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This recent news article (here is the original, in German) says that

Shouryya Ray, who moved to Germany from India with his family at the age of 12, has baffled scientists and mathematicians by solving two fundamental particle dynamics problems posed by Sir Isaac Newton over 350 years ago, Die Welt newspaper reported on Monday.

Ray’s solutions make it possible to now calculate not only the flight path of a ball, but also predict how it will hit and bounce off a wall. Previously it had only been possible to estimate this using a computer, wrote the paper.

What are the problems from this description? What is their precise formulation? Also, is there anywhere I can read the details of this person's proposed solutions?

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My suspicion is that this is yet another example of idiotic science journalism. I'm curious to know if I'm right though :) –  Colin K May 24 '12 at 22:50
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@Frank: Here is the full DailyMail article. –  Zev Chonoles May 27 '12 at 1:39
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I have sent an email to the organisers, asking them if the results are accessible somewhere. But the point is that this competition is a very well-reputed one, so it is likely that the student did something reasonable in a correct way. It's not so likely that it is the breakthrough that the newspapers make out of it or even worthy of a publication. (To give you an idea, the two runners-up in the category mathematics wrote a computer program to simulate the composition of fugues and a computer program for ray-tracing.) –  Phira May 27 '12 at 9:41
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I doubt that it is the student's fault that he is presented as the nerd genius solving the problems that have stumped centuries of mathematicians. Newspapers love this. "High school student shows a lot of promise and might be a very good researcher in 10 years." just doesn't cut it. –  Phira May 27 '12 at 9:43
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I agree with @Phira. This competition is organized in three stages: a regional stage, a state-wide stage and a nationwide stage. He made it to nationwide, but only scored second there. The nationwide winner was the runner-up from the second stage and his relativistic ray-tracer, so I doubt that Ray really solved an unsolved Math problem. –  mnemosyn May 28 '12 at 10:44
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up vote 19 down vote accepted

This thread(physicsforums.com) contains a link to Shouryya Ray's poster, in which he presents his results.

So the problem is to find the trajectory of a particle under influence of gravity and quadratic air resistance. The governing equations, as they appear on the poster:

$$ \dot u(t) + \alpha u(t) \sqrt{u(t)^2+v(t)^2} = 0 \\ \dot v(t) + \alpha v(t) \sqrt{u(t)^2 + v(t)^2} = -g\text, $$

subject to initial conditions $v(0) = v_0 > 0$ and $u(0) = u_0 \neq 0$.

Thus (it is easily inferred), in his notation, $u(t)$ is the horizontal velocity, $v(t)$ is the vertical velocity, $g$ is the gravitational acceleration, and $\alpha$ is a drag coefficient.

He then writes down the solutions

$$ u(t) = \frac{u_0}{1 + \alpha V_0 t - \tfrac{1}{2!}\alpha gt^2 \sin \theta + \tfrac{1}{3!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right) t^3 + \cdots} \\ v(t) = \frac{v_0 - g\left[t + \tfrac{1}{2!} \alpha V_0 t^2 - \tfrac{1}{3!} \alpha gt^3 \sin \theta + \tfrac{1}{4!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right)t^4 + \cdots\right]}{1 + \alpha V_0 t - \tfrac{1}{2!}\alpha gt^2 \sin \theta + \tfrac{1}{3!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right) t^3 + \cdots}\text.$$

From the diagram below the photo of Newton, one sees that $V_0$ is the inital speed, and $\theta$ is the initial elevation angle.

The poster (or at least the part that is visible) does not give details on the derivation of the solution. But some things can be seen:

  • He uses, right in the beginning, the substitution $\psi(t) = u(t)/v(t)$.

  • There is a section called "...öße der Bewegung". The first word is obscured, but a qualified guess would be "Erhaltungsgröße der Bewegung", which would translate as "conserved quantity of the motion". Here, the conserved quantity described by David Zaslavsky appears, modulo some sign issues.

  • However, this section seems to be a subsection to the bigger section "Aus der Lösung ablesbare Eigenschaften", or "Properties that can seen from the solution". That seems to imply that the solution implies the conservation law, rather than the solution being derived from the conservation law. The text in that section probably provides some clue, but it's only partly visible, and, well, my German is rusty. I welcome someone else to try to make sense of it.

  • Also part of the bigger section are subsections where he derives from his solution (a) the trajectory for classical, drag-free projectiles, (b) some "Lamb-Näherung", or "Lamb approximation".

  • The next section is called "Verallgemeneirungen", or "Generalizations". Here, he seems to consider two other problems, with drag of the form $\alpha V^2 + \beta$, in the presence of altitude-dependent horizontal wind. I'm not sure what the results here are.

