Preface

This section presents very important applications of ODFEs to pursuit problems.

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Glossary

Pursuit


Pursuit curve

People are familiar with chases and escapes since their childhood when every child played the game hide-and-seek.

On January 16, 1732, the French mathematician and hydrographer Pierre Bouguer (1698--1758) read his paper before the French Academy where he first formulated and solved the pursuit problem. Bouguer treated the case of a pirate ship pursuing a fleeing merchant vessel that he assumed to travel along straight line with constant speed vm. The pirate ship travels at constant speed vp along a curved path such that it is always moving directly toward the merchant, that is, the velocity vector of the pirate ship points directly at the merchant vessel at every instant of time. Bouguer's problem was to determine the equation \( y = y(x) \) of the curved path which is called the curve or ine of pursuit.

 To find the curve of pursuit for Bouguer's problem, let us denote by (x,y) the location of pirate ship in rectangualr coordinate system at arbitrary time \( t \ge 0 . \) We assume that initially, the pirate ship was at the origin, and the merchant vessel was at the point \( \left( x_0 , 0 \right) . \) At time t, the merchant vessel has sailed to the point \( \left( x_0 , v_m t \right) . \) The slope of the tangent line to the pursuit curve is

\[ \frac{{\text d}y}{{\text d}x} = \frac{v_m t -y}{x_0 -x} = \frac{y - v_m t}{x- x_0} . \]
We also know that, whatever the shape of the pursuit curve, the pirate ship has sailed along it at time t by distance \( v_p t . \) From calculus, we know taht this arc-length is also given by the integral
\[ v_p t = \int_0^x \sqrt{1 + \left( \frac{{\text d}y}{{\text d}s} \right)^2} \,{\text d}s , \]
where s is a dummy variable of integration. Using the expression for the derivative, we find
\[ \frac{1}{v_p} \int_0^x \sqrt{1 + \left( \frac{{\text d}y}{{\text d}s} \right)^2} \,{\text d}s = \frac{y}{v_m} - \frac{x- x_0}{v_m} \cdot \frac{{\text d}y}{{\text d} x} , \]
which, if we let \( {\text d}y /{\text d} x = p(x) , \) becomes
\[ \frac{1}{v_p} \int_0^x \sqrt{1 + p^2 (s)} \,{\text d}s = \frac{y}{v_m} - \frac{x- x_0}{v_m} \cdot p(x) . \]
Differentiating the above equation with respect to x, we arrive at
\[ \frac{1}{v_p} \,\sqrt{1 + p^2 (x)} = \frac{1}{v_m} \,\frac{{\text d}y}{{\text d}x} - \frac{x- x_0}{v_m} \cdot \frac{{\text d} p(x)}{{\text d} x} - \frac{1}{v_m} \, p(x) . \]
Using some algebra, we simplify the above equation to the following:
\[ \left( x - x_0 \right) \frac{{\text d}p(x)}{{\text d}x} = - \frac{v_m}{v_p} \cdot \sqrt{1 + p^2 (x)} = -n\, \sqrt{1 + p^2 (x)} , \qquad n= \frac{v_m}{v_p} . \]
Since it is assumed that the pirate ship travels faster that merchant vessel (otherwise, the problem has no solution), the constant n is less than 1. The n = 1 case is actually called tractrix and it was consided previously. The above equation is separable
\[ \frac{{\text d}p}{\sqrt{1+p^2}} = - \frac{n\,{\text d} x}{x- x_0} = \frac{n\,{\text d} x}{x_0 -x} . \]
Integrating both sides, we obtain
\[ \ln \left( p + \sqrt{1+p^2} \right) +C = - n\,\ln \left( x- x_0 \right) , \]
where the constant of integration is determined from the initial conditions \( p = {\text d}y / {\text d}x =0 \) at t = 0 when x = 0. Therefore, \( C = -n\, \ln x_0 , \) and we get
\[ \ln \left( p + \sqrt{1+p^2} \right) -n\, \ln x_0 = - n\,\ln \left( x- x_0 \right) . \]
Upon some pencil pushing, we arrive at
\[ \ln \left[ \left( p + \sqrt{1+p^2} \right) \left( 1 - \frac{x}{x_0} \right)^n \right] =0 , \]
which yields
\[ \left( p + \sqrt{1+p^2} \right) \left( 1 - \frac{x}{x_0} \right)^n =1 . \]
If we temporarily set \( q = p + \sqrt{1+p^2} \quad \Longrightarrow \quad p = \left( q^2 -1 \right) /(2q) , \) then
\[ p(x) = \frac{{\text d}y}{{\text d} x} = \frac{1}{2} \left[ \left( 1 - \frac{x}{x_0} \right)^{-n} - \left( 1 - \frac{x}{x_0} \right)^{n} \right] , \qquad n= \frac{v_m}{v_p} . \]
Once more integration yields
\[ y(x) +C = - \frac{x_0}{2} \, \frac{u^{1-n}}{1-n} + \frac{x_0}{2} \, \frac{u^{1+n}}{1+n} , \qquad u = 1 - \frac{x}{x_0} . \]
Factoring out the common term, we have
\[ y(x) +C = \frac{x_0}{2} \left( 1 - \frac{x}{x_0} \right) \left[ \frac{1}{1+n} \left( 1 - \frac{x}{x_0} \right)^n - \frac{1}{1-n} \left( 1 - \frac{x}{x_0} \right)^{-n} \right] . \]
Now we determine the value of constant of integration from the initial condition \( y=0 \) when x = 0:
\[ C = \frac{x_0}{2} \left( \frac{1}{1+n} - \frac{1}{1-n} \right) = - \frac{n}{1- n^2} \, x_0 . \]
Substituting into the formula for y(x), we get
\[ y(x) = \frac{n}{1- n^2} \, x_0 + \frac{x_0}{2} \left( 1 - \frac{x}{x_0} \right) \left[ \frac{1}{1+n} \left( 1 - \frac{x}{x_0} \right)^n - \frac{1}{1-n} \left( 1 - \frac{x}{x_0} \right)^{-n} \right] , \qquad n = \frac{v_m}{v_p} . \]
``Capture'' occurs when x = x0, that is, the pirate ship pursuit curve intersects the merchant's course. Of course, we have to assume that n is less than 1.
n = 0.66
x0 = 3
c = n*x0/(1 - n^2)
y[x_] = c + (x0 - x)/ 2*((1 - x/x0)^n /(1 + n) - (1 - x/x0)^(-n) /(1 - n))
Plot[y[x], {x, 0, x0}, PlotStyle -> Thick]
Pursuit path for \( n=0.66 \quad\mbox{and} \quad x_0 =3 . \)

 

 

Three bugs

 

  1. A. E. Dubinov, I. D. Dubinova, S. K. Saykov, The Lambert W-function and its Applications to Mathematical Problems of Physics (in Russian), Syrov (Russia), The Russian Federal Nuclear Center, 2006, pp. 1-160.
  2. S. G. Kazakova and E. S. Pisanova, Some Applications of the Lambert W-function to Theoretical Physics Education.
  3. Nahin, P.J., Chases and Escapes: The Mathematics Of Pursuit And Evasion (Princeton Puzzlers), Princeton University Press; Revised edition 2012.

 

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