From Ottawa Citizen (and all over, really):

An Indian-born teenager has won a research award for solving a mathematical problem first posed by Sir Isaac Newton more than 300 years ago that has baffled mathematicians ever since.

The solution devised by Shouryya Ray, 16, makes it possible to calculate exactly the path of a projectile under gravity and subject to air resistance.

This subject is of particular interest to me. I have been unable to locate his findings via the Internet. Where can I read his actual mathematical work?

So has he written an actual paper, and if so, will anyone get to read it?

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    $\begingroup$ My guesses: (1) The author did something highly worthy of a prize from a "Youth Research Foundation" (presumably a group whose purpose is to identify and encourage young talent, perhaps not specifically in mathematics). (2) This work has not been submitted, and probably was not intended to be submitted, to the math or physics research communities. (3) The flurry of media attention is a function of the enthusiasm of mathematically illiterate reporters for a good human interest story, not the importance of the work itself. (I would love to be wrong, but generally, that's what these things are.) $\endgroup$ Commented May 27, 2012 at 4:40
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    $\begingroup$ Online newspapers of India tend to trumpet achievements of those who live there or were born there. In one way it is annoying, when the description is really out of proportion to the facts. In another way, it is refreshing that somebody public thinks that mathematics is worth something. $\endgroup$
    – Will Jagy
    Commented May 27, 2012 at 4:43
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    $\begingroup$ I asked about this on physics.SE. $\endgroup$ Commented May 27, 2012 at 4:45
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    $\begingroup$ This is not the fault of Indian newspapers. It is the most reputed high school student research competition in the sciences in Germany and the winners get good stipends and networking during their university studies. The enthusiastic, but information-scarce articles are in German newspapers. I do think that it is correct to regard this as a mathematics problem, but physicists might be better situated to know what it is all about. $\endgroup$
    – Phira
    Commented May 27, 2012 at 9:32
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    $\begingroup$ Indeed, as shown in the reddit links in Zhen's comment, the differential equation is easily solved by separation of variables and very simple algebra. So it's hard to imagine what is supposedly so impressive. No doubt this was known long ago. It seems the achievement has been highly blown out of proportion by the media. One needn't be a "genius" to know this much calculus at age 16. Probably many readers here knew such. $\endgroup$ Commented May 28, 2012 at 21:50

4 Answers 4


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

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


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

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.

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    $\begingroup$ This accepted answer on physics.SE has been updated and provides the above link too. $\endgroup$ Commented Jun 7, 2012 at 22:04

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."


Knowing some German helps. When you go to this page and you click the picture, it enlarges it.

You can see some math symbols above his right shoulder. That may be all we can get until he puts the text of his presentation on the web somewhere.

Here's the full equation (also from the original site). It reads:

$$\frac{g^2}{2u^2}+\frac{\alpha g}2\left(\frac{v\sqrt{v^2+u^2}}{u^2}+\mathrm{arsinh}\,\left|\frac{v}u\right|\right)=\mathrm{const.}$$

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    $\begingroup$ Hmmm ... If the right-hand side is "const." anyway, then why not multiply through by $2$? $\endgroup$
    – Blue
    Commented May 28, 2012 at 21:21
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    $\begingroup$ Because this is physics $\endgroup$
    – Norbert
    Commented May 28, 2012 at 21:25
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    $\begingroup$ Physicits love division by $2$ :) Recall formula of kinetic energy, potential energy of expanded spring, distance for free fall and etc. $\endgroup$
    – Norbert
    Commented May 29, 2012 at 8:22
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    $\begingroup$ It's like a hint. $\endgroup$
    – jnm2
    Commented May 29, 2012 at 13:19
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    $\begingroup$ There are good reasons for all those factors of two. They arise from any equation described by Newton's second law and its generalizations. The law "(generalized) energy is the line integral of force with respect to (generalized) position" is only true if (generalized) energy is defined with a factor of one half in front of it. $\endgroup$ Commented Jun 7, 2012 at 16:28

Today I found this on arxiv.org Shouryya Ray Paper

It's the official paper from Shouryya Ray:

An analytic solution to the equations of the motion of a point mass with quadratic resistance and generalizations Shouryya Ray · Jochen Frohlich

why did he write it two years later from his original solution ?

  • $\begingroup$ Not two years -- the first version was posted in May 2013. $\endgroup$
    – user147263
    Commented Oct 19, 2014 at 5:54

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