  • The diagrams to the left seem to demonstrate the accuracy and convergence of his series solution by comparing them to Runge-Kutta. Though the text is kind of blurry, and, again, my German is rusty, so I'm not too sure.

  • Here's a rough translation of the first part of the "Zusammanfassung und Ausblick" (Summary and outlook), with suitable disclaimers as to the accuracy:

  • For the first time, a fully analytical solution of a long unsolved problem
  • Various excellent properties; in particular, conserved quantity $\Rightarrow$ fundamental [...] extraction of deep new insights using the complete analytical solutions (above all [...] perspectives and approximations are to be gained)
  • Convergence of the solution numerically demonstrated
  • Solution sketch for two generalizations

EDIT: Two professors at TU Dresden, who have seen Mr Ray's work, have written some comments:

Comments on some recent work by Shouryya Ray

There, the questions he solved are unambiguously stated, so that should answer any outstanding questions.

EDIT2: I should add: I do not doubt that Shouryya Ray is a very intelligent young man. The solution he gave can, perhaps, be obtained using standard methods. I believe, however, that he discovered the solution without being aware that the methods were standard, a very remarkable achievement indeed. I hope that this event has not discouraged him; no doubt, he'll be a successful physicist or mathematician one day, should he choose that path.

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Link to image of Shouryya Ray's poster is now dead. –  Qmechanic Jul 9 '12 at 20:00
    
Here is another link to the poster image. –  Qmechanic Jul 11 '12 at 20:23
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It is indeed quite difficult to find information on why exactly this project has attracted so much attention. What I've pieced together from comments on various websites and some images (mainly this one) is that Shouryya Ray discovered the following constant of motion for projectile motion with quadratic drag:

$$\frac{g^2}{2v_x^2} + \frac{\alpha g}{2}\left(\frac{v_y\sqrt{v_x^2 + v_y^2}}{v_x^2} + \sinh^{-1}\biggl|\frac{v_y}{v_x}\biggr|\right) = \text{const.}$$

This applies to a particle which is subject to a quadratic drag force,

$$\vec{F}_d = -m\alpha v\vec{v}$$

It's easily verified that the constant is constant by taking the time derivative and plugging in the equations of motion

$$\begin{align}\frac{\mathrm{d}v_x}{\mathrm{d}t} &= -\alpha v_x\sqrt{v_x^2 + v_y^2} \\ \frac{\mathrm{d}v_y}{\mathrm{d}t} &= -\alpha v_y\sqrt{v_x^2 + v_y^2} - g\end{align}$$

The prevailing opinion is that this has not been known before, although some people are claiming to have seen it in old textbooks (never with a reference, though, so take it for what you will).

I haven't heard anything concrete about how this could be put to practical use, although perhaps that is part of the technical details of the project. It's already possible to calculate ballistic trajectories with drag to very high precision using numerical methods, and the presence of this constant doesn't directly lead to a new method of calculating trajectories as far as I can tell.

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There is a discussion on Reddit about this subject, which describes the problem and a verification of the solution. See reddit.com/r/worldnews/comments/u7551/… –  jbatista May 28 '12 at 22:02
    
@jbatista yeah, that's one of the sources I was getting my information from. –  David Z May 28 '12 at 22:05
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So it sounds like a very neat result, and definitely impressive for a high school student; but not exactly worth a "Kid out-thinks Newton" headline. Junky science journalism, as always. –  Colin K May 28 '12 at 22:19
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The MathExchange cross-post math.stackexchange.com/q/150242 also copy+pastes the Reddit discussion; particularly salient is that it cites a result by G. W. Parker published in Am.J.Phys. 45 (1977) 606-610 discussing the same problem. This makes it even more interesting to find out about how Ray obtained his result. –  jbatista May 28 '12 at 22:19
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I) Here we would like to give a Hamiltonian formulation of a point particle in a constant gravitational field with quadratic air resistance

$$\tag{1} \dot{u}~=~ -\alpha u \sqrt{u^2+v^2}, \qquad \dot{v}~=~ -\alpha v \sqrt{u^2+v^2} -g. $$

The $u$ and $v$ are the horizontal and vertical velocity, 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 (1) become

$$\tag{4a} \dot{q}~=~ gp, $$ $$\tag{4b} \dot{p}~=~ \alpha \sqrt{1+q^2}. $$

III) Equation (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 (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

$$ H(q,p)~:=~T+V(q)~=~\frac{gp^2}{2}- \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$ $$\tag{12}~=~\frac{g}{2u^2} +\frac{\alpha}{2} \left( \frac{v\sqrt{u^2+v^2}}{u^2} + {\rm arsinh}\frac{v}{|u|}\right). $$

Since the Hamiltonian $H$ contains no explicit time dependence, the mechanical energy (12) is conserved in time, which is Shouryya Ray's first integral of motion.

