that in terms of the original velocity variables, the Poisson bracket reads
$$\tag{16} \{v(t_i),u(t_i)\}~=~u(t_i)^3.$$
We can write a formal solution to position, momentum, and force, as
$$ q(t_f) ~=~ e^{-\tau X_H}q(t_i) ~=~ q(t_i) - \tau X_H[q(t_i)]
+ \frac{\tau^2}{2}X_H[X_H[q(t_i)]]+\ldots \qquad $$
$$\tag{17a}\label{eq:17a} ~=~ q(t_i) + \tau g p(t_i) + \frac{\tau^2}{2}g F(q(t_i))
+\frac{\tau^3}{6}g \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))} +\ldots ,\qquad $$
$$ p(t_f) ~=~ e^{-\tau X_H}p(t_i) ~=~ p(t_i) - \tau X_H[p(t_i)]
+ \frac{\tau^2}{2}X_H[X_H[p(t_i)]]+\ldots\qquad $$
$$ ~=~p(t_i) + \tau F(q(t_i))
+\frac{\tau^2}{2}\frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))}$$
$$\tag{17b}\label{eq:17b} + \frac{g\alpha^2\tau^3}{6}
\left(q(t_i) + \frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right) +\ldots ,\qquad $$
$$ F(q(t_f)) ~=~ e^{-\tau X_H}F(q(t_i)) $$
$$~=~ F(q(t_i)) - \tau X_H[F(q(t_i))]
+ \frac{\tau^2}{2}X_H[X_H[F(q(t_i))]] + \ldots\qquad $$
$$ \tag{17c}\label{eq:17c}~=~ F(q(t_i)) + \tau \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))}
+\frac{g(\alpha\tau)^2}{2}\left(q(t_i)
+\frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right)
+\ldots ,\qquad $$
and calculate to any order in time $\tau:=t_f-t_i$, we would like.
(As a check, note that if one differentiates $\eqref{eq:17a}$ with respect to time $\tau$,
one gets $\eqref{eq:17b}$ multiplied by $g$, and if one differentiates $\eqref{eq:17b}$ with respect to time $\tau$, one gets $\eqref{eq:17c}$, cf. eqs. $\eqref{eq:4a}$-$\eqref{eq:4b}$.)
In this way, we can obtain a Taylor expansion in time $\tau$ of the form
$$ \tag{18} F(q(t_f)) ~=~\alpha\sum_{n,k,\ell\in \mathbb{N}_0}
\frac{c_{n,k,\ell}}{n!}\left(\tau\sqrt{\alpha g}\right)^n
\left(p(t_i)\sqrt{\frac{g}{\alpha}}\right)^k
\frac{q(t_i)^{\ell}}{(F(q(t_i))/\alpha)^{k+\ell-1}}. $$
The dimensionless universal constants $c_{n,k,\ell}=0$ are zero if either $n+k$ or $\frac{n+k}{2}+\ell$ are not an even integer. We have a closed expression
$$ F(q(t_f)) ~\approx~
\exp\left[\tau gp(t_i)\frac{\partial}{\partial q(t_i)}\right]F(q(t_i))
~=~ F(q(t_i)+\tau g p(t_i)) $$
$$ \tag{19} \qquad \text{for} \qquad
~ p(t_i)~\gg~\frac{ F(q(t_i))}{\sqrt{\alpha g}}, $$
i.e., when we can ignore the second term in the Hamiltonian vector field $\eqref{eq:15}$.
VII) The corresponding Lagrangian is
$$\tag{20} L(q,\dot{q})~=~T-V(q)~=~\frac{\dot{q}^2}{2g}+ \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$
with Lagrangian equation of motion
$$\tag{21} \ddot{q}~=~ \alpha g \sqrt{1+q^2}. $$
This is essentially Shouryya Ray's $\psi$ equation.
References:
- Herbert Goldstein, Classical Mechanics.
$^\dagger$ Note that if $u$ becomes zero at some point, it stays zero in the future, cf. eq. $\eqref{eq:1}$. If $u\equiv 0$ identically, then eq. $\eqref{eq:1}$ becomes
$$\tag{22}\label{eq:22} -\dot{v} ~=~ \alpha v |v| + g. $$
The solution to eq. $\eqref{eq:22}$ for negative $v\leq 0$ is
$$\tag{23} v(t) ~=~ -\sqrt{\frac{g}{\alpha}} \tanh(\sqrt{\alpha g}(t-t_0)) , \qquad t~\geq~ t_0, $$
where $t_0$ is an integration constant. In general,
$$\tag{24} (u(t),v(t)) ~\to~ \left(0, -\sqrt{\frac{g}{\alpha}}\right) \qquad \text{for} \qquad t ~\to~ \infty ,$$
while
$$\tag{25} (q(t),p(t)) ~\to~ (\infty,\infty) \qquad \text{for} \qquad t~\to~ \infty.$$
