Teenager solves Newton dynamics problem - where is the paper?

In the document Comments on some recentwork by Shouryya Ray by Prof. Dr. Ralph Chil and Prof. Dr. Jürgen Voigt (Technische Universität Dresden), dated June 4, 2012 it is written:

Conducting an internship at the Chair of Fluid Mechanics at TU Dresden, Shouryya Ray encountered two ordinary differential equations which are special cases of Newton's law that the derivative of the momentum of a particle equals the forces acting on it. In the first one, which describes the motion of a particle in a gas or fluid, this force is the sum of a damping force, which depends quadratically on the velocity, and the (constant) gravitational force. $$\begin{eqnarray*} \dot{u} &=&-u\sqrt{u^{2}+v^{2}},\qquad u(0)=u_{0}>0 \\ \dot{v} &=&-v\sqrt{u^{2}+v^{2}}-g,\quad v(0)=v_{0}. \end{eqnarray*}\tag{1}$$ Here, $u$ and $v$ are the horizontal and vertical velocity, respectively.

(...)

The second equation reads $$ \ddot{z}=-\dot{z}-z^{3/2},\qquad z(0)=0,\dot{z}(0)=z_{1},\tag{2} $$ and describes the trajectory of the center point $z(t)$ of a spherical particle during a normal collision with a plane wall. (...)

Let us come back to problem (1) which was the starting point of the media stories. In the context of Shouryya Ray's work it was an unfortunate circumstance, that a recent article from 2007$^8$ claims that no analytical solution of problem (1) was known, or that it was known only in special cases, namely falling objects$^9$. This might have misled Shouryya Ray who was not aware of the classical theory of ordinary differential equations. (...)

To conclude, Shouryya Ray has obtained analytic solutions of the problem (1), by transforming it successively to the problems (3)-(5), and by applying a recent result of D. Dominici in order to obtain a recursive formula for the coefficients of the power series representation of $\psi$. He then validated his results numerically. Given the level of prerequisites that he had, he made great progress. Nevertheless all his steps are basically known to experts and we emphasize that he did not solve an open problem posed by Newton. (...)

We hope that this small text gives the necessary information to the mathematical community, and that it allows the community to both put in context and appreciate the work of Shouryya Ray who plans to start a career in mathematics and physics.

The function $\psi$ is given by

$$\psi (t)=(v_{0}-g\Psi (t))/u_{0},$$

where

$$\Psi (t)=\int_{0}^{t}\exp \left[ \int_{0}^{\tau }\sqrt{u^{2}(s)+v^{2}(s)}ds \right] d\tau .$$

I've read about this text on this blog post.

PS. Also in Spanish the Francis (th)E mule Science's News post El problema de Newton y la solución que ha obtenido Shouryya Ray (16 años) discusses these problems.

Here's summmary of what I'm reading through the interwebs:

The original post is here

"The problem he solved is as follows:

Let $(x(t),y(t))$ be the position of a particle at time $t$. Let $g$ be the acceleration due to gravity and $c$ the constant of friction. Solve the differential equation:

$$(x''(t)^2 + (y''(t)+g)^2 )^{1/2} = c(x'(t)^2 + y'(t)^2 )$$

subject to the constraint that $(x''(t),y''(t)+g)$ is always opposite in direction to $(x'(t),y'(t))$.

Finding the general solution to this differential equation will find the general solution for the path of a particle which has a drag proportional to the square of the velocity (and opposite in direction). Here's an explanation of how this differential equation encodes the motion of such a particle:

The square of the velocity is:

$$x'(t)^2 + y'(t)^2$$

The total acceleration is:

$$( x''(t)^2 + y''(t)^2 )^{1/2}$$

The acceleration due to gravity is g in the negative y direction. Thus the drag (acceleration due only to friction) is [the preceding should probably read "the impedance (acceleration due to friction plus gravity) is"]:

$$\bigg( x''(t)2 + (y''(t)+g)2 \bigg)(1/2)$$

Thus path of such a particle satisfies the differential equation:

$$( x''(t)^2 + (y''(t)+g)^2 )^{1/2} = c(x'(t)^2 + y'(t)^2 )$$

Of course, we also require the direction of the drag $(x''(t),y''(t)+g)$ to be opposite to the direction of the velocity $(x'(t),y'(t))$. Once we find the intial position and velocity of the particle, uniqueness theorems tell us its path is uniquely determined."

The original post is here

"Here's a forward solution (found by reverse-engineering the answer):

Consider a projectile moving in gravity with quadratic air resistance. The governing equations are

$$u' = -a u \sqrt{ u^2 + v^2 }$$

$$v' = -a v \sqrt{ u^2 + v^2 } - g$$

where $a$ is the coefficient of air resistance defined by $|F| = ma|v|^2$.

Cross-multiply and rearrange to find

$$a \sqrt{ u^2 + v^2 } (uv'-vu') = gu'$$

Substitute $v = su$ and separate variables:

$$a \sqrt{ 1 + s^2} s' = g\frac {u'}{u^3}$$

Integrate both sides to get the answer:

$$\frac g {u^2} + a \left(\frac{v \sqrt{ u^2 + v^2 }}{u^2} + \sinh^{-1}\left|\frac v u\right| \right)= C"$$

"From what I can tell from the image the solution in the image is implicit and was derived by Parker at NCSU in 1977. It is still impressing for a 16-year-old. Here's the paper by Parker if anyone's is interested."

"Am I crazy... or is the equation not anywhere in the paper?"

"It isn't the same equation, but there are solutions using the logarithm definition of inverse hyperbolic functions. The solution isn't the same, but there's an implicit solution too. It's impressing because he found a way to solve the differential equation while Parker stopped at the hairy integral at the end of the "Exact Solutions" section. But yeah, my comment does look confusing, as the exact same equation isn't there."

"It was of little interest, thus nobody was really breaking their teeth on it. But it still is amazing, especially at such a young age."

"Exactly. This solution is implicit, therefore it has little use in actual calculations as you would need to numerically solve it in order to use it, you might as well solve the differential equation numerically directly. Exact solutions similar to the one presented here have been known since 1977 in a paper I posted in another thread. Anyway, the trick used to solve the ODE is quite clever, especially for a 16-year-old."