3
$\begingroup$

I have wondered the problem in http://www.helsinki.fi/~hmkokko/Stuff/Esdale/index.html for over year without success. If we try to minimize the functional equation T(\theta ) = \int_0^L\frac {dx}{v_0\cos \theta(x)} in the set W = {\theta \in C^1 [0,L]|C(\theta) = y} where C(\theta) = \int_0^L\frac {v_0\sin \theta (x) + v(x)}{v_0\cos \theta (x)}dx, can the solution be represented in a closed form or not? A solution or a proof that no closed form solution exists would be nice.

$\endgroup$
3

3 Answers 3

2
$\begingroup$

Here is some of your question properly translated.

$ T(\theta) = \int \frac{dx}{v_0\cos \theta(x)}dx $

$ W = {\theta \in C^1 [0,L]|C(\theta) = y} $

The definition of C(\theta) doesn't seem to want to work.

$\endgroup$
1
  • $\begingroup$ braces need help here, double the backslash... $W = \\{\theta \in C^1`[0,L] | C(\theta) = y\\}$ $\endgroup$ Nov 11, 2009 at 14:58
1
$\begingroup$

The straight line (with constant theta) given by:

y/x = (v0 sin(theta) + v)/(v0 cos(theta))

is the optimal solution. This can be readily seen from the question's formulation: The constrained functional:

int[0,L] dx/(v0 cos(theta)) + lambda int[0,L] dx(v0 sin(theta) + v)/(v0 cos(theta))

doesn't depend on the time derivative of theta, thus the Euler lagrange equation does not depend on the derivatives of theta, thus has a constant theta as a solution.

Another way to understand the solution is to describe it from the point of view of an observer moving with the current. In this case the problem reduces to the geodesic motion of a free particle on the two dimensional plane. It's solution is a straight line with constant velocity. The (Eucledian) inverse transformation to the original system transforms constant velocity straight lines to constant velocity straight lines.

Solutions of general dynamical problems of trajectory time minimization can be systematically formulated using Pontryagin's maximum principle (see the wikipedia article).

$\endgroup$
0
$\begingroup$

Sorry. The problem is that we start from the point $(0,0)$ and we swim to $(1,0)$. Now if you swim by straight line you won't necessary end to the point where you want to. Then you have to compute how the flow affects your swimming speed and the direction you are swimming to. I think this leads to the functional equation I got if I computed it correctly. Then you can probably find a solution curve which tells the direction you have to swim with respect to time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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