Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is from a problem seminar and I need help figuring out the solution.

Four bugs, $A,B,C,D$ are initially placed at the corners of a unit square. From a given initial moment, all four crawl simultaneously at one and the same speed $s>0$, $A$ towards $B$, $B$ towards $C$, $C$ towards $D$, and $D$ towards $A$, with each heading at every instant along the line joining it with its target.

Find a system of differential equations describing the trajectory of $A$.

Do the bugs eventually all meet at the center of the square? If so, how long did it take?

My attempt so far: $\def\grad{\mathbf\nabla}$ $\newcommand{\norm}[1]{\left\Vert #1 \right\Vert}$ Call ${\vec x}_1,{\vec x}_2,{\vec x}_3,{\vec x}_4$ the positions in the plane of $A,B,C,D$ respectively. Then $$\grad {\vec x}_1 = \frac{{\vec x}_2-{\vec x}_1}{\norm{{\vec x}_2-{\vec x}_1}}s$$ $$\grad {\vec x}_2 = \frac{{\vec x}_3-{\vec x}_2}{\norm{{\vec x}_3-{\vec x}_2}}s$$

$$\grad {\vec x}_3 = \frac{{\vec x}_4-{\vec x}_3}{\norm{{\vec x}_4-{\vec x}_3}}s$$

$$\grad {\vec x}_4 = \frac{{\vec x}_1-{\vec x}_4}{\norm{{\vec x}_1-{\vec x}_4}}s$$

with ${\vec x}_1(0) = (0,0),\ {\vec x}_2(0) = (0,1),\ {\vec x}_3(0) = (1,1),\ {\vec x}_4(0) = (1,0)$.

I have noticed that the sum of the gradients is 0, but I don't know if that is helpful or not. Can someone help me figure this problem out?

share|improve this question
    
This is related and has a nice animation in one of the answers: math.stackexchange.com/questions/44896/… –  joriki Nov 1 '11 at 8:43
    
Satisfied by an answer? –  Did Jun 28 '12 at 12:22
add comment

4 Answers

up vote 1 down vote accepted

Let the process take place in the square $|x|\leq 1$, $|y|\leq1$ of the complex $z$-plane. If $t\mapsto z(t)$ describes the movement of ant $A_0$ then the movement of the other ants is given by $$t\mapsto i^k\ z(t)\qquad(1\leq k\leq 3)\ .$$ By the given law of motion we have $$\dot z(t)\ =\ \lambda(t)\bigl(i\,z(t)-z(t)\bigr)$$ for some function $t\mapsto \lambda(t)>0$. As the velocity $v$ is given the factor $\lambda$ is determined by the condition $$v=|\dot z|=\lambda \sqrt{2}\,|z|\ .$$ Therefore $t\mapsto z(t)$ obeys the differential equation $$\dot z=v\ {i-1\over\sqrt{2}}\ {z\over|z|}\ .$$ We now write $z$ in the form $z(t)=r(t)\,e^{i\theta(t)}$, whereupon our differential equation becomes $$(\dot r + i r\dot\theta)e^{i\theta}\ =\ v\Bigl({i\over\sqrt{2}}-{1\over\sqrt{2}}\Bigr)\,e^{i\theta}\ .$$ Cancelling $e^{i\theta}$ and separating real and imaginary parts gives $$\dot r=-{v\over\sqrt{2}}\ ,\qquad \dot\theta ={v\over\sqrt{2}}\ {1\over r}\ ,$$ and we have the initial conditions $r(0)=\sqrt{2}$, $\ \theta(0)={\pi\over4}$.

