## A Random Pursuit

A missile is launched when an airplane is directly overhead, at a height $h$. The missile moves at a constant speed $v$ always heading directly towards the aircraft, which is moving along a straight line at constant velocity $u$. What is the shape of the missile’s trajectory? How far will the aircraft fly before it is hit by the missile? What if the plane takes evasive action by randomly changing its direction, but heading in the original direction on the average? What is the best strategy for the missile to maximize the probability of a hit?

### Pursuit of Uniform Motion

The deterministic version of this pursuit problem is well known. It is variously phrased as a dog chasing a rabbit, a peacock chasing a snake etc. and has been attributed to such ancients as Appolonius, Bhaskara . It has an elegant analytic solution. I am not sure how Bhaskara or Appolonius is supposed to have solved this problem originally, as the methods I know use calculus quite heavily. The answer is in the end algebraic. So may be there is a clever, purely algebraic method as well?

It is clear the outset that we need $v>u$ for a collision: the missile has farther to go.

Choose a co-ordinate system with origin at the launch site of the missile; the plane is moving along the $x$-axis at a value of $y$ equal to $h$. It turns out to be convenient to use $y$ as the independent variable, so we will try to determine $x$ as a function of $y$. The condition that the missile heads directly towards the plane gives the slope of this curve:

${dx\over dy}={x-ut\over y-h}.$
Now set

$p={dx\over dy},\quad (y-h)p=x-ut .$
The condition that the speed of the missile is $v$ becomes
$vdt=\sqrt{dx^2+dy^2}=\sqrt{1+p^2}dy$
Differentiating the earlier equation and using this one

$(y-h){dp\over dy}=-a\sqrt{1+p^2}$

where

$a={u\over v}<1.$ Rewriting, ${dp\over \sqrt{1+p^2}}=-a{dy\over y-h}.$ Integrating this, $p=-\sinh\left[\log\left({h-y\over h}\right)^a\right]$ In other words ${dx\over dy}={1\over 2}\left[\left({h-y\over h}\right)^{-a}-\left({h-y\over h}\right)^{a} \right]$ Solving $x={1\over 2}\left[{1\over 1+a}\left({h-y\over h}\right)^{1+a}-{1\over 1-a}\left({h-y\over h}\right)^{1-a}\right]+{ah\over 1-a^2}$ The collision happens when the value of $y$ reaches $h$; the distance the plane has moved by then is just the value of $x$ at this instant. That is, ${ah \over 1-a^2}.$ We see clearly why $a<1, u is needed. We can also get the time dependence of the missile trajectory: $vdt=\cosh\left[\log\left({h-y\over h}\right)^a\right]dy$ so that $t={h-y\over 2v}\left[{1\over 1-a}\left({h-y\over h}\right)^{-a}+{1\over 1+a}\left({h-y\over h}\right)^{a}\right].$

### Hamiltonian System

The differential equations above can also be thought as a Hamiltonian system. Indeed, if $p$ and $y$ are conjugate and

$H=v\ {\rm arcsinh}\ p +u\ \log {h-y\over h}$

we get

$\dot y={v\over \sqrt{1+p^2}},\quad \dot p={u\over h-y}.$

The conservation of the Hamiltonian gives the first integral of this system. With our initial conditions, it is just equal to zero. the time dependence is then determined by solving the equation for $\dot y$ after eliminating $p$ in favor of $y$.

### Randomness

What if the plane takes evasive action? It could change its direction randomly, keeping the average velocity the same as before. The missile would have to constantly change direction. Given a finite speed $v$ will it still catch the plane?

We get the stochastic differential system:

${d{\mathbf r}\over dt}=v{{\mathbf s}-{\mathbf r}\over | {\mathbf s}-{\mathbf r}|}$

where ${\mathbf s}$ is the path of the plane. One method would be to turn this into a linear Partial differential equation for the probability density of ${\mathbf r}$. It is unlikely to be exactly solvable though. Any ideas?

It could be convenient to make the change of variable

${\mathbf r}={\mathbf s}-[vt+w]{\mathbf n},\quad {\mathbf n}\cdot{\mathbf n}=1.$

so that

$\dot w=-\dot {\mathbf s}\cdot {\mathbf n},\quad {d{\mathbf n}\over dt}={(\dot{\mathbf s}\cdot{\mathbf n}){\mathbf n}-\dot{\mathbf s}\over w+vt}$

Then what?