$$\tag{13} \frac{dH}{dt}~=~ \frac{\partial H}{\partial t}~=~0. $$

VI) The Hamiltonian equations of motion are eqs. (4). 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

$$\tag{15} -X_H~:=~-\{H(q(t_i),p(t_i)), \cdot\} ~=~ gp(t_i)\frac{\partial}{\partial q(t_i)} + F(q(t_i))\frac{\partial}{\partial p(t_i)}. $$

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

$$ q(t_f) ~=~ e^{-\tau X_H}q(t_i) ~=~ q(t_i) - \tau X_H[q(t_i)] + \frac{\tau^2}{2}X_H[X_H[q(t_i)]]+\ldots \qquad $$ $$\tag{17a} ~=~ q(t_i) + \tau g p(t_i) + \frac{\tau^2}{2}g F(q(t_i)) +\frac{\tau^3}{6}g \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))} +\ldots ,\qquad $$ $$ p(t_f) ~=~ e^{-\tau X_H}p(t_i) ~=~ p(t_i) - \tau X_H[p(t_i)] + \frac{\tau^2}{2}X_H[X_H[p(t_i)]]+\ldots\qquad $$ $$ ~=~p(t_i) + \tau F(q(t_i)) +\frac{\tau^2}{2}\frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))}$$ $$\tag{17b} + \frac{g\alpha^2\tau^3}{6} \left(q(t_i) + \frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right) +\ldots ,\qquad $$ $$ F(q(t_f)) ~=~ e^{-\tau X_H}F(q(t_i)) $$ $$~=~ F(q(t_i)) - \tau X_H[F(q(t_i))] + \frac{\tau^2}{2}X_H[X_H[F(q(t_i))]] + \ldots\qquad $$ $$ \tag{17c}~=~ F(q(t_i)) + \tau \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))} +\frac{g(\alpha\tau)^2}{2}\left(q(t_i) +\frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right) +\ldots ,\qquad $$

and calculate to any order in time $\tau:=t_f-t_i$, we would like. (As a check, note that if one differentiates (17a) with respect to time $\tau$, one gets (17b) multiplied by $g$, and if one differentiates (17b) with respect to time $\tau$, one gets (17c), cf. eq. (4).) 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 (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:

  1. Herbert Goldstein, Classical Mechanics.

$^\dagger$ Note that if $u$ becomes zero at some point, it stays zero in the future, cf. eq.(1). If $u\equiv 0$ identically, then eq.(1) becomes

$$\tag{22} -\dot{v} ~=~ \alpha v |v| + g. $$

The solution to 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,

$$\tag{24} (u(t),v(t)) ~\to~ (0, -\sqrt{\frac{g}{\alpha}}) \qquad \text{for} \qquad t ~\to~ \infty ,$$

while

$$\tag{25} (q(t),p(t)) ~\to~ (\infty,\infty) \qquad \text{for} \qquad t~\to~ \infty.$$

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Interestingly, one may generalize to arbitrary power laws $\dot{u}= -\alpha u (u^2+v^2)^{\frac{r}{2}}$ and $\dot{v}= -\alpha v (u^2+v^2)^{\frac{r}{2}} -g$. Here $r$ is a real power with $r\neq 0,-1$. The canonical coordinates are then $q:=-\frac{v}{|u|}$ and $p:=|u|^{-r}$ with eoms $\dot{q}=\frac{g}{|u|}=gp^{\frac{1}{r}}$ and $\dot{p}=r\alpha(1+q^2)^{\frac{r}{2}}=:F(q)=-V^{\prime}(q)$. The potential $V(q)$ is (essentially) a hypergeometric function. The Hamiltonian reads $H(q,p):=\frac{g}{1+r^{-1}}p^{1+r^{-1}}+V(q)$. –  Qmechanic Feb 23 at 19:02
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The equations for this projectile problem were formulated by Jacob Bernoulli (1654-1705) and Gottfried Leibniz (1646-1716) developed a solution technique in 1696! The method develops an analytical solution for the velocity and angle of inclination (or equivalently the horizontal and vertical velocities). To obtain the horizontal and vertical displacements and the time by analytical methods from these intermediate results has not been successful since that then. It probably never will. However simple numerical techniques can yield solutions more efficiently without the use of power series representation. For example MATHEMATICA easily solves the equations from the intermediate results. I'm surprised that someone from the military ballistic community has not commented or even the aerospace guys. This must be very elementary to them. I happened upon this subject because I am reading an interesting book by Neville De Mestre called "The Mathematics of Projectiles in Sport". I recommend it. Although it was published in 1990 and is probably out of print it may be available through AMAZON BOOKS.

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