We immediately get $$r(t)=\sqrt{2}\Bigl(1- {v\over 2}\ t\Bigr)\ ,$$ from which we deduce that the process is over after time $T:={2\over v}$. Using this we find via a second integration $$\theta(t)\ =\ {\pi\over4} +\log{1\over 1-{v\over2} t}\qquad(0\leq t<T)\ .$$ This shows that the ants turn ever faster around the origin towards the end of the process.

share|improve this answer
add comment

Hints: Assume the square is centered at zero and the affix of the position of $A$ at time $t$ is $r(t)\mathrm e^{\mathrm i\theta(t)}$. Then:

  1. The affixes of the positions of $B$, $C$ and $D$ at time $t$ are $\mathrm ir(t)\mathrm e^{\mathrm i\theta(t)}$, $-r(t)\mathrm e^{\mathrm i\theta(t)}$ and $-\mathrm ir(t)\mathrm e^{\mathrm i\theta(t)}$ respectively.
  2. Hence, for example, the line $AB$ has the direction of $(\mathrm i-1)\mathrm e^{\mathrm i\theta(t)}$.
  3. At time $t$, the tangent of the curve that $A$ makes has the direction of $(r'(t)+\mathrm ir(t)\theta'(t))\mathrm e^{\mathrm i\theta(t)}$.
  4. The square of the speed at time $t$ is $(r'(t))^2+(r(t)\theta'(t))^2$.

The steps of the proof are as follows. Points 2. and 3. combined yield a first order differential equation characterizing the motion of $A$, hence of $B$, $C$ and $D$ through point 1. Solving this yields $\theta(t)$ as a function of $r(t)$. Finally, point 4. determines $r(t)$.

share|improve this answer
    
What is an affix? –  nullUser Oct 31 '11 at 22:57
1  
The affix of the point $(x,y)$ in the plane is the complex number $x+\mathrm iy$. –  Did Oct 31 '11 at 22:58
    
How do you make sure that the positions B,C,D remain as prescribed that is for instance B on $ir(t)e^{i\theta(t)}$? –  user17090 Oct 31 '11 at 23:13
    
@Ali, from the rotational symmetry of the problem. The initial position and the dynamics are symmetric under the 90 degree rotation, so this will remain the case for all later times. –  zyx Oct 31 '11 at 23:32
    
Yes $(r'(t)+\mathrm i\theta'(t))\mathrm e^{\mathrm i\theta(t)}$ is the derivative of $re^{i\theta}$ with respect to $t$, but this makes it the tangent in polar coordinates, I need the tangent in rectangular coordinates. This is $i\frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$, no? And I can't see whether or not these are equal. –  nullUser Nov 1 '11 at 0:16
show 3 more comments

I think I solved this. If bug 1 has coordinates $(x,y)$ then bug 2 has coordinates $(y,1-x)$ so since bug 1 moves into the direction of bug 2 the differential equation is

$$\frac{dy}{dx} = \frac{1-x-y}{y-x}.$$

Substitution of $x=p+1/2$ and $y=q+1/2$ leads to

$$\frac{dq}{dp}=\frac{p+q}{q-p}$$

and then with $f=q/p$ and $p=\exp(s)$ we get

$$\frac{df}{ds}=\frac{1+f^2}{1-f}$$

The solution starting at $(x,y)=(0,0)$ is the found through separation of variables

$$p = -\frac{\exp(\arctan(f)-\arctan(1))}{\sqrt{2(1+f^2)}}$$

from which as function of varying parameter $f$, $x$ and $y$ can be obtained.

share|improve this answer
add comment

From the previous result, changing to polar coordinates ( $p=r\cos(\Theta), q=r\sin(\Theta)$ ) one obtains a more transparent form of the solution:

$r= \exp(\Theta-5\cdot\frac{\pi}{4})/\sqrt{2} $

Letting theta decrease from zero to minus infinity, the path of the bug traced is seen to be be a logarithmic spiral.

A more elegant approach is perhaps using complex numbers. With the origin of the complex plane in the lower left corner of the square, and the two bugs are at $z$ and $i-iz$, the differential equation becomes

$\frac{dz}{dt} + (1+i)z = i$

for which the solution can be written immediately

$z = \frac{1+i}{2} (1-\exp(-(i+1)\cdot t))$ for $t\geq0$

which represents, of course, the same spiral. Since $|dz|=\exp(-t)dt$ it follows that total length of the path equals 1, the length of the side of the square.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.